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- TL;DR
- Help with getting the result in Schutz
Hi guys,
I can't seem to be able to get to
$$ (\rho + p) \frac {d\Phi} {dr} = - \frac {dp} {dr} $$
from
$$T^{\alpha\beta}_{\,\,\,\,;\beta} = 0$$
the only one of these 4 equations (in the case of a spherically symmetric static star) that does not identically vanish is that for ##\alpha=r##
Because ##T^{\alpha\beta}## is diagonal, that means ##T^{rr}_{\,\,\,\,;r}=0##.
We know that ##T^{rr}=p e^{-2\Lambda}## and that ##\Gamma^r_{\mu r} = \Lambda_{,r}##. So,
$$T^{rr}_{\,\,\,\,;r}=T^{rr}_{\,\,\,\,,r} + 2 \Gamma^r_{\mu r} T^{\mu r} = p_{,r}e^{-2\Lambda} - 2 p e^{-2\Lambda}\Lambda_{,r} + 2 \Lambda_{,r}p e^{-2\Lambda}=0 $$
And I get simply
$$ \frac {dp} {dr} = 0 $$
... which makes no sense! where did I go wrong? This is going to be embarrassing....
I can't seem to be able to get to
$$ (\rho + p) \frac {d\Phi} {dr} = - \frac {dp} {dr} $$
from
$$T^{\alpha\beta}_{\,\,\,\,;\beta} = 0$$
the only one of these 4 equations (in the case of a spherically symmetric static star) that does not identically vanish is that for ##\alpha=r##
Because ##T^{\alpha\beta}## is diagonal, that means ##T^{rr}_{\,\,\,\,;r}=0##.
We know that ##T^{rr}=p e^{-2\Lambda}## and that ##\Gamma^r_{\mu r} = \Lambda_{,r}##. So,
$$T^{rr}_{\,\,\,\,;r}=T^{rr}_{\,\,\,\,,r} + 2 \Gamma^r_{\mu r} T^{\mu r} = p_{,r}e^{-2\Lambda} - 2 p e^{-2\Lambda}\Lambda_{,r} + 2 \Lambda_{,r}p e^{-2\Lambda}=0 $$
And I get simply
$$ \frac {dp} {dr} = 0 $$
... which makes no sense! where did I go wrong? This is going to be embarrassing....