# Why this expression is time-reversal odd?

 P: 27 P and k are four-momentum of two particles. I read in a paper which said that $[\slashed{P},\slashed{k}]=\slashed{P}\slashed{k} - \slashed{k}\slashed{P}$ is time-reversal odd. Why?
 P: 754 could you please fix the Latex? What is P and k? momenta?
 Emeritus Sci Advisor PF Gold P: 29,239 ... and please make a full citation of the paper. Zz.
 Mentor P: 6,231 Why this expression is time-reversal odd? Unfortunately, MathJax does not support the "slashed" command from the amsmath package, which is used to implement Feynman's slash notation. Try $$\left[\gamma \cdot P , \gamma \cdot k \right] = \gamma \cdot P \ \gamma \cdot k − \gamma \cdot k \ \gamma \cdot P,$$ where ## \gamma \cdot P = \gamma^\mu P_\mu##. Or just use (the possible more useful in this case) ##\gamma^\mu P_\mu## and ##\gamma^\nu k_\nu##.[/edit]
 Thanks P: 1,948 Try [t e x]/ \!\!\!\! P[/ t e x]: $/ \!\!\!\! P$ and [t e x]/ \!\!\! k[/ t e x]: $/ \!\!\! k$
P: 27
 Quote by ZapperZ ... and please make a full citation of the paper. Zz.
http://arxiv.org/abs/hep-ph/0504130
Eq.(7), the ## A_4 ## term.

Regards
P: 27
 Quote by George Jones Unfortunately, MathJax does not support the "slashed" command from the amsmath package, which is used to implement Feynman's slash notation. Try $$\left[\gamma \cdot P , \gamma \cdot k \right] = \gamma \cdot P \ \gamma \cdot k − \gamma \cdot k \ \gamma \cdot P,$$ where ## \gamma \cdot P = \gamma^\mu P_\mu##. Or just use (the possible more useful in this case) ##\gamma^\mu P_\mu## and ##\gamma^\nu k_\nu##.[/edit]
Thanks a lot for the texing.
 P: 754 *wrong answer* Well, the reason why there is T-oddness I guess is because there is a $i$ in front, making that term complex..
P: 27
 Quote by ChrisVer Doesn't that say that $A_{4}$ is T-odd? no the [P,k]
But ##\Phi(P,k,S|n)## on the left of the Eq. must be T-even according to it's physical meaning, so, if ##A_4## is T-odd, ##[\gamma \cdot P, \gamma \cdot k]## must be T-odd.
Actually, I think it is ##[P, k]## be T-odd that infers ##A_4## be T-odd, not the opposite, because ##A_4## is an unknown coefficient function.
So I wonder why.
Thanks a lot!
 P: 754 lol, I changed my initial post because it was wrong, I guess that: "Well, the reason why there is T-oddness I guess is because there is a $i$ in front, making that term complex.." is your answer...
P: 27
 Quote by ChrisVer lol, I changed my initial post because it was wrong, I guess that: "Well, the reason why there is T-oddness I guess is because there is a $i$ in front, making that term complex.." is your answer...
Well, I think the ##i## in front is just for convenience. And the gamma matrices are not all real, for example, in Dirac representation, ##\gamma^2## is complex.
So, I think the problem is how does ##\left[\gamma \cdot P , \gamma \cdot k \right]##change under time reversal operation.

I know that, under time reversal, ##P=(P_0,\vec{P})##becomes ##(P_0,-\vec{P})##, and in QFT course I've learned that ##\gamma^{\mu}##becomes ##(-1)^{\mu}\gamma^{\mu}##(Peskin's book, Page71). But what about ##\left[\gamma \cdot P , \gamma \cdot k \right]##?
Thanks
P: 4,160
 Quote by Chenkb Well, I think the ##i## in front is just for convenience.
No, it's not there just for convenience. It's there because i[γ·P, γ·K] = 2 σμν Pμ Kν. And if you look at the table on p.71 of Peskin, you'll see how σμν transforms.
P: 27
 Quote by Bill_K No, it's not there just for convenience. It's there because i[γ·P, γ·K] = 2 σμν Pμ Kν. And if you look at the table on p.71 of Peskin, you'll see how σμν transforms.
Year, but for the terms in ##\sigma_{\mu\nu}P^{\mu}k^{\nu}## like ##\sigma_{02}P_0k_2##, is T-even, because ##\sigma_{02}## odd, ##P_0## even, and ##k_2## odd.
Regards!
Thanks
P: 4,160
 Quote by Chenkb Year, but for the terms in ##\sigma_{\mu\nu}P^{\mu}k^{\nu}## like ##\sigma_{02}P_0k_2##, is T-even, because ##\sigma_{02}## odd, ##P_0## even, and ##k_2## odd. Regards!
No, σ02 is even. In the sum, σ0i is even, σjk is odd, P0 and K0 are even, while Pj and Kj are odd. All the terms in the sum σμνPμKν wind up transforming the same way, namely they are all odd.
P: 27
 Quote by Bill_K No, σ02 is even. In the sum, σ0i is even, σjk is odd, P0 and K0 are even, while Pj and Kj are odd. All the terms in the sum σμνPμKν wind up transforming the same way, namely they are all odd.
Could you please explain why ##\sigma_{0i}## is T-even? isn't it ##-(-1)^{\mu}(-1)^{\nu}##?