Assignment 8, solutions
Problem 1:
HEPA is an asymmetric electron proton collider located near
the city of Hamburg in Germany. The energy of the electron beam is 26 GeV and
the energy of the proton beam is 820 GeV. Ignore baryon and lepton number
conservation and calculate
(a) the maximum number of neutral pions (mass of
p0 = 134.98 MeV) that can be
produced in one proton-electron collision.
(b) What momentum would a beam of electrons incident on
protons at rest need to have to produce the same number of pions as in part (a)
Solution:
 | Concepts, principles, relations
that apply to the problem:
Relativistic collisions, energy and momentum conservation, frame
transformations |
| Why do they apply?
In relativistic collisions between free particles energy and momentum are
always conserved. Often the physics is best visualized in the center of
momentum frame. |
| How do they apply?
(a) If in the CM frame the reaction products are
at rest, they have the largest mass M possible.
What is M?
[E2/c2 - P2]lab
before coll. = [E2/c2]CM before coll. =
[E2/c2]CM after coll.
invariance of norm of (P0,P)
energy conservation
lab, before:
(Ep + Ee)2/c2 – (Pp – Pe)2
= (Ep2 + Ee2 + 2EpEe)/c2
– Pp2 – Pe2 + 2PePp
=
85280.85 GeV2 /c2,
since
(Ep + Ee)2/c2 = 715716 GeV2
/c2
Ep2/c2 - mp2c2
= Pp2, Pp2 = (8202 -
0.9382) GeV2/c2 = 672399.1 GeV2/c2,
Ee2/c2 – me2c2
= Pe2, Pe2 = (262 –
(5.11*10-4)2) GeV2/c2 = 676 GeV2/c2.
ECM = 292 GeV
# of pions = 292/0.13498 = 2163
(b) Assume we want to create the same number of pions. Then we
want
(Ep + Ee)2/c2 – Pe2
= (Ep2+ Ee2 + 2EpEe)/c2
– Pe2 = (mp2c4 + 2mpc2Ee)/c2
+ me2c
= 85280.85 GeV2 /c2.
Ee = (85280.85 GeV2 - mp2c4
- me2c4)/(2mpc2) = 45458
GeV |
| Details of the calculation:
None |
Problem 2:
A pion (p) decays
at rest into a muon (m) and a neutrino (n).
In terms of the masses mp
and mm (use the approximation
mn = 0), find:
(a) the momentum, energy, and velocity of the outgoing
muon.
(b) In the rest frame of the outgoing muon, what is the
energy of the neutrino? What was the initial velocity of the pion in this
frame?
Solution:
| Concepts, principles, relations
that apply to the problem:
Conservation of energy and momentum, the Lorentz transformation |
| Why do they apply?
The decay of a particle is a
relativistic problem. In relativistic "collisions" energy
and momentum are always conserved. |
| How do they apply?
(a) In the rest frame of the decaying pion let the muon
move in the positive x-direction and the neutrino move in the negative
x-direction.
Energy conservation: Mpc2 = (Mm2c4
+ p2c2)1/2 + En.
Here p is the momentum of the muon.
Momentum conservation: pc = En.
Combine: (Mpc2 - En)2 = Mp2c4 + En2
- 2Mpc2En
= Mm2c4 + En2
En = (1/2)Mpc2
- (1/2)Mm2c2/Mp
= (c2/(2Mp))(Mp2 - Mm2)
p = En/c
= (c/(2Mp))(Mp2 - Mm2) = momentum of the outgoing muon.
Em = Mpc2
- En = (c2/(2Mp))(Mp2
+ Mm2) = energy of
the outgoing muon.
v = pc2/Em = c(Mp2
- Mm2)/ (Mp2
+ Mm2) = speed of
the outgoing muon.
(b) The muon moves with respect to the original rest frame
of the pion with speed v.
The original speed of the pion in the muons rest frame
therefore was v. It was moving in the negative x-direction. The relative speed
of the two frames is v.
b = v/c = (Mp2
- Mm2)/ (Mp2
+ Mm2)
g = (1-v2/c2)-1/2
= (Mp2 + Mm2)/(2MpMm)
Lorentz transformation of the (p0,p) =
(En/c, -i En/c)
4-vector from the original rest frame of the pion to the rest frame of the muon:
En’/c =
gEn/c
+ bg En/c = (Mp/Mm)En
is the energy of the neutrino in the rest frame of the muon. |
| Details of the calculation:
None |
Problem 3:
A f particle (mf = 1.020GeV/c2)
has a momentum of 3GeV/c along the z-axis in the laboratory frame.
It
decays f ® K+K- into two
charged kaons (mK+ = mK- = 0.494GeV/c2
).
(a) Calculate the momentum PK of each kaon in the
laboratory frame if the decay axis coincides with the z-axis.
(b) Calculate the momentum PK of each kaon in the
laboratory frame if the decay axis is perpendicular to the z-axis.
Solution:
| Concepts, principles, relations that apply to the
problem:
Relativistic "collisions", energy and momentum conservation |
| Why do they apply?
The decay of a particle is a
relativistic problem. In relativistic "collisions" energy
and momentum are always conserved. Often the physics is best visualized in the center of
momentum frame. |
| How do they apply?
In the rest frame of the mf, the
4-vector momentum of the f is (E/c,p) = (mfc,
0).
Each component of the 4-vector is conserved. After the decay we
therefore have for each particle:
Ek2 = (1/4)mf2c4
= mk2c4 + pk2c2,
pk2 = c2((1/4)mf2
- mk2).
Ek = 0.51 GeV, pk = 0.127 GeV/c.
The laboratory frame moves with respect to the rest frame in the negative
z-direction with speed v.
Ef' = gEf,
Ef'2 = g2Ef2,
mf2c4 + pf2c2
= g2Ef2,
g2 = [1.022 + 32]/1.022
, g = 3.1.
b = 0.947, v = 0.947c. |
| Details of the calculation:
We can now find the momentum of each kaon in the laboratory frame.
(a) Particle 1: E = Ek, pz = pk, px
= py = 0. pz' = 1.89 GeV/c.
Particle 2: E = Ek, pz = -pk, px
= py = 0. pz' = 1.11 GeV/c.
(b) Particle 1: E = Ek, pz = 0, px
= pk, py = 0. pz' = 1.5 GeV/c, px'
= 0.127 GeV/c.
Particle 2: E = Ek, pz = 0, px = -pk,
py = 0. pz' = 1.5 GeV/c, px' = -0.127
GeV/c.
For both particles: |p'| = (px'2 +pz'2)1/2
= 1.505 GeV/c. |
Problem 4:
In the following, two examples of a decaying system are
presented, where the decay products travel with velocities comparable to the
speed of light c.
1. Two electrons are ejected simultaneously in opposite
directions from an atom. Each electron has a speed as measured by a laboratory
observer of 0.5c. What is the speed of one electron as seen from the rest frame
of the other electron
(a) in the classical approach?
(b) in the relativistic approach?
Distinguish carefully the velocities in the respective
frames.
2. The neutral pi meson, p°,
has a rest mass of 135 MeV/c2. It decays into two photons (g
rays) of equal energy and opposite direction in its rest frame. In the
laboratory frame the p° is moving with a
total energy 25% larger than its rest energy.
(a) What are the energies of the g
rays, as measured in the laboratory, if the decay process causes them to be
emitted in opposite directions along the pion's original line of motion?
(b) What is the velocity of each
g rays as observed by the other?
Solution:
Problem 5:
A reference frame K' is moving with uniform velocity v = vi
with respect to reference frame K.
(a) In K, a plane wave with angular frequency w is
traveling in the i direction. What is its frequency in K'?
(b) In K, a plane wave with angular frequency w is
traveling in the j direction. What is its frequency in K'?
(c) In K, a plane wave with angular frequency w is
traveling in a direction making an angle of 45o with respect to the
i direction and the j direction. What is its frequency in K'?
Solution:
| Concepts, principles, relations that apply to the
problem:
The relativistic Doppler shift |
| Why do they apply?
Given the angular frequency w of a plane wave
in reference frame K we are asked to find the angular frequency w'
in reference frame K' moving with constant velocity with respect to K. |
| How do they apply?
(a)
w' = w[(1 - v/c)/(1 +
v/c)]1/2 if k and v are parallel to each other.
(b)
w' = gw if k and
v are perpendicular to each other. g
= (1 - v2/c2)-1/2.
There is a transverse Doppler shift, even if q =
p/2.
(c)
w' = g(w
- v×k) = gw(1 -
vcosq/c) = gw(1 -
v/(Ö2c)). |
|
Details of the calculation:
None |
Problem 6:
A, located on earth, signals with a laser pulse every six minutes. B
is on a space station that is stationary with respect to earth. C is in
a rocket traveling from A to B with a velocity v = 0.6c
relative to A.
(a) At what intervals does B receive signals from A?
(b) At what intervals does C receive signals from A?
(c) If C sends light pulses back to A, at what intervals does
A receive these pulses?
Solution:
| Concepts, principles, relations that apply to the
problem:
The Doppler shift |
| Why do they apply?
We are given the frequency of a signal in one inertial frame and are asked to
find it in another inertial frame.
f' = f[(1 - v/c)/(1 + v/c)]1/2 if the two frames recede from each
other. |
| How do they apply?
(a) B receives a signal every six minutes.
(b) f' = f[(1 - 0.6)/(1 + 0.6)]1/2 = 0.5f, T' = 2T, C
receives a signal every 12 minutes.
(c) f'' = 0.5f' = 0.25 f, since A recedes from C with speed v. A
receives a pulse every 24 minutes. |
| Details of the calculation:
None |
Problem 7:
A source emits electromagnetic waves with frequency f into a 4p
solid angle. What is the frequency f' of the waves observed by an observer
moving with speed v in a circular orbit around the source?
Solution:
| Concepts, principles, relations that apply to the
problem:
Transverse Doppler shift |
| Why do they apply?
The time interval between successive crests reaching the observer as
seen in the rest frame of the source is T.
In the frame of the observer it is the proper time interval
t = T/g.
We have t < T, f' > f. It is only the
instantaneous velocity which is important in calculating the time dilation. |
| How do they apply?
Let k be the wavevector and v the relative velocity;
f' = gf if k
and
v are perpendicular to each other. |
| Details of the calculation:
None |