Problems

Problem 1:

Reference frame K' moves with velocity vi with respect to reference frame K.  An electromagnetic plane wave is observed in K propagating in a direction -i + j with frequency n.   Find the frequency and direction of propagation of the plane wave when it is observed in K'.

Solution:

Concepts, principles, relations that apply to the problem:
The Doppler shift, the Lorentz transformation of the (k0,k) 4-vector
Why do they apply?
If we picture the wave as a series of crests and troughs moving through space, then the phase of any labeled point is a relativistically invariant quantity.  Observers in K and K' will agree on the phase of the a labeled point, i.e. they will agree if the point is on a crest, in a trough, etc.
We label the phase by f = k×r-wt = k'×r'-w't'.
Let  w = k0c, then k0ct - k×r = k0'ct' - k'×r',  (k0,k)×(ct,r) = ( k0',k')×(ct',r') for any 4-vector (ct,r).  The dot product of (k0,k) with any 4-vector yields a Lorentz invariant quantity.  Therefore (k0,k) is itself a 4-vector.
The angular frequency w of a sinusoidal electromagnetic wave with wave vector k (k = 2p/l = w/c) in a reference frame K is measured as w' in a reference frame K' moving with uniform velocity v with respect to K.
w' = gw(1 - (v/c)cosq), where q is the angle between the directions of k and v.
In frame K we have tanq  = ky/kx.  In frame K' we have tanq'  = ky'/kx'.  We find ky' and kx' by transforming the (k0,k) 4-vector.  Here k0 = w/c and |k| = 2p/l = w/c.
How do they apply?
q = 135o is the angle k makes with the x-axis.  The frequency of the plane wave in K' is n' = gn(1 + v/(21/2c)).
The Lorentz transformation of the (k0,k) 4-vector yields
.

tanq'  = ky'/kx' = -1/[g(bÖ2  + 1)] is the angle the wave vector makes with the x-axis in K'.  q' defines the direction of propagation of the plane wave in K'.

Details of the calculation:
None

Problem 2:

An energetic proton collides with a proton at rest.  What is the minimum energy the incident proton must have to make the reaction  p + p --> p + p + p + p possible?

Solution:

Concepts, principles, relations that apply to the problem:
Relativistic collisions, energy and momentum conservation, frame transformations
For each component pm of the 4-vector (p0,p1,p2,p3) we have
,

where i denotes the particles going into the collision and j denotes the particles emerging from the collision.

For transformations between reference frames we have
.

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?
The proton has minimum energy when the reaction products are at rest in the CM frame.
CM frame:

In the CM frame after the collision we have for the length of the momentum 4-vector
P02 - P2 = P02  = (Sp0)2 = (4mc)2 = 16m2c2.
Details of the calculation:
Lab frame:

The "length" of the momentum 4-vector is 16m2c2 before and after the collision.
Before the collision we have
P02 - P2 = (p0a + p0b)2 - (pa + pb)2 = p0a2 + p0b2 - pa2 + 2p0ap0b.
For any free particle we have p02- p2 = m2c2.  We therefore have:
P02 - P2 = 16m2c2 = 2m2c2 + 2p0ap0b,  p0ap0b = 7m2c2.  p0amc = 7m2c2.
Ea/c = 7mc.  mc2 = 938 MeV,  Ea = 6.6 GeV
The proton must have Ea - mc2 = 6mc2 of kinetic energy to make the reaction possible.

Problem 3:

A photon of energy E (massless) hits a proton of mass Mp at rest.  After the collision the photon is converted into an e+e- pair.  Assuming that the electron and positron each have rest mass me, what is the largest possible value of the recoil momentum of the proton?

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?
The proton will have the largest possible value of recoil momentum, if in the CM frame the electron and the positron move as a unit with minimum internal energy (positronium) in the opposite direction.

Neglect the binding energy of the positronium.  Let m1 = me+ + me- , m2 = Mp.
In the laboratory frame we have:
energy conservation:  E + m2c2 = (m12c4 + p12c2)1/2 + (m22c4 + p22c2)1/2,
momentum conservation:  (E/c)i = p1i + p2i.
Details of the calculation:
Let c = 1.  Then E + m2 = (m12 + p12)1/2 + (m22 + p22)1/2,  E = p1 + p2. 
We need to eliminate p1 and solve for p2 in terms of E, m1, and m2.
E2 + m22 + 2Em2 = m12 + p12 + m22 + p22 + 2(m12 + p12)1/2 (m22 + p22)1/2.
E2 = p12 + p22 + 2p1p2.
(2p1p2 + 2Em2 - m12)2 = 4(m12 + p12)(m22 + p22).
Write out all the terms, and substitute p1 = E - p2.  Obtain a quadratic equation for p2.
p22 + Ap2 + B = 0,  p2 = -(A/2) ± [(A/2)2 - B]1/2.  With m1 << m2 we have
A = E(m12 - 2m22 - 2Em2)/(m22 + 2Em2) » -E(2m2 + 2E)/(m2 + 2E) ,
B = m12(m22 + Em2 - m12/4)/(m22 + 2Em2) » m12(m2 + E)/(m2 + 2E)
p2max = -(A/2) + [(A/2)2 - B]1/2.

Problem 4:

A particle of mass 2m and kinetic energy 8mc2 collides with a particle of mass m at rest in the laboratory.  A particle of mass 3m and a particle of mass M emerge from the collision. What is the maximum possible value of M?

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?
M takes on its maximum value if in the CM frame the reaction products are at rest.
[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: 
(E2m + Em)2/c2 - P2m2 = (E2m2+ Em2 + 2E2mEm)/c2 - P2m2 

= 4m2c2 + P2m2 + m2c2 + 2E2mm - P2m2 = 5m2c2  + 2E2mm = 25m2c2
since E2m = 2mc2 + 8mc2 = 10m2c2

CM, after:
E2/c2 = 25m2c2,  E = 5mc2,  Mc2 = (5 - 3)mc2 = 2mc2

2m is the maximum possible value for M.

Details of the calculation:
None

Problem 5:

A particle of rest mass 1 MeV/c2 and kinetic energy 2 MeV collides with a stationary particle of rest mass 2 MeV/c2. After the collision, the two particles stick together.
(a) What are the energy, velocity and momentum of the incoming particle?
(b) What is the rest energy of the outgoing combined particle?

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)  Let mc2 = 1MeV.
Em = mc2 + 2mc2 = 3mc2
The energy of the incoming particle is 3 MeV.
g = 3,  1 - v2/c2 = 1/9,  v2/c2 = 8/9,  v = 0.9428c.
The velocity of the incoming particle is 0.9428c.
E2 = m2c4 + p2c2,  9m2c4 = m2c4 + p2c2, p2 = 8m2c2,  p = 2.828 MeV/c.
The momentum of the incoming particle is 2.828 MeV/c.
(b)  [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: 
(Em + E2m)2/c2 - Pm2 = (E2m2+ Em2 + 2E2mEm)/c2 - Pm2 

= 4m2c2 + Pm2 + m2c2 + 4Emm - Pm2 = 5m2c2  + 4Emm = 17m2c2
since Em = mc2 + 2mc2 = 3mc2

CM, after:
E2/c2 = 17m2c2,  E = 171/2mc2 = 4.123 MeV.

4.123 MeV is the rest energy of the outgoing particle.

Details of the calculation:
None

Problem 6:

A particle of rest mass M disintegrates at rest into a particle of rest mass m1, a particle of rest mass m2, and a high-energy photon (a gamma ray).  Find the maximum possible energy of the gamma ray in the rest frame of the initial particle of rest mass M.

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.
How do they apply?

The g-ray will have its maximum possible energy if after the disintegration the two particles have no relative kinetic energy.
Details of the calculation:
Energy conservation:  Mc2 = (m2c4 + p2c2)1/2 + Eg.
Here m = m1 + m2.
Momentum conservation:  pc = Eg.
Combine:  Eg = (1/2)Mc2 - (1/2)m2c2/M.
If m = M, then  Eg = 0.

Problem 7:

A p0 meson decays into two g-rays while at rest or in flight:  p0 ® g + g.

(a)  If the decaying p0 has velocity v and rest mass mp , and the g-ray is emitted at an angle q with respect to the original direction of the p0, find the g-ray energy as a function of mp , v, and q.
(b)  What is the maximum and minimum energy an emitted g-ray can have, and at what emission angles do these occur?
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.
How do they apply?
(a)  In the laboratory, we have from energy and momentum conservation:
gmc2 = hf1 + hf2,  gmv = hf1cosq1/c + hf2cosq2/c,  hf1sinq1/c = hf2sinq2/c.  Here m is the mass of the p0.
We want to find hf1 as a function of q1.
Details of the calculation:
Eliminate cosq2: (gmv -hf1cosq1/c)2 = (hf2/c)2(1-sin2q2) = (hf2/c)2 - (hf1/c)2sin2q1,
Eliminate hf2: (gmv -hf1cosq1/c)2 + (hf1/c)2sin2q1 = (hf2/c)2 = ( gmc- hf1/c)2.
Write out all the terms:  g2m2(v2 - c2) - 2gmhf1((v/c)cosq1 - 1) = 0.
 hf1 = mc2/[2g(1-(v/c)cosq1)].
(b)   hf1max = mc2/[2g(1-(v/c))].  cosq1 = 1, q1 = 0,  Emax = (mc2/2)[(1 + v/c)/(1 - v/c)]1/2.
 hf1min = mc2/[2g(1+(v/c))].  cosq1 = -1, q1 = 180,  Emin = (mc2/2)[(1 - v/c)/(1 + v/c)]1/2.
In the rest frame of the p0 the two g's are  emitted with energies (mc2/2) in the forward and backward directions.  In the lab frame they are Doppler shifted.

Problem 8:

Assume a photon with energy hn incident along the y-direction scatters off an electron with momentum p0 along the x-direction.  After the scattering the photon travels in the x-direction.

(a)  Find an expression for the energy of the scattered photon hn' in terms of hn and p0.
(b) What is the energy of the scattered photon if the energy of the incident photon is 5eV and the kinetic energy of the incident electron is 100eV?
(c)  What is the energy of the scattered photon if the energy of the incident photon is 5eV and the energy of the incident electron is 1GeV?

Solution:

Concepts, principles, relations that apply to the problem:
Energy and momentum conservation in relativistic collisions
Why do they apply?
In relativistic collisions between free particles energy and momentum are always conserved. 
How do they apply?
(a)
  before: after:
     
electron energy: E0 = (m2c4+p02c2)1/2     E = (m2c4+p2c2)1/2
electron momentum:    p0 i pcosq i + psinq j
photon energy: hn hn'
photon momentum: (hn/c) j (hn'/c) i

conservation of:

px    p0 = pcosq + hn'/c    (1)
py hn/c = psinq (2)
E E0 + hn = E + hn' (3)

squaring (1):  (p0 - hn'/c)2  = p2cos2q

squaring (2):  (hn/c)2 = p2sin2q

adding:  p2 = p02 + (hn'/c)2 - 2p0 hn'/c + (hn/c)2  (4)

squaring (3):  [E0 + h(n - n')]2  = E2 
h2(n - n')2 + E02  + 2E0h(n - n')  = m2c4 + p2c2
= m2c4 + p02c2 + (hn')2 - 2p0chn' + (hn)2 (from 4)

-h2nn' + E0 hn - E0 hn' + p0chn' = 0
hn' = E0hn/(hn + E0 - p0c)

Details of the calculation:
(b)  Let hn = 5eV, E0 = T + mc2 = 100eV + mc2.
p02c2 = E02 - m2c4 = m2c4(1-100eV/mc2)2 - m2c4 » 2mc2100eV.
hn' » 5eVmc2/(mc2 + 5eV - (mc2200eV)1/2) » 5eVmc2/{(mc2)1/2 [(mc2)1/2-(200eV)1/2]} » 5eV.
For a slow electron the photon energy does not increase appreciably.
Let hn = 5eV, E0 = T + mc2 = 1GeV + mc2 » 1GeV.
p02c2 = E02 - m2c4 » (1GeV)2.
hn' » 5eV 1GeV/(5eV + 1GeV -1GeV) » 1GeV.
The photon energy increases appreciably for a relativistic electron.

Link:
An interesting experiment that uses the high-energy photons created that way:
"Turning Light into Matter"

Problem 9:

A positron can be made by bombarding a stationary electron with a photon.
g + e- --> e- + e+ + e-
What is the minimum photon energy?

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?
The photon has minimum energy when the reaction products are at rest in the CM frame.
In the lab frame the positron and electrons move together as one particle with mass 3m,  (m = electron mass).
In the CM frame after the collision we have for the length of the momentum 4-vector
P02 - P2 = P02  = (Sp0)2 = (3mc)2 = 9m2c2.
The "length" of the total momentum 4-vector is 9m2c2 before and after the collision in the lab frame.
Before the collision we have
P02 - P2 = (hf/c + mc)2 - (hf/c)2 = m2c2 + 2hfm = 9m2c2.
hf = 4mc2 is the minimum photon energy.
Details of the calculation:
None

Problem 10:

At t = 0 in the lab frame a particle of mass 1kg has velocity (vx, vy) = (0.6c, 0) and acceleration (ax,ay) = (2,3) m/s2. What force is acting on the particle in the lab frame and in an inertial frame moving with velocity v = 0.6ci at t = 0, i.e. instantaneous inertial rest frame of the particle.  (You can assume that at t = 0 the origins of the two frames coincide and the particle is at the origin.)

Solution:

Concepts, principles, relations that apply to the problem:
F = dp/dt, 4-vectors, frame transformations
Why do they apply?
We are given dv/dt in the lab frame.  We must find the relationship between dv/dt and dp/dt to find F.
How do they apply?

In the lab frame:
Fx = dpx/dt,  Fy = dpy/dt,  px = gmvx, py = gmvy.
dpx/dt = gmax + (dg/dt)mvx,  dpy/dt = gmay + (dg/dt)mvy.
dg/dt = d(1 - vx2/c2 - vy2/c2)-1/2/dt = [(vxdvx/dt + vydvy/dt)/c2]g3.
at t = 0 we have dg/dt = [vx(dvx/dt)/c2]g3, since vy = 0.
Therefore
dpx/dt = gmax + g3max[vx2/c2] =  gmax[1+g2vx2/c2] = g3max.
dpy/dt = gmay.
(Fx, Fy) = m(g3ax, gay) , where g = [1/(Ö{1 - vx2/c2 - vy2/c2})] = 5/4 .
Thus (Fy, Fx) = ((5/4)3*2, (5/4)*3)N = (125/32, 15/4)N.

To transform to another inertial frame we use:
(p0, p) = (E/c, p) is a 4-vector.
At t = 0
px' = -g'b'E/c + g'px,  py' = py.
Here b' and g' are constants whose magnitude depends on the relative velocity of the two inertial frames.
dpx'/dt = d(-g'b'E/c + g'px)/dt,  dpy'/dt = dpy/dt.
The proper time interval is measured in the particles rest frame.

In the instantaneous inertial rest frame of the particle:
dt = dt', g'dt = g'dt' = dt.
Fx' = dpx'/dt' = g'd(-g'b'E/c + g'px)/dt.
-g'b'E/c + g'px = g'gm(vx - v'),
since E = gmc2, b'E/c = gmv', and px = gmvx.
Here g = (1 - v2/c2)-1/2, where v is the speed of the particle in the lab frame.
dg'gm(vx - v)/dt = mg'[(dg/dt)(vx - v) - g(dvx/dt - dv'/dt)]
At t = 0 we have (vx - v) = 0, and
dg'gm(vx - v)/dt = mg'gax = mg2ax,
Fx' = mg3ax = Fx.
For a frame such as the particle's, moving along with speed 0.6c with respect to the lab we have Fx¢ = Fx and Fy¢ = gFy . Thus the force in the instantaneous inertial rest frame of the particle is (Fx¢, Fy¢) = (125/32, 75/16)N = 25/16(5/2, 3)N.