# Transformation of coordinates for schwarzchild metric

1. Apr 1, 2013

### Storm Butler

the problem is as follows: Consider the following metric from the element of proper time,
(dτ)2 = (1 −M/r)^2(dt)^2 − (1 −M/r)^(−2)(dr)^2 − (r)^2[(d$\vartheta$)^2 + sin2 $\vartheta$(d$\phi$)^2].
There are singularities at r = M and r = 0. Change coordinates to (v, r,$\vartheta$ ,$\phi$ )
with the transformation t = v − f(r) so that grr = 0 and so removes the
singularity at r = M. Find df/dr, the new metric, and f such that v = 0 at
r = t = 0. Does this metric describe a black hole, that is can light originating
at r < M get to r > M.

I simply found the differential for t (dt=dv-f'(r)dr) and plugged that into the metric and squared it. i figured then that i just have to find and f(r) that will cancel the grr component. However if i do this i end up with a dvdr component. Is that correct? It seems odd only because we havent used non-diagonal metrics before.

thanks for any help

-Storm Butler