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.

  • 2
    $\begingroup$ Look at your first equation! $\tau = \int X d \lambda\implies d\tau = $?? Then, use $d\lambda=\frac{d\lambda}{d\tau}d\tau$. $\endgroup$
    – Danu
    Commented May 14, 2015 at 10:58
  • $\begingroup$ Related: physics.stackexchange.com/q/94348/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented May 14, 2015 at 12:28
  • $\begingroup$ 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. $\endgroup$
    – Peter4075
    Commented May 14, 2015 at 17:16

1 Answer 1


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


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})$ $d\lambda$


$\delta\tau = \int$ $\frac{d\lambda}{d\tau}$ $(-\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 $x^\sigma$.

  • $\begingroup$ 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 $d\lambda$s become $d\tau$s. Do I substitute $d\lambda=\frac{d\lambda}{d\tau}d\tau$ for every single $d\lambda$? $\endgroup$
    – Peter4075
    Commented May 14, 2015 at 14:09
  • $\begingroup$ No just the one at the end of the integral. It should work after using the chain rule. $\endgroup$
    – Horus
    Commented May 14, 2015 at 14:40
  • $\begingroup$ @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... $\endgroup$
    – tok3rat0r
    Commented May 14, 2015 at 16:55
  • $\begingroup$ (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)) $\endgroup$
    – tok3rat0r
    Commented May 14, 2015 at 16:59
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    $\begingroup$ 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. $\endgroup$ Commented Jun 26, 2015 at 15:12