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I am currently taking a course in General Relativity, and I've hit a bit of a roadblock with a homework assignment.

We are given the metric for Einstein's universe to be (forgive me, this is meant to be formatted for LaTeX, but I'm new at this whole stackexchange business) $ds^2 = c^2 dt^2 - \frac{1}{1-kr^2} dr^2 - r^2 d{\theta}^2 - r^2 \sin^2{\theta} d{\phi}^2$ and are asked to obtain the null geodesic equations. I understand that $ds = 0$ along null geodesics, and I am familiar with the equation forms involving the Christoffel symbols.

What confuses me is, how are we to specify what we take our $t, r, \theta$, and $\phi$ derivatives with respect to? If I treat it like when we worked through a non-null case (we used the Schwarzschild metric in class), then I feel like I'm either dividing by zero or doing some serious hand-waving, neither of which is appealing.

A couple of caveats: I have read in other textbooks that another method of working through this problem is using Killing vectors... we weren't taught that method, and I'd prefer to use the technique developed in class.

If anyone can point me in the right direction, that would be greatly appreciated.

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    $\begingroup$ @Jerry Schirmer, thanks for the equation fix! :) Hopefully I'll do it right next time. $\endgroup$ – user3421870 Mar 29 '14 at 21:37
  • $\begingroup$ you had it mostly right :) $\endgroup$ – Jerry Schirmer Mar 29 '14 at 21:50
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Rather than work out the answer for you, I'll work out the equivalent answer in Minkowski space, and leave it to you to develop the rest for the case of the Einstein static universe.

In Minkowski space, you have:

$$ds^{2} = - dt^{2} + dx^{2} + dy^{2} + dz^{2}$$

If one wants to timelike geodesics, one must find the path that is an extremum of:

$$ I = \frac{1}{2}\int ds^{2}\left(-\left(\frac{dt}{ds}\right)^{2} + \left(\frac{dx}{ds}\right)^{2} + \left(\frac{dy}{ds}\right)^{2} + \left(\frac{dz}{ds}\right)^{2}\right)$$

(henceforth, I write $\dot t$ to mean $\frac{dt}{ds}$) under an abitrary variation of $t(s), x(s), y(s), z(s)$, and which is also satisfies $ -1 = -{\dot t}^{2} + {\dot x}^{2} + {\dot y^{2}} + {\dot z^{2}}$

In this case, the variation is pretty trivial, and just gives you:

$$\delta I = \int \left( -{\dot t}\delta {\dot t} + {\dot x}\delta{\dot x} + {\dot y}\delta {\dot y} + {\dot z}\delta{\dot z}\right)$$

which just gives you $x^{a} = C^{a}(s - s_{0})$, where each of the $C^{a}$ are constants, and $s_{0}$ is an initial value for $s$.

Now, that takes care of the variation, but what about the timelike condition?

Well, then we just substitute this in, and we find that:

$$-1 = -\left(C^{t}\right)^{2} + \left(C^{x}\right)^{2} + \left(C^{y}\right)^{2} + \left(C^{z}\right)^{2}$$

and we find that the general solution is that we can choose the spatial components arbitrarily, and we have:

$$C^{t} = \sqrt{1 + C^{i}C_{i}}$$

OK, well, what if we wanted to find the null geodesics instead?

In this case, all we need to do is have the vectors satisfy a null condition:

$$ 0 = -{\dot t}^{2} + {\dot x}^{2} + {\dot y^{2}} + {\dot z^{2}}$$

And it should be pretty quick to see that the answer here is:

$$C^{t} = \sqrt{C^{i}C_{i}}$$

Now, you should be able to play the same game with your metric, and work out your geodesic equations.

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  • $\begingroup$ Thanks @Jerry Schirmer. I do understand what you've written here, but one unresolved part for me is: How can we "legally" take the derivatives with respect to ds, when ds is zero for the null case? (You'll have to forgive me, I was a mathematician before I was a physicist... old habits die hard!) $\endgroup$ – user3421870 Mar 29 '14 at 21:52
  • $\begingroup$ @user3421870: ds is not zero for the null case - null geodesics trace out "affine parameter". Think of it as "how many perpendicular null geodesics you cross as you trace out the null curve". Now, I WILL be zero, but that's no problem. $\endgroup$ – Jerry Schirmer Mar 29 '14 at 22:04
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    $\begingroup$ Ah! That's a good distinction! Thanks, @Jerry Schirmer. I think this puts me on the right track. I'll see what I come up with, and I'll be back if I haven't gotten it sorted. :) $\endgroup$ – user3421870 Mar 29 '14 at 22:08

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