# Carroll's derivation of the geodesic equations [duplicate]

In Carroll's derivation of the geodesic equations (page 69, http://preposterousuniverse.com/grnotes/grnotes-three.pdf), he starts with $$\tau=\int\left(-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}\right)^{1/2}d\lambda$$ and arrives at$$\delta\tau=\int\left(-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}\right)^{-1/2}\left(-\frac{1}{2}\partial_{\sigma}g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}\delta x^{\sigma}-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{d\left(\delta x^{\nu}\right)}{d\lambda}\right)d\lambda.$$ He then changes the curve parametrization from arbitrary $\lambda$ to proper time $\tau$ by plugging $$d\lambda=\left(-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}\right)^{-1/2}d\tau$$ into the above to obtain

$$\delta\tau=\int\left(-\frac{1}{2}\partial_{\sigma}g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}\delta x^{\sigma}-g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{d\left(\delta x^{\nu}\right)}{d\tau}\right)d\tau.$$

I cannot see how that substitution works. I've been told it uses the chain rule, but I just can't see it. Can anyone help? Thanks.

• Look at your first equation! $\tau = \int X d \lambda\implies d\tau =$?? Then, use $d\lambda=\frac{d\lambda}{d\tau}d\tau$.
– Danu
May 14, 2015 at 10:58
• Related: physics.stackexchange.com/q/94348/2451 and links therein. May 14, 2015 at 12:28
• Although my question has been flagged as being asked before and already having an answer, it wasn't answered at the sub-graduate level I could understand. Horus's answer, on the other hand, I could follow, and many thanks to him for persevering. May 14, 2015 at 17:16

Basically think of it this way. Take the original equation $$\tau = \int f(x) \,\mathrm{d}\lambda \tag{1}$$ which in differential form becomes

$$d\tau = f(x) \,\mathrm{d}\lambda \tag{2}$$

after a little rearranging gives

$\frac{d\lambda}{d\tau}$ = $(f(x))^{-1}$--------(3)

with the function $f(x)$ in this case being equal to

$f(x)$ = $(-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda})^{1/2}$--------(4)

as was demonstrated

EDIT:

Using eq (3)

$\frac{d\lambda}{d\tau}$ = $(f(x))^{-1}$ = $(-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda})^{-1/2}$

Substitute into

$\delta\tau = \int$ $(-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda})^{-1/2}$ $(-\frac{1}{2}$$g_{\mu\nu,\sigma}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}{\delta}x^{\sigma}-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{d({\delta}x^\nu)}{d\lambda}) d\lambda gives \delta\tau = \int \frac{d\lambda}{d\tau} (-\frac{1}{2}$$g_{\mu\nu,\sigma}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}{\delta}x^{\sigma}-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{d({\delta}x^\nu)}{d\lambda})$ $d\lambda$

$\delta\tau = \int$ $\frac{d\lambda}{d\tau}$ $(-\frac{1}{2}$$g_{\mu\nu,\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}{\delta}x^{\sigma}-g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{d({\delta}x^\nu)}{d\tau}) \frac{d\tau}{d\lambda}$$\frac{d\tau}{d\lambda}$ $d\lambda$

Use chain rule to get

$\delta\tau = \int$ $(-\frac{1}{2}$$g_{\mu\nu,\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}{\delta}x^{\sigma}-g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{d({\delta}x^\nu)}{d\tau}) d\tau Here I use , to represent the partial derivative with respect to x^\sigma. • Sorry, I realise I'm missing the blindingly obvious here (and being close voted to oblivion), but I can't still can't see how all those denominator d\lambdas become d\taus. Do I substitute d\lambda=\frac{d\lambda}{d\tau}d\tau for every single d\lambda? May 14, 2015 at 14:09 • No just the one at the end of the integral. It should work after using the chain rule. May 14, 2015 at 14:40 • @Horus - in your second-last step, where has the second \frac{d\tau}{d\lambda} come from in the integrand? I can see that you've multiplied through by \frac{d\lambda}{d\tau}\frac{d\tau}{d\lambda} to replace d\lambda with d\tau in each term. This leaves you with one \frac{d\tau}{d\lambda}, but I'm puzzled as to where the second one has come from? I looked at the other question on the same topic (linked), but found this step similarly opaque in that answer... May 14, 2015 at 16:55 • (The equivalent step in the top-rated answer here is the step from eq. (1) to eq (2), but there seems to be a totally arbitrary change from \frac{dx^\beta}{d\lambda} in the second term of eq. (1) to a corresponding \frac{dx^\beta}{d\tau} in the second term of eq. (2)) May 14, 2015 at 16:59 • I've fixed the first two equations here. Note the use of blocktset equations (use $$, not $), the use of \tag{} to get numbering, and the use \mathrm{} to force the typesetting of differentials in the conventional upright text, and finally the use of \, if you want to forcibly insert a thin space. Jun 26, 2015 at 15:12