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Stars as Laboratories for Fundamental Physics - MPP Theory Group

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122 Chapter 4<br />

Because the matrix element h<strong>as</strong> been <strong>as</strong>sumed to be a constant<br />

it can be pulled out of the integral which reduces to a ph<strong>as</strong>e-space<br />

volume. Nonrelativistically, d 3 p i /[2E i (2π) 3 ] = d 3 p i /[2m N (2π) 3 ] while<br />

E i = p 2 i /2m N in the energy δ function. The axion momentum is ignored<br />

according to Eq. (4.5). One uses CM momenta p 1,2 = p 0 ±p and p 3,4 =<br />

p 0 ± q where p 0 = 1 2 (p 1 + p 2 ) = 1 2 (p 3 + p 4 ) and defines u 2 ≡ p 2 /m N T<br />

and y ≡ v 2 ≡ q 2 /m N T . Then one finds explicitly<br />

Γ σ = 4 √ π απ 2 n B T 1/2 m −5/2<br />

N ,<br />

∫<br />

s(x) = 4 du dv u 2 v 2 e |x|−u2 δ(u 2 − v 2 − |x|)<br />

=<br />

∫ ∞<br />

0<br />

dy e −y ( |x|y + y 2) 1/2<br />

≈<br />

√1 + |x| π/4 . (4.7)<br />

The analytic <strong>for</strong>m is accurate to better than 2.2% everywhere; it h<strong>as</strong><br />

the correct <strong>as</strong>ymptotic behavior s(0) = 1 and s(|x|≫1) = (|x|π/4) 1/2 .<br />

We will see in Sect. 4.6.3 that <strong>for</strong> |x| ≫ 1 the nondegenerate s(x)<br />

must actually decre<strong>as</strong>e, in conflict with this explicit calculation. This<br />

problem reveals a pathology of the OPE potential which is too singular<br />

at short distances. For the present discussion this is of no concern so<br />

that I stick to the explicit OPE result in order to facilitate comparison<br />

with the existing literature.<br />

To determine the total emission rate one uses the first representation<br />

of s(x) and per<strong>for</strong>ms ∫ dx first to remove the δ function; the remaining<br />

integrals are e<strong>as</strong>ily done. Explicit results <strong>for</strong> s n ≡ ∫ ∞<br />

0 x n s(x) e −x dx are<br />

given in Tab. 4.1. The normalized axion energy spectrum dN a /dx =<br />

x s(x)e −x /s 1 is shown in Fig. 4.2 (solid line). The average axion energy<br />

is ⟨ω a ⟩/T = s 2 /s 1 = 16/7. The total nondegenerate energy-loss rate is<br />

Table 4.1. s n = ∫ ∞<br />

0<br />

x n s(x) e −x dx <strong>for</strong> nondegenerate (ND) and degenerate<br />

(D) conditions.<br />

n s n (ND) s n (D)<br />

1 8/5 2ζ 3 + 6ζ 5 /π 2<br />

2 128/35 31π 4 /315<br />

3 256/21 24ζ 5 + (180/π 2 ) ζ 7<br />

4 4096/77 82π 6 /315

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