homework and exercises - Carroll's derivation of the geodesic equations

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.

Answer

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 $x^\sigma$.