Lorentz invariance of an equation (metric)

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
felixphysics
Messages
3
Reaction score
0
soruboru.jpg
 
Physics news on Phys.org
I'm not sure what you mean by "metric is invariant." The metric transforms like a tensor, covariant or contravariant depending on whether the indices are up or down. The line element ##ds^2 = g_{\mu\nu} dx^\mu dx^\nu## is Lorentz invariant, but that's not what appears in your formula.

You should write down the explict transformation rules for ##V^\mu##, ##\partial/\partial x^\mu##, ##g^{\nu\sigma}##, and ##g_{\nu\sigma}##. You will need to combine them in your expression and carefully apply the derivatives with the appropriate product rules for differentiation.