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I want to show that the following quantity is a constant $$ g_{\mu\nu} \frac{d x^\mu }{d\tau} \frac{d x^\nu}{d \tau} = \epsilon, $$ where $\epsilon \equiv $ constant.

This can be achieved by formally integrating the geodesic equation. There is a neat trick involved which is performed by expressing the geodesic equation in the following way

$$ \frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha \beta} \frac{d x^\alpha }{d\tau} \frac{d x^\beta}{d \tau} \equiv \frac{d}{d\tau} \left( g_{\mu\nu} \frac{d x^\nu}{d\tau} \right) - \frac{1}{2} g_{\alpha \beta, \mu} \frac{d x^\alpha }{d\tau} \frac{d x^\beta}{d \tau} = 0. \tag{1} \label{eq:1} $$

So my question is regards Eq. (\ref{eq:1}) as I cannot see how this results obtained? Any suggestions?

For sake of completeness I will proceed to show that the above expression is indeed a constant.

Now, accepting (\ref{eq:1}) as gospel I went ahead with the following operations:

$$\frac{d}{d\tau} \left( g_{\mu\nu} \frac{d x^\nu}{d\tau} \right) - \frac{1}{2} g_{\alpha \beta, \mu} \frac{d x^\alpha }{d\tau} \frac{d x^\beta}{d \tau} = 0, \\ \frac{d g_{\mu\nu} }{d \tau} \frac{d x^\mu}{d\tau} \frac{d x^\nu}{d\tau} + g_{\mu\nu} \frac{dx^\nu}{d\tau} \frac{d^2 x^\mu}{d\tau^2} - \frac{1}{2} \frac{d g_{\alpha \beta}}{d\tau} \frac{d x^\alpha }{d\tau} \frac{d x^\beta}{d \tau} = 0, $$

where I've simply multiplied by $dx^\mu /d\tau$. Re-labelling $\mu \rightarrow \alpha, \mu \rightarrow \beta$ we get

$$g_{\alpha \beta} \frac{dx^\alpha}{d\tau} \frac{d^2 x^\beta}{d\tau^2} + \frac{1}{2} \frac{d g_{\alpha \beta}}{d\tau} \frac{d x^\alpha }{d\tau} \frac{d x^\beta}{d \tau} = 0,$$

which can be equivalently expressed as

$$\frac{d}{d\tau} \left( g_{\alpha \beta} \frac{d x^\alpha}{d \tau}\frac{d x^\beta}{d\tau} \right) = 0. $$

Of course we obtain a factor of two by differentiating the term $$ \frac{d}{d\tau}\left( \frac{d x^\alpha}{d \tau}\frac{d x^\beta}{d\tau} \right) = 2 \frac{d x^\alpha}{d \tau}\frac{d^2 x^\beta}{d\tau^2}.$$

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  • $\begingroup$ @Frobenius I'm wondering about the steps in Eq. (1). $\endgroup$ – Rumplestillskin Oct 28 '17 at 8:50
  • $\begingroup$ If a quantity is constant it's differential is zero? I didn't really test this out. $\endgroup$ – JMLCarter Oct 28 '17 at 9:52
  • $\begingroup$ @Frobenius I am obviously not being clear. I know what you are saying. However, this was just an exercise in a book on GR. I can do the exercise no problem and I understand that it is not a customary one. My question is, as stated above: for Eq. (1) they express the geodesic equation of motion in two equivalent ways and I can't see where the second one comes from. $\endgroup$ – Rumplestillskin Oct 29 '17 at 5:48
  • $\begingroup$ @Frobenius You are very correct with regards the indices... I must have been looking at the book for one and doing in my head for the other :)! $\endgroup$ – Rumplestillskin Oct 29 '17 at 6:50
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If I understand correctly, you want to show that the 4-velocity of a particle moving on a geodesic is constant. There is a neat way to do this using the "modern" differential geometry notation.

Consider a spacetime $(M,g)$ and denote by $X$ be the 4-velocity of the particle. Let $\nabla$ be the Levi-Civita connection associated to $g$ on $M$.

Suppose WLOG that the particle moves on an affinely parametrised geodesic, i.e. $$\tag{1}\label{geo}\nabla_{X} X=0.$$

Let $g(X,X)=\epsilon$. Then $$\nabla_{X}\epsilon=(\nabla_{X}g)(X,X)+2 g(\nabla_{X}X,X)$$ where the factor $2$ comes from the symmetry of $g$. Now, using the geodesic equation \eqref{geo} and the fact that the Levi-Civita connection is metric compatible $$\nabla g=0,$$ you obtain that $\epsilon$ is constant along geodesics.


Now, regarding the identity you mention, note that we can write ($X^{\mu}=\frac{dx^{\mu}}{d\tau}$) $$\frac{d}{d\tau}X^{\mu}=\frac{d}{d\tau}\left(g^{\mu\rho}g_{\rho\nu}X^{\nu}\right)=g^{\mu\rho}\frac{d}{d\tau}\left(g_{\rho\nu}X^{\nu}\right)+\left(\frac{d}{d\tau} g^{\mu\rho}\right)g_{\rho \nu} X^{\nu}.$$ Using $$\frac{d}{d\tau} g^{\mu\rho}=-g^{\mu\alpha}g^{\rho\beta}\frac{d}{d\tau}g_{\alpha\beta}=-g^{\mu\alpha}g^{\rho\beta}g_{\alpha\beta,\gamma} X^{\gamma}$$ we get $$\label{eq2}\tag{2}\frac{d}{d\tau}X^{\mu}=g^{\mu\rho}\frac{d}{d\tau}\left(g_{\rho\nu}X^{\nu}\right)-g^{\mu\alpha}g_{\alpha\beta,\gamma} X^{\gamma}X^{\beta}.$$ In a basis, the geodesic equation reads $$\frac{d}{d\tau} X^{\mu}+\Gamma^{\mu}_{\rho\sigma}X^{\rho}X^{\sigma}=0$$ and expanding the Christoffel symbols $$\frac{d}{d\tau} X^{\mu}+\frac{1}{2}g^{\mu\alpha}(2g_{\alpha(\rho,\sigma)}-g_{\rho\sigma,\alpha})X^{\rho}X^{\sigma}=0.$$ Note that we can drop the symmetrisation in the previous equation because that term is contracted with $X^{\rho}X^{\sigma}$, which is symmetric.

Finally, substituting \eqref{eq2} and relabelling the indices you get the identity $$\frac{d}{d\tau} X^{\mu}-\frac{1}{2}g^{\mu\alpha}g_{\rho\sigma,\alpha}X^{\rho}X^{\sigma}=0.$$

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  • $\begingroup$ great answer. Do you happen to have a reference for this? Is the process the same if you wanted to write the above expression in terms of coordinate time instead of proper time do you know? $\endgroup$ – Rumplestillskin Nov 24 '17 at 10:17

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