The other day, someone asked: What's a straightforward way to derive Schwarzschild celebrated metric?
The Schwarzschild solution of course is the spherically symmetric, static, vacuum solution of Einstein's general theory of relativity.
We assume a spherically symmetric, static, vacuum metric. That means that the metric can be written in the form,
\begin{align}
ds^2 = B dt^2 - A dr^2 - r^2 d\Omega^2,
\end{align}
with the (as yet unknown) quantities $A$ and $B$ being functions of the $r$ coordinate only.
The nonzero Christoffel-symbols associated with this metric are given by
\begin{align}
\Gamma_{tt}^r&=\frac{B'}{2A},&\Gamma_{tr}^t&=\frac{B'}{2B},&\Gamma_{rr}^r&=\frac{A'}{2A},\\
\Gamma_{r\theta}^\theta&=\frac{1}{r},&\Gamma_{r\phi}^\phi&=\frac{1}{r},&\Gamma_{\theta\theta}^r&=\frac{r}{A},\\
\Gamma_{\theta\phi}^\phi&=\frac{\cos\theta}{\sin\theta},&\Gamma_{\phi\phi}^r&=-\frac{r\sin^2\theta}{A},&\Gamma_{\phi\phi}^\theta&=-\sin\theta\cos\theta.
\end{align}
In a vacuum, the Einstein-tensor is zero, $G_{\mu\nu}=R_{\mu\nu}-\tfrac{1}{2}Rg_{\mu\nu}=0$ (with $R_{\mu\nu} = \Gamma_{\mu\nu,\alpha}^\alpha-\Gamma_{\mu\alpha,\nu}^\alpha+\Gamma_{\mu\nu}^\beta\Gamma_{\alpha\beta}^\alpha+\Gamma_{\mu\alpha}^\beta\Gamma_{\nu\beta}^\alpha$).
Calculating the mixed-index Einstein tensor is straightforward. We get, in particular,
\begin{align}
G_{tt} = \frac{A'r + A^2 - A}{A^2r^2}.
\end{align}
For a vacuum solution, this must be zero, hence $A'r + A^2 - A = 0,$ which is solved by $A = 1 / (1 - C/r)$ where $C$ is an integration constant.
Next, we take
\begin{align}
G_{rr}=-\frac{B'r+(1-A)B}{ABr^2}.
\end{align}
Again, in the vacuum the numerator of this expression must be zero. After substituting the solution for $A$ we can solve for $B$. We find that $B = K/A,$ where $K$ is another constant. We can set $K=1$ with impunity by simply rescaling the time coordinate.
To find the value of $C,$ we move on to the geodesic equations. Taking the Newtonian limit and radial motion, so $t \approx \tau$ (proper time); the angular coordinates are zero, $\phi = \theta = 0;$ and $C^2 \approx 0,$ as it is much smaller than the other quantities involved. Assuming zero initial velocity, so $\dot{r} = 0$ (the dot representing differentiation with respect to proper time), we get the equation
\begin{align}
\ddot{r} + C/2r^2 = 0.
\end{align}
comparing against Newton's gravity, we find that C = 2GM, so we finally get the Newtonian gravitational acceleration
\begin{align}
\frac{d^2r}{dt^2}=-\frac{GM}{r^2}.
\end{align}
This derivation can be automated. Here's a Maxima script that performs all the steps discussed above:
load(ctensor);
derivabbrev:true;
ct_coords:[t,r,p,q];
depends([A,B],r);
lg:diag_matrix(B,-A,-r^2,-r^2*sin(p)^2);
cmetric();
christof(false);
einstein(false);
ode2(num(ein[1,1]=0),A,r);
exp(lhs(%))=exp(rhs(%));
SA:solve(%,A)[1],exp(%c)=1/C,eval,factor;
ein[2,2],SA;
SB:ev(radcan(ode2(num(factor(%))=0,B,r)),%c=1,eval);
cgeodesic(true);
geod[2],p=0,diff(t,s)=1,diff(r,s)=0,SA,SB,diff,factor,eval;
%,C^2=0,expand;
The first six lines of this script set up the metric. Next, the Christoffel-symbols and the Einstein tensor are calculated. Then we solve for $G_{tt}=0$ using Maxima's built-in ODE solver, ode2, and obtain an expression for $A.$ Using this solution, we use $G_{rr}=0$ to solve for $B.$ Finally, we use Maxima to compute the geodesic equations of this metric and plug in the solutions for $A$ and $B$; after applying our approximation scheme, we get the final acceleration equation in the Newtonian limit.
