Spin statistic terms in absorption cross section

FunkyDwarf
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'Ello,

I have a question regarding the results in this paper (and another which I will mention later)

http://arxiv.org/abs/hep-ph/0212199

Now, I'm not so concerned about the 'braney' bit, but more their definition of the cross section in Eqn. (46). They have included the usual (2j+1) term (which is present even in non-relativistic physics) but it seems to me it should be the full g(j) = (2j+1)/((2a+1)(2b+1)) where a and b are the spins of the incident particle and the target (in this case one of them can be zero as they consider a black hole target which is modeled as a classical potential, in some sense).

A similar definition seems to be used in this paper by R. Fabbri:
http://prd.aps.org/abstract/PRD/v12/i4/p933_1

in this case in Eqn. (34)

In both cases they consider spin 1 and so one would expect a factor of (2j+1)/3, or no?

Any help would be appreciated :\

Cheers!
-Z
 
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Isn't the rule "average over initial, sum over final." Meaning that if the reaction probability is independent of the initial spin orientations, they don't contribute any weight. But the final spins are summed over, and do contribute a factor.
 
Well the transmission probability will certainly be spin dependent, but I'm talking about the spin term in the general cross section definition. I'm simply trying to work out why they include (2j+1) instead of the 'full' general term, given they are not considering a scalar incident particle.
 
Ok, slightly different question :)

The absorption cross section for spin 0 is defined as

<br /> \sigma =\frac{\pi}{\epsilon^2} \Sigma (2l+1) T_l<br />
for transmission coefficient $T_l$.

I understand the (2l+1) factor comes from the expansion of the plane wave to look at scattering. I don't suppose someone could point me to a similar expansion in the spin half case where the cross section is

<br /> \sigma =\frac{\pi}{\epsilon^2} \Sigma|\kappa| T_l<br />

Clearly the direct transformation between kappa and l doesn't allow you to go between these two equations (at least without a factor of 2).

Thanks,
-Z
 
Ok so apparently I'm a spazz and can't find the edit button, solved the above problems so thread can be closed/removed if needed :) Thanks guys!
 
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