3
$\begingroup$

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

Introduction

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

Question

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 :)

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.