D'Inverno derivation of Schwarzschild solution

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:
[itex]g_{00}= e^{h(t)}(1-2m/r)[/itex]
[itex]g_{11}=-(1-2m/r)^{-1}[/itex]
[itex]g_{22}=-r^2[/itex]
[itex]g_{33}=-r^2\sin^2\theta[/itex]
The final step is a time coordinate transformation that reduces [itex]g_{00}[/itex] to [itex]1-2m/r[/itex]. This is achieved by making [itex]e^{h(t')}=1[/itex], so [itex]h(t')=0[/itex]. He does this with the relation
[itex]t'=\int^t_c e^{\frac{1}{2}h(u)}du[/itex], c is an arbitrary constant
I suppose that, since c is arbitrary, I can assign whatever value to c to make [itex]h(t')=0[/itex], but why use this particular integral as the relation between t and t'? Is there something special about this integral?

Bill_K

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

PhyPsy

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