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Oct 15, 2020 at 19:21 comment added Prahar @SamapanBhadury - see comment above.
Oct 15, 2020 at 19:21 comment added Prahar Let us continue this discussion in chat.
Oct 15, 2020 at 19:21 comment added Samapan Bhadury can you kindly explain this statement or direct me to a reference.
Oct 15, 2020 at 19:17 comment added Prahar @SamapanBhadury No no. I mean that $\delta^\alpha_\beta$ is a Lorentz invariant tensor, but $\delta_{\alpha\beta}$ is NOT and we are only allowed to use Lorentz invariant tensors since the LHS is a Lorentz invariant tensor.
Oct 15, 2020 at 19:16 comment added Samapan Bhadury you mean to say $\delta^\alpha_\beta$ does not exist?
Oct 15, 2020 at 19:10 comment added Prahar @SamapanBhadury - $\delta$ has one upper index and one lower index. The LHS has all four indices lowered.
Oct 15, 2020 at 19:09 comment added Samapan Bhadury why not $\delta_{\alpha\beta} \delta{\gamma\delta}$? Why can't we construct with such $\delta$-functions?
Oct 15, 2020 at 18:54 comment added Prahar @SamapanBhadury - (1) what else could it be? I've already written down all the Lorentz invariant quantities for you. What remains is to construct something from those but with the right index structure. (2) ya, that sounds right. my metric convention is (-+++)
Oct 15, 2020 at 18:52 comment added Samapan Bhadury There is a difference of sign with your proof and the one of Peskin. Is it due to different metric choices?
Oct 15, 2020 at 18:50 comment added Samapan Bhadury @Pahar The one thing I don't understand is, why the only Lorentz invariant quantity possible is $\varepsilon_{\alpha \gamma} \varepsilon_{\dot{\beta} \dot{\delta}}$?
Jan 18, 2020 at 19:22 history answered Prahar CC BY-SA 4.0