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 = γmc2.  γ = 100 = (1 - v2/c2)-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 = N0exp(t/τγ) = N0exp(-2).  The fraction reaching the detector is N/N0 = 13.5%.

Problem 2:

In a proton-proton collision a π+ meson can be created through the reaction

p1 + p2 --> p + n + π+.

In the center of mass (CM) frame of reference each proton has an initial energy γmpc2, where mp is the mass of the proton and γ = (1 - β2)-1/2, with β = v/c.
Take the mass of the proton and the mass of the neutron to be 1837 me (me = electron mass) and the mass of the pion to be 273 me = 0.1486 mp.
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:2γmpc2 = 2.1486 mpc2.γ = 1.0743.v = 0.365 c = 1.1*108 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 xy-plane and makes an angle θ = 7.2o with the x axis?  When viewed from the spaceship, the structure is aligned with the x-axis.
(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.

 Event: coordinates in K coordinates in K' (1)  left edge at t = 0 (0, 0, 0, 0) (0, 0, 0, 0) (2)  right edge at t = 0 (0, Lcosθ, Lsinθ, 0) (-γβLcosθ, γLcosθ, Lsinθ, 0) (3)  left edge at t = t (ct, 0, ut, 0) (γct, -γβct, ut, 0) (4)  right edge at t = t (ct,  Lcosθ , Lsinθ + ut, 0) (γct - γβLcosθ, -γβct + γLcosθ, Lsinθ + ut, 0)

From events 1 and 3 we find vx' = ∆x'/∆t' = -γβct/γt = -βc = -v.
From events 1 and 3 we find vy' = ∆y'/∆t' = ut/γt = u/γ.
In K' at t' = 0 the left edge of the stick is at x' = 0, y' = 0.
At t' = -γβLcosθ/c the right edge of the stick is at x' = γLcosθ, y' = Lsinθ.
The stick travels with velocity v' = -v i + (u/γ) j.
Therefore  at t' = 0 the right edge of the stick is at  x' = γLcosθ - vγβL/c,  y' = Lsinθ + (u/γ)γβL/c,
or x' = γL(cosθ - β2),  y' = L(sinθ + uv/c2).
For y’ = 0 at t’ = 0 we need sinθ + uv/c2 = 0,  v = -sinθ c2/u = -2 sin7.2o c = -0.25 c.
The ship is traveling in the negative x direction with velocity v = -0.25c i with respect to the station.

(b)  At t’ = 0 the right edge of the stick is at x’ = 0, y’ = 0, the right edge of the stick is at
x' = γL(cosθ - β2), y’ = 0, the length of the stick is L' = γL(cosθ - β2) = 0.96 L = 9.6 km.

Problem 4:

A pion (mass Mp) decays into a muon (mass Mμ) 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 τ1 in the pion's rest frame.  Find the lifetime t0 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:  Mpc2 = (Mμ2c4 + p2c2)1/2 + En.Here p is the momentum of the muon.Momentum conservation:  pc = En.Combine:  (Mpc2 - En)2 = Mp2c4 + En2 - 2Mpc2En = Mμ2c4 + En2.En = ½Mpc2 - ½Mμ2c2/Mp.Eμ = Mpc2 - En = ½Mpc2 + ½ Mμ2c2/ Mp.T = Eμ - Mμc2 = (Mp - Mμ)2c2/(2Mp). (b)  τ0 = τ1/γ.  We need to find γ.T = (Mp - Mμ)2c2/(2Mp) = (γ - 1)Mμc2. (γ - 1) = (Mp - Mμ)2/(2MμMp),  γ = ( Mp2 + Mμ2)(/2MμMp), τ0 = τ12MμMp/( Mp2 + Mμ2).

Problem 5:

An excited nucleus of 57Fe formed by the radioactive decay of 57Co emits a gamma ray of 1.44 *104 eV.  In the process, there is conservation of energy and m0c2 = γma0c2 + hf,  where m0c2 is the initial mass of the nucleus and ma0c2 is its mass after the emission of the gamma ray.  There is also conservation of momentum, hf/c = γma0u, where u is the recoil velocity of the iron nucleus.  The energy released by the reaction is Er = (m0 - ma0)c2.
(a)  Show that  hf = Er(m0 + ma0)/(2m0) = (1 - Er/(2m0c2))Er.
Thus hf < Er: part of Er goes to the photon, and the other part supplies kinetic energy to the recoiling nucleus.

(b)  Set m0 = 57*1.7*10-27 kg, and show that Er/(2m0c2)) ~ 1.3*10-7.
Thus the fraction of the available energy Er 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 57Fe 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 57Fe moves in the direction of a sample of normal 57Fe, 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 1013?  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 gamma-ray detector placed behind the sample of normal 57Fe when the source of activated Fe moves
(i) toward the normal 57Fe,
(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 = mc2 + ½mv2 + O(v4) = mc2 + mgh + O(v4).
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 = mc2/(mc2 + mgh + O(v4)) = (1 + gh/c2 + O(v4))-1 = 1 - gh/c2 + O(v4).]

(g)  A normal 57Fe 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:  m0c2 = γma0c2 + hf.    (1) momentum conservation:  hf/c = γma0u.            (2) Define:  Er = (m0 - ma0)c2.                                (3) Show:  hf = Er(m0 - ma0)/(2m0). m0c2 - hf = γma0c2.                                          (1) m02c4 + h2f2 - 2m0c2hf = γ2ma02c4.                    square (1) -2m0hf = γ2ma02c2 - m02c2 - h2f2/c2 = γ2ma02c2 - m02c2 - γ2ma02u2 = γ2ma02(c2 - u2) - m02c2 = ma02c2 - m02c2.         insert (2) 2m0hf = (ma0 - m0)(ma0 + m0)c2 = Er(ma0 + m0).   rearrange hf = Er(ma0 + m0)/(2m0) = Er(ma0 - m0 + m0 + m0)/(2m0) = Er(1 - Er/(2m0c2)). (b)  Set m0 = 57*1.7*10-27 kg,  hf = 1.44*104 eV. Er - Er2/(2m0c2)) - hf = 0.  Er2 - 2m0c2Er + 2m0c2hf = 0. Er = m0c2 - ((m0c2)2 - 2m0c2hf)1/2. m0c2 = 8.72*10-9 J,  hf = 2.3*10-15 J,  Er = 2.3 *10-15 J. Er/(2m0c2) ~ 1.3*10-7. (c)  If m0 = 10-3 kg, then m0c2 = 9*1013 J,  Er/(2m0c2) ~ 1.27*10-29.The fraction that appears as recoil energy is now negligible.The gamma ray energy is now hf = Er.  It has increased by Er2/(2m0c2)) = 1.3*10-7 Er.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 gamma-ray detector increases for both directions, because the sample 57Fe absorbs less. (f)  A  14.4 keV gamma emitted from 57Fe ray travels vertically upward in a uniform gravitational field.We have Δf/f = -gΔh/c2 from E2/E1 = 1 - g(h2 - h1)/c2.  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.