My question is pretty simple, but has some annoying build-up beforehand:


If one wants to find the geodesic equation starting from extremizing the proper time along a path, it comes down to finding extrema of $$\tau =\int \sqrt{-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}} d\tau$$ Which, as shown in Carroll's GR book, is equivalent to finding the extrema of $$I=\frac{1}{2}\int g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}d\tau$$ Now, what we want is for the variation of this integral, $\delta I$, to vanish. To find $\delta I$, we have to vary each part individually: $$\delta I = \frac{1}{2}\int \delta (g_{\mu\nu})\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}+g_{\mu\nu}\delta(\frac{dx^{\mu}}{d\tau})\frac{dx^{\nu}}{d\tau}+g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\delta(\frac{dx^{\nu}}{d\tau})$$


At this point in the derivation, I attempted a shortcut with the last term in the integral: $$g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\delta(\frac{dx^{\nu}}{d\tau})=g_{\nu\mu}\frac{dx^{\mu}}{d\tau}\delta(\frac{dx^{\nu}}{d\tau})=g_{\nu\mu}\delta(\frac{dx^{\nu}}{d\tau})\frac{dx^{\mu}}{d\tau} $$

I used the fact that $g$ is symmetric (and multiplication is commutative :D). Interchanging the summation indices on the last expression, $\mu\to\nu$, and $\nu\to\mu$ (this is valid since they're just dummy indices; their names don't matter), we have: $$g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\delta(\frac{dx^{\nu}}{d\tau})=g_{\mu\nu}\delta(\frac{dx^{\mu}}{d\tau})\frac{dx^{\nu}}{d\tau}$$

So luckily, our variation becomes a little simpler:

$$\delta I = \frac{1}{2}\int \delta (g_{\mu\nu})\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}+2g_{\mu\nu}\delta(\frac{dx^{\mu}}{d\tau})\frac{dx^{\nu}}{d\tau}$$

However, when working with this variation (integrating by parts, doing all the usual business), I don't arrive at the correct geodesic equation... So I'm kind of forced to conclude that what I did is incorrect. Can anybody help me by pointing out where I go wrong in my "shortcut"? Thanks in advance :)

  • $\begingroup$ Everything you've done looks perfectly correct; I would just double-check your algebra. Are you making sure to note that $\delta g_{\mu\nu} = \delta x^\sigma \partial_\sigma g_{\mu\nu}$ and $dg_{\mu\nu}/d\tau = (dx^\sigma/d\tau) \partial_\sigma g_{\mu\nu}$? $\endgroup$
    – Sebastian
    Feb 5 '17 at 19:30
  • 1
    $\begingroup$ Could you edit the answer with the rest of your calculation? It looks good to me so far. One other trick you may be missing is that $2\dot{x}^\mu \dot{x}^\nu \partial_\nu g_{\mu \rho} = \dot{x}^\mu \dot{x}^\nu \partial_\nu g_{\mu \rho} + \dot{x}^\mu \dot{x}^\nu \partial_\mu g_{\nu \rho} $. $\endgroup$
    – gj255
    Feb 5 '17 at 19:30

Sorry everyone, it seems that what I did was correct! I just didn't apply the handy trick mentioned by gj255 in the comments (to be used a few lines after what I had): $$2\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}\partial_{\nu}g_{\mu\rho}=\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}\partial_{\nu}g_{\mu\rho}+\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}\partial_{\mu}g_{\nu\rho} $$ Using this, the result comes out :)

Ironically enough, this trick is sort of the reverse of my shortcut (it "fixes" that I changed the order a bit), so it was kind of useless to try to be clever here!

  • $\begingroup$ Yup, that's perfectly right! Of course, your original result would have given the right answer all along (i.e. the correct motion); it's just that we prefer the connection to be symmetrized in its upper indices. $\endgroup$
    – knzhou
    Feb 6 '17 at 1:14

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