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In the derivation of the BGV theorem paper:

https://arxiv.org/abs/2307.10958

There is the following relation (numbered (8)):

$\dfrac{d}{d s}\left[a^2(t) \dfrac{d \mathbf{x}}{d s}\right]=0$

It's very likely dumb but i don't see how they get that start from the FLWR metric and geodesic equation...

If anyone can help.

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1 Answer 1

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This can be seen from the variational principle $$ \delta \int\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}=0 $$ and eq.(7) in the paper, i.e. $u_\mu u^\mu=-1$. From the Lagrange equations $$ \frac{d}{ds}\frac{\partial L}{\partial \dot x_\mu}=\frac{\partial L}{\partial x_\mu}, $$ where the dot means derivative with respect to $s$, it is seen immediately that $$ \frac{d}{ds}\left(\frac{1}{\sqrt{\dot x_0^2-a^2(t)|\dot{\bf x}|^2}}a^2(t)\dot{\bf x}\right)=0, $$ giving the expected result.

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  • $\begingroup$ Thanks. I will try to fill-in the gaps in the next three weeks and understand why you have a square root in th LHS of the final result that doesn't appear on the paper. I will come back after these 3 weeks if i fail. $\endgroup$ Mar 28 at 13:28
  • $\begingroup$ But the square root is 1 due to $u_\mu u^\mu=-1$ so, it is exactly the result in the paper. $\endgroup$
    – Jon
    Mar 28 at 13:40
  • $\begingroup$ Indeed the square root is 1. Try however to give me the time to check all LoL.... You are faster than your shadow and i'm slow as a snail... ;-) $\endgroup$ Mar 28 at 13:46

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