If one is constrained to the $xt$ plane, one can define the intersection with that plane of the null hypersurfaces originating at some point $P$ as

$$ g_{tt} \frac{d P^t}{d \lambda}\frac{d P^t}{d \lambda} + g_{xx}\frac{d P^x}{d \lambda}\frac{d P^x}{d \lambda} = 0,\tag{1}$$

$$ \sqrt{ \frac{g_{xx}}{-g_{tt}}} \frac{d P^x}{d \lambda} = \frac{d P^t}{d \lambda}. \tag{2}$$

It is not clear that a curve satisfying this equation will also satisfy the Geodesic equation:

$$ \frac{d^2 P^{\alpha}}{d \lambda^2} = - \Gamma^{\alpha}_{\beta \gamma} \frac{d P^{\beta}}{d \lambda} \frac{d P^{\gamma}}{d \lambda}.\tag{3}$$

What would be a reasonable approach to show that the first equation is also a geodesic?

  • $\begingroup$ Why do think this is a geodesic? $\endgroup$
    – Ryan Unger
    Dec 4 '15 at 13:25

Comments to the question (v3):

  1. In 3+1D, the 1D-intersection of a null-hypersurface with the constraints $y=0=z$ does not need to locally be a geodesic, as simple counterexamples show.

  2. The analogous 1+1D question is more interesting: In 1+1D, the eq. (1) is locally always a non-affine parametrized geodesic. It does not have to satisfy the affinely parametrized geodesic eq. (3). For a proof, see my Phys.SE answer here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.