Soal Dan Penyelesaian

8/12/2019 Soal Dan Penyelesaian

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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 . The particles proper acceleration, , is defined as its accelerationmeasured in its instantaneous rest frame. Specifically, suppose at time t the particlehas velocityv. Then it is instantaneously at rest in the frame moving with velocityv relative to . Then = dv /dt.

Show that the acceleration observed in is related to by

dvdt

=3(v)

where(v) = 1/

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

If the frame is moving with velocity vx relative to , then

t= (t + x/c)

dt= dt(1 + uxv/c2)and,

ux= u

x+ v1 + uxv/c2

dux = du

x

2(1 + uxv/c2)2

.

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

ax=dux

dt =

ax3(1 + uxv/c

2)3.

Now if is the instantaneous rest frame of the particle, ux = 0, v =ux and a

x . So,

ax=

3.

1


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8.20 Special Relativity IAP 2008 2

2. Constant proper acceleration (5 points)

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

(a) Use the result of the previous problem to show that its velocity (as measured in

) is given by

v(t) = 0t1 + ( 0tc )

2

>From problem 1,dv

dt =0(1 v2/c2)3/2

d(1 2)3/2 =

0c

dt

1 2 =

0tc

v(t) = 0t1 + (0t/c)2

.

(b) Let0= g = 9.8m/sec2. How long would it take the particle to reach v = 0.99c?,

v= 0.999c? according to an observer in ?

>From part (a),

t= c

0

1 2.

So, time taken to reach = 0.99 is about 6.8 years.And time taken to reach = 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 = d

= t

0

dt1 2=

t0

dt 1

1 + 20t2/c2

= c

0sinh1

0t

c


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So, the proper time taken to reach = 0.99 is about 2.6 years andthe proper time taken to reach = 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) = 0t1 + ( 0tc )

2

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

and show that you reproduce the result obtained in class:

X(T) =

T2 + 1 1whereT is in years and X in light-years.

Using v(t) =dx/dt,

dx= 0t1 + (0tc )

2dt.

Substituting 0t/c= sinh :

dx= c2

0

sinh cosh d1 + sinh2

= c2

0sinh d.

Integrating:

x(2) x(1) = c2

0 (cosh 2 cosh 1)=

c2

0(

1 + sinh2 2

1 + sinh2 1)

x(t2) x(t2) = c2

0(

1 + (

0t2c

)2

1 + (0t1

c )2).

As 0 =g = 10m/s, c/0 = 3 107s= 1 yr. Then T =0t/c is in yearunits and X=x0/c

2 is light year units. This gives the result obtained

in class:

X(T) =

T2 + 1 1.

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

T= sinh


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again both times in years.

>From the value of v in (a),

1

v(t)2 =

1

1 + (0t

c )2

= 1

(v)

.

But d=dt/. Again substituting 0t/c= sinh ,

d= c

0d.

This gives T= sinh , where 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 then

negative. Thus the velocity is

v(t) = 0t

1 + (0tc )2

, 0 t T /2

= 0(T t)

1 + (0(Tt)

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= Tt, for the second integration youcan easily see that the itegral will be equal to the integration

of 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)) =

T2 + 42;

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


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=

1 + (

0(t)

c )2, 0 t T /2

= 1 + ( 0(T t)c

)2, T /2

t

T.

= T/20

1dt +

TT/2

1dt

= 2

T/20

11 + T2

dT

= 2 s inh1T

2

T= 2 sinh2

.

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 earths 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 0, we derived the followingresult in Problem 2 :

v(t) = 0t1 + (0t/c)2

,

where v, t are measured in the earth frame.We integrate this to obtain a hyperbolic equation for the earth-measured

distance of travel, d, as a function of time:

d(t) = c2

0

1 + (0t/c)2 c

2

0.

We can solve for t to get

t=

d

d

c2+

2

0

.

Convert the earth time to the astronauts time using

= c

0sinh1

0t

c


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derived in Problem 2.

So, for travel halfway to the moon with 0= 9.8m/s2,

d=1

2 3.82 108m

t= 1.734 hrs = 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

0= 1.734 hrs

which gives a total time of

3.47 hrs .

For such a short journey, there is no difference between Galilean, earth

and 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=1

24.5

1012m

t= 7.843 days = 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 earthsreference frame, at the half way point of the journey?


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For travel halfway to Alpha Centauri,

d=1

2 4.3 lyrs

t= 2.96 yrs = 1.78 yrs

So the total travel time is

3.56 yrs .

At the halfway point,

= 0t/c1 + (0t/c)2

= 0.95

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

of4.08 yrs .

Now the earth frame travel time is larger than the Galilean time which

in turn is larger than the astronauts time.

(d) What is the value ofthat 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

4.3c5.92

= 0.726 c

and a rocket traveling at that constant speed will complete the journey

in 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, they

travelled 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 ofG; or start with aparticular value forGand iterate; or use a symbolic manipulation package like Matlab


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or Mathematica, both available on Athena.)

In problem 3 we solved for x(t), with acceleration 0. In this problem theacceleration is G. Suppose the whole trip takes time T, then the midpointis reached at T /2. Thus

x(1

2T) =

c2

G(

1 + (

GT

2c )2 1)

Gc2

x(1

2T) =

1 + sinh2

G

2c 1

= coshG

2c 1

= 2 sinh2G

4c

Gggx(T /2)

c2 = 2 sinh2

G

gg

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 = sinh G

eG

2

G = 32

ln 2 + 3.5 ln 10 +1

2ln G.

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 functions

of 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) =m(v)c2.


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(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 Earths motion around the Sun?

The exact relativistic correction to the earths mass is given by

mex = m m= m

1

22 +

3

84 +

5

166 +

35

1288 + . . .

.

In approximation used in part (b) we just took first two terms into

account i.e. m= m(0) + m(1). Percentage error due to this isgiven by

mex mm

100% =

5

166 +

35

1288 + . . .

3.1 1023

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 0.99cto0.999c?

The proton mass is mp= 939 MeV/c2. The work done in increasing its

speed from v1 to v2 is

W =

11 21

11 22

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)


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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 1027 kg.me = 9.1094 1031 kg.m 4He = 6.6419 1027 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 1012J .A kilogram of protons contains

4 1.4947 1026 protonswhich release

1.4947 1026 4.5290 1012J= 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 of

methane. The usual units are KCal/mole).

We can calculate that 1000 kg of methane burns to produces how much

energy by looking up the values in CRC. The reaction is

CH4+ 2O2 CO2+ H2Owhich 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 energy

as that produced by fusing 1 kg of protons:

1000 6.7694 1014

5.586 1010 kg= 1.2119 107 kg .


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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 (0C) to boiling (100C)?

The amount of heat energy transfered to 1000 kg of water to raise its temperatur

from 0 C to 100 C is

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

This is equivalent to a mass of

4.67 109 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 1016 kg/s 1

2 1030 smu / kg (360024365) 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, then

classically it would require

1

2mec

2 = 4.1 1014 J .


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

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

T = (

1)mec

2 =1

2

mec2

= 32

= 0.745 .

12. A useful approximation (4 points)

(a) Show that for an extremely relativistic particle, the particle speedu differs from

the speed of light c by

u= c u= c2

m0c

2

E

2

wherem0 is the rest mass and E is the energy.

E= m0c2

=

1 m0c2

E 2

1 12

m0c

2

E

2for a particle with E >> m0c

2

u= c u c2

m0c

2

E

2

(b) Find ufor electrons produced by

i. MITs Bates Accelerator Center, whereE= 900 MeV.

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

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

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


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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)

French6, 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) + Mf(c, v)Where E is the energy of photons.

>From conservation of four-momentum:

P =P

we get two equations:

Mic= E/c + Mfc

0 =

E/c + Mfv

Eliminate E from above:

Mic= Mfv + Mfc

Mi/Mf=(1 + ) =

1 +

1

French6, Problem 6-9, page 201: A laser beam

(a) Energy radiated by laser per second

P0= 1020 s1

hc

= 33.1 W

Since laser is at rest w.r.t earth therefore this is also the rate at which

observer on earth observes the power radiated.


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(b) Let E be the total energy radiated by laser in form of photons. Byenergy conservation in earths frame

M c2 =(MM)c2 + E,and by momentum conservation

0 = Ec

+ (MM)v.

Eliminating E from above two equation we get

M c2 =(MM)c(c + v)which simplifies to

MMM

=

1 1 +

.

Expanding in powers of beta upto leading term we get

= MM v= M

M c.

Now

M= 1020 hc 10 yrs/c2 = 1.16 107 kg.

This gives

v= 3.48 m/s.

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

E=

1 1 +

E

where E is energy of photon in laser frame. Similarly rate at which photonare observed on earths frame is given by

r=

1 1 +

r

where r = 1020 s1. Therefore observed power on earth as a function ofbeta is given by

P =Er=1 1 +

Er =

1 1 +

P0.

Taking P =P0 P and expanding in power of we getP = 2P0= 2 33.1 1.16 108 W = 7.7 107 W .

So the rate seen on earth after 10 years has decreased by 7.7 107 W.

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


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(a) of the 0 is:

=1 GeV+ 135 MeV

135 MeV = 1135/135 = 8.4 = 0.9929

Before the decay we have the four-momentum:

mc(1, )

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 =mc2

E E =mc2

E=1

2mc2(1 + ) = 1131 MeV

E =1

2mc2(1 ) = 4 MeV

(b) This time well have:

E/c(1, cos , sin ) + E/c(1, +cos , sin ) =mc(1, , 0)

The last component of the above equation will give:

E= E

>From the zeroth component (energy conservation) well get:

2E/c = mc

>From the 1st component well get:

2E/c cos = mc

After deviding the above two equations we get:

cos = = 0.1192 Radians= 6.8

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

13.6.


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French6, Problem 6-14, page 202: Impossible processes

Factors of c are suppressed in this calculation.

(a) A single photon (with four-momentum p= (E, E, 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,

p+ pi= pf.

Squaring both sides,

m2e =m2e+ 2ppi

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

(b) A single photon (with four-momentum p= (E, E, 0, 0)) in empty spaceis transformed into an electron and a positron with four-momenta p1and p2 respectively. So,

p=p1+ p2

0 = 2m2e+ 2p1 p2 p1 p2= m2e

which is impossible because the scalar product of four-momenta corresponding

to 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 in

the center of mass frame of the electron-positron pair. Then the photon

must have zero momentum to conserve momentum but that would mean that

energy cannot be conserved.

(c) A fast positron and a stationary electron annihilate, producing only

one photon. This is the reverse process of (b) (Lorentz transformed

to 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 = (mp+K)/mp = 1.46 which givesv= 0.731 c. From energy conservation we get

2mp+ K= 2E


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

mpv= 2Ev cos

cos =(mp+ K)v

2Ev = 0.7426which gives = 42.045. Therefore inclusive angle is 2= 84.09.

French6, Problem 7-1, page 225: Kaon decay

Lets denote P1 the four-momentum of the K meson, P2 the moving and P3the at rest.

P1= mKKc(1, K)

P2 = mc(1, )

P3= mc(1, 0)

We have:

P1 = P2+ P3 (P1 P3)2 =P22m2Kc

2 + m2c2 2mc2mKK=m2c2

K= mK2m

= 1.8

EK=mKKc2 = 889 MeV

KK= 889 494 = 395 MeV

from Energy conservation:

E = 889 137 = 752 MeVK = 752 137 = 615 MeV

French6, 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 look

like (3mec, 0) since P2 is invariant under lorentz transformation well have:

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

= 9m2ec

2

0 + m2ec2 + 2meE= 9m

2ec

2

E= 4mec2 = 2.04 MeV

French6, Problem 7-4, page 226: elastic scattering


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

(a) A proton (with rest mass m) accelerated in a proton synchrotron to a kineticenergyKstrikes 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 the

kinetic 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 the

two protons is:

Ei= K+ 2mc2

pi=

K2 + 2Kmc2/c

After the collision we can treat all the products as a single particle

with mass M and velocity v. having energy and momentum:

Ef=M c2

pf = Mv

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

K+ 2mc2 = Mc2

K2 + 2Kmc2/c= M v

DEviding the two equations gives ,

= v/c=

K2 + 2Kmc2

K+ 2mc2 =

K

K+ 2mc2

We can then find :

=

1 +

K

2mc2

This gives us M c2, the total energy available for making particlesand endowing them with kinetic energy (i.e. the total energy in the

frame where the momenum is zero),

M c2 =K+ 2mc2

= 2mc2 =

1 +

K

2mc2


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(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 T eV

16. Center-of-mass collisions (RH) (5 points)(a) In modern experimental high-energy physics, energetic particles are made to

circulate 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= 2mc21 + K

2mc2 .

Four-momenum before the collision:

P=mc(1, ) + mc(1,) = 2mc(1, 0)After the collision the four-momenum is P with P2 =M2c2. But weknow that P2 =P2:

M2c2 = 4m22c2

>From the relation

K= (

1)mc2

you can find as a function of K:

= 1 + K

mc2

E=M c2 = 2mc2 = 2mc2(1 + Kmc2

)


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(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?

ForE= 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 the

advantages of storage rings.

This means that head-on collision has more advantages because we need

less Kinetic energy to produce a given mass or with a given Kinetic

energy we can produce more massive particles.

17. Compton scattering (5 points)

Following the discussion in lecture, consider Compton scattering, in which a photonwith wavelength 0 scatters from an electron at rest. After the collision, a photon ofwavelength is emitted at an angle relative to the direction of the initial photonand the electron recoils with the velocity and direction required by momentum and

energy conservation. Derive the relationship between the scattering angle and thewavelength change = = 0 for Compton scattering:

= h

mc(1 cos )

(a) Use invariants as in lecture.

A photon with four-momentum p= (E, E, 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 p = (E , Ecos , Esin , 0) where denotes the angle that the scattered photon makes with the direction of

the incident photon. The electron is scattered to some four-momentum pf.

p+ pi= p+ pf

p2f = (p+ pi p)2

m2e =m2e+ 2me(EE) 2EE(1 cos )


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E = mec2E

mec2 + (1 cos )E (with factors of c restored).

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

E=hc

, E =

hc

.

So,

= + h

mc(1 cos )

= hmc

(1 cos )where 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 ,p

sin , 0).Energy conservation gives

E+ me = E

+ E

e

Ee = E+ me EMomentum conservation along x-direction gives

E = E

cos +p cos

p cos = EEcos

and along y-direction

p sin = Esin .

Squaring and adding last two equation we get

(p)2 =E2+ (E

)2 2EEcos .

Subtracting the last equation from square of energy conservation equation

gives

(Ee)2 (p)2 = m2e+ 2me(E E) 2EE(1 cos )

me(EE) = EE(1 cos )

1E 1

E

= 1

me (1 cos )

= hmec

(1 cos ) (substituting factors of c back)

= hmec

(1 cos ).


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18. Production of hyperons at rest (7 points)

TheK meson and the 0 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 0 + 0can be used to make 0s 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, mKc

2 = 494 MeV,m0c

2 = 135 MeV, m0c2 = 1116 MeV.

(a) Find the energy of the incidentK beam required to just produce 0 hyperonsat rest in the lab.

Let pK, pp, p, p be the four-momenta of the K, p, 0, 0 particles respectivel

We know thatpK= (EK, |pK|) ,

pp = (mp, 0) ,

p = (m, 0) .

>From conservation of four-momentum,

pK+ pp= p+p

p2 = (pK+ pp p)2

m2 =m

2K+ m

2p+ m

2+ 2(mp

m)EK

2mpm

EK= m2Km2+ (m mp)2

2(m mp) 726 MeV .

(b) What is the 0 energy for this magicK energy?

E =EK+ Ep E= EK+ mp m 549 MeV .(c) Check momentum conservation.

|pK| =

E2Km2K 533MeV ,

|p| =

E2 m2 533MeV .The hyperon and proton have no 3-momentum. Hence, momentum is conserved

as long as the K and the are directed in the same direction.


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(d) Could the process be run the other way? That is, could a0 beam (assuming onewas available) be used to make a K+ at rest by the reaction 0 +p 0 + 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 the

following substitutions:

K , K , .

So,

E ==m2 m2+ (mKmp)2

2(mKmp) 1156 MeV> m,

E = E+ Ep EK=E+ mp mK 1601 MeV> m .Thus, the process can be run the other way.

19. Invariant Product (5 points)

Lorentz transformation for A and B are given by

a0 = a0cosh axsinh ax= axcosh a0sinh ay,z =ay,zb0 = b0cosh bxsinh bx= bxcosh b0sinh by,z =by,z

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

definition

A B = a0b0 axbx ayby azbz.

Using the transformation we get

A B = (a0cosh axsinh )(b0cosh bxsinh )(axcosh a0sinh )(bxcosh b0sinh ) ayby azbz

= a0b0cosh2 + axbxsinh

2 (a0bx+ axb0)cosh sinh axbxcosh2 a0b0sinh2 + (axb0+ a0bx)cosh sinh ayby azbz

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

= a0b0 axbx ayby azbz= A B

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

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