A compilation of undergraduate physics problems

Thomas GarelService de Physique Théorique, CEA/DSM/SPhT

Unité de recherche associée au CNRS91191 Gif-sur-Yvette cedex, France

This compilation of physics exercises and comments may be useful for undergraduates. It contains seven chapters(Introduction, Mechanics, Central forces, Vibrations, Optics, Electromagnetism, Thermodynamics), and does not aimat being exhaustive or original. Many exercises can be found in specialized Journals (Am. J. Phys., Eur. J. Phys.,....),or books. Some post-exercises comments suggest further studies for the interested student.

2

Introduction

I1 Diameter of a fuse

A fuse is a cylindrical wire, designed to melt and disconnect an electric circuit, if the current exceeds a certain valueIc. For a given material, what is the dependence of Ic on the length L and on the diameter D of the fuse?

The fuse melts when the Joule heat cannot be evacuated through its surface, so that its temperature exceeds themelting point. The Joule power is PJ ∼ RI2, where R ∼ 4ρLπD2 is the resistance of the fuse. The surface evacuatedpower is PS ∼ f0πDL, where f0, like the resistivity ρ, depends on the nature of the fuse material. One gets

PJPS

∼ A0I2

D3

where A0 ∼ 4ρπ2f0 only depends on the material. This shows that the critical current Ic (i) does not depend on thefuse length L (ii) depends on the fuse diameter as Ic ∼ D3/2.

I2 Some results on the Earth

(2-a) Small rocks are not round, but large planets, such as Jupiter, are. Compare the gravitational and electrostaticenergies of cohesion for the case of the Earth.

One expects that the gravitational energy EG of a spherical body (mass M , radius R) is given by EG ∼ −βGM2

R ,where β = O(1). To calculate the electrostatic energy Eel of this body, we imagine that it contains 2N charged

particles (N with charge q and N with charge −q), on a lattice of periodicity a. We get Eel ∼ −α Nq2

4πǫ0a, with an

Ewald constant α = O(1). One has the relation R ∼ N1/3a, up to numerical prefactors, leading to

EGEel

∼ ( NNc

)2/3

where Nc ∼ ( q2/4πǫ0Gm2 )

3/2, with m the mass of the particles. One can then show that an iron-made earth is notcompletely dominated by gravitation. Jupiter, on the contrary has N ≫ Nc. Note that both forces are long-ranged, but charge neutrality has no equivalent in gravitation. The gravitational energy is therefore not extensive:(EG ∼ N5/3), contrarily to the electrostatic energy Eel ∼ N . Hence the existence of a critical Nc.

(2-b) A celestial body has a rotation period of 110 second. Find its composition.

Suppose that the cohesive energy of the rotating body is due to gravitation. This requires m(ω2r−g(r)) < 0, whereg(r) is the gravitational field at distance r from the center. Using Gauss theorem leads to ω2 < 4π3 ρG, where ρ is the

specific density of the star. One gets ρ > ω2

G , showing that this star has the density of a nucleus.

(2-c) Estimate the pressure at the center of the Earth.

Dimensionally, one has EG ∼ GM2

R ∼ PR3, leading to P ∼ GM2

R4 . Putting rough numbers gives P of the order of

1012 Pa.

(2-d) What is the maximum size of an asteroid from which you can escape by jumping?

We consider an earth-like asteroid of density ρA = ρE , and that your muscles are as efficient there as they areon the earth, with a maximum output of ∼ mgEh (h being something like your height). One must therefore havemgEh ∼ GMAmRA . Using gE =

GMER2E

and MAR3A

= MER3E

leads to RA ∼√REh of the order of a few kilometers.

(2-e) Estimate the thickness of an isothermal atmosphere.

3

For an isothermal atmosphere, Boltzmann distribution yields h ∼ kBTmg . Since the earth atmosphere is 75 % nitrogen(N2;m = 28mp), we get h ∼ 10kms.More generally, defining δ = GMm/RkBT , one has h ∼

kBTmg ∼ Rδ since g = GMR2 . For Pluto, one finds h ∼ 64kms for an

atmosphere of methane (CH4;m = 16mp).The lifetime of the isothermal atmosphere is considered in the Thermodynamics section (problem T5).

I3 Surface and bulk

(3-a) A naive view of the melting transition in an infinite solid is the following: when the thermal energy of anatom is of the order of magnitude of the cohesive energy of this atom in the solid, this solid melts. What is then thevariation of the melting temperature in a finite system (sphere, cylinder,...)?

In a finite system, the atoms close to the surface are a priori different from their bulk counterparts. For instance,if the interactions are short-ranged, the surface atoms have a smaller number of atoms to interact with than the bulkatoms. Hence the internal energy

E = Nbǫb −Nsǫs

where Nb (resp. Ns) are the number of atoms in the bulk (resp. at the surface) and ǫb,s the atom interaction energies.

For a sphere of radius R, one has Ns ∼ N2/3b , so that the melting temperature reads

Tf (R) = Tf (∞)−a

R= Tf (∞)(1−

δ

R)

where δ is a microscopic length (possibly depending on the orientation [hkl] of the surface) A similar reasoning forthe cylinder yields Tf (R,L) = T (∞)− aR − bL .For Coulomb forces, which are long-ranged and submitted to a global neutrality constraint, we have mutatis

mutandis the same result (see e.g. the calculation of the Ewald constant). For gravitational forces on the contrary,the distinction between bulk and surfaces does not apply (one gets E ∼ N5/3ǫ for a homogenous star, where N is thetotal number of atoms).

(3-b) Explain why some insects can walk on water.

If a leg on the insect exerts pressure on the water surface, it will tend to increase this surface. This means thatthe number of atoms less bound than in the bulk water will increase. The water will in turn try to react, exertingan upward force on the leg of the insect. This upward (capillary) force is to be balanced against gravity. If γ is thesurface tension, the balance of forces reads γL ∼ mg ∼ ρL3g. Surface phenomena dominate bulk phenomena belowthe capillary length Lc ∼

√γρg , of the order of 3 mm for water.

I4 Examples of scaling

(4-a) One multiplies all dimensions of a resistor, capacitor, inductance by the same factor k. Variation of thesequantities?.

For the resistor, one has initially R = ρLS . so that Rk = ρkLk2L2 = k

−1R. One similarly finds Ck = kC and Lk = kL.

(4-b) An horizontal magnetized needle oscillates at the end of a torsionless wire with a period T . One cuts it intotwo, and make one half oscillate in the same conditions. The period is T ′. Relation between T and T ′?.

If B is the horizontal component of the earth magnetic field, J the moment of inertia of the neeedle and M itsmagnetization, we have

T = 2π

√J

MB

Since for a needle, J is proportional to L3 and M to L, one finds T ′ = T2 .

4

(4-c) How long and how far can an animal of size L walk in the desert?

Assuming water reserves proportional to the volume of the animal and evaporation proportional to its surface, themaximum time should scale as L.There is also a famous argument stating that its speed v should scale as L0 (on a flat ground) and L−1 uphill. It

assumes that the power of the animal scales like L2. On a flat ground, the animal must overcome the power of the airresistance v× v2L2 ∼ v3L2, hence v ∼ L0. Uphill, the main force is gravity, so that mgv ∼ L3v ∼ L2, hence v ∼ L−1.Similar ideas suggest (i) that the maximal jumping height for an animal is independent of its size (ii) the terminal

velocity of a falling animal of mass M is proportional to L1/2 i.e. to M1/6.

(4-d) On the 1945 atomic bomb: suppose that the radius of the mushroom at time t depends only on t, on the airdensity ρ and on the energy E released. Find the relation between these quantities.

This story supposedly tells how G.I. Taylor estimated the power released by an atomic blast just by looking at somephotographs, showing how the mushroom expands in time. We have [E] =ML2T−2 and [ρ] =ML−3. To cancel themass M , we consider [Eρ ] = L

5T−2, from which we get

r(t) = C(Et2

ρ)1/5

where C is a dimensionless constant. Measuring r(t) at various times yields an estimate of E.

I5 Some static problems

(5-a) One wishes to hang a painting on the wall. This painting is idealized as a very thin homogeneous half-circleof radius R. An inextensible massless string of length L (L > 2R) is hooked on the painting at both extremities ofdiameter AB (Figure). This string can slide with no friction on a nail K driven in the wall. One also neglects thefriction between the painting and the wall. Discuss the equilibrium position(s).

��������

Knail

A

B

O

G

AK +KB = L

AB = 2R

A) Quick solution

For equilibrium to occur, the string tensions at A and B must be equal (in modulus), and their moments w.r.t

center of gravity G of the half-circle should be equal as well. This implies sinK̂AG = sinK̂BG, which in turn yields

• K̂AG = K̂BG (horizontal position for AB)

5

• K̂AG = π − K̂BG (two slanted solutions, symmetrical w.r.t to the vertical KG) The quadrilateral KAGB isthen inscribable and one has ÂKB = π − ÂGB

It easily seen that the horizontal position is unstable: a slight increase in the angle K̂AG creates a torque thatfurther increases this angle. The two slanted positions are stable. One may also show that KG is maximum (implyingthat the potential energy is minimum ) if the quadrilateral KABG is inscribable.

B) More detailed solution

Since OG = 4R3π by elementary geometrical considerations, one may build triangle AGB, and find the supplementary

angle ÂKB. One must therefore build a triangle AKB, knowing side AB, the opposite angle ÂKB, and the sum ofthe two other sides AK +KB = L.

The problem has a solution iff Lsin ÂKB2 ≤ AB < L. Furthermore, sin ÂKB2 = cosÂGO, with tgÂGO = 3π4 .Putting everything together, one finds that the stable slanted positions exist for 4L√

9+16π2≤ 2R < L. The unstable

solution exist for 2R < L.

(5-b) A lawn sprayer consists of a small spherical cap, with a non-uniform density of holes through which water isejected at speed v0. Find the density of holes ρ(α) to get a uniform watering around point R (see figure).

M

O

z

R

(Oz,OM) = α

From Newton’s equations in the gravity field, one has R =v20g sin2α, and the watering around point R is proportional

to RdR. This water comes from the holes between α and α + dα. Their number is proportional to ρ(α)sinαdα. Wetherefore have

ρ(α)sinαdα ∝ RdR ∝ sin4αdα

that is ρ(α) ∝ sin4αsinα .

(5-c) A solid block (height l, density µs, unit cross section ) floats in an infinite pool of liquid (density µl > µs).Consider two different states (i) the reference state where the block just rests above the water (ii) a second state wherethe immersed part of the block has height u. Calculate their potential energy difference and “deduce” Archimedes’principle

µl

(i) (ii)

u

l

µs

In state (i) the immersed part of the block has u = 0. The center of gravity of the block has therefore dropped byu in state (ii). The displaced liquid has its center of gravity at −u/2 in (i), and at u = 0 in (ii) since the liquid poolis infinite. We therefore have

∆E = Eii − Ei = Slµsg(−u+ l/2− l/2) + Sµlu(0− (−u/2))

6

where solid and liquid contributions have been detailed. Extremizing this difference, d∆Edu = 0, we obtain ueq =µslµl

,

expressing Archimedes’ principle.

(5-d) Find the shape of a road where a square wheel may roll without slipping. It is also required that each point ofthe road is a point of static equilibrium of the wheel.

a

x

y

O

C

D

G

E

B

α

αE

G

B

Let C, with coordinates (x, y) be the contact point of the square wheel (side length 2a) with the road. If eachpoint of the road is a point of static equilibrium, the center of mass of the wheel G moves along a horizontal line, andremains directly above the contact point C (see figure). Pure rolling implies that the line segment BC has the samelength as the arc OC.The horizontality condition reads GB=GC+CD so that

a√2 =

a

cosα+ y

Since dydx = tanα, one easily gets the shape of the road as

y(x) = a√2(1− 1√

2cosh(k − x

a)) (1)

where k is found by requiring y(0) = 0. The road is catenary shaped, and remains so for any polygon-shaped wheels.

Comment A sinusoidal road is obtained for an elliptic wheel.

7

Mechanics

M1 Oscillators

One considers an infinite quasi-planar conductor, whose thickness e is small. The surfaces of this conductor carrya uniform surface charge density σ > 0. A negatively charged particle is constrained to move perpendicularly to theconductor, and may go through it via a small hole which does not perturb the conductor. Initially, the particle isreleased with zero velocity at a distance d (d≫ e) from the conductor. If τ is the period of the particule motion, showthat τ ∼ dα and give the value of the exponent α. One may compare this result with other periodic motions (planetrevolving around the sun, harmonic oscillator,..)

d

e

σ

Let us denote by z the axis of the motion. The electric field can be calculated by Gauss’theorem; it is along z forsymmetry reasons. Taking into account the fact that the conductor has two surfaces, one finds

~E =σ

ε0sign(z) ~ez

The force on the charge q < 0 is attractive ~F = q ~E, so that the equation of motion read

mz̈ =qσ

ǫ0; z > 0

mz̈ = −qσǫ0

; z < 0

Solving these equations give a periodic motion with period τ ∼ d1/2 yielding α = 1/2.This result can be obtained by writing the potential at point z as V (z) ∼ zβ . Newton’s equation mz̈ = F = −dVdz

can be written in a dimensional way

[z]

[t]2= [z]β−1

yielding a time T -length L relation T ∼ Lα, avec α = 1− β/2. The above case correspond to β = 1. Kepler’s thirdlaw corresponds to β = −1 and the harmonic oscillator to β = 2 (period independent of the amplitude).

Note that a similar dimensional analysis for the diffusion equation

∂c

∂t= D

∂2c

∂x2

shows that time T and space L are related by L2 ∼ DT , in marked contrast to wave propagation where L ∼ cT . Ina traditional (non-microwave) oven , the roasting time of a mass M is proportional to L2, i.e. to M2/3.

M2 Equilibrium configurations of current carrying wires

(2-a) In classical mechanics a standard method for calibrating the orbits of charged particules with momentummodulus p in a static magnetic field is to determine the configuration in the same magnetic field assumed by aperfectly flexible wire, carrying current I and under a mechanical tension (modulus T ). Why?

8

Newton’s equation of motion reads

d~p

dt= ~F = q~v × ~B (2)

where ~p = m~v is the particle momentum, and ~F is Lorentz force.

Since ~F · ~v = 0, one has v = |~v| = cst. One therefore writes

~v = v~t = vd~r

ds; ~p = p~t = p

d~r

ds

where ~t is tangent to the trajectory. Since ds = vdt, eq(2) reads

vd

ds(pd~t

ds) = qv~t× ~B

that is

d2~r

ds2=q

p

d~r

ds× ~B (3)

For a flexible wire, the tension is directed along the tangent to the wire, so that the equilibrium equation of thewire reads

Id~r × ~B + T (d~rds

)2 − T (d~r

ds)1 = ~0 (4)

so that one gets

d2~r

ds2= − I

T

d~r

ds× ~B (5)

which has formally the same structure as eq(3) with the correspondence qp → − IT .

(2-b) Derive the shape of a flexible rope of length L and linear mass density µ in the gravity field

Equilibrium for an element ds of the rope reads

~T (s+ ds)− ~T (s) + µ~gds = ~0 (6)

that is

d~T

ds+ µ~g = ~0

Projecting this equation onto the x and y axis, and using ds2 = dx2+dy2, one finally gets the familiar catenary shape

y(x) =T0µg

(cosh(µgx

T0)− cosh(µgx0

T0)) (7)

where we have chosen y(±x0) = 0 and where T0 is given by the implicit equation

2T0sinh(µgx0T0

) = µgL

Comment 1 This catenary shape is also of interest for the calculation of a soap film stretched between two coaxialrings at a distance 2x0, since the soap film may have an equilibrium surface known as the catenoid, generated by therotation of a catenary around the rings axis.

9

Comment 2 Several generalizations of the above calculations are possible, and we mention only two (i) In equation

6, on may add a Lorentz force I ~ds × ~B if the rope carries a current I in the presence of a magnetic field ~B. Forvery large fields, one expects a circular shape of the rope (maximum flux rule). For intermediate field, one can solveexactly the problem (ii) One may also study the problem of an hanging rope at an air-water interface. The abovecalculation applies to the part of the rope hanging in the air. For the part hanging in water, one cannot neglect theArchimedes thrust in equation 6. At the air-water interface, the rope (i.e. its tangent vector) is continuous, but the(radius of) curvature jumps.

Comment 3 The problem of the equilibrium shape of a flexible rope in a given force field has often an opticalgeometric analogy (see e.g. the problem of the mirage explained by catenary-like trajectories of the rays). We illustratethis analogy below.

(2-c) Tension of a rotating flexible rope of linear mass density µWe consider a rotating rope (constant angular velocity ω, and assume that it is in equilibrium in the rotating frame.

Neglecting the weight we have

~T (s+ ds)− ~T (s) + ~fds = ~0 (8)

where the centrifugal force is given by ~f = −~∇V and V = − 12µω2r2, r being the distance to the rotation axis. For aflexible rope, one has ~T = T~t, where ~t = d~rds is the tangent unit vector. Multiplying equation (8) by

~t, we get

T − V = Constant

Note that the Constant is to be understood along the rope. We thus have

T (r) = T (0)− 12µω2r2

This example shows that the rope equation (dT~t)

ds =~∇V is directly linked to the optical geometry ray equation

d(n~t)ds =

~∇n derived from Fermat’s principle. One expects direct analogies between the shape of the rope in potentialV and the ray trajectories in a medium of index n.

M3 Closure of a car’s door

One considers a Galilean frame of reference R, and a stopped car with its door open. The angle between the carmain body and the door is a right angle. The door is square shaped (of side d) and homogeneous (mass m). Itsmoment of inertia JI with respect to the hinge I is known. The car starts to move with a constant acceleration γ.Study the motion of the door until it is closed (final rotation speed, closure time,...).

������

������

θ

γ

dI

Ghinge

~i

~j

~k

10

Let us consider in R an orthonormal trihedron (~i,~j,~k) where ~i is parallel to the initial position of the door, ~j isparallel to the direction of the acceleration ~γ, (with ~k =~i×~j). Point G is the instantaneous center of gravity of thedoor and point I denotes the hinge.Denoting by O the origin of R, we have for the system ”door”

Md2−−→OG

dt2= ~F =M~g + ~N + ~T ;

d~LGdt

=−→GI × ~T (9)

where we have introduced the normal ( ~N) and tangential (~T ) components of the hinge reaction on the door. Since

one has clearly ~N +M~g = ~0, eq.(9) can be rewritten as

Md2−−→OG

dt2= ~T ;

d~LGdt

=−→GI × ~T (10)

Since d~LGdt = −JGθ̈~k, and

−−→OG =

−→OI +

−→IG (implying that ~VG = ~VI +

d−→IGdt ), we get

−JGθ̈~k =−→GI ×M(d

~VIdt

+d−̇→IG

dt)

Using (−→GI = −d2cosθ, d2 sinθ, 0) and ~γ =

d~VIdt , we have

−JGθ̈~k =M−→GI × ~γ +M(d

2)2θ̈~k

Projecting on the ~k axis, and noting that JI = JG +M(d2 )

2, one gets

JI θ̈ =1

2Mγdcosθ (11)

From eq.(11) and the initial conditions θ(t = 0) = θ̇(t = 0) = 0, the final rotation speed of the door reads θ̇f =MγdJI

,

(for the corresponding angle θf =π2 ). The closure time is given by

tf =

√JIMγd

∫ π/2

0

dθ√sinθ

The numerical value of the integral is 2.62...

M4. The charge-monopole problem

One postulates the existence of a magnetic charge g (also called magnetic monopole). This magnetic charge is fixed

at the origin O of a coordinate system. At a point M, such that−−→OM = ~r, this charge creates a magnetic field

~B = gµ04π

~r

r3

Study the motion of a particle of charge q and mass m in the magnetic field of this monopole fixed at point O.

Introducing the angular momentum of the particle with respect to point O, ~L = m~r ∧ ~v, one will show in particularthat the vector

~J = ~L+ α~r

r

with α = −qg µ04π , is an invariant vector of the motion.

The equation of motion reads

md~v

dt= q~v × ~B = α

~L

mr3(12)

11

where α = −gq µ04π and ~L = m~r × ~v.Using eq(12), it is readily seen that d

~Jdt =

~0 with

~J = ~L+ α~r

r(13)

Since ~L · ~r = 0, we have ~J · ~r = Jrcos( ~J,~r) = αr implying that

cos( ~J,~r) =α

J

The orbit of the particle therefore lies on the surface of a cone, with point O as summit, axis parallel to the direction

of ~J , and with half-angle cos−1 αJIt is possible to show that

• |~L| is time invariant but not ~L

• r2(t) = r20 + v20t2 with appropriate initial conditions

• denoting ~̂r = ~rr , we have

d~̂r

dt= ~ω(t)× ~̂r

where ~ω =~J

mr2(t) . This shows that the particle rotates at non-constant angular speed on the surface of the cone.

Comment 1 Since ~r is a polar vector and ~L a pseudovector (also called an axial vector), the homogeneity of eq.(12) requires the magnetic charge g to be a pseudoscalar (moreover odd under time inversion).

Comment 2 It can be shown that a static electromagnetic field may have a momentum density ~pem =~Sc2 , where

~S = 1µ0~E × ~B is the Poynting vector of the field. Dimensionnally, one indeed has [pem] = ǫc , where ǫ is an energy per

unit volume. The second term on the r.h.s of eq (13) is the integral over space of the field angular momentum density~r × ~pem and may be called the field angular momentum. Equation (13) then expresses the conservation of the totalangular momentum (mechanical + field).

M5 Comments on the virial theorem.

(5-a) Statement of the theorem

Take a point mass m and let ~F be the applied force. Newton’s equation reads

md~v

dt= ~F

We further assume that the motion of the particle is bounded at all times (finite position ~r, finite velocity ~v,..).Multiplying both sides of the equation by ·~r, we have

md~v

dt· ~r = m(d(~v · ~r)

dt− ~v2) = ~F · ~r (14)

Denoting by < A > the following time average

< A >= limT→∞1

T

∫ t+T

t

A(u)du (15)

12

we get from equation (14) and the boundedness of the motion

−2 < T >=< ~F · ~r > (16)

where T is the kinetic energy of the particle.

Comment 1 If the motion is not only bounded, but periodic, one can show that equation (16) holds, with theaverage symbol < .. > meaning now average over one period of the motion.

Comment 2 Applications to a bounded and interacting system of particules is straightforward. If ~Fi is the forceon particle i at position ~ri, we have

−2 < T >=<∑

i

~Fi · ~ri > (17)

where 2T =∑imiv

2i is the kinetic energy of the system.

(5-b) Application to the Kepler problem

As mentionned above, we are concerned here with the case of E < 0, i.e. the particle (planet) follows an elliptic

trajectory. In this case, the force ~F derives from the potential energy U = −GMmr and the virial theorem, asinterpreted in the periodic case, reads

2 < T >= − < ~F · ~r >=< −→∇U · ~r = − < U > (18)

Comment 1 The same relation applies to a system of N particles interacting through gravitational forces (seeequation (17)). If one adopts the ergodic hypothesis stating that time averaging is identical to ensemble averaging,one may link the time averaged kinetic energy to its canonical ensemble averaging, and write < T >= 32NkBτ , wherekB is Boltzmann’s constant, and τ the temperature. Knowing an order of magnitude for U may give an estimate of

τ . A naive guess for the sun would be 3NkBτS =35GM2SRS

, yielding τS ≃ 107 K for an hydrogen-made sun.

Comment 2 Due to friction on the upper atmosphere, a satellite orbiting around the earth will see its total energydecrease (remember that E < 0 and the satellite loses altitude). Through the virial theorem, this means that thesatellite increases its kinetic energy, and therefore its speed.

(5-c) Application to the harmonic oscillator

In this case, the force ~F derives from U = mω2

2 r2 and the virial theorem, as interpreted in the periodic case, reads

2 < T >= − < ~F · ~r >=< −→∇U · ~r = 2 < U > (19)

that is < T >=< U >.

CommentMore generally , if U is a homogeneous function of degree n, the virial theorem reads 2 < T >= n < U >.The cases n = 2 and n = −1 respectively correspond to the harmonic oscillator and the Kepler motion.

(5-d) The case of the Lorentz force

An interesting case concerns an electron around the nucleus, in the presence of a magnetic field ~B. The force onthe electron reads

~F = q( ~E + ~v × ~B)

where ~E is the electric field created by the nucleus. Performing the same steps as above leads to

−2 < T >=< ~F · ~r >=< U > +q < ~r × ~v > · ~B (20)

13

that is

−2 < T >=< U > +2 < ~M > · ~B

where U is the electrostatic energy and < ~M >= 12q < ~r × ~v > the average magnetic moment of the electron.

(5-e) Application to a gas of particules

(5-e-1) Non-interacting particles

We consider now a gas of N particles, where the virial theorem reads

−2 < T >=<∑

i

~Fi · ~ri > (21)

where the average kinetic energy of the gas is linked to the temperature τ through < T >= 32NKBτ (ergodichypothesis for space dimension d = 3). In equation (21), the only force on the particles is the force exterted by thewalls of the vessel when their position ~ri corresponds to the positions of these walls (there are no other interactionsbetween the molecules of the gas). The time averaged force per unit area of the wall is (minus) the pressure p, so that

<∑

i

~Fi · ~ri >= −∫

wall

p−−−−→dSwall · ~r (22)

where−−−−→dSwall is oriented along the outward normal to the wall, and the integral is evaluated over the walls. Since

∫~r · −→dS =

∫div~r dV = 3V

where V is the volume of the vessel, the virial theorem finally reads

2 < T >= 3pV (23)

that is pV = Nkbτ .

Comment 1 Equation (23) can also be found by simple dimensional arguments applied to quantum mechanics.

The kinetic energy for classical particule reads T =∑ip2i2m . Now de Broglie’s relation implies that a wave of wave

length λ is associated to a particule through pi ∼ hλ . For non-interacting particles, the only natural length is thevessel ’s linear dimension L ∼ V −1/3 (in three dimensions). So we expect λ ∼ L, yielding

T ∝ V −2/3

. The pressure of the gas is given by p = − dTdV = 2T3V , in agreement with eq. (23).

Comment 2 If one deals with relativistic particles, one can also deduce a virial theorem. It is simpler for photonswhere the energy-momentum relation is linear, as opposed to the quadratic dependence for classical massive particles.Since T =

∑i pic (c being the speed of light in the vacuum), de Broglie’s relation implies now that T ∝ V −1/3, so

that the pressure of the gas of photons reads p = T3V .

(5-e-2) Repulsive interactions A/rn

Let us suppose that there is a two-body cental repulsion of the form

u(rij) =A

(|~ri − ~rj |)n

between molecules i and j of the gas. The total energy of the gas reads

U =∑

i

14

and equation (21) becomes

−2 < T >=<∑

i

~Fi · ~ri >= −3pV + n < U > (24)

i.e. pV = NkBτ +n3 < U >, where we have also assumed that the gas is three-dimensional.

(5-e-3) Logarithmic interaction in two space dimensions

We consider that molecule i carries a charge qi = ±q and that the gas of N molecules is globally neutral. This caseis of interest for the melting of two-dimensional crystals, where the melting may be linked to the collective behaviourof a gas of dislocations, interacting through a two dimensional Coulomb-like logarithmic interaction.

u(rij) = −Aqiqj ln(|~ri − ~rj |)

a

where A is an appropriately dimensioned constant, and a a microscopic length. The total energy reads

U =∑

i=<∑

i

~Fi · ~ri >= −2pV −N

2Aq2

where V is the “volume” of the two-dimensional vessel. Since in two dimensions, one has < T >= NkBτ , the equationof state of the gas reads

pV = NkB(τ − τ0)

with τ0 =Aq2

4kB. A more detailed theory shows that the above results apply only for τ >> τ0.

M6 On slowly varying parameters in some oscillators

(6-a) Spring constant

Consider a point mass attached to a spring with a slowly varying stiffness constant (the variation time scale is muchlarger than the period of the oscillation). Find how this variation affects the amplitude of the oscillation.

The energy of the oscillator is E = 12mẋ2 + 12kx

2. If k varies , we have

dE =1

2dkx2 =

dk

k(1

2kx2) (25)

Since k varies on time scales much longer than the period, we average equation (25) on one period and get

dE =dk

k<

1

2kx2 >av (26)

For a harmonic oscillator, the virial theorem tells us that

<1

2kx2 >av=

E

2

which, together with equation (26), yields

E = cst√k

15

Since E = 12ka2, we finally get

ak1/4 = cst (27)

Comment This result can also be obtained directly through the use of the adiabatic invariant Eω where ω =√

km .

(6-b) Vertical bouncer

A point mass m bounces elastically from the floor. Due to the fall of meteorites, the gravity acceleration g varieswith time on a time scale much larger than the period of the motion. Find how this variation affects the amplitude ofthe motion.

Following closely the steps of exercise (6-a), we have

dE = mxdg =dg

g(mgx) (28)

Since g varies on time scales much longer than the period, we average equation (28) on one period and get

dE =dg

g< mgx >av (29)

For oscillation in a linear potential, the virial theorem tells us that

< mgx >av=2E

3

which, together with equation (29), yields

E = cstg2/3

Since E = mgh, we finally get

hg1/3 = cst (30)

Comment 1 This result can also be obtained directly through the use of the adiabatic invariant Eω where ω ∼√

gh .

Comment 2 Some subtle points may arise if the slowly varying parameter is the mass of the particle m.

M7 Pendulum with nail

Consider the figure below, where a mass m hanging from a massless rope of length l is released with zero velocityfrom point A0. A nail is located at point A

′ on the vertical OA, such that AA′ = l′. Study the possible motions ofmass m in the cases (i) l = 5l′ (ii) l = 2l′ and α0 =

π2 .

There are a few natural possibilities to be inquired

• asymmetric oscillations

• wrapping of the taut rope around the nail

• vanishing of the rope tension

16

l

l′

α0

O

nail

A

A’A0

m

m

β

Asymmetric oscillationsDue to energy conservation, the maximum angle β0 on the left part of the figure would be given by

mgl′(1− cosβ0) = mgl(1− cosα0) (31)

Writing β0 <π2 leads to the conditions

1− cosα0 <l′

l(32)

Note that the period of oscillation is shorter than for the symmetric pendulum. The angular amplitude is larger onthe left side of the figure.

Wrapping of the taut rope around the nail

If the initial potential energy is high enough, the taut rope may wrap around the nail so that the mass m may passbetween O and A′ with a velocity v′ which we can find from the equilibrium condition

mv′2

l′= mg + T

where T > 0 is the tension of the rope (which in this position has the same direction as the weight).This implies

v′2 > gl′

The initial potential energy must therefore satisfy

mgl(1− cosα0) > 2mgl′ +1

2mv′2

that is

1− cosα0 >5l′

2l(33)

Vanishing of the rope tension

We have assumed in the two cases above that the rope remains under tension. Let us know consider in more detailthe equilibrium conditions involving the tension T of the rope. At a general angle β on the left part of the motion,this equilibrium condition reads

mv2

l′= T −mgcosβ

Together with the conservation of energy

1

2mv2 +mgl′(1− cosβ) = mgl(1− cosα0)

17

this shows that T = 0 for

cosβ =2

3(1− l

l′(1− cosα0)) (34)

Note that equation (34) gives back equation (32) if β < π2 , and equation (33) if T > 0 for β = π.

The two cases to be studied are (i) l = 5l′ (ii) l = 2l′; α0 =π2 .

In the first case, one may have asymmetric oscillations (1− cosα0 < 15 ), wrapping of the taut rope around the nail(1− cosα0 > 12 ) and vanishing of the tension in between (the trajectory of mass m is then a parabola until the ropegets taut again)

In the second case, one has only a vanishing of the tension for cosβ = − 23 .

M8 Oscillations of a rectangular plate

A pendulum is made out of a thin homogeneous rectangular plate, which oscillates around an axis K perpendicularto the plate and passing through the middle of a side of length a (Figure).

A) Axis K being fixed, one considers several such plates, of fixed mass M , but with different ratios r = ab . What isthe qualitative variation of the oscillation period TA(r) with r ?

B) One fixes now axis K in the middle of a side of length b and one considers several such plates, of fixed mass M ,but with different ratios r = ab . What is the qualitative variation of the oscillation period of the plates TB(r) with r ?Can one find a ratio r0 such as TA(r0) = TB(r0)?

K

b

a

This problem is borrowed from the book by JM Lévy-Leblond ,La Physique en questions: Mécanique, Vuibert ed.,1998.

Since the mass M is fixed and the plate is homogeneous, one has in all cases ab = l20 = cst. The oscillation periodis of the general form

T (a, b) ∼√

J

Mgd

where the values of J and d depend on the choice of axis K and/or the geometry of the plate.

A) one has d = b2 .

For a≫ b (r ≫ 1), one has J ∼Ma2 leading to TA ∼√

a2

b ∼ r3/4 since ab = cst.For a≪ b (r ≪ 1), one has J ∼Mb2 leading to TA ∼

√b ∼ r−1/4 since ab = cst.

B) In this case, one exchanges the role of a and b so that one has to substitute in the analysis of A r by 1r .We thus haveFor a≪ b (r ≪ 1) , TB(r) ∼ r−3/4For a≫ b (r ≫ 1), TB(r) ∼ r1/4

It is easily seen that the minimum rAm (resp rBm =

1rAm

) of TA(r) (resp TB(r)) is distinct from the point r = 1. The

curves TA(r) and TB(r) have therefore three points of intersection r = 1, r0, r−10 (Figure).

18

TB(r)

TA(r)

r

T (r)

r = 1 rAm r0r−10 r

Bm

Comment More exact calculations would be as follows. The moment of inertia of the plate w.r.t to its center of

mass G is JG =M(a2+b2)

12 . In geometry (A), we have J = JG +Mb2

4 , so that, setting a = l0√r, b = l0√

r, we get

TA(r) ∼√

J

Mg b2∼

√l0gfA(r)

with

fA(r) =

√r

32

6+

2

3r12

This yields rAm =√

43 .

M9 A classical potential well

A point particle (mass m) slides without friction along the x axis, on a table where a well of length 2a and heighth has been hollowed out (Figure). The collisions on the (horizontal and vertical) walls conserve the particle energy.The same experiment is done many times, with different initial velocities ~v0 = v0~x. Study the motion of the particleand find the probability that it escapes from the well.

~v0

m

2a

h

x

The crux of the problem is to notice that there are two length scales in the problem, the well length 2a and thebounce length l (defined as the horizontal distance between two successive contacts). One can in some sense unfoldthe well and consider trajectories that are symmetrical with respect to the vertical walls.The bounce length l is given by the classical parabolic trajectory and reads

l = 2v0(2h

g)1/2

We consider that the particle is incident from the left.

19

• If (2p+ 1)2a = nl, with (p, n) integers, the particle is transmitted• If (2p)2a = (2n+ 1)l, with (p, n) integers, the particle is reflected

It is readily seen that all rational ratios 2al do lead to either reflection or transmission. Transmission in particularrequires that the energy of the particle is of the form

Epn =1

2mv20 =

mga2

4h

(2p+ 1)2

n2

Performing many experiments with different v0, i.e. different l will amount to test how many l’s lead to a rationalratio 2al . The escape probability is therefore zero.

M10 Coupled oscillators

(10-a) A magnetic example

Two identical magnetic dipoles ~µ1 and ~µ2 can freely rotate around two fixed axis perpendicular to the line joiningthem (Figure). The distance between the dipoles is a and their moment of inertia with respect to their rotation axisis denoted by J . Study the equilibrium positions and the small oscillations around them.

α β

~µ2~µ1a

Solution

The interaction energy of two dipoles reads

E =µ04π

(~µ1 · ~µ2r3

− 3(~µ1 · ~r)(~µ2 · ~r)r5

)(35)

Setting W0 =µ04π

µ2

a3 , we get

E =W0(cos(α− β)− 3 cosα cosβ

)

A traditional study shows that the stable equilibrium position is α = β = 0. Other configurations, includingα = π2 = −β, are unstable. Expanding E to second order, we get

E(α, β) = E(0, 0) +W0(α2 + β2 + αβ)

from which we get the small oscillation equations

Jα̈ = −∂E∂α

= −W0(2α+ β) (36)

Jβ̈ = −∂E∂β

= −W0(α+ 2β)

The eigenmodes of eq(36) are either symmetrical (α = β) or antisymmetrical (α = −β) with respective pulsationsω2sym =

3W0J and ω

2antisym =

W0J .

(10-b) Two neutral atoms

Two hydrogen atoms are separated by a distance R, large in comparison with their size. The electron of each atom isbound to its nucleus by an elastic force, and we assume its motion to be along the x-axis. Calculate the eigenfrequenciesof the system.

The electrostatic energy of the system reads (see Figure)

U =1

4πǫ0(e2

R+

e2

R+ x1 − x2− e

2

R+ x1− e

2

R− x2) (37)

20

R

+ +− −

x2

x1

Expanding equation (37) for x1, x2 ≪ R yields

U =1

4πǫ0(−2e

2x1x2R3

)

so that the total energy of the system reads

E =m

2(ẋ1

2 + ω20x21) +

m

2(ẋ2

2 + ω20x22)−

1

4πǫ0(2e2x1x2R3

) (38)

By symmetry, we introduce the symmetric (xs =x1+x2√

2) and antisymmetric (xas =

x1−x2√2

) modes, and we obtain

their eigenfrequencies as

mω2s = mω20 −

e2

4πǫ0R3; mω2as = mω

20 +

e2

4πǫ0R3(39)

Comment This classical model is also of interest in quantum mechanics (Van der Waals- London interactions).Thefact that the coupling tends to push apart the frequencies of the coupled oscillators is quite a general phenomenon inclassical and quantum physics (where it is called “level repulsion”).

M11. On time of ascent and time of descent

Compare, in the presence of the drag force of the air, the time of ascent and the time of descent of a stone thrownupwards with initial velocity v0

On the way up, the drag force adds to the weight, contrary to the situation on the way down. The acceleration istherefore larger on the way up, hence a smaller time of ascent (Tascent < Tdescent). One may also consider a pointon the way up or down: the gravity potential energy is the same at this point, but some energy is lost in-between tothe drag force. The velocity at this point is therefore less on the way down than on the way up. Since this is true atevery point, it is true on average, hence a longer time of descent.

Compare for the same initial speed v0 the height and time of ascent with and without the drag force of the air.

Let us called Hascent (resp H0ascent) the height reached with the air drag (resp. without the air drag). The

respective time of ascent are Tascent and T0ascent. Let F denote the air drag and P the weight of the stone. The energy

conservation reads

1

2mv20 =

∫ H0ascent0

Pdz =

∫ Hascent

0

(F + P )dz

that is

PH0ascent = PHascent +

∫ Hascent

0

Fdz

yielding Hascent < H0ascent, as expected.

Let us now use the conservation of momentum

mv0 =

∫ T 0ascent0

Pdt =

∫ Tascent

0

(P + F )dt

21

yielding

PT 0ascent = PTascent +

∫ Tascent

0

Fdt

Thus Tascent < T0ascent.

We know that the time of descent Tdescent with air friction is longer than Tascent, but, at this stage, we have noway to compare it to T 0ascent = T

0descent.

M12 A variable mass pendulum

A pendulum consists of a long hollow tube (mass M0, length H) that can be filled with water at a height h, with0 < h < H. The pendulum oscillates around its point of support A (Figure). Qualitatively draw the curve T (x), whereT is the period of the small oscillations and x = hH .

������

h

H

A

The general formula for the small oscillations of a compound pendulum is

T = 2π

√JA

Mtotgd(Gx, A)(40)

where JA is the total (pendulum and water) moment of inertia w.r.t to the point of support A, Mtot = M0 +m(x)the total mass , and d(Gx, A) is the distance between the center of gravity Gx of the total system and point A. Wefirst get

JA = J0 + J(x) =1

3M0H

2 + (1

3m(x)(

h

2)2 +m(x)(H − h

2)2)

where we have applied Huyghens’theorem to calculate J(x) for the water. The distance d(Gx, A) is defined by

(M0 +m(x))d(Gx, A) =M0H

2+m(x)(H − h

2)

Finally, we define the total mass of water m1 = m(x = 1) when the tube is completely filled, and write

m(x) = m1x

with x = hH .It is easily seen that we have T (1) = T (0), and therefore the curve T (x) has some maximum in-between.

22

Comment This behavior is qualitatively the same as the one found in a leaky pendulum.

M13. Elastic bounce of a rough rotating ball

Take a rotating ball bouncing in a perfectly elastic way on the ground. The ball is also prefectly rough, i.e there isno slip at the ball-ground contact point. Write the final state of the ball as a function of initial state.

We denote by J = αmR2 the moment of inertia of the ball (mass m, radius R) w.r.t an axis passing through itscentre of gravity G (α = 25 for a homogeneous sphere). The final and initial physical quantities will be characterizedby subscripts f and i. Newton’s law reads

md~vGdt

= ~F (41)

where ~vG is the the velocity of point G and ~F the total force on the ball, that we write as

~F = ~N +m~g + ~T

where the reaction of the ground has been decomposed into normal ( ~N) and tangential (~T ) components. The angularmomentum theorem w.r.t point G reads

Jd~ω

dt=

−−→GK × ~F = −−→GK × ~T (42)

since the normal force ( ~N +m~g) has zero torque.

~VG

~T

z

G

K

~ω

Restricting equation (41) to the tangential plane, we therefore have

Jd~ω

dt=

−−→GK ×md

~VGdt

(43)

where ~VG is the tangential component of ~vG.Equation (43) can be written just before and just after the collision as

[J~ω −−−→GK ×m~VG]f = [J~ω −−−→GK ×m~VG]i (44)

which expresses the conservation of the angular momentum w.r.t the contact point K in states f and i. Projectionon the axis out of the page yields

(Jω −mRVG)f = (Jω −mRVG)i (45)

So far, we have not considered the “perfectly elastic” character of the collision. The conservation of kinetic energyjust before and just after implies

23

[1

2Jω2 +

1

2m(V 2G + v

2z)]f = [

1

2Jω2 +

1

2m(V 2G + v

2z)]i (46)

where we have introduced rotational and translational components. We have also introduced the vertical componentvz of ~vG. It is clear that the particular case (ω = 0; VG = 0) implies that

(vz)f = −(vz)iWe will admit that this is true in the more general situation, so that equation (46) reads

[1

2Jω2 +

1

2mV 2G]f = [

1

2Jω2 +

1

2mV 2G]i (47)

Equations (45) and (47) then give

(ωR+ VG)f = −(ωR+ VG)iand, using J = αmR2

ωf =α− 1α+ 1

ωi −2

α+ 1

ViR

Vf = −2α

α+ 1ωiR−

α− 1α+ 1

Vi (48)

where we have dropped the subscript G for VG.

Comment 1 Equations (48) can be put in matrix form (f) = Mα(i), where the matrix Mα obeys M2α = I, the

matrix unity. This identity is linked to the time reversal invariance of the collision.

Comment 2 Equation (45) shows that the ball may oscillate between two contact points if Vi = αωiR. In general,equation (48) shows that “angle of incidence=angle of reflection” is wrong for a rotating ball. A nice example isprovided by the case α = 13 , where a ball thrown without spin returns precisely along its initial path after a series ofthree collisions (floor-underside of the table-floor).

Comment 3 This model may be useful to understand the importance of “backspin” for basket-ball players.

M14. Rolling without slipping

The laboratory frame (K) is a Galilean frame of reference. An experimentalist has there a (rather old) record player,rotating at a constant angular speed (Ω) around a vertical axis going through its centre U (Figure). One considers aspherical homogeneous ball (mass m, radius R, moment of inertia J = 25mR

2 with respect to an axis going throughits centre G). At time t = 0, the ball is put on the rotating record player tray. One assumes that the ball rolls withoutslipping (perfectly rough tray). Study the trajectory of the ball in the laboratory frame (K).

��������������������������������

��������������������������������

U

ball

x

rotating tray

view from above

~Ω

24

Let us denote by G the center of gravity of the ball and by I its contact point with the rotating tray. The relevant

forces on the ball are (i) its weight m~g (ii) the tray’s reaction with normal and tangential components ~R = ~T + ~N .Newton’s equations read

md~vGdt

= ~T ; ~N +m~g = ~0

Jd~ω

dt= ~GI × (m~g + ~R) = ~GI × ~T

We now have to express the rolling without slipping condition, namely that the speed of point I considered asbelonging to the ball is the same as the speed of point I considered as belonging to the tray, namely

~vI = ~vG + ~IG× ~ω = ~Ω× ~UI (49)

Taking the derivative of eq(49) we have

d~vGdt

+ ~IG× d~ωdt

= ~Ω× ~vG

since d~UIdt = ~vG. Putting everything together and using J =

25mR

2 leads to

d~vGdt

=2

7~Ω× ~vG (50)

so that the motion of point G is a circle.

Comment 1 In the rotating frame, the trajectory of point G is an epicycloid.

Comment 2 Equation (50) shows that this problem is similar to the problem of a charged particle in a magneticfield. If the rotating tray is not exactly horizontal, there is an extra drift force on the ball similar to an electric field.The trajectory of the ball is then cycloidal.

Comment 3 An interesting variation on the above theme is as follows: if on sends a ball (which rolls withoutslipping ) with initial velocity ~v0, on a rotating tray , show (i) that the velocity of the ball as it leaves the tray is ~v0(ii) the final trajectory of the ball is aligned with its initial rectilinear trajectory (with a circular trajectory in-betweenon the rotating tray).

M15 Larmor’s theorem

Consider the electron of the hydrogen atom describing a circular trajectory. One applies a small magnetic field ~Bperpendicular to the plane of the orbit. Show that one can find a rotating frame where the equation of motion of theelectron is the same as the equation of motion before the introduction of the magnetic field, up to terms of order B2.

In the lab frame the equation of motion before the introduction of the magnetic field ~B reads

m~alab = −e ~E

where ~E = e4πǫ0~rr3 is the electric field created by the hydrogen atom nucleus. In the presence of the magnetic field,

we have

m~alab = −e( ~E + ~vlab × ~B) (51)

In a rotating frame (angular velocity ~ω), the usual velocity and acceleration composition laws imply that

m~arot = −e( ~E + ~vlab × ~B)− 2m~ω × ~vrot −m~ω × (~ω × ~r) (52)

25

where ~vlab = ~vrot + ~ω × ~r. Choosing

~ω =e

2m~B

equation (52) reads

m~arot = −e ~E − e(~ω × ~r)× ~B −m~ω × (~ω × ~r) = −e ~E +O(B2) (53)

which is the sought for result.

Comment 1 Traditionnally, one shows that a consequence of equation (53) is diamagnetism, namely that thevariation of magnetic moment induced by the field is

∆~µ = −e2r2

4m~B

This result can also be shown directly, provided one assumes that the radius of the orbit r does not change at orderB. It will take some time to establish the magnetic field, so that an e.m.f given by

E = −dΦdt

= −d(πr2B)

dt

will create an extra electric field ~E′ such that E =∫~E′ · ~dl.

In the lab frame, the change in angular momentum is

d~L

dt= −e~r × ( ~E + ~E′) = −e~r × ~E′ = +1

2

d(er2 ~B)

dt

so that ~L− er2 ~B2 is a constant vector. Since ~µ = −e~L

2m , we indeed get

∆~µ = −e2r2

4m~B

Comment 2 It is not too surprising to find a relation between angular rotations and magnetic fields since both arepseudo-vectors (and odd under time reversal). A deeper connection exists between inertial forces and electromagneticforces as shown by the following expressions

~Fem = q( ~E + ~v × ~B −∂ ~A

∂t)

~Fin = m(−~ω × (~ω × ~r) + 2~v × ~ω −∂~ω

∂t× ~r) (54)

where we have allowed for time-dependent fields and angular rotations. It is clearly possible to mimic the inertialforces by the appropriate electromagnetic forces.

(i) q ~E ↔ −m~ω × (~ω × ~r)(ii) q ~B ↔ 2mω(iii) q ~A↔ m~ω × ~rNote that the only parameter of the rotation being the vector ~ω, the corresponding electromagnetic fields must be

constrained by ( ~E · ~B = 0 and ~A = 12 ~B × ~r)

Comment 3 The Zeeman effectWe consider, in the lab frame, a harmonic oscillator in the xOy plane, and a magnetic field ~B = B~ez. The equation

of motion reads

~̈r = −ω20~r +q

m~̇r × ~B

In the Larmor rotating frame, we get back an harmonic oscillator of pulsation (±ω0). In the lab frame, we haveeigenpulsations (ω0 ± qB02m ).

26

M16. The Foucault pendulum

Study the small oscillations of a linear pendulum (point mass m attached to a string of length L) at a point oflatitude λ on the rotating Earth.

We consider the rotating Earth and the motion of the pendulum in the tangent planeXY (Figure).

East

North

zenith

Earth

λ

~Ω

m

~gX

Z

Y~T

~Ω

L

λ

A

A Z

Y

Xα

λ

Neglecting terms of order Ω2, the equation of motion reads

m~̈r = m~g + ~T − 2m~Ω× ~̇r

where ~T is the tension of the string. Since ~r is constrained to be in the XY plane, the only active component of ~Ωis the vertical one Ωz = Ωsinλ. Instead of studying the Foucault pendulum, we will study the Bravais variant, wherethe mass m describes a circle of radius R, either clockwise or counterclockwise. The equation of motion now reads

−mω2r = −T sinα± 2m(Ωz)(ωR) (55)

where the + sign corresponds to counterclockwise motion. Since the angle α is small we have T = mgcosα ≃ mg andsinα ≃ α ≃ RL . Equation (55) now reads

ω2 ± 2ωΩz −g

L= 0 (56)

so that ω± ≃√

gL ∓ Ωz. The superposition of two such circular motions, of amplitude R is a sinusoidal motion of

amplitude 2R, with angular frequency√

gL , whose direction of oscillation changes clockwise with angular frquency

Ωz = Ωsinλ. This is the traditional Foucault pendulum.

Comment An optical version of the Foucault pendulum has been experimentally tested by Michelson et al. (1925).The first part of the experiment is to divide a beam into two to beams of light (one clockwise and one counterclockwise)around a very small circuit, before the beams recombine to give an interference pattern. The same experiment isdone now with the two beams encircling a very large area (of the order of a square kilometer). One measures ashift of the interference pattern compared to the small circuit experiment. To evaluate this shift, we use a classicalpicture. We identify the two beams with the two circular motions of the Bravais-Foucault pendulum with speedv∓ = v ±R(ω− − ω+) and get a shift

δl = tδv = t(ω− − ω+)R = (2πR

v)2ΩzR =

4ΩzA

c0

where A = πR2 is the surface of the large circuit and we have boldly set v = c0 (speed of light).

27

M17 Motion of a charged particle close to a current carrying wire.

Study the motion of a charged particle in the magnetic field of a long straight wire

The wire is chosen as the z axis of the system (Figure), and cylindrical coordinates are used. The particle charge

is denoted by q. The magnetic field at point M is ~B(M) = µ0I2πr~eθ, so that the equations of motion are

r̈ − rθ̇2 = −α żr

z̈ = α ṙr

rθ̈ + 2ṙθ̇ = 1rd(r2θ̇)dt = 0 (57)

where α = µ0Iq2πm .

θ

M

I

z

~eθ

~ez

~er

r

Integrating the second and third equations above, with initial conditions (r0, ż0, ...) gives

r2θ̇ = r20 θ̇0 = β; ż = ż0 + αlnr

r0

so that the first equation may be cast in the form

r̈ = −dVeffdr

(58)

where

Veff (r) =β

2r2+α2

2ln2

r

r0+ αż0ln

r

r0

is an effective radial potential for the charged particle. For representative initial conditions, , it is clear that thepotential Veff (r) has a minimum for finite r, hence the possibility of a stable trajectory of the particle around thewire.

Comment For atoms with a permanent magnetic moment ~µ, the energy reads

E =m

2~v2 − ~µ · ~B = m

2(ṙ2 + (rθ̇)2 + ż2)− µθ

µ0I

2πr

If ~µ is aligned with ~B, one has again a stable trajectory around the wire,with an effective radial potential

Veff (r) =β

r2− γr

where γ > 0. If ~µ is antialigned with ~B, the effective potential is repulsive (γ < 0). This device is used to filter (andguide) atoms of a given orientation along a wire.

28

Central forces

C1 Some central force motions

(1-a) One considers an infinite straight wire, carrying a uniform linear charge density λ. In the presence of the

electric field ~E created by the wire, a particle of mass m acquires an electric dipole moment ~p = α~E, where α is apositive dimensionful constant. Study the motion of the particle as a function of the initial conditions.

����

(λ)

~p = α~E

The (radial) electric field created by the wire can be calculated with Gauss’ theorem and reads

~E(~r) =λ

2πǫ0r~er (59)

where r is the distance to the wire.The potential energy of the particle reads

U(~r) = U(r) = −12~p · ~E = −A

r2

where A = α2 (λ

2πǫ0)2

The motion of the particle is two-dimensional (i.e. in the plane orthogonal to the wire); motion along the wire isunaffected (no force). We have therefore to study the central force motion described by the Hamiltonian

E = 12m(~̇r)2 + U(r) =

1

2m(ṙ)2 +

~L2

2mr2− Ar2

=1

2m(ṙ)2 + Ueff (r) (60)

with ~L = m~r × ~v being a constant of the motion.Equation (60) shows that, depending on the initial conditions, there is a critical value L2c = 2mA generating different

trajectories

• L > Lc: the effective potential Ueff (r) is positive (repulsive). The minimum distance approach (ṙ = 0) is givenby r20 =

L2−L2c2mE . The particle escapes to infinity.

• L = Lc: the effective potential Ueff (r) vanishes. The particle reaches the wire with a non-zero velocity

• L < Lc: the effective potential Ueff (r) is attractive. The particle again reaches the wire with a non-zero velocity.

(1-b) A point particle (mass m) is sent from infinity (with velocity ~v∞ and impact parameter b) towards a target-

particle of mass M (Figure). The interaction energy between the particule and the target is Ep(r) = −A b2

r2 where ris the distance between the particle and the target and A a positive dimensionful constant. Explain how this problemcan be reduced to a one-body problem and study the various possibles trajectories.

���

���

���

���

M

m b

−→v∞

29

The reduction from a two-body problem to a one-body problem works as follows. Consider the equations of motionfor particles 1 and 2

m1−→̈r1 = ~F2→1 (61)

m2−→̈r2 = ~F1→2 = −~F2→1

Adding and substrating these equations gives respectively the motion of the center of mass (CM) and the relativemotion (R). We get

m1−→̈r1 +m2

−→̈r2 = 0 = (m1 +m2)

−−−→¨RCM

and

µ−−−−−−→

¨(r1 − r2) = µ−→̈rR = ~F2→1 = ~F (~rR)

with the reduced mass µ = m1m2m1+m2 .

The (CM) has therefore a constant velocity (see the initial conditions), and may be taken as the origine of a newgalilean frame.

In the problem at hand, we are dealing with a central force ~F (~r) = −−→∇Ep(r), and the relative motion canbe described as the usual one-body central force problem. Introducing the radial coordinate r, the energy of the(relative) problem reads

E =1

2µ(−→̇r )2 + Ep(r) =

1

2µ(ṙ)2 +

~L2

2µr2− Ab

2

r2=

1

2µ(ṙ)2 + Ueff (r)

with the angular momentum with respect to the (CM) ~L = µ~r × ~v being a constant of the motion (|~L| = µr2θ̇) andUeff (r) =

~L2

2µr2 − Ab2

r2

We are therfore back to the previous problem: depending on the initial conditions, there is a critical value L2c =2µAb2 generating different trajectories

• L > Lc: the effective potential Ueff (r) is positive (repulsive). The minimum distance approach (ṙ = 0) is givenby r20 =

L2−L2c2µE . The particle escapes to infinity.

• L = Lc: the effective potential Ueff (r) vanishes. The particle reaches the wire with a non-zero velocity

• L < Lc: the effective potential Ueff (r) is attractive. The particle again reaches the wire with a non-zero velocity.

Comment 1 It is important to note that the angular momentum with respect to the (CM) is the same in thelaboratory frame and in the (CM) frame.

Comment 2 One may also study the two-body problem in a rotating frame. If a common origin on the rotationaxis is chosen for both the inertial laboratory and rotating frames, one can show that the decomposition in a (CM)and relative motions still works.

Comment 3 For charged particles, this decomposition works at zero magnetic field. If the magnetic field isnon-zero, it works only if q1m1 =

q2m2

.

Comment 4 One considers an isolated sytem of two charged particles with charges (q1, q2) and masses (m1,m2).

Take the origin of coordinates at (CM) of the system (i.e. m1~r1 +m2~r2 = ~0). The total angular momentum reads

~L = ~r1 × ~p1 + ~r2 × ~p2 = (~r1 − ~r2)× ~p1

since ~p1 + ~p2 = ~0. The magnetic moment of the system reads

~m =1

2(q1~r1 × ~v1 + q2~r2 × ~v2)

30

It is easily checked that

~m =1

2

( q1m21

+q2m22

) m1m2m1 +m2

~L

(1-c) One considers a point particle (mass M) at the origin O of a Galilean frame (K). Another point particlewith mass m (m ≪ M) is sent from infinity with vector velocity ~v∞ and impact parameter b (Figure). Gravitation(constant G) is the interaction energy of the problem and mass M is fixed. If (~r,~v, ~L) respectively represent theposition, velocity and angular momentum of particle m with respect to point O, show that the Runge-Lenz vector

~A = ~v × ~L−GMm~rr

is an invariant vector of the motion. Give the expression of the angle D -called the deviation- between final and initialdirections of particle m.

One reminds the reader of the identity

~A× ( ~B × ~C) = ~B( ~A · ~C)− ~C( ~A · ~B)

������

������

��������������������������������������

Db

M

m,−→v∞

Since mass M is fixed, the equation of motion of mass m is given by

m~̈r = ~F = −GMm ~rr3

(62)

implying among other that the angular momentum ~L = ~r×m~v of mass m w.r.t point O is constant. The Runge-Lenzvector reads

~A = ~v × ~L−GMm~rr

We therefore have

d ~A

dt=d~v

dt× ~L+ ~v × d

~L

dt−GMm d

dt(~r

r) (63)

Using eq(62) and d~Ldt =

~0 , we have

d ~A

dt=

~F

m× ~L−GMm(~v

r− v~rr2

)

Using d~r2

dt = ~r · ~v = dr2

dt = rv, it is easily shown that

~r × (~r × ~v) = ~r(rv)− r2~v

yielding

d ~A

dt= ~0

The initial and final Runge-Lenz vectors thus satisfy

31

~Ai = v∞~i× ~L−GMm(−~i) = ~Af = vf × ~L−GMm~uf (64)

where we have used ~vi = v∞~i. Taking an orthonormal basis (~i,~j,~k =~i×~j), we have

~vf = v∞~uf = v∞(~icosD −~jsinD)

and ~L = −mv∞b~k. Plugging these values in eq.(64) we obtain

tgD

2=GM

bv2∞

independently of mass m. For small deviations (i.e. for large b), we have D(b) ∼ 2GMbv2∞ . A hand-waving argument forthis result is as follows. For large b, the gravitational force is weak, and if we consider that this force applies only

around the minimal distance approach r0 ∼ b for a small time δt ∼ bv∞ , we have D ∼δpypx

∼ F (b)δtmv∞ ∼GMbv2∞

.

(1-d) One considers a spherical homogeneous star (radius R, mass M). The center O of the star is at the originof a Galilean frame (K). A point particle with mass m (m ≪ M) is sent from infinity with vector velocity ~v∞ andimpact parameter b ≪ R (Figure). Gravitation (constant G) is the interaction energy of the problem and mass M isfixed (M ≫ m). All collisions of particle m with the particles of the star are neglected. Write down the equation ofmotion of the particle m inside the star (one notices that a time scale T0 shows up in this equation). If the particlespends a time τ ≪ T0 in the star, show that the star deviates its trajectory by an angle D(b) ∼ bf , where f is a lengththe physical meaning of which will be explained. What can you say of D(b) for b≫ R ? Draw qualitatively the curveD(b) for all b

�������

�������

m,−→v∞

b

M

R

O D

The equation of motion inside the star reads

m~̈r = m~g(~r) = −mg(r)~er

where, according to Gauss’ theorem

g · 4πr2 = 4π3ρr3

with M = 4π3 ρR3 We therefore get an harmonic oscillator equation

~̈r = −ω20~r

with ω0 =2πT0

= GMR3 .

If the particle spends a time τ ≪ T0 in a star, this means that the gravity force mg(r) will be exerted for a shorttime τ . This means that the initial particle trajectory will be very little deflected. A rough calculation is as follows.The particle moves along the x axis with speed v∞ and enters the star with an impact parameter b small. Thedeviation D(b) can be calculated as

D(b) ∼ δpypx

∼ mg(b)τmv∞

with τ ∼ Rv∞ and g(b) ∼GMR3 b.

32

One finally gets

D(b) ∼ bf, f ∼ R

2v2∞GM

Since the deviation is proportional to b, this means that particles with different impact parameters will be focussedat the same point at distance f from the star, which can be called the focal length of the system. This calculationmakes sense iff f ≫ R. For large impact parameter b, b ≫ R, one expects that the deviation D(b) decreases withb. This means that the full curve D(b) is a non-monotonic function of b, with a maximum around b ∼ R. Thisnon-monotonicity has important implications for gravitational lenses.

(1-e) Analogies

A) A parallel light beam is sent on a spherical glass lens (index n > 1, radius R). Calculate, in the framework ofgeometrical optics, the focal length of the lens (Figure).

B) Sound waves are sent in the same conditions (Figure) on a balloon filled with a perfect gas (G). Discuss thetwo cases (G)=CO2, He.

C) A beam of particles with the same mass m and velocity v∞ is sent in the same conditions (Figure) on a sphericalhomogeneous star (mass M , radius R). The beam radius is b, with b

33

Now for perfect gases, the sound velocity is proportional to M−1/2 where M is the molar mass (dimensionally, oneexpects Mc2 ∼ RT ). For Helium , one has cHe > cair, so that the helium balloon is expected to be a divergingacoustic lens. For CO2, one has cCO2 < cair, resulting in an convergent acoustic lens of focal length

nR2(n−1) with

n = caircCO2> 1.

C)

The equation of motion inside the star reads

m~̈r = m~g(~r) = −mg(r)~er

where, according to Gauss’ theorem

g · 4πr2 = 4π3ρr3

with M = 4π3 ρR3 We therefore get an harmonic oscillator equation

~̈r = −ω20~r

with ω0 =2πT0

= GMR3 .

If the particle spends a time τ ≪ T0 in a star, this means that the gravity force mg(r) will be exerted for a shorttime τ . This means that the initial particle trajectory will be very little deflected. A rough calculation is as follows.The particle moves along the x axis with speed v∞ and enters the star with an impact parameter b small. Thedeviation D(b) can be calculated as

D(b) ∼ δpypx

∼ mg(b)τmv∞

with τ ∼ Rv∞ and g(b) ∼GMR3 b.

One finally gets

D(b) ∼ bf, f ∼ R

2v2∞GM

Since the deviation is proportional to b, this means that particles with different impact parameters will be focussedat the same point at distance f from the star, which can be called the focal length of the system. This calculationmakes sense iff f ≫ R.

(1-f) Derive Newton’s gravitation law between two homogeneous bodies A and B residing in a material medium Fof mass density ρF

The simplest (and non-rigorous) idea is to invoke the linearity of Poisson’equation and to write (see Figure)

= +A B

F

F

A−F B−F

0

where A (resp B,F ) stands for the first body with density ρA (resp ρB , ρF ) and A − F (resp B − F ) for anhypothetical body of density ρA − ρF (resp ρB − ρF ). We therefore have on the left the situation we are trying to

34

solve and on the far right two bodies of volume VA (resp VB) and density ρA − ρF (resp ρB − ρF ) interacting in thevacuum.Newton’s law in the medium F reads

~F = −GVA(ρA − ρF )VB(ρB − ρF )~err2

= −GmAmB(1−ρFρA

)(1− ρFρB

)~err2

Note the analogy with Archimedes principle, which can be put on a more rigorous general relativity-like basis.

(1-g) Describe what is meant by a geostationary satellite. Extends the discussion to a geostationary rope, of uniformmass per unit length µ, with one free end barely touching the earth.

Equilibrium with respect to the rotating earth first requires that the satellite orbit is in the equatorial plane, wheregravity and centrifugal forces are collinear. One may then writes

0 = m(ω2rs −GMer2s

)

yielding a circular orbit approximately 36000 kms above the earth equator (and following of course the sense of theearth rotation)For a rope to be geostationary, it must also lie in the equatorial plane and follow the earth rotation. The equilibrium

condition for a portion between r and r + dr reads

dT = µdr(GMer2

− ω2r)

where T is the tension at position r. With free ends, one has T (re) = T (re + L) = 0, where re is the radius of theearth and L the length of the rope. Integrating the previous equation with the boundary conditions leads to

T (r) = µGMe(1

re− 1r)− µω

2

2(r2 − r2e)

and L = re2 ((1 +8GMeω2r3e

)1/2 − 3), of the order of 144000kms. Note that the tension is maximal at r = rs.

C2 Lagrange points

One considers the isolated Earth-Sun system (respective masses M2,M1), and assume a circular motion of the E-Ssystem. Discuss the equilibrium positions of a small mass m w.r.t the E-S system on the E-S axis. Show that thereare two more equilibrium positions in the plane of the E-S motion, such that the triangle m-E-S is equilateral

Let G be the center of mass (CM) of the isolated E-S system. The relative motion equation reads

µ~̈r = −GM1M2~r

r3

with the reduced mass µ = M1M2M1+M2 . The angular rotation around G is thus ω2 = G(M1+M2)d3 , if d denotes the E-S

distance. In this rotating frame, the force on a small mass m at point P read

m−̈−→GP = ~FE + ~FS − 2m~ω × ~vP −m~ω × (~ω × ~GP ) (66)

where we have included the inertial forces as well as the gravitational forces ~FE , ~FS . Equilibrium then implies

~0 = −GM2−−→EP

EP 3−GM1

−→SP

SP 3− ω2−−→GP (67)

35

On the E-S axis, we choose the origin at the (CM), xG = 0. Projecting equation (67) successively for x < xS ,xS < x < xE and x > xE , we obtain the Lagrange points L3, L1, L2. The calculations become simpler if M1 >> M2,namely

L3 : x3 ≃ −d(1 +5M212M1

); L1 : x1 ≃ d(1− (M23M1

)1/3); L2 : x2 ≃ d(1 + (M23M1

)1/3)

Off the E-S axis, equation (67) together with the value ω2 = G(M1+M2)d3 implies that there are two more pointsL4, L5, symmetrical w.r.t the S-E axis such that the triangle PES is equilateral

L4 : (d

2(M1 −M2M1 +M2

), d

√3

2); L5 : (

d

2(M1 −M2M1 +M2

),−d√3

2)

The stability analysis around the Lagrange points is tricky, due to the presence of the Coriolis force in equation(66),which does not derive from a potential. For small displacement, it is found that the points L1, L2, L3 are unstable,whereas points L4, L5 may be stable if

M1M2

> 24.9599..... Note that “unstable” does not imply that the trajectoryof mass m is unbounded. Some “horseshoe” orbits are known, where mass m wanders periodically around pointsL3, L4, L5.

Comment The Lagrange points L1, L2 are used for satellite purposes (e.g. the solar observatory SOHO at L1and the Planck Microwave Probe at L2). Point L3 is a favorite of science fiction writing (Planet X). Note that TheSun-Jupiter system has large asteroids around its L4, L5 points.

L3

L5

L4

L1 L2

y

x

S E

G

C3 Estimating the Roche limit

Consider a spherical homogeneous body (C) orbiting around the earth (E). Write down the equilibrium conditionfor a small part of (C) located at a finite distance x from the center of (C) along the (CE) axis.

(C)

(E)

d x

36

Let d being the distance between the center of the two bodies. The distance between the earth and the test massis d− x. We write the equilibrium condition in the non-inertial frame rotating with (C)

0 = −ω2(d− x) + GM(d− x)2 −

4π

3Gρx (68)

where the third force represents, with the help of Gauss’ theorem, the gravitational attraction of (C) on the massunity. In equation (68), the rotation speed ω is given by mω2d = GMmd2 . Expending to first order in x/d yields

d = r(3M

m)

13 (69)

where r is the radius of (C).

Comment 1 Equation (69) can alternatively be written as d = RE(3ρEρC

)13 , where the ρ’s are the densities of the

two bodies, and RE the radius of the earth. If the distance is smaller than the Roche limit, the earth attractionbecomes too large for the self-gravitating term and the orbiting body (C) breaks up. This mecanism may play a rolein the formation of rings around the outer planets.

Comment 2 The full calculation is rather complicated and we just quote the final answer d = 2.44 r(Mm )13 (as

compared to a factor 313 ∼ 1.44 in equation (69)). Note that we assumed that gravitation is the cohesive force holding

together the orbiting body (C): this is only true for a large enough body (the cohesion of smaller bodies is mostlyelectrostatic in nature).

C4. The dumbbell satellite

Consider a dumbbell-shaped satellite of two point masses m connected by a massless rod of length 2d, where the rodlies in the plane of the orbit. The orientation of the satellite w.r.t to the direction of the earth is given by an angle θ(Figure). Discuss the small oscillations of the rod around its stable equilibrium position.

earth

m

m

θ

The potential energy of the satellite is

Ep = −GMm(1

R++

1

R−)

where R2± = R2 + d2 ± 2Rdcosθ (Figure). Expanding to second order in dR yields

Ep ≃ −2GMm

R[1 +

(3cos2θ − 1)2

(d

R)2]

The equilibrium positions are (θ = 0, stable) and (θ = π2 , unstable). The kinetic energy term reads Ec = 212mx

2θ̇2 =

mx2θ̇2. The small amplitude oscillations around θ = 0 have a period T = T0√3, where T0 =

2π√GM

R3/2 is the period of

revolution of the satellite around the earth.

C5 More on the gravidyne

37

The gravidyne is a dumbbell-shaped satellite, which may change its orbit through periodic shape changes (Beletsky).We follow here a pedagogical path to Beletsky’s ideas.

(5-a) Motion of a point particle in a gravitation field

We will restrict ourselves to bound states (E < 0). The force on a point mass m reads

~F = −−→∇U = −GMm ~rr3

where U(r) = −GMmr is the gravity potential energy. Well-known facts about central force motions include (i)the conservation of the angular momentum ~L = ~r × m~v (ii) the conservation of the energy E = 12m

−→v 2 + U(r) =12mṙ

2+Ueff (r), with Ueff (r) =L2

2mr2 +U(r) (Figure (a)). For the specific case of gravitation, the Runge-Lenz vector~A = ~v × ~L−GMm~rr is also conserved.

The precise derivation of the equation reads as follows. Conservation of ~L implies θ̇ = Lmr2 . Together with

ṙ =√

2m (E − Ueff (r)), this explicitly yields

dθ

dr=

L

mr21√

( 2Em − 2m ( L2

2mr2 − GMmr ))(70)

Setting u = 1r allows one to integrate equation (70) as

r =p

1 + ecosθ(71)

with

p =L2

GMm2; e =

√1− 2|E|L

2

m(GMm)2

where we have taken E < 0. Equation (71) shows that the point mass m describes an ellipse with origin at a focalpoint (where mass M sits), semi-latus rectum (or parameter) p and eccentricity e < 1 (Figure (b)).

Ueff (r)

r

E < 0

rmin rmax

rminrmax

m

M

r θp

2a

2b

(a) (b)

The ellipse is further characterized by (rmin, rmax) = (p

1+e ,p

1−e ), so that its major a and minor b axis, related by

p = b2

a , are given by

a =p

1− e2 ; b =p√

1− e2=

L√2m|E|

38

These results imply that the total energy does not depend on the minor axis b

E = −GMm2a

Other relations of interest include

L = mωab; ω2a3 = GM

where T = 2πω is the period of the motion.

(5-b) Motion of a dumbbell satellite (dumbbell perpendicular to the plane of the orbit)

We now consider that the satellite has the shape of dumbbell, with two point masses linked by a massless rigidrod of length 2d. The rod is constrained to be perpendicular to the plane of the orbit, i.e. parallel to direction Mz(Figure).

The position of the masses read ~r± = r~er ± d~ez. Introducing the tensions ~T± in the rod, the equilibrium conditionsin the rotating frame read

~0 = −m~ω × (~ω × ~r±) + ~T± −GMm~r±r3±

where r+ = r− =√r2 + d2 and ~T+ + ~T− = ~0 (equilibrium of the rod). Adding the two equations yields ω2 =

GMr3 , as

for a point mass. Substracting gives ~T = T~ez, with T = GMmd

(r2+d2)3/2≃ mω2d for r >> d.

It is easily shown that the angular momentum ~L with respect to point M is conserved, since ~L = ~r+×m~v++~r−×m~v−implies

d~L

dt= ~r+ × ~F+ + ~r− × ~F− = ~r × (~T+ + ~T−) = ~0

The dumbbell satellite with the rod constrained to be perpendicular to the plane of the orbit behaves almost asa point mass 2m located at its center of mass. However, comparing the strength of the gravitational forces givesf = GM 2mr2 for the point mass and f̃ = GM

2mr(r2+d2)3/2

for the dumbbell satellite.The force f̃ is slightly smaller than

f by an amount of order (dr )2. It is precisely this small difference which lies at the heart of the gravidyne.

α

m

m

2dMr

z

(5-c) The gravidyne

We give below a heuristic treatment of Beltsky’s ideas. We have seen above that the angular momentum L was aconserved quantity for the dumbbell satellite in the perpendicular geometry. Let us write down the conservation ofthe energy as

E =1

2mv2 − GMm√

r2 + d2=

1

2mv2 − GMm

r+GMm

r− GMm√

r2 + d2= −GMm

2p(1− e2) + GMm

r− GMm√

r2 + d2(72)

39

where we have used previous results. Equation (72) shows that for non-zero d, the eccentricity e depend on r since itcan be rewritten as

e2 +2p

r[1− r√

r2 + d2] = Cst (73)

In equation (73), the parameter p = L2

GMm2 is constant since L is constant. See the Figure. At the perigee (r = rmin),the eccentricty is minimal, and the contrary at the apogee (r = rmax).

e2

r/p

1

1

3

5

7

2

4

68

If the gravidyne keeps its rod-like shape, the orbit is periodic (see trajectory 1-2-1-2...on the Figure). If at position2 (the apogee), some appropriate motor can quickly brings the two masses together, the point-like satellite will followtrajectory 2-3 (constant excentricity for a point mass).Remember that the gravity force is larger for a point mass thanfor the extended shape: the perigee corresponding to point 3 is therefore smaller than the initial perigee 1. Now atperigee 3, the same motor quickly expands the point mass satellite to its original rod configuration. The eccentricitywill then increase with r and a new apogee 4 is reached..... The gravitational energy will increase so that the motionbecomes unbound (e → 1). The variation of the eccentricity of the orbit is of order ( da )2 at each cycle, suggestingthat a number of cycles of order (ad )

2 is necessary to get an unbound orbit. For a = 7000 kms. and d = 100 kms, thegravidyne will escape to infinity in a year or so.

Comment 1 One can calculate the work done by the motor during one extension-contraction cycle.

Comment 2 The same idea may be applied, mutatis mutandis, to a rod-like satellite, where the rod lies in theplane of the orbit. In this case, the gravity force f̃ is larger for the extended shape than the gravity force f in thecontracted shape: the extension must therefore take place at the apogee and the contraction at the perigee.

C6 On the relation between the Kepler and harmonic oscillator problems.

(6-a-1) The Laplace-Runge-Lenz vector and the symmetry of the Kepler problem

For the Kepler problem, the Laplace-Runge-Lenz (LRL) vector

~A = ~v × ~L−GMm~rr

(74)

is invariant along the particle’s trajectory since

d ~A

dt=d~v

dt× ~L+ ~v × d

~L

dt−GMm d

dt(~r

r) (75)

Using m~̈r = ~F = −GMm ~rr3 and d~Ldt =

~0 , we have

d ~A

dt=

~F

m× ~L−GMm(~v

r− v~rr2

)

40

Using d~r2

dt = ~r · ~v = dr2

dt = rv, and ~r × (~r × ~v) = ~r(rv)− r2~v yields

d ~A

dt= ~0

One may therefore evaluate it at the perigee of the orbit (~rmin = rmin~er; rmin =p

1+e ). In polar coordinates

(~er, ~eθ), we get

Ar = ~A · ~er = rmin(L2

mr2min− GMm

rmin) (76)

and

Aθ = ~A · ~eθ = −(mr2θ̇ṙ)rmin = 0

since ṙ = 0 at the perigee. Equation (76) can be rewritten as

Ar = ~A · ~er = rmin(E +L2

2mr2min) = (GMm)e (77)

where we have used E = −GMm2a and a =p

1−e2 .

So, at any point of the orbit, ~A is parallel to the major axis, with length | ~A| = (GMm)e.

Comment 1 At this point, symmetry considerations are relevant. First of all, the (LRL) vector is even w.r.t. timeinversion. As for spatial symmetry , the Kepler problem has an axis of symmetry along the apogee- force center-perigee direction. It is therefore not totally unexpected to find the invariant Laplace vector, which is a true (or polar)vector along this direction.

Comment 2 From the invariance of ~L and ~A, one may build an extra invariant of the motion, called the Hamilton

vector ~H such that ~A = ~H × ~L (or ~H = ~L× ~AL2 ), directed along the minor axis b of the ellipse (the Hamilton vector isodd under time inversion). It is easily shown that the eccentricity of the bounded Kepler motion can be reexpressedas e = HLGMm .

Comment 3 Since the Kepler problem is analogous tho the motion of an electron around the nucleus, it may be

of interest to study the effect of a constant electric field ~E on the Kepler results. For the Hamiltonian

H = m2~v2 − k

r+ e~r · ~E

the angular momentum ~L and Hamilton vector ~H are no longer invariant vectors. They are “replaced” by the followinginvariant scalar quantities

Γ = ~L · ~E; ∆ = ~A · ~E + e2(~r × ~E)2

(6-a-2) The Laplace-Runge-Lenz tensor and the symmetry of the harmonic oscillator

The two-dimensional harmonic oscillator is described by an applied force ~F on a point mass m, such that

~F = −−→∇U ; U = mω2

2(~r)2 =

mω2

2(x2 + y2)

with force center at point O, (0, 0). With appropriate initial conditions, the trajectory is easily shown to be an ellipseof center the force center. This implies that there are now two perpendicular axis of symmetry (see Figure), and thisis a major difference with the Kepler problem. In particular, we do not expect to find a single invariant vector such

as the (LRL) vector ~A.

41

x

y

~r

OA’A

B’

B

θ

OA = a; OB = b

In fact, one may try to find two perpendicular (LRL) vectors ~A and ~B respectively parallel to the x and y axis,with the same characteristic as in the Kepler problem (polar and even under time inversion). These vectors can then

be considered as the eigenvectors of a symmetric (polar, even) tensor (or matrix) ¯̄A. More detailed calculations showthat

¯̄A =m

2v̄v̄ +

mω

2r̄r̄ (78)

Due to the higher symmetry of the motion, the eigenvectors ~A and ~B of the (LRL) tensor are not invariant along

the whole trajectory: ~A for instance, is invariant only for x > 0, and reverses its direction as x becomes negative (at

point B of the figure). The angular momentum ~L = ~r ×m~v is conserved and is equal to ±mω ~A× ~B.

Comment 1 One may also define a couple of Hamilton vectors ~HA and ~HB respectively parallel to ~B and ~A.

Comment 2 In polar coordinates, the equation of the trajectory reads

1

r2=

1

2(1

a2+

1

b2) +

1

2(1

a2− 1b2)cos2θ

where a and b are the semi- major and minor axis.

(6-b) Connexion between the two problems

Let us write equation (70) for the Kepler motion.

dθKeplerdr

=L

mr21√

(2EKepler

m − 2m ( L2

2mr2 − GMmr ))(79)

and change the variable to u =√r. We have

dθKeplerdu

= 2L

mu21√

(2GM − 2m ( L2

2mu2 − EKepleru2))(80)

Let us now write the equivalent of equation (70) for the harmonic oscillator

dθoscdr

=L

mr21√

( 2Eoscm − 2m ( L2

2mr2 +mω2r2

2 ))(81)

Since EKepler < 0 for a bound state, we have a complete equivalence between the two motions. Because of thefactor 2 in in the transformed Kepler motion (80), two Kepler orbits are performed for every