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Problem 7.
Obtain the Lorentz transformation in which the velocity is at an infinitesimal angle dθ counterclockwise from the z axis, by means of a similarity transformation applied to Eq. (7-18). Show directly that the resulting matrix is orthogonal and that the inverse matrix is obtained by substituting −v for v.
We can obtain this transformation by first applying a pure rotation to rotate the z axis into the boost axis, then applying a pure boost along the (new) z axis, and then applying the inverse of the original rotation to bring the z axis back in line with where it was originally. Symbolically we have L = R−^1 KR where R is the rotation to achieve the new z axis, and K is the boost along the z axis. Goldstein tells us that the new z axis is to be rotated dθ counterclockise from the original z axis, but he doesn’t tell us in which plane, i.e. we know θ but not φ for the new z axis in the unrotated coordinates. We’ll assume the z axis is rotated around the x axis, in a sense such that if you’re standing on the positive x axis, looking toward the negative x axis, the rotation appears to be counterclockwise, so that the positive z axis is rotated toward the negative y
axis. Then, using the real metric,
L =
0 cos dθ sin dθ 0 0 − sin dθ cos dθ 0 0 0 0 1
0 0 γ −βγ 0 0 −βγ γ
0 cos dθ − sin dθ 0 0 sin dθ cos dθ 0 0 0 0 1
0 cos dθ sin dθ 0 0 − sin dθ cos dθ 0 0 0 0 1
0 cos dθ − sin dθ 0 0 γ sin dθ γ cos dθ −βγ 0 −βγ sin dθ −βγ cos dθ γ
0 cos^2 dθ + γ sin^2 dθ (γ − 1) sin dθ cos dθ −βγ sin dθ 0 (γ − 1) sin dθ cos dθ sin^2 dθ + γ cos^2 dθ −βγ cos dθ 0 −βγ sin dθ −βγ cos dθ γ
Problem 7.
A rocket of length l 0 in its rest system is moving with constant speed along the z axis of an inertial system. An observer at the origin observes the apparent length of the rocket at any time by noting the z coordinates that can be seen for the head and tail of the rocket. How does this apparent length vary as the rocket moves from the extreme left of the observer to the extreme right?
Let’s imagine a coordinate system in which the rocket is at rest and centered at the origin. Then the world lines of the rocket’s top and bottom are
xt μ′ = { 0 , 0 , +L 0 / 2 , τ } xb μ′ = { 0 , 0 , −L 0 / 2 , τ }.
where we are parameterizing the world lines by the proper time τ. Now, the rest frame of the observer is moving in the negative z direction with speed v = βc relative to the rest frame of the rocket. Transforming the world lines of the rocket’s top and bottom to the rest frame of the observer, we have
xtμ = { 0 , 0 , γ(L 0 /2 + vτ ), γ(τ + βL 0 / 2 c)} (1) xbμ = { 0 , 0 , γ(−L 0 /2 + vτ ), γ(τ − βL 0 / 2 c)}. (2)
Now consider the observer. At any time t in his own reference frame, he is receiving light from two events, namely, the top and bottom of the rocket moving past imaginary distance signposts that we pretend to exist up and down the z axis. He sees the top of the rocket lined up with one distance signpost and the bottom of the rocket lined up with another, and from the difference between the two signposts he computes the length of the rocket. Of course, the light that he sees was emitted by the rocket some time in the past, and, moreover, the
Problem 7.
Two particles with rest masses m 1 and m 2 are observed to move along the observer’s z axis toward each other with speeds v 1 and v 2 , respectively. Upon collision they are observed to coalesce into one particle of rest mass m 3 moving with speed v 3 relative to the observer. Find m 3 and v 3 in terms of m 1 , m 2 , v 1 , and v 2. Would it be possible for the resultant particle to be a photon, that is m 3 = 0, if neither m 1 nor m 2 are zero?
Equating the 3rd and 4th components of the initial and final 4-momentum of the system yields
γ 1 m 1 v 1 − γ 2 m 2 v 2 = γ 3 m 3 v 3 γ 1 m 1 c + γ 2 m 2 c = γ 3 m 3 c
Solving the second for m 3 yields
m 3 =
γ 1 γ 3
m 1 +
γ 2 γ 3
m 2 (5)
and plugging this into the first yields v 3 in terms of the properties of particles 1 and 2:
v 3 =
γ 1 m 1 v 1 − γ 2 m 2 v 2 γ 1 m 1 + γ 2 m 2
Then
β 3 =
v 3 c
γ 1 m 1 β 1 − γ 2 m 2 β 2 γ 1 m 1 + γ 2 m 2
1 − β^23 =
γ 12 m^21 + 2γ 1 γ 2 m 1 m 2 + γ^22 m^22 − [γ 12 m^21 β^21 + γ 22 m^22 β 22 − 2 γ 1 γ 2 m 1 m 2 β 1 β 2 ] (γ 1 m 1 + γ 2 m 2 )^2
=
γ 12 m^21 (1 − β 12 ) + γ 22 m^22 (1 − β 22 ) + 2γ 1 γ 2 m 1 m 2 (1 − β 1 β 2 ) (γ 1 m 1 + γ 2 m 2 )^2
=
m^21 + m^22 + 2γ 1 γ 2 m 1 m 2 (1 − β 1 β 2 ) (γ 1 m 1 + γ 2 m 2 )^2
and hence
γ 32 =
1 − β^23
(γ 1 m 1 + γ 2 m 2 )^2 m^21 + m^22 + 2γ 1 γ 2 m 1 m 2 (1 − β 1 β 2 )
Now, (5) shows that, for m 3 to be zero when either m 1 or m 2 is zero, we must have γ 3 = ∞. That this condition cannot be met for nonzero m 1 , m 2 is evident from the denominator of (6), in which all terms are positive (since β 1 β 2 < 1 if m 1 or m 2 is nonzero).
Problem 7.
A meson of mass π comes to rest and disintegrates into a meson of mass μ and a neutrino of zero mass. Show that the kinetic energy of motion of the μ meson (i.e. without the rest mass energy) is
(π − μ)^2 2 π
c^2.
Working in the rest frame of the pion, the conservation relations are
πc^2 = (μ^2 c^4 + p^2 μc^2 )^1 /^2 + pν c (energy conservation) 0 = pμ + pν (momentum conservation).
From the second of these it follows that the muon and neutrino must have the same momentum, whose magnitude we’ll call p. Then the energy conservation relation becomes
πc^2 = (μ^2 c^4 + p^2 c^2 )^1 /^2 + pc −→ (πc − p)^2 = μ^2 c^2 + p^2
−→ p =
π^2 − μ^2 2 π
c.
Then the total energy of the muon is
Eμ = (μ^2 c^4 + p^2 c^2 )^1 /^2
= c^2
μ^2 +
(π^2 − μ^2 )^2 4 π^2
c^2 2 π
4 π^2 μ^2 + (π^2 − μ^2 )^2
c^2 2 π (π^2 + μ^2 )
Then subtracting out the rest energy to get the kinetic energy, we obtain
K = Eμ − μc^2 =
c^2 2 π
(π^2 + μ^2 ) − μc^2
c^2 2 π (π^2 + μ^2 − 2 πμ)
c^2 2 π
(π − μ)^2
as advertised.
rest in the COM system). Then the momentum conservation relation becomes simply
pπ = pK + pλ (8)
and the energy conservation relation is (with c = 1)
(m^2 π + p^2 π )^1 /^2 + mn = (m^2 K + p^2 K )^1 /^2 + (m^2 Λ + p^2 Λ)^1 /^2. (9)
The problem is to find the minimum value of pπ that satisfies (9) subject to the constraint (8). To solve this we must first resolve a subquestion: for a given pπ , what is the relative allocation of momentum to pK and pΛ that minimizes (9)? Minimizing
Ef = (m^2 K + p^2 K )^1 /^2 + (m^2 Λ + p^2 Λ)^1 /^2.
subject to pK + pΛ = pπ , we obtain the condition
pK (m^2 K + p^2 K )^1 /^2
pΛ (m^2 Λ + p^2 Λ)^1 /^2
=⇒ pK =
mK mΛ
pΛ (10)
Combining this with (8) yields
pΛ =
mΛ mK + mΛ
pπ pK =
mK mK + mΛ
pπ. (11)
For a given total momentum pπ , the minimum possible energy the final system can have is realized when pπ is partitioned between pK and pΛ according to (11). Plugging into (8), the relation defining the threshold momentum is
(m^2 π + p^2 π )^1 /^2 + mn =
m^2 K +
mK mK + mΛ
p^2 π
m^2 Λ +
mΛ mK + mΛ
p^2 π
Solving numerically yields pπ ≈ 655 MeV/c, for a total pion energy of about 670 MeV.
Problem 7.
A photon may be described classically as a particle of zero mass possessing never- theless a momentum h/λ = hν/c, and therefore a kinetic energy hν. If the photon collides with an electron of mass m at rest it will be scattered at some angle θ with a new energy hν′. Show that the change in energy is related to the scattering angle by the formula λ′^ − λ = 2λc sin^2
θ 2
where λc = h/mc, known as the Compton wavelength. Show also that the kinetic energy of the recoil motion of the electron is
T = hν
( (^) λc λ
sin^2 θ 2 1 + 2
( (^) λc λ
sin^2 θ/ 2
Let’s assume the photon is initially travelling along the z axis. Then the sum of the initial photon and electron four-momenta is
pμ,i = pμ,γ + pμ,e =
h/λ h/λ
mc
h/λ mc + h/λ
.^ (12)
Without loss of generality we may assume that the photon and electron move in the xz plane after the scatter. If the photon’s velocity makes an angle θ with the z axis, while the electron’s velocity makes an angle φ, the four-momentum after the collision is
pμ,f = pμ,γ + pμ,e =
(h/λ′) sin θ 0 (h/λ′) cos θ h/λ′
pe sin φ 0 √^ pe^ cos^ φ m^2 c^2 + p^2 e
(h/λ′) sin θ + pe sin φ 0 (h/λ′) cos θ + pe cos φ (h/λ′) +
m^2 c^2 + p^2 e
Equating (12) and (13) yields three separate equations:
(h/λ′) sin θ + pe sin φ = 0 (14) (h/λ′) cos θ + pe cos φ = h/λ (15) h/λ′^ +
m^2 c^2 + p^2 e = mc + h/λ (16)
From the first of these we find
sin φ = −
h λ′pe
sin θ =⇒ cos φ =
[
h λ′pe
sin^2 θ
] 1 / 2
spherical polar angles θ and φ = 0). Then the 4-momenta are
p 1 =
E
c
E
c
p 2 =
E
c sin θ, 0 ,
E
c cos θ,
E
c
pt = p 1 + p 2 =
E
c sin θ, 0 ,
E + E cos θ c
E + E
c
It’s convenient to rotate our reference frame to one in which the space portion of the composite four-momentum of the two photons is all along the z direction. In this frame the total four-momentum is
p′ t =
c
E^2 + E^2 + 2EE cos θ,
E + E
c
At threshold energy, the two produced particles have the same four-momenta:
p 3 = p 4 =
0 , 0 , p, (m^2 c^2 + p^2 )^1 /^2
and 4-momentum conservation requires that twice (19) add up to (18), which yields two conditions:
2 p = (^1) c
E^2 + E^2 + 2EE cos θ −→ p^2 c^2 = 14 (E^2 + E^2 + 2EE cos θ) 2
m^2 c^2 + p^2 = E+ c E −→ m^2 c^4 + p^2 c^2 = 14 (E^2 + E^2 + 2EE)
Subtracting the first of these from the second, we obtain
m^2 c^4 =
EE
(1 − cos θ)
or
E =
2 m^2 c^4 E(1 − cos θ)
as advertised.
