Deriving the Schwarzschild solution

[Deadstar]

Right so there's this part in my notes where we begin to derive the schwarzchild solution. There's a substitution part I don't understand fully (I think) but I'll start from the beginning...The Schwarzschild Solution.

The solution corresponds to the metric corresponding to a static, spherically symmetric gravitational field in the empty spacetime surrounding a central mass (like the Sun).

Choosing coordinates (t, r, θ, ϕ), it can be shown that a metric of this type is of the
form:

$$ds^2 = -e^{A(r)}dt^2 + e^{B(r)}dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)$$ (6.8)

where A(r) and B(r) describe deviation of the metric from Minkowski spacetime. Note
that for constant t and r the metric reduces to the standard metric for the surface of a
sphere. As one is dealing with vacuum, one is poised to solve

$$R_{ab} = 0$$ (6.9)

Where (I'm pretty sure) $$R_{ab}$$ is the Ricci tensor.

Substituting (6.8) in (6.9) we have, after some algebra...

--------

Then there's expressions for $$R_{tt}$$, $$R_{rr}$$ etc...

But I don't know exactly how to sub 6.8 into 6.9!

Would I be right in thinking we just calculate... (When ab = tt)

$$R_{tt} = R^c_{tct} = 0$$ where $$R^c_{tct}$$ is the Riemann curvature tensor.

Then we can use that,

$$\Gamma^c_{ab} = \frac{1}{2}g^{cd}(\partial_a g_{bd} + \partial_b g_{ad} - \partial_d g_{ab})$$

to calculate $$R^c_{tct}$$ where we have...

$$g_{tt} = -e^{A(r)}$$
$$g_{rr} = e^{B(r)}$$

and so on...

Is this the right method? Seems like it's going to take a loooong while to do this though.

 

[betel]

Yes, this is the right method and Yes this will take a long while to do it. But it is worth it. Btw. you will only need $$R_{tt}, R_{rr}$$.