# Integral curves and null geodesics

Let be $(M^{n+1},g)$ a spacetime (Lorentz manifold, connexe and time-oriented), $n\ge 2$, and $S\subset M$ a null hypersurface (codim $S=1$ and the restriction of $g$ to each tangent space $T_p S$ is degenerate).

If $K$ is a null vector field of $S$, show that the integral curves of $K$ are null geodesics of $S$.

I would like understand why that problem is equivalent to show that $$\nabla_K K=\lambda K,$$

where $\lambda\in C^\infty(S)$ is a smooth function. Cause, by $\nabla_K K=\lambda K$, if $\alpha$ is a integral curve of $K$, $\frac{d\alpha}{dt}=K(\alpha(t))$, then

$$\frac{D}{dt}\Big(\frac{d\alpha}{dt}\Big)=K(\alpha (t)),$$

and I don't understand why the right-hand side is zero.

Everyone can help me? Thanks.

## 1 Answer

Actually, by setting $K'=\alpha K$ and calculating the covariant derivative with respect to $K'$ we have $$\nabla_{K'}K'=\alpha K(\alpha)K+\alpha^2\nabla_K K=(\alpha K(\alpha)+\lambda\alpha^2)K.$$ So if $\alpha$ satisfies $\alpha K(\alpha)+\lambda\alpha^2=0$ we have desired vector field which correspond integral curve is a geodesic.