0

I have two short questions in the steps of doing Lorentz transformations with angular momentum.

First, the infinitesimal form of the Lorentz transform is set up as $\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + w^{\mu}_{\nu}$, where $w^{\mu}_{\nu}$ is infinitesimal. I understand up to the part where the condition which makes $\Lambda$ a Lorentz transformation would be $$(\delta^{\mu}_{\sigma} + w^{\mu}_{\sigma})(\delta^{\nu}_{\tau} + w^{\nu}_{\tau})\eta^{\sigma\tau} = \eta^{\mu\nu}.$$

However, I am having a trouble getting how the condition $ w^{\mu \nu}+w^{\nu \mu} = 0$ would ensure the above equations to hold.

After a few steps away, I understood up to the part $\delta L = -w^{\mu}_{\nu} x^{\nu} \partial_{\mu}L$, where $L$ is a Lagrangian density.

But I am not exactly understanding how the fact that $w^{\mu}_{\mu} = 0$ due to the antisymmetry makes $-w^{\mu}_{\nu} x^{\nu} \partial_{\mu}L = -\partial_{\mu}(w^{\mu}_{\nu}x^{\nu}L)$.

Qmechanic
  • 220,844
Ellen Lee
  • 11

1 Answers1

0

First, $\partial_\mu \Lambda^\mu_\nu = 0$ since the $\Lambda^\mu_\nu$s are just numbers. This implies $\partial_\mu w^\mu_\nu = 0$. Then, you just have to compute $\partial_\mu(w^\mu_\nu x^\nu L) = w^\mu_\nu (\partial_\mu x^\nu) L + w^\mu_\nu x^\nu (\partial_\mu L)$. The first term may be rewritten $w^\mu_\nu \delta^\nu_\mu$, which vanishes.

Theosics
  • 1