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0
votes
How to prove $ g^{\mu\nu}\Lambda^{\rho}{}_{\mu}\Lambda^{\sigma}{}_{\nu}=g^{\rho\sigma} $ for...
Let me use condition
$${\Lambda^{-1}}^{\rho}_{}{\nu}= \Lambda_{\nu}{}^{\rho}$$
We can raise $\nu$ index left side as well as right side,
$${\Lambda^{-1}}^{\rho\nu}= \Lambda^{\nu\rho}$$
Now contract ab …
2
votes
Derivative of line element in general relativity is zero?
Actually you don't need $\frac{d}{d\lambda} \sqrt{-g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu} =0$.
Since $$L=\sqrt{-g_{\mu\nu}\dot{x}^ \mu \dot {x}^\nu} $$, $L^2$ also satisfies Euler-Lagrange equations.
So …