So, I was watching this video https://www.youtube.com/watch?v=cj-QpsZsDDY&t=3968s , where at some point the prof explains that a divergence can be ontained by "exploiting the antisymmetry" of an infinitesimal Lorentz transformation.

What it boils down to is this: Let $x^\mu$ be some inertial coordinate system in flat 4D spacetime with Minkowstki metric $\eta_{\mu\nu}$. Consider the infinitesimal Lorentz transformation $$ x'^\mu=\Lambda^\mu_{\;\nu}x^\nu=(\delta^\mu_{\;\nu}+\omega^\mu_{\;\nu})x^\nu $$ with $|\omega^\mu_{\;\nu}| \ll 1$. It is easy to show (see e.g. this question) that this requires that $\omega^{\mu\sigma}=\eta^{\sigma\nu}\omega^\mu_{\;\nu}$ be antisymmetric.

Now here's my question: apparently this antisymmetry allows one to conclude, that $$ \partial_\mu(\omega^\mu_{\;\nu} x^\nu)=0 $$ Yet, I can't manage to prove that. Here's what I came up with sofar $$ \partial_\mu(\omega^\mu_{\;\nu} x^\nu)=\partial_\mu(\omega^{\mu\sigma}\eta_{\sigma\nu}x^\nu) \\ =\frac{1}{2}\eta_{\sigma\nu}\partial_\mu(\omega^{\mu\sigma}x^\nu)-\frac{1}{2}\eta_{\sigma\nu}\partial_\mu(\omega^{\sigma\mu}x^\nu) \\ =\frac{1}{2}\eta_{\sigma\nu}\partial_\mu(\omega^{\mu\sigma}x^\nu)-\frac{1}{2}\eta_{\mu\nu}\partial_\sigma(\omega^{\mu\sigma}x^\nu) $$

none of which seems to bring me any closer though... Am a missing something trivial?


Ok, never mind. I did indeed miss something completely trivial. In my scenario, the $\omega^\mu_{\;\nu}$'s are constant; I've just been doing far too much general relativity lately, where everything depends on position ;)

Anyway, remembering that $\partial_\sigma \omega^\mu_{\;\nu}=0$, the proof is indeed trivial:

$$ \partial_\mu(\omega^\mu_{\;\nu} x^\nu)=\omega^\mu_{\;\nu}\partial_\mu x^\nu=\omega^{\mu\sigma}\eta_{\sigma\nu}\delta^\nu_{\;\mu}=\omega^{\mu\sigma}\eta_{\sigma\mu}=0 $$ since $\eta_{\sigma\mu}$ is symmetric. My bad.

By the way, anyone interested in the original issue, which is finding the conserved currents in a scalar field, generated by Lorentz symmetries, can also look at this question, which I just found. It confirms, among other things, my solution here.

  • $\begingroup$ Moral: stop doing general relativity? :p $\endgroup$ – QuantumDot Mar 8 at 20:19
  • $\begingroup$ Absolutely not. I love general relativity :) $\endgroup$ – James Bates Mar 11 at 14:53

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