Phys_QUIZ_

Physics Problems

Physics Questions/Problems

This quiz has been accessed times since December 1995.For faster access in Europe see this MIRROR site.

Exercise your mind! One challenging problem is posed each month. Students (undergraduate, graduate, high-school,...) and researchers (from academy, industry,...) from all over the world are welcome to send their solutions, discussions and suggestions. Solutions should be mailed to [email protected]. The best (and, hopefully, correct) solution will be published on this WWW server. [Please, ask your webmaster or equivalent computer-wiz to include a reference (link) to this page in the appropriate web page of your organization.]

What's new - a list of latest changes.

D=Discussion, A=Answer, A-=Incomplete answer

2004/2005

# Title of the Question D A

01/04 Isotropic universe A-

02/04 Balancing plank A

03/04 Frozen Earth

04/04 Nonuniform Archimedes

05/04 Solar sails A

06/04 Thin ice D

07/04 Neutral capacitors

08/04 Falling magnet D

09/04 Frictionless rotation D A

http://star.tau.ac.il/QUIZ/ (1 of 7)11.7.2005 9:20:37

http://quiz.thphy.uni-duesseldorf.de/

mailto:[email protected]

Physics Problems

10/04 Wire capacitor

11/04 Gas of rotators

12/04 High-viscosity motion

01/05 Rope between inclines A

02/05 Negative index

03/05 Against the wind

04/05 Linear molecules

2003Many problems given that year were not solved. We are still waiting for solutions/discussions/comments


01/03 Bricks and sticks A

02/03 Reference frames A

03/03 Plank on a log A

04/03 Electric outlet D

05/03 Leiden jar

06/03 Meteorites D

07/03 Hula hoop D

08/03 Pillow A-

09/03 Delicate balance A

10/03 Three-sided coin A

11/03 Immortal bacteria

12/03 Three-phase currents A-


Physics Problems

2002Several problems given that year were not solved. We are still waiting for solutions/discussions/comments


01/02 Equipotential surfaces A

02/02 Maximal gravity A

03/02 Soap bubble A

04/02 Scattering dipoles A

05/02 Line of charges A

06/02 Isotropic radiator D A

07/02 Current in a wire A

08/02 Electrostatic machine A

09/02 Polymer at a wall D A

10/02 Least action A-

11/02 Inviscid motion

12/02 Charged drop D

2001Several problems given that year were not solved. We are still waiting for solutions/discussions/comments


01/01 Floating sphere A

02/01 Limited pendulum A

03/01 Incandescent light A

04/01 Ball in a box D A-

05/01 Dipole in a shell A


Physics Problems

06/01 Minimal container A

07/01 Conducting chessboard A

08/01 Dandelion seed

09/01 The power of dimension D A

10/01 Area maximization D

11/01 Resistor networks A

12/01 Bad weather A

2000Solutions of two problems are still incomplete. We are still waiting for solutions/discussions/comments


01/00 Climbing an iceberg A

02/00 Halley's planet A

03/00 Leaning tower of bricks A

04/00 Bike ride A

05/00 The first digit D A

06/00 Hydrogen atom A-

07/00 Energetics of walk A

08/00 Radiating sphere A

09/00 Perpetuum mobile A

10/00 Charges on a circle D A-

11/00 Frying in oil A

12/00 Coaxial cable A


Physics Problems

1999Some yet unsolved problems were given that year, and some solutions were incomplete. We are still waiting for solutions/discussions/comments


01/99 Mind reading

02/99 Interstellar travel A

03/99 Half-empty bottle A

04/99 Cup-a-tea D

05/99 Circuit of batteries A

06/99 Stretched molecule A

07/99 Missing energy D A

08/99 Zeroth order rainbow D A

09/99 Mean deflecting force A-

10/99 3D billiard A

11/99 Ohm's law D A

12/99 Creatures great and small A

1998Some yet unsolved problems were given that year, and some solutions were incomplete. We are still waiting for solutions/discussions/comments


01/98 Delayed weather D

02/98 A floating triangle A

03/98 Weighing a fly A

04/98 Knotted fields A-

05/98 Unequal halves A


Physics Problems

06/98 Floating bodies of equilibrium D A-

07/98 Slippery slope A

08/98 Body heat A-

09/98 Missing momentum A

10/98 Falling clouds A

11/98 Bouncing brick A-

12/98 Coupled pendulums A

1997 Some yet unsolved problems were given that year, and some solutions were incomplete. We are still waiting for solutions/discussions/comments


01/97 What causes A/r potential? D A

02/97 Does voltmeter measure EMF? - A

03/97 Solvable quantum Hamiltonians - A

04/97 StatMech in x2+1/x2 potential - A

05/97 At the lake-shore... D A

06/97 Breaking windows A

07/97 Enhancing a jump A

08/97 Quantifying fog A

09/97 Lunar Olympics - A-

10/97 Inertial capillarity - A

11/97 Variable capacitor - A

12/97 Once in a blue moon - A


Physics Problems

1996 Some yet unsolved problems were given that year. We are still waiting for solutions/discussions/comments


01/96 Running on water D A-

02/96 Floating faster than a stream D A

03/96 Candle at night D A

04/96 Burning Roman ships D A

05/96 Motion of hockey puck D A

06/96 Fish versus spermatozoa D

07/96 Disintegrating chimney A

08/96 French flag - A-

09/96 Avoiding flat tire D A

10/96 Splashing water

11/96 Wet spots A

12/96 Coffee stains D A

List of other sites with problems/puzzles

If you want to contribute a problem, send it to [email protected]



What's New

What's New in Physics Questions/Problems

In our main page we keep posting old problems. Some of them have not yet been solved. Sometimes we receive a solution or discussion of an old problem. To make sure that the changes in the old problem do not go unnoticed, we publish this list of latest additions and changes in our Quiz.

06/05 A new problem (04/05) has been published.06/05 A comment/discussion was added to 07/03.06/05 A discussion of 12/02 has been published.05/05 A new problem (03/05) has been published.05/05 A solution of 09/04 has been published.05/05 A discussion of 08/04 has been published.05/05 A new problem (02/05) has been published.03/05 A solution of 01/05 has been published.03/05 An comment has been added to the solution of 05/04.02/05 A new problem (01/05) has been published.12/04 New discussion has been added to the solution of 02/01.12/04 A solution of 04/01 has been published.12/04 A new problem (12/04) has been published.11/04 A discussion of 09/04 has been published.11/04 A new problem (11/04) has been published.11/04 A new problem (10/04) has been published.11/04 A new problem (09/04) has been published.10/04 A solution of 01/04 has been published.10/04 A solution of 08/03 has been published.10/04 A discussion item about (06/03) has been published.09/04 A solution of 05/04 has been published.09/04 A discussion item about (06/04) has been published.08/04 A new problem (08/04) has been published.07/04 A new problem (07/04) has been published.07/04 An additional partial answer of 12/03 was published.07/04 A new problem (06/04) has been published.07/04 A new problem (05/04) has been published.06/04 A new problem (04/04) has been published.05/04 A solution of 02/04 has been published.05/04 A partial answer of 12/03 was published.05/04 A new problem (03/04) has been published.05/04 A new problem (02/04) has been published.05/04 A new problem (01/04) has been published.04/04 Problem 09/03 was solved and its solution published.

http://star.tau.ac.il/QUIZ/whatsnew.html (1 of 6)11.7.2005 9:20:39

What's New

04/04 Problem 08/02 was solved and its solution published.04/04 Avery old problem 08/97 has been solved.03/04 A new problem (12/03) has been published.03/04 A new problem (11/03) has been published.01/04 Problem 10/03 was solved and its solution published.12/03 A new problem (10/03) has been published.12/03 A new problem (09/03) has been published.12/03 A new problem (08/03) has been published.12/03 A solution of problem s12/98 has been published.11/03 A solution of problem 09/02 has been published.10/03 Problem 03/96 has been finally solved, and its solution published.10/03 An additional discussion of problem 07/03 has been published.09/03 A discussion of problem 07/03 has been published.08/03 A discussion of problem 10/01 has been published.08/03 A discussion of problem 09/02 has been published.07/03 A new problem (07/03) has been published.07/03 A solution of problem 01/03 has been published.07/03 A comment has been added to problem 08/99.06/03 A new problem (06/03) has been published.06/03 A new problem (05/03) has been published.06/03 A solution of problem 02/03 has been published.06/03 Dicussion of problem 04/04 has been published.05/03 A solution of problem 03/03 has been published.04/03 A new problem (04/03) has been published.03/03 A new problem (03/03) has been published.03/03 A solution of problem 07/02 has been published.02/03 A solution of problem 05/97 has been published.02/03 A solution of problem 04/02 has been published.01/03 A new problem (02/03) has been published.01/03 An additional example has been added to the solution of 10/02.01/03 Partial solution of problem 10/02 has been published.01/03 A solution of problem 10/00 has been published.01/03 A new problem (01/03) has been published. Happy New Year!12/02 Solution of 06/02 has been published.12/02 A new problem (12/02) has been published.11/02 A new problem (11/02) has been published.09/02 A new problem (10/02) has been published.09/02 Problem 07/01 has been solved and its solution published.09/02 Problem 05/02 has been solved in its solution published.09/02 A new problem (09/02) has been published.08/02 A discussion about 06/02 has been published.08/02 A new problem (08/02) has been published.08/02 Problem 03/02 has been solved in its solution published.


What's New

07/02 Problem 01/02 has been solved in its solution published.07/02 A new problem (07/02) has been published.06/02 A new problem (06/02) has been published.05/02 Additional part of problem 06/98 has been solved, and solution published.05/02 Problem 03/01 has been solved in its solution published.05/02 Problem 02/02 has been solved in its solution published.05/02 A new problem (05/02) has been published.04/02 A new problem (04/02) has been published.03/02 Problem 12/01 has been solved in its solution published.03/02 A new problem (03/02) has been published.02/02 Problem 11/01 has been solved in its solution published.02/02 A new problem (02/02) has been published.01/02 A new problem (01/02) has been published. Happy New Year.12/01 Solution of 09/01 has been published.12/01 A new problem (12/01) has been published.11/01 A new problem (11/01) has been published.10/01 A new problem (10/01) has been published.09/01 A discussion of 09/01 has been published.09/01 Solution of 01/01 has been published.08/01 A new problem (09/01) has been published.08/01 A discussion of 04/01 has been published.07/01 A new problem (08/01) has been published.07/01 Solution of 06/01 has been published.07/01 Solution of 05/01 has been published.06/01 A new problem (07/01) has been published.06/01 A new problem (06/01) has been published.06/01 Solution of 02/01 has been published.05/01 Important information was added to 02/98 and 06/98.05/01 A post-post-script was added to problem 01/00.05/01 Solution of 12/00 has been published.05/01 A new problem (05/01) has been published.04/01 Expansion of the solution of 06/00 has been published.04/01 A new problem (04/01) has been published.03/01 Solution of 08/00 has been published.03/01 A very old problem (06/97) has been finally solved.03/01 A new problem (03/01) has been published.02/01 A new problem (02/01) has been published.12/00 A new problem (01/01) has been published. Happy New Year.12/00 A solution of 11/00 was published.12/00 Remarks added to solution 08/98.12/00 Remarks added to solution 02/00.12/00 A new problem (12/00) has been published.11/00 Remarks added to solution 11/98 and it has been "re-opened" for further discussions.


What's New

11/00 A solution of 11/99 was published.11/00 A solution of 02/00 was published.11/00 Correction+additional material were added to problem 08/96.11/00 Discussion of problem 10/00 has been updated.11/00 A solution of 09/00 was published.11/00 Discussion of problem 04/99 has been updated.11/00 A new problem (11/00) has been published.10/00 Partial solution of 06/00 was published.09/00 A new problem (10/00) has been published.09/00 Prob. 07/00 has been solved and the solution published.09/00 A new problem (09/00) has been published.08/00 The main part of the problem 06/98 has been solved. One part remains unsolved.08/00 A new problem (08/00) has been published.07/00 Prob. 04/00 has been solved and the solution published.07/00 A new problem (07/00) has been published.06/00 A new problem (06/00) has been published.05/00 Prob. 05/00 has been solved and the solution published.05/00 Prob. 01/00 has been solved and the solution published.05/00 A discussion of 05/00 has been published.05/00 A new problem (05/00) has been published.05/00 A postscript email has been added to an old problem (04/96).04/00 A new problem (04/00) has been published.03/00 Additional comments have been added to an old problem (12/97).03/00 Prob. 03/00 has been solved and the solution published.03/00 Prob. 12/99 has been solved and the solution published.03/00 A new problem (03/00) has been published.02/00 Prob. 09/99 has been solved and the solution published.02/00 Prob. 02/99 has been solved and the solution published.02/00 Discussion of problem 11/99 has been published.02/00 Additional comment/insight to an old solved problem (03/99) has been published.01/00 An illustration has been added to an old solved problem (04/96).01/00 A new problem (02/00) has been published.01/00 Additional material regarding solution (10/98) was published.12/99 A new problem (01/00) has been published. Happy New Year!11/99 A new problem (12/99) has been published.11/99 Prob. 10/99 has been solved and the solution published.11/99 Prob. 07/99 has been solved and the solution published.11/99 A new problem (11/99) has been published.10/99 Prob. 08/99 has been solved and the solution published.10/99 A new problem (10/99) has been published.09/99 Discussion of the problem 04/99 has been published.09/99 Prob. 06/99 has been solved and the solution published.08/99 A new problem (09/99) has been published.


What's New

08/99 Discussion of the problem 08/99 has been published.07/99 A new problem (08/99) has been published.07/99 Discussion of the problem 07/99 has been published.07/99 Prob. 09/98 has been solved and the solution published.07/99 A new problem (07/99) has been published.06/99 A new problem (06/99) has been published.06/99 Prob. 05/99 has been solved and the solution published.04/99 Prob. 03/99 has been solved and the solution published.04/99 A new problem (05/99) has been published.03/99 Partial solution was published in the discussion of the problem (05/96).03/99 A new problem (04/99) has been published.03/99 Additional solutions of (08/98) have been published.03/99 A new problem (03/99) has been published.02/99 An new discussion of the problem 01/98 has been published.02/99 An old problem (07/96) has been solved.02/99 Prob. 11/98 has been solved and the solution published01/99 A new problem (02/99) has been published.12/98 First problem for the new calendar year (01/99) has been published. Happy Hollidays!12/98 Prob. 10/98 has been solved and the solution published11/98 A new problem (12/98) has been published.11/98 Additional comment has been added to the discussion of 05/97.11/98 A new problem (11/98) has been published.10/98 One possible method of solution of 08/98 has been published.9/98 An old problem (07/97) has been solved.9/98 Prob. 07/98 has been solved.9/98 A new problem (10/98) has been published.8/98 A new problem (09/98) has been published.7/98 Solution of old prob. (09/96) has been recieved and published.7/98 A follow-up discussion about the validity of the solution 04/98 has been published.7/98 A new problem (08/98) has been published.6/98 Discussion of the problem (01/98) was updated.6/98 A new problem (07/98) has been published.6/98 Discussion of the problem 06/98 has been published.5/98 A new problem (06/98) has been published.5/98 Prob. 02/98 has been solved.5/98 Prob. 03/98 has been solved.5/98 Prob. 05/98 has been solved.5/98 Solution of old prob. (12/96) has been published.4/98 A new problem (05/98) has been published.4/98 Solution of the problem (04/98) and a follow-up question have been published.


What's New

Back to "front page"


http://star.tau.ac.il/QUIZ/quiz_body.html


Question 01/04

Question 01/04

ISOTROPIC UNIVERSE

Observer O is located in an infinite space and is bombarded by particles (shown in blue) coming from infinity and moving along straight lines in random directions. This "rain" of particles is isotropic in its directions. At large distance from the observer there is a region of scatterers S shown in red. The region contains many scatterers of the blue particles. The scatterers are not isotropic. However, their orientation is random. Can the observer O detect the presence of the region S in the space by simply observing the distribution of particles arriving at O from different directions? Note: You may assume that the scatterers are "dilute", i.e. a particle is not scattered more than once.


http://star.tau.ac.il/QUIZ/04/Q01.04.html11.7.2005 9:20:41



Answer 01/04

Answer to the Question 01/04

ISOTROPIC UNIVERSE

The question was:

Observer O is located in an infinite space and is bombarded by particles (shown in blue) coming from infinity and moving along straight lines in random directions. This "rain" of particles is isotropic in its directions. At large distance from the observer there is a region of scatterers S shown in red. The region contains many scatterers of the blue particles. The scatterers are not isotropic. However, their orientation is random. Can the observer O detect the presence of the region S in the space by simply observing the distribution of particles arriving at O from different directions?

http://star.tau.ac.il/QUIZ/04/A01.04.html (1 of 2)11.7.2005 9:20:42

Answer 01/04

Note: You may assume that the scatterers are "dilute", i.e. a particle is not scattered more than once.

(10/04) A solution to the problem has been submitted (7/7/04) by John G. Florakis, from University of Athens, Greece (e-mail [email protected]). You can see the solution in the PDF file. However, we feel that a simpler solution, less specific, and relying on basic principles of mechanics should be possible.

The answer: The region should be undetectable.




http://star.tau.ac.il/QUIZ/04/Isotropic2.pdf



Question 02/04

Question 02/04

BALANCING PLANK

A thick plank is placed on a log of semicircular cross section. What is the thickest plank such that the plank on log has stable oscillations?

This follow-up question (to 03/03) has been suggested by David J.C. MacKay.





Answer 02/04


BALANCING PLANK

The question was:

A thick plank is placed on a log of semicircular cross section. What is the thickest plank such that the plank on log has stable oscillations?

(6/04) The problem has been solved (17/4/04) by Chetan Mandayam Nayakar, an undergraduate student at Indian Institute of Technology, Madras (e-mail [email protected]) (his solution is presented below), and by several other people.

The answer: The plank becomes unstable if its thickness exceeds the diameter of the log.

The solution: Let the radius of the log be R and the thickness of the plank be t. At equilibrium, the height of the plank is equal to R+(t/2). When the plank undergoes an infinitesimal angular displacement of {theta} from equilibrium, the height of the plank above the log axis becomes (R+(t/2))*cos({theta})+(R*{theta}*sin({theta})).



Answer 02/04

For stable equilibrium, the new height should be greater than R+(t/2). Applying the limit {theta} tending to zero and solving the inequality, the result is t < 2R.

p.s. Actually the solution can be found in the answers to problem 03/03.





Question 03/04

Question 03/04

FROZEN EARTH

What would happen to the magnetic field of Earth if the planet suddenly freezes? (What will be the eventual value of the magnetic field and how long will it take to get there?)





Question 04/04

Question 04/04

NONUNIFORM ARCHIMEDES

A planet with VERY inhomogeneous mass distribution may have a very non-uniform gravitational field on its surface, and equipotential surfaces of the field will not be flat as depicted by black lines in the picture above. Consider a body of water (lake) on the surface of such a planet. Discuss flotation of bodies in such a liquid. Can you formulate a generalized Archimedes law? (As an example consider flotation of a solid sphere.)

This this problem was inspired by discussions with Gyozo Egri, Marton Kormos and Bence Kocsis.





Question 05/04

Question 05/04

SOLAR SAILS

Consider a cosmic sail-boat moving under the influence of the pressure of sun-light at a distance from the Sun similar to that of Earth. What kind of accelerations you can expect? How do those light-pressure forces compare with the gravitational pull of the Sun?





Answer 05/04


SOLAR SAILS

The question was:

Consider a cosmic sail-boat moving under the influence of the pressure of sun-light at a distance from the Sun similar to that of Earth. What kind of accelerations you can expect? How do those light-pressure forces compare with the gravitational pull of the Sun?

(9/04) The problem has been solved (16/9/04) by Iavor Veltchev (e-mail [email protected]). His solution can be found in the following PDF file.

The answer: Both gravitational forces and the light pressure decrease as inverse squared distance from the Sun, and therefore the answer is independent of the distance. Since gravitational force is proportional to the mass of the sail, while the light pressure is proportionl to the area, by equating these two we can find the maximal allowed thickness of the sail. Unfortunately, it comes out to be smaller than a micron. So much for solar sailing...

(3/05) Y. Kantor: Not all is lost: While we cannot use the solar wind to move in arbitrary directions like a sailboat in the sea, we nevertheless can use it for a gradual propulsion in space. On 30/3/05 Anthony O'Brien (e-mail [email protected]) wrote us the following:

The most efficient way to utilise the solar pressure is direct the photons perpendicular to the direction of the sun. If you put a craft into space, by putting it into orbit around the sun, you can utilise the solar wind purely for acceleration. Then you can spiral the craft out from the sun. This is how most interplanetary missions are being proposed, even for conventional rocket craft.


http://star.tau.ac.il/QUIZ/04/A05.04.html11.7.2005 9:20:46


http://star.tau.ac.il/QUIZ/04/sail.pdf

mailto:



Question 06/04

Question 06/04

THIN ICE

Why can a person run on thin ice (over a pond), but cannot stand on it?

This problem appears without solution in "Physics Problems" by P. L. Kapitza.





Discussion 06/04

Discussion of the Question 06/04

THIN ICE

The question was:

Why can a person run on thin ice (over a pond), but cannot stand on it?

(9/04): Sam Crown from Gresham, Oregon, USA (e-mail [email protected]) sent us an email correctly identifying the reasons for thin ice supporting the runner. He wrote:

The non-compressible water beneath the ice has to flow away from the area under the runner's feet before the ice can bend. Given the inertial and viscous nature of the water, this flow process isn't instantaneous; if the load applied beneath the runner's boots is quickly applied and released, the ice won't deflect enough to fail. When the runner stops, the water beneath the ice flows out from under his feet and the ice bends and breaks.

Can we make this solution more quantitative?


http://star.tau.ac.il/QUIZ/04/D06.04.html11.7.2005 9:20:48




Question 07/04

Question 07/04

NEUTRAL CAPACITORS

In electric circuits it is usually assumed that the charges on the two plates of each capacitor are equal in size and have opposite signs. E.g., in the piece of a circuit shown above it is usually assumed that: Q1=-Q2=Q3=-Q4. What are the conditions for such relations to be (approximately) correct?





Question 08/04

Question 08/04

FALLING MAGNET

Bar magnet falling in a vertical copper pipe quickly reaches "terminal" velocity and stops accelerating. Find that velocity as a function of the features of the magnet and the pipe.





Discussion 08/04


FALLING MAGNET

The question was:

Bar magnet falling in a vertical copper pipe quickly reaches "terminal" velocity and stops accelerating. Find that velocity as a function of the features of the magnet and the pipe.

(5/05) Y. Kantor: We did not get a complete solution of the problem yet. As a hint, we publish a solution of a closely related problem ("Pitching pennies into a magnet") by Kirk T. McDonald from Princeton University, NJ, USA. A copy of the solution can be found in the following PDF file.

http://star.tau.ac.il/QUIZ/04/D08.04.html (1 of 2)11.7.2005 9:20:50

http://star.tau.ac.il/QUIZ/04/pennies.pdf

Discussion 08/04





Question 09/04

Question 09/04

FRICTIONLESS ROTATION

A man is standing straight on the surface of ice facing east. The ice is so slippery that there is no friction between the man's shoes and the ice. Is there a sequence of moves that can be made, such that at the end of it the man will be standing straight facing west?





Discussion 09/04




(11/04) Y. Kantor:


Discussion 09/04

We recieved a surprisingly large amount of e-mails containing the statement that angular momentum conservation implies a "direction conservation." This prejudice is extremely common among physicist, and is particularly visible in answers that they give to a question "How does a cat fall on its feet." In various web-sites you can find answers ranging from "the cat is rotating its tail at high speed" to "the cat always(!) has an initial rotational velocity which it then regulates by spreading its feet like a rotating ice-skater does with his/her hands". In fact a body which can be deformed in its shape does not have to preserve its orientation. T.R. Kane and M.P. Scher, in Int. J. Solids Struct. 5, 663 (1969) "mapped" the motions of a cat on two cylinders which can rotate and change the angles between their axes. They showed by detailed analysis of equations of mechanics, how a cat performs the "impossible rotation." [See more in R. Montgomery, Fields Inst. Comm. 1, 193 (1993) and J. Marsden, in Motion, Control and Geometry (Nat. Acad. Press, Washington, D.C., 1997).] In fact, already in 1894 Etienne-Jules Marey and Georges Demeny made a movie that shows in great detail the way a cat thrown upside-down (without any initial rotation) turns over during a short fall.





Answer 09/04



The question was:


(5/05) Y. Kantor: We received a very large amount of proposed solutions. (We will not list the names of the numerous contributors, but we wish to thank them all.) Roughly, the solutions fall into three groups: a. Using external objects. E.g.: Take off your shoe and rotate it above your head. b. Actively use the ice as a provider of normal force. E.g.: Lie down on the ice, and, by pushing with hands turn around on your back; then get up. c. Use body motions to perform the rotation. (The fact that the ice supports you is used to the "minimal extent.")

The solution presented below does not follow exactly any of the proposed solutions. Nevertheless, it follows the "spirit" of some of the proposals mentioned in the item (c) above.

The solution: Since it is a bit difficult to explain the nature of the moves, I hope that the demonstration by my son in the pictures below, will help you to visualize the required motion.

The rotation is accomplished by repeated sequence of hand motions as described below: Bend your hand and the stretch it out forward. Swing it sideways and as a result your body will rotate in the opposite direction. Bend your hand back.

The idea of the entire exercise is that when you swing your hand, the moment of inertia of the entire body is different from the moment when you bring the hand back to the original position. Such change in the moment of inertia is a necessary condition for performing this rotation.

Comment: It should be noted that the act of stretching the hand forward on a frictionless ice is not a trivial accomplishment. The


Answer 09/04

fact that one does not fall down (due to rotation of the body in the opposite direction), is due to the small momentum of force that can be generated by the normal forces of ice acting on our feet.





Question 10/04

Question 10/04

WIRE CAPACITOR

A capacitor consists of two very long parallel conducting circular wires almost touching each other (see the cross section in the picture). Find the electric field of the capacitor. Does it have a finite capacitance? Explain!





Question 11/04

Question 11/04

GAS OF ROTATORS

Consider a very dilute gas of thin rodlike molecules. The kinetic energy of such a gas is significantly larger than kinetic energy of an ideal gas of point-like objects, because each rotational degree of freedom has (on the average) kinetic energy (1/2)kT. Nevertheless, the pressure of the gas is the same as of point-like molecules. Show from dynamical considerations that the rotational degrees of freedom do not contribute to the pressure.





Question 12/04

Question 12/04

HIGH-VISCOSITY MOTION

Consider two long cylinders kept parallel at fixed distance from each other and rotated by built-in motors in opposite directions, as shown in the figure. The system is immersed in a very viscous liquid. Find the relation between the parameters of the system and the velocity of its propulsion. (Neglect gravity. Assume low Reynolds number. Solve a two-dimensional problem. Assume that the cylinders are extremely long, and disregard the influence of contraption that holds them together.)





Question 01/05

Question 01/05

ROPE BETWEEN INCLINES

A rope rests on two platforms which are both inclined at an angle θ (which you are free to pick), as shown. The rope has uniform mass density, and its coeffciient of friction with the platforms is 1. The system has left-right symmetry. What is the largest possible fraction of the rope that does not touch the platforms? What angle θ allows this maximum value?





Answer 01/05


ROPE BETWEEN INCLINES

The question was:

A rope rests on two platforms which are both inclined at an angle θ (which you are free to pick), as shown. The rope has uniform mass density, and its coeffciient of friction with the platforms is 1. The system has left-right symmetry. What is the largest possible fraction of the rope that does not touch the platforms? What angle θ allows this maximum value?

(3/05) This problem has been solved correctly (25/2/05) by Chetan Mandayam Nayakar, a student at India Insttitute of Technology, Madras, India (e-mail [email protected]), (26/2/05) by Sebastian Popescu, a lecturer at the Physics Department at "Al. I. Cuza" University in Iasi, Romania (e-mail [email protected]), (5/3/05) by Ioannis Florakis, from University of Athens, Greece (e-mail [email protected]). This problem and its solution appear at the web site of David Morin from the Physics Department of Harvard University, Cambridge, Mass., USA. You can see the solution in the PDF file.

The answer:


mailto:<[email protected]



http://star.tau.ac.il/QUIZ/05/sol_rope.pdf

Answer 01/05

The maximal possible fraction of the rope that is hanging is (√2-1)2 ≈ 0.172, which is obtained for θ = 22.5 degrees.





Question 02/05

Question 02/05

NEGATIVE INDEX

A ray of light is incident on a boundary between vacuum and material that (at that particular frequency) has dielectric constant ε=-1 and magnetic permeability μ=-1. (Stricly speaking, these quantities must also have imaginary parts. We will assume that they are very small.) The incidence angle is θ. What can you say about the direction of the reflected and the transmitted rays.





Question 03/05

Question 03/05

AGAINST THE WIND

A boat carries a wind motor of a windmill type, which drives the propeller screw. Discuss the possibility that such a boat can sail against the wind.






Question 04/05

Question 04/05

LINEAR MOLECULES

Consider a two-dimensional gas of molecules inside a rectangular box of small width d and very large length. If the box is filled by point-like molecules with density n, then the pressure is p=nkBT. (kB is the Boltzmann constant, and T is the temperature.) If the point-like molecules are replaced by thin linear molecules of length L, then the pressure will slightly increase. (a) Consider a case when L is smaller (but not much smaller) than d, and the molecules do not interact with each other, but they cannot penetrate the walls of the box. (See picture (a).) Calculate the exact pressure in the box. (b) Consider a case when the molecules of lengths L form rigid circles of diameter L/{pi}. (See picture (b).) Calculate the exact pressure in the box. (c) Note that your answers to (a) and (b) produced corrections to the leading term (nkBT) that are simply related to each other. What is the reason for that simple relation?





Question 01/03

Question 01/03

BRICKS AND STICKS

Two bricks are separated by three parallel wooden sticks of circular cross sections. The sticks are positioned as shown in the picture, separated from the ends of the brick and from each other by distances l1, l2, l3, l4. The weight of the top brick is P. What are the forces applied by each stick on the brick.





Answer 01/03


BRICKS AND STICKS

The question was:

Two bricks are separated by three parallel wooden sticks of circular cross sections. The sticks are positioned as shown in the picture, separated from the ends of the brick and from each other by distances l1, l2, l3, l4. The weight of the top brick is P. What are the forces applied by each stick on the brick.

(6/03) The problem has been solved (21/1/2003) by Victor Ivanov, a teaching assistant at the Faculty of Physics of Sofia University (Bulgaria) (e-mail [email protected]) (the solution below follows (to large extent) his derivation), (9/3/2003) by Gareth Pearce a Ph.D. student at U. Sydney (e-mail [email protected]), (22/3/2003) by Regis Lachaume, a PhD student at Grenoble Observatory (UJF, Grenoble, France) [email protected]) (see his solution at the following postscript file, and (16/6/2003) by Ansgar Esztermann, from Heinrich-Heine-Universitaet Duesseldorf (Germany) (e-mail [email protected]).





http://star.tau.ac.il/QUIZ/03/lach_brck.ps


Answer 01/03

The answer:

For the sake of simplicity we introduce the coordinates Xi (i=1...3) of the points, where the rods touch

the surface of the upper brick. They are defined as the horizontal positions of those points with respect to the center of the brick. The Y-axis is directed downwards from the brick's center. If we assume that the bricks and the rods are perfect rigid bodies then the conditions for equilibrium read: (1) SUM(Fi) = P (force balance), and:

(2) SUM(Fi*Xi) = 0 (torques balance),

where Fi are the forces of normal reaction exerted from the rods on the upper brick. It is evident that if

the number of rods is 3 or more, the two equations are insufficient to determine F's unequivocally (the problem is statically undefined). To cope with the problem we must go beyond the rigid body approximation. Let us to assume that the rods acquire some small deformation when the upper brick is put on them. We will neglect the deformations of the surfaces of the bricks since as a rule the ceramic material of the bricks is rather stiff compared to the wood. It can be shown that the deformation of the rod is a linear function of the force, despite the fact that it has only a "point contact" with the brick. (This is explained in "Comment 1" at the bottom of this page.) Thus we can treat the rods as elastic springs of equal stiffness K, and assume that the Hook's law is satisfied: Fi = K*Ui, where Ui is the deformation of the i-th rod. We assume that when the upper brick is

laid onto the rods, due to their deformations, it's center is lowered by a small amount Y, and its surface tilts a little bit at an angle A with respect to the horizontal (the later effect takes place if the rods are placed asymmetrically with respect to the center of the brick). Then the deformation of the i-th rod is: Ui = Y + A*Xi.

By substituting the forces Fi obtained from the Hook's law in Eqn. (1) and (2) we obtain the values of Y

and A: Y = (P/(3K))*D/(D - M2); A = - (P/(NK))*M/(D - M2); where D = SUM(Xi

2)/3 and M = SUM(Xi)/3. Finally we obtain the forces of reaction:

(3) Fi = (P/3)*(D - Xi*M)/(D - M2).

It is evident from the last expression that if the rods are placed symmetrically with respect to the center of the brick, then the weight of the brick is spread uniformly among them with equal portions of P/3. A*Xi.

Comment 1: Contacts involving cylinders or spheres were treated in 19th century by H. Hertz. You can find this treatment in most elasticity textbooks (e.g., the book by Landau and Lifshitz). What is special in this case is that we have a point contact, which upon application of force broadens into a contact with finite area. For the case of a contact between two spheres, or a contact between a sphere and infinitely rigid plane, this leads to conclusion the the displacement (deformation of the sphere) is proportional to the force in the power 2/3. (The prefactor in this relation depends both on the elastic properties of the


Answer 01/03

sphere and on its radius.) The situation is quite different in the "two-dimensional" case of cylinder. It can be derived from the general formalism that treats contact of ellipsoids, or can be considered directly as was done by H. Poritsky in J. Appl. Mech. 17, p. 191 (1950). In this paper it has been shown that the displacement is simply proportional to the force applied to the cylinder. Thus the simple linear relation between the force and deformation used in the above derivation is justified. The only pathology caused by the presence of the point contact is related to the dependence of the proportionality constant on the properties of the cylinder: As in the case of a "regular" contact the force constant is proportional to the elastic constant (Young modulus) of the stick. However, for the point the force constant very weakly (logarithmically) depends on the diameter of the stick.

Comment 2: Regis Lachaume, a PhD student at Grenoble Observatory (UJF, Grenoble, France) also considered (23/1/2003) a very different case, in which the bricks and the sticks are rigid enough so that their deformation is small. However, the "brick" is so long that its transverse deformations become significant. This situation is not suited, for our "brick and sticks" case, but probably correctly describes a long board supported by sticks. See his solution in postscript file.



http://star.tau.ac.il/QUIZ/03/lach_brck.ps



Question 02/03

Question 02/03

REFERENCE FRAMES

A positively charged particle starts moving with velocity V0 parallel to a uniformly positively charged infinite plate. The plate generates field E perpendicular to it. The motion of the particle can be described by a parabola; after a while it will acquire velocity V1 in the vertical direction and will continue to move with velocity V0 in the horizontal direction, as depicted in Fig. (a) Now let us look at this motion in the reference frame moving with velocity V0 to the right. As depicted in Fig. (b), the particle moves with increasing velocity in the vertical direction, but does not move in the horizontal direction. However, in this reference frame, the infinite plate moves to the left with constant velocity and generates a magnetic field of size V0E/c in the direction perpendicular to the plane of the picture. As a result, there is a force acting on the particle in the horizontal direction, and consequently it must accelerate! Find the error in the above reasoning, and explain the apparent paradox.

This problem has been suggested by Lev Vaidman.





Answer 02/03


REFERENCE FRAMES

The question was:

A positively charged particle starts moving with velocity V0 parallel to a uniformly positively charged infinite plate. The plate generates field E perpendicular to it. The motion of the particle can be described by a parabola; after a while it will acquire velocity V1 in the vertical direction and will continue to move with velocity V0 in the horizontal direction, as depicted in Fig. (a) Now let us look at this motion in the reference frame moving with velocity V0 to the right. As depicted in Fig. (b), the particle moves with increasing velocity in the vertical direction, but does not move in the horizontal direction. However, in this reference frame, the infinite plate moves to the left with constant velocity and generates a magnetic field of size V0E/c in the direction perpendicular to the plane of the picture. As a result, there is a force acting on the particle in the horizontal direction, and consequently it must accelerate! Find the error in the above reasoning, and explain the apparent paradox.

(5/03) The problem has been solved (4/2/2003) by Srikant Marakani a graduate student the University of Chicago (e-mail [email protected]), (10/2/03) by Dimitrios Vardis (e-mail vardis@ote.




Answer 02/03

net), (8/3/03) by Alexander Shpunt (e-mail [email protected]), (12/3/03) by Yoga Divayana an undergraduate Student of School of Electrical and Electronics Engineering Nanyang Technological University (Singapore) (e-mail [email protected]), (30/5/03) by Ivan Sirakov from Laboratoire de Rheologie des Matieres Plastiques at CNRS, St. Etiene, France (e-mail [email protected]) - see his very detailed solution in the following PDF file, (24/6/2003) jointly by Roberto D'Agosta (e-mail [email protected]) and Martino De Prato (e-mail [email protected]), and (26/06/03) by Carlos Soria from Sevilla (Spain) (e-mail [email protected]).

The answer:

The answer to the problem is that the velocity V0 is not constant in the original reference frame since its

relativistic momentum mV0/sqrt(1-V2/c2), where V is the total velocity, would then be increasing.

Hence, V0 must be decreasing which would give the same direction for the acceleration as the force due

to the magnetic field in the second frame. One can directly verify this statement quantitatively.

Dimitrios Vardis sent us the following detailed description of the process as is seen from each of the reference frames:

Let K be the inertial frame relative to which the plate is at rest and K' the inertial frame relative to which the plate is moving with velocity - Vo. The relativistic equation of motion dp/dt = F is invariant, i.e. it has the same form

in all inertial frames of reference. It is more convenient in our case to write this equation in the form. dp/dt = F or d(mu)/dt = F or mdu/dt + dm/dt u = F or ma = F - dm/dt u but mc2= E or c2dm/dt = dE/dt = F.u so the last of the above equations takes the form ma = F- (F u/c2)u. In the K frame the projection of the above equation along the direction of the relative motion of the two frames (x-axis) is

max = Fx - (qEV1/c2)ux

or

max = - (qEV1/c2)Vo at the moment the velocity along x-axis is Vo)

In the K' frame relative to which the charged plate is moving it generates an electrical field perpendicular to the plate and a magnetic field parallel to the plate and at right angle to the page. Taking the y-axis perpendicular to the plate, the electric and magnetic field vectors in the K' frame are

E'y = gy

B'y=-uEg/c2 (g = Lorentz factor)

The equation of motion in K' is max = -qVoEg/2 (along x-axis at the moment the velocity of charged particle

relative to K is Vo) etc. Therefore there is an acceleration of the particle along the x-axis in both frames as

requires the principle of relativity...







http://star.tau.ac.il/QUIZ/03/0203sol.pdf





Answer 02/03

Alexander Shpunt performed a detailed calculation of the actual motion described in the following postrscript file. In his derivation you can actually see that horizontal velocity is decreasing due to increasing mass of the particle.



http://star.tau.ac.il/QUIZ/03/refframe_sol.ps

http://star.tau.ac.il/QUIZ/03/refframe_sol.ps



Question 03/03

Question 03/03

PLANK ON A LOG

A thin plank is placed on a log of semicircular cross section. If the plank is slightly tilted it will start oscillating. Find the frequency of the oscillations. Assume that all masses and dimensions are known.





Answer 03/03


PLANK ON A LOG

The question was:

A thin plank is placed on a log of semicircular cross section. If the plank is slightly tilted it will start oscillating. Find the frequency of the oscillations. Assume that all masses and dimensions are known.

(5/03) The problem has been solved (12/3/03) by Alex Smolyanitskiy (e-mail [email protected]), (12/3/03) by Chetan Mandayam Nayakar (e-mail [email protected]), (1/4/03) by Zoran Hadzibabic (e-mail [email protected]), and (7/4/03) by Luca Visinelli (e-mail [email protected]). Alex Smolyanitskiy treated such a problem long before it was published in our QUIZ. Thus, he submitted and extremely detailed solution which even accounts for details that we neglected and can be seen in this postscipt file. A nice solution of the problem also has been submitted (12/5/03) by Matthias Punk (e-mail [email protected]) - see his solution in this postscipt file.

The answer: The angular frequency of oscillation will be sqrt{12gR/L2}, where L is the length of ther plank, R is the radius of the log, and g is the acceleration of free fall.

The solution: We first notice that once the plan is deflected by angle {theta} the restoring torque (to the lowest order in {theta} is MgR{theta}, where M and L are the mass and the length of the plank. This should be equated with the moment of interia ML2/12 of the plank around its center multiplied by angular acceleration. This immediately leads to the desired answer. However, the outlined solution disregards the fact that the center of mass of the plank is moving (left-right and up-down), and that the contact point is moving relative to the center of mass. Nevertheless, a detailed analysis of all the approximations, as explained in the solutions of Smolyanitskiy and Punk, shows that for small angle oscillations these "complications" can be neglected.






http://star.tau.ac.il/QUIZ/03/ruler_can_sol.ps


http://star.tau.ac.il/QUIZ/03/matthias_sol.ps

Answer 03/03





Question 04/03

Question 04/03

ELECTRIC OUTLET

Electric outlet has two "current carrying holes", one of which is the "phase" while the other one is "zero". (In addition, many outlets have a third hole for the "ground".) After many years of usage with high-power devices (ovens, heaters, etc.) one may notice that the "phase" side of the outlet shows some signs of overheating. Sometimes (especially in low quality outlets or plugs, and in the cases when the plugs are repeatedly inserted/removed from the outlets) burning signs can be detected on the plug or the outlet. (These are more frequent in 220/230 Volt outlets.) Obviously, the same current is flowing through both holes of the outlet. Why do these signs appear on the "phase" hole/pin more frequently than on the "zero" side?





Discussion 04/03


ELECTRIC OUTLET

The question was:

Electric outlet has two "current carrying holes", one of which is the "phase" while the other one is "zero". (In addition, many outlets have a third hole for the "ground".) After many years of usage with high-power devices (ovens, heaters, etc.) one may notice that the "phase" side of the outlet shows some signs of overheating. Sometimes (especially in low quality outlets or plugs, and in the cases when the plugs are repeatedly inserted/removed from the outlets) burning signs can be detected on the plug or the outlet. (These are more frequent in 220/230 Volt outlets.) Obviously, the same current is flowing through both holes of the outlet. Why do these signs appear on the "phase" hole/pin more frequently than on the "zero" side?

(6/03): We got (25/5/2003) the following email from Carlos Soria from Sevilla, Spain (e-mail [email protected]):

Burning signs appear in the ''phase'' hole because a little spark is produced each time something is plugged or unplugged, because there is a potential difference between the ''phase'' and the plug, not present in the earthed side.

As the plug is approaching to the outlet, a electric field is forming, and charge is being accumulated at both sides. The spark join these two 'electrodes' carrying charge and weakening the electric field, that is to say, by generating 'displacement current' in the opposite direction. So, total current may be null during the contact on the ''phase'' side and, consequently, on the earthed side, with the difference that in the phase side spark conduction current is liberating heat.

This could explain why phase side is burnt but not earthed side. Of course, the higher the voltage, the stonger the effect.



Discussion 04/03

(10/03): We got (5/10/2003) an interesting email from an anonymous correspondent (e-mail [email protected]). We do not think that this email solves the problem, but nevertheless it is an interesting observation. Here is what he wrote:

I had an oven that I plug and unplug often because the switch doesn't work. The ground side is the side that is pitted. The answer is that if you are right handed; you are going to put the plug in with the right side going into the outlet first. The left side is the side which will draw the arc and get pitted.



mailto:

mailto:



Question 05/03

Question 05/03

LEIDEN JAR

What fraction of energy of "Leiden jar" is transfromed into sound during its discharge.

This problem appears without solution in "Physics Problems" by P. L. Kapitza. A nice web-site describing condensers in general (and Leiden jar in particular) can be found here.



http://physics.kenyon.edu/EarlyApparatus/Electricity/Condenser/Condenser.html



Question 06/03

Question 06/03

METEORITES

What is the cumulative effect of all the meteorites that fell on the Earth during the last billion years on the length of a day? Did it become longer or shorter? By how much?





Discussion 06/03


METEORITES

The question was:

What is the cumulative effect of all the meteorites that fell on the Earth during the last billion years on the length of a day? Did it become longer or shorter? By how much?

(10/04): A proposed solution to the problem has been submitted (22/10/04) by John G. Florakis, from the University of Athens, Greece (e-mail [email protected]). You can see the solution in the PDF file. However, we feel that the proposed solution considers only one aspect of the problem,- the increase in the mass of the Earth. It completely disregards net angular momentum that can be imparted by meteorites and makes no attempt to estimate the magnitudes of various contributions.



http://star.tau.ac.il/QUIZ/03/Meteorites.pdf

Discussion 06/03





Question 07/03

Question 07/03

HULA HOOP

What movement should be performed by a person rotating a hoop round his body?






Discussion 07/03


HULA HOOP

The question was:

What movement should be performed by a person rotating a hoop round his body?

(9/03) Y. Kantor: We got quite a large amount of emails considering the kinematics in its simplest form. Since we are dealing with quite a complicated system, let us split the "general" question into a sequence of simpler questions. Stage A: Assume that the body has circular cross-section. Its radius r may depend on height. The hoop has radius R. See the following picture, depicting top and side views (top and bottom of the picture). If we completely disregard energy losses, there is no need to move the body, and the center of mass of the hoop will perform a circular motion around the center of the body and the radius of that motion will be Rcos(a)-r, where a is the angle between the plane of the hoop and a horizontal plane.


Discussion 07/03

In the idealized situation described above, what will be the force between the body and the hoop? How the frequency of the rotation of the hoop is related to parameters of the problem (including "body shape", i.e. dependence of r on height)? Is there an optimal frequency? What is the angle a?

Stage B: How will the above answers change if the body does not have a circular cross section. (Consider a case of ellipse.)

Stage C: In the presence of losses, what kind of motion should be performed to "inject" energy into the hoop.

(6/05) Y. Kantor: In March 2004 issue of Biological Cybernetics was published a paper entitled "Coordination Modes in the Multi-Segmental Dynamics of Hula-Hooping" (see it in the pdf format). It was later awarded the jokular Ig Nobel award. Some interesting data on hula-hooping can be found there.



http://star.tau.ac.il/QUIZ/03/hulahoop_paper.pdf



Question 08/03

Question 08/03

PILLOW

Why is pillow soft?





Answer 08/03


PILLOW

The question was:

Why is pillow soft?

(10/04) The problem has been solved (25/8/04) by Somak Ray (e-mail [email protected]). Here is his (slightly edited) answer:

Soft matters are deformable. A pillow is soft because when we press it, it moulds around the head and so the reactive force is distributed on a large surface area and so exerts less pressure giving a 'soft' feeling.

On the contrary when we press a hand against a concrete wall say, the force exerted by the wall acts on fewer points on the hand like on the knuckles and joints, thereby exerting more pressure and so the wall feels "hard'.

(10/04) Y. Kantor: The above answer essentially answers the question. Can we make the concept of softness more "quantitative"? We expect additional correspondence on the subject.






Question 09/03

Question 09/03

DELICATE BALANCE

Consider a lead pencil balanced on its point. Estimate how long it will remain standing if quantum effects (uncertainty principle) are taken into account. Compare that answer with the time it takes for thermal effects (air molecules, thermal radiation (if it's in vacuum)) to topple it.

This question was suggested by Mario Cesar from Colima, Mexico.





Answer 09/03


DELICATE BALANCE

The question was:

Consider a lead pencil balanced on its point. Estimate how long it will remain standing if quantum effects (uncertainty principle) are taken into account. Compare that answer with the time it takes for thermal effects (air molecules, thermal radiation (if it's in vacuum)) to topple it.

(4/04) The problem has been solved (7/3/04) by John G. Florakis (e-mail [email protected]). His solution can be found in the following PDF file. He has shown that in each of the cases the time is few seconds.




http://star.tau.ac.il/QUIZ/03/Equilibrium2.pdf



Question 10/03

Question 10/03

THREE-SIDED COIN

What should be the dimensions of a coin that has probability of 1/3 to fall on its obverse ("heads") side, 1/3 - on its reverse ("tails") side, and 1/3 - on its edge.





Answer 10/03


THREE-SIDED COIN

The question was:

What should be the dimensions of a coin that has probability of 1/3 to fall on its obverse ("heads") side, 1/3 - on its reverse ("tails") side, and 1/3 - on its edge.

(1/04) The problem has been solved (1/1/04) by Shiva (e-mail [email protected]).

The answer: The width of the coin should be 2R/sqrt(3), where R is the radius of the coin.

The solution: For the probability to be the same, since the coin is round, a vertical line through the center of gravity should pass through a corner when the coin is tilted by 30 degrees from the "standing on the edge" position. This way, out of possible 360 degrees, one third (120 degrees) will correspond to the coin eventually resting on the edge.






Question 11/03

Question 11/03

IMMORTAL BACTERIA

Bacteria are almost immortal. (Apollo 12 crew recovered bacteria that were accidentally brought to the Moon 3 years earlier by Surveyor probe, and those bacteria were revived. There are claims of revival of 30 million year-old bacteria.) How long will it take for the cosmic radiation to destroy the DNA of a bacterium.

This question was inspired by "Fermi question" site that can be found here.



http://www.physics.uwo.ca/science_olympics/events/puzzles/fermi_questions.html



Question 12/03

Question 12/03

THREE-PHASE CURRENTS

Why do power companies produce three-phase currents. What are the advantages over single-phase, two-phase, four-phase,... currents?

This question was suggested by D. J. Bergman.





Answer 12/03


THREE-PHASE CURRENTS

The question was:

Why do power companies produce three-phase currents. What are the advantages over single-phase, two-phase, four-phase,... currents?

Solution:

Since there are numerous aspects to this problem we will publish partial answers as they arrive. (Credits to people that provided that particular answer will be given at the bottom of this page.)

1. If the number of phases N is greater or equal 3, then the total instantaneous power is constant, i.e. the generator runs under a constant load. (This assumes that all phases are equally loaded.) To prove that we note that instantenous load (up to a numerical prefactor) is I=SUMn=0

N-1cos2(wt+2{pi}n/N)

Note that dI/dt=2 SUMn=0

N-1cos(wt+2{pi}n/N)sin(wt+2{pi}n/N)=

SUMn=0N-1sin(2wt+4{pi}n/N)=0.

for N equal to or greater than 3.

2. If the potential difference (amplitude) between 0 and one of the "phase wires" is Vo, then for 2-phase

current the potential difference between the wires of two phases will be 2Vo. For a larger number of

phases (N=3,4,...) the potential difference between two phases depends on which phase we choose. For the case of even N there will be pairs of phases that have potential difference 2Vo. For N=3, the potential

difference between any pair will be 31/2Vo. For other even Ns, there will always be a pair that has a

potential difference larger than 31/2Vo! In fact for large N the maximal potential difference will always

be close to 2Vo. Thus, in the three-phase case we have the smallest maximal potential difference between the wires. This enables, to use less insulation when the wires are close together. Only, three-phase system has this advantage!

3. It should be noted that 3-phase motors (as opposed to motors using 1-phase current) are both simpler, because it is easy to create a "rotating magnetic field" with 3-phases available, and do not need auxillary device to start the motor.


Answer 12/03

Credits: 1. This item was solved (in part) by (28/3/04) by Luca Pappalardo, Mechanical Engineering student at the University of Salerno (Italy) (e-mail [email protected]).

2. This observation was made by D. J. Bergman.

3. This observation was made (29/7/2004) an anonymous correspondent (e-mail [email protected])




mailto:

mailto:



Question 01/02

Question 01/02

EQUIPOTENTIAL SURFACES

Let the equationF(x,y,z)=crepresent a set of nonintersecting surfaces. (Each c corresponds to a different surface.) What is the condition that this is a set of equipotential surfaces? In other words, can we define such potential V(c) that its Laplacian vanishes?





Answer 01/02


EQUIPOTENTIAL SURFACES

The question was:

Let the equationF(x,y,z)=crepresent a set of nonintersecting surfaces. (Each c corresponds to a different surface.) What is the condition that this is a set of equipotential surfaces? In other words, can we define such potential V(c) that its Laplacian vanishes?

(7/2002) The problem has been solved (26/6/2002) by Thomas Garel from Service de Physique Theorique, CE-Saclay, Gif-sur-Yvette (Francei) (e-mail [email protected]); the solution below follows (to large extent) his derivation.

The solution: Derivatives of V with respect to space coordinates, can be taken by first taking the derivative of V with respect to c and then the derivative of c with respect to, say, x. Then grad2V=V''(c)(gradc)2+V'(c)grad2c=0 where grad2 denotes Laplacian, and prime denotes derivative with respect to c. This leads to the condition [grad2c]/[(grad c)2=-V''(c)/V'(c)=G(c) i.e. the ratio on the l.h.s. of the equation must be a function of only c. This is the true and the only requirement of the set of surfaces determined by F. The arbitrary function G(c) determines the potential V: V=A {\int} exp[-{\int}G(c) dc] dc + B where {\int} denotes the integral sign.

The solution and its applications can be found in the book of W. R. Smythe Static and Dynamic Electricity (McGraw-Hill, 1950).



http://www-spht.cea.fr/pisp/garel




Question 02/02

Question 02/02

MAXIMAL GRAVITY

Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?





Answer 02/02


MAXIMAL GRAVITY

The question was:

Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

(5/2002) The problem has been solved (9/2/2002) by Eva Sandlokk (e-mail [email protected]), (18/2/2002) by Kirk T McDonald from Henry Laboratories, Princeton, NJ (e-mail [email protected]) - see his solution in PDF format, as well as by Mattew Horton (1/4/2002) (e-mail [email protected]), and (2/4/2002) Daniel Gulotta from Illinois Mathematics and Science Academy (e-mail [email protected]). While direct analytical solution of the problem is possible, a very interesting approach was suggested (2/4/2002) by Horst Schneider from the Physics Department in Martin-Luther-Universitat Halle-Wittenberg in Germany (e-mail [email protected]); the approach begins with numerical optimization, and eventually arrives to an analytical solution - see files 1 and 2. It was also partially solved (26/3/2002) by Jared Daniel Kaplan from Stanford (e-mail [email protected]).

The answer: The optimal body has axial symmetry and has the shape depicted at the picture below. The green dot denotes the point of maximal free-fall acceleration. The body can be described by relating distance r from the point at which the maximal acceleration is found as a function of the angle {theta} from the axis of symmetry (vertical black line).

r2=R2cos({theta}).

The acceleration at the green point will be g=(4/5)(15/4)1/3{pi}2/3M/V2/3. This is only 2.6% larger than the gravity on the surface of a spherical planet.





http://star.tau.ac.il/QUIZ/02/A0202_KMD.pdf




http://star.tau.ac.il/QUIZ/02/A0202_HS1.ps

http://star.tau.ac.il/QUIZ/02/A0202_HS2.ps


Answer 02/02

The solution: Assume for a moment that we already know that the body should have axial symmetry. Consider a narrow ring of matter lying on the surface of the body. (Center of the ring coincides with the axis of symmetry and its plane is perpendicular to that axis. It is depicted on the figure below.) If the mass of the ring is dM and the distance between the points of the ring and the "green point" is r, then the contribution of the ring to free-fall acceleration will be G*{dM}cos({theta})/r2.


Answer 02/02

If we remove this ring of material from the surface and place it somewhere else on the surface, it will have different r and different angle {theta}. Since the shape is optimal, it is "stationary", i.e. infinitesimal changes should not change the acceleration. This will happen only if cos({theta})/r2 is constant. Consequently, r2=R2cos({theta}). The constant R denotes the maximal diameter of the body. Once the shape of the body is given, we can easily find (by an elementary integration*) that its volume V=(4/15){pi}R3, while the acceleration at point g is given** by the expression which has been presented above.

Using similar argument you can now show that the body must have axial symmetry. Imagine that is not so, i.e. the ring which was described above is not complete - then you can easily see that the free-fall acceleration can be increased by relocating some of the surface material.

For a more detailed solution see PDF format.

Comments:



http://star.tau.ac.il/QUIZ/02/A0202_KMD.pdf



Question 03/02

Question 03/02

SOAP BUBBLE

Calculate the time of disappearance of a soap bubble connected with the atmosphere via a capillary.

This question appears (without solution) in Physics Problems by P.L. Kapitza.





Answer 03/02


SOAP BUBBLE

The question was:

Calculate the time of disappearance of a soap bubble connected with the atmosphere via a capillary.

(6/2002) The problem has been solved (13/5/02) by Alex Smolyanitskiy MS student in in Electrical Engineering, Columbia University (NY) (e-mail [email protected]) (the solution below closely follows his solution), and (22/6/02) by Chetan Mandayam Nayakar (e-mail [email protected]).

The solution: In this solution we assume that the process is sufficiently slow that we can (a) disregard the effects of viscosity of the soap bubble and the times required for liquid in the wall of the bubble to readjust its shape. (b) assume that the process is almost static despite constantly changing volume of the bubble. E.g., below we use Laplace equation and assume that the pressure is constant everywhere inside the bubble. (c) assume that the air flow in the capillary is laminary.

The pressure difference between inside and outside of a spherical bubble is given by the Laplace equation p=4s/r,

where s is the surface tension of soapy water (approximately 0.025N/m), r is the radius of the bubble. In the above equation we introduced an additional prefactor 2 to account for the fact that the bubble has two surfaces. If we disregard the viscosity of the air and assume laminar flow, Bernoulli relation can be used to relate this pressure difference with the flow velocity v inside the capillary.

p=D·v2/2,

where D is the density of the air (1.3 kg/m3). If the cross section area of the capillary is A then we can easily relate the outflow of air via capillary with the decrease in the radius of the bubble

v·A·dt=-4{pi}·r2·dr

The above three equations can be simply combined into a single differential equation, which determines the rate of change of the radius of the bubble:

r5/2·dr=-(A/{pi})·(s/2D)1/2dt

This equation can be directly integrated by assuming that during time T the radius of the bubble changes from its initial value R to zero, leading to




Answer 03/02

T=(2{pi}/7A)·(2D·R/s)1/2·R3

As an example, let us consider a bubble of diameter R=0.05m, connected to a capillary of 1mm radius, i.e. A=3·10-6 m2. By substituting all those numbers into the above result we find T=80 sec.

Alex Smolyanitskiy experimentally verified the above theoretical expression for several values of soap bubble diameters.

This result has been obtained by assuming that air flow inside the capillary has everywhere the same velocity v. A more careful analysis will show the velocity does depend on the distance from the central axis of the capillary. This effect can be effectively taken into account by using smaller (effective) cross section of the capillary.

One can directly estimate the Reynolds number for typical flow velocities predicted by this treatment, and find the for buble radius ranging between 1 and 10 cm, and capillary diameters of few milimeters, the Reynolds number will be between 1 and 10, i.e. low enough to ensure laminar flow that was assumed in this derivation.





Question 04/02

Question 04/02

SCATTERING DIPOLES

A system consists of two ideal dipoles placed at positions (0,0,0) and (0,0,a). Dipole moment p of each dipole is related to electric field E on that dipole via relation p=gE. (g is so small that you can neglect the interaction between the dipoles, i.e. electric field created by one dipole at the position of the other dipole is negligible.) An incoming electromagnetic wave of wavelength L=2a is scattered by the system. Consider two cases: (a) the incoming wave is in x-direction, and (b) the incoming wave is in z-direction. Estimate the ratio between the total scattering cross sections between those two cases. (You may neglect dimensionless prefactors of order unity.)

This problem was given at an exam by Y. Kantor





Answer 04/02


SCATTERING DIPOLES

The question was:

A system consists of two ideal dipoles placed at positions (0,0,0) and (0,0,a). Dipole moment p of each dipole is related to electric field E on that dipole via relation p=gE. (g is so small that you can neglect the interaction between the dipoles, i.e. electric field created by one dipole at the position of the other dipole is negligible.) An incoming electromagnetic wave of wavelength L=2a is scattered by the system. Consider two cases: (a) the incoming wave is in x-direction, and (b) the incoming wave is in z-direction. Estimate the ratio between the total scattering cross sections between those two cases. (You may neglect dimensionless prefactors of order unity.)

(2/2003) We did not receive any solutions of the problem, and decided to publish the following solution and comments.

Solution:

(Comment: All expressions are presented in Gaussian electromagnetic units.)

When electromagnetic field propagates x direction, both dipoles will will feel the same field at the same time, and the total electric dipole of the system will be p=2gE. Since the dipoles are not close to each other, the system will also have higher moments (e.g. quadrupole), and its radiation will not be pure dipole radiation. However, the order of magnitude estimate of dipole radiation will suffice in an estimate of the cross section. We note the the power radiated by a dipole must be proportional to p2, since the electric field must be proportional to charge (and thus to dipole moment), and the power is proportional to squared field. From dimensional considerations, the radiated power must be of order cp2/L4, where c is the speed of light. Substituting, expression for p, and dividing the result by the flux of the incoming wave, we find that the cross section is of order g2/L4. (This expression could be obtained directly from dimensional analysis.)

When electromagnetic field propagates z direction, both dipoles will feel field pointing in opposite directions, and consequently the total dipole moment will vanish. However, one can easily convince himself, that non-vanishing elements of quadrupole moment are of order Q=pa=gaE. From dimensional considerations, we can also establish that the radiation of quadrupole is of order cQ2/L6. Substituting, expression for Q, and dividing the result by the flux of the incoming wave, we find that the cross section is of order g2a2/L6. Since L=2a, this expression is of the same order as cross section obtained in the i previous ("dipole radiation").

Those results really should not be surprising: Once we established that cross section must be


Answer 04/02

proportional to squared charges (and thus to g2, and keeping in mind that the units of g are [length]3, we must divide g2 by something with dimensions [length]4 to obtain a cross section (that has units [length]2). Since a and L are of the same order, we can use one of them to get the correct result g2/L4.





Question 05/02

Question 05/02

LINE OF CHARGES

Consider N identical "charges" placed along a straight line. A pair of charges at positions xi and xj repel each other via potential -ln|xi-xj|. Each charge is attracted to the origin of coordinates by a

parabolic potential xi2/2. Show that the equilibrium positions of the charges are given by the

zeroes of the Hermite polynomial HN.

This problem was brought to our attention by M. Fogler.





Answer 05/02


LINE OF CHARGES

The question was:

Consider N identical "charges" placed along a straight line. A pair of charges at positions xi and xj repel each other via potential -ln|xi-xj|. Each charge is attracted to the origin of coordinates by a

parabolic potential xi2/2. Show that the equilibrium positions of the charges are given by the

zeroes of the Hermite polynomial HN.

(9/02) The problem was originally considered by T.J. Stieltjes, Comp. Rend. Acad. Sci., Paris, 100, 439 and 620 (1885). (It can be found the book: T. J. Stieltjes, Oevres Complés, published by Groningen, P. Noordhoff (1914-18).)

The problem has been solved (22/5/02) by Oded Farago from Materials Research Laboratory in University of California at Santa Barbara (e-mail [email protected]), and (16/8/02) by Markus Walser (e-mail [email protected]) (see his solution in the following postscript file), as well as (2/9/02) by Armin Rahmani (e-mail [email protected]).

A very useful ('classical') reference on properties of orthogonal polynomials is G. Szegö Orthogonal Polynomials, American Math. Soc. Colloquium Publications, vol. XXIII (1959). Some recent developments in the area of "electrostatic analogies" of orthogonal poynomials can be found in the paper of M.E.H. Ismail, Pacific J. Math., 193, 355 (2000) (here is a copy in PDF format). It also contains many useful references.

The solution:

Hermite polynomial HN(x) of order N satisfies differential equation

HN"- 2 x HN'+2 N HN=0,

where ' and " denote the first an second derivatives with respect to x. If x1 is one the zeroes of the

polynomial then we immediately see from the above equation that

HN"(x1)/[2HN'(x1)= 1/x1.

Now consider an arbitrary polynomial G(x) of order N-1, with zeroes at x2, x3,... , xN, and an additional

function F(x)=(x-x1)G(x). For such polynomials it is true that

1/(x1-x2)+ 1/(x1-x3)+...+ 1/(x1-xN)=G'(x1)/G(x1)= F"(x1)/2F'(x1).




http://star.tau.ac.il/QUIZ/02/walser.ps


http://star.tau.ac.il/QUIZ/02/Ismail_PJM.pdf

Answer 05/02

If function F happens to be HN(x), and xi are the zeroes of the polynomial, then the above relation will

have the form

1/(x1-x2)+ 1/(x1-x3)+...+ 1/(x1-xN)=1/(2x1).

Obviously, x1 in the above relation can denote any zero of the polynomial. However, the above relation

is exactly the equilibrium condition of a "charge" at position x1, that is attracted to the origin via

potential x12, and is repelled by any other charges by potential -Sumi=2,Nln|x1-xi|. The equilibrium

position is determined by equating the derivative of the potential with respect to x1 and equating the

result to zero. This coincides with the equation above!

Thus we proved that the condition of equilibrium of each charge coincides with the condition satisfied by the zero of Hermite polynomial.





Question 06/02

Question 06/02

ISOTROPIC RADIATOR

Is it possible to build and antenna that will radiate coherent electromagnetic radiation with intensity independent of direction?





Discussion 06/02


ISOTROPIC RADIATOR

The question was:


(8/02) Y. Kantor: We got simultaneously (2/8/2002) two almost identical remarks from Jared Kaplan from Stanford University, US (e-mail [email protected]), and from Ben Zickel from Rafael, Israel (e-mail [email protected]). Both comments point out that according to a topological theorem, frequently called as "the hairy ball theorem" it is impossible to cover the surface of a sphere with vectors of fixed size but continuously varying direction. At some point, such coverage must have discontinuity, i.e. jump in the direction of the vector field.

So, if we would attempt to create an antenna that produces radiation such that on a remote sphere we have fields oscillating with the same phase, this would violate the above theorem, since configuration of the fields at the moment of their (simultaneous) maximum would violate the theorem. Notice, however, that the problem required coherency, which implies fixed phase differences between various parts of the field, but not identical phases.







Answer 06/02


ISOTROPIC RADIATOR

The question was:


(12/02) It was definitely a difficult problem, and it seems that no solutions can be expected. Therefore, we publish the known solution of the problem. This problem was originally considered and solved by H. Matzner, M. Milgrom and S. Shtrikman in their article Magnetoelectric symmetry and electromagnetic radiation which was published in Ferroelectrics, 161, 213-219 (1994).

The solution:

A direct calculation of the vector potential, electric field and radiation intensity produced by the following current shows that it produces direction independent radiation intensity. The current density is:

J(x,y,z)=-D(x)D(y)D(z+1/8)x+D'(x)D(y)cos({pi}/4+2{pi}z)z

for -1/8 < z < 1/8, where all distances are measured in wavelengths, D denotes Dirac delta-function, and D' denotes its derivative. Such current distribution is produced by U-shaped antenna (depicted in the following figure) when h-> 0. For a detailed calculation see the paper mentioned above.


Answer 06/02

It should be noted that the polarization of the radiation emitted by such an antenna depends on the direction and changes from linear to elliptic to circular (right and left polarized) depending on the direction.





Question 07/02

Question 07/02

CURRENT IN A WIRE

In freshman physics we are told that a current in an "infinite" cylindrical wire flows with constant density J independent of this distance from the center of the wire. But we also learn how to use Ampere's Law to calculate the magnetic field in the wire. Doesn't the magnetic field act to make the current density non-uniform?

This problem has been suggested by B. Svetitsky.





Answer 07/02


CURRENT IN A WIRE

The question was:

In freshman physics we are told that a current in an "infinite" cylindrical wire flows with constant density J independent of this distance from th e center of the wire. But we also learn how to use Ampere's Law to calculate the magnetic field in the wire. Doesn't the magnetic field act to make the current density non-uniform?

(3/2003) The problem has been solved (15/7/02) by Jared Kaplan from Stanford University, California (e-mail [email protected]) (below we present his (slightly edited) solution), (18/11/02) by Eric T. Lane from Physics Department, University of Tennessee at Chattanooga (e-mail [email protected]), and by (4/2/03)Srikant Marakani a graduate student the University of Chicago (e-mail [email protected]). On 24/8/02 Regis Lachaume from UJF, Grenoble (France) ((e-mail [email protected]) produced of very simple estimate demonstrating that density variation must be a very small number (see PDF file).

The solution: [This is an edited solution of J. Kaplan. It is presented in MKSA (SI) unit system.] Simplifying assumptions: -Current is carried exclusively by moving, positive charges, all moving with the same speed -Negative charges are stationary and uniformly distributed in the wire -the wire is cylindrical, and current/charge density is a function of the radius only

Let d=density of negative charge, r=radius, e=permitivity, u=permeability, p(r)=density of moving positive charge as a function of r, and v=velocity of positive charge. Thus the current density = p(r)*v. Then we have: E from negative charge E-=d*r/(2e)

E from positive charge, E+= Integral(p(x)*x*dx,0,r)/(e*r)

B = uv*Integral(p(x)*x*dx,0,r)/r

Using the integral relations for E and B with the symmetry of the setup. In equilibrium, the force on the positive charge should be zero. We allow a force on the negative charge for the moment, because we are assuming it's rigidly held in place. F on + charge = 0 =(E-+E+)*p(r) + v*p(r)*B

with the convention that a force outwards has positive sign, we get: I(p(x)*x*dx,0,r) = d*r2/(2e*(1/(e) - u*v2))

An appropriate p(r) is evident at this point. Formally, we can differentiate with respect to r and then







http://star.tau.ac.il/QUIZ/02/Lachaume.pdf

Answer 07/02

solve, yielding: p(r) = d/(1 - e*u*v2) Although the precision of the model is questionable, we are led to the result that p(r) is independent of r, thus the current should be uniformly distributed. It is interesting that we sort of see the necessity of special relativity since e*u = 1/c2 means that p(r) = d/(1-(v/c)2), meaning that the total positive charge does not equal the total negative charge exactly. Note that if the positive charge was stationary while the negative charge moved the model would give the same prediction.

Srikant Marakani wrote:

The solution is that the current density is in fact in equilibrium when it is uniform despite the magnetic field in the wire which is proportion to the radius from Ampere's law. Due to the transformation of the current density four vector (or, if you prefer it, the Lorentz contraction) the charge density of the carriers is increased giving rise to an electric field which also increases radially. It is not difficult to see from Maxwell's equations that E=-vxB where v is the drift velocity of the charges and there is no net force on the charge carriers. Another way to solve it is to look at two fields, both for uniformly charged cylinders of equal but opposite charge density. Lorentz boost one of those fields and then superpose them with the other to get the full field. We start with only electric fields before the boost but end up with both electric and magnetic fields after the boost and superposition which have the required form. [...] Finally, the best argument (though I am not quite sure it is watertight) I can think of is to use Lorentz invariance. We transform to the reference frame where the background charges are moving with velocity -v so that the original charge carriers are now stationary and the other way around. If there is any change in the radial charge density (except due to the Lorentz contraction) of charge carriers in the original frame, then it should occur for the background charges as well since in the new frame, they are the charge carriers. If the overall mechanical stress is zero, there is no electrical force that will make both the background and carriers move in the same way, so there should be no variation in the current density.

Eric T. Lane also remarked that the presence of confining wire is essential for obtaining the solution (described above). However, "in a plasma, moving charge sets up a magnetic field that generates an attraction among the parts of the current, giving rise to the magnetic pinch effect."





Question 08/02

Question 08/02

ELECTROSTATIC MACHINE

Electrostatic machine depicted in the figure consists of an insulated metallic spherical container into which drops charged with a certain potential are dripping from a faucet above. Determine the dependence of the maximal potential of the spherical container on the height from which the drops are falling.






Answer 08/02


ELECTROSTATIC MACHINE

The question was:

Electrostatic machine depicted in the figure consists of an insulated metallic spherical container into which drops charged with a certain potential are dripping from a faucet above. Determine the dependence of the maximal potential of the spherical container on the height from which the drops are falling.

(4/2004) The problem has been solved (16/4/04) by Sebastian Popescu (e-mail [email protected]) from Complex Systems Group in the Faculty of Physics of "Al. I. Cuza" University, Iasi, Romania. The solution can be found here.




http://star.tau.ac.il/QUIZ/02/KelvinGenerator.pdf



Question 09/02

Question 09/02

POLYMER AT A WALL

One of the simplest idealizations of a flexible polymer chain consists of replacing it by a random walk on a cubic lattice in three-dimensional space or on a square lattice in two-dimensional space. (This walk is allowed to self-intersect, and it has no "bending energy", i.e. each step is independent of the previous one.) An example of such "polymer" is depicted by a red line in Fig. (a). Assume that one end of the polymer is tied to a surface (denoted by brown color in the Figure). The surface will be considered adsorbing, i.e. every time the polymer (or the random walk) touches the surface its energy decreases by V. E.g., the energy of the configuration in Fig. (a) is -3V, because the walk touches the surface three times.

A very long polymer will be either localized near the surface as depicted in Fig. (c), or will be delocalized as in Fig. (b), depending on the temperature T. (Figs. (b) and (c) have different scale from Fig. (a) and therefore the lattice is not shown.)

http://star.tau.ac.il/QUIZ/02/Q09.02.html (1 of 2)11.7.2005 9:21:46

Question 09/02

Describe the temperature-dependence of the "localization" of the polymer.





Discussion 09/02


POLYMER AT A WALL RADIATOR

The question was:




Discussion 09/02


(8/03) Y. Kantor: Nenad Vukmirovic, an undergraduate student at the University of Belgrade (Serbia, Yugoslavia) (e-mail [email protected]) submitted (6/8/03) the following numerical solution of the problem. The numerical solution has several problems ((a) the method is not proper for low temperatures,- a true Monte Carlo technique would be a proper replacement of the numerical methods in the proposed solution; (b) the definition of the localization width through the maximal separation of the polymer from the wall is somewhat problematic, because it considers the extreme rather than a typical point). Nevertheless, the numerical solution provides us with a qualitative picture of the transition, and an order of magnitude estimate of the transition point - the critical T is of order of 0.1V/kB.

We are waiting for more accurate solutions.




http://star.tau.ac.il/QUIZ/02/dl.ps



Answer 09/02


POLYMER AT A WALL

The question was:




Answer 09/02


(11/2003) The problem has been solved (6/10/03) by Yoram Burak from Tel Aviv University, Israel (e-mail [email protected]) (see his solution in PDF format), and (17/10/03) by Oleg Semenov (Moscow, Russia) (e-mail [email protected]) (see his solution in PDF format). The solutions approach the problem from slightly different angles, but they both arrive at the same answer: the critical temperature is V/akB, where the numerical constant a=ln(4/3)=0.29.




http://star.tau.ac.il/QUIZ/02/polymer.pdf


http://star.tau.ac.il/QUIZ/02/polymer1.pdf



Question 10/02

Question 10/02

LEAST ACTION

In analytical mechanics Newton's laws are reformulated as Hamilton's Principle of Least Action. The actual derivation of Newton's laws (via the Euler-Lagrange equations) requires only that the classical path be a stationary point of the action functional. But the term "Least Action" seems to imply that the path must be a minimum of the functional.

Can you think of examples where the classical path is a local maximum of the action? If not, how about a saddle point?

This question has been suggested by B. Svetitsky.





Answer 10/02


LEAST ACTION

The question was:

In analytical mechanics Newton's laws are reformulated as Hamilton's Principle of Least Action. The actual derivation of Newton's laws (via the Euler-Lagrange equations) requires only that the classical path be a stationary point of the action functional. But the term "Least Action" seems to imply that the path must be a minimum of the functional.

Can you think of examples where the classical path is a local maximum of the action? If not, how about a saddle point?

(1/2003) On 29/12/2002 Gleb Gribakin from the Department of Applied Mathematics and Theoretical Physics Queen's University Belfast, U.K. (email [email protected]) suggested the following example:

Suppose the particle is confined to the surface of a sphere and the potential energy on this surface is constant, i.e. the particle moves freely on the surface. The path of the particle then lies on a circle whose diameter is equal to that of the sphere. For any two points on such a circle that are not diametrically opposite, say 1 and 2, there are two possible paths which start at 1 and end at 2. The shorter arc is a local (and global) minimum of the action integral. The longer one must be a saddle point. A variation of the path, which keeps the midpoint of the longer arc fixed, increases the action. On the other hand, if we allow the midpoint to move in the direction perpendicular to the original long arc, the action integral can be made smaller (together with the length of the path).

Benjamin Svetitsky who originally suggested this problem offered an additional example of saddle point, which can be found in the following file.

David Augier (email [email protected] suggested (28/1/03) the following example in which and extremum (in Fermat's principle of least time in optics) does not correspond to minimum:

Consider an ellipsoidal (concave) mirror. As everybody know, its two foci, A and B, verify AM+BM=constant for every M on the surface of the ellipsoid. As a consequence, every AMB is a light ray. Let M be a certain point on the surface of the ellipsoid. Now, forget this ellipsoidal mirror (just keep its shape in mind) and consider a concave mirror, which is tangent to the ellipsoid at M, and strictly included inside the ellipsoid (except M of course). As this new mirror is tangent to the ellipsoid, AMB is still a light ray (Snell-Descartes laws are still valid). However, every ANB (N belongs to the surface of the concave miror), with N quite near to M, is clearly smaller than AMB, and is not a light ray (still because of Snell-Descartes laws). AMB is a local maximum.



http://star.tau.ac.il/QUIZ/02/least_action.pdf


Answer 10/02

If the concave mirror is stricly included outside the ellipsoid, the light ray is a local minimum. And if you consider a mirror with a shape like x3 (still tangent to the ellipsoid), a part of the mirror is inside the ellipsoid, and another part is outside. Then AMB is only stationnary (saddle point).

Y. Kantor: We are waiting for more examples and discussions.





Question 11/02

Question 11/02

INVISCID MOTION

A solid sphere of radius R and density d is suspended on a spring with spring constant k in an inviscid liquid of (smaller) density do. The spring is slightly stretched and the sphere starts small oscillations. Will those oscillations ever stop? Why?





Question 12/02

Question 12/02

CHARGED DROP

The shape of a freely suspended liquid drop is kept spherical (with radius R) by the surface tension g. [For simplicity we assume weightlessness.] Assume that the liquid is conducting and it is being gradually charged. What will happen as the charge Q increases?





Discussion 12/02


CHARGED DROP

The question was:

The shape of a freely suspended liquid drop is kept spherical (with radius R) by the surface tension g. [For simplicity we assume weightlessness.] Assume that the liquid is conducting and it is being gradually charged. What will happen as the charge Q increases?

(6/05) Y. Kantor: Finally we have some progress on the problem. We got (2/8/2002) a suggested solution from J.I.I. de la Torre (e-mail [email protected]). (His solution can be found in the following PDF file.) He compares the total energy (electrostatic energy+surface energy) for two cases: a single spherical drop, and two spherical drops with total volume equal to the volume of the original (single) drop. He finds that when the charge Q exceeds certain value that is defined by the volume of the original drop and the surface tension, the system would prefer to be split into two spherical drops.

What does it mean? It means that single drop solution is NOT a global minimum if the charge is too large. At this point, however, it is not clear whether for even smaller charges one can find geometrical configurations that have a lower energy than a single drop. Moreover, it has been shown that for certain charges the single drop geometry is not a global minimum. But is it a local minimum? I.e., is the spherical configuration locally stable?

We are waiting for additional attempts to solve the problem.




http://star.tau.ac.il/QUIZ/02/ChargedDrop_part.pdf



Question 01/01

Question 01/01

FLOATING SPHERE

A sphere of uniform density floating in a glass of water (left figure) can be at any distance from the axis of the glass. Where will be the sphere located if the glass is rotating with a constant angular velocity (right figure)?





Answer 01/01


FLOATING SPHERE

The question was:

A sphere of uniform density floating in a glass of water (left figure) can be at any distance from the axis of the glass. Where will be the sphere located if the glass is rotating with a constant angular velocity (right figure)?

(9/01) We received a surprisingly large number of wrong solutions of the problem. Indeed it is a bit more complicated than appears at the first glance. So we decided to publish the solution without further delays... A nice solution of this problem can be found in G.K. Batchelor, An Introduction to Fluid Dynamics, par.1.4 (Cambridge University Press, 1967 (and later)).

Answer: This sphere will float at the center of the surface (i.e. on the axis of symmetry/rotation).

The solution:

The necessary condition for equilibrium of a fluid is the equilibrium between the external ("long-range")


Answer 01/01

force density F and the gradient of the pressure P:

F=grad P.

If the long-range force is be expressed as a product of density d and a (minus) gradient of some potential U then:

-d grad U=grad P.

By taking curl of both sides we get:

(grad d) x (grad U)=0.

Thus levels of constant density and constant potential coincide. In a reference frame rotating with angular velocity w the potential U is given by:

U=gz-(1/2)w2(x2+y2),

where the second term accounts for the centrifugal force. Thus, the equipotential surfaces, and consequently also the surfaces of constant density are paraboloids of rotation, as depicted in the following figure:

Of course, the surface of the liquid is also a paraboloid of rotation. Now, look at the sphere and at the liquid which it displaced. Imagine that the sphere is not there and its place is filled with a liquid. This liquid would be at equilibrium. However, it would NOT be of constant density - it's density would be slightly higher on the right Now if we place a uniform density sphere it's mass distribution is closer to the axis of rotation, and consequently the centrifugal force is smaller that the force from the liquid it expelled (which is equal to the centrifugal force that would apply to the liquid if it would be there), and therefore it would NOT be at equilibrium, but would be driven towards the center. Consequently, at the


Answer 01/01

equilibrium the sphere would float in the center of the surface.





Question 02/01

Question 02/01

LIMITED PENDULUM

Period of a pendulum depends only on its length only for very small oscillations. For large oscilations the period depends on the amplitude. Such amplitude-dependence can be eliminated by making the string of the pendulum (shown in red) to rap around a limiting curve (shown in blue). What is the shape of this curve?

p.s. Huygens tried to use such curve in a clock, but the idea created technical problems and was eventually abandoned.





Answer 02/01


LIMITED PENDULUM

The question was:

Period of a pendulum depends only on its length only for very small oscillations. For large oscillations the period depends on the amplitude. Such amplitude-dependence can be eliminated by making the string of the pendulum (shown in red) to rap around a limiting curve (shown in blue). What is the shape of this curve?

(6/2001) An elegant and short solution of the problem has been submitted (28/4/2001) by Szabolcs Galambosi a Ph.D. student at Department of Physics at University of Helsinki (e-mail [email protected])). Derivation of the equation of motion below, closely follows his solution. [Arvind Murugan from Caltech (e-mail [email protected]) pointed out (11/2/2001) that this is a well known problem solved by Huygens and it initiated the study of involutes and evolutes. We received some correct answers but no detailed solutions from several people (Itzhak Shapir (e-mail [email protected]), Chethan Krishnan (e-mail [email protected]))]

Answer: The curve is cycloid.

The solution:

DEFINITIONS







Answer 02/01

Cycloid (ordinary cycloid) is the locus of a point of a circle that rolls along the straight line.

Thus, if the blue circle in the above picture rolls along the horizontal line and in its leftmost position a point has been marked on its lowest part, then that point will trace the red cycloid. (Cycloid was first studied by Cusa, and later was properly defined by Mersenne in 1599. Galileo also studied this curve and gave it its name. Roberval found that the area below the cycloid is 3{pi}R2, where R is the radius of the circle.) The equation of cycloid is:

x=R[f-sin(f)]y=R[1-cos(f)]

where f is the angle by which the generating circle has been rotated.

Evolute is a curve defined by the loci of the centers of curvature of a given curve.

Involute is a curve traced by a point on straight line rolling without slipping along a given curve. Involution is the inverse operation of evolution.

There are two curves (cycloid and logarithmic spiral) whose involutes are exactly the same curves. Evolute of the red cycloid above is the green cycloid, and, consequently, the red curve is the evolute of the green curve! We do not prove these properties (they can be found in books of analytical geometry). We will use those properties below.

PHYSICS

Now consider the above picture inverted (upside down), so that the heigh h=-y. A bead of mass m


Answer 02/01

sliding along a cycloidal curve (the inverted red curve) will have potential energy change (relative to initial height ho): mg(h-ho). This decrease in potential energy should be equal the kinetic energy:

(1/2)mv2=(1/2)m[(dx/dt)2+(dy/dt)2 =mR2[1-cos(f)](df/dt)2

From the equality between the loss of potential energy and the kinetic energy we find:

dt=[(R/g)(1-cos(f))/(cos(fo)-cos(f))]1/2df

where fo is the value of the angle (of the generating circle) corresponding to ho. This equation can be

integrated from f=fo to f={pi} to give the time of quarter period (time from uppermost point to

lowermost point). Direct integration shows that this integral is independent of fo!!! The oscillation

period comes out to be identical with the period of of small-amplitude oscillations of regular pendulum of length 2R. Thus the period of the bead sliding on an (inverted) cycloid is independent of the amplitude.

Szabolcs Galambosi suggested an alternative proof of time independent behavior. He suggested to form Lagrangian L=T-U from the kinetic energy T and the potential energy U that have been calculated above. Everything in this Lagrangian will be expressed in terms of f and its first time derivative f'. The resulting Euler-Lagrange equation will be:

2f"Rcos(f/2)-sin(f/2)f'2+g sin(f/2)=0,

which after a substitution a=Rsin(f/2) becomes

a"+(g/2R)a=0,

which is a simple harmonic equation for a, corresponding to a small-oscillations-pendulum of length 2R.

Now consider a bead hanging on string of length 2R. If its swinging is limited by the green cycloid (see picture above - inverted) then the bead will trace out a curve which is involute of the green cycloid, i.e. it will move along the red cycloid. Consequently, the period of such pendulum will be independent of the amplitude.

This property (that is called tautochrone or isochrone) was discovered by Huygens in his Horologium Oscillatorium in 1673.

(4/2004): A very straightforward and short solution of the problem has been suggested (16/4/04) by Sebastian Popescu (e-mail [email protected]) from Complex Systems Group in the Faculty of Physics of "Al. I. Cuza" University, Iasi, Romania. The solution can be found here.

(12/2004): A very interesting additiona has been provided (17/12/04) by Ioannis Florakis (e-mail



http://star.tau.ac.il/QUIZ/01/cycloidal_pendulum.pdf

Answer 02/01

[email protected]) from University of Athens, Greece. He has shown that harmonic oscillator (in one dimension) and cycloid (in two dimensions) are the only possible symmetric isochrone solutions of the problem. His discussion of the problem can be found here.




http://star.tau.ac.il/QUIZ/01/Isochrone2.pdf



Question 03/01

Question 03/01

INCANDESCENT LIGHT

The "usual" incandescent light bulb operates in AC current. What is the size of the temperature oscillations of the filament of the bulb.






Answer 03/01


INCANDESCENT LIGHT

The question was:

The "usual" incandescent light bulb operates in AC current. What is the size of the temperature oscillations of the filament of the bulb.

(5/02) The problem was solved (3/4/2002) by Regis Lachaume a Ph.D. student at Grenoble Observatory (France) (e-mail [email protected])). His solution can be found in the PDF file. A more "empirical" approach has been suggested (5/5/2002) by Szabolcs Galambosi from Department of Physics of University of Helsinki (Finland) (e-mail [email protected])).

Answer: The temperature fluctuations will be about 80 K.

The solution:

A filament of a light bulb is formed by drawing tungsten metal into a very fine wire. The following figures show the close-up view of filament, with horizontal bar indicating the dimensions. Typical diameter of the wire is few tens of microns.

Relative ammount of radiation emitted at different wavelengths depend on the temperature of the body. The following graph shows the relative amounts of radiation at various wavelengths depending on the temperature of the filament. Notice, that even at the highest temperature depicted in this graph, most of the intensity is wasted in the invisible (heating) range of infra-red and larger wavelengths. Usually the tungsten light bulbs operate at 2800K temperature. From the point of view of producing visible light, even higher temperatures are needed, but at higher temperatures the filament evaporates too fast.



http://star.tau.ac.il/QUIZ/01/bulb.pdf


Answer 03/01

For the purpose of our calculation we will consider a L=0.7 meter long circular wire, D=30 microns in diameter. At (r.m.s.) voltage V=220 Volts, the power dissipated by such wire is P=V2/R=V2{pi}D2/[4*L*{rho}=60 Watts, where the resistance R has been calculated using specific resistance {rho}=8*10-7 Ohm*m at the temperature To=2800K.

We note that the power radiated by such a wire at this temperature is given by the Stefan-Boltzmann radiation law Pr=p*S*sB*To

4=60 Watts,

where we used tungsten emissivity p=0.28, and the Stefan-Boltzmann constant sB=5.64*10-8 Watt/

[m2*K4], and the surface area S was calculated from the dimensions of the cylinder. [The fact that P=Pr

indicates consistency of our choice of dimensions, although it should not be taken too literally, because the emissivity of the filament maybe actually smaller due to its geometry, and some of the heat is lost via heat conduction.] If we increase the length of the filament by factor b, and its diameter by factor b2, then the resistance will decrease by factor b3, while the power will increase by factor b3. Simultaneously, the surface area of the filament (to which the emission of light is proportional) will also increase by factor b3.

Let us assume that the only source of energy loss is radiation. Then, the change in the heat content (given by the l.h.s. of the equation below) will be equal to the difference between the heat produced by the current (first term on the r.h.s) and radiated heat (second term on r.h.s.).

C(dT/dt)=2(V2/R)cos2({omega}t)-p*S*sB*T4

where the heat capacity of the wire C can be calculate by multiplying the volume of the wire, by the mass density of tungsten 19.3*103kg/m3, and by specific heat 133 J/[kg*K] of tungsten. By averaging the above equation over time we can define the mean temperature of the filament To via relation


Answer 03/01

V2/R=p*S*sB*To4

If we assume that temperature fluctuations are very small, then we can neglect the fact that the second term in the r.h.s. includes temperature that is time dependent, and replace T by To. The equation, then

becomes easily solvable, and its solution has the form

T=To+G*sin(2{omega}t),

where G is the amplitude of temperature fluctuations. Its value can be found by direct substitution into the equation, leading to

G=To/[2{omega}to], where to=C/p*S*sB*T03.

By substituting, the values of the parameters mentioned above we find that the "timescale" to=0.056 sec,

and 2{omega}to=35. [We used the European current frequency of 50Hz.] Consequently, temperature

fluctuation G=2800/35=80K.

Both Lachaume and Galambosi suggested an alternative approach to the problem. The approach began by an estimate of the time it takes to a bulb to cool off (assuming that it is approximately exponential decay in temperature). Galambosi convinced himself that the time is of order of 1 sec, since the afterglow of a lamp can be seen about that amount of time. From there he assume, that during quarter-period of oscillation of current (0.005 sec) the temperature will drop by 0.5%, i.e. by about 10K. Lachaumeme used similar argument, with much shorter estimate of decay time and concluded that the temperature oscillation should be larger than 50K. Note that our "timescale" in the above solution was 0.056 sec - a bit shorter than these intuitive estimates. Maybe somebody can actually perform this measurement, and tell us what is the experimental answer?!





Question 04/01

Question 04/01

BALL IN A BOX

A hollow cylinder, with its both ends closed, is filled with a fluid and is at rest in the space. Inside the cylinder there is a small hard ball with a density equal to the density of the fluid. The ball is initially at rest and is close to the center of one of the lids (let's call it a "front lid"). The cylinder suddenly gains an acceleration a and then moves with that constant acceleration (the motion is non-relativistic although the magnitude of the acceleration can be large). The direction of the acceleration is along the axis of the cylinder pointing from the "rear lid" to the "front lid". If viewed from the reference frame of the cylinder at the very first moment after the acceleration appears the ball will, of course, start gaining speed in the direction toward the "rear lid". The question is: will the ball hit the rear lid?

This question was created by Princeton Ph.D. students Akakii Melikidze, Ilya Nemenman and Lesik Motrunich.





Discussion 04/01


BALL IN A BOX

The question was:


(17/8/01) Y. Kantor: We are getting many replies in which people attempt to deduce the position of the sphere from the assumption that the density of the liquid does not change under acceleration and consequently the sphere will always be at a neutral buoyancy. Clearly, such assumption of absolutely incompressible fluid only leads to a conclusion of completely "indifferent" equilibrium, and cannot serve as a guide of the final position of the sphere. If, however, the fluid is even slightly compressible, and the initial temperatures and pressure were such that the acceleration will not cause a phase separation (into, say, liquid and gaseous states) we may assume that the density will be lower at the front lid and higher at the back lid. If the changes in density are not large (that depends on the compressibility and the acceleration), then we can assume that we will have an almost linear profile of density along the axis of the cylinder. This should lead us to the conclusion regarding the final position of the sphere. However, one needs to do more careful dynamical analysis to decide what will happen before the sphere settles into its final position.





Answer 04/01


BALL IN A BOX

The question was:


(1/03) Armin Rahmani (5/9/02) (e-mail [email protected]) supplied a partial solution. We also got many qualitative solutions of the problem. However, the closest to a complete quantitative solution came (25/11/04) Eduardo Aoun Tannuri from University of Sao Paulo, Brazil (e-mail [email protected]) that supplied a rather detailed answer that can be found in PDF file.

Answer: The the rear lid will not be reached. However, the solution of the problem is much more interesting that the final answer, because one needs to consider many interesting effects. Not all effects have been considered and we are waiting for additional emails on the subject.

The solution:

On the most elementary level the solution appears to be very simple. If we for a moment assume that as a result of acceleration the fluid density has settled into a linear profile, with the original density maintained in the center of the cylinder, then the difference between the apparent "weight" ma of the ball and the Archimedes force will create a force that increases linearly in distance from the center. Thus in the absence of friction we have a "harmonic oscillator". If the ball start at the front lid, then the amplitude of its oscillations will be half-length of the cylinder, and therefore it will reach the rear lid after half-period of its oscillation. If some friction (fluid viscosity) is introduced, the ball will not reach the rear lid, and its oscillations around the center will be damped. Eventually, it will settle at the center of the cylinder. However, the problem is slightly more difficult: it is even not trivial to identify all kinds of forces that act on the ball.

The solution of Tannuri addresses most of the aspects of the problem: when the ball is moving it is experiencing the "inertia" force (due to acceleration of the reference frame - the box), the force due to




http://star.tau.ac.il/QUIZ/01/bib.pdf

Answer 04/01

acceleration of displaced fluid (this would be present even in the absence of viscosity), the Archimedes force, and small viscous forces. One point still needs to be clarified: How does the liquid go from the initial uniform density state to the final state where the density is a linear function of the position? It is clear that a wave will be created in the beginning of acceleration. That wave will be damped by viscosity, How long will it take? Is that time negligible compared to time scale over which the ball performs its motion?





Question 05/01

Question 05/01

DIPOLE IN A SHELL

A dipole of size p has been placed inside a conducting grounded spherical shell or radius R at a distance a from the center of the shell (a < R). The dipole points in radial direction. (The dimensions of the dipole are small enough, so it can be treated as an ideal dipole.) Find the electrical potential inside the shell.

This question appeared in an Analytical Electromagnetism exam given at Tel Aviv University by Y. Kantor in 2001.





Answer 06/01


DIPOLE IN A SHELL

The question was:

A dipole of size p has been placed inside a conducting grounded spherical shell or radius R at a distance a from the center of the shell (a < R). The dipole points in radial direction. (The dimensions of the dipole are small enough, so it can be treated as an ideal dipole.) Find the electrical potential inside the shell.

(7/2001) The problem has been solved by Gerome Gosset (16/5/2001) (e-mail [email protected]).

Answer: The potential is a sum of potentials created by the dipole p described in the problem, and an image dipole P=p(R/a)3 located at distance l=R2/a from the center, and a charge (monopole) Q=pR/a2 located at the same position as the image dipole. (p and P have the same direction.)

The solution:

This question can be solved using the "method of images", i.e. by considering a potential created by charges inside the shell and imaginary charges outside the shell. (Since we are only interested in solution




Answer 06/01

for the potential inside the shell, the charges outside the shell can be added at will and adjusted in such a way that correct boundary conditions are satisfied at the boundary of the shell.)

It is quite difficult, to find proper images for a dipole sitting inside the shell. Therefore we shall start with a simpler problem. If a charge q (red) is placed at distance a from the center of the sphere then (as is well known from elementary textbooks), we can find an image charge q' (green) at distance l from the center of the shell such that potential created by both charges vanishes on the shell. To determine the values of q' and l it is enough to write down the potential at two points of the shell (the closest, and the most remote to the charge):

q/(R-a)+q'/(l-R)=0; q/(R+a)+q'/(l+R)=0;

By solving these equations we immediately find that:

q'=-qR/a; l=R2/a

By substituting these values into a general expression for a potential created by two charges, we can easily verify that the potential vanishes everywhere on the spherical shell.

Now consider two charges, q and -q, at positions a and a-d, inside the shell when d is small and qd=p. The images of these two charges will be q', mentioned above and q"=qR/(a-d) located at l"=R2/(a-d).

Now take the limit of vanishing d, diverging q, while their product (p) is fixed. First we notice that the sum q'+q" does not vanish in that limit but rather approaches Q=pR/a2. Thus we have an image charge Q at position l. In addition, the dipole moment of the images (=q"(l"-l)) approaches P=p(R/a)3. Thus the image of the dipole p consists of a dipole and a charge located at distance l from the center.





Question 06/01

Question 06/01

MINIMAL CONTAINER

A given amount of gas is held in a spherical container. For what gas pressure the weight of the container will be minimal?






Answer 06/01


MINIMAL CONTAINER

The question was:

A given amount of gas is held in a spherical container. For what gas pressure the weight of the container will be minimal?

(7/01) The problem has been answered correctly by Fred Goesmann (6/6/2001) from Max-Planck-Institut fuer Aeronomie in Katlenburg-Lindau (Germany) (e-mail [email protected]) - see remarks at the bottom of the page, and by Larry Weinstein (5/7/2001) from Old Dominion University (Norfolk, VA, USA) (e-mail [email protected])

Answer: The weight of the container is almost independent of gas pressure.

The solution:

The weight of the container is proportional to HR2, where H is the thickness of the wall and R is the radius of the sphere. (This assumes that the thickness of the walls is much smaller than the radius of the sphere.) The thickness should be such that the container does not rupture under pressure p. If the critical stress that can be sustained by the metal from which the container is made is S, then the critical thickness is proportional to (p/S)R. This result can be obtained from a direct analysis of the forces, since the stress in the sphere must create a resulting force opposing the internal pressure.* Alternatively, the same answer can be obtained from pure dimensional analysis: Abviously, the H must be proportional to pressure since all the forces are proportional to it; the only combination of pressure, critical stress and radius that has dimensions of length is (p/S)R. Since the pressure is inversely proportional to the volume of the gas (i.e. proportional to 1/R3), the critical thickness H is simply proportional to 1/R2. Consequently, the weight of the container is independent of R ! As we saw, the decrease of thickness with increasing radius is exactly compensated by increasing area, and consequently the weight of the container is really independent of the radius of the container, or of the gas pressure. (Of course, if you do not neglect the ratio between the thickness and the radius, then you will discover that larger radia (smaller pressures) are slightly more advantageous.)

* Comment: A simple way to get that relation is as follows: Consider half of the sphere exactly at a critical pressure; the pressure p applies force p{pi}R2 to the hemisphere; it is opposed by the stress S which is applied to the the container walls, which have cross section area 2{pi}RH, creating force 2{pi}RHS. By equating the two forces, you will get the desired relation.




Answer 06/01

Remark: Fred Goesmann wrote us (slightly edited): In our department we are building a gas-chromatograph/mass-spectrometer for space applications and we need to take carrier gas with us. And as always one has to minimise mass, volume, power, everything. A description can be found here. The surprising result is that the mass of a spherical container is independent of the pressure. ... The result holds also for long cylindrical containers.



http://roland.mpae.gwdg.de/cosac/



Question 07/01

Question 07/01

CONDUCTING CHESSBOARD

Small square metal sheets are made from two different metals (green and yellow in the picture) and combined into a single very large "chessboard". The two-dimensional conductivity of one metal is S1 and of the other - is S2. Calculate the effective two-dimensional conductivity of this "chessboard" by following steps:

(a) Show that the electrostatic potential inside every, say, green square is given by the same harmonic function f1(x,y) (up to a constant which needs to be added depending on the detailed position of the square). Similar function g1(x,y) describes yellow squares.

(b) Functions f1(x,y) and g1(x,y) can be treated as real parts of ANALYTIC functions F(x,y)=f1(x,y)

+if2(x,y) and G(x,y)=g1(x,y)+ig2(x,y) of a complex variable z=x+iy. Show that f2(x,y) and g2(x,y) also solve (a different) chessboard conductivity problem.

(c) By comparing the solutions of the conductivity problems described in (b) find the effective conductivity of this "chessboard".


Question 07/01





Answer 07/01


CONDUCTING CHESSBOARD

The question was:

Small square metal sheets are made from two different metals (green and yellow in the picture) and combined into a single very large "chessboard". The two-dimensional conductivity of one metal is S1 and of the other - is S2. Calculate the effective two-dimensional conductivity of this "chessboard" by following steps:

(a) Show that the electrostatic potential inside every, say, green square is given by the same harmonic function f1(x,y) (up to a constant which needs to be added depending on the detailed position of the square). Similar function g1(x,y) describes yellow squares.

(b) Functions f1(x,y) and g1(x,y) can be treated as real parts of ANALYTIC functions F(x,y)=f1(x,y)

+if2(x,y) and G(x,y)=g1(x,y)+ig2(x,y) of a complex variable z=x+iy. Show that f2(x,y) and g2(x,y) also solve (a different) chessboard conductivity problem.

(c) By comparing the solutions of the conductivity problems described in (b) find the effective conductivity of this "chessboard".


Answer 07/01

Answer (to part (c)): The effective conductivity of this "chessboard" is S=sqrt{S1*S2}.

Historical note: The problem of conductivity of two-dimensional composites was originally considered by A.M. Dykhne, JETP 7, 110 (1970). In particular, using his theory it can be shown that a symmetric mixture of two components has conductivity which is geometrical average of the conductivities of constituent materials. This applies also to the chessboard problem, i.e. the effective conductivity can be found without actually finding the potentials! Nevertheless, the potentials themselves are known: they have been found by V.L. Berdichevskii and published in Vestnik Moskovskogo Universiteta, 40(4), 56 (1985). For additional (more recent) references see the pdf file mentioned in the discussion of a modified problem by K. T. McDonald below.

(9/98) The problem has been solved (19/9/02) by Jared Kaplan (e-mail [email protected]), an undergraduate student at Stanford, CA (USA), and by Dan Gulotta (e-mail [email protected]), a high school senior at Illinois Mathematics and Science Academy in Aurora, IL (USA).

The solution:

The detailed solution provided by Kaplan and Gulotta can be found in the following pdf file. Here, we only mention few essential points:

(a) In potentials of the problem in each type of the squares are assumed to be real parts of complex functions, then the imaginary parts represent field rotated by 90 degrees.

(b) By examining the boundary conditions between the grains one notices that the fields generated by the imaginary parts are currents in a conductivity problem in which conductivity Si of each square has

been replaced by 1/Si.

(c) By writing down the expressions for the effective conductivities of the original and the "rotated" problems, we notice that they are simply related (one the the inverse of the other). By further using the isotropy of the problem, homogeneity of the solutions, and the symmetry between the two components, one arrives at the expression for conductivity.

This problem is highly non-trivial, and you are advised to read the detailed solution mentioned above. It should be noticed that the solution is valid not only for checkerboard, but also for random two-dimensional 50%-50% mixtures which are symmetric (i.e. both components have the same type of geometry). Over the years, the solutions were slightly extended. In particular a three component mixture can be solved, under some (very restrictive) conditions on geometry and conductivity.

(9/2002) Y. Kantor: If you are sure that you understood the solution of the continuous problem here is another challenge for you: show that the same result is valid for a two-dimensional square lattice whose bonds are randomly chosen resistors of two types. Each bond has a 50%-50% chance to have either of




http://star.tau.ac.il/QUIZ/01/chess_KG.pdf

Answer 07/01

two selected values. The effective resistivity of such random lattice is given by the geometrical mean of two resistor values. While the solution of this discrete problem resembles, in its spirit, the continuous solution, you will discover many challenging obstacles that did not appear in continuum problem.

Below we present some additional approaches and related problems.

"Heuristic" approach: Now that we know what the answer is, we can try to see if we can get it easier, by using some general properties of conductivity and some guesses. The effective conductivity S is a function of constituent conductivities S1 and S2: S=H(S1,S2). Obviously (because of linearity of

equations of conductivity), the increase of both constituent conductivities by factor k, will increase the effective conductivity by the same factor. Therefore H(ka,kb)=k*H(a,b). By choosing k=1/S2, we get (S/

S2)=H((S1/S2),1). So now we see that (S/S2) is only a function of (S1/S2). What kind of function it can

be? The function is equal to 1, when (S1/S2)=1. The function obviously vanishes if the argument

vanishes (i.e. if one of the material is non-conducting), and the function is infinite when the argument is infinite (i.e. first material has infinite conductivity, while the second material has finite conductivity). [The latter property is not self-evident; one really needs to look deeper into a problem of a contact between two corners, to understand what is happening to currents.] All these properties lead us to a guess that the function is a simple power law. [We do not know simple rigorous argument that this should be so.] Consequently,

(S/S2)=(S1/S2)p,

or

S=S1p*S2

1-p.

However, the value of S cannot change if we exchange locations of green and yellow squares, i.e. exchange S1 and S2 in the above formula. Consequently, p=1/2, and the answer S=sqrt{S1*S2} follows.

We should note that this "heuristic" approach involved guesses regarding the shape of the function, and explicitly used the symmetry of the system (interchange two types of materials). In three dimensions, the simple power law assumption is incorrect, and, consequently, there the solution is not known.

Kirk T. McDonald from Princeton (e-mail [email protected] suggested (22/10/2001) a related simpler problem, which allows an exact solution which includes finding the fields! He considered a circular (two-dimensional) region, divided into four parts by two straight lines passing through its center and forming pre-defined angle between them. The conductivities of 4 regions alternate between two possible conductivities. The conductance of such system can be found exactly. The



Answer 07/01

modified problem and its solution are presented in the following pdf file.



http://star.tau.ac.il/QUIZ/01/chess_McD.pdf



Question 08/01

Question 08/01

DANDELION SEED

Dandelion (Taraxacum Officinale) is a nice yellow flower (see left picture) which produces a seed-head (center picture) full of seeds (right picture) which are "well adapted" to be flown by wind to large distances. Give a quantitative evaluation of the "design" of the seed. In what sense it is "optimal"? (Or, maybe, it is not optimal?)





Question 09/01

Question 09/01

THE POWER OF DIMENSION

Physical quantities always have dimensions that are products of powers of basic units. E.g., the energy is measured in joules and 1 J = kg*m2/sec2 Why aren't there any quantities which are NOT powers of elementary units, but rather are more complicated functions?





Discussion 09/01



The question was:


(27/9/01) Yevgeny Kats (e-mail [email protected]) made a valid remark that some units, such as decibel, pH, and others, do not really represent product of powers of dimensions. Indeed the term "units" frequently represents a method of measurement rather than actual dimensionality. E.g., decibel is really a dimensionless unit which represents "10 times logarithm of a ratio between some measured and some standard quantity of energy flux". Similarly, other dimensionless units (angle degrees, radian) denote a method of measurement. Our question, of course, does NOT consider such usage of the term "unit".






Answer 09j/01



The question was:


(12/01) Most of the answers submitted to us concentrated on "non-power-law" examples which might lead to absurd result. A more constructive approach to the question has been submitted by Jared Daniel Kaplan (9/2001) from Stanford (e-mail [email protected]). However, we feel that we did not receive a really general proof of the case that details all the assumptions involved. An excellent exposition and in depth consideration of this question can be found in the book Scaling, self-similarity, and intermediate asymptotics by G.I. Barenblatt (Cambridge U. Press, 1996). (We are grateful to Jerome Gosset (e-mail [email protected]) for bringing this book to our attention.) Our discussion below largely follows that book.

Answer: Only power laws (and products of power laws) satisfy the principle underlying the concept of "unit" as defined below.

Detailed explanation

A set of fundamental units that is sufficient to measure properties of a class of phenomena is called a system of units. E.g., gram, centimeter and second can form a system for a broad range of phenomena. Similarly, kilogram, meter and second can form another system of units. Both systems rely on mass M, length L and time T units, and we shall say that they belong to one class of sytems of units. On the other hand, kilogram-force, speed-of-light and minute, also form a system. However, they belong to a different class (force, velocity, time).

The function which determines factor by which the numerical value of a physical quantity changes upon passage between two systems of units within the same class is called dimension function. E.g., the dimension function for "mass density" is M/L3. It tells us that if mass unit has been increased by a factor of 2 and length unit has been increased by a factor of 4, the number of the same quantity will decrease by factor of 32.

Can there be a dimension function sinM*logT? No! We are going to show that the dimension function




Answer 09j/01

must be a power-law monomial. This is a consequence of a simple principle:

All systems within a single class are equivalent, and there is no single distinguished or somehow preferred system of units

Without loss of generality, let us consider some specific system of units, e.g. the system built on L, M and T. Consider a mechanical quantity A that depends on all these units and the dimension function is [A]=F(L,M,T). Now let us consider systems of units 1 and 2 which have been obtained from an original system L,M,T by decreasing the units by factors L1,M1,T1 and L2,M2,T2, respectively. If the value of our

mechanical quantity in the original units was A, then in the new units it will be A1=A*F(L1,M1,T1) and

A2=A*F(L2,M2,T2), respectively, and therefore:

A1/A2=F(L1,M1,T1)/F(L2,M2,T2).

However, we could have used system 1 as our basic system, and consequently treated system 2, as modification (decrease) of system 1 units by factors L2/L1, M2/M1, T2/T1, and therefore

A2=A1F(L2/L1, M2/M1, T2/T1).

By comparing the above expressions we see that

F(L1,M1,T1)/F(L2,M2,T2)= F(L2/L1, M2/M1, T2/T1)

If we now differentiate both sides with respect to L2, and then set L2=L1=L, M2=M1=M, T2=T1=T, then

we get:

DLF(L,M,T)/F(L,M,T)=(1/L)DLF(1,1,1)=c/L,

where c simply denotes DLF(1,1,1), and DL denotes partial derivative with respect to L. Integrating the

last equation with respect to L, we find that

F(L,M,T)=LcG(M,T),

where G is some (as yet) undetermined function of two variables. By repeating the above procedure with M and with T we will arrive to the conclusion that the remaining variables must also appear as power laws, and consequently,

F=g Lc Md Tf,

where g, c, d, f are constants. This proves that dimension function must be a power-law monomial.





Question 10/01

Question 10/01

AREA MAXIMIZATION

A flexible electric wire-loop (of fixed length) (black line in the picture) is placed on a horizontal plane and is free to slide on it. At certain points rings (shown in red) are anchoring the wire to the plane. The wire can slide through the rings. A battery ensures that a constant current is flowing through the ring. Strong magnetic field B is applied perpendicular to the plane.

(a) Show that the wire will assume shape maximizing the area surrounded by the wire (consistent with the constraints imposed by the rings).

(b) What happens if the rings are placed in such a way that they force the current loop to cross itself?

This problem appears (in a more "commercial" form: maximization of area surrounded by a fence of given length, which must pass through given points) in the book Laws of Electromagnetism by A.A. Borovoi, E.B. Finkel'shtein and A.N. Kheruvimov ("Nauka", Moscow, 1970).





Question 10/01


Discussion10/01


AREA MAXIMIZATION

The question was:

A flexible electric wire-loop (of fixed length) (black line in the picture) is placed on a horizontal plane and is free to slide on it. At certain points rings (shown in red) are anchoring the wire to the plane. The wire can slide through the rings. A battery ensures that a constant current is flowing through the ring. Strong magnetic field B is applied perpendicular to the plane.

(a) Show that the wire will assume shape maximizing the area surrounded by the wire (consistent with the constraints imposed by the rings).

(b) What happens if the rings are placed in such a way that they force the current loop to cross itself?

(8/03) Y. Kantor: From the various submitted solutions it became clear to us that the problem was poorly formulated, i.e. the general solution is more complicated than implied in the question. The treatment of a simple "convex polygon" case has been presented (6/6/2002) by Chetan Mandayam Nayakar (e-mail [email protected]). His solution was as follows:

Area A of a regular (non-selfintersecting) curve can be calculated using the formula below, where n is the normal to the surface.



Discussion10/01

A=(1/2){\int}rxdr·n= (1/(2B)){\int}rxdr·B= (1/(2BI))I{\int}rxdr·B= (1/(2BI))M·B In the second part of the formula we replace n by B/B, where B is the magnetic field, which by the conditions of the problem is perpendicular to the plane. In the third part we multiplied and divided by current I, and noted that the integral is simply the definition of the magnetic moment (in MKSA units) M. We also note, that the expression for the area can be rewritten as A=(1/(2BI))I{\int}Bxdr·r= (1/(2BI))I{\int}dF·r Thus, the area is proportional to the integral of the radial component of the force multiplied by the radius.

We should, however, notice, that the change of the sign of magnetic field would change the above derivation, and actually lead to minimization (rather than maximization) of the area.

Ivan Sirakov Laboratoire de Rheologie des Matieres Plastiques of CNRS, St Etienne (France) (e-mail [email protected]) sent us (22/7/03) a detailed solution of the problem which can be seen here in PDF format. The solution considers directly the forces, and demonstrates that equilibrium positions can be minima, maxima or even saddle points. While this not a complete solution (yet), it indicates the complexity of the problem, and nice examples. We do not know whether more general statement about the problem can be made.




http://star.tau.ac.il/QUIZ/01/Solution_Max_area.pdf



Question 11/01

Question 11/01

RESISTOR NETWORKS

Consider and network built of wire segments having unit resistance. Find the resistance between two adjacent nodes of the network (denoted by blue and red in the figures) if the network is: (a) infinite simple cubic network; (b) infinite square network; (c) semi-infinite "ladder".

We welcome suggestions of additional analytically solvable infinite networks.





Answer 11/01


RESISTOR NETWORKS

The question was:

Consider and network built of wire segments having unit resistance. Find the resistance between two adjacent nodes of the network (denoted by blue and red in the figures) if the network is: (a) infinite simple cubic network; (b) infinite square network; (c) semi-infinite "ladder".

(3/02)The problem was solved correctly by many people. Loic Turban from Universite Henri Poincare (Nancy, France) (e-mail [email protected]) both solved (19/12/01) the general case of z-coordinated periodic lattice, and suggested addition solvable geometry (see below). Predrag Lazic a Ph.D. student in Zagreb, Croatia (e-mail [email protected]) both solved (26/1/2002) the proposed questions and suggested an additional solvable geometry.

Answer: (a) 1/3; (b) 1/2; (c) 1+sqrt{3}.

The solution:

(a) and (b) Square, cubic and general hypercubic lattices in D space dimensions can be solved as follows: Assume that current I is inserted from outside into the "red" junction and withdrawn at infinity. Since the system is symmetric around the injection points the current will split equally into 2D currents in bonds connected to the junction. In particular, along the bond connecting blue and red junctions there will be




Answer 11/01

current I/(2D). Now consider a different situation: current I is removed from blue junction and is injected from infinity. At each of the bonds entering the blue junction, there will be a current I/(2D). Now consider a linear superposition of two above situations: Current I is injected into the red junction and current I is withdrawn and the blue junction, and nothing happens at infinity. The current on the bond connecting the red and blue junctions will be I/(2D)+I/(2D)=I/D. Therefore, the voltage between the junctions will be V=I/D, and consequently the effective resistance will be R=V/I=1/D. Thus, the resistance will be 1/3 for cubic, and 1/2 for square lattice.

(c) Assume that the resistance of the semi-infinite ladder is R. Then we can view the system as consisting of three unit resistors, when in parallel to the center resistor we connected an effective resistor R, as depicted in the following figure:

From simple sequential/parallel resistor rules we find:

R=2+1/(1+1/R)

which can be solved and gives R=1+sqrt{3}.

There are additional analytically solvable infinite networks. E.g., consider the following triangular network:

It can be solved in the same way as the square network. Convince yourself that the resistance between adjacent sites is 1/3. In general, if coordination number (number of nearest neighbors of a site) on a periodic (symmetric) lattice is z, then the resistance will be 2/z. E.g., for honeycomb lattice, z=3, and the resistance will be 2/3.

Loic Turban suggested to look at the resistance between the points of a honeycomb lattice separated by two bonds. If current I is inserted a point A then current (1/3)I will flow in a bond connecting A with its nearest neighbor B. That current will split, and (1/6)I will flow from B to next-nearest-neighbor C. The


Answer 11/01

voltage drop between A and C will be (1/2)I. Similar argument can be repeated for current I being extracted from C; however, this time there will be current (1/3)I in BC, and (1/6)I and AB. Superposing two cases, we get current I inserted at A and removed at C, causing voltage drop of I, and consequently the equivalent resistance is unity. Similar "symmetry arguments" can prove that the resistance between second neighbors in a diamond lattice is 2/3. Also, one can show that the second neighbor resistance in a dice lattice is 1/2. The dice lattice is a two-dimensional lattice (despite its "3D appearance") shown in the following figure:

Predrag Lazic suggested to look at the resistance between the points of Sierpinski gasket. This geometry is obtained by iterative procedure. There are many simple fractal geometries that can be solved.





Question 12/01

Question 12/01

BAD WEATHER

One winter day a weatherman in New York said that the snow was melted by the recent rainfall and that's why there were floods. There were about 30 inches of snow and about 1 inch of rain. Is the weatherman correct in his assertion?

This question was asked by Robert Ristinen at the 1996 oral exam at University of Colorado at Boulder.





Answer 12/01


BAD WEATHER

The question was:

One winter day a weatherman in New York said that the snow was melted by the recent rainfall and that's why there were floods. There were about 30 inches of snow and about 1 inch of rain. Is the weatherman correct in his assertion?

(03/02) The problem has been answered correctly by Will Brunner (6/12/2001) from the University of Colorado at Boulder (US) (e-mail [email protected]).

Answer: The weatherman was wrong.

The solution:

If you ever looked at a snowflake, or kept snow in your hand - you surely know how tenuous it is. The density of snow is approximately 0.1 g/cm3, although it depends on temperature and other weather conditions and maybe as large as 0.2 and as small as 0.02. Consequently, most of the snow is simply "air", and 30" of snow corresponds to about 3 inches of compressed material (ice). Consequently, even if 1" of rain could melt all the snow we would be faced with 4" of water. However, we need to examine whether the rain could actually melt all the snow. Let us assume that it was really warm rain (in the middle of the winter) of say 20 degrees C. Each cm3 of rain the would cool to 0 degrees and release 20 calories of heat. The latent heat of ice is 80 cal/g, and consequently 1" or rain could melt 1/4" of ice, leading to amount of water equivalent to 1.25". Certainly not enough for a flood.

Comment: Bill Bruml (13/1/2002)) (e-mail [email protected]) pointed out two important points that may modify the above solution or even make it invalid: 1. Snowflakes have a very large surface (relative to their bulk). This may effectively decrease the latent heat. 2. If the snow was really not snow, but "slush", i.e. matrix of frozen material holding water - then the heat will only be required to melt the matrix, and thus much more water can be generated.

Comment: (26/12/01) Luca Visinelli - an Italian student (e-mail [email protected]) pointed to an additional potential source of energy: the kinetic energy of falling drops is converted to heat. However, for rain drops falling at say 10 m/s velocity, the kinetic energy of 1 gm of rain is 0.05J=0.01cal. Completely negligible amount! (The velocity of the drop is a result of compensation between its weight and air friction and, consequently, depends on its radius, but for any reasonable value of drop size, the kinetic energy is still too small.)





Answer 12/01





Question 01/00

Question 01/00

CLIMBING AN ICEBERG

A mountaineer climbs an iceberg - icy cone-shaped mountain. He has made a loop with non-sliding knot and thrown it over the top of the mountain as shown in the figure. If the mountain is steep enough, the climber will reach the summit; if it is not, the loop will slip of the top. Find the critical slope of the mountain.

D. Khmelnitskii told us this problem. He heard it from M. Spector in 1970. [5/2001: See also a postscript in the answer.]





Answer 01/00


CLIMBING AN ICEBERG

The question was:

A mountaineer climbs an iceberg - icy cone-shaped mountain. He has made a loop with non-sliding knot and thrown it over the top of the mountain as shown in the figure. If the mountain is steep enough, the climber will reach the summit; if it is not, the loop will slip of the top. Find the critical slope of the mountain.

The problem has been solved (18/1/2000) by Jhinhwan Lee a Ph.D. student at Center for Science in Nanometer Scale at Seoul National University, Korea (e-mail [email protected]), by Sumit Banerjee (13/2/2000) from Indian Institutes of Science and Astrophysics (e-mail [email protected]), and (26/2/2000) by Andrew Wiggin (e-mail [email protected]).

D. Khmelnitskii told us both the problem and its solution which is presented below. (See also the p.p.s. at the end!)

The answer: The critical angle A between the generatrix of the cone and the vertical (see Fig.(a) below) is 30o.

The solution:

The solution is given by a simple lemma, consisting in the following: Consider a rope on a convex surface with two equal weights at its ends. It is obvious that the length must be the shortest possible, otherwise the potential energy of the weights would not be minimum. So, according to the lemma, a rope on a convex surface lies along a geodesic line, i.e. the shortest curve between two points lying wholly on the surface.

Now, if we cut our cone along the generatrix, passing through the knot of the loop (green line on Fig.(a) below), and then roll it out on the plane, we will get a sector, which will look either like Fig.(b) or Fig.(c) below depending on the angle A. Green lines on the resulting sectors are just the line along which the cone was cut. Here the thick lines are the projections of the loop on the plane. A geodesic on a plane is a straight line. So, the straight line is exactly what we need. Moreover, we can see that the loop does not slip off the top of the mountain only if the angle B at the top of the sector does not exceed 180o as depicted on Fig.(b). Otherwise (see Fig.(c)), the line lies outside the sector and the loop has no equilibrium on the cone.

To find the critical angle A of the cone, we note that if L is the generatrix of the cone, then the length of the arc, which






http://www.tcm.phy.cam.ac.uk/~dek12/

Answer 01/00

subtends the angle B, equals the circumference of the base of the cone 2{pi}r, where r is the radius of the base: B L = 2 {pi} r. As r = L sin(A) and the critical angle B = {pi}, sin(A) = 0.5, and therefore A= 30o.

P.S. An interesting related problem can be found in the book Riemannian Geometry by S.Gallot, D.Hulin and J.Lafontaine (Springer Verlag, 1987 (1st edition), 1990 (2nd edition)).

P.P.S. (5/2001) Pavel Novikov from Institute of Semiconductor Physics, Russian Academy of Science (Siberian Branch) (e-


Answer 01/00

mail [email protected]) wrote us the following: In 1982, when I studied in the Novosibirsk State University (Russia), Dr. Spector lead seminars for us on Continious Media Physics and I heard the same problem from him. He gave us only 5 minutes to solve the problem. Nobody did and he told us an answer... By the way only three my friends solved this beautiful problem. Now two of them work in Institute of Semiconductor Physics in the Novosibirsk Academgorodok.






Question 02/00

Question 02/00

HALLEY'S PLANET

Not so many years ago there was no order of magnitude estimate of the age of the planet Earth. Edmund Halley (yes, that's the "comet guy") wrote an article in Philosophical Transactions of the Royal Society of London, 29, 296 (1714), where he observed that lakes that emit no rivers (such as Dead Sea, or Caspian Sea) are very salty, and thus it is reasonable to assume that rivers bring salt into the lakes. He further claimed that "'tis not improbable that the ocean it self is become salt from the same cause." Consequently, by measuring the salinity of the oceans and by determining the amount of salt brought by rivers every year, one could estimate the age of the Earth. Halley lacked the data to actually perform an estimate of the age of the Earth using his method. Halley's assumptions are wrong, but nevertheless it was a nice try for his time. What would be the age of Earth according to Halley's method?

This problem was suggested by Y. Kantor.





Answer 02/00


HALLEY'S PLANET

The question was:

Not so many years ago there was no order of magnitude estimate of the age of the planet Earth. Edmund Halley (yes, that's the "comet guy") wrote an article in Philosophical Transactions of the Royal Society of London, 29, 296 (1714), where he observed that lakes that emit no rivers (such as Dead Sea, or Caspian Sea) are very salty, and thus it is reasonable to assume that rivers bring salt into the lakes. He further claimed that "'tis not improbable that the ocean it self is become salt from the same cause." Consequently, by measuring the salinity of the oceans and by determining the amount of salt brought by rivers every year, one could estimate the age of the Earth. Halley lacked the data to actually perform an estimate of the age of the Earth using his method. Halley's assumptions are wrong, but nevertheless it was a nice try for his time. What would be the age of Earth according to Halley's method?

(11/2000) Y. Kantor: The problem was solved (9/2000) by Marakani Srikant from the National University of Singapore (e-mail [email protected]). He suggested a nice "back of the envelope" solution. (E.g., he estimated the total river discharge as 5 times Amazon river, etc. While some of his numbers were by an order of magnitude off the target, the final answer is rather close to what we present below.)

Answer: About 40 million years.

Solution:

The total amount of water in the oceans is 1.37 109 km3. The total river inflow per year is 37,300 km3. 1 kilogram of sea water contains 10 grams of Na+ ions, or 35 grams of salts (mostly sodium and magnesium chlorides) per kilogram of water. It is difficult to define what exactly is called "fresh water" and what is the mean salinity of rivers. A nice reference on salinity is here. It says that there is about factor 1000 between salinity of soft river water and the sea.

From this data we deduce that "the age of the earth" is 1.37 109 km3/37,300 * 1000=4 107years. Not bad for an estimate of geological time, although we are more than two orders magnitude off. Now, can you tell the reasons for this number being such a strong underestimate of an actual answer?

(12/2000) Paul N. Taylor from Oxford, England (e-mail [email protected]) sent us the following e-mail regarding the question asked above:



http://www-ib.berkeley.edu/IB/instruction/IB150/material/lectures/lecture22/OsmoticEnvironments.html


Answer 02/00

In the answer to the problem of the age of the earth from the salt in the sea you pose the supplementary question of why that method produces such an underestimate for the age of the Earth. The key to the problem is that the ocean basins constitute a rather leaky container. The rivers go on and on feeding salt into the system, but it is perhaps not quite so clear how the salt is then taken out of the ocean water system.

Over the last few decades our understanding of the workings of the oceans have been greatly enhanced by the development of the theory of plate tectonics and sea floor spreading. New ocean floor is being created at the mid-ocean ridges by the frequent eruption of basaltic lavas (MORB: Mid Ocean Ridge Basalts). When freshly erupted, these basalts have a rather low sodium concentration. However, another very important process taking place at and near the mid-ocean ridges is the hydrothermal interaction of seawater with the newly erupted lavas. In this process massive volumes of ocean water are processed through the basalt lava piles in convection cells which penetrate deep into the lava pile. The seawater is strongly heated by this process and very extensive chemical exchanges take place between the seawater and the basalts. The composition of hydrothermally altered basalts typically show sodium concentrations greatly increased relative to fresh MORB. The fate of most of this highly modified basalt, on a timescale of ca. 200 Million Years is to be transported away from the mid-ocean ridge by the "conveyor belt" process of sea-floor spreading eventually to be consumed back into the mantle at a subduction zone. Much of the sodium and the water loaded onto the "conveyor belt" by the hydrothermal activity at the ridge is probably released from the descending subducted slab as it warms up, and these materials contribute significantly to a second round of volcanic activity in the island arcs or Andean-type volcanic chains which presently surround the Pacific [the oldest of the Earth's present oceans]. These issues are all dealt with very effectively in most of the modern textbooks of Igneous Petrology or Plate Tectonics.

The "age of the Earth" by the salt in the sea method has found important applications in modern geochemistry. The "age" given by this method is now understood better in terms of a residence time in the ocean water mass for each chemical species - given by (the mass of the species in the ocean water mass)/(the rate of supply of the species to the ocean mass). The matter is dealt with in "Inorganic Geochemistry" by Paul Henderson, Pergamon 1982; and also in "The Continental Crust: its composition and evolution" by SR Taylor & SM McClennan, Blackwell 1985.

p.s. It appears that creationists just love this way of measuring age of earth, since it ensures "young" earth (maybe not young enough for their taste, but "it's at least something"). If you want to have some "creationist fun" take a look at the Answers in Genesis. (We thank Ido Golding from Tel Aviv University for this reference.)



http://www.answersingenesis.org/docs/3910.asp



Question 03/00

Question 03/00

LEANING TOWER OF BRICKS

By placing brick on top of another brick, slightly shifted as shown in the figure, we can create a situation that the top brick in such a pile is displaced horizontally by h relative to the bottom brick. Can this be done for an arbitrary h? What is the optimal way to do it?





Answer 03/00


LEANING TOWER OF BRICKS

The question was:

By placing brick on top of another brick, slightly shifted as shown in the figure, we can create a situation that the top brick in such a pile is displaced horizontally by h relative to the bottom brick. Can this be done for an arbitrary h? What is the optimal way to do it?

(3/2000) The problem has been solved by Avi Nagar (e-mail [email protected]), by Lluis Batet from Technical University of Catalonia, Barcelona, Spain (e-mail [email protected]), by Andrew




Answer 03/00

Wiggin (e-mail [email protected]), and by Robert White (e-mail [email protected]).

The solution: If the topmost brick is shifted by 1/2 brick relatively to the brick below it, and the second-from-the-top brick is shifted by 1/4 brick relatively to the brick below it, ... , n-from-the-top brick is shifted by 1/(2n) brick relatively to the brick below it, - then there is no limit to the displacement of the bricks, because the series Sum{n=1 to n=infinity}(1/n) diverges(*). At the same time one can directly verify the the center of mass of any k top-most bricks is exactly above the edge of the brick just below them: Let us calculate the horizontal position of the center-of-mass of k top-most bricks relative to the brick just below them. The displacement of the top-most brick in such a situation is (1/2)[1+1/2+1/3+...+1/k], the location of the second-from-the top is (1/2)[1/2+1/3+...+1/k], etc., and consequently position of the center of mass

(1/k)*(1/2)*[1+2*(1/2)+3*(1/3)+k*(1/k)]=1/2

i.e. it is exactly above the edge of the brick below them.

Since at every level the bricks above a given brick are shifted to their extreme positions, the solution presented here is optimal for any number of bricks.

* Note: For finite sum (up to a large integer N) Sum{n=1 to n=N}(1/n)=ln(N)+Gamma, where Gamma is Euler's constant. Thus, for a particular displacement h expressed in the units of single brick length we will need exp(2*h-Gamma) (rounded to the next integer) bricks.







Question 04/00

Question 04/00

BIKE RIDE

Suppose your bike has no rear fender. How slowly do you have to ride on a wet road to avoid getting mud on your bottom?

This question was contributed by T.A. Witten.



http://physics.uchicago.edu/t_cond.html#Witten



Answer 04/00


BIKE RIDE

The question was:

Suppose your bike has no rear fender. How slowly do you have to ride on a wet road to avoid getting mud on your bottom?

(7/2000) The problem has been solved by Jared D. Kaplan, - a high school student from the Illinois Mathematics and Science Academy (e-mail [email protected]). Partial answers have been also provided by Avi Nagar (e-mail [email protected]), by Regis Lachaume, a PhD student at Grenoble Observatory (France) (e-mail [email protected]), and by Ido Golding, a PhD student at Tel Aviv University (Israel) (e-mail [email protected]).

The answer: About 12 km/hour (3.3 m/sec).

The solution: Below we present a (slightly edited) version of the solution submitted by Kaplan.

For this problem I set up a coordinate system, where the center of the rear wheel is at the origin, the back of the bike seat is at (d,h), and the radius of the rear wheel is r. The way I worked the problem was to express the height of a mud particle at a distance d from the origin in terms of its velocity upon leaving the wheel. The only tricky thing was dealing with what point on the wheel the mud leaves from. I expressed this as the angle A that the radius to the point of departure makes with the negative x-axis in






Answer 04/00

my coordinate diagram, knowing the mud would be flying off perpendicular to this radius. Also, let the velocity of the bike be v:

vx = v*sin(A)

vy = v*cos(A)

Let the function H(t) denote the height of the mud particle at time t, and let D(t) denote the horizontal distance covered at time t:

H(t) = v*cos(A)*t - g/2*t2 + r*sin(A) D(t) = v*sin(A)*t - r*cos(A)

So when D(t) = d, if H(t) > h, you get mud on your back. D(t) = d --> v*sin(A)*t - r*cos(A) = d --> t = (d+r*cos(A))/(v*sin(A))

Plugging this time into our H function we get:

H(d/v*csc(A) + r/v*cot(A)) > h cot(A)*(d+r*cos(A)) - g/2*(d/v*csc(A) + r/v*cot(A))2 + r*sin(A) > h

we can solve it for v and get:

v^2 < g(d+r*cos(A))2/(d*sin(2A)+2r*sin(A)-2h*sin(A)2) v < sqrt(g/(d*sin(2A)+2r*sin(A)-2h*sin(A)2))*(d + r*cos(A))

What this all means is that if the velocity is less than this function at all A's, then we won't get dirty. I don't think this function can be minimized in general, but I applied it to my particular bicycle with measurements of: h = .58 meters d = .20 meters r = .33 meters

I found numerically the minimum point of the above function to be approximately 3.28 m/s, meaning that as long as the velocity that the mud splatters off at is less than this, I won't get dirty. This is really nice because the this velocity is exactly the same as the maximum velocity of the bike. I didn't take drag with the air or any other more subtle features into account, but I hope this was at least a good approximation.





Question 05/00

Question 05/00

THE FIRST DIGIT

Back side of the front cover of the Handbook of Chemistry and Physics lists 18 important fundamental (dimensional) constants and non-standard units such as speed of light, charge of an electron, Stephan-Boltzmann constant, electron volt, etc. Values of SEVEN out of 18 constants begin with digit 1.

Why is digit 1 so prevalent in that list?

Will the situation change if instead of MKSA (SI) units we will use, say, inch, second, pound and electrostatic charge unit?

This question was suggested by Y. Kantor.





Discussion 05/00


THE FIRST DIGIT

The question was:




(5/00) Y. Kantor: we got several enlightening e-mails on the subject. Below we present edited extracts from these e-mails.

(3/5/2000)Dr. Fred Goesmann from Max-Planck-Institut fuer Aeronomie (Katlenburg-Lindau) (e-mail [email protected]) wrote: If numbers are equidistributed on a logarithmic scale, which is not unreasonable to assume, since they vary over so many orders of magnitude, numbers starting with a 1 should account for about one third of all numbers. (see, for example log graph paper). 7 out of 18 is not far out. Hence this phenomenon should really not depend on the system of units employed.

(3/5/2000)Mike Salem from Case Western Reserve University (e-mail [email protected]) wrote: Last year, I wrote a program which multiplied each of the numbers from 1 to 1000 by each other, and each of the resulting products by each number from 1 to 1000, for a total of a billion products. Fractional multiples of arbitrary units don't make any difference since we are only concerned with the leading (or first non-zero) digit. With this simple test, I found that 31-33% (I can't remember the actual fraction) of the products had one as a leading digit, confirming what my roommate suggested. Idealy, one would find the product of all numbers, an infinite number of times (I suppose), however my experiment, which was also conducted with the numbers from 1 to 100 and with 1 to 1000 only multiplied by eachother once, suggested that the fraction converges to a value somewhere in the low 30 percentile.

(5/5/2000)Ansgar Esztermann from University of Duesseldorf (e-mail [email protected]) wrote: It all comes down to chosing the correct distribution function. If we pick some numbers at random (without any constraints, such as order of magnitude), their relative distances should be about constant, so (x1-x2)/x1=const=:1/

C Since there is one number within a "length" of x1-x2, the density of numbers is






Discussion 05/00

rho(x)=1/(x1-x2)=C/x

The probability of a number starting with 1 is then obtained by integrating: p=int12 rho(x) dx / int110 rho(x) dx

=(ln(2)-ln(1))/(ln(10)-ln(1)) =log10(2) =~ 0.3

Thus, a logarithmic distribution (as proposed by F. Goesmann) is obtained.

(4/5/2000)Andy Frohmader from Case Western Reserve University (e-mail [email protected]) sugested the following argument (which was send to us by his friend Mike Salem): The definitions of the units are entirely arbitrary... Thus, the probability that any constant is any particular value should be equal to the probability that the constant is twice that value... Let the probability that the mantissa of a constant is on the interval [m, n) be p([m, n)). Then p([1, 2)) = p([2, 4)) = p([4, 8)) = p([8, 10) U [1, 1.6)), and so on. Adding n of these terms gives np([1, 2)) = p([1, 2)) + p([2, 4)) + ... + p([mantissa(2(n-1)), mantissa(2n)) (or the last term may have to be broken up into a union of 2 intervals as before). Note that the mantissa must be somewhere on the interval [1, 10), so p([1, 10)) = 1. The above sum covers the entire interval [1, 10) [log(2n)] times (where [x] is the greatest integer which is not greater than x; also, the logarithms are in base 10) and part of an additional time. Hence, [log(2n)] < np([1, 2)) < [log(2n)] + 1. Rearranging this gives [nlog(2)] < np([1, 2)) < [nlog(2)] + 1, or [nlog(2)]/n < p([1, 2)) < ([nlog(2)] + 1)/n. Taking the limit as n goes to infinity here gives that p([1, 2)) = log(2). But p([1, 2)) is just the probability that the mantissa starts with a 1, which is, of course, the probability that any physical constant begins with a 1. Therefore, the probability of any constant beginning with a 1 is log(2) = .301...






Answer 05/00


THE FIRST DIGIT

The question was:




(5/2000) Many people contributed discussions/comments/references. Here are some names: Dr. Fred Goesmann (e-mail [email protected], Mike Salem (e-mail [email protected]), Rasmus C. Egeberg (e-mail [email protected]), Ansgar Esztermann (e-mail [email protected]), Andy Frohmader from Case Western Reserve University (e-mail [email protected]), Carlos Ungil Gutierrez-Rave (e-mail [email protected]). Each of the emails contained a different aspect of the solution, or told us where such derivation can be found. Below we present a (mathematically non-rigorous) solution of the question. (You are also advised to look at the discussion of this problem where additional insights and alternative derivations can be found.)

The answer: The prevalence of 1 as a first (significant) digit is dictated by the broad range of numbers in the list, and has nothing to do with physics and the choice of units.

The solution:

Prevalence of digit 1 is not specific to the list mentioned in this question. Many lists of physical properties will have 1 as their one of the most common first digit. The only real requirement is - the list must include a very broad range, i.e the quantities should range over several orders of magnitude. (Physical constants listed in the Handbook range over many orders of magnitude (from 10-27 to 1023) in MKSA units.) Approximately we can say that the list has no typical scale. Without knowing what exactly is included in the list we can only guess, i.e. talk about probability distribution of the numbers. We expect the probability distribution to be scale invariant, i.e. probability density p(kx) should be









Answer 05/00

proportional to probability p(x). (This means that the distribution will be independent of the choice of units!) One can easily convince himself that only p(x)~1/x has this property*. (We note that this probability density is not normalizable. In reality the are some very remote cutoffs (minimal and maximal numbers) which will determine the normalization prefactor.) Now we observe, that probability P1 to find a number that begins with digit D=1, is a sum of probabilities of number being in the

intervals: 1< x< 2, 0.1< x< 0.2, 10< x< 20, 0.01< x< 0.02, ... Similar, relation can be written for other digits. By performing such summations we find that

PD=log10(1+(1/D))

In particular, this means that P1=0.30103, i.e. almost one third of numbers begin with digit 1. Histogram

of the probabilities is presented below:

From the derivation it must be clear that the law has nothing to do with physics, - it it as valid for stock prices, numbers appearing in a newspaper, etc. It was first noted by S. Newcomb (in 1881 in Amer J. Math. 4, 39(1881)) and investigated in detail by Frank Benford a physicist at the General Electric Company (in Proc. Amer. Phil. Soc. 78, 551(1938)), and, hence, is known as Benford's Law. Only recently, the law received rigorous formulation and proof. For detailed description and references see description of the law by Eric W. Weisstein. (The above histogram was taken from there.) See also the article by Malcolm W. Browne on the applications to economics and fraud detection.

Some additional references are http://www.newscientist.com/ns/19990710/thepowerof.html, http://mathworld.wolfram.com/ZipfsLaw.html, http://linkage.rockefeller.edu/wli/zipf/.

* This property can be proven as follows: (1) Assume that p(kx)=f(k)p(x); then from normalization condition follows that f(k)=1/k. (2) Since p(kx)=p(x)/k for any k, we can take k=1/x, and get p(x)=p(1)/x.


http://mathworld.wolfram.com/BenfordsLaw.html

http://courses.nus.edu.sg/course/mathelmr/080498sci-benford.htm

http://www.newscientist.com/ns/19990710/thepowerof.html

http://mathworld.wolfram.com/ZipfsLaw.html

http://mathworld.wolfram.com/ZipfsLaw.html

http://linkage.rockefeller.edu/wli/zipf/

Answer 05/00





Question 06/00

Question 06/00

HYDROGEN ATOM

Is it possible that two different transitions in hydrogen atom give the same frequency of radiation?

Comment 1: Disregard degeneracies due to different orbital and magnetic numbers ("l" and "m"). Consider only different atomic levels "n".

Comment 2: Consider a "simple" model of hydrogen atom, i.e. disregard complications introduced by spin, Lamb shift, etc.

This question was asked by Walter Wyss at the 1997 oral exam at University of Colorado at Boulder.





Answer 06/00


HYDROGEN ATOM

The question was:

Is it possible that two different transitions in hydrogen atom give the same frequency of radiation?

(10/00) We did not recieve a general solution of the problem. However, a satisfactory solution has been submitted by Jared D. Kaplan (9/7/2000) (e-mail [email protected]) and his friend Howard Liu, both of them high school students from the Illinois Mathematics and Science Academy. Their solution is presented at the bottom of this page. Iddo Ussishkin (27/7/2000) from the Department of Condensed Matter Physics at the Weizmann Institute of Science in Israel (e-mail [email protected]) has correctly identified the "lowest possible" transition pair. A subset of solutions has also been identified by Ganesh Sundaram (28/9/00) (e-mail [email protected]).

The answer: There are infinitely many such transitions.

The solution:

Since the energy of the n'th level is propotional to -1/n2 the transition frequency for transiiton from m'th level to n'th level is proportional to (1/n2-1/m2). So we need to find different pairs of integers for which the difference in inverse squares is the same. And this can beeasily done: E.g., the transition from n=35 to n=7 has the same energy difference as the transition from n=7 to n=5. Similarly, the transition from n=90 to n=9 has the same energy difference as the transition from n=6 to n=5. Below we present a large sequence of such pairs for n<200. (This sequence has been found "by inspection" of all possible transitions with relatively small n's. We do not know a general solution. However, see below.)

( 6 -> 5)==( 90 -> 9) ( 7 -> 5)==( 35 -> 7) ( 9 -> 5)==( 90 -> 6) ( 8 -> 6)==( 72 -> 9) ( 9 -> 6)==( 72 -> 8) ( 8 -> 7)==( 56 -> 14) ( 14 -> 7)==( 56 -> 8) ( 11 -> 10)==( 55 -> 22) ( 12 -> 10)==(180 -> 18) ( 14 -> 10)==( 70 -> 14) ( 18 -> 10)==(180 -> 12) ( 22 -> 10)==( 55 -> 11) ( 16 -> 12)==(144 -> 18) ( 18 -> 12)==(144 -> 16)





Answer 06/00

( 16 -> 14)==(112 -> 28) ( 18 -> 14)==( 63 -> 21) ( 21 -> 14)==( 63 -> 18) ( 28 -> 14)==(112 -> 16) ( 21 -> 15)==(105 -> 21) ( 22 -> 20)==(110 -> 44) ( 28 -> 20)==(140 -> 28) ( 44 -> 20)==(110 -> 22) ( 24 -> 21)==(168 -> 42) ( 42 -> 21)==(168 -> 24) ( 35 -> 25)==(175 -> 35) ( 36 -> 28)==(126 -> 42) ( 42 -> 28)==(126 -> 36) ( 33 -> 30)==(165 -> 66) ( 34 -> 30)==( 85 -> 51) ( 51 -> 30)==( 85 -> 34) ( 66 -> 30)==(165 -> 33) ( 44 -> 33)==(176 -> 48) ( 48 -> 33)==(176 -> 44) ( 42 -> 35)==(120 -> 56) ( 56 -> 35)==(120 -> 42) ( 45 -> 36)==( 80 -> 48) ( 48 -> 36)==( 80 -> 45) ( 40 -> 38)==(152 -> 95) ( 95 -> 38)==(152 -> 40) ( 42 -> 39)==(182 -> 91) ( 45 -> 39)==(117 -> 65) ( 65 -> 39)==(117 -> 45) ( 91 -> 39)==(182 -> 42) ( 54 -> 42)==(189 -> 63) ( 63 -> 42)==(189 -> 54) ( 70 -> 55)==(154 -> 77) ( 77 -> 55)==(154 -> 70) ( 68 -> 60)==(170 -> 102) (102 -> 60)==(170 -> 68) ( 75 -> 65)==(156 -> 100) (100 -> 65)==(156 -> 75) ( 90 -> 72)==(160 -> 96) ( 96 -> 72)==(160 -> 90)

Here is what Kaplan and Liu wrote (7/00) about this problem (slightly edited):

The energy (and, consequently, the radiation frequency) released when electrons jump between energy levels a and b (integers) is proportional to (1/a2-1/b2). So the problem is equivalent to finding solutions to the diophantine equation: 1/a2 - 1/b2 = 1/c2 - 1/d2, where a != c and b != d.


Answer 06/00

Which we can simplify to: (b2-a2)/(ab)2 = (d2-c2)/(cd)2 Our method of finding solutions here was to find solutions to: b2-a2 = d2-c2 and then to multiply a and b by some number x and c and d by some number y so that the original expression works. We noticed that (52-12)=(72-52)=24, so we just need to find the constants x and y: x2(52) = y2(5272) So obviously we can let x = 7 and y = 1, to get the solution: (a,b,c,d) = (7,35,5,7), meaning that a jump from energy level 35 to energy level 7 is the same as a jump from 7 to 5. There are obviously an infinite number of solutions of the form (7k, 35k, 5k, 7k), where k is a positive integer. I see no reason why there shouldn't be an infinity of other solutions, but so far I haven't proven this.

Later (8/00) Kaplan expanded the solution he suggested. His line of reasoning is quite complicated and tedious, so we only will give you the final answer which you can verify directly: a = (k3+k6)/2 b = (k6-k3)/2 c = k2(k+1)(k4+k2+1)/2 d = k2(k-1)(k4+k2+1)/2

Ussishkin made few interesting and important remarks (7/00):

1. The l and m quantum numbers should be chosen to agree with the transition selection rules. 2. On the practical level, observing these transitions may be very difficult, as they involve high lying levels, and various deviations from the "ideal" hydrogen may become important.

(4/0) We recieved quite a general method for generating pairs, triples, etc. of pairs of states producing identical frequencies. The solution has been sent by L.F.Perondi - a researcher at Instituto Nacional de Pesquisas Espaciais - SP, Brazil (e-mail [email protected]). While his method does not produce all the possible states, it certainly very broad. The method is as follows:

We give a procedure for finding a set of pairs (x1,y1;x2,y2;.....;xn,yn), with xi and yi integers, which satisfy the

equation

(1) ... 1/x12 - 1/y1

2 = 1/x22 - 1/y2

2 = ... = 1/xn2 - 1/yn

2 = R,

where R is a rational number.

For a particular pair of integers (xi,yi), we may write

(2) ... 1/xi2 - 1/yi

2 = (yi-xi) * (yi+xi) / (xi2 * yi

2) = m1 * m2 * ... * mk / D2,

D = d1* d2* ... * dl,

where mi and di denote the prime factors of the numerator and the denominator of Eq. (2), respectively. Let {mi}

and {mj} denote a partition of {m1,m2,...,mk} into two groups, such that if C1 and C2 are the integers formed by

the multiplication of all elements in {mi} and {mj}, respectively, then C1*C2 = m1*m2*...*mk. Making use of

these definitions, Eq. (2) may be rewritten as



Answer 06/00

(3) ... m1 * m2 * ... * mk / D2 = ( (C1 + C2)/(2 * D) )2 - ( (C1 - C2)/(2 * D) )2.

From Eq. (3), we observe that if D is chosen so as to contain the prime factors appearing in either (C1+C2)/2 and

(C1-C2)/2, then this equation assumes the form given on the left-hand side of Eq. (2), with

(4) ... xi = 2 * D / ( C1 + C2 ) , yi = 2 * D / ( C1 - C2) .

Hence, departing from a given set of integers {m1,m2,..., mk}, we may "construct" as many solutions to Eq. (2) as

there are partitions of {m1,m2,...,mk} in the sense described above.

Without any loss of generality , it may be easily shown that one may drop the factor 2 appearing in Eqs. (3) and (4) (it will appear multiplying m1 * m2 * ... * mk) and still get correct solutions to Eq. (1) through the same

procedure described above.

Taking what has been described so far, we may summarise the procedure for generating solutions to Eq. (1), as follows:

- choose a set {m1,m2,...,mk}; for each partition, find the prime factors in the corresponding numbers ( C1 + C2 )

and ( C1 - C2 );

- form D in such a way that it contains all prime factors emerging from the above step, (D is given by the minimum common multiple of the set of the ( C1 + C2 ) and ( C1 - C2 ) factors);

- compute the pairs (xi,yi) from xi = D / ( C1 + C2 ) and yi = D / ( C1 - C2).

A few examples will help illustrating the procedure outlined above.

Let m1=3, m2 =5 (evidently, 1 is always a member of the set); consider the partitions:

2,3 (C1 + C2) = 5

(C1 - C2 ) = 1

6,1 (C1 + C2) = 7

(C1 - C2) = 5 ;

hence, D = 5 * 7, and

x1 = D / 5 = 7

y1 = D / 1 = 35

x2 = D / 7 = 5

y2 = D / 5 = 7.

Now let us consider an example with three pairs. Let m1 = 22, m2 =3

4,3 (C1 + C2) = 7

(C1 - C2 ) = 1


Answer 06/00

6,2 (C1 + C2) = 8 = 2^3

(C1 - C2 ) = 4 = 22

12,1 (C1 + C2) = 13

(C1 - C2 ) = 11;

hence, D = 23 * 7 * 11 * 13, and

x1 = D / 7 = 1,144

y1 = D / 1 = 9,152

x2 = D / 23 = 1,001

y2 = D / 22 = 2,002

x3 = D / 13 = 616

y3 = D / 11 = 728.

From the above example, by taking only the first two partitions we find another pair of solutions: now D = 23 * 7 and the corresponding two pairs are (8, 56) and (7, 14).

Concluding, we may say the number of pairs that satisfy equations of the form of Eq. (1) may be made as large as one pleases, through a suitable choice of R. The procedure given above allows one to build solutions with arbitrary number of pairs.

Y. Kantor: We are still waiting for a general solution of the problem.





Question 07/00

Question 07/00

ENERGETICS OF WALK

Energy expenditure (power) for walking on a level for speed v smaller than 2m/s is approximately given by the expression: P=M+K*v. (At speeds higher than 2m/s the walk is naturally transformed into running. "Race-walking" has higher speeds but involves a very different kind of body motion.) In the expression for power, M represents basal energy spent even when we are standing, which is about 80W (see problem 08/98).

Why do we need to use energy in order to walk on a horizontal plane?

What is the coefficient K in the above expression?





Answer 07/00


ENERGETICS OF WALK

The question was:

Energy expenditure (power) for walking on a level for speed v smaller than 2m/s is approximately given by the expression: P=M+K*v. (At speeds higher than 2m/s the walk is naturally transformed into running. "Race-walking" has higher speeds but involves a very different kind of body motion.) In the expression for power, M represents basal energy spent even when we are standing, which is about 80W (see problem 08/98).

Why do we need to use energy in order to walk on a horizontal plane?

What is the coefficient K in the above expression?

(7/2000) The problem has been solved by Victor Ivanov from Faculty of Physics at Sofia University, Bulgaria (e-mail [email protected]). His solution can be read here in postscript format, It also was solved by Travis Brooks from Stanford (e-mail [email protected]), and by Lihi Goldsmith from Comverse Network Systems (e-mail [email protected]).

The answer and solution:

This is what Philip and Phylis Morrison have to say about walking in the March 1999 issue of Scientific American:

Walking resembles the motions of a pendulum. Consider the walker at the moment when one leg is slanted behind, the other slanted ahead, making an inverted V. Soon the lagging leg rises to pass the other. When they pass more or less straight beneath the torso, both feet are in ground contact, and the walker's center of gravity is raised a little, given that the legs, nearly straight and close to vertical, hold the body's center of gravity higher off the floor than the slanted inverted V did. The inverted V forms again, only now with the legs exchanged. The total mechanical energy alternates in form, changing from the kinetic energy of swinging legs and rising body to gravitational potential energy near maximum body height, and back again. A pendulum handles energy similarly, exchanging kinetic energy near the midpoint of its arc for gravitational energy around the point of least height, then up again. The energy so stored is largely recovered and serves to reduce the energy cost of walking by about half of what would be needed were the swinging legs not also stores of gravitational energy. The walker rolls up and over the high midpoint of a step on near- straight legs, then smoothly down again under gravity.

The work that has to be done to raise the center of mass of the body is the most important energy expenditure. However, it is not the only energy required for walking: We also need to maintain some



http://star.tau.ac.il/QUIZ/00/a07_sol.ps



Answer 07/00

(varying amount) of kinetic energy of the center of mass, as well as some kinetic energy spent on relative (translational and rotational) motion of various limbs relative to the center of mass. The latter energy is relatively small (see, e.g., a very detailed discussion in the paper by P.A. Willems, G.A. Gavanga and N.C. Heglund in The Journal of Experimental Biology 198, 379-393 (1995)). The kinetic energy of the center of mass is rather significant and should be taken into account. On the other hand, the kinetic energy is partially used as to modify the potential energy of the center of mass. Thus, a very crude approximation would be by considering the work needed to raise center of mass. If the inverted V description is close to reality, then assuming the length of the legs of that V is about 1 m while the size of a step is l=0.5 m we find that the height of the center of mass oscillates between "position" /\ and "position" || by about 3 cm, which for 70 kg person would mean that the potential energy changes by A=20 J. If the walking speed is v, then the number of steps per unit time is v/l, and the power required for such walk is P'=A*v/l. Therefore, A/l=40 N is the coefficient K which we needed to find.

This is, of course, a very crude (under)estimate. Real measurements of energy expenditure show that the coefficient K=160 N (approximately).

P.S. We recommend the book Energies, by V. Smil (MIT Press, Cambridge, Mass., 1998). It has some interesting facts about energetics of walking, as well as many useful facts about energy in general.





Question 08/00

Question 08/00

RADIATING SPHERE

Consider a thin charged spherical shell with time dependent radius R(t)=Ro+A*cos(wt) and total charge Q. How much power will be radiated by this shell?

This question has been suggested by M. Schwartz.





Answer 08/00


RADIATING SPHERE

The question was:

Consider a thin charged spherical shell with time dependent radius R(t)=Ro+A*cos(wt) and total charge Q. How much power will be radiated by this shell?

(3/01) We received various solutions of the problem from Jürgen Holetzeck (14/9/2000) from Wiesbaden, Germany (e-mail [email protected]), Nadav Shnerb (20/9/00) from Department of Physics in Judea and Samaria College, Israel (e-mail [email protected]), Michael Cohen (2/10/00) from Department of Physics & Astronomy at University of Pennsylvania (e-mail [email protected]), Kirk T. McDonald (5/10/00) from Princeton University (e-mail [email protected]), Zoran Hadzibabic (8/12/00) from MIT (e-mail [email protected]), Luis F. Rodriguez (27/1/01) from Institute of Astronomy of the National Autonomuos Institute of Mexico (e-mail [email protected]), Sylvain Wolf (5/3/01) from University of Lausanne, Switzerland (e-mail [email protected]), and Itzhak Shapir (18/3/01) (e-mail [email protected]).

The answer: There will be no radiation.

The solution:

There are several ways to see that there will be no radiation from the sphere:

(a) The radiation consists of multipoles (dipole (magnetic or electric) radiation, quadrupole radiation, etc.) The symmetry of the source is "monopole", since clearly there is no directional dependence. Since there is no such thing as "monopole radiation", the spherical shell will not radiate. One can also say that by symmetry both electric and magnetic fields must be radial, while far away from the source the electromagnetic wave must be transverse!

(b) Let us to calculate the electromagnetic fields more carefully. The (particular Fourier component (frequency) of the) vector potential A at position x can be found by integrating the expression (1/c)J(x')exp[(w/c)|x-x'|]/|x-x'| over the positions x' of the source of radiation which is the (Fourier component of the) current density J(x'). Due to symmetry of the source the results of this integration will be of the form f(r)r where f(r) is some function of the distance from the center of the sphere and r is the radius-vector. [See, e.g.., Ch.9 in J.D. Jackson, Classical electrodynamics, Wiley, NY, 2nd edition (1975). The equations above are written in Gaussian units.] Since curl of radial vector A vanishes, the magnetic field will be zero everywhere and therefore there will be no radiation.











Answer 08/00

(c) Gauss law is applicable also in dynamic circumstances. If we draw a large sphere around our source charges, and assume that the electric field is isotropic, i.e. is in radial direction, and depends only on the distance from the center, then we reach a conclusion that the electric field is is constant in time. Consequently, we do not expect any radiation.

(d) Because of the symmetry the magnetic field B must point in radial direction and depend only on the distance from the center. This, together with equation div B=0, leads to the conclusion that B=0 everywhere. Consequently, there is no radiation.

Comment: Michael Cohen told us that this problem appears as a problem in Sec. 14 of the Classical Electricity and Magnetism by W.K.H. Panofsky and M. Phillips.





Question 09/00

Question 09/00

PERPETUUM MOBILE

The perpetual motion machine depicted in this figure was described on the pages of the journal "Power" in the beginning of 20th century. It consists of a well oiled rope hanging on a wheel and passing through a tube filled with water. The Archimedes force makes the right side of the rope lighter than the left side and therefore it will move and rotate the wheel counterclockwise. Or will it?...





Answer 09/00


PERPETUUM MOBILE

The question was:

The perpetual motion machine depicted in this figure was described on the pages of the journal "Power" in the beginning of 20th century. It consists of a well oiled rope hanging on a wheel and passing through a tube filled with water. The Archimedes force makes the right side of the rope lighter than the left side and therefore it will move and rotate the wheel counterclockwise. Or will it?...

(10/2000) The problem has been solved by Cao Zhi Heng (24/9/00) a freshman at University of Tokyo (e-mail [email protected]), by Sahal Yacoob (9/10/00) from the University of Cape Town (South Africa) (e-mail [email protected]), by Keith Selbo (12/10/00) (e-mail [email protected]), by John Moore (26/10/00) from Bio-Imaging Research (e-mail [email protected]).

The solution:

Archimedes law in its simple form (force is equal to the weight of displaced water) applies only when






Answer 09/00

the body is either completely surrounded by water, or is floating on a surface, but is surrounded by water from all other sides. When this is not true one must directly analyze the pressures on the body and calculate the net force. Calculation of this type will convince you that the water has no effect on the motion of the rope. There are many ways to do it: for instance, you can mentally "divide" the rope into tiny disk-like slices and ask yourself what an effect the water has on that particular slice; obviously there will be no effect in the direction along the centerline of the rope. (The water pressure difference will try to push the bottom bend of the rope upwards towards the bend of the tube; this however, has no effect on overall rotation of the rope.) Consequently, the perpetuum mobile will not work.

Heng wrote: In accordance with Archimedes' principle the weight of an object in water is reduced by the weight of displaced water. The buoyancy is created the water pressure difference between the top and the bottom of that object. Thus, an object on the bottom of a sea does not "feel" buoyancy. For the same reason, there will be no "net" buoyancy force in the geometry presented in the picture.

Alternatively, as noted by Selbo, one can say that the force of the water pressure is always (locally) perpendicular to the rope, and therefore will have no contribution along the rope.

A very pleasant and informative book on the history of perpetual engines is Perpetual Motion: The History of Obsession by Arthur W.J.G. Ord-Hume (George Alen & Unwin Ltd., London).

Some related web-sites: History of Perpetual Motion.



http://www.voicenet.com/~eric/dennis4.html



Question 10/00

Question 10/00

CHARGES ON A CIRCLE

Several identical point charges (shown in red) are placed on a circular disk (shown in blue) so as to minimize the electrostatic energy of the system. Obviously, single charge can be placed anywhere as depicted in Fig. (a), while a pair of charges will occupy opposite sides of the diameter of the circle, as in Fig. (b). Similarly, three charges will form an equilateral triangle, as in Fig. (c). What can you say about the geometric arrangement of 4, 5, 6, ... charges?





Discussion 10/00


CHARGES ON A CIRCLE

The question was:


(6/11/00) Y. Kantor: We are getting many replies claiming that the optimal placement of the charges is the equidistant placement along the boundary of the disk, i.e. on the circle. This seems to be true for number of charges n=2 or 3. But is it true for larger n? It is possible that equally spread charges represent a local energy minimum. But is it a global minimum? As an example, let us compare configuration of equally spaced charges on the boundary with configuration in which one charge is in the center of the circle while the remaining are equally spaced on the boundaries. Two such configurations are depicted for n=7 below.

The following table presents the energies of those two configurations for various values of n. (It is assumed that those are unit charges and the radius of the circle is also unity.)

n with charge in center without charge in center 3 2.500000000000000 1.732050807568877


Discussion 10/00

4 4.732050807568877 3.828427124746190 5 7.828427124746190 6.881909602355869 6 11.88190960235587 10.96410161513776 7 16.96410161513776 16.13335409673741 8 23.13335409673741 22.43892676967297 9 30.43892676967297 29.92344919779823 10 38.92344919779824 38.62449897970962 11 48.62449897970962 48.57567511970017 12 59.57567511970017 59.80736151791218 13 71.80736151791218 72.34728957471518 14 85.34728957471515 86.22096479601028 15 100.2209647960103 101.4519980160739 16 116.4519980160739 118.0623677300199 17 134.0623677300200 136.0726314193666 18 153.0726314193665 155.5020983001627 19 173.5020983001626 176.3689723517624 20 195.3689723517624 198.6904720782230

You immediately see that for n=12 it is already worthwhile to put one of the charges in the center, and therefore the configuration where all the charges are on the boundaries is not the global minimum!

Obviously, we did not intend to find the global minimum. We just tried to demonstrate that equally spaced charges on the boundary, are not necessarily the optimal placement of the charges...





Answer 10/00


CHARGES ON A CIRCLE

The question was:


(1/2003) We did not receive an answer to the question. However, Karthik Tadinada brought to our attention a paper by K.J. Nurmela (a postscript file of this work can be found here), where a detailed study of cases containing up to 80 charges has been performed. The paper also contains numerous references to other works on the subject.

The answer as we know it so far:

For number of charges n<12, the minimum energy configuration consists of charges spread out on the boundary. For n=12,13,14,15 and 16, we get one charge in the center and the remainder on the boundary. For n=17 and 18 already two charges are inside the circle. As the number of charges continue to increase, the positions of the charges resemble more and more Wigner crystal (triangular lattice) of charges. As the number of charges approaches infinity, the mean density of the charge distribution approaches the known density of continuous charge distribution on a circle (density ~ 1/sqrt(1-r2), where r is the distance from the center of the disk).



http://star.tau.ac.il/QUIZ/00/charges_on_circle.ps



Question 11/00

Question 11/00

FRYING IN OIL

The boiling point of olive oil is higher that the melting point of tin. How can you explain that it is possible to fry in tinned skillet.

This question was asked by Edoardo Amaldi during a teaching session in Enrico Fermi's room. (Good Italian skillets were made of tin-lined copper.)





Answer 11/00


FRYING IN OIL

The question was:

The boiling point of olive oil is higher that the melting point of tin. How can you explain that it is possible to fry in tinned skillet.

(24/12/2000) The problem has been solved correctly by by Itzhak Shapir (5/11/2000) (e-mail [email protected]), by John Moore (20/11/2000) (e-mail [email protected]), by James Montaldi (29/11/2000) (e-mail [email protected]), by Will Brunner (10/12/2000) (e-mail [email protected]), and by Lior Fainshil (15/12/2000) from Tel Aviv University (e-mail [email protected]).

The answer: Oil does not boil, when frying. It is water in the food that boils! And the boiling water keeps the temperature of the oil and pan almost fixed. (However, when the food is left for too long time to fry most of its water is lost and then the food may be burnt, as well as the oil and the pan.)

This question was asked by Edoardo Amaldi during a teaching session in Enrico Fermi's room. It was answered by a student Miss Ginestra Giovene who later became Mrs. Amaldi. The story is related by Laura Fermi in the book Atoms in the Family (My Life with Enrico Fermi) [Tomash Publishers (American Institute of Physics), 1987 (Vol. 9 in the series "The History of Modern Physics 1800-1950"); Originally published by Univ. of Chicago Press (1954)]











Question 12/00

Question 12/00

COAXIAL CABLE

A coaxial cable is built of two parallel metal cylinders of arbitrary cross-section. If current is flowing in the inner cylinder and an equal and opposite current is flowing in the outer cylinder, then magnetic field will be present inside the cable, and one can define inductance L per unit length of the cable. Similarly, one can charge the inner and outer cylinders with opposite charges, and view it as a capacitor with capacitance C per unit length of the cable. Show that expression (LC)1/2 is independent of the shape of the cross-sections and find its value.

This was a homework question in the course of Applied Electromagnetism given by Y. Kantor in 1996-99.





Answer 12/00


COAXIAL CABLE

The question was:

A coaxial cable is built of two parallel metal cylinders of arbitrary cross-section. If current is flowing in the inner cylinder and an equal and opposite current is flowing in the outer cylinder, then magnetic field will be present inside the cable, and one can define inductance L per unit length of the cable. Similarly, one can charge the inner and outer cylinders with opposite charges, and view it as a capacitor with capacitance C per unit length of the cable. Show that expression (LC)1/2 is independent of the shape of the cross-sections and find its value.

(5/2001) The problem has been solved correctly by Victor Ivanov (5/12/2000) from the Faculty of Physics, Sofia University, Bulgaria (e-mail [email protected]), by Zilong Chen (8/12/2000) from MIT (e-mail zilong), by Zoran Hadzibabic (8/12/2000) from the School of Science at MIT (e-mail [email protected]), by Ee Hou Yong (14/1/2001) from Stanford University (e-mail [email protected]), and by Itzhak Shapir (18/3/2001) (e-mail [email protected]).

The answer: 1/c, where c is the speed of light.

The solution: There are several possible approaches to the problem. (The solutions are presented in Gaussian units.)

Solution No. 1:








Answer 12/00

The problems of calculation of C and L (per unit length) are essentially two-dimensional problems. The equations above prove the required statement. Below we explain the proof:

Calculation of capacitance requires solution of two-dimensional electrostatic equation, i.e. finding a harmonic two-dimensional function F(x,y) which has a constant value (say, V1) on the inner surface, and

a different constant value (say, V2) on the outer surface. ["Harmonic" means a function with vanishing

Laplacian.] The capacitance per unit length of the cable will be determined by Q/(V1-V2), where the

charge per unit length Q is determined by the the integral of the normal component of the electric field (grad F) along, say, inner surface, i.e. the (two-dimensional) electric flux, divided by 4{pi}.

In two dimensions any harmonic function F can be treated as a real part of an analytic function. The imaginary part G of the same analytic function will also be a harmonic function representing a potential of a field which is everywhere perpendicular to the field created by the potential F. [See the proof of this property in Ch. 4, of W.K.H. Panofsky and M. Phillips Classical Electricity and Magnetism, Addison-


Answer 12/00

Wesley, 2nd ed. (1972). Of course, the existence of potential G can be also proven directly without introduction of analytic functions.] Moreover the two functions will satisfy Cauchy-Riemann relations: x-derivative of F coincides with y-derivative of G, x-derivative of G coincides with minus y-derivative of F. From these relations, we can directly see that if we choose two points in a two-dimensional plane, the potential difference between the two points calculated using one of the potentials, i.e. integral of a field along some line joining the points, coincides with a flux of the other field through a line joining the points.

One can easily convince himself that grad G represents the magnetic field which we are looking for the calculation of the inductance: In the empty space between the cylinders divergence of the magnetic field is supposed to vanish, which will happen if it is a gradient of a harmonic function. One can also convince himself that this solution satisfies the correct boundary conditions.

Now we note, that according to the Biot-Savart's law the line integral of the magnetic field is [4{pi}/c]I, where I is the current in the inner cylinder. However, as we mentioned in the second paragraph such line integral should be equal to the electric flux of the electrostatic problem. Alternatively, the potential difference in the electrostatic problem is equal the flux of the magnetic field in the magnetic problem. It is this flux which enters the definition of the inductance. [Inductance (per unit length) is flux divided by the current and by c.]

Now we can do a sequence of substitutions: inductance is expressed in terms of magnetic flux and current. Flux can be expressed in terms of voltage difference in electrostatic problem which can be expressed in terms of capacitance and electric flux. The electric flux coincides with line integral of magnetic field, and consequently can be expressed in terms of current. If we execute this sequence of substitutions we see that everything cancels out and we are left with an expression relating the product of capacitance and inductance to the inverse of c2.

Solution No. 2: Solution suggested by Zoran Hadzibabic essentially coincides with the solution described above. However, he chooses a simpler approach. He first considers a case of linear currents and linear charges for which both the electrical and magnetic fields can be easily found and related to each other. In particular, one can easily see that electric and magnetic fields are perpendicular to each other. Now, he "builds" the solution of the problem in question by construction electrostatic problem from linear charges and magnetostatic problem from linear currents and noticing that correct boundary conditions (requiring minimization of energy) are identical for both problems. Consequently, he finds that relation between the product of capacitance and inductance and the speed of light in the general case remain the same as in trivial cases.

Solution No. 3: Solution suggested by Victor Ivanov was very original and different from most of the submitted solutions. First he assumed that an electrostatic problem has been solved and expressed the capacitance via electrostatic energy (i.e., an integral of squared electric field). Then he looked at the same problem from a reference frame moving with small velocity v along the axis of the of the cylinder.


Answer 12/00

To the lowest order in in v/c the relativistic effects do not change the charge density. Thus the current will be simply the charge density multiplied by v. On the other hand, the magnetic field B in the new reference frame will be related to electric field E in the old reference frame by B=vE/c. The transverse dimensions of the cable will not be changed by relativistic effects. We can now use the knowledge of the new magnetic field and the new current to calculate the inductance (e.g., via energy relations) and relate it to the capacitance of the original problem, and to obtained the desired answer.

Solution No. 4: This solution is "a little bit cheating". Nevertheless, it is a valid solution. We first think of the couple of cylinders as a wave-guide. Consider a transverse electromagnetic wave propagating through the guide. For detailed desciption of such waves see, e.g., chapter 8 in J.D. Jackson Classical Electomagnetism, 2nd ed., Wiley, NY (1975). It can be shown that the speed of propagation of such wave is c independently of the cross section of the guide. On the other hand we can view the same problem as a "classical" transmission line problem, of a series of capacitors and inductances, and show that the speed of propagation of the wave is the inverse of the square root of LC. This leads to the desire result. Of course, strictly speaking the "transmission line" treatment is "quasistatic"; however, it will be correct at low frequency, and it is sufficient for us to prove it at any frequency.





Question 01/99

Question 01/99

MIND READING

Thought is an electrical activity of the brain and therefore it should be possible (at least in principle) to read your mind from a distance by monitoring the electromagnetic radiation. Unfortunately, the useful signal will probably drown in the sea of noise. What is the minimal time required to receive one bit of information (one bit of thought!) transmitted by your mind at a distance of one meter?

This question was considered in 1970s by D.S. Chernavskii. We will appreciate any info and references on works regarding this specific question.





Question 02/99

Question 02/99

INTERSTELLAR TRAVEL

In 1960 Bussard proposed an interstellar vehicle that would collect its fuel from interstellar medium. A large scoop would pick up protons present in space, and the protons would be fed into a fusion reactor where part of their mass would be converted into kinetic energy. How big would the scoop have to be in order to produce a constant 1g acceleration for a 3500-ton vehicle?





Answer 02/99


INTERSTELLAR TRAVEL

The question was:

In 1960 Bussard proposed an interstellar vehicle that would collect its fuel from interstellar medium. A large scoop would pick up protons present in space, and the protons would be fed into a fusion reactor where part of their mass would be converted into kinetic energy. How big would the scoop have to be in order to produce a constant 1g acceleration for a 3500-ton vehicle?

(02/2000) The problem has been solved correctly by Lenny Eusebi, Physics student at Wesleyan University in Middletown, Connecticut (e-mail [email protected]). A detailed discussion of the problem can be found in the article by R.W. Bussard, Astronautica Acta 6, 179 (1960). Short solution of the problem is presented below.

The answer: The radius of the scoop should be of order of 10,000 km.

The solution:

The acceleration a (=10 m/sec2) of a vehicle of mass M=3500 ton will be

a=nmc2Afk / M

where n is the density of protons in the interstellar space (=0.1 cm-3, or slightly above that value), m is the mass of the proton, c is the speed of light, A is the area of the scoop, f fraction of mass converted into energy (it can be estimated by comparing, say, binding energy of He or deuterium with its mass; this number is smaller than 0.01). k efficiency with which the fusion energy is converted into thrust. (Of order 1(?).) This equation is obtained by equating the required force Ma with energy which can be released by protons in space in a volume which is multiplication of the area of the scoop and unit length. From this expression we can extract the area (and, thus, the radius) of the scoop.






Question 03/99

Question 03/99

HALF-EMPTY BOTTLE

You are riding in a train and have just opened a bottle of beer. How much of the beer should you drink so that after putting your partially empty bottle on a table (which is shaking) it would be as stable as possible?

This question was suggested by Sergei Flach.(3/99) This question appears in the physics problem book by H. Vogel. (We thank Dr. Fred Goesmann for bringing this to our attention.)





Answer 03/99

Answer to Question 03/99

HALF-EMPTY BOTTLE

The question was:

You are riding in a train and have just opened a bottle of beer. How much of the beer should you drink so that after putting your partially empty bottle on a table (which is shaking) it would be as stable as possible?

(3/99) The problem has been solved by Eran Toledo (e-mail [email protected]), a Ph.D. student at Tel Aviv University, and by Andreas Wendler (e-mail [email protected]) from JENOPTIK SYSTEMHAUS, Jena, Germany. We have also been told that the problem and its solution appear as problem 2.3.2 in Probleme aus der Physik by H. Vogel (Springer-Verlag).

The solution:

(3/99) The most stable condition of the bottle is when its center of mass is lowest. Let the mass of the empty bottle be M, the height of its center of mass H, and the cross-sectional area A, while the density of liquid is d and the height of the liquid is h. Then the height of the center of mass of the bottle with the liquid is

(M H+d A h2/2)/(M+d A h),

since the mass of the liquid is d A h and the height of the center of mass of the liquid is h/2. By taking the derivative with respect to h and equating it to zero we find the condition for minimizing the height of the joint center of mass. The solution is

h=[M/(d A)] (sqrt(1+2 d A H/M)-1)

If the bottle is heavy (M >> d A H) this expression has a very simple approximate form: The optimal height h is simply equal to H, i.e., the level of the liquid should be at the position of the center of mass of the bottle.

It is interesting to note that whatever the relative weight of the bottle and the liquid, when the level of the beer h is optimal it coincides with the height of the center of mass of the entire system (bottle+beer).

(2/2000) Peter W. Martin from Bureau International des Poids et Mesures (e-mail [email protected]) suggested a very nice and simple argument proving that the total center of mass will be at minimum when the level of the beer coincides with the level of the total center of mass. His agument is independent of the shape of the bottle and goes as follows:





Answer 03/99

Imagine filling an empty bottle. Clearly the center of gravity of the system will start to decrease in height because we are adding matter below the total center of gravity. This will continue until the center of gravity reaches the level of the liquid. At that point the addition of more liquid will then cause the center of gravity to rise.





Question 04/99

Question 04/99

CUP-A-TEA

When you finish stirring sugar into your cup of tea the water comes to rest in a few seconds. Is the decay of the water's rotation caused primarily by the walls of the cup or by its bottom?





Discussion 04/99


CUP-A-TEA

The question was:

When you finish stirring sugar into your cup of tea the water comes to rest in a few seconds. Is the decay of the water's rotation caused primarily by the walls of the cup or by its bottom?

(9/99) Andrew Wiggin (e-mail [email protected]) sent us the following e-mail (slightly edited):

There's no order of magnitude difference between the two effects (bottom vs. walls), but I'm still fairly confident which way that factor of 2 or 3 goes. We have definite evidence for convection (piling up of sugar at the center of the cup), and we know that takes up energy, so the bottom definitely has a thing going for it. On the other hand the walls do have a bigger area, but if we get into the dependence on the dimensions of the cup, things will get nasty... Also, I tried another experiment; if you shake the cup, the oscillations do seem to go on for longer than the rotation when you stir it. Admittedly, it's hard to do this experiment in an unbiased way, and the geometry complicates it a bit. But think about shaking a rectangular aquarium, or just disturbing water in a baby's bath tub, the way I remember such situations, there is definitely less damping there

(10/00)John Moore from Bio-Imaging Research (e-mail [email protected]) sent us the follwoing e-mail (slightly edited):

The local frictional force (to a first approximation) depends on the velocity, which is probably constant at its peripheral value V along the sidewall of height H, but decreases linearly from the outer radius R towards the center on the bottom. Integrating the local force times the area, we have V x 2R x H on the sidewalls, and V x 2(R2)/3 on the bottom. Thus, if R > 3H, the bottom will have more influence. But few cups are six time wider than they are high, so the snswer favors the walls. Note : the flow pattern won't be as simple as assumed above, but it's unlikely to change the conclusion for reasonable cups. Note: An exact solution will depend on how the rotational flow was set up: was the cup rotated as a whole (and for how long?), or was it stirred (and from the center or the outside, and with a spoon or something else?)? However it is started, the three-dimensional flow field will be continuously altering, which means that the sidewall-to-bottom force ratio will also be changing. If one insists on an exact answer, the initial conditions must also be specified exactly.







Question 05/99

Question 05/99

CIRCUIT OF BATTERIES

Twelve identical 1 volt batteries are connected into an electrical circuit as shown in the figure. What will the voltmeter show?

This question was created by Y. Kantor and A. Palevski.





Answer 05/99


CIRCUIT OF BATTERIES

The question was:

Twelve identical 1 volt batteries are connected into an electrical circuit as shown in the figure. What will the voltmeter show?

(5/99) The problem has been solved by Eitan Federovsky (e-mail [email protected]), by Andy Tan (e-mail [email protected]), physics tutor at the National Junior College, Singapore, and his colleagues, and by D. Augier (e-mail [email protected]).

The answer: The voltmeter will show zero voltage.

The solution:

Assuming that the voltmeter has infinite resistance, the current in the loop will be equal to the total electromotive force, 12 volts, divided by the total resistance, 12 times the internal resistance R of one battery. So the current will be (1 V)/R. The internal voltage drop in each battery (current times internal resistance) will therefore equal R (1 V)/R=1 V which is the electromotive force on that battery. The voltage drop across a single battery will be equal to the difference between the electromotive force and the internal voltage drop and therefore it will be equal to zero. The same is true for any group of batteries.








Question 06/99

Question 06/99

STRETCHED MOLECULE

A simple 3-atom molecule can be modeled as two infinitely rigid rods ("bonds") of length l which are freely jointed, i.e., the angle between those two rods can vary freely (see the picture). The molecule is in thermal equilibrium with its surrounding at temperature T. With what force F must the ends of the molecule be pulled apart in order that the distance between the end-atoms be h? (Obviously h < 2l.) What is the origin of the resistance of the molecule to stretching?





Answer 06/99


STRETCHED MOLECULE

The question was:

A simple 3-atom molecule can be modeled as two infinitely rigid rods ("bonds") of length l which are freely jointed, i.e., the angle between those two rods can vary freely (see the picture). The molecule is in thermal equilibrium with its surrounding at temperature T. With what force F must the ends of the molecule be pulled apart in order that the distance between the end-atoms be h? (Obviously h < 2l.) What is the origin of the resistance of the molecule to stretching?

(5/99) The problem has been solved (partially) by Andrew Wiggin (e-mail [email protected]).

The solution:

If the end atoms are held in place then the central atom can move in a circle of radius r=sqrt{l2-(h/2)2}. Thus the entropy of the system will be S=k ln(2{pi}r), so the configurational part of the free energy will be -TS. Taking a derivative with respect to h we find the force:

F=kTh/(4l2-h2)

While we usually say that "the force has entropic origin," the true mechanical reason is quite simple. At temperature T, the center atom will have mean kinetic energy (1/2)mv2=(1/2)kT. Since it is moving along a circle of radius r, it will need a centripetal force mv2/r =kT/r applied to it. A simple geometric calculation shows that the force F needed to create such centripetal force is exactly what we got above.



Answer 06/99

(By the way, this is exactly the principle on which the mechanical device called an inertial regulator is based: The change in speed of rotation changes the force F which is used to regulate the fuel supply to, say, a steam engine.)





Question 07/99

Question 07/99

MISSING ENERGY

A body starts sliding (from rest) without friction down a hill of height h. In the beginning the body has potential energy and no kinetic energy; at the end the body has only kinetic energy. It follows from energy conservation, mgh=½mv2, that the velocity of the body at the bottom of the hill will be v=(2gh)1/2.

Now consider the same process from a reference frame moving with velocity v

(to the right) relative to the ground. In this reference frame, the body in the beginning has potential energy mgh and kinetic energy (1/2)mv2, while at the end of the process both energies vanish. Where did the energy disappear? Give a detailed account of missing energy!





Discussion 07/99


MISSING ENERGY

The question was:A body starts sliding (from rest) without friction down a hill of height h. In the beginning the body has potential energy and no kinetic energy; at the end the body has only kinetic energy. It follows from energy conservation, mgh=½mv2, that the velocity of the body at the bottom of the hill will be v=(2gh)1/2.



(7/99) Y. Kantor: Probably, it is our fault, that many people misunderstood the question. Many people sent in solutions which are correct but are NOT what we wanted to hear. Roughly speaking, many solutions were as follows:

Consider the system of Earth and the body in the moving reference frame. In this system, we can use the momentum conservations to determine the velocity of Earth, and from there we can verify that the energy of the system Earth+body is conserved.

We received several solutions of this kind. Some of them used the momentum conservation directly, while others did it less directly, i.e. used the dynamics laws in a form which would lead to momentum conservation. Solutions of this kind were received from G. Fedel (e-mail [email protected]), A. Esztermann (e-mail [email protected]), Y. Kats (e-mail [email protected]), and D. Augier (e-mail [email protected]).

There is something strange in solutions of this kind: momentum conservation is used to derive the law of energy conservation. Clearly, this is not possible. Nevertheless, people succeeded doing it because they assumed that they know that the body will be at rest (relative to the moving frame of reference) when it reaches the bottom of the hill. However, this assumption is a consequence of energy conservation, and, therefore, cannot be used in a closed consistent derivation.






Discussion 07/99

What we really want, is a solution which uses dynamical laws in the moving frame of reference and proves energy conservation. In particular, we want a detailed account where and when does the energy of the body go. Let's say it explicitly: the energy of the body decreases because it performs work = negative work is performed on the body; when and what amount of work is performed?





Answer 07/99


MISSING ENERGY

The question was:

A body starts sliding (from rest) without friction down a hill of height h. In the beginning the body has potential energy and no kinetic energy; at the end the body has only kinetic energy. It follows from energy conservation, mgh=½mv2, that the velocity of the body at the bottom of the hill will be v=(2gh)1/2.



(7/99-11/99) The problem has been solved by Gerrit Danker (e-mail [email protected]), by Fernando Ferreira (e-mail [email protected]), and by Adi Armoni (e-mail [email protected]).

The solution:

Before we begin the solution it is important to understand several basic facts about this problem. [At the very end we will present a solution of a more general problem, but we start with the problem in the form in which it was posed.] The following figure depicts the horizontal and vertical components of the velocity in the reference frame of the ground. Both velocities are linear functions of time t as long as the body is on the slope. However, when the body reaches the bottom of the hill the direction of its velocity suddenly changes. At that moment there is no change in the magnitude of the velocity, but both vertical and horizontal components change suddenly.






Answer 07/99

Now let us look at the same events from the reference frame which moves relative to the ground with velocity v to the left. In this reference frame, the discontinuities in the velocity components are still present, and therefore the total velocity undergoes a sudden change from a finite value to zero, when the body reaches the bottom of the slope and goes into the horizontal plane. Clearly, in what follows we will need to consider both what happens on the slope and what happens at the last moment of the slide - during the transfer from the slope to the horizontal plane.


Answer 07/99

In the moving frame of reference the initial energy of the body is mgh+½mv2=2mgh. All our calculations will be performed in the moving frame of reference. In this frame (which moves with velocity v relative to the ground) the body is no longer moving in the direction parallel to the direction of the slope of the hill. Nevertheless, we can consider the components of the velocity which are parallel and perpendicular to the slope. Since the body remains in contact with the slope, its velocity perpendicular to the slope does not change, and therefore there is no acceleration in that direction. Therefore the component of the weight mg cos A, where A is the angle of inclination of the hill, is compensated by the normal force of equal size (and opposite direction). The acceleration of the body in the vertical direction is a=g-g cos2 A=g sin2 A, which is the difference between the weight and the vertical component of the


Answer 07/99

normal force of the hill divided by m. Since the vertical component of the initial velocity vanishes, the time required to travel from height h to the bottom of the hill is

t=sqrt{2h/a}=sqrt{2h/(g sin2 A)}.

The work performed by the normal force of the hill will be equal to its product with the distance that the body travels in direction perpendicular to the slope of the hill (with velocity v sin A). The distance traveled by the body in the direction perpendicular to the hill is v sin A t=v sqrt{2h/g}. Thus the work performed by the hill on the body is:

W1=-mg cos A v sqrt{2h/g}= -2mgh cos A.

Note that this work depends on the slope and does not compensate for all of the energy loss 2mgh, i.e., the body reaches the bottom of the slope with kinetic energy 2mgh(1-cos A). This, however, is not the end of the story. At the very bottom of the slope the body moves from the slope to the horizontal plane. For a very short time it moves on a highly curved path. In the reference frame of the Earth, this last stage modifies the momentum, because the direction of motion changes; however, the energy remains unchanged. In the moving reference frame, work is performed by and on the body during this last stage of motion.

There are several ways to calculate the work performed on the body at this last stage. A clear intuitive understanding of the process can be obtained if one replaces the sharp corner (the transition between the slope and the horizontal plane) by an arc of a small circle which interpolates between the slope and the horizontal plane. The angle of the arc as viewed from the center of that circle is A. Now one needs to evaluate the work performed on the body while it moves along the arc. Of course, the force depends on the radius of curvature R of the arc, but the final answer does not depend on that radius.

By analyzing the motion of the body in the moving reference frame, it can be shown (we do not give the explicit derivation) that the force that the surface applies on the body is given by mU2/R, where U is the velocity in the reference frame of the Earth. Moreover, the reader can convince himself that U does not change during the motion along the arc and is equal to v, although in the moving reference frame the velocity decreases with time. When the radius vector drawn from the center of curvature to the position of the body changes its angle B from the vertical by dB, then the work performed on the body is mv2sin B dB. As angle B changes from A to 0, by integrating the expression for the work we find that the total work performed is exactly -2mgh(1-cos A), i.e., it eliminates the remaining energy of the body. In this paragraph we did not actually calculate the work performed on the arc, but only indicated how it can be done.

All the above remarks were somewhat specific to the given shape of the hill. Adi Armoni solved the problem for a hill of any shape. The postscript file of his solution is HERE. Note that his solution is very general. Read it and compare with the solution for the particular shape of the hill presented above.



http://star.tau.ac.il/QUIZ/99/hill_sol.ps



Question 08/99

Question 08/99

ZEROTH ORDER RAINBOW

A ray of light coming from the sun (A) enters a water drop, is reflected internally once, and exits in direction B. The angle of maximal intensity of this reflected light depends on the index of refraction of the water and thus is slightly different for different colors. This gives rise to the primary rainbow at about 42 degrees opposite to the direction of sun. (The secondary rainbow corresponds to two internal reflections.) Question: Why can't we see a zeroth order rainbow by looking at rays like C that exit the drop without internal reflections?

This problem was suggested by D. J. Bergman





Discussion 08/99



The question was:


(8/99) Dr. Fred Goesmann from Max-Planck-Institut fuer Aeronomie (Katlenburg-Lindau) (e-mail [email protected]) wrote:

Are you sure such a rainbow is never seen? A few weeks back I saw something which would fit. It was near sunset and coloured bands were visible to the right and to the left of the sun (I do not recall the angle, but I would guess it to be about 20 degrees to either side, maybe less). The coloured patterns were only visible near the horizon and on the same height as the sun. Of course one would probably need a highly transparent thin curtain of droplets through which the sun is seen, but is this outright impossible? I do not think so.

Y. Kantor: I am not sure what is described in the above observation. However, I am guessing that these



Discussion 08/99

were coronae - concentric colored rings repeating around the sun or moon several times(!) with colors going from white to blue to green, yellow and red and repeating. This is caused by interference of light diffracted in drops of water. This effect should be distinguished from (much stronger) rainbows caused by geometrical optics effects. Beautiful pictures of coronae can be found in the book Color and Light in Nature, by D.K. Lynch and W. Livingston (Cambridge Univ. Press, 1995).

We welcome comments on the subject.





Answer 08/99



The question was:


(9/99) The problem has been solved by Yevgeny Kats (e-mail [email protected]), by David Augier (e-mail [email protected]), and by Yinon Arieli (e-mail [email protected]).

The solution:

Using Snell's law one can easily show that an incoming ray that has an incidence angle (angle between the ray and the radius of the drop, i.e., the normal to the surface) denoted by a will be deflected by an angle b=2[a-arcsin((sin a)/n)], where n is the index of refraction of water (approximately 1.33).





Answer 08/99

Different rays are refracted in different directions. The intensity of the observed light depends on how much light is scattered in a given direction and therefore it depends on the rate at which the angle of deflection changes as the impact parameter of the incoming ray changes. The slower the change, the larger the intensity.

In ordinary (i.e., first or second order) rainbows the derivative db/da vanishes for some value of a. At that (stationary) point the intensity of the light will be maximal. Since that optimal angle depends on n, and the index of refraction depends on color, each color will have slightly different angle of maximal intensity. This is how a regular rainbow is created. (Actually it would be more correct to measure the deflection angle as a function of impact parameter, rather than incidence angle. However, the point at which db/da=0 is obviously also the point at which the derivative with respect to impact parameter vanishes.)

In the zeroth order rainbow the derivative never vanishes. The equation db/da=0 leads to the condition n=1, which cannot be satisfied. Therefore there will be no zeroth order rainbow!

Comment (28/12/01): The mathematics of regular rainbow is nicely explained here.

Comment (14/7/03): Tobias Hackstock brought to our attention that the entire discussion above relies on the fact that we are considering spherical drops. Ice crystals, naturally, produce optical phenomena (such as "sundogs", "halos") detrmined by the fact that the crystal has flat faces, and the direction of transmission depends on the orientation of the crystal. (Consequently, the tranmission will be maximum when transmission angle is stationary as a function of the orientation of the ice crystal.) This web site provides a detailed explanation of several related phenonomena.



http://www.webnexus.com/users/billv/rainbow/rainbowmath.html

http://www.sundog.clara.co.uk/halo/parhelia.htm



Question 09/99

Question 09/99

MEAN DEFLECTING FORCE

Consider a bead sliding without friction on a wire placed in a horizontal plane and forming a closed loop. The force acting on the bead is always perpendicular to the direction of its motion, to the right or to the left depending the direction the wire is curving. A force to the right or to the left, relative to the direction of the bead's motion, is defined as positive or negative, respectively. Let the mass of the particle be m and its (constant) velocity v, while the length of the wire loop is L.

(a) If the loop doesn't intersect itself, what is the time average of the perpendicular force, and how does it depend on the shape of the loop?

(b) What if the wire does cross itself? (This means that parts of the loop are slightly out of horizontal plane.)

This question was suggested by Y. Kantor. It was inspired by Newton's first attempt to calculate centripetal force as it appears in his "Waste Book" (dated 1664-1665).





Answer 09/99


MEAN DEFLECTING FORCE

The question was:

Consider a bead sliding without friction on a wire placed in a horizontal plane and forming a closed loop. The force acting on the bead is always perpendicular to the direction of its motion, to the right or to the left depending the direction the wire is curving. A force to the right or to the left, relative to the direction of the bead's motion, is defined as positive or negative, respectively. Let the mass of the particle be m and its (constant) velocity v, while the length of the wire loop is L.

(a) If the loop doesn't intersect itself, what is the time average of the perpendicular force, and how does it depend on the shape of the loop?

(b) What if the wire does cross itself? (This means that parts of the loop are slightly out of horizontal plane.)

(9/99-2/00) The problem has been solved by Andrew Wiggin (e-mail [email protected]) (his solution is presented below), by Christian von Ferber from University of Duesseldorf in Germany (e-mail [email protected]), by Baruch Meerson from Hebrew University in Jerusalem (e-mail [email protected]), by Jhinhwan Lee a Ph.D. student at Center for Science in Nanometer Scale at Seoul National University, Korea (e-mail [email protected]), and by Sumit Banerjee from Indian Institutes of Science and Astrophysics (e-mail [email protected]).







Answer 09/99

The answer:

The mean deflecting force for non-self-intersecting loop is 2{pi}m*v2/L, - just like for a circle. It does not depend in the details of the shape. For a general figure the prefactor 2{pi} should be replaced by the total angle by which the direction of bead motion was rotated.

The solution:

First we assume that the loop has no crossings. The positive/negative direction of the force depends on the positive/negative curvature radius R, corresponding to deflection to the right/left. The instantaneous acceleration is always

a = v2/R = F/m

Since the speed is constant, change in the velocity dv directly maps into the deflection angle d(theta).

a dt = dv = v d(theta)

Hence, integrating

Fav = 1/T int[F dt], where T = L/v trivially gives that the average force depends only on the total deflection angle, delta(theta)

Fav = (m v2/L) delta(theta)

For a simple loop delta(theta) = +/- 2{pi}. So ignoring the overall sign, Fav = 2{pi}m*v2/L, same as for the circle.

In (b), one needs a careful evalution of the total angle by which the velocity was rotated. It is quite obvious that this will be |delta(theta)| = 2{pi}*N, where N is an integer that tells how many complete turns are made by the loop. It has to be determined from the geometry of the loop.

We do not know a simple rule relating the total angle of rotation and the shape of the curve. (Of course, one can simply integrate the d(theta) and find the angle analytically.) There are some simple relations: e.g., if the number of self-intersections of the projection of the curve is even (odd), then the total number of rotations is odd (even). Unfortunately, this is insufficient... We received many suggestions how to calculate the number of rotations. The suggestions, were either as complicated as direct evaluation of the angle by "following the curve" or they included poorly-defined rules which needed to be refined every time a more complicated shape was considered.





Question 10/99

Question 10/99

3D BILLIARD

Can a billiard ball inside a cube bounce off every wall once and return to its starting point?

Comment: This is a "mathematical" problem. Disregard gravitation, etc.





Answer 10/99


3D BILLIARDS

The question was:

Can a billiard ball inside a cube bounce off every wall once and return to its starting point?

(10/99) The problem has been solved by Bill Schwennicke (e-mail [email protected]), by Andrew Wiggin (e-mail [email protected]), by Avi Nagar (e-mail [email protected]), and by Heinz Kabelka from the Inst.f.Experimentalphysik, Vienna, Austria (e-mail [email protected]).

The answer: Such a path exists. It was originally found by Hugo Steinhaus. (See M. Gardner's Sixth book of mathematical games from Scientific American, Freeman, San Francisco, 1971. It describes several problems of this type.) Below is one possible solution. It is followed by solutions submitted to us, which expand upon the solution and offer an intuitive argument.

One solution:

Say the size of the cube is 3a, and its corners are located at coordinates(0,0,0,), (3a,0,0), (0,3a,0), (0,0,3a), (3a,3a,0), (0,3a,3a), (3a,0,3a), (3a,3a,3a).A path satisfying the conditions of the problem connects the points(0,a,2a) -> (a,2a,3a) -> (2a,3a,2a) -> (3a,2a,a) -> (2a,a,0) -> (a,0,a) -> (0,a,2a). The length of each straight segment of the path is sqrt{3}a. The projection of this path onto any face of the cube is a rectangle, which proves that at any impact point the angle of impact is equal to the angle of reflection. The following figure depicts the path:







Answer 10/99

Bill Schwennicke noted that there are infinitely many solutions of this type. Here is his solution of the problem (slightly edited):

Assume a unit cube with opposite corners at (0,0,0) and (1,1,1). An infinitesimal ball heading out from (0,0,0) at an equal angle from each axis would hit the cube at (1,1,1) and then bounce straight back. Offsetting the starting point a little, but heading out at the same angle, the ball would proceed as in the following example: (0,1/8,2/8), (6/8,7/8,1), (7/8,1,7/8), (1,7/8,6/8), (2/8,1/8,0), (1/8,0,1/8), (0,1/8,2/8), etc., thereby making the required circuit. A finite-sized ball with radius r can be substituted for the infinitesimal one by expanding the cube by an amount r in each direction, and having the above coordinates be for the center of the ball.

Andrew Wiggin presented a nice argument:

On every collision, the normal component of the velocity just reverses, and the other two are conserved. Hence the directions are decoupled, and the answer doesn't depend on the number of dimensions. All that matters is that the period of the motion in each direction is the same, i.e., that the absolute value of the velocity is the same in all directions. So the starting velocity should be of the form V = c(±1, ±1, ±1), and the starting point can be anything. The ball will always move parallel to one of the body diagonals of the cube.

P.S. (11/99) Avi Nagar and Heinz Kabelka made the following observation (we present A. Nagar's message, slightly edited): I just wanted to point out the similarity of this problem to a corner-reflector which is desciribed as an experiment in one of the pages of the Melbourne University Physics Department (http://www.ph.unimelb.edu.au/lecdem/oa6.htm). [Before you continue, take a look at this experiment by clicking here.] The problem can be viewed as a combination of two corner reflectors. Since each reflects light (or a billiard ball) parallel to its original direction, we need to find a direction that will ensure that after two corner reflections we'll be back at the exact starting direction/point. The


http://star.tau.ac.il/QUIZ/99/oa6.html

Answer 10/99

symmetry of the cube ensures that any direction parallel to the big diagonals is a good solution (because not only is the beam reflected from a corner reflector parallel to the incoming beam, but it also has the same offset from the corner).





Question 11/99

Question 11/99

OHM'S LAW

Nowadays high school and university students use voltmeters and ammeters to demonstrate Ohm's law. These devices were not available in Ohm's time! (Moreover, these devices rely on laws (including Ohm's law) not known at the time.) How did Ohm do it? Suggest experiments (using only equipment available in 1826, or similar present-day equipment) showing that: (a) "voltage" is proportional to "current"; (b) for fixed "voltage", the "current" is proportional to the cross-sectional area of the wire and inversely proportional to its length.





Discussion 11/99


OHM'S LAW

The question was:


(2/2000) Ahmet Uysal (e-mail [email protected]) sent us the following e-mail:

There are lots of interesting points in your question. Lets have a look:

* In 1820 Oersted showed that a current can affect a magnet. It was possible that to make a galvanometer in 1826. And Ohm produced his tools by himself.

* More important point is a voltmeter has same principle with ammeter. And both of them stand on Ohm's law. How can you prove a law by using itself?

So, this is very interesting point. But I think you have to change the way of your question. It is far from the reality. You can look at these documents to get more real information:

1- Electromagnetics: History, Theory and Applications ; Robert Selliot. 2- Intellectual Mastery of Nature: Physicists from Ohm to Einstein. 3- American Journal of Physics, 1963, sh.544






Answer 11/99


OHM'S LAW

The question was:


(11/00) No good complete solution has been recieved. A partial solution of the problem has been provided (8/8/00) by Javier Groshaus (e-mail [email protected]) from the Physics Faculty in Technion (Haifa, Israel). Below we present our "editorial" solution of the problem. Some of Groshaus' suggestions appear in the comment at the bottom of this page.

The solution:

As a voltage source we can use batteries. Ohm first used "chemical batteries" but those had very short life-time, i.e their electromotive force (EMF) varied during the experiment. Eventually, he used a thermocouple as a voltage source. The voltage, or more correctly the EMF, was therefore N*E where N is the number of batteries connected in series and E is EMF of a single battery. If the resistance of wire was X, and internal resistance of the battery was r, then the current I was given by

I=N*E/(N*r+X)

Current can be measured by placing a magnetized needle hanging on a string at certain fixed distance from the wire. Since we are not supposed to know exactly the angular dependence of the magnetic force, we will simply for each current twist the string on which the needle hangs until the needle returns in its position before the current began flowing. The angle by which the string was twisted is proportional to the force momentum, and thus measures the strength of the current. (Of course we assume that forces are indeed proportional to the current.)

We will begin the experiment by keeping the same wire (the same X) and changing N. By plotting 1/I versus 1/N, we will establish r/E, and will establish the "voltage-current" relation of the Ohm's law. Now, by lengthening the wire of making its cross section larger we can investigate dependence of resistance on geometry of the wire. (Of course we cannot measure the actual absolute values of the



Answer 11/99

resistance, but all we need is the dependence on length and cross section area...) Ohm published his results on geometry dependence of X in Journal fur Chemie und Physik, 46, p. 160 (1826).

Nice short description of Ohm's work in its historic context can be found in the book History of Physics (Storia della Fisica) by Mario Gliozzi.

Comment: Groshaus suggested to measure the voltage by attaching one of the leads of the battery to electroscope. Can it be done with sufficient accuracy?





Question 12/99

Question 12/99

CREATURES GREAT AND SMALL

Why is it that creatures of such different sizes as a flea, a grasshopper, a human, and a lion can jump (i.e., raise their center of gravity) 1 meter in the air at most (give or take a factor of two)?





Answer 12/99


CREATURES GREAT AND SMALL

The question was:

Why is it that creatures of such different sizes as a flea, a grasshopper, a human, and a lion can jump (i.e., raise their center of gravity) 1 meter in the air at most (give or take a factor of two)?

(3/00) The problem has been solved (6/12/99) by Ansgar Esztermann (e-mail [email protected]), who remarked that got his idea about scaling laws from a book by Isaac Asimov. (Additional list of people who suggested correct, but slightly different solution can be found below.)

The solution:

The height of a jump is proportional to the ratio between the energy which can be supplied to the body E and the weight of the body W.

The weight W is proportional to L3, where L is the linear dimension of the solid. The energy E is proportional to the product of the force produced by the foot multiplied by the distance along which this force acts. The latter distance is proportional to the foot size and therefore proportional to L, while the force is proportional to the cross-section of the muscles and therefore is proportional to L2. Thus E is proportional to L3, and the ratio E/W is independent of the size of the body L !

Mark de Graaf from Technische Universiteit Eindhoven in Netherlands (e-mail [email protected]), Dan Peleg (e-mail [email protected]), and Louis Zammit Mangion from the Department of Physics of University of Malta (e-mail [email protected]) also correctly solved the problem. However, they directly claimed that the work performed by a muscle is proportional to its volume (which is correct - since the force is proportional to cross-section and distance is proportional to the length of the muscle). However, it seems to us that the argument presented above is "cleaner".

Comment: This and similar problems are discussed in the book of J.M. Smith Mathematical Ideas in Biology (Cambridge University Press, 1968). See also the discussion by F. Felber from Jaycor, San Diego, California in a letter to Physics Today, p.11, March 1999. See also the lecture: "Scaling: why giants don't exist" by Michael Fowler from Physics Department of University of Virginia. (We thank L.Z. Mangion for bringing this reference to our attention.)







http://www.phys.virginia.edu/classes/109N/lectures/scaling.html

http://www.phys.virginia.edu/classes/109N/lectures/scaling.html

Answer 12/99





Question 01/98

Question 01/98

DELAYED WEATHER Summer (winter) solstice in the Northern hemisphere, i.e. the date when the sun is highest (lowest) in the sky, is on June 21 (December 21). Yet the hottest (coldest) weather is a month or two later. Give a quantitative explanation of this delay.

This question appeared in the New Scientist, and received a qualitative answer in the 22 Nov. 97 issue.





Discussion 01/98


DELAYED WEATHER

The question was:

Summer (winter) solstice in the Northern hemisphere, i.e. the date when the sun is highest (lowest) in the sky, is on June 21 (December 21). Yet the hottest (coldest) weather is a month or two later. Give a quantitative explanation of this delay.

Since a qualitative answer has already been published in the New Scientist (22 Nov. 97 issue) we may as well give it to you. A. Westwood and S. Collins from the University of Leeds said that "the Earth has certain heat capacity, which leads to a thermal time lag." Now we know the reason! All we need is a (semi)quantitative estimate of the time lag.

(6/98) Michael Bertschik (e-mail [email protected]), a Ph.D. sudent at University of Heidelberg suggested the following solution:

Mass of atmosphere to heat (thickness ~10km, constant pressure) is M=5*1018m3*density(air) = 6,5*1021g. Difference of extraterrestrial solar constant and "ground" solar constant: (1400-970) W/m2 = 430 W/m2; hence absorbed energy E (irradiated surface of earth * solar constant) is E = Pi*R(earth)2 * 430 W/m2 = 5,5*1016 W = 5,5*1016 J/s. Heat capacity of air: cv = 0,73 J/(g*K). Heating per second in Kelvin: H = E/(cv*M) = 1,16 * 10-5 K/s. So the heating per day is ~ 1 Kelvin.

Y. Kantor: This solution disregards daily fluctuations of temperature, which could be explained using similar arguments. In addition the energy flux in a calculation of this type must be positive or negative depending on the season. The "jury" was not convinced by the argument...

(2/99) Bill Bruml (e-mail [email protected]), wrote the following message which may give us some hints regarding the correct solution:

Very crudely: Model solar input as varying from some average value by (1+a*sin(t)) over the course of the year. Model heat losses as constant and ignore any heat transfer between the latitudes. If the earths long term average temperature stays the same, the heat loss is the same as the average heat input so the net heat input is just a*sin(t) and the integrated excess heat (which is the temperature) is just (-1/a)*cos(t). At the winter solstice sin(t) (solar input) is at its minimum value and -cos(t) (temperature) is at its mid point. At the vernal equinox, sin(t) (solar input) is at its mid point and -cos(t) (temperature) is at its minimum. So, I get (semi)quantitative (about as semi as it can get) information about the date of the coldest day without needing needing to know the magnitude of the heat flows. With this first order model, the question for more sophisticated models is not, "Why is the coldest day after the winter solstice?" but, "Why is the coldest day




Discussion 01/98

before the vernal equinox?"





Question 02/98

Question 02/98

A FLOATING TRIANGLE

A wooden log (density 0.5 gm/cm3) is floating in the water. The cross section of the log is an equilateral triangle. Which side up will it float?

This question was created by Y. Kantor





Answer 02/98


A FLOATING TRIANGLE

The question was:

A wooden log (density 0.5 gm/cm3) is floating in the water. The cross section of the log is an equilateral triangle. Which side up will it float?

(3/98) Christian von Ferber (e-mail [email protected]), a post-doc at Tel Aviv U., correctly solved the problem.

The answer is: The triangle will float as depicted in Fig. A, i.e one of its medians will be exactly at the waterline.

Here we present some main points of the solution.

Since the density of the body is half the density of the water, the waterline will divide the triangle into two halves. (If the density (relative to the water) is rho then only the fraction rho of the area of the triangle will be under water.) The point of application of weight of the triangle is its center of mass (c.m.). One can easily convince himself that the c.m. of a triangle is at the intersection of the medians. The direction of the force of weight is vertical (down). The point of application of the Archimedes force is the center of mass of the underwater part of the triangle (center of buoyancy). Its direction is vertical (up) and its size is equal to the weight. At the equilibrium, the points of application of the weight and the Archimedes force must be exactly one above the other.

In order to determine the stability of the equilibrium it is convenient to look at the potential energy of the system which is proportional to W=(vertical coordinate of the c.m.)-(vertical coordinate of the center of buoyancy)



Answer 02/98

A direct calculation of the quantity W shows that the system has minimum of energy at the position depicted in Fig. A, while the positions of a corner pointing up or down correspond to maxima of energy. Note that due to the symmetry there are 6 stable equilibrium positions.

If rho is NOT equal to 0.5, the equilibrium orientation depends on rho. Only a fraction rho of the triangle will be under the water. It is interesting to note that there is the is a relation between cases with density rho and 1-rho. In these cases the underwater and above-water parts of the triangle simply exchange roles. For low densities [rho < (7/16)] the triangle will float his corner pointing exactly up as in Fig. C. (There are 3 such positions.) For high densities [rho > (9/16)] the triangle will float his corner pointing exactly down as in Fig. D. (There are 3 such positions.) At the intermediate densities [(7/16) < rho < (9/16)] the solutions are not symmetric, and the position of flotation is as in Fig. B. (There are 6 such positions.) The following figure depicts the rho-dependence of the angle between one of the sides of the triangle and the water surface.

(5/2001) Y. Kantor: In 2/2001 we received a letter from Paul Erdos (e-mail [email protected]) Professeur honoraire at The Institute of Theoretical Physics of University of Lausanne, Switzerland (not to be confused with the late mathematician Paul Erdos (1913-1996)). He has drawn our attention to his papers on "Floating equilibrium of symmetrical objects and the breaking of symmetry" [P. Erdos and G. Schilbler, Am.J.Phys. 60, 335 (1992) (Part 1), Am.J.Phys. 60, 345 (1992) (Part 2)]. In Part 1 of these papers the floating equilibrium for the current case of a triangular prism is treated in detail and in



Answer 02/98

particular the above curve for the tilt angle as function of the density rho is derived. The same analysis for the square prism yields a similar result. The symmetry breaking behavior of floating symmetric bodies is observed in Part 2 of this study also for the cube, the octahedron and the tetrahedron.





Question 03/98

Question 03/98

WEIGHING A FLY

An open bottle lies on a precise weighing scale. A fly then enters the bottle and flutters around in futile attempt to escape. How will the weight reading of the scale change?

A lazier fly, instead of using its wings to fly, uses a small helium filled balloon in order to remain airborne. What happens if it happens to wander into the same bottle?

This question was created by D. J. Bergman





Answer 03/98


WEIGHING A FLY

The question was:

An open bottle lies on a precise weighing scale. A fly then enters the bottle and flutters around in futile attempt to escape. How will the weight reading of the scale change?

A lazier fly, instead of using its wings to fly, uses a small helium filled balloon in order to remain airborne. What happens if it happens to wander into the same bottle?

The problem has been solved by K. MacArthur (e-mail [email protected]). The solution is presented below.

Solution:

In the first case, the reading of the scale will increase approximately by the same amount as if the fly was sitting on the bottom of the bottle, i.e. it will increase by the weight of the fly: Since the fly "flutters around" it exerts a force on the air equal to its weight (and the air exerts the same force on the fly), while the air in its turn acts with the same force on the bottom of the bottle.

In the second case the lazy fly + balloon displace an amount (weight) of air which is exactly equal to their weight (that's why it remains airborne) and thus will have no effect on the scale.






Question 04/98

Question 04/98

KNOTTED FIELDS

Field lines of magnetic fields are always closed loops. Does there exist a configuration of currents which creates a KNOTTED line of a magnetic field.

This problem is a reformulation of a problem posed by Stanislaw Ulam in "The Scottish Book" in 1935 (Problem No. 18), where it appears without a solution (see Ref.[*]).

[*]The Scottish Book ed. by R. Daniel Mauldin, Birkhauser, Boston (1981).



http://star.tau.ac.il/QUIZ/98/scottish.html



Answer 04/98


KNOTTED FIELDS

The question was:

Field lines of magnetic fields are always closed loops. Does there exist a configuration of currents which creates a KNOTTED line of a magnetic field.

(4/98) It has been been observed by several people, that the question, in the form in which it was presented, has a trivial solution:

Consider a knotted line in three-dimensional space (trefoil knot, granny knot, or knot of any topology you like). Take a long narrow solenoid and fold it into the shape of the knot. Connect the ends of the solenoid (and, of course, insert a power source somewhere along the line). Inside the solenoid there will be magnetic field lines parallel to its axis, and therefore those lines will form a knot.

So we see that the solution is simple. HOWEVER, we notice the current which formed the knotted magnetic field line is by itself knotted. Therefore, we would like to ask the follow-up question:

Is it possible to create a knotted field line by a linear current loop which is NOT KNOTTED?

(6/98) Christian von Ferber (e-mail [email protected]), a post-doc at Tel Aviv U., correctly solved the follow-up problem. He had shown that one by slight modification of the construction which was explained above it is possible to create knotted field by unknotted currents. Here is what he wrote:

Assume we have formed a trefoil knot with our solenoid with the knot topology given by the sketch below. The magnetic field lines are then obviously knotted.

We will now unknot the wire without changing the topology of the magnetic field line along the center of the solenoid. For this note that we only have to change one 'bridge' in the sketch above to a 'tunnel'.

Imagine a blow-up of one bridge. We pull apart a little bit the two crossing parts of the solenoid to illustrate the bridge formed by the thin wire:



Answer 04/98

We may then unknot the wire while not moving other parts of the solenoid:

Finally we push together the solenoid again keeping the changed topology of the wire. Thus the centerline of the solenoid is unchanged in the form of a trefoil knot, while the wire itself is of zero-knot topology.

(7/98) Y. Kantor: After publication of the solution we received an e-mail from Michael Trott (e-mail [email protected]) from Wolfram Research, Inc. claiming that "It is wrong belief that magnetic field lines are closed! Typically they are not closed." The discussion between Trott, and v. Ferber on the subject followed, and is presented here. Since the argument was partially numerical, I do not feel that a firm proof of either position has been presented. I therefore present the entire argument to you. Further comments are welcome!




http://star.tau.ac.il/QUIZ/98/knots.html



Question 05/98

Question 05/98

UNEQUAL HALVES

Why is the time from autumn to spring equinox in the Northern hemisphere three days shorter than the time from the spring to autumn equinox?





Answer 05/98


UNEQUAL HALVES

The question was:

Why is the time from autumn to spring equinox in the Northern hemisphere three days shorter than the time from the spring to autumn equinox?

The problem has been solved qualitatively by K. MacArthur (e-mail [email protected]) and by O. Melinger. A slightly more quantitative solution is presented below.

Solution:

The orbit of the Earth around the Sun is almost, but not quite, circular. In January the distance to the Sun is 147 million km, while in July it is 152 million km. According to Keplers laws, this causes more than 3 percent difference in maximal and minimal speeds of the Earth, and more than one percent difference in the mean speeds in the two half-year periods. At the same time the distance traveled by the earth in the "winter half-year" is slightly shorter. Those two effects together cause the "winter half-year" to be about two percent shorter than the "summer half-year".






Question 06/98

Question 06/98

FLOATING BODIES OF EQUILIBRIUM

Stanislav Ulam asked (in "The Scottish Book") whether a sphere is the only solid of uniform density which will float in water in any position. We ask a simpler, two-dimensional, question: Consider a long log of circular cross-section; it will, obviously, float in any position without tending to rotate. (Of course, the axis of the log is asumed to be parallel to the water surface.)

(a) Are there any additional cross-section shapes such that the log will float in any position if the density of the floating body is 0.5 gm/cm3?

(b) Are there any shapes for some other (predefined) values of the density? (Of course, we assume that the given density is smaller than the density of water.)

(c) Are there any shapes, besides the circle, which will float in any position independent of the density of the material?

Partial answers to the question are welcome.A related question appeared in Feb. 98 and was solved.






Discussion 06/98



The question was:

Stanislav Ulam asked (in "The Scottish Book") whether a sphere is the only solid of uniform density which will float in water in any position. We ask a simpler, two-dimensional, question: Consider a long log of circular cross-section; it will, obviously, float in any position without tending to rotate. (Of course, the axis of the log is assumed to be parallel to the water surface.)




(6/98) Y. Kantor: Many people noted that part (c) of the question can be easily solved. Omer Edhan (e-mail [email protected]) was the first to answer correctly. Most of the solutions provided only a partial argumentation. The complete and correct argument for part (c) is as follows:

Since the question refers to floating body of arbitrary density it must be also correct in the limit of vanishing density. A body is in stable equilibrium when the difference between the height of the center of mass and the center of buoyancy (center of mass of the part which is under the water surface) is minimal. In the vanishing density limit, the center of buoyancy is at the water surface, and the question is reduced to minimization of the height of the center of mass. Since the body is supposed to be in equilibrium independently of its orientation the height of the center of mass must be independent of the orientation, and therefore the distance between the center of mass and the water must be independent of orientation. Therefore the external boundary of the cross section of the log must be a circle. (L. Montejano, proved this statement in Studies in Applied Mathematics 53, p. 243 (1974).) (Of course, one can also use a log with a hole - log of circular cross section from which a smaller concentric circle has been removed.)







Discussion 06/98



The question was:

Stanislav Ulam asked (in "The Scottish Book") whether a sphere is the only solid of uniform density which will float in water in any position. We ask a simpler, two-dimensional, question: Consider a long log of circular cross-section; it will, obviously, float in any position without tending to rotate. (Of course, the axis of the log is assumed to be parallel to the water surface.)




Answers and solutions

PART (a)

(5/2000) Ch.v.Ferber: It took some time, but finally part (a) of the problem has been solved by Franz Wegner from the Inst. für Theoretische Physik in Ruprecht-Karls-Universitat Heidelberg (e-mail [email protected] ). His elegant derivation can be found in the postscript file .

Outline:Wegner first proves a remarkable property of the `waterline', the line that the water level draws on the cross section of the log: This line divides the cross section in two areas, one above, one below the water level. For a log that solves the problem not only the areas above and below are constant, i.e. independent of the position of the log, but also the length of the waterline itself is constant. This condition holds independent of the density of the log. For density 1/2 Wegner explicitly constructs curves that comply with this condition and shows that these indeed solve the original problem.

History: This problem (in response to Ulam's question) was originally solved by H. Auerbach. The solution was published in Studia Math. 7, 121-142 (1938) as cited in `The Scottish Book'. Auerbach for his solution he had reinvented curves that had been discussed by K. Zindler even before Ulam had posed his problem Monatsh. Math. Phys. 31, 25-57 (1921). Auerbach's solution, written in French and using a




http://star.tau.ac.il/QUIZ/98/wegner.ps

Discussion 06/98

more geometrical approach, is equivalent in most part to that of Wegner. The only difference is that the condition for convexity as also suspected by Wegner is too restrictive. In fact one may allow the solution to be non-convex at some points, convexity `nearly everywhere' is enough. In this way the following shape is shown to be a solution:

In Wegner's notation this corresponds to

s(phi)=cot(alpha+phi) for 0<phi<pi/2

s(phi)=-cot(alpha+pi-phi) for pi/2<phi<pi with tan(alpha/2)=0.54037..

PART (b)

(5/2000) Ch.v.Ferber: We do not yet have a full answer to the part of the problem when density is not equal 0.5. However, it should be mentioned that the property of the `waterline' to be constant and to divide the shape at a constant ratio remains true as proven by Wegner. Auerbach furthermore proves the following condition to be necessary and sufficient: The waterline chord is constant and divides the perimeter into arcs of constant length. So it remains to be shown that the corresponding generalizations of the Zindler curves exist.

(5/2001) Y. Kantor: In 2/2001 we received a letter from Paul Erdos (e-mail [email protected]) Professeur honoraire at The Institute of Theoretical Physics of University of Lausanne, Switzerland (not to be confused with the late mathematician Paul Erdos (1913-1996)). In his letter, Erdos pointed out that Auerbach, in his 1938 paper, suspected that for densities different from 0, 1/2 and 1 the shape should be circle. Auerbach noted that E. Salkowski claimed to have proved this property (in Sb. D. Heidelb. Acad. D. Wiss., Math.-nat (1934), 57-62) but admitted that he does not understand the proof. Erdos pointed out



Discussion 06/98

to us that Salkowski's proof is erroneous. Erdos also sent us his own (yet unpublished) proof that at for specific non-trivial, different from 0, 1/2, 1, densities the cross section must be a circle.

(5/2002) Y. Kantor: Recently (3/2002 and 5/2002) Franz Wegner from the Inst. für Theoretische Physik in Ruprecht-Karls-Universitat Heidelberg (e-mail [email protected] ) found equations (and shapes) of non-circular logs that will float in arbitrary orientations, for specific sets of densities. See details in the following papers in PDF format: 1 and 2.

PART (c)

(6/98) Y. Kantor: Many people noted that part (c) of the question can be easily solved. Omer Edhan (e-mail [email protected]) was the first to answer correctly. Most of the solutions provided only a partial argumentation. The complete and correct argument for part (c) is as follows:

Since the question refers to floating body of arbitrary density it must be also correct in the limit of vanishing density. A body is in stable equilibrium when the difference between the height of the center of mass and the center of buoyancy (center of mass of the part which is under the water surface) is minimal. In the vanishing density limit, the center of buoyancy is at the water surface, and the question is reduced to minimization of the height of the center of mass. Since the body is supposed to be in equilibrium independently of its orientation the height of the center of mass must be independent of the orientation, and therefore the distance between the center of mass and the water must be independent of orientation. Therefore the external boundary of the cross section of the log must be a circle. (L. Montejano, proved this statement in Studies in Applied Mathematics 53, p. 243 (1974).) (Of course, one can also use a log with a hole - log of circular cross section from which a smaller concentric circle has been removed.)

(5/2001) Y. Kantor: In a letter by P. Erdos (see par.(b)) it was brought to our attention that H. Steinhaus proved (Bull. Acad. Poloaise des Sc. Math., Astr. et Phys. XI No.4 (1963) 173-174) that for a three dimensional body on a horizontal support must be a sphere with it center of gravity coinciding with the center of the sphere. So, L. Montejano essentially proves the same thing (in two dimensions) apparently unaware of Steinhaus' work.




http://star.tau.ac.il/QUIZ/98/Wegner0.pdf

http://star.tau.ac.il/QUIZ/98/Wegner1.pdf




Question 07/98

Question 07/98

SLIPPERY SLOPE

A body is sliding along the curve which is shown to-scale in the above figure. It starts at point A, an slides with friction coefficient k. Describe a method to determine the point B at which the body stops.

This question appeared in Physics Olympiad - Oslo 1996.





Answer 07/98


SLIPPERY SLOPE

The question was:

A body is sliding along the curve which is shown to-scale in the above figure. It starts at point A, an slides with friction coefficient k. Describe a method to determine the point B at which the body stops.

(7/98) The problem has been solved correctly by Omer Edhan (e-mail [email protected]) and Itzhak Shapir (e-mail [email protected].

The answer: Starting from point A draw a straight line of slope -k. The point of intersection of the line with the curve is the point B where the body stops.

The solution:

The friction force F=k*N, where N is the normal force between the body and the slope. N on a straight slope is simply equal to the weight P of the body multiplied by cos(alpha), where alpha is the angle




Answer 07/98

between the tangent to the slope and the horizontal direction. Since the slope is curved an additional component should be added because of the presence of centripetal force. However, if we assume that the radius of curvature R is always large while the velocity V is small (more accurately, that V2/R is much smaller than the acceleration of free fall) then we may neglect the effects of curvature.

The element of work performed by the friction force when the body is displaced by amount ds along the curve isdW=-k*P*ds*cos(alpha)=-k*P*dx,where dx is the horizontal displacement element. Thus the total work performed by friction forces is -k*P*X where X the total horizontal displacement. At point B this work must be equal to the change in potential energy P*Y, where Y is the vertical displacement. Thus, k=-Y/X, and consequently the position of B is determined by the intersection of the curve with a straight line with slope -k.





Question 08/98

Question 08/98

BODY HEAT

How much heat is produced by a human body (in Watts)?

p.s. We expect to receive many different (independent) methods of estimating this quantity.





Answer 08/98


BODY HEAT

The question was:

How much heat is produced by a human body (in Watts)?

(27/7/98) Itzhak Shapir (e-mail [email protected]) solved the problem by considering the food intake by a man, and reached the conclusion that it is 58Watts.

Solution:

1. An average human consumes about 1200 food calories (1,200,000 calories) which are about 5,000,000 Joules per day (24 hours).

2. Most of that energy is processed in the body and results in heat. In that case, the heat power is 5,000,000/(24*3600) J/sec ~ 58W

(3/99) Tal Kuzniz (e-mail [email protected]) from Tel Aviv University (Israel) and Mark Davison (e-mail [email protected]) from University of Paisley (Scotland) suggested to look at thermal radiation balance between the body and the surrounding. The solutions used different estimates of the total area of the body, as well as different estimates of the normal temperature difference between the body and the air. (We need such temperature difference that a (naked) person does not feel hot (i.e. does not sweat) or does not feel cold.) A little opinion poll which was conducted converged on (or rather averaged at) 25-26C as ideal temperature.

Solution:

Body area A=1.5m2

Body temperature: T=310KAir temperature: t=299KStefan-Boltzmann constant: S=5.67*10-8W/(m2K4)Skin emissivity close to unity.Power=S*A*(T4-t4).This expression leads to estimate of 105W, which looks like a nice estimate.

Y. Kantor: Here is my personal "solution" of the problem, which is based on heat conductance:





Answer 08/98

Solution:

Person (or rather, naked person) feels well at temperature of about 25C. At temperatures below that one needs clothing to reduce the heat loss while at higher temperatures one sweats to increase heat loss. So, we may assume that when the temperature difference between the body and the environment is 12C the heat flow by thermal conductivity is equal to heat production. If we assume that the human body is a sphere (of course) then the temperature field around it will be T=const + A/r, and the total heat current will be 4{pi}Ak, where k=2.6J/m*sec*deg is the heat conductivity of the air. (The latter expression is simply 4{pi}r2 multiplied by conductivity k and multiplied by the temperature gradient A/r2. The prefactor A in the expression for temperature is determined by assuming that the temperature difference is 12C when r is the radius of the sphere R. Thus A=12*R, and the final result for heat loss is 4{pi} k *12*R. Assuming that the "equivalent sphere" representing the human being has radius somewhere between 0.2m and 0.6m, we find that the heat production is between 80W and 240W.

For those who are not happy representing a human being as a sphere, let us assume that it is a prolate ellipsoid with larger semi-axis a=0.8m and the minor semi-axis b=0.2m. In this case the coefficient A in the above derivation should be the temperature difference (12C) multiplied by the capacitance of the ellipsoid (see Landau and Lifshitz, Electrodynamics of Continuous Medium). That capacitance is sqrt(a^2-b^2)/Arch(a/b)=0.4m. Substituting this number we get heat current of 160W.

Y. Kantor: We note that radiation and heat conduction together remove more heat than is produced by human body. Maybe our estimates are wrong by a factor of 2 or more... We are still waiting for comments and more quantitative estimates.

(12/2000) Lawrence Weinstein (e-mail [email protected]) from the Physics Department of the Old Dominion University, Virginia, wrote us the following e-mail:

(1) 1200 calories per day is much too few. It is typically more like 2000-2500 (at least for well fed Americans). This will double the estimated heat available from 58 W to between 100 and 120 W. (The rule of thumb is that one person = 100 W light bulb.)

(2) Skin temperature is not 37 C; core temperature is 37 C. This is especially true for the extremities. The average skin temperature will vary significantly to maintain the heat balance. This temperature is controled by varying the blood flow to the surface and to the extremities. Presumably, 25 C is the lowest temperature at which people are comfortable. This implies that skin and extremity temperatures will be much lower than 37 C at this point. A better reference would be the highest comfortable temperature in a high humidity environment (to minimize the cooling effects of sweating). I would guess this to be about 30-31 C. (This implies a comfort zone (without sweating) of about 5 C.)

If we use 33 C for the body's surface temperature (may be too low for the torso but too high for the extremities) and 25 C for the ambient temperature, then we get 65 W for the radiation output and about 110 W for conductive output (using the prolate elipsoid) for a total of 175 W. If we use 37 C for the body's surface temperature at an ambient temperature of 31 C, then we get 60 W for the radiation output and 80 W for the prolate elipsoid for a total of 140 W.


http://www.physics.odu.edu/~weinstei


Answer 08/98

These numbers are reasonable because your body is less metabolically active and cools at night, emitting significantly less heat for at least 8 hours.





Question 09/98

Question 09/98

MISSING MOMENTUM

Consider the creation of a phonon by the scattering of a neutron from a crystal. Let the initial and final momentum of the neutron be p and p'. The neutron interacts only weakly with matter, and is essentially a free particle. The phonon which is produced has a pseudomomentum K=p-p'.

(a) Prove that momentum of a phonon vanishes.

(b) Where does the missing momentum go?

This is a reformulation of a question asked by R. Peierls.





Answer 09/98


MISSING MOMENTUM

The question was:

Consider the creation of a phonon by the scattering of a neutron from a crystal. Let the initial and final momentum of the neutron be p and p'. The neutron interacts only weakly with matter, and is essentially a free particle. The phonon which is produced has a pseudomomentum K=p-p'.

(a) Prove that momentum of a phonon vanishes.

(b) Where does the missing momentum go?

(6/99) The problem has been solved by Groshaus Javier (e-mail [email protected]). This problem was discussed in great detail by Rudolf Pierls in his book Surprises in Theoretical Physics, ch.4.2 (Princeton University Press, NJ, 1979). Short solution of the problem is presented below.

The solution:

In a phonon instantaneous displacement of an atom is proportional to cos(Q x - w t) where Q=K/{hbar}, x is the position of the atom, w the relevant frequency, and t is time. Thus the velocity of an atom (and, consequently, its momentum) will also oscillate sinusoidally and on the average will be zero. Moreover, if we sum over all positions of atoms on the lattice, and Q is such that the phonon forms a standing wave the summation over all atoms vanishes identically at every moment! Thus a phonon has vanishing momentum.

The above argument is not valid for Q=0 mode. Thus, the momentum of the neutron is transferred to the entire lattice. This means that the entire solid starts moving with velocity K/M, where M is the mass of the entire body.

The actual situation is a little more complicated, since the neutron is somewhat localized and therefore creates not a single phonon but rather a localized packet of phonons. Read the book of Pierls about this question as well as other related questions.






Question 10/98

Question 10/98

FALLING CLOUDS

Clouds are made of water droplets, and water is about 800 times denser than air. Why don't clouds fall like a rock to the ground?

This question was "borrowed" from the "challenge questions" of H.S. Greenside



http://www.phy.duke.edu/%7Ehsg/physics-challenges/challenges.html



Answer 10/98


FALLING CLOUDS

The question was:

Clouds are made of water droplets, and water is about 800 times denser than air. Why don't clouds fall like a rock to the ground?

(11/98) The problem has been solved with varying degrees of accuracy and completness, by Oren Melinger (e-mail [email protected]), Kerry MacArthur (e-mail [email protected]), Dmitriji Podolsky from Ukraine, and William Chess (e-mail [email protected]). Short solution of the problem is presented below.

The answer: The clouds do fall, but they fall very slowly...

The solution:

A falling drop feels a frictional force from the air's viscosity that opposes gravity (the drop has to push its way through the air), the drop reaches a constant terminal velocity, and this terminal velocity is smaller the smaller the radius of a fluid particle or ice crystal. It turns out that this terminal velocity is quite small for drops the size of a few millimeters and so, although the drops in a cloud fall, they fall so slowly that only careful measurements would reveal this fact.

More quantitatively we can say the following: The weight of a drop of air of radius R is 4{pi}R3{rho}g/3 where {rho} is the density of the water (=103 kg/m3, and g (=9.8 m/sec2) free-fall acceleration. If the drop is falling through air with velocity v, the viscosity of the air n (=1.75*10-5 N*sec/m2) will cause friction (Stokes law) 6{pi}nRv. [Here we neglect the density of the air and assume that the viscosity of the drop is much larger than viscosity of the air; we also asume that the drop is approximately spherical.] The terminal velocity of the falling drop is obtained by equating the weight and the friction force, and is given by

v=2{rho}gR2/9n .

Thus 1 micron drop will have velocity of 0.13 mm/sec, or 11 m/day. Larger, 10 micron drops will still fall slowly, - 1.1km/day. Such fall rates can be neglected, especially since the motion of the air itself can be faster than that.

Drops significantly smaller than 1 micron are not visible and will not be percieved as clouds, while 0.1 mm drops will fall with velocity of about 1 m/sec, i.e. it will rain. Larger drops will fall with even larger





Answer 10/98

velocities; however, the air-friction starts increasing faster than than v, i.e. the velocity will not increase as fast with increasing drop size. At such large speeds the weight of the drop is balanced by the drag force (1/2){rho}airC{pi}R2v2, where {rho}air is the density of the air (=1.2kg/m3) and C=1.2 is the drag

coefficient for sphere. According to this equation 1 mm drop will fall with velocity of 4.3 m/sec and 10 mm drop will fall with velocity of 13.6 m/sec. However, drops larger than 5 mm are usually broken into smaller drops that fall more slowly.

Comment 1: This problem is related to the Milliken oil-drop experiment in which the absolute magnitude of the electric charge was measured for the first time.

Comment 2: There is a nice book that discusses the question of falling clouds and other meteorological phenomena: W. E. Knowles Middleton, A History of the Theories of Rain and Other Forms of Precipitation, Franklin Watts, Inc. (NY), 1965.

We thank Henry S. Greenside for his help in formulation of this solution.





Question 11/98

Question 11/98

BOUNCING BRICK

A heavy brick is falling from a height of 1 meter, hits a tennis ball and jumps back to the height of (almost) 1 meter. (We assume that the mass of the tennis ball is negligible compared to the mass of the brick. The collision with the ball is assumed to be elastic, i.e. there is no energy loss.) To what height will the tennis ball jump?

This was one of the Physics Olympics questions in the Soviet Union (R.I.P.)





Answer 11/98


BOUNCING BRICK

The question was:

A heavy brick is falling from a height of 1 meter, hits a tennis ball and jumps back to the height of (almost) 1 meter. (We assume that the mass of the tennis ball is negligible compared to the mass of the brick. The collision with the ball is assumed to be elastic, i.e. there is no energy loss.) To what height will the tennis ball jump?

(2/1999) The problem has been solved correctly by William Chess (e-mail [email protected]), by Eitan Federovsky (e-mail [email protected]), by Bill Bruml (e-mail [email protected]), by Age Sjastad (e-mail [email protected]), and by Javier Groshaus (e-mail [email protected]) M.Sc. student from Technion (Haifa, Israel).

The answer: The tennis ball will jump to the height of 1/4 meter.

The solution:

The brick will fall and strike the tennis ball, which will compress symmetrically and then return to its original shape. During this compression, the center of mass of the tennis ball moves exactly half the







Answer 11/98

distance that the topmost point of tennis ball moves (the point first contacted by the brick). Since the distances moved by these two points on the ball occur during the same time period, the velocity of the center of mass of the ball is exactly one-half the velocity of the top of the ball. During the rebound, at the moment the tennis ball returns to its original shape, the brick (and the top of the tennis ball) will have reached its maximum upward velocity, which, according to the problem statement, is sufficient to propel the brick to a maximum height of almost 1 m. Since the maximum height of a projectile is proportional to the square of the vertical component of its initial velocity (H=v2/(2g)), and since the center of mass of the tennis ball was moving at one-half the velocity of the brick at the moment of restoration, the tennis ball will reach a maximum height equal to one-quarter of the maximum height of the brick.

(11/2000) A very interesting and important remark has been made by Bill Unruh (e-mail [email protected]). He noticed that, although it is claimed that everything in the problem is elastic, the solution does not seem to be reversible in time: Consider the brick at the lower-most point with the ball maximally compressed. We claimed that if the time "runs forward" then both the brick and the ball will jump. However, if we "run the time backward" then the brick returns its it original position, while the ball is still on the ground. So the solution is not "time reversible". (Actually the time reversibility should be slightly more complicated, since there is a third body ("the ground") involved. However, the conclusion regarding ireversibility is not changed.) Obviously, both the formulation of the problem and its solution neglected many things; some of them are more important, while the others are negligible. E.g., the fact that the "sound" or compression wave in the ball propagates at a finite speed seems to be not particularly relevant, provided this speed is much larger than the velocity of the brick. However, certain "inelastic" aspects may be important. We welcome a more detailed analysis of the interaction between the ball and the brick.






Question 12/98

Question 12/98

COUPLED PENDULUMS

Two pendulums with equal masses m are coupled by a spring with spring constant k as shown in the figure. Length of the left pendulum is l, while the length of the right pendulum is different: l+x. The length of the right pendulum can be varied. Assume that at the initial moment the right pendulum is shorter than the left one (x<0). We deflect the left pendulum and release it. The system starts oscillating. Now we gradually increase the length of the right pendulum until it is longer than the left one. How will the system behave at the end of this process?

This question was suggested by S. Nussinov.





Answer 12/98


COUPLED PENDULUMS

The question was:

Two pendulums with equal masses m are coupled by a spring with spring constant k as shown in the figure. Length of the left pendulum is l, while the length of the right pendulum is different: l+x. The length of the right pendulum can be varied. Assume that at the initial moment the right pendulum is shorter than the left one (x<0). We deflect the left pendulum and release it. The system starts oscillating. Now we gradually increase the length of the right pendulum until it is longer than the left one. How will the system behave at the end of this process?

(12/2003) The problem has been solved correctly by Sergei Flach from Max Planck Institute for the Physics of Complex Systems, Dresden, Germany (e-mail [email protected]) almost immediately after publication of the problem (12/98). We expected that this problem will generate many attempts at the solution, but recieved only few partial solutions. John G. Florakis, a Physics student at the University of Athens (Greece) (e-mail [email protected]) came (7/2003) very close to the desired solution. After 5 years, it seems that we should publish the solution. Flach's (slightly editted) solution is presented below.

The solution:

There is an ambiguity in the problem, because the system is nonlinear and nonintegrable, so in general an irregular motion is generated, with any outcome you wish. But I assume that the pendula are weakly excited, and therefore we can linearize the equations of motion. Further I assume that 'gradually' means adiabatically slowly, i.e. when we solve at any fixed length l+x, the eigenvalue problem and prepare the system in a mixture of the two possible eigenstates (f1,f2) c1*f1+c2*f2 where f1,f2 are the eigenstates and c1,c2 are the weight factors, then upon changing the length l+x we do not change c1,c2, but simply gradually change f1,f2. Due to the nonzero coupling we generically won't run through a degeneracy




Answer 12/98

(eigenvalues e1,e2 are different for all x), what we find is an avoided crossing, i.e. for weak coupling e1(x) and e2(x) will come very close if x=0. Simple calculation of the tuning of a 2x2 matrix by a parameter close to a degeneracy shows that after the avoided crossing f1 --> f2 and f2 --> -f1. This implies that after the crossing (after the change of sign in x) the system is in a state c1*f2 - c2*f1.

Suppose that when at the beginning x was positive, our initial condition was such that that the solution was not permutational invariant (i.e. e.g. one oscillator oscillates with large amplitude and the other one with small amplitude). That can in general happen because nonzero x breaks permutational symmetry. Now after the experiment x --> -x and consequently now we have the permuted solution - i.e. the right oscillator is having large amplitudes, while the left one does not.

w2=sqrt[0.5(g/l1+g/l2+2k/m) +- sqrt[0.25*(g/l1-g/l2)2+(k/m)2]]

The graph is for the case when (k/m)=0.05(g/l2).





Question 01/97

Question 01/97

An electrostatic potential has been measured everywhere outside a sphere of radius R. It was found that the potential is spherically symmetric, i.e. depends only on the distance r from the center of the sphere, and is given by the expression A/r, where A is some constant. No measurement of the potential inside the sphere has been performed. What can you say about the charge distribution inside the sphere.

This question was suggested by M. Schwartz.





Discussion 01/97


The question was:An electrostatic potential has been measured everywhere outside a sphere of radius R. It was found that the potential is spherically symmetric, i.e. depends only on the distance r from the center of the sphere, and is given by the expression A/r, where A is some constant. No measurement of the potential inside the sphere has been performed. What can you say about the charge distribution inside the sphere.

Once the question was published, I immediately received several responses. Everybody stated, correctly, that the total charge inside the sphere should be A, since this is the only way to produce outside the sphere the potential A/r (this can be proven using Gauss law). However, regarding the charge distribution inside the sphere we did not receive satisfactory answers. Some people claimed (and even attempted a mathematical proof) that the charge distribution inside the sphere should be spherically symmetric. This is obviously wrong - consider the following example: Take a uniformly charged sphere (figure below - left), cut out a spherical cavity inside the sphere (as shown on the figure below - right), and concentrate all the charge which was in the piece which was cut into a single point which will be placed in the center of the cut. Obviously, the point charge produces a field outside the cut identical to the field which was produced by the spherical area which was cut-out. Therefore, the field, and potential outside the large sphere was unchanged by this cut-and-concentrate procedure. However, the resulting charge distribution is NOT spherically symmetric. Therefore, we conclude that the charge distribution does not have to be spherically symmetric. Nevertheless: Can we say something intelligent (and constructive) about possible charge distributions?





Answer 01/97


The question was:An electrostatic potential has been measured everywhere outside a sphere of radius R. It was found that the potential is spherically symmetric, i.e. depends only on the distance r from the center of the sphere, and is given by the expression A/r, where A is some constant. No measurement of the potential inside the sphere has been performed. What can you say about the charge distribution inside the sphere.

(5/97) Many people noted correctly that the total charge of the sphere must be A (in Gaussian units). This can be proven directly using Gauss law. At the same time it was demonstrated (see Discussion) that the charge distribution does not have to be spherically symmetric.

(6/97) Itamar Borukhov from Tel Aviv U. wrote that presenting the potential at point r as a volume integral over rho(r')/|r-r'| where rho(r') is the density of the charge, and by expanding 1/|r-r'| in powers of r and integrating over r', we get a multipole expansion of the potential. Since we know that the resulting potential is spherically symmetric, all the multipoles must vanish. Thus the statement which can be made is: the charge distribution is such that all multipoles (except the monopole) vanish.

(6/97) Y. Kantor: So now we know that the multipoles vanish. It may look strange, since the charge density is not constant. However, we must keep in mind that multipoles are "kind-of" expansion of the charge density into functions rl Ylm(theta,phi), where r is the distance from the center, theta,phi are the

polar angles and Y is the spherical-harmonic function. This is not a complete set of functions within a sphere. (It is only a complete set for harmonic functions, and the density is not a harmonic function.) So now we somewhat reduced the list of functions which can create the potential A/r. Can we rephrase the restrictions in a more constructive form, i.e. say something like: "the most general charge density producing such a potential is..."?

(9/97) Koenraad Audenaert from University of Gent (RUG), Belgium (email: [email protected]) gave a constructive solution: He has shown that the charge density must be an expansion of terms of the form Ylm(theta,phi)Pk

(0,2)(2r-1) where Pk(0,2)(2r-1) is Jacobian polynomial. For detailed

derivation see the following postscript file.




http://star.tau.ac.il/QUIZ/97/A01_97_potential.ps



Question 02/97

Question 02/97

Flux of the magnetic field B through a metalic ring is changing with time. It creates 12 Volt electromotive force in the ring. A voltmeter is connected to the contacts shown in the picture, i.e. the contacts are are separated by 1/4 of a circle. What will be the reading of the voltmeter?

This question was created by Y. Kantor and A. Palevski





Answer 02/97


The question was:Flux of the magnetic field B through a metalic ring is changing with time. It creates 12 Volt electromotive force in the ring. A voltmeter is connected to the contacts shown in the picture, i.e. the contacts are are separated by 1/4 of a circle. What will be the reading of the voltmeter?

The problem was solved correctly by Itamar Borukhov, Eilon Brenner, Boaz Kol and Udi Fuchs from Tel Aviv U. (4/6/97). The answer: if the voltmeter is positioned in the lower-right corner outside the ring, then it will show 3V, while if it is positioned in the upper-left corner outside the ring, then it will show 9V.

The outline of the solution is as follows:1. Since this is NOT an electrostatic problem it is impossible to define potential differences (voltages). Neverthless the integral of the electric field E along a certain segment (electromotive force - EMF) of a ring plays a very similar role: Since the Ohm's law is still valid (current is proportional to the local field, therefore the sum of resistances of segments of circuit multiplied by currents must be equal to the EMF in that circuit. However, the EMF is equal to the rate of change of flux within that specific loop.2. Voltmeter has high internal resistance an measures the current flowing through it. It displays on its scale the current multiplied by internal resistance (that's what we call "reading of the voltmeter").3. The current is conserved at every junction.4. One needs to choose two loops in the electric scheme and apply the above laws.

Here is what Borukhov, Brenner, Kol and Fuchs actually wrote:The Voltmeter in the picture will read 3V.

The answer depends on the geometry in which the Voltmeter is connected to the circuit. For example, if the Voltmeter was connected to the same points, but this time from the other side of the ring - it would read 9V.


Answer 02/97

The subtle point is that the voltage is not a local quantity but rather a global one depending on the amount of flux circled by the current loop. Therefore one has to solve for the current passing through the Voltmeter. One way to solve the problem is to look at closed current loops using Kirhoff's law (see Figure). [Note: the "power sources" in the figure are "symbolic" - they denote integrals of electric field.] For example, suppose we want to solve for the currents using the top and bottom loops. Then the equations would be: top: 12V = 3/4*R*I1 + 1/4*R*(I1-Ir)bottom: 0 = Ir*r + 1/4*R*(I2-I1)which gives in the limit r>>R: Ir*r = 3V If we want to use the top circuit and the external one then we get: top: 12V = 3/4*R*I1 + 1/4*R*(I1-Ir)external: 12V = 3/4*R*I1 + Ir*r which gives the same result.





Question 03/97

Question 03/97

1. It is known that a real positive function (x) is a groundstate of a one-dimensional quantum

Hamiltonian H. Find the potential V(x) of that Hamiltonian.

2. Suggest a way for construction of potential V(x) and two functions (x) and (x) (i.e. all

three functions can be chosen by you) in such a way that the functions correspond to the ground state and the first excited state of the quantum Hamiltonian.

The first part of this question appeared as a qualifying exam question at MIT, while the second part was suggested by M. Schwartz





Answer 03/97


The question was:1. It is known that a real positive function (x) is a groundstate of a one-dimensional quantum

Hamiltonian H. Find the potential V(x) of that Hamiltonian.

2. Suggest a way for construction of potential V(x) and two functions (x) and (x) (i.e. all

three functions can be chosen by you) in such a way that the functions correspond to the ground state and the first excited state of the quantum Hamiltonian.

(8/97) The problem was solved correctly by Adi Armoni from TAU (e-mail: [email protected]).

The first part of the problem is simple: simply write down the Schrodinger equation and express V(x) in terms of everything else. The second part is slightly more complicated: One assumes that the ratio between the eigenfunction of the first excited state and the ground state is some known function f(x). (E.g., f(x)=xn.) By writing down Schrodinger equation for (x) and =f(x) (x) we can get a first order differential equation for

which can be readily integrated.

For a detailed solution see the following postscript file.




http://star.tau.ac.il/QUIZ/97/schrod.ps



Question 04/97

Question 04/97

A particle is placed in a one-dimensional potential

2 BU(x)=A*x + --- 2 x

at temperature T. Its motion is restricted to x>0. Find its mean energy.

The origin of the problem is unknown.





Answer 04/97


The question was:A particle is placed in a one-dimensional potential

2 BU(x)=A*x + --- 2 x

at temperature T. Its motion is restricted to x>0. Find its mean energy.

The problem was solved correctly by Oded Farago from Tel Aviv U. (5/15/97). The answer: The mean total energy is kT+2*sqrt(A*B)The solution:

1. The mean kinetic energy of the particle is < p2/2m> = kT/2 (Equipartition theorem) 2. The mean potential energy can be split into two parts:< A*x2+B/x2> = < A*x2-B/x2> + 2*B*< 1/x2>.a. The first part can be evaluated using virial theorem:< A*x2-B/x2> = 1/2*< x*dU/dx> = kT/2b. The second part can be evaluated as follows: First perform a change in variable y=sqrt(sqrt(A/B))x. In this new variable the potential is brought to the form U=sqrt(AB)*(y2+1/y2). Thus the average< 1/x2> =sqrt(A/B) int (1/y2) exp[-(sqrt(AB)/kT)(y2+1/y2)] dy / int exp[-(sqrt(AB)/kT)(y2+1/y2)] dyNow in the first integral perform variable change z=1/y and it will become identical with the second integral, leaving us with the answer < 1/x2> =sqrt(A/B). Collecting all the above results we finally get the total mean energy: kT+2*sqrt(A*B).

Y. Kantor: Note something interesting: the potential has a minimum at x=sqrt(sqrt(B/A)). The potential energy at that point is U=2*sqrt(A*B). At low T the potential can be treated as approximately harmonic (parabolic), i.e. the potential energy due to the temperature will be kT/2. If we add similar term for kinetic energy, we get the total energy kT+sqrt(A*B). The surprising fact is that this relation holds also when the temperatures are not low and the potential cannot be approximated as harmonic...





Question 05/97

Question 05/97

When you throw a stone at a glancing angle into the water it skips several times before sinking. Can you increase the number of skips by modifying the angle or velocity at which the stone is thrown? Can you increase it by choosing a stone of special weight or shape? Is there a limit to the number of skips?

p.s. This problem is suggested towards your summer vacation. You may try an experimental approach...

This problem was suggested by R. Mints.





Discussion 05/97


The question was:When you throw a stone at a glancing angle into the water it skips several times before sinking. Can you increase the number of skips by modifying the angle or velocity at which the stone is thrown? Can you increase it by choosing a stone of special weight or shape? Is there a limit to the number of skips?

(11/96) Y. Tsori from Tel Aviv U. wrote (slightly edited):The number of skips n depends upon the following factors:1) The angle of glancing theta.2) mg (where m is the mass of the stone).3) mu - the friction coefficient between the stone and the water. I assume friction with the air is negligible.4) The initial speed of the stone, v.I assumed that the shape of the stone is given, and does not change. From dimensional analysis, it is clear that... n can only depend on the angle theta.

(11/96) Y. Kantor: The "friction coefficient" must be defined carefully. Let's say that friction force depends on velocity of the stone, viscosity (of the water) and size of the stone. Whatever the combination, clearly we can create a dimensionless quantity from the weight of the stone and the friction force. Thus the angle theta is not the only dimensionless quantity in the problem.

(11/98) Christian von Ferber from Duesseldorf U. (e-mail [email protected]) wrote:When the Rhine flooded our garden at the time, I had the opportunity to test the question about the jumping stone. An important ingredient is the angular momentum of the stone. Without spinning it will never jump. Also the jumps follow a curved line. I suspect that the spin stabilizes the angle between the (flat) stone and the water surface. It also results in 'violating' the law of reflection when the stone jumps off the surface again. As the stone will not be reflected any more, when it has lost the angular momentum, the question seems to be connected with 5/96 "motion of the hockey puck" and at least as complicated, as in addition hydrodynamics comes into play.






Answer 05/97


The question was:When you throw a stone at a glancing angle into the water it skips several times before sinking. Can you increase the number of skips by modifying the angle or velocity at which the stone is thrown? Can you increase it by choosing a stone of special weight or shape? Is there a limit to the number of skips?

(2/2003) We waited for many years for the solution of this problem. Finally, we saw a paper by Lyderic Bocquet from Departement de Physique des Materiaux, Universite Lyon that has been published on xxx.lanl.gov preprint server (physics/0210015). You can find a copy of the paper here.



http://star.tau.ac.il/QUIZ/97/Bocquet.pdf



Question 06/97

Question 06/97

At what speed will a tennis ball be able to break a glass window?

This question appears in the "Problems in Physics" by P.L. Kapitza





Answer 06/97


The question was:At what speed will a tennis ball be able to break a glass window?

(3/2001) Finally, after more than 3 years we have a solution of the problem. Larry Weinstein from Old Dominion University, Norfolk, VA, USA (e-mail [email protected]) suggested a solution presented below. The solution includes a number of assumptions and approximations. We will appreciate comments on the solution.

The answer: the speed is between 20 and 30 m/s. The solution is presented in the following postscript file




http://star.tau.ac.il/QUIZ/97/tennis.ps



Question 07/97

Question 07/97

By how much can an upward jump of an acrobat be increased using a trampoline once.






Answer 07/97


The question was:By how much can an upward jump of an acrobat be increased using a trampoline once.

(9/98) This problem has been solved correctly by Martin Doerr (e-mail [email protected]) from Universite Libre de Bruxelles.

The answer: the height of the jump will be doubled. The solution as was sent in by Doerr:

The acrobat jumps twice (once onto the trampoline and the second time from the trampoline). Since the trampoline does not give or take energy by itself (well, its an ideal trampoline, of course), the max kinetic energy he can have at the end is twice the one without trampoline. It is actually a little less, particularly if the trampoline is too hard, since his legs can't take the total acceleration (like for a steel trampoline, which works well with steel balls). If the tramp is too soft, he will also not be able to gain more than his initial (single jump) kinetic energy (try it). The period of the tramp must be matched to the artists optimal jump time (wouldn't you have guessed).






Question 08/97

Question 08/97

How many drops are there in a cubic centimeter of fog if the visibility is 100m and the fog disappears within an hour.






Answer 07/97


The question was:How many drops are there in a cubic centimeter of fog if the visibility is 100m and the fog disappears within an hour.

(4/04) This problem has been solved correctly (8/4/04) by Sebastian Popesku (e-mail [email protected]) from Complex Systems Group in the Faculty of Physics of "Al. I. Cuza" University, Iasi, Romania. The solution can be found here.

The answer: About 7 drops/cm3.




http://star.tau.ac.il/QUIZ/97/fog_visibility.pdf



Question 09/97

Question 09/97

Lunar Olympics will be conducted in a closed stadium located on the Moon. The stadium will be filled with air, pressurized to one atmosphere. What will be the Lunar-Olympic records in various fields (race, long jump, shot put, high jump, pole vault, weight lifting). The table at the answer page details several sports and their records on Earth. (The numbers are approximate and slightly rounded.) Find the equivalent records on the Moon. The answer page will be filled-in gradually as the answers arrive.

Oversimplified (and sometimes completely wrong) answers to this question (or parts of it) can be found in many elementary physics textbooks).





Answer 09/97


The question was:Lunar Olympics will be conducted in a closed stadium located on the Moon. The stadium will be filled with air, pressurized to one atmosphere. What will be the Lunar-Olympic records in various fields. The table below details several sports and their records on Earth. (The numbers are approximate and slightly rounded.) Find the equivalent records on the Moon.

The following table will be filled out gradually as the answers arrive. Please send your answers, and suggestions.

Sport Earth record Lunar record Comment

100m/10000m race 9.8 sec/27 min

long jump 9 m

shot put 23 m

high jump 2.5 m

pole vault 6.1 m

weight lifting-snatch/jerk/press

148kg/188kg/? 1450N/1840N/?

900kg/1140kg/? 1450N/1840N/?

6a,6b

6a. The snatch and clean-and-jerk world records belong to N. Suleymanoglu who is 150cm tall and weighs 64 kilograms 6b. (9/97) (Y. Kantor) It is resonably clear that at least approximately the weight which can be lifted on the Moon is equal to the weight on Earth (i.e. the athlete applies the same force). Consequently the mass which can be lifted on the Moon is larger than the mass on Earth by the ratio between the free-fall acceleration on Earth (=9.8m/sec2) and Moon (=1.62m/sec2). However, if you try to be a bit more accurate there must be some difference.


Answer 09/97





Question 10/97

Question 10/97

A small vertical tube (radius smaller than capillary length) is put in contact with the surface of a (perfectly) wetting liquid of very low viscosity. Describe the motion of the liquid in the tube. What is the ratio between the maximal height and the final height reached by the liquid surface in the tube?





Answer 10/97


The question was:A small vertical tube (radius smaller than capillary length) is put in contact with the surface of a (perfectly) wetting liquid of very low viscosity. Describe the motion of the liquid in the tube. What is the ratio between the maximal height and the final height reached by the liquid surface in the tube?

3/98 We received several answers. The answers included correct qualitative discussions, but none of them actually solved the problem quantitatively. It doesn't seem, that we are going to receive a complete and correct answer. Therefore we publish the solution.

The problem was considered theoretically and experimentally by D. Quere. Detailed explanation as well as nice experimental results can be found in Europhys.Lett. 39, 533 (1997). When the tube is inserted into liquid, the liquid will start raising due to surface tension (capillary) forces. Eventually (after a long time) the height of the liquid will be such that the capillary forces will be balanced by the weight of the liquid column. However, before that final height is reached the liquid will oscillate (due to inertia) around its final height. The maximal height reached by the liquid column (due to the oscillations) is 1.5 times the final height of the column.

The following are the main points of the solution:

The final height H of the column of liquid is determined by the balance between capillary force F=2{pi}r{gamma} and the weight Mg={pi}r2 H {rho}g, where {gamma} is the surface tension, {rho} is the density, r is the radius of the tube. Thus the final height is H=2{gamma}/({rho}gr).

If viscosity can be neglected then the motion of the liquid column can be described by the equation d(Mv)/dt = F - Mg where v=dh/dt, and h is the height of the liquid at time t, while M is the mass of the column at time t. The solution of this equation is a parabola, with its maximum h=1.5 H. (This solution means that after reaching the maximal point the liquid interface swings back to h=0. This will not happen when viscosity is taken into account. The height will simply perform damped oscillations around H.)





Question 11/97

Question 11/97

A variable capacitor is connected to two terminals of a battery of electromotive force E. The capacitor initially has a capacitance C(0) and charge q(0). The capacitance is caused to change with time so that current I is constant. Calculate the power supplied by the battery, and compare it with the time rate-of-change of the energy of the capacitor. Account for any difference.

This was a Candidacy Exam question at the University of Chicago.





Answer 11/97


The question was:A variable capacitor is connected to two terminals of a battery of electromotive force E. The capacitor initially has a capacitance C(0) and charge q(0). The capacitance is caused to change with time so that current I is constant. Calculate the power supplied by the battery, and compare it with the time rate-of-change of the energy of the capacitor. Account for any difference.

(11/97) This problem has been solved correctly by Oded Farago (e-mail [email protected]) from Tel Aviv University and by Tom Snyder (e-mail [email protected]) from Lincoln Land Community College, Springfield, Illinois. This problem was published in the "Graduate Problems in Physics" by J.A.Cronin, D.F. Greenberg and V.L. Telegdi, where a short solution can be found.

Here is (a slightly editted version of) the solution submitted by Snyder:

The battery supplies the constant power P1=EI. The electrostatic energy of the capacitor, U=qE/2, is changing at a rate dU/dt = (E/2)(dq/dt)=EI/2. Thus the battery is doing twice as much work as being stored in the capacitor. The difference is a work done by the capacitor on external agent that is causing the capacitance to change. To find an expression for the rate at which the agent delivers energy to the capacitor we imagine the situation in which the capacitor has been given a charge Qo and then isolated from the source. Thus Q will remain fixed at Qo. Now suppose the external agent varies the capacitance. Writing the potential energy U of the capacitor as Qo2/(2C) we find dU/dt = - (Qo/C)2(C'/2) = - (C'/2)V2 where V is the instantaneous value of the potential difference between the plates and C' is the rate of change of capacitance. Since the capacitor is isolated from the source of emf, dU/dt here must represent the rate at which energy is supplied to the capacitor by the external agent alone. Denoting this rate by P2, we have P2 = -C'V2/2. Although derived for the case where the capacitor is isolated, this expression for P2 must be true in general since it involves only the instantaneous value of V and the instantaneous rate at which the agent is varying the capacitance, C'. So we may apply this expression for P2 to the original situation. In this case V = E, a constant. Then P2 = -C'E2/2 = -d(CE)/dt * E/2 = -dQ/dt * E/2 = -IE/2 The total rate at which energy is delivered to the capacitor is then P1 + P2 = IE - IE/2 = IE/2 which is just the rate at which energy is being stored in the capacitor, dU/dt.







Question 12/97

Question 12/97

Once in a blue moon the moon looks blue or green or purple. Why?





Answer 12/97


The question was:Once in a blue moon the moon looks blue or green or purple. Why?

(12/97) Oren Melinger correctly noted that the change in the color of the moon is due to scattering of the light (reflected from the moon) by drops or particles. A slighly more detailed solution is provided below:

We are used that the light scattering in the atmosphere is more effective for shorter wavelengths, and therefore the color of, say, sun is shifted in the "red direction". However, sometimes the size of the water drops is of the same order as the visible light wavelength. In such case the scattering cross section is very sensitive to wavelength. In particular 600 to 1000 nanometer drops will scatter stronger the red light than the blue light. Thus the white light reflected from the moon, passing through a cloud will have red scattered out, and will become blue or green. For detailed explanation see H.C. van de Hulst Light scattering by small particles, Wiley, NY (1957). See also D.K. Lynch and W. Livingston Color and light in nature, Cambridge U. Press (1995).

p.s. O. Melinger also remarked that the current usage of the term Blue Moon means appearance of two full moons in a single calendar month.

p.p.s. The common usage of the term "Blue Moon" as something extremely rare or nonexistent can be traced as far back as Shakespeare (1528):

Yf they say the mone is bleweWe must believe that it is true.

p.p.p.s. (3/2000) Few years after publication of our "QUIZ" question about blue moon, a similar question has been asked and answered in "The Last Word" section of the New Scientist - see the last page of the 11 March 2000 No.2229 issue. In addition of mentioning various recorded case of blue moon due to atmospheric dust, some of the answers describe a case of clearly blue SUN! In particular, D. Harper mentions a report of R. Wilson in Monthly Notices of Royal Astronomical Society, 111, p. 477 (1951) about "deep indigo blue sun" which was observed in Edinburgh on Sept. 26, 1950 few days after extensive forest fires in Canada. (Spectrogram confirmed that this is NOT an optical illusion - red color was missing from the spectrum.) Apparently, the same case was reported by A. Watson (a pilot); RAF checked and found that at 11 km height there was a


http://www.skypub.com/sights/moonplanets/9905bluemoon.html

http://www.newscientist.com/

Answer 12/97

layer of smoke, above which sun looked normal. See all those questions and answers in the on line edition of the New Scientist. A local copy of those questions and answers can be found here.



http://www.newscientist.com/lastword/lastword.jsp?id=lw1159

http://star.tau.ac.il/QUIZ/97/lastword.html



Question 01/96

Question 01/96

What is the minimal speed with which a person can run on the surface of the water.

(Please, no remarks about the gentelman who walked on the water (lake of Galilee) two thousand years ago.) This question appears in the "Problems in Physics" by P.L. Kapitza





Discussion 01/96


The question was:What is the minimal speed with which a person can run on the surface of the water.

R. Mints (1/96): It is easy to set an upper bound - the escape velocity (8 km/sec) is certainly sufficient to walk on water. Actually, it is sufficient "to walk on air"...Y. Kantor(1/96): It is easy to set a better bound. If I will run at the speed of sound in the water (1500m/sec=5400km/hour) then the water will not react (i.e. will behave as a solid). This is certainly an upper bound. However, I think that the actual answer will be around 100km/hour - just think about swimming "butterfly" style...





Answer 01/96


The question was:What is the minimal speed with which a person can run on the surface of the water.

(2/96)The most reasonable answer was submitted by Deniz Ertas (post-doc at Harvard).The shortened (and maybe slightly distorted) version of his answer can be expressed as follows: If water with density rho is pushed with velocity v by a foot of area A, then a force rho*A*v^2 is generated (check the dimensionality!). This force must be of order of the weight W of the person. Thus an order of magnitude estimate should be v=sqrt{W/rho*A}=10 m/sec. There is however a catch: The question was about the horizontal velocity, while the argument considered velocity of vertical treading. Can we actually move with arbitrarily slow velocity provided we are treading with sufficient speed? [Ertas' detailed explanation appears at the bottom of this page.]

(12/96) Few months after we finished the discussions, our question recieved an unexpected answer from the works of people ( J.W. Glasheen and T.A. McMahon) investigating lizards running on the water. See the abstracts of two papers on the subject: 1, 2 and the figure at the bottom of this paragraph. (See also Sept. 1997 issue of Scientific American.) From these works we can deduce the following: (1) The force rho*A*v^2 has a prefactor of about 1/3. The prefactor depends on the shape of the foot, but we have an answer only for a foot of circular shape (of course!). There are also some corrections: (a) there is a dependence of the force on the depth as a result of hydrostatic pressure; it is negligible in our case; (b) there is an impact force during the entrance of the foot into the water - also negligible in our case. The presence of the prefactor increases our velocity estimate to velocities closer to 20 m/sec. (2) The the papers mentioned above also tell us about an additional problem: As the foot enters the water, the water separates and the foot must be withdrawn before the cavity in the water closes. However, we note that a person running on the water will not push his foot into the water more than 0.5 m, which will last for about 0.02 sec. From the graphs in the papers it is clear that the person has enough time to withdraw the foot before the cavity closes. (3) It is not clear how far forward should a person place his foot in the next step, but it seems reasonable that he must move at least 0.5 m (the depth of the cavity created in the previous step). In this case the horizontal running velocity will also be about 20 m/sec. (4) The most important problem raised in the articles mentioned above is the question of power: the person is supposed to apply a force of 600N at speed of 20 m/sec, i.e. the power will be 12000 Watts. This is way beyond the human ability. So now we know that it is impossible to run on water barefoot. Can we bring all the numbers into a reasonable range by wearing BIG shoes?


http://star.tau.ac.il/QUIZ/96/F01.96a.html

http://star.tau.ac.il/QUIZ/96/F01.96b.html

Answer 01/96

Here is what D. Ertas actually wrote:

Of course, it all depends on how you get your thrust; if you have a jet propeller attached, you could hover over the water and walk as slowly as you like. But that's cheating. Well, consider the honcho waterskiing person who refuses to wear skis and goes barefoot. People certainly do this, and although an auxiliary power is employed, the scenario can be changed such that the person himself provides the thrust, with additional factors of order one involved, perhaps: The lagging foot pushing against the water at a large attack angle and the leading foot gliding over the water at a small attack angle. No need to even approach the speed of sound, this requires relatively small velocities. The key is to create a lifting force that counteracts the gravitational force; the actual motion over the water can be adjusted at will. Assume that my leading foot makes an angle Theta with the horizontal; this is the "attack" angle I referred to earlier. The foot is moving with velocity v w.r.t. the water. Go the the coordinate frame where the foot is stationary. Then, the water exerts a normal pressure P=2*Sin(theta)*rho*v^2, where rho is the specific gravity of water. The vertical component of this pressure P*Cos(theta) should balance my weight pressure, which is given by W/A, where W is my weight and A is the contact area (say half the area of my foot). Then, the typical velocity involved is given by v=sqrt(W/(Sin(2*theta)*rho*A)). Plugging in typical values: W=600N, theta=pi/12, rho=1000 kg/m3, A=0.01 m2, the result is v=11 m/s or 21 knots; which seems plausible when the waterskier is considered.

Let me elaborate a little more on how you get the thrust. Let's asssume that we aren't really walking, but treading water. How fast would we need to tread the water with our feet in order to stay on top of the water? The answer should be essentially the same as the previous answer (think of two feet pushing outward at an angle and creating a lift that keeps the body up. Of course, an additional factor of 2 or so is needed so that we don't sink while bringing the feet closer) Thus, in fact the velocity I calculated is the minimum velocity needed for at least some part of my body to move. If you can tread on top of water, you sure must be able to walk at arbitrarily small velocity. (Of course, each foot still needs to move at about the originally calculated velocity.)





Question 02/96

Question 02/96

Surface of a river has a slope. Can a body freely float in the water with velocity exceeding the velocity of the river.






Discussion 02/96


The question was:Surface of a river has a slope. Can a body freely float in the water with velocity exceeding the velocity of the river.

Some people responded with the following argument: Obviously a body can slide on a frozen water (ice) provided the slope is steep enough; thus we have an example of motion faster than the stream. This answer is wrong because in the problem it was stated that the body floats in the water. Think what flotation implies...





Answer 02/96


The question was:Surface of a river has a slope. Can a body freely float in the water with velocity exceeding the velocity of the river.

No correct answers have been submitted. The answer is NO for the following reasons:Body floating in the water expels amount of water equal to its weight. Thus, we can simply replace the body by water which will fill its place. This replacement water will not flow faster than the rest of the water. In slightly different terms: If the body moves downstream faster than the water, it actually moves the same amount of water upstream; therefore, there is no energetic advantage (no gain of potential energy) in moving the body downstream.





Question 03/96

Question 03/96

What is the maximal distance from which you can see (with your naked eye) a burning candle in the middle of the night?

This question was derived from an exam question at Cambridge U.





Discussion 03/96


The question was:What is the maximal distance from which you can see (with your naked eye) a burning candle in the middle of the night?

You deserve few hints: the luminous intensity of one candle is about one candela (surprise!). Thus, the luminous flux is about 4{pi}=10 lumens, which correspond to about 0.03 Watts.





Answer 03/96


The question was:What is the maximal distance from which you can see (with your naked eye) a burning candle in the middle of the night?

I admit: it was a difficult question... It can be solved in one of two ways:

1. Eric M. Bram suggested a solution which seems to set the true maximal distance from which the candle can be practically seen. He assumed that we can see the stars of 6th magnitude, and by comparing their illuminance with that of a candle he arrived to the conclusion that the candle can be seen from 10 km distance. (See details of his solution below.) Later, assuming that even 8th magnitude stars are visible he revised his estimate to 60 km.

2. There is also a different approach: Since the luminous flux of candle is about S=0.03 Watts (see discussion of the problem), it emits approximately N=0.03 Watts/(3*10^{-20})Joules=10^{17} photons/second. (The number by which it was divided is the energy of a photon of yellow light.) From distance R the number of photons per second which reach the eye is N*A/(4{pi}R^2), where A is the area of the pupil (A is about 40 mm^2 for the dark-adapted eye). In order to see a continuous spot of light we need about 20 photons/second entering our eye (this is the "memory" of an eye; think about the speed of the movies!). Thus, N*A/(4{pi}R^2)=20, leading to R of about 140 km. (Some estimates in this solution were approximate and their modification may increase the number by, maybe, factor 2 or 3.)

Comment: The second solution assumed absolute darkness (no noise from surrounding light sources) and no atmospheric interference. We should probably conclude that away from a city on a really dark night the maximal distance is about 10 km.

Here is what Eric M. Bram actually wrote regarding the solution in part No. 1:

From C. W. Allen, "Astrophysical Quantities" (3d Ed, 1973) I see that a magnitude 0 star gives (outside the Earth's atmosphere) an illuminance (light received per unit surface) of 2.54(10 Exp-10) phot ==> [1 phot = 1 lumen/cm^2]. The faintest stars one is supposed to be able to see with the unaided eye is 6'th magnitude (of course we're inside the earth's atmosphere, so this might throw off the calculation, but forget that). Since the difference in magnitude m1-m2 between two stars with fluxes S1 and S2 is -2.5(log(S1/S2)), the ratio S1/S1 = 10 Exp ((-0.4)(m1-m2)). The six magnitude difference between magnitude 6 and 0 is therefore 10 Exp ((-0.4)(6)) or 10 Exp -2.4, so a magnitude 6 star (assumed to be the dimmest that can be seen) gives an illuminance of 2.54(10 Exp-10)(10 Exp -2.4) phot = 2.54(10 Exp -12.4) phot. Since 1 meter-candle (which I think I recall is the illuminance given by a candle at 1 meter) is 1 lux = 1 lumen/


Answer 03/96

m^2 = 10 Exp -4 phot, the candle at 1 meter looks (10 Exp -4)/((2.54)(10 Exp -12.4)) = (10 Exp 8.4)/(2.54) as bright as the faintest star that can be seen. Brightness varying inversely as distance squared, of course, in order to be as faint as that faintest star that can be seen with the unaided eye, the candle would therefore have to be Sqrt((10 Exp 8.4)/(2.54)) = 9944.5 meters away. HOWEVER: 6'th magnitude is the traditional limit among amateur astronomers for "naked eye" objects, but I and most experienced observers can see 7'th magnitude stars in a really good sky. And you're right, the night sky is never totally dark with only one star in it, so that should depress the "practical" value even more. Also, the initial illuminence value for the 0 magnitude star I used is for outside the Earth's atmosphere. Without the atmosphere the stars would be quite brilliant, and steady! So the faintest then might be more like 8'th magnitude. And every magnitude fainter the eye can see multiplies the "candle distance" answer by a factor of about 2.5. So if 8'th magnitude is the actual "naked eye star" limit with absolutely dark sky and no atmospheric absorption, perfect seeing, etc., the answer would be 6.31 times as large.





Question 04/96

Question 04/96

The legend has it that Greeks, on the advice of Archimedes, have burned the Roman ships by focusing the sunlight (reflected from their shields) on the ships. Is it possible?

This question was derived from an entrance exam question to the Moscow Physical-Technical Institute.





Discussion 04/96


The question was:


When I was a child I was able to use an ordinary lense (3cm in diameter) with a reasonable focal distance (10cm) to light a piece of wood. Therefore,...





Answer 04/96


The question was:


The answer is YES, but just barely...

Dry wood can be lighted by focusing the sunlight by a lens of diameter d=3cm and focal length f=10cm (the "regular" variety of a lens...). The angular diameter of the sun is a=0.01rad. The flux of light through such a lens is:

F=E({pi}/4)d^2,

where E is the flux of the sunlight on the surface of earth. The area of the image of sun created by such a lens is

S=({pi}/4)(af)^2.

Thus the illumination required to light the wood is

I=F/S=Ed^2/(af)^2=900E.

If the shields of Greek soldiers were flat and the ships were close to shore we can neglect the widening of the reflected beams of light. If we further assume 50% reflectivity of the shields, we can see that 1800 soldiers can create illumination of 900E required to light the wood, provided they all direct their shields in such a way that the reflected light falls at the same spot.


Answer 04/96

Comment: There is a beautiful and very informative web-site dedicated to Archimedes:http://www.mcs.drexel.edu/~crorres/Archimedes/contents.html

p.s. (4/2000) We received an interesting email from Jim Hunt detailing the history of the inquiry into the subject of ship burning. See here.



http://www.mcs.drexel.edu/~crorres/Archimedes/contents.html

http://star.tau.ac.il/QUIZ/96/brs.txt



Question 05/96

Question 05/96

When hockey puck is sliding on the ice it usually performs both linear motion (translation) and rotation. One notices, that the rotation stops exactly at the same moment as the center of mass comes to rest, i.e. the translation ends. Why?

This question was raised by several high-school, college and university teachers on physics discussion list at [email protected] in 1989. It is related to a question which appears in Resnick and Halliday book. However, even before that it has been considered by a number of physicists.





Discussion 05/96


The question was:When hockey puck is sliding on the ice it usually performs both linear motion (translation) and rotation. One notices, that the rotation stops exactly at the same moment as the center of mass comes to rest, i.e. the translation ends. Why?

We may assume (approximately) that the motion is slowed down by the "usual solid friction", i.e. the element of puck with mass (dm) which weighs g*(dm) experiences the friction force k*g*(dm) in the direction opposite to the direction of motion of that element.

(10/97) Gil Ariel and Shai Machness from Tel Aviv University wrote down the correct equations describing the motion of puck. They, however, were able to analyze the solution only in the limit of very small linear velocity or very small angular velocity.

(3/99) Eitan Federovsky (e-mail [email protected]) considered a simpler case and reached results which are slightly more susceptible to analysis. Below we present an edited version of his solution. (His equations are essentially the same as equations obtained by Ariel and Machness, but simplified geometry enables to make "one extra step".)

Instead of a disk we consider a ring of radius r and mass m sliding with linear velocity (of the center of mass) v and rotation with angular velocity w. The velocity of each point of the ring is sum of the velocity of center of mass v and velocity of rotation around center of mass (equal to the vector product of angular velocity vector and radius vector. Assume that the center of mass is moving in x-direction. The x component of the force will be

-k*m*g/(2{Pi}) int ((v-w*r*sin(alpha))/sqrt(v2 +(w*r)2 - 2*w*r*v*sin(alpha))

where the integral is performed over all angles alpha. Note that the integrand is simply an x-component of the total velocity of an element of a ring divided by the absolute value of the total velocity. Obviously this force is equal to m*(dv/dt).

Similarly we can calculate the torque. Instead of projecting force on x-axis, we now project it on direction tangential to the circle. The result is

-k*m*g*r/(2{Pi}) int ((w*r-v*sin(alpha))/sqrt(v2 +(w*r)2 - 2*w*r*v*sin(alpha))

which is equal to m*r2(dw/dt). By comparing the expressions for torque and force we immediately see that they are given by function of the same type:

F(x) = int (x-sin(alpha))/sqrt(1+x2-2*x*sin(alpha))



Discussion 05/96

dv/dt=-k*g/(2{Pi})*F(v/w*r)

d(w*r)/dt=-k*g/(2{Pi})*F(w*r/v).

Behavior of the function F is depicted in the following figure both as function of v/w*r and its inverse.

We see that if v is small compared to w*r the force acting to stop v is very small, while the moments on w*r are high, and also in the other way. This is physically explained by noticing that the friction force is only affected by the mass and the direction of the motion!. If w*r is very high and v is small, the total velocity vector direction is influenced mainly by w*r, and when adding we get from symmetry that the total force is zero, while the moments are high. If v is high and w*r is small, the direction of total velocity vector is mainly derived from v, so total velocity vector ~ v and the total force is high, while from symmetry the total moments are small.

(3/99) Y. Kantor: In addition we notice that if we start with v=w*r then d(w*r)/dt=dv/dt and therefore the ratio v/(w*r) will not change with time. Therefore, the rotation and translation will stop simultaneously. It remains to be shown, that whatever the initial ratio (except for trivial cases v=0 or w=0) the final ratio will approach the fixed value v/(w*r)=1! It seems that we passed the conceptually difficult stages of the solution. Nevertheless, there is some hard work ahead. We are waiting for continuation of the solution.





Answer 05/96


The question was:When hockey puck is sliding on the ice it usually performs both linear motion (translation) and rotation. One notices, that the rotation stops exactly at the same moment as the center of mass comes to rest, i.e. the translation ends. Why?

(10/03) Ivan Sirakov from CNRS, St Etienne, France (e-mail [email protected]) submitted (28/8/03) solution that apears in the following pdf file, as well as ilustration of the process that can be found in the zipped avi file. He essentially provides a complete solution of the problem.

In fact this problem was considered in the literature, quite some time ago. K. Volyenli and E. Eriksen analysed the motion in the article apearing in the American Journal of Physics 53, 1149 (1985).



mailto:<[email protected]

http://star.tau.ac.il/QUIZ/96/Hockey_puck.pdf

http://star.tau.ac.il/QUIZ/96/hockey.zip



Question 06/96

Question 06/96

Fish (left figure) are swimming by waving their tail left-and-right, while the spermatozoa (right figure) are spinning their tail. Why can't spermatozoa swim like fish?





Discussion 06/96


The question was:

Fish (left figure) are swimming by waving their tail left-and-right, while the spermatozoa (right figure) are spinning their tail. Why can't spermatozoa swim like fish?

Y. Kantor (3/97): Half-a-year after publication of the question we still did not get a satisfactory answer. It was correctly noted by several people that the size of spermatozoa is much smaller than that of the fish, and consequently the Reynolds number must be very different. Thus hydrodynamic equations describing the swimming must be quite different in those cases. So we are on the right track, and it seems that only few steps are missing in the solution...





Question 07/96

Question 07/96

A very tall chimney (width w, height h) starts falling. At some point it will break as shown in the Fig.(b). When will this happen and what will be the height of the breaking point?

This was one of the qualifying exam questions at MIT.





Answer 07/96


The question was:A very tall chimney (width w, height h) starts falling. At some point it will break as shown in the Fig.(b). When will this happen and what will be the height of the breaking point?

(1/99) The problem has been solved correctly by Chethan Krishnan (e-mail [email protected]) from Indian Institute of Technology, Madras, India. Below we present his (slightly edited) solution.

The answer: The chimney will break at 1/3 of its height.

The solution:

If the angle made by the chimney while it is falling is A (at a particular instant) then the torque at the base will be (1) . . . T1 = 0.5*m*g*L*cosA.



Answer 07/96

The moment of inertia of the chimney (relative to its base) is I1=m*L2/3. From the basic dynamical

equation T1 = I1*a we get that angular acceleration

(2) . . . a = 1.5*g*cosA/L.We see that the angular acceleration is inversely proportion to the length. The entire chimney is rotating with the same angular acceleration. Thus if we for instance look at only at some specific top part of the chimney, it will be rotating "slower than it should" under its own weight. It is "slowed down" because not only its weight is exercising a torque, but also the bottom part acts on it, and slows it down. That additional torque will eventually tear the chimney into parts. The torque on a particular point on the rod - at a distance x from the top - caused by the weight of the top part will be (3) . . . T2 = 0.5 *(m*x/L)*g*x*cosA=m*x2*a/3,

where we substituted the result of Eq.(2) into Eq.(3). This torque plus the torque T3 (which acts in the

opposite direction) equal to the moment of inertia of the top part alone I2 multiplied by the same angular

acceleration a. (4) . . . I2*a = (m*x/L)*x2*a/3.

From Eqs.(2),(3), and (4), we get that (5) . . . T3=T2-I2*a = m*x2*a/3-(m*x3/L)*a/3

This torque vanishes at the base and at the top of the chimney, and has a maximum at x=2L/3. Thus the chimney will break at a at height equal to 1/3 of its total height.





Question 08/96

Question 08/96

What is the ratio of the red, white and blue bars of the French tricolore? Why?

This question is from Physics book by Gerthsen, Kneser, Vogel





Answer 08/96


The question was: What is the ratio of the red, white and blue bars of the French tricolore? Why?

(1/1997) No correct answers have been recieved. Even not from France... Below we give you the solution. We would, nevertheless, like to have a detailed calculation.

The answer: Chromatic aberration in the eye makes red objects near blue objects appear closer. This effect has for long been known and used by artists. To correct for this and evoke the impression of three equal bars the tricolore is divided into 37:33:30 red:white:blue.

(11/2000) Bill Unruh (e-mail [email protected]) brought to our attention an interesting web page detailing the history and proportions of French flag (see here). The main points (as conveyed by Zeljko Heimer and Armand Noel du Payrat) are as follows: 1. The "tricolore" was created during French revolution, officially adopted in 1794, and re-adopted in 1830. The proportions of colors were specified as "equal". The 1958 Constitution does not specify the proportions. 2. "The proportions of vertical stripes on the French flag when used at sea as the civil or naval ensign are 30:33:37, to give a good visual effect when flying". (What does it mean?) 3. The Tricolore ensign was adopted by decree dated 27 pluviose an II (15 February 1794) and by decree dated 7 March 1848. The proportions 30:33:37 were decided by regulation dated 17 May 1853.




http://fotw.digibel.be/flags/fr.html



Question 09/96

Question 09/96

A car participates in a race. Its tire explodes. What should be the the speed of the car, so that the tire does not become "flat"?






Discussion 09/96


The question was:

A car participates in a race. Its tire explodes. What should be the the speed of the car, so that the tire does not become "flat"?

Ryan Werstuik ([email protected]) suggested (3/30/97) that in order to prevent deflation of the tire one needs to achieve the speed at which "there will not be enough time while the wheel is off the ground for any air to escape". He did not provide a detailed calculation but estimated that the speed will be close to the speed of sound.

Y.K.: It seems to me that the excess air will escape the tire at any speed, and that we do not really need to maintain air-pressure inside the wheel in order to prevent flat tire. Keep in mind that there are "centrifugal" forces acting on the tire...





Answer 09/96


The question was:A car participates in a race. Its tire explodes. What should be the the speed of the car, so that the tire does not become "flat"?

(7/98) The problem has been solved correctly by Itzhak Shapir (e-mail [email protected]). The answer is about 150 km/hour.

The solution:

Quarter of the weight of a car is supported by each wheel. The spped of the car must be large enough so the the centrifugal force on the part of the wheel which is in contact with the ground be equal to that part of the weight of the car. Here is the detailed solution by Shapir:

a. Assumptions on the car are that it has a mass M=1000kg, equally shared among wheels (i.e. 250kg per wheel)

b. Assumptions on the wheel are as follows: (1) outer diameter is 0.5m; (2) inner diameter is 0.3m; (3) tire mass is 5kg, equally spread around its outer diameter; (4) effective vertically oriented centripetal force is equally spread on 10cm of the tires footprint under its axes.

c. Method of calculation:

(1) A segment having mass m rotating around the axes with angular speed w at a distance r from the center of movement will apply the force : f=m*r*w**2o or f=m*v**2/r, where v=w*r is the linear velocity. In case of that tire v is the car's velocity.

(2) The mass of the effective tire footprint (10cm) is:m=5*0.1/(3.14*0.5) ~0.32kg

(3) In order to balance the 250kg weight (~2500N force) the equation will be:

2500 = 0.32*v**2/0.15 (0.15m is the flat position equals inner radius)

or v**2 = 2500*0.15/0.32 ~ 1250

so v = sqrt(1250) ~ 35 m/sec, which is about 120km/h

This is the speed to start recovering from flat tire.



Answer 09/96

d. Further calculation may be of the speed to maintain original tire inflation. Same assumptions on 10cm footprint and 250kg car mass on the wheel, but radius will be about 0.25m instead 0.15. In that case speed is calculated to be about 150km/h.





Question 10/96

Question 10/96

A stone falls into the water and water drops are splashed. Why do the water drops fly upwards? Does the maximal height reached by the drops depend (primarily) on the size of the stone or on its speed? What is the maximal height?






Question 11/96

Question 11/96

Why wet spots are dark?





Answer 11/96


The question was:Why wet spots are dark?

No correct answers have been received.

There are two explanations for the darkness of wet surfaces (see D.K. Lynch and W. Livingston, Color and Light in Nature, Cambridge U. Press, 1995):

1. If a thin layer of water overlays a non-porous substance then the light which in the absence of water was reflected from the surface of the substance to the eye of the observer (i.e. was not absorbed by the surface) reaches the the water-air interface and part of it is again reflected towards the surface, where it can be absorbed again.

2. If the substance is porous then part of the light is reflected from the surface (and has the typical color of the material) while another part of light is repeatedly scattered from the irregular features of the surface (some of which are of the size of the wavelength) and eventually reaches the eye of the observer. The latter part is essentially "white". This white light "dilutes" the "true color" of the substance. When the surface is covered by a thin layer of water the scattering is reduced, because now the light does not go from air (with refraction coefficient n=1) to material (with large n), but first goes to the water (with n=1.3) and only then to the material. The reduced contrast decreases the scattering. Thus reflected "true color" is no longer "diluted" by the white light. The substance looks darker and its color is "juicier".





Question 12/96

Question 12/96

When a coffee drop dries out, the coffee doesn't stay uniform; instead it concentrates in a ring around the boundary. The same thing happens when almost any drop dries. What makes the coffee become so nonuniform?

In case you have never seen a stain of a dried drop, R. Deegan provided for you the pictures of dried coffee (left) and dried latex (right):

This question was asked by R.D. Deegan, O. Bakajin, T.F. Dupont, G. Huber, S.R. Nagel and T.A. Witten





Discussion 12/96


The question was:When a coffee drop dries out, the coffee doesn't stay uniform; instead it concentrates in a ring around the boundary. The same thing happens when almost any drop dries. What makes the coffee become so nonuniform?

Jonathan Baugh ([email protected]) wrote on 3/19/97:If the surface that the drop sits on is not too smooth, then the diameter of the drop will remain relatively fixed over the duration of evaporation. Since the rate of evaporation is essentially uniform over the entire surface of the drop, the boundary can only remain fixed if there is a flow of liquid from the center outward. This flow tends to collect particulates at the edge, giving the familiar ring shaped stain. Anyway, that's my short, hand-waving theory.

Y. Kantor: I think that Jonathan is moving the right direction: the boundaries of the drop are indeed fixed during the process of evaporation. One needs, however, to get some more quantitative estimate regarding the evaporation rates from the different parts of the drop, and estimate what would be the current inside the drop. It seems that for small drops one can assume that the shape of the surface of the drop is parabolic.





Answer 12/96


The question was:When a coffee drop dries out, the coffee doesn't stay uniform; instead it concentrates in a ring around the boundary. The same thing happens when almost any drop dries. What makes the coffee become so nonuniform?

No complete unswer was submitted. Here is a brief explanation, by Rob Deegan (the original appears on the web):

Why do drying drops form a ring?

The image to the left is a common household sight: a dried coffee stain. It is immediately appearent from the ring-like appearance of the stain that the majority of the residue is concentrated along the perimeter. This is intriguing because it indicates that some physical or chemical process is at work in drying drops which is able to segregate the solute from the solvent. At the University of Chicago, we attempted to answer the questions of how general is this phenomenon, what is driving mechanism responsible for it, and are there practical technological implications?

We have determined that ring formation is a ubiquitous and robust phenomenon. It doesn't depend on the solute, the solvent, or the substrate so long as the solvent is partially wetting and volatile, and the contact line is pinned.

Traditional mechanism of solute transport--Marangoni flow, Rayleigh-Benard convection, diffusion, electrostatic and electrokinetic effects, and wetting phenomenon--do not account for ring formation. The phenomenon is due to a previously unexplored form of capillary flow: the contact line of the drying drop is pinned such that liquid evaporating from the edge must be replenished by liquid from the interior.

This mechanism is illustrated in the schematic to the right. Pictures a and b show a cross section of a drop during an infinitesimal increment of evaporation. The green hatched area represents the volume of liquid removed by evaporation in a single time step. The liquid removed from the edge must be replenished by an outward flow (indicated in blue).


http://mrsec.uchicago.edu/MRSEC/Nuggets/Coffee/

http://wcssex.ucsd.edu/alp/rb.html

Answer 12/96

We are also able to qualitatively alter the morphology of the ring deposit. Since evaporation is the driving mechanism, we can produce different deposits by altering the evaporation profile.

In the figure to the left, we indicate schematically three evaporation profiles by the amount of liquid removed at a given radius (red hatched lines) and a corresponding photograph of the resulting deposit. Interestingly, when the evaporation profile is such that no water leaves from the edge (bottom profile), no ring is formed. This qualitatively demonstrates the validity of the above mechanism.

As an experimental system, we used a colloidal suspension of polystyrene spheres in water. Fluorescent video microscopy was used to image the spheres, and image analysis techniques were used to extract information about the concentration and flow of the spheres in the bulk and their accumulation in the ring. We are able, using the simple mathematical model of the mechanism introduced above, to accurately predict these quantities (see pre-print).

Ring deposition is a potentially useful technology impacting printing, washing and coating processes. We are pursuing two application possibilities: fine wire assembly and DNA stretching. Using a colloidal suspension of gold particles it might be possible to assemble a fine wire with dimensions comparable to those achieved with modern lithographic methods. It might also be possible to stretch DNA using the strong shear flow that develops in ring forming drops when the evaporation rate is enhanced at the edge.

"Capillary flow as the cause of ring stains from dried liquid drops"Robert D. Deegan, Olgica Bakajin, Todd F. Dupont, Greg Huber,


http://arnold.uchicago.edu/~tten/Nagel.7.4.pdf

http://arnold.uchicago.edu/~tten/Nagel.7.4.pdf

Answer 12/96

Sidney R. Nagel, and Thomas A. Witten, was published in Nature.



http://www.nature.com/



List of sites with puzzles/problems


Site Comments Location Rating

Los Alamos Mega-Math project

Los Alamos G

Naor's puzzler Moni Naor's collection of Math puzzles

Weizmann Inst.

X

The MathSoft Puzzle Page

Cambridge, MA, USA

R

Physics Quiz Germany PG

Challenge Physics Challenges

Duke University, USA

PG

NRICH math online club

Cambridge University, England

G

BrainTeasers by Quantum Journal

Nat.Sc. Teachers Assoc. (US)

G

Physics Prob. Collection by A. Melikidze

Princeton University, NJ, USA

R

http://star.tau.ac.il/QUIZ/MISC/puz_list.html (1 of 2)11.7.2005 9:24:26

http://www.cs.uidaho.edu/~casey931/mega-math/

http://www.wisdom.weizmann.ac.il/~naor/puzzler.html

http://www.mathsoft.com/puzzle.html

http://www.jgiesen.de/Quiz/QuizEnglish.html

http://www.phy.duke.edu/%7Ehsg/physics-challenges/challenges.html

http://www.nrich.maths.org.uk/

http://www.nsta.org/quantum/

http://pupgg.princeton.edu/~melikidze/problems.htmlx


"Ponder this" - math puzzles

IBM (T.J. Watson) Res.Labs, NY,USA

R

"Physics of the outdoors" by S.M. van Roode

The Netherlands

R

Ortvay Yearly Physics Competition

Hungary R

The rating refers to the difficulty (prerequisite knowledge required) to solve a typical problem in that site. It is vaguely defined as follows: G=high-school; PG=high-school/undergraduate students; R=university students; X=graduate student/researcher.

If you know an address of an additional site with physics/math problems or puzzles, please mail the address to [email protected]

[Please, ask your webmaster or equivalent computer-wiz to include a reference (link) to our page (http://star.tau.ac.il/QUIZ/) in the appropriate web page of your organization.]

http://star.tau.ac.il/QUIZ/MISC/puz_list.html (2 of 2)11.7.2005 9:24:26

http://www.research.ibm.com/ponder/

http://home.hetnet.nl/~smvanroode/nvhvv/index.html

http://ortvay.elte.hu/main.html


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