Thanks for your help all.
I suppose my problem now boils down to this. In Feynman diagrams we have internal lines and external lines. We call external lines real particles with on shell momentum, and internal lines virtual particles with no such on-shell constraint. Now if we have a one vertex...
Thanks.
I don't really understand the on/off-shell business. I understand what it means,
p^{\mu}p_{\mu}=m^2
but where does that condition come from, and why doesn't it apply to intermediate particles.
Many thanks,
Gordon.
I didn't mean to say that, it's just that I'm only thinking in terms of lowest order diagrams, which I would hope are the leading order in behaviour anyway.
Regards,
Gordon.
Hello,
Thanks for the help.
Just to clarify, what you have said is that all photons, are intermediate states between charged particles and therefore the distinction between real and virtual particles is based on their proximity to the mass shell. All photons real and virtual can be detected...
According to what I have read, virtual particles are those which are intermediate photons in interactions, which cannot be directly observed. If my understanding is correct that would mean that they consist of (in low order perturbation theory), (among other things) u and t channel photons. But...
Yes that is what I mean, commute the first gamma all the way to the end and then bring it back to the start using the cyclic property, then commute it through the gamma 5 to leave it back where it started. However, in this case, this process does not result in an overall factor of -1 on the big...
On reflection it may not be zero. There are six indices with four possibilities for each. Therefore some must be equal. Remember that unequal indices anticommute. The possibilities are:
1. 5 indices are equal. Four can be squared and removed, leaving
tr(\gamma^{5}\gamma^{\mu}\gamma^{\nu})=0...
yes, your strange feeling is correct. after pushing the \gamma^{\mu} all the way back to where it started you end up with a plus sign on the big trace meaning it cancels with the one on the far side leaving you with the funny identity...
The \phi integral just adds a factor of 2\pi, change variables in the \theta integral to x=\cos(\theta) and then it becomes a lot nicer.
Good luck.
Eoin Kerrane.
Hello,
I think I need the following for a QFT problem.tr(\gamma^{5}\gamma^{\mu}\gamma^{\alpha}\gamma^{\beta}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma})
I know that...