# D'Inverno derivation of Schwarzschild solution

1. Dec 29, 2011

### PhyPsy

If you happen to have D'Inverno's Introducing Einstein's Relativity, this is on page 187. He has reduced the metric to non-zero components:
$g_{00}= e^{h(t)}(1-2m/r)$
$g_{11}=-(1-2m/r)^{-1}$
$g_{22}=-r^2$
$g_{33}=-r^2\sin^2\theta$
The final step is a time coordinate transformation that reduces $g_{00}$ to $1-2m/r$. This is achieved by making $e^{h(t')}=1$, so $h(t')=0$. He does this with the relation
$t'=\int^t_c e^{\frac{1}{2}h(u)}du$, c is an arbitrary constant
I suppose that, since c is arbitrary, I can assign whatever value to c to make $h(t')=0$, but why use this particular integral as the relation between t and t'? Is there something special about this integral?

2. Dec 30, 2011

### Bill_K

This is pretty easy. You want (1 - 2m/r) dt'2 = eh(t)(1 - 2m/r) dt2, so take dt' = eh(t)/2 dt

3. Dec 30, 2011

### PhyPsy

Ah, OK...but why would the relation not be $t'=\int e^{h(t)/2}dt$? Instead, they have it as a definite integral from c to t.