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, 2015 at 13:25

1 Answer 1


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.


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