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Stars as Laboratories for Fundamental Physics - MPP Theory Group

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104 Chapter 3<br />

by n e = p 3 F/3π 2 . This also implies that |q| 2 ≈ |p 1 − p 2 | 2 ≈ 2p 2 F(1 − c 12 )<br />

where c 12 is the cosine of the angle between p 1 and p 2 . With these<br />

approximations and the velocity at the Fermi surface β F ≡ p F /E F =<br />

p F /(m 2 e + p 2 F) 1/2 one finds<br />

where<br />

∫<br />

F =<br />

Q = π2 α 2 α ′<br />

15<br />

dΩ2<br />

4π<br />

∫<br />

dΩa<br />

4π<br />

T 4 ( ∑<br />

m 2 e j<br />

n j Z 2 j<br />

)<br />

F, (3.33)<br />

(1 − β 2 F) [2 (1 − c 12 ) − (c 1a − c 2a ) 2 ]<br />

(1 − c 1a β F ) (1 − c 2a β F ) (1 − c 12 )(1 − c 12 + κ 2 )<br />

(3.34)<br />

with κ 2 ≡ k 2 S/2p 2 F.<br />

For a single species of nuclei with charge Ze and atomic weight A<br />

the energy-loss rate per unit m<strong>as</strong>s is<br />

ϵ = π2 α 2 α ′<br />

15<br />

Z 2<br />

A<br />

T 4<br />

F<br />

m u m 2 e<br />

= α ′ 1.08×10 27 erg g −1 −1 Z2<br />

s<br />

A T 8 4 F, (3.35)<br />

where again T 8 = T/10 8 K. Because F is of order unity <strong>for</strong> all conditions,<br />

the bremsstrahlung rate mostly depends on the temperature and<br />

chemical composition, and ϵ is not suppressed at high density.<br />

Expanding Eq. (3.34) in powers of β F <strong>for</strong> nonrelativistic or partially<br />

relativistic electrons one finds<br />

F = 2 ( ) [ ( ) 2 + κ<br />

2 2 + 5κ<br />

2 2 + κ<br />

2<br />

3 ln + ln − 2 ]<br />

β<br />

κ 2 15 κ 2 F 2 + O(β 4<br />

3<br />

F).<br />

(3.36)<br />

There<strong>for</strong>e, in contr<strong>as</strong>t to the nondegenerate calculation this expression<br />

would diverge in the absence of screening.<br />

Another approximation can be made if one observes that Coulomb<br />

scattering is mostly <strong>for</strong>ward, i.e. the main contribution to the integral<br />

is from c 12 ≈ 1 which implies c 1a ≈ c 2a . With c 2a = c 1a only in the<br />

denominator one obtains<br />

F = 2 ( ) [ ( ) ]<br />

2 + κ<br />

2 2 + 3κ<br />

2 2 + κ<br />

2<br />

3 ln + ln − 1 f(β<br />

κ 2 6 κ 2 F ) (3.37)<br />

with<br />

− 3 (1 − β2 F)<br />

2β 3 F<br />

( ) 1 + βF<br />

. (3.38)<br />

1 − β F<br />

f(β F ) = 3 − 2β2 F<br />

βF<br />

2 ln<br />

The function f(β F ) is 0 at β F = 0 and rises monotonically to 1 <strong>for</strong>

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