Stars as Laboratories for Fundamental Physics - MPP Theory Group

Two-Photon Coupling of Low-Mass Bosons 171
is from decay; it involves a Bose stimulation factor [1+(e ωL/T −1) −1 ] =
(1−e −ωL/T ) −1 for the final-state γ L . For the axions, a stimulation factor
is not included because they are assumed to escape immediately.
Because a longitudinal excitation has no magnetic field, the electric
field in the matrix element corresponding to Eq. (5.1) must be
associated with a longitudinal, the magnetic field with a transverse excitation.
In general B = ∇ × A and E = −∇A 0 − ∂ t A with the vector
potential A. For a propagating mode A ∝ e −i(ωt−k·x) ϵ where ϵ is a
polarization vector. This implies that B ∝ k T × ϵ T where ϵ T is a polarization
vector transverse to k T with ϵ 2 T = 1, and E ∝ ω L ϵ L − k L ϵ 0 L
where ϵ L = (kL, 2 ω L k L )kL
−1 (ωL 2 − kL) 2 −1/2 according to Eq. (6.27) so that
E ∝ ˆk L (ωL 2 − kL) 2 1/2 . Then one finds from the usual Feynman rules
|M| 2 = g 2 aγ (ω 2 L − k 2 L) |ˆk L · (k T × ϵ T )| 2 . (5.12)
Note that |ˆk L · (k T × ϵ T )| 2 = |(ˆk L × k T ) · ϵ T | 2 . Averaging over the two
transverse polarization states yields 1 2 |ˆk L × k T | 2 .
If one writes Z L = ˜Z L ω 2 L/(ω 2 L − k 2 L) as in Sect. 6.3 and performs the
d 3 k L integration in Eq. (5.11) one finds
Γ γT →a =
g2 ∫
aγ
16 (2π) 2
×
dΩ a dω a Z T ˜Z L
ω a ω L
ω T
|ˆk L × k T | 2
[ δ(ωT + ω L − ω a )
e ω L/T
− 1
+ δ(ω T − ω L − ω a )
1 − e −ω L/T
]
. (5.13)
In a nondegenerate, nonrelativistic plasma typically ω P ≪ T . Because
ω L = ω P one may expand the exponentials so that (e ω L/T −1) −1 → T/ω L
and also (1 − e −ω L/T ) −1 → T/ω L . If ω T = O(T ) ≫ ω P one may ignore
ω L in the δ functions which are then trivial to integrate,
Γ γT →a =
g2 ∫
aγT
8 (2π) 2
dΩ a Z T ˜Z L |ˆk L × k T | 2 . (5.14)
In this limit a has the same energy as γ T while γ L has only provided
momentum.
To finish up, note that in the nondegenerate, nonrelativistic limit
Z T = ˜Z L = 1. Because ω P ≪ T and ω T = O(T ) we have ω P ≪ ω T
so that k T ≈ ω T . Moreover, k L = k T − k a yielding |ˆk L × k T | 2 =
|k a ×k T | 2 /|k a −k T | 2 = ω 2 T(1−z 2 )/2(1−z) = 1 2 ω2 T(1+z) where z is the