- #1

- 463

- 39

[tex]M = M^{\mu} \epsilon^{*}_{\mu}}[/tex]

then

[tex] \sum_\epsilon |\epsilon^{*}_\mu M^\mu |^2 = \sum_\epsilon \epsilon^{*}_\mu epsilon\nu M^\mu M^{* \nu}[/tex]

and you can replace the sum over polarizations with a [tex]-g_{\mu \nu}[/tex]

But what if you cannot separate it out? Say your M is of the form:

[tex]M=\epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma + A\epsilon^{*}_\mu[/tex]

Do you square it out, but then the first term will be a [tex]-g^{\alpha \alpha'}[/tex] so each term gets summed over different indices?

[tex]\sum_\epsilon |M|^2 =\left( \epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma + A\epsilon^{*}_\mu\right)\left(\epsilon_{\nu \alpha' \beta' \sigma'} \epsilon^{\alpha'} q^{\beta'} p^{\sigma'} + A\epsilon_\nu \right)[/tex]

[tex]\sum_\epsilon |M|^2 =\left( \epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma \epsilon_{\nu \alpha' \beta' \sigma'} \epsilon^{\alpha'} q^{\beta'} p^{\sigma'} + A\epsilon^{*}_\mu\epsilon_{\nu \alpha' \beta' \sigma'} \epsilon^{\alpha'} q^{\beta'} p^{\sigma'} + \epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma A\epsilon_\nu+ A\epsilon^{*}_\mu A\epsilon_\nu\right)[/tex]

So my question is, for every pair of polarization vectors do I make the replacement to the metric tensor? Or do I multiply the entire thing by g mu nu?