# Conformal invariance of null geodesics

1. Sep 24, 2011

### PhyPsy

Hi, folks. I hope this is the right forum for this question. I'm not actually taking any classes, but I am doing self-study using D'Inverno's Introducing Einstein's Relativity. I have a solution, and I want someone to check it for me.

1. The problem statement, all variables and given/known data
Prove that the null geodesics of two conformally related metrics coincide.

2. Relevant equations
Conformally related metrics: $\overline{g}$ab = $\Omega$2gab
Null geodesics: 0 = gab(dxa/du)(dxb/du)

3. The attempt at a solution
I define the parameter u = $\frac{1}{2}$$\Omega$2. Thus $\frac{du}{d\Omega}$ = $\Omega$.

Now, I use the chain rule on the null geodesics equation:
0 = $\Omega$2gab(dxa/d$\Omega$)$\frac{d\Omega}{du}$(dxb/d$\Omega$)$\frac{d\Omega}{du}$
0 = $\Omega$2gab(dxa/d$\Omega$)(dxb/d$\Omega$)($\frac{du}{d\Omega}$)-2
0 = $\Omega$2gab(dxa/d$\Omega$)(dxb/d$\Omega$)$\Omega$-2
0 = gab(dxa/d$\Omega$)(dxb/d$\Omega$), which is the null geodesics equation with the new parameter $\Omega$.

So is this a legitimate proof of the coinciding of null geodesics of conformally related metrics?

2. Sep 29, 2011

### PhyPsy

Bump; can anyone help me?

3. Sep 29, 2011

### diazona

Hmm... is u a parameter for the geodesic? If so, I don't think you can relate it to Ω like that, because the proof should work for arbitrary conformal transformations including (for example) those where Ω is a constant.

4. Sep 29, 2011

### Dick

No, it's not right. 0 = gab(dxa/du)(dxb/du) isn't the geodesic equation. It's just says that it's a null curve. And those are obviously conformally invariant. Not all null curves are null geodesics. If you want to see the right way to do it look at Appendix D in Robert Wald's book General Relativity.

5. Sep 30, 2011

### PhyPsy

OK, I see now that I should be using the equation:
d2xa/ds2 + $\Gamma$abc(dxb/ds)(dxc/ds) = 0

Unfortunately, I am coming up with
d2xa/ds2 + {$\Gamma$abc + ($\delta$ac$\partial$b$\Omega$ + $\delta$ab$\partial$c$\Omega$ - gadgbc$\partial$d$\Omega$) / $\Omega$}(dxb/ds)(dxc/ds) = 0
as the transformation. I don't see how that is invariant unless d$\Omega$ = 0, and I don't see any reason to make that assumption.

I think I will try looking for that Wald book you mentioned.

Update: Wow, thanks for that tip, Dick. The explanation in the Wald book really cleared things up. It starts by using the affine geodesic equation,
Xa$\nabla$aXb = 0,
and uses a relation between covariant derivatives that I did not find in the Inverno book:
$\overline{\nabla}$aXb = $\nabla$aXb + TbacXc

I find how Tbac transforms conformally and get this equation:
Xa$\nabla$aXb = 2XaXb$\nabla$a(ln $\Omega$) - gbdgacXaXc$\nabla$d(ln $\Omega$)

Since it is a null geodesic, gacXaXc = 0, so the equation reduces to:
Xa$\nabla$aXb = 2XaXb$\nabla$a(ln $\Omega$)

The non-affine geodesic equation is Xa$\nabla$aXb = $\lambda$Xb, so I just define $\lambda$ = 2Xa$\nabla$a(ln $\Omega$), and the geodesic equation is satisfied. This is why the Inverno book gave the hint that both equations need not be affinely parametized.

Last edited: Sep 30, 2011