Assignment 7, solutions
Problem 1:
A monochromatic particle beam consist of particles whose
total energy is 100 times their rest energy. The rest lifetime of the particles
is 0.10 ns. In the laboratory, the distance between the point where the
particles are generated and the detector is 6.0 m.
(a) What is the
speed of the particles?
(b) What fraction of the generated particles
reaches the detector?
Solution:
Concepts: Proper time, mean life  
Reasoning: We are given the proper time interval τ (mean lifetime) in frame 1 and the energy in frame 2. We must find the speed and mean lifetime in frame 2.  
Details of the calculation: (a) E = γmc^{2}. γ = 100 = (1  v^{2}/c^{2})^{1/2}, v = 0.99995 c ~ c. (b) In the laboratory frame the lifetime of the particles is τγ = 10 ns. If they do not decay, they reach the detector after t = (6 m)/c = 20 ns. The number of particles left after 20 ns is N = N_{0}exp(t/τγ) = N_{0}exp(2). The fraction reaching the detector is N/N_{0} = 13.5%. 
Problem 2:
In a protonproton collision a π^{+} meson can be created through the reaction
p_{1} + p_{2} > p + n + π^{+}.
In the center of mass (CM) frame of reference each proton
has an initial energy γm_{p}c^{2}, where m_{p} is the
mass of the proton and γ = (1  β^{2})^{1/2}, with β = v/c.
Take the mass of the proton and the mass of the neutron to be 1837 m_{e}
(m_{e} = electron mass) and the mass of the pion to be 273 m_{e}
= 0.1486 m_{p}.
Determine the minimum value of the initial velocity v
for which π^{+} creation is possible.
Solution:
Concepts: Energy and momentum conservation  
Reasoning: In relativistic collisions between free particles energy and momentum are always conserved.  
Details of the calculation: Momentum conservation: In the CM frame the total momentum is zero before and after the collision. For the minimum value of the initial velocity v the three particles are at rest after the collision. Energy conservation: 2γm_{p}c^{2} = 2.1486 m_{p}c^{2}. γ = 1.0743. v = 0.365 c = 1.1*10^{8} m/s. 
Problem 3:
A spaceship travels with velocity v = vi
with respect to a space
station. In the frame of the space station a linear structure of
length L = 10 km moves with velocity u = (c/2)j in the positive
y direction. The structure lies in the xyplane and makes an
angle θ = 7.2^{o} with the x axis?
When viewed from the spaceship,
the structure is aligned with the xaxis.
(a) What is the velocity v of the spaceship (magnitude and direction)?
(b) What is the length of the structure in the frame of the spaceship'?
Solution:
Concepts: Relativistic kinematics  
Reasoning: We are asked to relate observations in different inertial frames.  
Details of the calculation: (a) Let us look at the positions of the left and the right ends of the structure at time 0 and at time t in frames K (space station) and K' (spaceship). To transform the coordinates of an event from K to K' we use the matrix below.

Problem 4:
A pion (mass M_{p}) decays into a muon (mass M_{μ}) and a
neutrino (massless).
(a) Find the kinetic energy of the muon in the
rest frame of the pion.
(b) The muon is unstable, with a lifetime τ_{1}
in the pion's rest frame. Find the lifetime t_{0} of the muon in
its own rest frame.
Solution:
Concepts: Relativistic "collisions", energy and momentum conservation  
Reasoning: The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved.  
Details of the calculation: (a) Energy conservation: M_{p}c^{2} = (M_{μ}^{2}c^{4} + p^{2}c^{2})^{1/2} + E_{n}. Here p is the momentum of the muon. Momentum conservation: pc = E_{n}. Combine: (M_{p}c^{2}  E_{n})^{2} = M_{p}^{2}c^{4} + E_{n}^{2}  2M_{p}c^{2}E_{n} = M_{μ}^{2}c^{4} + E_{n}^{2}.^{}E_{n} = ½M_{p}c^{2}  ½M_{μ}^{2}c^{2}/M_{p}. E_{μ} = M_{p}c^{2}  E_{n} = ½M_{p}c^{2} + ½ M_{μ}^{2}c^{2}/ M_{p}. T = E_{μ}  M_{μ}c^{2} = (M_{p}  M_{μ})^{2}c^{2}/(2M_{p}). (b) τ_{0} = τ_{1}/γ. We need to find γ. T = (M_{p}  M_{μ})^{2}c^{2}/(2M_{p}) = (γ  1)M_{μ}c^{2}. (γ  1) = (M_{p}  M_{μ})^{2}/(2M_{μ}M_{p}), γ = ( M_{p}^{2} + M_{μ}^{2})(/2M_{μ}M_{p}), τ_{0} = τ_{1}2M_{μ}M_{p}/( M_{p}^{2} + M_{μ}^{2}). 
Problem 5:
An excited nucleus of ^{57}Fe
formed by the radioactive decay of ^{57}Co emits a gamma ray of 1.44 *10^{4} eV. In the process, there is conservation of energy and m_{0}c^{2}
= γm_{a0}c^{2}
+ hf, where m_{0}c^{2}
is the initial mass of the nucleus and m_{a0}c^{2}
is its mass after the emission of the gamma ray. There is also
conservation of momentum, hf/c
= γm_{a0}u,
where u is the recoil velocity of the iron nucleus. The energy released by
the reaction is E_{r} = (m_{0}
 m_{a0})c^{2}.
(a) Show that
hf
= E_{r}(m_{0}
+ m_{a0})/(2m_{0})
= (1  E_{r}/(2m_{0}c^{2}))E_{r}.
Thus hf
< E_{r}:
part of E_{r}
goes to the photon, and the other part supplies kinetic energy to the recoiling
nucleus.
(b) Set m_{0} = 57*1.7*10^{27} kg, and show that E_{r}/(2m_{0}c^{2}))
~ 1.3*10^{7}.
Thus the fraction of the available energy E_{r}
that appears as recoil is small.
(c) Mössbauer discovered in 1958 that, with solid iron, a significant fraction
of the atoms recoil as if they were locked rigidly to the rest of the solid.
This is the Mössbauer
effect. If the sample has a mass of 1 gram, by what fraction is the gamma
ray energy shifted in the recoil process?
(d) A sample of normal ^{57}Fe
absorbs gamma rays of 14.4 keV by the inverse recoilless process much more
strongly than it absorbs gamma rays of any nearby energy. The excited
nuclei thus formed reemit 14.4 keV radiation in random directions some time
later. This is resonant scattering. If a sample of activated ^{57}Fe
moves in the direction of a sample of normal ^{57}Fe,
what must be the value of the velocity v that will shift the frequency of the
gamma rays, as seen by the normal nuclei, by 3 parts in 10^{13}?
This is one line width.
(e) A Doppler shift in the gamma ray results in a much lower absorption by a
nucleus if the shift is of the order of one line width or more. What
happens to the counting rate of a gammaray detector placed behind the sample of
normal ^{57}Fe
when the source of activated Fe moves
(i) toward the normal ^{57}Fe,
(ii) away from it?
(f) If a 14.4 keV gamma ray travels 22.5 meters vertically upward, by what
fraction will its energy decrease?
[Gravitation redshift, a thought experiment: Suppose
a particle of rest mass m is dropped from the top of a tower and falls freely with
acceleration g. It reaches the ground with a velocity v = (2gh)^{1/2},
so its total energy E, as measured by an observer at the foot of the tower is E
= mc^{2} + ½mv^{2} + O(v^{4}) = mc^{2} + mgh
+ O(v^{4}).
Suppose an observer has some magical method of converting all this energy into a
photon of the same energy. Upon its arrival at the top of the tower with
energy E the photon is again
magically changed into a particle of rest mass
m' =
E'/c. Energy conservation requires
that m' = m.
Therefore
E'/E = mc^{2}/(mc^{2}
+ mgh + O(v^{4})) = (1 + gh/c^{2} + O(v^{4}))^{1}
= 1  gh/c^{2} + O(v^{4}).]
(g) A normal ^{57}Fe absorber located at this height must move in
what direction and at what speed in order for resonant scattering to occur?
Solution:
Concepts: Relativistic decay, energy and momentum conservation, the Doppler shift  
Reasoning: The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. The Doppler shift can compensate for the gravitational redshift to produce optimum absorption.  
Details of the calculation: (a) energy conservation: m_{0}c^{2} = γm_{a0}c^{2} + hf. (1) momentum conservation: hf/c = γm_{a0}u. (2) Define: E_{r} = (m_{0}  m_{a0})c^{2}. (3) Show: hf = E_{r}(m_{0}  m_{a0})/(2m_{0}). m_{0}c^{2}  hf = γm_{a0}c^{2}. (1) m_{0}^{2}c^{4} + h^{2}f^{2}  2m_{0}c^{2}hf = γ^{2}m_{a0}^{2}c^{4}. square (1) 2m_{0}hf = γ^{2}m_{a0}^{2}c^{2}  m_{0}^{2}c^{2}  h^{2}f^{2}/c^{2} = γ^{2}m_{a0}^{2}c^{2}  m_{0}^{2}c^{2}  γ^{2}m_{a0}^{2}u^{2} = γ^{2}m_{a0}^{2}(c^{2}  u^{2})  m_{0}^{2}c^{2 }= m_{a0}^{2}c^{2}  m_{0}^{2}c^{2}. insert (2) 2m_{0}hf = (m_{a0}  m_{0})(m_{a0} + m_{0})c^{2} = E_{r}(m_{a0} + m_{0}). rearrange hf = E_{r}(m_{a0} + m_{0})/(2m_{0}) = E_{r}(m_{a0}  m_{0 }+ m_{0 }+ m_{0})/(2m_{0}) = E_{r}(1  E_{r}/(2m_{0}c^{2})). (b) Set m_{0} = 57*1.7*10^{27} kg, hf = 1.44*10^{4} eV. E_{r}  E_{r}^{2}/(2m_{0}c^{2}))  hf = 0. E_{r}^{2}  2m_{0}c^{2}E_{r} + 2m_{0}c^{2}hf = 0. E_{r} = m_{0}c^{2}  ((m_{0}c^{2})^{2}  2m_{0}c^{2}hf)^{1/2}. m_{0}c^{2} = 8.72*10^{9} J, hf = 2.3*10^{15} J, E_{r} = 2.3 *10^{15} J. E_{r}/(2m_{0}c^{2}) ~ 1.3*10^{7}. (c) If m_{0} = 10^{3} kg, then m_{0}c^{2} = 9*10^{13} J, E_{r}/(2m_{0}c^{2}) ~ 1.27*10^{29}. The fraction that appears as recoil energy is now negligible. The gamma ray energy is now hf = E_{r}. It has increased by E_{r}^{2}/(2m_{0}c^{2})) = 1.3*10^{7 }E_{r}. Δf/f = 1.3*10^{7}. (d) We want Δf/f = 3*10^{13}. Δf/f = (f'  f)/f = [(1 + v/c)/(1  v/c)]^{1/2}  1 if the samples approach each other. We can solve for v/c ~ 3*10^{13}. v ~ 10^{4} m/s. (e) The counting rate of a gammaray detector increases for both directions, because the sample ^{57}Fe absorbs less. (f) A 14.4 keV gamma emitted from ^{57}Fe ray travels vertically upward in a uniform gravitational field. We have Δf/f = gΔh/c^{2} from E_{2}/E_{1} = 1  g(h_{2 } h_{1})/c^{2}. Δf/f = 2.45*10^{15}. Such a shift can also be produced if the absorber moves away from the source. We would need Δf/f = [(1  v/c)/(1 + v/c)]^{1/2} 1 = 2.45*10^{15}, v/c = 2.45*10^{15} away from source. (g) The absorber has to move towards the source with v/c = 2.45*10^{15} in order to increase, in its own frame, the frequency for optimum absorption, because the frequency was decreased by the gravitational redshift. If the absorber is at rest we still have absorption, since Δf/f_{g} << 3*10^{13}. 