Casimir's Trick/Evaluating Cross Sectionsby Spriteling Tags: casimir, cross, sections, trick or evaluating 

#1
Jul1811, 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 4d Levi Civita tensor contracted with the metric tensor. I'm working on finding the crosssection for neutrinoproton 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..) 



#2
Jul1811, 02:41 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

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.




#3
Jul1811, 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 4vectors. How would you be able to evaluate this? 



#4
Jul1811, 06:35 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

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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