Proving Geodesic Length Unchanged by Small Curve Changes

[Mmmm]

Homework Statement

Prove that the proper length of geodesic between two points is unchanged to first order by small changes in the curve that do not change its endpoints.

Homework Equations

Length of curve = \int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{1}{2}d\lambda

where g_{\alpha\beta} is the metric for the particular coordinate system, and
\frac{dx^\alpha}{d\lambda}= U^\alpha is the gradient of the curve.

On a geodesic, g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda} is constant.

The Attempt at a Solution

The equation of the curve is
x^\alpha = x^\alpha(\lambda)

For a small change in the curve, the equation becomes
x^\alpha = x^\alpha(\lambda) + \delta x^\alpha(\lambda)
where
\delta x^\alpha(\lambda_{2})=\delta x^\alpha(\lambda_{1}) = 0
to ensure that the ends of the curve are unchanged.

So

Length of curve = \int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{d}{d\lambda}(x^\alpha + \delta x^\alpha)\frac{d}{d\lambda}(x^\alpha + \delta x^\alpha)\right|^\frac{1}{2}d\lambda

Multiplying out:
=\int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|^\frac{1}{2}d\lambda

Expanding to first order:

\approx \int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{1}{2}+\frac{1}{2}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{-1}{2} \left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|d\lambda

The first term here is the length of the original curve again so the second term is the change in length to first order given a small change in the curve. I must prove that this is 0.

\Delta l = \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{-1}{2} \left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|d\lambda

Now on a geodesic g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda} is constant, so i will just call this factor C.

so
\Delta l = C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|d\lambda

Rename some indices to get a common factor of \frac{d\delta x^\gamma}{d\lambda}:

= C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|g_{\alpha\gamma}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\gamma}{d\lambda}+g_{\gamma\beta}\frac{d\delta x^\gamma}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\gamma}{d\lambda}\right|d\lambda

Factorise and rename more indices:

= C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|\frac{d\delta x^\gamma}{d\lambda}\right|\left|g_{\alpha\gamma}\frac{d x^\alpha}{d\lambda}+g_{\gamma\alpha}\frac{dx^\alpha}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\right|d\lambda

use symmetry of g to make first two terms equal:

= C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|\frac{d\delta x^\gamma}{d\lambda}\right|\left| 2 g_{\alpha\gamma}\frac{d x^\alpha}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\right|d\lambda

Now integrate by parts:

= C \left[ \frac{1}{2}\left|{\delta x^\gamma}\left( 2 g_{\alpha\gamma}\frac{d x^\alpha}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\right) \right| \right] ^{\lambda_{2}}_{\lambda_{1}} - C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|\delta x^\gamma}\left( 2 \frac{d}{d \lambda} \left( g_{\alpha\gamma} U^\alpha \right) + \frac{d}{d \lambda} \left( g_{\alpha\gamma} \frac{d \delta x^\alpha}{d\lambda} \right) \right)\right|d\lambda

The first term vanishes as \delta x^\alpha(\lambda_{2})=\delta x^\alpha(\lambda_{1}) = 0

= C \int ^{\lambda_{2}}_{\lambda_{1}} \left|\delta x^\gamma}\left[ - \frac{d}{d \lambda} \left( g_{\alpha\gamma} U^\alpha \right) - \frac{1}{2} \frac{d}{d \lambda} \left( g_{\alpha\gamma} \frac{d \delta x^\alpha}{d\lambda} \right) \right]\right|d\lambda

This is where I get a bit stuck. The expression in square brackets should come out as the geodesic equation and it is nearly there. The problem is the second term, it should be

\frac{1}{2}g_{\alpha \beta , \gamma} U^\alpha U^\beta

(which I got from the answer in the back of the book)
to give

\Delta l = C \int ^{\lambda_{2}}_{\lambda_{1}} \left|\delta x^\gamma}\left[ - \frac{d}{d \lambda} \left( g_{\alpha\gamma} U^\alpha \right) + \frac{1}{2}g_{\alpha \beta , \gamma} U^\alpha U^\beta \right]\right|d\lambda

Then, indeed with a bit of indices tweaking you do get the geodesic equation and everything vanishes.

I have an extra \frac{d \delta x^\alpha}{d\lambda} and I just have no idea how to get rid of it.

 

[Dick]

It's easy to get rid of. It gets rid of itself when the variation goes to zero. It shouldn't have been there to begin with. It came from a term that's second order in the variation. You should have thrown that away at the beginning. The term you are missing comes from the variation of the g, g(x(\lambda)+\delta x(\lambda)).

 

[Mmmm]

Aaaaahhhhhhh... of course. after a taylor expansion of g right at the beginning and a removal of that 2nd order term it all comes out beautifully.

Thanks so much Dick... Now I can sleep at night ;)