PHYS 4D
Solution to HW 9
March 6, 2011
Problem Giancoli 36-64 (II) A certain galaxy has a Doppler shift given by f0 − f = 0.697f0 , How fast is it
moving away from us?
Solution:
f
f0 − f = 0.697f0 ⇒
= 0.303.
f0
The galaxy is moving away from the Earth. Eq. 36-15b ⇒
√
f = f0
1−
c−v
⇒v=
c+v
1+
f2
f02
f2
f02
c = 0.83c.
Problem Giancoli 36-66 (II) Starting from Eq. 36-15a, show that the Doppler shift in wavelength is
v
∆λ
=
λ
c
if v ≪ c.
Solution: Eq. 36-15a:
√
λ = λ0
λ − λ0
λ
√
=
≈
c−v
1−
=1−
c+v
(
v) v
1− 1−
= .
c
c
√
c+v
c−v
1−
1+
v
c
v
c
(
v )− 21
v ) 12 (
1+
=1− 1−
c
c
Problem Giancoli 36-68 (III) A certain atom emits light of frequency f0 when at rest. A monatomic gas
composed of these atoms is at temperature T . Some of the gas atoms move toward and others away from an observer
due to their random thermal motion. Using the rms speed of thermal motion, show that the fractional difference
between the Doppler-shifted
frequencies for atoms moving directly toward the observer and directly away from the
√
observer is ∆f /f0 ≈ 2 3kT /mc2 ; assume mc2 ≫ 3kT . Evaluate ∆f /f0 for a gas of hydrogen atoms at 550K. [This
”Doppler-broadening” effect is commonly used to measure gas temperature, such as in astronomy.]
Solution: The difference between the Doppler-shifted frequencies for atoms moving directly toward the observer
and directly away from the observer is
√
√
c+v
c−v
∆f = f0
− f0
c−v
c+v
2v
2v
= f0 √ c
= f0 √
.
2
(c − v) (c + v)
1 − vc2
1
Using the rms speed of thermal motion,
√
v
∆f
3kT
v
⇒ =
m
c
√
3kT
= vrms =
⇒
mc2
√
3kT
2 mc
2
2
= f0 √
= f0 √
3kT
mc2
1 − mc
2
3kT − 1
assume mc2 ≫ 3kT ,
√
3kT
.
mc2
Problem Griffiths 12-30 Suppose you have a collection of particles, all moving in the x direction, with energies
E1 , E2 , E3 , . . . and momenta p1 , p2 , p3 , . . .. Find the velocity of the center of momentum frame, in which the total
momentum is zero.
Solution:
∆f /f0 ≈ 2
ET
=
E1 + E2 + E3 + . . . ;
pT
p̄T
=
=
p1 + p2 + p3 + . . . ;
γ (pT − βET /c) = 0 ⇒ β = v/c = pT c/ET .
v
=
pT c2 /ET = c2 (p1 + p2 + p3 + . . .) / (E1 + E2 + E3 + . . .) .
Problem Griffiths 12-33 A neutral pion of (rest) mass m and momentum p = (3/4)mc decays into two photons.
One of the photons is emitted in the same direction as the original pion, and the other in the opposite direction.
Find the energy of each photon.
Solution: Pion’s energy:
E 2 = p2 c2 + m2 c4 =
9 2 4
25 2 4
5
m c + m2 c4 =
m c ⇒ E = mc2 .
16
16
4
Conservation of energy:
5 2
mc = EA + EB .
4
Conservation of momentum:
EA
EB
3 2
mc = pA + pB =
−
4
c
c
⇒
2EA = 2mc2 .
⇒
1 2
mc .
4
Problem Griffiths 12-35 In a pair annihilation experiment, an electron (mass m) with momentum pe hits a
positron (same mass, but opposite charge) at rest. They annihilate, producing two photons.
(a) Why couldn’t they produce just one photon?
(b) If one of the photons emerges at 60◦ to the incident electron direction, what is its energy?
Solution:
(a) One photon is impossible, because in the center of momentum frame, we’d be left with a photon at rest,
whereas photons have to travel at speed c.
(b) Conservation of energy:
√
p2e c2 + m2 c4 + mc2 = EA + EB .
EA = mc2 , EB =
Conservation of momentum:
horizontal:
pe =
EB
1
EA
cos 60◦ +
cos θ ⇒ EB cos θ = pe c − EA ,
c
c
2
2
vertical:
0=
√
EA
EB
3
sin 60◦ −
sin θ ⇒ EB sin θ =
EA .
c
c
2
Square and add ⇒
(
⇒
1
pe c − EA
2
(√
)2
3
EA
2
)2
2
EB
=
2
EB
2
= p2e c2 − pe cEA + EA
[√
]2
p2e c2 + m2 c4 + mc2 − EA
=
√
(
)
2
= p2e c2 + m2 c4 + 2 p20 c2 + m2 c4 mc2 − EA + m2 c4 − 2mc2 EA + EA
+
√
√
mc2 + p20 c2 + m2 c4
mc2 mc + 2pe + p20 + m2 c2
√
EA = mc
=
2
mc + 34 p0
mc2 + p20 c2 + m2 c4 − pe c/2
2
3