## Question

Show that conformal transformations leave null geodesics invariant, that is, that the null geodesics of ##g_{\mu\nu}## are the same as those of ##\omega^2g_{\mu\nu}##. (We already know that they leave null curves invariant; you have to show that the transformed curves are still geodesics.) What is the relationship between the affine parameter in the original and conformal metrics?

## Answer

The answer to this is a bit feeble I think - so feeble that I forgot to post it for ten days. It relies on Carroll's assertion in section 3.4 on the properties of geodesics that from some kind of equation like his 3.58 you can always find an equation that satisfies the geodesic equation and he gives us the relationship of the affine parameter to the magic equation. It would have been nice to prove the assertion but I think that was out of scope.

It's at: Ex G1 Conformal Null Geodesics.pdf (a mere two pages).

Thank you for providing this answer, it helped me get unstuck!

ReplyDeleteActually the claim (3.58) is not too hard to prove. Essentially we want to solve the differential equation (3.59). If we write λ=g(α), (3.59) becomes f = -(1/g) (1/g)' (1/g)^(-2) = g'/g. Integrate on both sides, we have ln g = c + \int_0^α f(x) dx, or equivalently, g = c' exp(\int_0^α f(x) dx).

ReplyDeletesupplying delimiters:

DeleteIf we write ##λ=g(α)##, (3.59) becomes $$f = -(1/g) (1/g)' (1/g)^(-2) = g'/g$$ Integrate on both sides, we have $$ln g = c + \int_0^α f(x) dx$$ or equivalently, $$g = c' exp(\int_0^α f(x) dx)$$.

For posterity, I am also posting my solutions to the other exercises in Appendix G (or reference to existing solutions). For Ex. 2, refer to Exercise I.5.17 in A. Zee’s Einstein Gravity in a Nutshell. In that book a solution is provided.

ReplyDeleteFor Ex. 3. (a), see https://physics.stackexchange.com/a/747035/269697. (b) is trivial, since a photon geodesics in \tilde{g} is also a photon geodesics in g. (c) is messed up. There is no way for this claim to be true in the general case. In fact, if we take the metric to be (3.72), we can calculate the Christoffel symbols and then the LHS of (G.20). The result is that (G.20) is true only if R(t) = a(t). For (d), compare (3.173) to (G.20), we see that when p^μ is light-like, a conformal Killing vector will also lead to the conservation of K^μ p_μ. In the original coordinate, ξ^μ = (a, 0, 0, 0). Let p^μ = (E, E, 0, 0). We see aE must be a constant.

ReplyDelete