Physics 161
Homework 2 - Solutions
Wednesday August 31, 2011
Make sure your name is on every page, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them.
The homework is due at the beginning of class on Wednesday, September 7th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. Draw the Feynman diagram representing electron-electron scattering to first order in
the coupling constant (note: there are two such diagrams). Draw three distinct second
order-diagrams.
————————————————————————————————————
Solution
A first-order Feynman diagram (in electrodynamics) involves the exchange of a single
photon. There are two diagrams involving the exchange of the photon between two
electrons. The first is seen in the first diagram below, where the two electrons move
towards each other, exchange a photon, then move apart. However, since the electrons
are identical particles, and indistinguishable, we can’t tell which electron ends up
where. So, the second diagram differs by the exchange of the final electrons, as seen in
the diagram to the right in the figure.
e−
e−
e−
e−
γ
γ
−
e
e
−
e−
e−
A second-order diagram involves the exchange of two photons. There are many such
diagrams, four of which are seen below.
The first diagram involves the emission of a photon by an electron, before it exchanges
one with the other electron. The first electron then interacts with the second electron,
and only then reabsorbs the original photon. This diagram has the net effect of renormalizing the electron’s magnetic moment. The second diagram has the first electron
e−
e−
e−
e−
e−
γ
γ
γ
e−
e−
e−
γ
e−
e−
e−
e−
emitting and reabsorbing a photon, before it interacts with the other electron. This
diagram renormalizes the mass of the electron. In both diagrams, the electron interacts
with its own field.
e−
e−
γ
e−
e−
e−
e−
γ
e−
e−
γ
γ
e−
e−
e−
e−
The third diagram above simply shows the interaction between two electrons via the
exchange of two photons. The final diagram shows the interaction between the electrons
via the exchange of one photon, which then forms a virtual electron-positron pair. The
pair recombines back into a photon, only then being absorbed by the second electron.
This final diagram leads to a renormalization of the electron’s charge. Note, of course,
that there are plenty of variations on these diagrams differing by the exchange of the
final electrons, or by putting the photon in the second diagram on a different electron,
etc.
2. (a) How many different meson combinations can you make with 1, 2, 3, 4, 5, or 6
different quark flavors (recalling that there are three different colors for each quark
and that any observable particle must be colorless)? What’s the general formula
for n flavors?
(b) How many different baryons combinations can you make with 1, 2, 3, 4, 5, or 6
different quark flavors? What’s the general formula for n flavors?
————————————————————————————————————
Solution
(a) Suppose that we start with one photon, say u (which, of course, includes its
antiparticle, ū). There is only one meson that can be made from this combination,
¯ ūd, and dd.
¯
uū. With two quarks (u and d), we can make four mesons, uū, ud,
¯ us̄, dū, dd,
¯ ds̄,
With three mesons (u, d, s), then the possibilities are uū, ud,
¯ and ss̄, for a total of nine. It’s clear that the general case for n different
sū, sd,
quarks is n2 , and so four quarks gives 16 possibilities, five gives 25, and six gives
36 mesons.
(b) With one quark, we can only have one baryon, uuu, where each u is a different
color. With two quarks, u and d, then we can form uuu, uud, udd, and ddd, or
four baryons. With three quarks (u, d, and s), then we can form uuu, uud, uus,
udd, uds, dus, ddd, dds, dss, and sss. This gives a total of ten baryons.
This pattern is harder to simply see, so we’ll figure it out. For n quarks, for all
three different, there are n different ways that we can make a baryon (uuu, ddd,
etc.). For two the same and one different there are n(n − 1) different ways, uud,
uus, dds, etc. (n ways of picking the first one, and n − 1 ways of picking the
second, since the first quark is used up). Finally, for all three different, there are
n ways of picking the first, n − 1 of picking the second, and n − 2 of picking the
third. However, because dsb is the same as bsd, for example, we need to divide by
all six different permutations, for a grand total of n(n − 1)(n − 2)/6 possibilities.
So, the total number is the sum
n + n(n − 1) +
n(n−1)(n−2)
6
=
=
=
=
1
(6n + 6n(n − 1) + n(n − 1)(n −
6
1
(6n + 6n2 − 6n + n3 − 3n2 + 3)
6
1
(n3 + 3n2 + 3n)
6
n
(n + 1) (n + 2) .
6
2))
As a check, for n = 1, then the total is 61 (2)(3) = 1, while for n = 2, then we
have 26 (3)(4) = 4, and for three we get, 36 (4)(5) = 10, which all check out. So, for
four, five, and six quarks we have 64 (5)(6) = 20, 56 (6)(7) = 35, and 66 (7)(8) = 56
possibilities, respectively.
3. Sketch the lowest-order Feynman diagram representing Delbruck scattering, γ + γ →
γ + γ. (This process, the scattering of light by light, has no analog in classical electrodynamics.)
————————————————————————————————————
Solution
Photons do not interact with photons directly (since they carry no electric charge), and
so we can’t simply draw a diagram with three photons coming together at a vertex, like
we can with electrons and a photon. The only vertex that we really have in quantum
electrodynamics is an incoming electron, emitting (or absorbing) a photon, giving a
final electron. The simplest diagram that can account for light scattered off of light is
seen in the diagram below.
γ
γ
e−
e−
e−
e−
γ
γ
In this diagram, two photons interact via the creation of virtual electron-positron pairs,
which then reannihilate forming final photons. This is a second-order process, and so
is suppressed at low energies.
4. Draw all the lowest-order diagrams contributing to the process e+ + e− → W + + W − .
(One of them involves the direct coupling of Z 0 to W ± s, and the other the coupling of
γ to W s, so if a positron-electron collider is ever built with sufficient energy to make
two W s, these interactions will be directly observable.)
————————————————————————————————————
Solution
There are three lowest-order diagrams for this process. In the first diagram, the electron
and positron interact and directly produce the W ± pair. In the second case, the
electron-positron pair mutually annihilate and form a virtual Z 0 particle, which then
transforms into a W ± pair.
W−
W
−
W
W+
+
Z0
e−
e−
e−
e−
e−
Finally, the last diagram has a similar process to the second diagram, only now the
electron-positron pair annihilates to a virtual photon before decaying to a W ± pair.
W−
W+
γ
e−
e−
5. Examine the following processes, and state for each one whether it is possible or impossible, according to the Standard Model of Particle Physics (which does not include
GUTs, with their potential violation of the conservation of lepton number and baryon
number). If the reaction is possible, state which interaction is responsible - strong,
electromagnetic, or weak; if it is impossible, cite a conservation law that prevents it
from occurring. Look up any unfamiliar particles.
(a) p + p̄ → π + + π 0
(b) η → γ + γ
(c) e− + e+ → µ+ + µ−
(d) µ− → e− + ν̄e
(e) ν̄e + p → n + e+
(f) p → e+ + γ
(g) p + p → p + p + p + p̄
(h) π 0 → γ + γ
————————————————————————————————————
Solution
(a) This reaction is impossible since it violates charge conservation. The reactants
begin with zero charge, but the products carry a net positive charge.
(b) This reaction is possible via electromagneticeffects. The η particle is a mixture
of u, d, and s quarks and antiquarks, η = √16 uū + dd¯ − 2ss̄ . Each of these pairs
can interact electromagnetically.
(c) This reaction is possible via electromagnetic effects. The electron-positron pair
annihilates, forming a virtual photon, which then decays into a µ± pair.
(d) This reaction is impossible since it violates muon number conservation. The
reactants have a net muon number of 1, but the products have a net muon number
of zero, since none of the products are muons (or muon-type neutrinos).
(e) This reaction is possible via weak effects. It is a variation of neutron decay,
n → p + e− + ν̄e , with the electron moved to the left-hand side of the reaction,
changing it to a positron.
(f) This reaction is impossible since it violates baryon and lepton number conservation. The reactant starts with a baryon number of one, and lepton number of
zero, while the reactants have baryon number of zero, and lepton number of -1.
(g) This reaction is possible via strong effects. The extra energy imparted by slamming the protons together can form quark-antiquark pairs, which can form the
proton-antiproton pair.
(h) This reaction is possible via electromagnetic
effects.
The neutral pion is a combi
1
0
¯
¯
√
nation of a uū and dd pair, π = 2 uū − dd , which can all interact electromagnetically, giving two photons.
6. Muons can be created by cosmic rays in the upper atmosphere. A muon at rest has
a mean lifetime of about 2.2 microseconds. Suppose that a muon is created 10 km in
the air, and is produced with a velocity of 0.99c.
(a) According to classical physics, how far does the muon travel before decaying
(according to an observer on the ground)?
(b) According to Special Relativity, how far does the muon travel before decaying
(according to an observer on the ground)?
(c) In it’s own rest frame, the muon still lasts only 2.2 microseconds. How can it
make it to the point given in part (b)?
————————————————————————————————————
Solution
(a) According to classical physics, the muon travels a distance d = vτ , where v is the
velocity of the muon (0.99c), and τ is the lifetime of the muon. So,
dclassical = vτ = 0.99 3.00 × 108 2.2 × 10−6 ≈ 654 meters.
The muons only make it about 650 meters.
(b) According to Special Relativity the muons last longer, due to time dilation effects.
Their lifetime, as measured by stationary observers on the ground, is
τ0 = p
2.2 × 10−6
=√
= 1.6 × 10−5 seconds.
2
2
2
1 − 0.99
1 − v /c
τ
During this time, the muons can move a distance dSR = vτ 0 , where τ 0 is the new
lifetime. Thus, dSR = 0.99 (3.00 × 108 ) (1.6 × 10−5 ) = 4630 meters, which is more
than seven times further than the classical result.
(c) From the muon’s point of view, it is stationary, and the Earth is rushing up
towards it. This means that the p
muon sees the distance between it and the Earth
contracted by an amount d0 = 1 − v 2 /c2 d. This is contracted by exactly the
same factor by which the lifetime was dilated from the stationary observer’s point
of view. Thus, both the observers on the Earth and the muon agree on the location
that the muon reaches.
7. Show that the interval, ds2 , is invariant under Lorentz transformations.
————————————————————————————————————
Solution
The interval reads
ds2 = −c2 dt2 + dx2 + dy 2 + dz 2 .
Under a Lorentz transformation
2
t0 = √t−vx/c
2
x
0
=
1−v /c2
√ x−vt2 2
1−v /c
y0 = y
z 0 = z,
then the interval changes to ds02 = −c2 dt02 + dx02 + dy 02 + dz 02 . Now, taking the
differentials of the Lorentz transformations gives
2
dt−vdx/c
dt0 = √
2 2
1−v /c
dx0 = √dx−vdt
2 2
1−v /c
dy 0 = dy
dz 0 = dz,
and so
dt02
dx02
dy 02
dz 02
2
2
2
+v dx
= dt −2vdxdt/c
1−v 2 /c2
2
2 dt2
= dx −2vdxdt+v
1−v 2 /c2
= dy 2
= dz 2 .
2 /c4
Thus,
2
2 +v 2 dx2 /c4
2
2 dt2
+ dx −2vdxdt+v
ds02 = −c2 dt −2vdxdt/c
+ dy 2 + dz 2
2
2
1−v /c
1−v 2 /c2
= 1−v12 /c2 (−c2 dt2 + 2vdxdt − v 2 dx2 /c2 + dx2 − 2vdxdt + v 2 dt2 ) + dy 2 + dz 2
= 1−v12 /c2 (−c2 (1 − v 2 /c2 ) dt2 + (1 − v 2 /c2 ) dx2 ) + dy 2 + dz 2
= −c2 dt2 + dx2 + dy 2 + dz 2
= ds2 ,
which shows the invariance of the interval under the Lorentz transformations.
8. A pion at rest decays into a muon and a neutrino (π − → µ− + ν̄µ ). On the average,
how far will the muon travel (in vacuum) before disintegrating?
————————————————————————————————————
Solution
The muon is ejected with some velocity, v (that we need to determine), with respect
to the rest frame of the pion. By time dilation the muon lasts a time τ = γτ0 , where
−1/2
γ = (1 − v 2 /c2 )
, and τ0 = 2.2 µs is the average lifetime of the muon. So, in this
time the muon can travel a distance d = vτ = γvτ0 . Thus, we just have to determine
v, which we can do using energy and momentum conservation.
We could get our answer using four-momenta, but we’ll simply use energy and momentum conservation and a bit of extra algebra. Energy conservation says that
Ei = Eπ = Eµ + Eµ = Ef . Now, since the pion is initially at rest, it’s initial energy
is simply its rest-mass energy, Ei = mπ c2 . The energy of the muon is Eµ = γmµ c2 ,
while the neutrino has Eν = pν c, since it is assumed massless. Now, by momentum
conservation, since the pion is at rest we have 0 = pµ + pν . Thus, pν = −pµ = −γmµ v,
which gives for our energy conservation expression,
s
1 − v/c
v
= m µ c2
,
mπ c2 = γmµ c2 − γmµ vc = γmµ c2 1 −
c
1 + v/c
or
s
mπ
1 − v/c
=
.
1 + v/c
mµ
So, we just need to solve this expression for the velocity, v. Squaring both sides and
rearranging gives
2
2
2 !
mπ mπ
v
v
mπ
v
⇒1−
.
1− =
1+
= 1+
c
mµ
c
mµ
mµ
c
Solving gives
v=
m2π − m2µ
m2π + m2µ
c.
From this we find that
γ = √ 12 2
1−v /c
2 2 2 −1/2
m −m
=
1 − mπ2 +mµ2
π
µ
2 2 2 −1/2
m2π +m2µ 2
m −m
=
− m2π +m2µ
m2π +m2µ
π
µ
−1/2
m4π +2m2π m2µ +m4µ −m4µ +2m2µ m2π −m4µ
=
2
(m2π +m2µ )
−1/2
4m2π m2µ
=
2
(m2π +m2µ )
2
2
m +m
= 2mπ π mµµ .
Thus, the total distance that the muon travels will be
2
2
2
mπ + m2µ
mπ − m2µ
mπ − m2µ
d = γvτ =
c τ0 =
cτ0 .
2mπ mµ
m2π + m2µ
2mπ mµ
Plugging in the numbers gives
d=
m2π − m2µ
2mπ mµ
cτ0 =
(139.6)2 − (105.7)2
2 (139.6) (105.7)
!
3 × 108
2.2 × 10−6 = 186 meters.
9. Show that the Minkowski metric is invariant under Lorentz transformations.
————————————————————————————————————
Solution
The Minkowski metric reads (with our choice

−1 0
 0 1
ηµν = 
 0 0
0 0
of metric signature)

0 0
0 0 
,
1 0 
0 1
while the Lorentz transformation matrix reads

γ
−γ vc
 −γ v
γ
c
Λµ ν = 
 0
0
0
0
0
0
1
0

0
0 
.
0 
1
Under a Lorentz transformation the Minkowski metric becomes
0
ηµν
= Λα µ Λβ ν ηαβ = ΛT ηΛ,
where the last expression is written in matrix notation (and the superscript T denotes
the transpose of the matrix). Thus we just have to multiply the three matrices together,




−1 0 0 0
γ
−γ vc 0 0
γ
−γ vc 0 0
v


 −γ v
γ
0 0 
γ
0 0 
c
  0 1 0 0   −γ c

ΛT ηΛ = 
 0
0
1 0  0 0 1 0  0
0
1 0 
0
0 1  0 0 0 1
0
0 1
 0
 0
v
v
γ
−γ c 0 0
−γ γ c 0 0
 −γ v
  −γ v γ 0 0 
γ
0
0
c
c


= 
 0
0
1 0  0
0 1 0 
0
0 1
0
0 0 1
 02
2 2
−γ (1 − v /c )
0
0 0
2
2 2

0
γ (1 − v /c ) 0 0 

= 

0
0
1 0 
0
0
0 1


−1 0 0 0
 0 1 0 0 

= 
 0 0 1 0 ,
0 0 0 1
−1/2
where we have used the fact that γ = (1 − v 2 /c2 )
. Thus, we see that the Lorentz
transformation has returned the same Minkowski metric; in other words, the Minkowski
metric is invariant under Lorentz transformations.
10. A photon of wavelength λ collides elastically with a charged particle of mass m. If the
photon scatters at an angle θ, find its outgoing wavelength, λ0 .
————————————————————————————————————
Solution
We will solve this problem using fourvectors. The photon is incident along the
x axis, and strikes the stationary electron.
The photon scatters upwards at an angle
θ, with respect to it’s original direction.
The electron is scattered downwards by the
same angle, in order to conserve momentum, as seen in the figure to the right. The
initial four-momentum of the photon is pµγ ,
while that of the electron is pµe . The final
four-momenta we’ll call pµγ0 and pµe0 .
γ0
γ
e−
From the picture, and noting that Eγ = pγ c for a photon, we can write
pµγ = Ecγ (1, 1, 0, 0)
E 0
pµγ0 = cγ (1, cos θ, sin θ, 0)
pµe = (m
e c, 0, 0, 0)
Ee0
µ
pe 0 =
, pe0 cos θ, −pe0 sin θ, 0 .
c
Conservation of momentum and energy gives pµγ + pµe = pµγ0 + pµe0 , which, upon rearranging gives
pµγ − pµγ0 = pµe0 − pµe .
Now, squaring both sides gives
p2γ + p2γ 0 − 2pµγ pµγ 0 = p2e0 + p2e − 2pµe0 pµe .
Recalling that p2 = −m2 c2 for a particle of mass m, and that a photon is massless,
then we have
−2pµγ pµγ 0 = −2m2e c2 − 2pµe0 pµe ⇒ pµγ pµγ 0 = m2e c2 + pµe0 pµe .
Now, from our expressions for the four-momenta, then
Eγ E
0
pµγ pµγ 0 = − c2 γ (1 − cos θ)
pµe0 pµe = −me Ee0 ,
giving
−
Eγ Eγ 0
(1 − cos θ) = m2e c2 − me Ee0 .
c2
Now, by energy conservation, Ee0 = Eγ + Ee − Eγ 0 = Eγ − Eγ 0 + me c2 , and so
−
Eγ Eγ 0
2
2 2
0 + me c
.
c
−
m
E
−
E
(1
−
cos
θ)
=
m
e
γ
γ
e
c2
Canceling off the m2e c2 terms and dividing through by −me Eγ Eγ 0 gives
1
1
1
−
.
(1 − cos θ) =
2
me c
Eγ 0
Eγ
Now, recalling that the energy of a photon is E = hν =
hc
,
λ
λ0
λ
1
−
=
(1 − cos θ) .
hc hc
me c2
Finally, solving for λ0 we get our final answer,
λ0 = λ +
h
(1 − cos θ) .
me c
then