Physics 8.20 Special Relativity IAP 2008Physics 8.20 Special Relativity IAP 2008 Problem Set # 4...

Physics 8.20 Special Relativity IAP 2008

Problem Set # 4 Solutions

1. Proper acceleration (4 points)

Note: this problem and the next two ask you to work through the details ofresults derived in lecture. Let a particle be moving along the x-axis when viewedin the frame Σ. The particle’s proper acceleration, α, is defined as its accelerationmeasured in its instantaneous rest frame. Specifically, suppose at time t the particlehas velocity v. Then it is instantaneously at rest in the frame Σ′ moving with velocityv relative to Σ. Then α = dv′/dt′.

Show that the acceleration observed in Σ is related to α by

dv

dt= γ−3(v)α

where γ(v) = 1/√

1− v2/c2 as usual, and v is the instantaneous velocity.

If the frame Σ′ is moving with velocity vx̂ relative to Σ, then

t = γ(t′ + βx′/c)

⇒ dt = γdt′(1 + u′xv/c2)

and,

ux =u′x + v

1 + u′xv/c2

⇒ dux =du′x

γ2(1 + u′xv/c2)2.

So the acceleration of a particle in the two frames is related by

ax =dux

dt=

a′xγ3(1 + u′xv/c2)3

.

Now if Σ′ is the instantaneous rest frame of the particle, u′x = 0, v =ux and a′x ≡ α. So,

ax =α

γ3.

1

8.20 Special Relativity IAP 2008 2

2. Constant proper acceleration (5 points)

Suppose a particle experiences constant proper acceleration, α = α0. Suppose itstarts out at rest in Σ at t = 0.

(a) Use the result of the previous problem to show that its velocity (as measured inΣ) is given by

v(t) =α0t√

1 + (α0tc )2

>From problem 1,dv

dt= α0(1− v2/c2)3/2

⇒ dβ

(1− β2)3/2=

α0

cdt

⇒ β√1− β2

=α0t

c

⇒ v(t) =α0t√

1 + (α0t/c)2.

(b) Let α0 = g = 9.8m/sec2. How long would it take the particle to reach v = 0.99c?,v = 0.999c? according to an observer in Σ?

>From part (a),

t =c

α0

β√1− β2

.

So, time taken to reach β = 0.99 is about 6.8 years.And time taken to reach β = 0.999 is about 21.7 years.

(c) How long would it take to reach these speeds according to an observer on theparticle?

The observer on the particle measures proper time.

dt = γdτ

⇒ τ =∫ t

0dt̃√

1− β2

=∫ t

0dt̃

1√1 + α2

0t̃2/c2

=c

α0sinh−1 α0t

c


So, the proper time taken to reach β = 0.99 is about 2.6 years andthe proper time taken to reach β = 0.999 is about 3.7 years (usingthe results of part (b)).

3. Hyperbolic space travel I (7 points)

(a) Take the result of the previous problem,

v(t) =α0t√

1 + (α0tc )2

and integrate v(t) = dx/dt to obtain an expression for x(t). Take α = g =10m/sec2

and show that you reproduce the result obtained in class:

X(T ) =√

T 2 + 1− 1

where T is in years and X in light-years.

Using v(t) = dx/dt,

dx =α0t√

1 + (α0tc )2

dt.

Substituting α0t/c = sinh χ:

dx =c2

α0

sinhχ coshχdχ√1 + sinh2 χ

=c2

α0sinhχdχ.

Integrating:

x(χ2)− x(χ1) =c2

α0(coshχ2 − coshχ1)

=c2

α0(√

1 + sinh2 χ2 −√

1 + sinh2 χ1)

⇒ x(t2)− x(t2) =c2

α0(

√1 + (

α0t2c

)2 −√

1 + (α0t1

c)2).

As α0 = g = 10m/s, c/α0 = 3× 107s = 1 yr. Then T = α0t/c is in yearunits and X = xα0/c2 is light year units. This gives the result obtainedin class:

X(T ) =√

T 2 + 1− 1.

(b) Likewise confirm the result from lecture that relates the passage of proper time(experienced by the astronauts), τ , and the passage of time in the rest framefrom which the rocket originated:

T = sinh τ


— again both times in years.

>From the value of v in (a),

1− v(t)2 =1

1 + (α0tc )2

=1

γ(v).

But dτ = dt/γ. Again substituting α0t/c = sinh χ,

dτ =c

α0dχ.

This gives T = sinh τ, where τ is now measured in yrs.

(c) Now suppose that the space travellers take a more “realistic” journey where theyaccelerate at g for the first half of the trip and decelerate at g for the secondhalf. Find expressions for

i. How far can they travel in a time T (measured in the originating frame)?

Here, for the first half, the accelaration is positve, and thennegative. Thus the velocity is

v(t) =α0t√

1 + (α0tc )2

, 0 ≤ t ≤ T/2

=α0(T − t)√

1 + (α0(T−t)c )2

, T/2 ≤ t ≤ T.

Using the integrated equation for x in part (a) with t2 = T/2,t1 = 0 and then integrating the same way for t2 = T, t1 = T/2.By a change of variable x = T−t, for the second integration youcan easily see that the itegral will be equal to the integrationof v(t)dt for the part of the trip we get:

x(T )−x(0) = x(T )−x(T/2)+x(T/2)−x(0) = 2×(x(T/2)−x(0)) =√

T 2 + 4−2;

ii. How much proper time, τ , passes on a journey that takes T years (as ob-served in the originating frame).


γ =

√1 + (

α0(t)c

)2, 0 ≤ t ≤ T/2

=

√1 + (

α0(T − t)c

)2, T/2 ≤ t ≤ T.

⇒ τ =∫ T/2

0γ−1dt +

∫ T

T/2γ−1dt

= 2∫ T/2

0

1√1 + T 2

dT

= 2 sinh−1 T

2⇒ T = 2 sinh

τ

2.

4. Hyperbolic space travel II (7 points)

Assume that a rocket can produce an acceleration g0 = 9.8m/sec2. [As describedin lecture, this acceleration is approximately the same as that due to gravity at thesurface of the earth. Astronauts will be able to live comfortably in this spaceship, asif in earth’s gravity.) Assume, in addition, that in travelling to any destination therocket will accelerate half the way and decelerate during the second half of the journey.

(a) Calculate the travel time as measured by the space traveller to the moon. (As-sume the moon is at a distance of 382,000 km.) Compare with the Galileananswer.

For a rocket with constant proper acceleration α0, we derived the followingresult in Problem 2 :

v(t) =α0t√

1 + (α0t/c)2,

where v, t are measured in the earth frame.We integrate this to obtain a hyperbolic equation for the earth-measureddistance of travel, d, as a function of time:

d(t) =c2

α0

√1 + (α0t/c)2 − c2

α0.

We can solve for t to get

t =

√d

(d

c2+

2α0

).

Convert the earth time to the astronaut’s time using

τ =c

α0sinh−1 α0t

c


derived in Problem 2.So, for travel halfway to the moon with α0 = 9.8m/s2,

d =12× 3.82× 108m

⇒ t = 1.734 hrs

⇒ τ = 1.734 hrs

Twice the halfway travel time gives the total travel time of

3.47 hrs .

A Galilean calculation would give a halfway travel time (in all frames)of √

2d

α0= 1.734 hrs

which gives a total time of

3.47 hrs .

For such a short journey, there is no difference between Galilean, earthand astronaut travel times.

(b) Answer the same questions for travel to Neptune, assumed to be at a distanceof 4.5× 109 km.

For travel halfway to Neptune,

d =12× 4.5× 1012m

⇒ t = 7.843 days

⇒ τ = 7.842 days

So the total travel time is

15.7 days .

A Galilean calculation would give a total travel time (in all frames)of

15.7 days .

So once again, the journey is short enough to not produce any significantdifference between the Galilean, earth and astronaut travel times.

(c) Answer the same questions for travel to Alpha Centauri, assumed to be at adistance of 4.3 light years away. What is the velocity of the rocket, in the earth’sreference frame, at the half way point of the journey?


For travel halfway to Alpha Centauri,

d =12× 4.3 lyrs

⇒ t = 2.96 yrs

⇒ τ = 1.78 yrs

So the total travel time is

3.56 yrs .

At the halfway point,

β =α0t/c√

1 + (α0t/c)2= 0.95

A Galilean calculation would give a total travel time (in all frames)of

4.08 yrs .

Now the earth frame travel time is larger than the Galilean time whichin turn is larger than the astronaut’s time.

(d) What is the value of β that would enable a second astronaut, travelling at con-stant speed, to travel from the earth to Alpha Centauri in the same travel timeas that taken by the rocket described above?

The rocket described above takes 5.90 yrs (as measured on earth) totravel to Alpha Centauri 4.3 lyrs away. This corresponds to an averagespeed of

4.3c

5.92= 0.726 c

and a rocket traveling at that constant speed will complete the journeyin the same amount of time.

5. Hyperbolic space travel III (8 points)

Aliens arrive on Earth and tell us that they have come from very far away. They hadbeen watching light emitted from the primitive Earth and saw that life was evolving.They took a risk and headed off for Earth. Their journey took them a distance of100,000,000 light-years. After getting our units right, we figured out that only 5 yearsof proper time elapsed for them during their great journey. Being smart aliens, theytravelled hyperbolically — they accelerated at the natural acceleration of gravity ontheir planet, G, for the first half, and then decelerated for the second half of thejourney.

How strong is gravity on the surface of their planet? (Hint: To solve the transcen-dental equation for G, you could graph both sides as a function of G; or start with aparticular value for G and iterate; or use a symbolic manipulation package like Matlab


or Mathematica, both available on Athena.)

In problem 3 we solved for x(t), with acceleration α0. In this problem theacceleration is G. Suppose the whole trip takes time T, then the midpointis reached at T/2. Thus

x(12T ) =

c2

G(

√1 + (

GT

2c)2 − 1)

⇒ G

c2x(

12T ) =

√1 + sinh2 Gτ

2c− 1

= coshGτ

2c− 1

= 2 sinh2 Gτ

4c

⇒ G

g× gx(T/2)

c2= 2 sinh2 G

g× gτ

4c.

As c/g ≈ 1 yr, we can now insert the values given in lt years and years.Call 2.5G/2g = G′. Then, inserting the values

2G′ × 107 = sinh2 G′

⇒√

2G′ × 103.5 = sinhG′

≈ eG′

2

⇒ G′ =32

ln 2 + 3.5 ln 10 +12

lnG′.

Solving by iteration: Choose on right hand side

G′ = 1 ⇒ G′ = 9.098;G′ = 9.098 ⇒ G′ = 10.202;

G′ = 10.202 ⇒ G′ = 10.261;G′ = 10.261 ⇒ G′ = 10.262;G′ = 10.262 ⇒ G′ = 10.263.

Thus G′ = 10.26, which gives G = 80.5 m/s2.

We could also plot the left and right hand sides of the equations as functionsof G′ and look at their point of intersection.

6. Non-relativistic limit (3 points)

The correct relativistic formula for the energy of a particle of rest mass m and speedv is E(v) = mγ(v)c2.


(a) Expand E(v) to order v4. Identify the first correction to the usual non-relativisticformula for the kinetic energy, T = 1

2mv2.

The relativistic expression for the energy of a particle of rest massm and speed v is

E(v) = γ(v)mc2

= mc2

(1 +

12β2 +

38β4 + . . .

)= mc2 +

12mv2 +

38m

v4

c2+ . . .

So the first relativistic correction to the kinetic energy is

∆T =38m

v4

c2.

(b) How large is the correction of part a) for the Earth relative to the Sun? In otherwords, approximately how much heavier is the Earth on account of its motionaround the Sun? You will need the Earth’s mass and its orbital speed to com-plete this estimate.

The earth’s mass is m = 6×1024 kg and its orbital speed is v = 30 km/s.The leading order correction to mass is given by

∆m(0) =mv2

2c2

=12mβ2

= 3× 1016kg

This is correction by(∆m

m× 100

)or 5× 10−7% .

The correction to galilean kinetic energy calculated in part (a) contributesa mass of (next to leading order correction)

∆m(1) =∆T

c2

=38mβ4

= 2.25× 108kg

This is a correction by(∆m

m× 100

)or 4× 10−15% .


(c) How large a percentage error did you make by using the approximation of parta) instead of the exact result in the case of the Earth’s motion around the Sun?

The exact relativistic correction to the earth’s mass is given by

∆mex = γm−m

= m

(12β2 +

38β4 +

516

β6 +35128

β8 + . . .

).

In approximation used in part (b) we just took first two terms intoaccount i.e. ∆m = ∆m(0) +∆m(1). Percentage error due to this isgiven by

∆mex −∆m

m× 100% =

(516

β6 +35128

β8 + . . .

)≈ 3.1× 10−23

7. Hard work (2 points)

A proton initially moves at v = 0.99c in some reference frame.

(a) How much work must be done on the proton to increase its speed from 0.99c to0.999c?

The proton mass is mp = 939 MeV/c2. The work done in increasing itsspeed from v1 to v2 is

W =

(1√

1− β21

− 1√1− β2

2

)× 939 MeV .

(a) If v1 = 0.999c and v2 = 0.99c, then

W ≈ 14.3 GeV .

(b) How much work must be done to increase its speed from 0.999c to 0.9999c?

If v1 = 0.9999c and v2 = 0.999c, then

W ≈ 45.4 GeV .

8. Solar Power (5 points)


The net result of the fusion reaction that fuels the sun is to turn four protons andtwo electrons into one helium nucleus,

4p+ + 2e− → 4He++.

Other particles are given off (neutrinos and photons), but you can assume they even-tually show up as energy. The masses of the relevant nuclei are as follows:

mp = 1.6726× 10−27 kg.me = 9.1094× 10−31 kg.m 4He = 6.6419× 10−27 kg.

(a) How much energy is released when a kilogram of protons combines with justenough electrons to fuse completely to form helium?

4 protons fuse with 2 electrons to release

(4mp + 2me −mHe)c2 = 4.5290× 10−12J .

A kilogram of protons contains

4× 1.4947× 1026 protons

which release

1.4947× 1026 × 4.5290× 10−12J = 6.7694× 1014 J .

(b) How many kilograms of methane would you have to burn to produce the sameamount of energy? ( Go to your favorite source of chemistry information (theWeb, perhaps, or the CRC Handbook) and determine the reaction by whichmethane burns and how much energy is released per gram molecular weight ofmethane. The usual units are KCal/mole).

We can calculate that 1000 kg of methane burns to produces how muchenergy by looking up the values in CRC. The reaction is

CH4 + 2O2 → CO2 + H2O

which releases an energy of 212.8 Kcal/mole. So for 1000 kg, this gives:

212.8× 103cal/mole × 10000.016

moles × 4.2 J/cal = 5.586× 1010J

Dividing by c2, the mass equivalent of the above energy is

0.6207 micro-grams .

So the amount of methane that will produce the same amount of energyas that produced by fusing 1 kg of protons:

1000× 6.7694× 1014

5.586× 1010kg = 1.2119× 107 kg .


9. E = mc2 (2 points)

Assume the heat capacity of water is equal to 4.2 joules/g-K, and constant as a func-tion of temperature. How much does the mass of a ton of water increase when it isheated from freezing (0◦C) to boiling (100◦C)?

The amount of heat energy transfered to 1000 kg of water to raise its temperaturefrom 0◦ C to 100◦ C is

106 g × 4.2 J/(◦ g) × 100◦ = 4.2× 108J .

This is equivalent to a mass of

4.67× 10−9 kg .

10. Enormous energies (2 points)

Quasars are the nuclei of active galaxies in the early stages of their formation. Atypical quasar radiates enery at the rate of 1041 watts. At what rate is the mass ofthis quasar being reduced to supply this energy? Express your answer in solar massunits per year, where one solar mass unit, 1 smu = 2 × 1030 kg, is the mass of oursun.

The mass of the quasar is reduced at a rate of

1041

9× 1016kg/s × 1

2× 1030smu / kg ×(3600×24×365) s/yr = 17.52 smu/yr .

11. Classical physics and the speed of light (2 points)

(a) How much energy would it take to accelerate an electron to the speed of lightaccording to “classical” (before special relativity) physics?

If an electron is accelerated from rest to the speed of light, thenclassically it would require

12mec

2 = 4.1× 10−14 J .


(b) With this energy what would its actual velocity be?

If the above amount of energy is given to an electron then

T = (γ − 1)mec2 =

12mec

2

⇒ γ =32

⇒ β = 0.745 .

12. A useful approximation (4 points)

(a) Show that for an extremely relativistic particle, the particle speed u differs fromthe speed of light c by

∆u = c− u =c

2

(m0c

2

E

)2

where m0 is the rest mass and E is the energy.

E = γm0c2

⇒ β =

√1−

(m0c2

E

)2

⇒ β ≈ 1− 12

(m0c

2

E

)2

for a particle with E >> m0c2

⇒ ∆u = c− u ≈ c

2

(m0c

2

E

)2

(b) Find ∆u for electrons produced by

i. MIT’s Bates Accelerator Center, where E = 900 MeV.

E = 900 MeV at Bates Accelerator Center gives ∆u = 48.4 m/s .

ii. The Jeffreson Lab (in Newport News, Virginia), where E = 12 GeV.

E = 12 GeV at Jefferson Lab gives ∆u = 0.272 m/s .


iii. The Stanford Linear Accelerator Center (in Palo Alto, California), whereE = 50 Gev.

E = 50 GeV at SLAC gives ∆u = 0.0157 m/s .

13. Pressure of light (5 points)

French §6, Problem 6-7, page 201: A photon rocket

Four momentum of the system BEFORE collision:

P = Mi(c, 0)

Four momentum of the system AFTER collision:

P ′ =E

c(1,−1) + Mfγ(c, v)

Where E is the energy of photons.

>From conservation of four-momentum:

P = P ′

we get two equations:Mic = E/c + Mfγc

0 = −E/c + Mfγv

Eliminate E from above:

Mic = Mfγv + Mfγc

⇒ Mi/Mf = γ(1 + β) =

√1 + β

1− β

French §6, Problem 6-9, page 201: A laser beam

(a) Energy radiated by laser per second

P0 = 1020 s−1 hc

λ= 33.1 W

Since laser is at rest w.r.t earth therefore this is also the rate at whichobserver on earth observes the power radiated.


(b) Let Eγ be the total energy radiated by laser in form of photons. Byenergy conservation in earth’s frame

Mc2 = γ(M −∆M)c2 + Eγ ,

and by momentum conservation

0 = −Eγ

c+ (M −∆M)γv.

Eliminating Eγ from above two equation we get

Mc2 = γ(M −∆M)c(c + v)

which simplifies to

M −∆M

M=

√1− β

1 + β.

Expanding in powers of beta upto leading term we get

β =∆M

M⇒ v =

∆M

Mc.

Now

∆M = 1020 × hc

λ× 10 yrs/c2 = 1.16× 10−7 kg.

This givesv = 3.48 m/s.

(c) Incident energy of each photon when laser is moving with velocity βc

Eγ =

√1− β

1 + βE′

γ

where E′γ is energy of photon in laser frame. Similarly rate at which photon

are observed on earth’s frame is given by

r =

√1− β

1 + βr′

where r′ = 1020 s−1. Therefore observed power on earth as a function ofbeta is given by

P = Eγr =1− β

1 + βE′

γr′ =1− β

1 + βP0.

Taking P = P0 −∆P and expanding in power of β we get

∆P = 2P0β = 2× 33.1× 1.16× 10−8 W = 7.7× 10−7 W .

So the rate seen on earth after 10 years has decreased by 7.7× 10−7 W.

(d) Loss in incident power is accounted for by energy used in acceleratingthe laser.


14. Relativistic collisions and decays (17 points)

Note that many of these problems are easiest if you use invariants.

French §6, Problem 6-5, page 200: Three particle decay

(a) Four momentum Before the decay:

P = M0c(1, 0, 0)

Four momentum after the decay:

P ′ = m0γ1c(1,−β1, 0) + m0γ2c(1, 0,−β2) + m0γ3c(1, β3 cos θ, β3 sin θ)

Where we know:β1 = 4/5

γ1 = 5/3

β2 = 3/5

γ2 = 5/4

>From the conservation of four-momentum P = P ′ we’ll get three equations:

M0 = m0(γ1 + γ2 + γ3) (1)

γ1β1 = γ3β3 cos θ (2)

γ2β2 = γ3β3 sin θ (3)

Devide (2) and (3):

tan θ =γ2β2

γ1β1= 9/16

(2) ⇒ γ3β3 =γ1β1

cos θ= γ1β1

√1 + tan2 θ =

√33712

⇒ β23 =

337337 + 144

β3 = 0.84

(b) >From (1) we have:

M0/m0 = γ1 + γ2 + γ3 = 1.67 + 1.25 + 1.84 = 4.76

French §6, Problem 6-11, page 201: Pion decay


(a) γ of the π0 is:

γ =1 GeV + 135 MeV

135 MeV= 1135/135 = 8.4 ⇒ β = 0.9929

Before the decay we have the four-momentum:

mγc(1, β)

After the decay we have the two photons:

(E/c,E/c) + (E′/c,−E′/c)

>From the law of conservation of four-momentum we get:

E + E′ = mγc2

E − E′ = mγβc2

E =12mc2γ(1 + β) = 1131 MeV

E′ =12mc2γ(1− β) = 4 MeV

(b) This time we’ll have:

E/c(1, cos θ, sin θ) + E′/c(1,+cos θ,− sin θ) = mγc(1, β, 0)

The last component of the above equation will give:

E = E′

>From the zeroth component (energy conservation) we’ll get:

2E/c = mγc

>From the 1st component we’ll get:

2E/c cos θ = mγcβ

After deviding the above two equations we get:

cos θ = β ⇒ θ = 0.1192 Radians = 6.8◦

The angle between the two gamma rays is twice the above angle , i.e.13.6◦.


French §6, Problem 6-14, page 202: Impossible processes

Factors of c are suppressed in this calculation.

(a) A single photon (with four-momentum pγ = (Eγ , Eγ , 0, 0)) strikes a stationaryelectron (pi = (me, 0, 0, 0)) and gives up all its energy to the electron(which has four-momentum pf). From conservation of four-momentum,

pγ + pi = pf .

Squaring both sides,m2

e = m2e + 2pγ · pi

⇒ meEγ = 0

which is impossible unless Eγ = 0 (i.e. there is no photon).

(b) A single photon (with four-momentum pγ = (Eγ , Eγ , 0, 0)) in empty spaceis transformed into an electron and a positron with four-momenta p1

and p2 respectively. So,

pγ = p1 + p2

⇒ 0 = 2m2e + 2p1 · p2

⇒ p1 · p2 = −m2e

which is impossible because the scalar product of four-momenta correspondingto massive particles has to be greater than the product of the masses(work it out in the center of mass frame to convince yourself of this).Another way to deduce that this process is impossible is to work inthe center of mass frame of the electron-positron pair. Then the photonmust have zero momentum to conserve momentum but that would mean thatenergy cannot be conserved.

(c) A fast positron and a stationary electron annihilate, producing onlyone photon. This is the reverse process of (b) (Lorentz transformedto the rest frame of the electron) and so is impossible because (b)is impossible.

French §6, Problem 6-15 (a), page 202: Proton Collision

Lorentz factor for incident proton is γ = (mp+K)/mp = 1.46 which givesv = 0.731 c. From energy conservation we get

2mp + K = 2E′


where E′ is energy of each final proton. This gives E′ = 1158.5 MeV . Thereforeγ′ = 1158.5/940 = 1.23 which gives v′ = 0.585 c. From momentum conservation

γmpv = 2E′v′ cos θ

⇒ cos θ =(mp + K)v

2E′v′= 0.7426

which gives θ = 42.045◦. Therefore inclusive angle is 2θ = 84.09◦.

French §6, Problem 7-1, page 225: Kaon decay

Let’s denote P1 the four-momentum of the K meson, P2 the moving π and P3

the π at rest.

P1 = mKγKc(1, βK)

P2 = mπγπc(1, βπ)

P3 = mπc(1, 0)

We have:P1 = P2 + P3 ⇒ (P1 − P3)2 = P 2

2

m2Kc2 + m2

πc2 − 2mπc2mKγK = m2πc2

⇒ γK =mK

2mπ= 1.8

EK = mKγKc2 = 889 MeV

KK = 889− 494 = 395 MeV

from Energy conservation:

Eπ = 889− 137 = 752 MeV

Kπ = 752− 137 = 615 MeV

French §6, Problem 7-2, page 225: Pair production

For the minimum energy of the gamma ray, the two e− and the e+ will be atrest in their CENTER of MASS frame so their total four-momentum will looklike (3mec, 0) since P 2 is invariant under lorentz transformation we’ll have:

[E/c(1, 1) + mec(1, 0)]2 = 9m2ec

2

0 + m2ec

2 + 2meE = 9m2ec

2

E = 4mec2 = 2.04 MeV

French §6, Problem 7-4, page 226: elastic scattering


15. Fixed-target collisions (RH) (5 points)

(a) A proton (with rest mass m) accelerated in a proton synchrotron to a kineticenergy K strikes a second (target) proton at rest in the laboratory. The collisionis entirely inelastic in that the rest energy of the two protons, plus all of thekinetic energy consistent with the law of conservation of momentum, is availableto generate new particles and to endow them with kinetic energy. Show that theenergy available for this purpose is given by

E = 2mc2

√1 +

K

2mc2.

Before the collision, the total energy and the total momentum of thetwo protons is:

Ei = K + 2mc2

pi =√

K2 + 2Kmc2/c

After the collision we can treat all the products as a single particlewith mass M and velocity v. having energy and momentum:

Ef = γMc2

pf = γMv

By energy and momentum conservation, E − i = Ef and pi = pf:

K + 2mc2 = γMc2√K2 + 2Kmc2/c = γMv

DEviding the two equations gives β,

β = v/c =√

K2 + 2Kmc2

K + 2mc2=

√K

K + 2mc2

We can then find γ:

γ =

√1 +

K

2mc2

This gives us Mc2, the total energy available for making particlesand endowing them with kinetic energy (i.e. the total energy in theframe where the momenum is zero),

Mc2 =K + 2mc2

γ= 2mc2 =

√1 +

K

2mc2


(b) How much energy is made available when 100-GeV protons are used in this fash-ion?

E0 = 2(0.938 GeV )

√1 +

1002× 0.938

= 13.6 GeV

(c) What proton energy would be required to make 100 GeV available? This shouldbe compared with the result in the next problem.

For E0 = 100 GeV ,

100 GeV = 2(0.938 GeV )

√1 +

K

2× 0.938⇒ K = 5.33 TeV

16. Center-of-mass collisions (RH) (5 points)

(a) In modern experimental high-energy physics, energetic particles are made tocirculate in opposite directions in so-called storage rings and are permitted tocollide head-on. In this arrangement, each particle has the same kinetic energyK in the laboratory. The collisions may be viewed as totally inelastic, in thatthe rest energy of the two colliding protons, plus all the available kinetic energy,can be used to generate new particles and to endow them with kinetic energy.Show that, in contrast to the previous problem, the available energy in thisarrangement can be written in the form

E = 2mc2

(1 +

K

2mc2

).

Four-momenum before the collision:

P = mγc(1, β) + mγc(1,−β) = 2mγc(1, 0)

After the collision the four-momenum is P ′ with P ′2 = M2c2. But weknow that P 2 = P ′2:

M2c2 = 4m2γ2c2

>From the relationK = (γ − 1)mc2

you can find γ as a function of K:

γ = 1 +K

mc2

⇒ E = Mc2 = 2mγc2 = 2mc2(1 +K

mc2)


(b) How much energy is made available when 100-GeV protons are used in this fash-ion?

for K = 100 GeV ,

E = 2(0.938 GeV )(1 + 100/0.938) = 202 GeV

(c) What proton energy would be required to make 100 GeV available?

For E = 100 GeV ,

100 GeV = 2(0.938 GeV )(1 + K/0.938)

K = 43.0 GeV

(d) Compare these results with those of the previous problem and comment on theadvantages of storage rings.

This means that head-on collision has more advantages because we needless Kinetic energy to produce a given mass or with a given Kineticenergy we can produce more massive particles.

17. Compton scattering (5 points)

Following the discussion in lecture, consider Compton scattering, in which a photonwith wavelength λ0 scatters from an electron at rest. After the collision, a photon ofwavelength λ is emitted at an angle θ relative to the direction of the initial photonand the electron recoils with the velocity and direction required by momentum andenergy conservation. Derive the relationship between the scattering angle θ and thewavelength change ∆λ = λ = λ0 for Compton scattering:

∆λ =h

mc(1− cos θ)

(a) Use invariants as in lecture.

A photon with four-momentum pγ = (Eγ , Eγ , 0, 0) strikes a stationary electronwith pi = (me, 0, 0, 0). We have temporarily suppressed factors of c. Thephoton is scattered to a four-momentum pγ′ = (Eγ′ , Eγ′ cos θ, Eγ′ sin θ, 0) whereθ denotes the angle that the scattered photon makes with the direction ofthe incident photon. The electron is scattered to some four-momentum pf.

pγ + pi = pγ′ + pf

⇒ p2f = (pγ + pi − pγ′)2

⇒ m2e = m2

e + 2me(Eγ − Eγ′)− 2EγEγ′(1− cos θ)


⇒ Eγ′ =mec

2Eγ

mec2 + (1− cos θ)Eγ(with factors of c restored).

Now the initial and final wavelengths (λ and λ′) are obtained from

Eγ =hc

λ,Eγ′ =

hc

λ′.

So,

λ′ = λ +h

mc(1− cos θ)

⇒ ∆λ =h

mc(1− cos θ)

where ∆λ is the final wavelength minus the initial wavelength.

(b) Use energy conservation, momentum conservation in the x-direction, and momen-tum conservation in the y-direction, and eliminate the appropriate variables.

Let four momentum of final electron be denoted by (E′e, p

′ cos φ,−p′ sinφ, 0).Energy conservation gives

Eγ + me = E′γ + E′

e

⇒ E′e = Eγ + me − E′

γ

Momentum conservation along x-direction gives

Eγ = E′γ cos θ + p′ cos φ

⇒ p′ cos φ = Eγ − E′γ cos θ

and along y-directionp′ sinφ = E′

γ sin θ .

Squaring and adding last two equation we get

(p′)2 = E2γ + (E′

γ)2 − 2EγE′γ cos θ .

Subtracting the last equation from square of energy conservation equationgives

(E′e)

2 − (p′)2 = m2e + 2me(Eγ − E′

γ)− 2EγE′γ(1− cos θ)

⇒ me(Eγ − E′γ) = EγE′

γ(1− cos θ)

⇒(

1E′

γ

− 1Eγ

)=

1me

(1− cos θ)

⇒ λ′ − λ =h

mec(1− cos θ) (substituting factors of c back)

⇒ ∆λ =h

mec(1− cos θ).


18. Production of hyperons at rest (7 points)

The K− meson and the Λ0 hyperon are two commonly encountered unstable particles.For example, they are commonly produced in air showers by cosmic rays and severalof each have whizzed through you during the time you have been working on thisproblem set. The reaction

K− + p → Λ0 + π0

can be used to make Λ0’s at rest in the laboratory by scattering K− mesons off astationary proton (hydrogen) target.

The rest energies of all the particles involed are: mpc2 = 939 MeV, mK±c2 = 494 MeV,

mπ0c2 = 135 MeV, mΛ0c2 = 1116 MeV.

(a) Find the energy of the incident K− beam required to just produce Λ0 hyperonsat rest in the lab.

Let pK , pp, pΛ, pπ be the four-momenta of the K−, p,Λ0, π0 particles respectively.We know that

pK = (EK , |~pK |) ,

pp = (mp, 0) ,

pΛ = (mΛ, 0) .

>From conservation of four-momentum,

pK + pp = pΛ + pπ

⇒ p2π = (pK + pp − pΛ)2

⇒ m2π = m2

K + m2p + m2

Λ + 2(mp −mΛ)EK − 2mpmΛ

⇒ EK =m2

K −m2π + (mΛ −mp)2

2(mΛ −mp)≈ 726 MeV .

(b) What is the π0 energy for this “magic” K− energy?

Eπ = EK + Ep − EΛ = EK + mp −mΛ ≈ 549 MeV .

(c) Check momentum conservation.

|~pK | =√

E2K −m2

K ≈ 533MeV ,

|~pπ| =√

E2π −m2

π ≈ 533MeV .

The hyperon and proton have no 3-momentum. Hence, momentum is conservedas long as the K and the π are directed in the same direction.


(d) Could the process be run the other way? That is, could a π0 beam (assuming onewas available) be used to make a K+ at rest by the reaction π0 + p → Λ0 +K+?

If a beam of pions strikes stationary protons and makes kaons at rest,then all the results of the previous calculation will hold with thefollowing substitutions:

K → π ,Λ → K , π → Λ .

So,

Eπ ==m2

π −m2Λ + (mK −mp)2

2(mK −mp)≈ 1156 MeV > mπ ,

EΛ = Eπ + Ep − EK = Eπ + mp −mK ≈ 1601 MeV > mΛ .

Thus, the process can be run the other way.

19. Invariant Product (5 points)

Lorentz transformation for A′ and B′ are given by

a′0 = a0 cosh η − ax sinh η a′x = ax cosh η − a0 sinh η a′y,z = ay,z

b′0 = b0 cosh η − bx sinh η b′x = bx cosh η − b0 sinh η b′y,z = by,z

Let us explicitly evaluate the dot product and check the invariance. Bydefinition

A′ ·B′ = a′0b′0 − a′xb′x − a′yb

′y − a′zb

′z .

Using the transformation we get

A′ ·B′ = (a0 cosh η − ax sinh η)(b0 cosh η − bx sinh η)−(ax cosh η − a0 sinh η)(bx cosh η − b0 sinh η)− ayby − azbz

= a0b0 cosh2 η + axbx sinh2 η − (a0bx + axb0) cosh η sinh η

−axbx cosh2 η − a0b0 sinh2 η + (axb0 + a0bx) cosh η sinh η − ayby − azbz

= a0b0(cosh2 η − sinh2 η)− axbx(cosh2 η − sinh2 η)− ayby − azbz

= a0b0 − axbx − ayby − azbz

= A ·B

where in the second last step we have used identity cosh2 η − sinh2 η = 1.This verifies the invariance of the product.

Physics 8.20 Special Relativity IAP 2008Physics 8.20 Special Relativity IAP 2008 Problem Set # 4...

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