Math 3181
Name:
Dr. Franz Rothe
February 10, 2016
16SPR\4080_spr16h2a.tex
2
Solution of Homework
Theorem 1. The four-momentum of a particle moving in any field of forces is
pµ = [E/c, p~ ]
The total energy E and the rest mass m are related by
q
(2.1)
E = m2 c4 + c2 p~ 2
The momentum and the velocity by
(2.2)
p~ =
E
~v
c2
These formulas retains their meaning for m = 0, as does occur for a photon, or other
massless particle.
Problem 2.1. A particle is non-relativistic if c|~p | mc2 , or equivalently β 1. Use
the power expansion
√
x x2
1+x=1+ −
± ...
2
8
to get the approximation of a kinetic energy for a non-relativistic particle.
Answer. From the relation (2.1), the total energy E = T + mc2 and the kinetic energy
T are
r
q
p2
E = m2 c4 + c2 p~ 2 = mc2 1 + 2 2
mc
#
"r
2
2
p
p4
p
2
2
1 + 2 2 − 1 = mc
T = mc
−
± ...
mc
2m2 c2 8m4 c4
T '
p2
2m
as well known from the basics of classical mechanics.
Problem 2.2. Check that the kinetic energy of the scattered electron is
T 0 = mc2
λ2C
(1 − cos θ)
λλ0
Take as an example the monochromatic Mo-Kα -rays from A.H. Compton’s experiment
of 1922. The incident wavelength is λ = 70 pm. Calculate the kinetic energy of the
scattered electron.
1
Answer. From the 0-component of the four-momentum conservation, the kinetic energy
of the electron is T 0 = cq 0 − cq 0 0 and converted to wavelengths becomes
λC
1
1
λC
0
0
00
0
2
T = cq − cq = ~ω − ~ω = hc
−
= mc
− 0
λ λ0
λ
λ
The equation (??) for the shift of the wave length yields
T 0 = mc2
λ2C
(1 − cos θ)
λλ0
With the data from Compton’s experiment of 1922, one obtains T 0 = (1−cos θ)635.9 eV.
Problem 2.3. Determine the direction of the scattered electron, from the y-component
of the four-momentum conservation.
(a) Write down the y-component of the four-momentum conservation.
(b) Get sin φ in terms of sin θ, and the wave length λ0 and relativity parameter γ 0 for
the scattered electron.
(c) From the kinetic energy T 0 obtained in the previous problem see that
γ0 − 1 =
λ2C
(1 − cos θ)
λλ0
Use this expression to simply. After a bid lengthy calculation one gets sin φ =
−A cos(θ/2) with a factor A < 1 which is approaching one for non-relativistic
electron.
(d) Express the result sin φ = − cos(θ/2), to be expected in the experiments, in simple
geometric terms.
Answer. (a) The y-components ~q 0 y = (~ω 0 /c) sin θ and p~ 0 y = mcγ 0 β 0 sin φ add up to
zero. One gets cp0 sin φ + ~ω 0 sin θ = 0.
(b)
sin φ = −
~ω 0
h
h
λC
p
sin
θ
=
−
sin
θ
=
−
sin
θ
=
−
sin θ
cp0
λ0 p0
mcλ0 β 0 γ 0
λ0 (γ 02 − 1)
(c)
√
λλ0
λC
λC
sin θ
√ 0
p
sin φ = − 0 √ 0
sin θ = − 0 √ 0
λ γ +1 γ −1
λ γ + 1 λC
(1 − cos θ)
r r
λ
2
=−
cos(θ/2)
0
0
λ γ +1
2
(d) The non-relativistic electron is scattered along the ray opposite to the angle bisector
of the incident and scattered photon. The deviation from this rule is of first order
in v/c.
Problem 2.4. Rotate the lab-system by the angle φ such that the scattered electron
moves along the positive x-axis. Transform the components of the four-momenta for the
incoming and scattered photon and electron to this system. Convince yourself that
ω
sin(φ − θ)
λC
= 0 =1+
(1 − cos θ)
sin φ
ω
λ
Conclude that for an experiment with λ λC , one gets indeed φ ≈ θ/2 − 90◦ .
Answer. One applies the rotation matrix

1
0
0
0
 0 cos φ sin φ 0
R=
 0 − sin φ cos φ 0
0
0
0
1




to get transformed components of four-momenta. For the incident photon and electron
(~ω/c)R[1, 1, 0, 0]T = (~ω/c)[1, cos φ, − sin φ, 0]T
and
(mc)R[1, 0, 0, 0]T = (mc)[1, 0, 0, 0]T
and for the scattered photon and electron
(~ω 0 /c)R[1, cos θ, sin θ, 0]T = (~ω 0 /c)[1, cos(θ − φ), sin(θ − φ), 0]T
and
(E 0 /c)R[1, β 0 cos φ, β 0 sin φ, 0]T = (E 0 /c)[1, β 0 , 0, 0]T
Only the photon has a momentum along the new y-axis. By conservation of momentum
one gets −~ω sin φ = ~ω 0 sin(θ − φ). Together with the formula for the shift of the wave
length
sin(φ − θ)
ω
λC
= 0 =1+
(1 − cos θ)
sin φ
ω
λ
For an experiment with λ λC , one gets indeed sin(φ − θ) ≈ sin φ. Hence φ − θ ≈
180◦ − φ , φ ≈ θ/2 + 90◦ and |φ| ≈ 90◦ − |θ|/2.
Problem 2.5. Inverse Compton scattering 1 occurs whenever a photon scatters off a
particle moving with almost speed of light. Suppose that a particle with rest mass M and
total energy E collides head on with a photon of energy Eγ . For simplicity, assume that
the scattered particles move back along the same axis.
1
See Hobson General Relativity p.132 problem 5.14
3
(i) Convince yourself that the conservation of the four-momentum implies
Eγ (E + cp) = Eγ0 (E − cp) + 2Eγ Eγ0
(ii) Solve for Eγ0 and use that M c2 E and hence E +cp ≈ 2E holds with any accuracy
needed. Show that under this assumption the scattered photon has the energy
Eγ0 =
4E 2 · Eγ
M 2 c4 + 4E · Eγ
(iii) Such a situation may occur for example for a collision of an ultra-relativistic cosmic
ray with a photon from the cosmic background radiation. How much energy can
such a cosmic ray proton transfer to a microwave background photon?
Energies up to E = 1020 eV may occur in cosmic rays. The proton has rest energy
M c2 = 938.272 MeV. A typical energy of photon from the cosmic background
radiation is Eγ = 2.7◦ K · 8.617 · 10−5 eV/K.
Answer. The four-momenta of the incident photon and proton are
cq = Eγ [1, 1, 0, 0] and cp = [E, −cp, 0, 0]
the four-momenta for the scattered photon and proton are
cq0 = Eγ0 [1, −1, 0, 0] and cp0 = [E 0 , −cp0 , 0, 0]
From the conservation of four-momentumq+p = q0 +p0 , we eliminate the less interesting
quantity p0 . I prefer to take advantage of the invariant scalar products and get, with
the same calculation as above
q · p = q0 · p0 = q0 · (p + q − q0 ) = q0 · (p + q)
We put the assumed vectors into the last equation and obtain
Eγ (E + cp) = Eγ0 (E − cp) + 2Eγ Eγ0
Eγ0 =
Eγ (E + cp)2
Eγ (E + cp)2
Eγ (E + cp)
= 2
=
E − cp + 2Eγ
E − c2 p2 + 2Eγ (E + cp)
M 2 c4 + 2Eγ (E + cp)
We use that M c2 E and hence E + cp ≈ 2E holds with any accuracy needed to
simplify
Eγ0 =
4Eγ · E 2
M 2 c4 + 4Eγ · E
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Definition 1 (Center of mass system). For any set of colliding particles, the center
of mass system or CMS, is the inertial system for which the total momentum of the
incoming particles is zero. Thus their four-momentum is [M c, 0, 0, 0] where M is the
total rest mass.
Problem 2.6. Consider the Compton scattering process in the CMS system, with incoming photon moving in the positive x-direction. Write down the four-momenta of the
incident photon and electron, and the possible four-momenta for the scattered photon
and electron.
Answer. The four-momenta of the incident photon and electron are
q = (~ω/c)[1, 1, 0, 0] and p = (γmc)[1, −β, 0, 0]
with ~ω = mc2 βγ, and the possible four-momenta for the scattered photon and electron
are
q0 = (~ω/c)[1, cos θ, sin θ, 0] and p0 = (γmc)[1, −β cos θ, −β sin θ, 0]
with any angle θ.
Problem 2.7. The Bevatron at Berkeley was built with the idea of producing antiprotons, 2 by the reaction p + p → p + p + p + p. Thus one intended to let a high-energy
proton strike a proton at rest;—at that point of history, it was not yet possible to send
two proton rays against each other. By known conservation laws, it was clear that only
an additional pair of proton and antiproton could be expected. Find the threshold for the
energy E, and kinetic energy T , of the incoming proton at which this reaction becomes
possible.
(i) Calculate the total energy-momentum four-vector ptot of the incoming particles in
the lab system.
(ii) At the threshold, the created four particles cannot have any additional kinetic energy. Thus they need to be at rest in the CMS-system. Calculate the total energymomentum four-vector p0tot of the scattered particles in the CMS-system.
(iii) From the conservation of the four-momentum and Lorentz invariance, we know
that
ptot · ptot = p0tot · p0tot
(iv) Determine E and c|~p | from the two equations
(E + M c2 )2 − c2 p~ 2 = (4M c2 )2
E 2 − c2 p~ 2 = M 2 c4
2
See also D. Griffiths Introduction to Elementary Particles, p.106
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(v) Determine the kinetic energy T and the speed of the incoming protons at threshold.
Answer. (i) ptot = [(E + M c2 )/c, p, 0, 0] is the total four-momentum for the incoming
two protons, in the lab system.
(ii) p0tot = [4M c, 0, 0, 0] is the total energy-momentum four-vector of the four scattered
particles, in the CMS system.
(iii) The Lorentz invariant of the total four-momenta are
ptot · ptot = (E/c + M c)2 − p~ 2
p0tot · p0tot = 16M 2 c2
(iv) From the conservation of the four-momentum and Lorentz invariance, we get the
first equation. The second one refers to the incoming proton from the ray.
(E + M c2 )2 − c2 p~ 2 = (4M c2 )2
E 2 − c2 p~ 2 = M 2 c4
√
One gets E = 7M c2 and c|~p | = 48M c2 .
(v) T = E − M c2 = 6M c2 . Indeed the antiprotons were discovered when the machine
√
reached about 6000 MeV. At threshold, the speed of the proton is β = ( 48)/7
times the speed of light.
Problem 2.8 (Twin paradox or travelling keeps young). For this problem, I put
c = 1 and consider only one space dimension. We do a series of simplifying calculations,
and prove the conjecture at first in 1 + 1 dimensions. Confirm for any two time-like
vectors x = (t, x) ∈ Future and p = (E, p) ∈ Future, the reversed triangle inequality
|x + p| ≥ |x| + |p|
holds—even with proper inequality > unless they are proportional.
Answer. All the following inequalities are equivalent:
|x| + |p| ≤ |x + p|
|x| + |p| + 2|x||p| ≤ |x + p|2
p
t2 − x2 + E 2 − p2 + 2 (t2 − x2 )(E 2 − p2 ) ≤ (t + E)2 − (x + p)2
p
2 (t2 − x2 )(E 2 − p2 ) ≤ 2tE − 2xp
2
2
(t2 − x2 )(E 2 − p2 ) ≤ [tE − xp]2
−t2 p2 − x2 E 2 ≤ −2tExp
0 ≤ (tp − xE)2
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