5.10 Example: The Photon-Electron-Proton Fluid
123
−
1
tu
2 p a
1 p b
1 + p a
3
p b
1 − p b
4
+ 2 p
[a
1 p
b]
4
,
(5.10.66)
Z := 1 +
m 2
p + m 2
e + 2 p 1 · q 2
m 2
e + p 1 · p 2
+
2( p 1 · q 2 ) 2
m 2
e + p 1 · p 2
2 ,
(5.10.67)
˜
Z :=
2m p m e
m 2
e + p 1 · p 2
,
(5.10.68)
Z ab :=
p a
1 p b
1 + 2q a
2 q b
1
m 2
e + p 1 · p 2
+
m 2
p
p a
1 − p b
2
p b
1 + 2
p a
1 − p a
2
q b
2 − q a
2 p b
1
p 1 · p 2
m 2
e + p 1 · p 2
2
,
(5.10.69)
˜
Z ab :=
m p p a
1 q b
2
m e
m 2
e + p 1 · p 2
+
m p m e
p a
1 − p a
2
q b
2
m 2
e + p 1 · p 2
2 ,
(5.10.70)
W a := m 3
e
1
k 1 · p 1
−
1
k 1 · p 2
2
k a
2 − m e
1
k 1 · p 1
−
1
k 1 · p 2
p a
2 + 2k a
1
+
m e
k 1 · p 1
k 1 · p 2
k 1 · p 1
− 1
+
1
m e
k 1 · p 1
k 1 · p 2
−
k 1 · p 2
k 1 · p 1
p a
1 ,
(5.10.71)
W ab :=
1 − B
m 2
e
k a
1 k b
1 + Bp a
1 p b
1
+ (1 + B)
k a
1 p b
1
k 1 · p 1
−
p a
1 − p a
2
k b
1
k 1 · p 2
,
we have made use of the Mandelstam variables
t := ( p 1 − p 3 ) 2 ≡ −2
m 2
e + p 1 · p 3
,
u := ( p 1 − p 4 ) 2 ≡ −2
m 2
e + p 1 · p 4
,
(5.10.72)
in the e + e → e + e case and again the remaining terms can be obtained by interchanging electrons and protons. The “TT” superscript in Eq. (5.10.30) stands for
“transverse-traceless part”
X
TT
ab := X ab −
1
2
ab X
c
c ,
(5.10.73)
so the right-hand side expression is consistently transverse with respect to k
a
1 . Similarly, we also check that the integrands of (5.10.34), (5.10.35) and (5.10.36) are
consistently transverse with respect to p
a
1 . As mentioned at the end of Sect. 5.5, the
integrands of the collision matrices are not necessarily hermitian, only their integrals
are, which for the real variables employed here translates into imaginary contributions. For instance, ˜
C
e contains an imaginary term of the form
˜
C
e
⊃ i ˜
P ab,1
d
3 k 2
(2π) 3 2k 2
d
3 p 1
(2π) 3 2E p 1
d
3 p 2
(2π) 3 2E p 2
K
ab
(
k 1 ,
k 2 ,
p 1 ,
p 2 ) . (5.10.74)
To see that this is zero, we note that, whatever the K
ab function, the result of the
integral must be a tensor distribution by Lorentz invariance. Consequently, given the
available dependencies, the integral can only be of the form
d 3 k 2
(2π) 3 2k 2
d 3 p 1
(2π) 3 2E p 1
d 3 p 2
(2π) 3 2E p 2
K ab (
k 1 ,
k 2 ,
p 1 ,
p 2 ) = f 1 (k 1 ) η ab + f 2 (k 1 ) k a
1 k b
1 + f 3 (k 1 ) ab
1 ,
(5.10.75)
123
−
1
tu
2 p a
1 p b
1 + p a
3
p b
1 − p b
4
+ 2 p
[a
1 p
b]
4
,
(5.10.66)
Z := 1 +
m 2
p + m 2
e + 2 p 1 · q 2
m 2
e + p 1 · p 2
+
2( p 1 · q 2 ) 2
m 2
e + p 1 · p 2
2 ,
(5.10.67)
˜
Z :=
2m p m e
m 2
e + p 1 · p 2
,
(5.10.68)
Z ab :=
p a
1 p b
1 + 2q a
2 q b
1
m 2
e + p 1 · p 2
+
m 2
p
p a
1 − p b
2
p b
1 + 2
p a
1 − p a
2
q b
2 − q a
2 p b
1
p 1 · p 2
m 2
e + p 1 · p 2
2
,
(5.10.69)
˜
Z ab :=
m p p a
1 q b
2
m e
m 2
e + p 1 · p 2
+
m p m e
p a
1 − p a
2
q b
2
m 2
e + p 1 · p 2
2 ,
(5.10.70)
W a := m 3
e
1
k 1 · p 1
−
1
k 1 · p 2
2
k a
2 − m e
1
k 1 · p 1
−
1
k 1 · p 2
p a
2 + 2k a
1
+
m e
k 1 · p 1
k 1 · p 2
k 1 · p 1
− 1
+
1
m e
k 1 · p 1
k 1 · p 2
−
k 1 · p 2
k 1 · p 1
p a
1 ,
(5.10.71)
W ab :=
1 − B
m 2
e
k a
1 k b
1 + Bp a
1 p b
1
+ (1 + B)
k a
1 p b
1
k 1 · p 1
−
p a
1 − p a
2
k b
1
k 1 · p 2
,
we have made use of the Mandelstam variables
t := ( p 1 − p 3 ) 2 ≡ −2
m 2
e + p 1 · p 3
,
u := ( p 1 − p 4 ) 2 ≡ −2
m 2
e + p 1 · p 4
,
(5.10.72)
in the e + e → e + e case and again the remaining terms can be obtained by interchanging electrons and protons. The “TT” superscript in Eq. (5.10.30) stands for
“transverse-traceless part”
X
TT
ab := X ab −
1
2
ab X
c
c ,
(5.10.73)
so the right-hand side expression is consistently transverse with respect to k
a
1 . Similarly, we also check that the integrands of (5.10.34), (5.10.35) and (5.10.36) are
consistently transverse with respect to p
a
1 . As mentioned at the end of Sect. 5.5, the
integrands of the collision matrices are not necessarily hermitian, only their integrals
are, which for the real variables employed here translates into imaginary contributions. For instance, ˜
C
e contains an imaginary term of the form
˜
C
e
⊃ i ˜
P ab,1
d
3 k 2
(2π) 3 2k 2
d
3 p 1
(2π) 3 2E p 1
d
3 p 2
(2π) 3 2E p 2
K
ab
(
k 1 ,
k 2 ,
p 1 ,
p 2 ) . (5.10.74)
To see that this is zero, we note that, whatever the K
ab function, the result of the
integral must be a tensor distribution by Lorentz invariance. Consequently, given the
available dependencies, the integral can only be of the form
d 3 k 2
(2π) 3 2k 2
d 3 p 1
(2π) 3 2E p 1
d 3 p 2
(2π) 3 2E p 2
K ab (
k 1 ,
k 2 ,
p 1 ,
p 2 ) = f 1 (k 1 ) η ab + f 2 (k 1 ) k a
1 k b
1 + f 3 (k 1 ) ab
1 ,
(5.10.75)