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I have found this equality that is the derivative of the energy for a local observer in GR, with energy defined as $E=-p^\mu u_\mu$:

$$dE/d\tau = -p^\mu p^\nu\nabla_{(\mu}u_{\nu)}$$

trying to derive it myself, what I did is (asssuming $m=1$ in $p^\mu = mdx^\mu/d\tau$):

$$\frac{dE}{d\tau} = \frac{dp^\mu}{d\tau}\frac{\partial E}{\partial p^\mu} = -p^\mu p^\nu\nabla_\mu u_\nu - p^\mu u_\nu\nabla_\mu p^\nu$$

and the first term is $-p^\mu p^\nu\nabla_{(\mu}u_{\nu)}$, but what about the second term? It seems like a divergence of the 4-momentum and perhaps that should be zero, but the indices don't match (it's not of the form $\nabla_\mu p^\mu$).

EDIT/ANSWER: As @Cham below has explained, the $u$ in $E = -p^\mu u_\mu$ is the velocity of the observer, while $p$ is the momentum of the particle, so it is incorrect to do $p^\mu=mdx^\mu /d\tau$.

On the other hand, the correct expression for the derivative is $$\frac{d E}{d \tau}=-u^{\mu} u^{\nu} \nabla_{\nu} p_{\mu}-p_{\mu} u^{\nu} \nabla_{\nu} u^{\mu}$$

where the second term is 0 (equation of a geodesic) and the first one is the desired term (we can perform the symmetrization).

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    $\begingroup$ Is $\tau$ the proper time of the observer, or of the particle? I guess it should be the proper time of the observer to make sense of that equation. $\endgroup$ – Cham Sep 10 at 14:49
  • $\begingroup$ Yes, it is for the observer. $\endgroup$ – David Sep 10 at 14:55
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    $\begingroup$ Then I think you mixed up that parameter with the particle's proper time in your second equation. If $\tau$ is the observer's proper time, then I get $$\frac{dE}{d\tau} = -\, u^{\mu} \, u^{\nu} \, \nabla_{\nu} \, p_{\mu} - p_{\mu} \, u^{\nu} \, \nabla_{\nu} \, u^{\mu}.$$ The second term vanishes if the observer is in free fall. The first term could be "symetrized" and we get your first equation (with $u^{\mu}$ and $p^{\mu}$ reversed). $\endgroup$ – Cham Sep 10 at 15:01
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    $\begingroup$ In general, $p^{\mu} \ne m \, u^{\mu}$ ! If $p^{\mu} = m \, u^{\mu}$, then $E = -\, p_{\mu} \, u^{\mu} = m$, which is trivially a constant and $dE/d\tau = 0$. Don't confuse the oberver's 4-velocity and the particle's 4-momentum. $\endgroup$ – Cham Sep 10 at 15:25
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    $\begingroup$ To make things clearer, I would write $u_{\mathcal{O}}^{\mu}$ for the observer and $p^{\mu} = m \, u_{\mathcal{P}}^{\mu}$ for the particle, so $$E_{\mathcal{PO}} = -\, p_{\mu} \, u_{\mathcal{O}}^{\mu} = -\, m \, g_{\mu \nu} \, u_{\mathcal{P}}^{\mu} \, u_{\mathcal{O}}^{\nu}$$ is the $\mathcal{P}$ particle's energy as seen by the observer $\mathcal{O}$. $\endgroup$ – Cham Sep 10 at 15:28

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