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PhyPsy
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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?
[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?