Null Hypersurfaces: Questions on Kruskal Diagram, Eqn 2.65-2.75, and Geodesic

[latentcorpse]

So, if you look at these notes
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf

(i) On page 27, there is the Kruskal diagram at the top of the page, how do we know what direction to put the arrows on the U and V axes?

(ii) Going from eqn (2.65) to (2.66), why does the $\frac{\partial S}{\partial v} \frac{\partial}{\partial r}$ term vanish? Surely since $S=r-2M$ and $v=t+r^*$ where $r^*$ is the Reggae Wheeler radial coordinate that is a function of $r$, we would expect $S$ to vary with $v$ also and hence expect $\frac{\partial S}{\partial v} \neq 0$

(iii) In going from (2.74) to (2.75), how does $l^\rho D^\mu l_\rho = \frac{1}{2} l^{2, \mu}$ where $D$ represents covariant derivative.

(iv) At the bottom of page 28, he explains how he gets (2.75) into teh form of something proportional to $l^\mu$ in order to prove it is a geodesic (see eqn (2.8) for definition). Why does the third term in (2.75) vanish? And can someone explain what he does to the 2nd term in (2.75) to get it proportional to $l^\mu$?

Thanks a lot!

 

[NateD]

(i) On the Kruskal diagram, the arrows on the U and V axes indicate the direction of increasing values of the coordinates. The arrows should point towards larger values of U and V.(ii) The term \frac{\partial S}{\partial v} \frac{\partial}{\partial r} vanishes because the partial derivative of S with respect to v is zero. Since S=r-2M, its partial derivative with respect to v is 0 since v is a function of r, not of S.(iii) In going from (2.74) to (2.75), the l^\rho D^\mu l_\rho term can be simplified by using the definition of covariant derivatives. The covariant derivative of a vector is defined as D^\mu l_\rho = \partial^\mu l_\rho + \Gamma^\mu_{\alpha \rho} l^\alpha, where the second term is known as the Christoffel symbol. By substituting this definition into the l^\rho D^\mu l_\rho term, we get l^\rho \partial^\mu l_\rho + l^\rho \Gamma^\mu_{\alpha \rho} l^\alpha = \frac{1}{2} l^{2, \mu}.(iv) The third term in (2.75) vanishes because l^\mu l_\nu is the same as l^\mu l^\nu. The second term can be simplified by noting that l^\mu l^\nu \nabla_\mu l_\nu can be expressed as (l^\mu \nabla_\mu l_\nu) l^\nu, which is equal to l^\mu l_\nu l^\nu = l^2 l^\mu.