Generators of a null hypersurface

[WannabeNewton]

For a hypersurface \Sigma ,defined by f(\mathbf{x}) = const., the vector field \zeta^{\mu } = \triangledown ^{\mu }f will be normal to \Sigma. If \boldsymbol{\zeta } is null - like then \Sigma will be a null hypersurface. My question is on the justification of the statement that the set of all null geodesics on \Sigma are the generators of \Sigma. The geodesic equation can be written as \zeta ^{\mu }\triangledown _{\mu }\zeta_{\nu } = \upsilon (\alpha )\zeta _{\nu } where \upsilon (\alpha ) = 0 if \alpha is affine. Using \zeta _{\nu } = \triangledown _{\nu }f, once can arrive at \zeta ^{\mu }\triangledown _{\mu }\zeta_{\nu } = \frac{1}{2}\triangledown _{\nu }(\zeta ^{\mu }\zeta _{\mu }). So, apparently this is where the problem comes because even though \zeta ^{\mu }\zeta _{\mu } = 0 on \Sigma we can't be sure it vanishes off of it so we don't know that \triangledown _{\nu }(\zeta ^{\mu }\zeta _{\mu }) = 0. So if you specify the null hypersurface by \zeta ^{\mu }\zeta _{\mu } = 0 then the normal vector field will be \triangledown _{\nu }(\zeta ^{\mu }\zeta _{\mu }) = g\triangledown _{\nu }f for some scalar function g(x). So here is where I am confused: you have \zeta ^{\mu }\triangledown _{\mu }\zeta _{\nu } = \frac{1}{2}g\zeta _{\nu } and you can rescale by setting \xi ^{\mu } = h(x)\zeta ^{\mu } so that \xi ^{\mu }\triangledown _{\mu }\xi ^{\nu } = 0 but how do you actually rescale the original normal vector field so that with the rescaled vector field you have the right - side vanishing? Because if this is true then the rescaled normal vector field acts as the generator of the integral curves that, obeying that equation, turn out to be null geodesics whose union is \Sigma but I don't get how you can just rescale the original normal vector field to make the right side of the aforementioned equation vanish.

 

[Kevin_Axion]

Why you so smart?

Don't know what to say, maybe look into Killing vectors.