# Integrating along a null geodesic

I was studying the details of integration on null hypersurfaces through A Relativist's Toolkit by Eric Poisson.

If we consider a null hypersurface in 4-dimensional spacetime, then the surface element, according to Poisson, is given by$$\mathrm{d}\Sigma_\alpha=-k_\alpha \sqrt{\sigma} \, \mathrm{d}^2\theta \, \mathrm{d}\lambda\,,$$where $k$ is a future pointing vector tangent to the surface along the null generators, $\theta$ are the coordinates transverse to the null generators and $\lambda$ parametrizes the null generators.

If we are in two dimensional spacetime, the null hypersurfaces are simply null geodesics. In this case we don't have transverse directions to the generators and hence I expect that integrating any smooth function along a null geodesic should yield zero. Do you agree with this statement?

$\newcommand{\d}{\mathrm{d}}$No, the integral will not generally yield zero, because you are only integrating over one dimension. But probably more relevant: that's not the right volume element to use when integrating in a different number of dimensions. You are using fallacious logic, analogous to saying that the circumference of a circle is zero because the area element of the sphere looks like $\sin\theta\, \d\theta\, \d\phi$ but you're only integrating over $\phi$.
In this case, Poisson was talking about the volume element for null cones in 3+1 relativity, so you can't expect his result to apply in any other number of dimensions. If you want to integrate on a null cone in 1+1 relativity, you'll need a different volume element. So take a step back, and think about what Poisson was actually doing. Follow his logic, but for the 1+1 case. You should find that there is no need to introduce the other two coordinates, and therefore you'll end up with something like $\d\Sigma_a = -k_a\, \d\lambda$. Another good (if more concise) reference for this sort of result is section 2.8 of Hawking & Ellis, in which they basically say that the volume element on a hypersurface with normal $k^a$ is just the "contraction" of the full manifold's volume form with $k^a$ — or essentially the Hodge dual of $k_a$.