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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.

Qmechanic
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Peter4075
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    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 '15 at 10:58
  • Related: http://physics.stackexchange.com/q/94348/2451 and links therein. – Qmechanic May 14 '15 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. – Peter4075 May 14 '15 at 17:16

1 Answers1

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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 $x^\sigma$.

dmckee --- ex-moderator kitten
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Horus
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  • 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 $d\lambda$? – Peter4075 May 14 '15 at 14:09
  • No just the one at the end of the integral. It should work after using the chain rule. – Horus May 14 '15 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... – tok3rat0r May 14 '15 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)) – tok3rat0r May 14 '15 at 16:59
  • Well, I think I get it now. Very well explained. Thanks. – Peter4075 May 14 '15 at 17:20
  • @tok3rat0r - hasn't Horus multiplied through twice by $\frac{d\lambda}{d\tau}\frac{d\tau}{d\lambda}$? That gives the $\frac{d\lambda}{d\tau}\frac{d\lambda}{d\tau}d\lambda$ term at the end, which cancels to give $d\tau$. – Peter4075 May 14 '15 at 17:55
  • @Peter4075 - just realised this a few minutes ago; I'd completely missed the fact that the terms in brackets are both double derivatives! All finally becomes clear... – tok3rat0r May 14 '15 at 18:19
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    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. – dmckee --- ex-moderator kitten Jun 26 '15 at 15:12
  • And while you're at it, we have revision histories, so you don't need to add "edit" to your post. You might want to consider reformatting the post slightly to eliminate the need for "edit". – Kyle Kanos Jun 26 '15 at 15:30