1

LECTURE # 28RELATIVITY V

MASS ENERGY EQUIVALENCEEXAMPLES

PHYS 270-SPRING 2010

Dennis Papadopoulos

APRIL 23, 2010

Relativistic EnergyRelativistic EnergyThe total energy E of a particle is

This total energy consists of a rest energy

and a relativistic expression for the kinetic energy

This expression for the kinetic energy is very nearly ½mu2 when u << c.

€

K = dW =0

u

∫ Fdx0

u

∫ =dp

dtdx

0

u

∫

If → dp /dt = mdv /dt

K =1

2mv 2

Otherwise

K = mc 2(γ −1)€

Eo=mc2 Invariant

Energy conservation requires that M=2m+2K/c2

Mass is not conserved

K=2mec2

Law of conservation of total energy

E E i

i

(p

i

)i mic2

e-(fast)+e-(at rest)e-+e-+e-+e+

M=2m+2K/c2

M

Pair creation

Conservation of EnergyConservation of EnergyThe creation and annihilation of particles with mass, processes strictly forbidden in Newtonian mechanics, are vivid proof that neither mass nor the Newtonian definition of energy is conserved. Even so, the total energy—the kinetic energy and the energy equivalent of mass—remains a conserved quantity.

Mass and energy are not the same thing, but they are equivalent in the sense that mass can be transformed into energy and energy can be transformed into mass as long as the total energy is conserved.

• Energy : The measure of a system’s capacity to do work

• Units of Energy: Joule = Nt x m, eV= 1.6 x 10-19 J, Cal = 4.2 x 103 J

• Examples : It takes 100 Joules to lift 10 kg by 1 meter against the Earth’s gravity (g= 10 m/sec2); It takes .4 MJ to accelerate a 1000 kg car to 30 m/sec (105 km/hr); It takes about 1010J to accelerate a missile to 5 km/sec. [E=1/2 M(kg) v2 (m/sec) J]

• Chemical Energy Storage : Chemical energy is stored in the chemical bonds of molecules. As an order of magnititude a few eV per bond. A 1 kg steak store approximately 1000 Cal or approximately 4 MJ. This is the amount (4-10 MJ/kg)stored in one kg of chemical explosives (TNT). Also a typical battery has few MJ of stored energy.

• Energy Transformations: Energy has many forms, e.g. potential, kinetic, chemical, acoustic, radiation, light etc. Each can be transformed to the other, but overall energy is conserved.

• Explosion : In ordinary, e.g. TNT, explosion chemical energy istransformed in kinetic energy of the fragments, acoustic energy ofthe snow-plowed air (shock or blast wave) and some radiation fromthe heated air. In an explosion we deliver the energy fast.

Power: P= Energy/time. Units are Watt = J/sec. Hair dryer 1 kW, Light bulb 100 W.

Power worldwide 1.3 Twatts, In a year multiply by 2x107 secs to get 3x1019 J or 104 MT (MT=4x1015 J)

CHEMICAL BINDING ENERGY

TWO OXYGEN ATOMS ATTRACT EACH OTHER TO FORM O2 WHILE RELEASING 5 eV OF ENERGY. THEREFORE 2 OXYGEN ATOMS ARE HEAVIER THAN AN OXYGEN MOLECULE BY

m= 5 eV/c2 =9x10-36kg

MASS OF OXYGEN MOLECULE IS 5x10-26kg.m/m=2x10-10

FORM 1 GRAM OF O2 AND GET 2x104 JOULES

All matter is anAssembly of Atoms

Atomic number vs. Mass number

Radioactivity alpha decayRa(226,88)->Rn(222,86)+He(4,2)U(238,92)->Th(234,90)+He(4,2)

Beta decayC(14,6)->N(14,7)+e-+

p n

E=mc2 1 kg has the potential to generate 9x1016 Joules Could provide electric power to city of 800000 for 3 years

Efficiency= mc2

Chemical reactions (0il, coal, etc)

Nuclear power – Fission

Nuclear power – Fusion

1010 tons/year

300 tons/year

300 kg/year

30 tons/year

EXAMPLES OF CONVERTING MASS TO ENERGY

• Nuclear fission (e.g., of Uranium)– Nuclear Fission – the splitting up of atomic nuclei– E.g., Uranium-235 nuclei split into fragments when

smashed by a moving neutron. One possible nuclear reaction is

– Mass of fragments slightly less than mass of initial nucleus + neutron

– That mass has been converted into energy (gamma-rays and kinetic energy of fragments)

BaKrnnU 14489235 31

FISSION

One case of the fission of 236U. The net mass of the initial neutron plus the 235U nucleus is 219,883 MeV/c2. The net mass of the fission products (two neutrons, a 95Mo nucleus and a 139La nucleus) is 219,675 MeV/c2 - smaller because of the stronger binding of the Mo and La nuclei. The "missing mass'' of 208 MeV/c2 goes into the kinetic energy of the fragments (mainly the neutrons), which of course adds up to 208 MeV.

Fusion

• Nuclear fusion (e.g. hydrogen)– Fusion – the sticking together of atomic nuclei– Much more important for Astrophysics than fission

• e.g. power source for stars such as the Sun.• Explosive mechanism for particular kind of supernova

– Important example – hydrogen fusion.• Ram together 4 hydrogen nuclei to form helium nucleus• Spits out couple of “positrons” and “neutrinos” in process

22 4 41 eHeH

Fusion

– Mass of final helium nucleus plus positrons and neutrinos is less than original 4 hydrogen nuclei

– Mass has been converted into energy (gamma-rays and kinetic energy of final particles)

• This (and other very similar) nuclear reaction is the energy source for…– Hydrogen Bombs (about 1kg of mass converted into

energy gives 20 Megaton bomb)– The Sun (about 4109 kg converted into energy per

second)

Annihilation

• Anti-matter– For every kind of particle, there is an antiparticle…

• Electron anti-electron (also called positron)• Proton anti-proton• Neutron anti-neutron

– Anti-particles have opposite properties than the corresponding particles (e.g., opposite charge)… but exactly same mass.

– When a particle and its antiparticle meet, they can completely annihilate each other… all of their mass is turned into energy (gamma-rays)!

EXAMPLES OF CONVERTING ENERGY TO MASS

• Particle/anti-particle production– Opposite process to that just discussed!– Energy (e.g., gamma-rays) can produce particle/anti-

particle pairs

– Very fundamental process in Nature… shall see later that this process, operating in early universe, is responsible for all of the mass that we see today!

PAIR PRODUCTION

Electron-positron PAIR PRODUCTION by gamma rays (above) and by electrons (below). The positron (e+) is the ANTIPARTICLE of the electron (e-). The gamma ray ( ) must have an energy of at least 1.022 MeV [twice the rest mass energy of an electron] and the pair production must take place near a heavy nucleus (Z) which absorbs the momentum of the .

• Particle production in a particle accelerator– Can reproduce conditions similar to early universe in

modern particle accelerators…

A real particle creation event

Space-time Diagrams

Space-time diagrams• Because space and time are “mixed up” in

relativity, it is often useful to make a diagram of events that includes both their space and time coordinates.

• This is simplest to do for events that take place along a line in space (one-dimensional space) – Plot as a 2D graph– use two coordinates: x and ct

• Can be generalized to events taking place in a plane (two-dimensional space) using a 3D graph (volume rendered image): x, y and ct

• Can also be generalized to events taking place in 3D space using a 4D graph, but this is difficult to visualize x

ctlight

Stationary object

Moving objects

World lines of events

Time, the fourth dimension?

“Spacetime”

x

ct

In x,y space the two space dimensions are interchangeable if they have the same units. A similar relationship can be used to express the relationship between space and time in relativity.

Light propagating in one dimension in a spacetime coordinate system as viewed from a frame S. The distance traveled is equal to the speed of light times the time elapsed.

45°

x x cparticlevelocity c

t c t slope

v=c

v>c

v<c

t in years distance in lightyearst in secs distance in lighteseconds

v/c=tan

€

2 =1

2 − sec2 θ

SIMULTANEITY

Spacetime diagrams in different frames• Changing from one

reference frame to another…– Affects time coordinate

(time-dilation)– Affects space coordinate

(length contraction)– Leads to a distortion of the

space-time diagram as shown in figure.

• Events that are simultaneous in one frame are not simultaneous in another frame

ct

x

A

ctJ ctM

xJ

xM

-1

Light cone for event “A”

“LightCone”

Different kinds of space-time intervals

“LightCone” “time like”“light like”

“Space like”

Time-like: s2>0

Light-like: s2=0

Space-like: s2<0€

s2 = (cΔt)2 − (Δx)2 = inv

Past, future and “elsewhere”.

“Future of A” (causally-connected)

“Past of A” (causally-connected)

“Elsewhere”(causally-disconnected)

ct

x

x=ctx=-ctA B

C

O

Could an event at O cause A?

Yes, because a “messenger” at O would not have to travel at a speed greater than the speed of light to get there.

Could an event at O cause B?

A light signal sent from O could reach B.

Could an event at O cause C?

No, the spacetime distance between O and C is greater than could be covered by light. It would require time travel.

xtc

xtc

xtc

here,now

where light that is here now may have been in the past

where light that is here now may go in the future

The twin paradox• Suppose Andy (A) and Betty (B) are twins.• Andy stays on Earth, while Betty leaves Earth, travels (at a

large fraction of the speed of light) to visit her aunt on a planet orbiting Alpha Centauri, and returns

• When Betty gets home, she finds Andy is greatly aged compared her herself.

• Andy attributes this to the time dilation he observes for Betty’s clock during her journey

• Is this correct? • What about reciprocity? Doesn’t Betty observe Andy’s

clock as dilated, from her point of view? Wouldn’t that mean she would find him much older, when she returns?

• Who’s really older?? What’s going on???

Andy’s point of view• Andy’s world line, in his own frame,

is a straight line• Betty’s journey has world line with

two segments, one for outbound (towards larger x) and one for return (towards smaller x)

• Both of Betty’s segments are at angles 45 to vertical, because she travels at vc

• If Andy is older by t years when Betty returns, he expects that due to time dilation she will have aged by t/ years

• Since 1/ = (1-v2/c2)1/2 1, Betty will be younger than Andy, and the faster Betty travels, the more difference there will be

ct

x

A

B (outbound)

B (return)

€

(cΔtA /2)2 − L2 = (cΔtB /2)2

ΔtB < ΔtA

L = vΔtA /2

(ΔtA )2 = (ΔtB )2 /[1− (v /c)2]

ΔtA = γΔtB

L

04/18/23 40

Betty’s point of view• Consider frame moving with

Betty’s outbound velocity• Andy on Earth will have straight

world line moving towards smaller x

• Betty’s return journey world line is not the same as her outbound world line, instead pointing toward smaller x

• Both Andy’s world line and Betty’s return world line are at angles 45 to vertical (inside of the light cone)

• Betty’s return world line is closer to light cone than Andy’s world line

ct

x

A

B (outbound)

B (return)

ct

x

A

B (outbound)

B (return)

For frame moving with Betty’s return velocity, situation is similar

04/18/23 41

Solution of the paradox• From any perspective,

Andy’s world line has a single segment

• From any perspective, Betty’s world line has two different segments

• There is no single inertial frame for Betty’s trip, so reciprocity of time dilation with Andy cannot apply for whole journey

• Betty’s proper time is truly shorter -- she is younger than Andy when she returns

ct

x

A

B (outbound)

B (return)

04/18/23 42

Different kinds of world lines

• Regardless of frame, Betty’s world line does not connect start and end points with a straight line, while Andy’s does

• This is because Betty’s journey involves accelerations, while Andy’s does not

ct

x

A

B (outbound)

B (return)

ct

x

A

B (outbound)

B (return)

QuestionsQuestions

Which of these is an inertial reference frames (or a very good approximation)?

A. A rocket being launchedB. A car rolling down a steep hillC. A sky diver falling at terminal speedD. A roller coaster going over the top of a hillE. None of the above

A. A rocket being launchedB. A car rolling down a steep hillC. A sky diver falling at terminal speedD. A roller coaster going over the top of a hillE. None of the above

Which of these is an inertial reference frames (or a very good approximation)?

Ocean waves are approaching the beach at 10 m/s. A boat heading out to sea travels at 6 m/s. How fast are the waves moving in the boat’s reference frame?

A. 4 m/sB. 6 m/sC. 16 m/s D. 10 m/s

A. 4 m/sB. 6 m/sC. 16 m/s D. 10 m/s

Ocean waves are approaching the beach at 10 m/s. A boat heading out to sea travels at 6 m/s. How fast are the waves moving in the boat’s reference frame?

A carpenter is working on a house two blocks away. You notice a slight delay between seeing the carpenter’s hammer hit the nail and hearing the blow. At what time does the event “hammer hits nail” occur?

A. Very slightly after you see the hammer hit.B. Very slightly after you hear the hammer hit.C. Very slightly before you see the hammer hit.D. At the instant you hear the blow.E. At the instant you see the hammer hit.

A. Very slightly after you see the hammer hit.B. Very slightly after you hear the hammer hit.C. Very slightly before you see the hammer hit.D. At the instant you hear the blow.E. At the instant you see the hammer hit.

A carpenter is working on a house two blocks away. You notice a slight delay between seeing the carpenter’s hammer hit the nail and hearing the blow. At what time does the event “hammer hits nail” occur?

A tree and a pole are 3000 m apart. Each is suddenly hit by a bolt of lightning. Mark, who is standing at rest midway between the two, sees the two lightning bolts at the same instant of time. Nancy is at rest under the tree. Define event 1 to be “lightning strikes tree” and event 2 to be “lightning strikes pole.” For Nancy, does event 1 occur before, after or at the same time as event 2?

A. before event 2B. after event 2C. at the same time as event 2

A tree and a pole are 3000 m apart. Each is suddenly hit by a bolt of lightning. Mark, who is standing at rest midway between the two, sees the two lightning bolts at the same instant of time. Nancy is at rest under the tree. Define event 1 to be “lightning strikes tree” and event 2 to be “lightning strikes pole.” For Nancy, does event 1 occur before, after or at the same time as event 2?A. before event 2

B. after event 2C. at the same time as event 2

A tree and a pole are 3000 m apart. Each is suddenly hit by a bolt of lightning. Mark, who is standing at rest midway between the two, sees the two lightning bolts at the same instant of time. Nancy is flying her rocket at v = 0.5c in the direction from the tree toward the pole. The lightning hits the tree just as she passes by it. Define event 1 to be “lightning strikes tree” and event 2 to be “lightning strikes pole.” For Nancy, does event 1 occur before, after or at the same time as event 2?

A. before event 2B. after event 2C. at the same time as event 2

A tree and a pole are 3000 m apart. Each is suddenly hit by a bolt of lightning. Mark, who is standing at rest midway between the two, sees the two lightning bolts at the same instant of time. Nancy is flying her rocket at v = 0.5c in the direction from the tree toward the pole. The lightning hits the tree just as she passes by it. Define event 1 to be “lightning strikes tree” and event 2 to be “lightning strikes pole.” For Nancy, does event 1 occur before, after or at the same time as event 2?

A. before event 2B. after event 2C. at the same time as event 2

ct

x1 2

t2

t1

.5 c

tNancy

Molly flies her rocket past Nick at constant velocity v. Molly and Nick both measure the time it takes the rocket, from nose to tail, to pass Nick. Which of the following is true?

A. Nick measures a shorter time interval than Molly.B. Molly measures a shorter time interval than Nick.C. Both Molly and Nick measure the same amount of

time.

A. Nick measures a shorter time interval than Molly.B. Molly measures a shorter time interval than Nick.C. Both Molly and Nick measure the same amount of

time.

Molly flies her rocket past Nick at constant velocity v. Molly and Nick both measure the time it takes the rocket, from nose to tail, to pass Nick. Which of the following is true?

Uses one clock

(ctN )2 (ctM )2 l2 (ctM )2

Beth and Charles are at rest relative to each other. Anjay runs past at velocity v while holding a long pole parallel to his motion. Anjay, Beth, and Charles each measure the length of the pole at the instant Anjay passes Beth. Rank in order, from largest to smallest, the three lengths LA, LB, and LC.

A. LA = LB = LC

B. LB = LC > LA

C. LA > LB = LC

D. LA > LB > LC

E. LB > LC > LA

A. LA = LB = LC

B. LB = LC > LA

C. LA > LB = LC

D. LA > LB > LC

E. LB > LC > LA

Beth and Charles are at rest relative to each other. Anjay runs past at velocity v while holding a long pole parallel to his motion. Anjay, Beth, and Charles each measure the length of the pole at the instant Anjay passes Beth. Rank in order, from largest to smallest, the three lengths LA, LB, and LC.

Proper length longest possible

An electron moves through the lab at 99% the speed of light. The lab reference frame is S and the electron’s reference frame is S´. In which reference frame is the electron’s rest mass larger?

A. Frame S, the lab frameB. Frame S´, the electron’s frameC. It is the same in both frames.

A. Frame S, the lab frameB. Frame S´, the electron’s frameC. It is the same in both frames.

An electron moves through the lab at 99% the speed of light. The lab reference frame is S and the electron’s reference frame is S´. In which reference frame is the electron’s rest mass larger?

In relativity, the Galilean transformations are replaced by the

A. Einstein tranformations.B. Lorentz transformations.C. Feynman transformations.D. Maxwell transformations.E. Laplace tranformations.

In relativity, the Galilean transformations are replaced by the

A. Einstein tranformations.B. Lorentz transformations.C. Feynman transformations.D. Maxwell transformations.E. Laplace tranformations.

Which of these topics was not discussed in this chapter?

A. TeleportationB. SimultaneityC. Time dilationD. Length contraction

Which of these topics was not discussed in this chapter?

A. TeleportationB. SimultaneityC. Time dilationD. Length contraction

What is energy? The capability to do WORK ->W

What is work ?

€

dW = Fdx =dp

dtdx = vdp = vd(γmv) = mv(vdγ + γdv)

dcdv

dc

vdv

dvc

vdv

cv

cvd

)11(

)/1(

/2

2

1

222

2

2

2

3

2/322

2

€

dW = mc 2dγ

K = dW = mc 2 dγ1

γ

∫0

v

∫ = mc 2(γ −1)€

v /c <<1 → γ =1/ 1− v 2 /c 2 ≈1+1

2

v 2

c 2

K ≈1

2mv 2

OPTIONAL

€

K = dW = mc 2 dγ1

γ

∫0

v

∫ = mc 2(γ −1)

Where did the 1 on the right hand side come from? It's the starting value of the integral. Now all of the terms in this equation are energies. When > 1, we have non-zero kinetic energy. So, if we think of moc2 as the total energy of body, and write

mc2= mc2 +K

then ( 1)mc2 is the kinetic energy, and moc2 is an energy that a body has when v = 0 and = 1.

Remember the proper time and proper length. These were the time and length of a body measured in its own frame. So we could write

E=Eo+K

where E0 = mc2 is the proper energy

of a body - the energy that it has, even when it is not moving.

OPTIONAL

€

K = dW =0

u

∫ Fdx0

u

∫ =dp

dtdx

0

u

∫

If → dp /dt = mdv /dt

K =1

2mv 2

Otherwise

K = mc 2(γ −1)

€