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Casimir's Trick/Evaluating Cross Sections

by Spriteling
Tags: casimir, cross, sections, trick or evaluating
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Spriteling
#1
Jul18-11, 01:36 PM
P: 36
Hi all.

I'm working on a project at the moment, and I've run into some questions regarding the 4-d Levi Civita tensor contracted with the metric tensor.

I'm working on finding the cross-section for neutrino-proton scattering. While ccontracting the hadronic and leptonic tensors, I end up with a term like [itex]i \epsilon_{\alpha \mu \beta \nu} k'_{\alpha}k_{\beta}g^{\mu \nu}[/itex] but I am uncertain as to how to evaluate this term. I've looked all through Griffiths and Halzen and Marten and I've googled but I can't find a definitive answer. Can anyone help?
(I'm sorry if this is in the wrong section; I wasn't sure if it should go here, as it is related to HEP, or int he homework help thing, which it isn't really, but it is a question..)
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fzero
#2
Jul18-11, 02:41 PM
Sci Advisor
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PF Gold
P: 2,603
Symmetry of the metric ([itex]g^{\mu \nu}=g^{\nu \mu}[/itex]) and total antisymmetry of [itex] \epsilon_{\alpha \mu \beta \nu}[/itex] imply that [itex]\epsilon_{\alpha \mu \beta \nu}g^{\mu \nu}=0[/itex] (together with all possible permutations of indices). So all terms of that form vanish.
Spriteling
#3
Jul18-11, 02:58 PM
P: 36
Ah, that makes quite a lot of sense.

What if, on the other hand, you have a term like [itex]i \epsilon_{\alpha \mu \beta \nu} k'_{\alpha}k_{\beta}P^{\mu} P^{\nu}[/itex]? k, k' and P are all 4-vectors. How would you be able to evaluate this?

fzero
#4
Jul18-11, 06:35 PM
Sci Advisor
HW Helper
PF Gold
P: 2,603
Casimir's Trick/Evaluating Cross Sections

The product [itex]P^\mu P^\nu[/itex] is symmetric under [itex]\mu\leftrightarrow\nu[/itex], so that term vanishes by symmetry as well. If there were a prime on one of the [itex]P[/itex]s, it would not.

To anticipate the next logical question, we can ask how to express [itex]i \epsilon_{\alpha \mu \beta \nu} k'^{\alpha}k^{\beta}P^{\mu} P'^{\nu}[/itex]. There really isn't any simpler form to reduce it to, but we can note that

[itex]i \epsilon_{\alpha \mu \beta \nu} k'^{\alpha}k^{\beta}P^{\mu} P'^{\nu} = i\left( k'^0k^1P^2 P'^3 \pm \text{permutations} \right).[/itex]


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