Physics 253a
1
Problem Set 3 Solutions
October 6, 2010
1.
(a) In the center of mass frame, the
e
+
and
e

approach each other with opposite
momenta and total energy
E
cm
:
p
e
+
=
E
cm
2
(1
,
1
,
0
,
0) and
p
e

=
E
cm
2
(1
,

1
,
0
,
0).
The muon and antimuon look the same, but with their momenta rotated by
θ
:
p
μ
+
=
E
cm
2
(1
,
cos
θ,
sin
θ,
0) and
p
μ

=
E
cm
2
(1
,

cos
θ,

sin
θ,
0).
Now,
s
=
(
p
e
+
+
p
e

)
2
=
(
E
cm
,
0
)
2
=
E
2
cm
t
=
(
p
μ


p
e

)
2
=

2
p
μ

·
p
e

=

E
2
cm
2
(1

cos
θ
)
u
=
(
p
μ
+

p
e

)
=

2
p
μ
+
·
p
e

=

E
cm
2
(1 + cos
θ
)
(b) Note that
s
+
t
+
u
= 0.
(c) In class we found
dσ
d
Ω
=
e
4
64
π
2
E
2
cm
(1 + cos
2
θ
)
Note that
4
tu
s
2
= 1

cos
2
θ
. So we can rewrite the above as
dσ
d
Ω
=
e
4
32
π
2
s
1

2
tu
s
2
=
e
4
32
π
2
s
3
(
t
2
+
u
2
)
where we’ve used
s
2
= (
t
+
u
)
2
=
t
2
+ 2
tu
+
u
2
.
(d) We have
s
+
t
+
u
=
(2
m
2
e
+ 2
p
e

·
p
e
+
) + (
m
2
e
+
m
2
μ

2
p
e

·
p
μ

)
+(
m
2
e
+
m
2
μ

2
p
e

·
p
μ
+
)
=
4
m
2
e
+ 2
m
2
μ

2
p
e

·
(
p
μ
+
+
p
μ


p
e
+
)
=
4
m
2
e
+ 2
m
2
μ

2
p
e

·
p
e

=
2(
m
2
e
+
m
2
μ
)
Where we’ve used conservation of momentum in the last line. In general, we’d
have
s
+
t
+
u
=
∑
i
m
2
i
.
Physics 253a
2
2.
(a) (Almost) straight off Wikipedia:
dσ
d
Ω
=
(2
Z
)
2
α
2
mv
2
2
1
sin
4
(
θ/
2)
where
m
is the mass of the
α
particle,
v
is its incoming velocity,
Z
is the nucleon
charge, and
α
=
e
2
4
π
is the fine structure constant (the Wikipedia article forgets
the product of the
α
particle and nucleon charge 2
·
Z
). This formula treats the
nucleus
N
as having infinite mass, so that the
α
particle scatters with the same
kinetic energy
E
kin
=
1
2
mv
2
as it started with. It also assumes the
α
particle is
a nonrelativistic point particle moving in a classical Coulomb potential.
(b)
i. Let the
α
particle have initial momentum
p
i
= (
E,
p
i
) and final momentum
p
f
= (
E,
p
f
). Note that the initial and final energies are the same in the limit
that the nucleon mass is infinite. Also,

p

=
√
2
mE
kin
in the nonrelativistic
limit. The relevant Feynman diagram is
γ
N
α, p
i
N
α, p
f
By momentum conservation, the virtual photon has
k
=
p
f

p
i
=
(0
,
p
f

p
i
)
=
p
2
mE
kin
(0
,
1

cos
θ,

sin
θ,
0)
ii. & iii. Rewriting
m
e
→
m
, and taking
e
2
→
2
Ze
2
, we have
dσ
d
Ω
=
(2
Z
)
2
e
4
4
π
2
m
2
k
4
=
(2
Z
)
2
e
4
4
π
2
m
2
4
m
2
E
2
kin
((1

cos
θ
)
2
+ sin
θ
2
)
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 SCHWARTZ
 Electron, Center Of Mass, Energy, Mass, Momentum, Fundamental physics concepts, Feynman diagram