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The Schwarzschild Solution
\bigskip

Physically, the Schwarzschild solution represents the geometry of an
``isolated, nonrotating star, which has settled down to equilibrium''.
What properties would we expect such a solution to have?
Firstly, we would expect the solution to be static, \ie,
we would expect to have a timelike, hypersurface-orthogonal Killing
vector $t^a$.
Secondly, we would expect the solution to be spherically
symmetric, \ie, we would expect to have Killing vectors
$\xi_{\!1}^a$, $\xi_{\!2}^a$, and $\xi_{\!3}^a$, which are
spacelike, linearly dependent, and have the commutation relations
(\ref{RotGroup}).
Finally, we would expect that the time-translation and rotations
commute, \ie, we would expect to have the additional commutation relations
\begin{equation}
      [t,\xi_{\!1}]^a=[t,\xi_{\!2}]^a=[t,\xi_{\!3}]^a=0
      \label{TimeT-Rot}
\end{equation}
To summarize, we are concerned with space-time having four Killing
vectors, with the commutation relations (\ref{RotGroup}) and
(\ref{TimeT-Rot}).  For the matter composing the star, we have a fluid.
Thus, we have the density $\rho$, pressure $p$, and (unit) velocity field
$\eta^a$.  Since the star is supposed to have ``settled down to equilibrium'',
we suppose that the fluid does not ``move relative to static observers'',
\ie, we take $\eta^a$ to be a multiple of $t^a$.


To summarize, the Schwarzschild solution is a space-time with four Killing
vectors, $t^a$ (timelike, hypersurface-orthogonal), and
$\xi_{\!1}^a$, $\xi_{\!2}^a$, and $\xi_{\!1}^a$ (spacelike, linearly
independent), subject to (\ref{RotGroup}) and (\ref{TimeT-Rot}),
where the matter is a fluid with four-velocity field proportional to $t^a$.
We now discuss the geometry of the Schwarzschild solution.


Set $\lambda=t^a t_a$ (so, $\lambda<0$, since $t^a$ is timelike).
We begin with the physical interpretation of this $\lambda$.  Since $t^a$
is hypersurface-orthogonal, we have, from (\ref{D-HypOrth}),
\begin{equation}
     \grad_b t_c = -\lambda^{-1} t_{[b}\grad_{c]}\lambda
     \label{D-TimeT}
\end{equation}
Hence,
\begin{eqnarray*}
     R^m{}_{abc}t_m&=&\grad_a\grad_b t_c\\
                   &=&\grad_a(-\lambda^{-1} t_{[b}\grad_{c]}\lambda)\\
                   &=&\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                         -\lambda^{-1}(\grad_a t_{[b})\grad_{c]}\lambda
                         -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
                   &=&\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                         -\12\lambda^{-1}[(\grad_a t_b)\grad_c\lambda
                                         -(\grad_a t_c)\grad_b\lambda]
                         -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
                   &=&\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                       -\12\lambda^{-1}[
                          (-\lambda^{-1} t_{[a}\grad_{b]}\lambda)\grad_c\lambda
                         -(-\lambda^{-1} t_{[a}\grad_{c]}\lambda)\grad_b\lambda]
                       -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
                   &=&\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                       -\12\lambda^{-2}[
                          2(-t_a\grad_b\lambda+t_b\grad_a\lambda)\grad_c\lambda
                         -2(-t_a\grad_c\lambda+t_c\grad_a\lambda)\grad_b\lambda]
                       -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
                   &=&\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                       -\12\lambda^{-2}[
                          2(t_b\grad_a\lambda)\grad_c\lambda
                         -2(t_c\grad_a\lambda)\grad_b\lambda]
                       -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
                   &=&\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                       -\12\lambda^{-2}[
                          (\grad_a\lambda)(t_{[b}\grad_{c]}\lambda]
                       -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
                   &=&\12\lambda^{-2}(\grad_a\lambda) t_{[b}\grad_{c]}\lambda)
                       -\lambda^{-1}(\grad_a \grad_{[c}\lambda) t_{b]}\\
\end{eqnarray*}
where, in the first step, we have used (73\ref{73}),
in the second we have substituted (\ref{D-TimeT}),
in the third we have expanded using the Leibnitz rule,
and, in the fifth, we have again used (\ref{D-TimeT}).
Contracting the previous equation ``a'' and ``c'',
\begin{eqnarray*}
     R^m{}_bt_m &=& \12\lambda^{-2}(\grad^c\lambda) t_{[b}\grad_{c]}\lambda)
                       -\lambda^{-1}(\grad^c \grad_{[c}\lambda) t_{b]}\\
                &=& \12\12\lambda^{-2}\grad^c\lambda) (t_b\grad_c\lambda
                                                       -t_c\grad_b\lambda)
                       -\12\lambda^{-1}[(\grad^c\grad_c\lambda) t_b
                                        -(\grad^c\grad_b\lambda) t_c]\\
82
\end{eqnarray*}
where, in the second step, we have expanded the antisymmetrizations,
and, in the third step, we have used $t^c\grad_c\lambda=0$ [since $t_a$
is a Killing vector] and written ``$\grad^2$'' for ``$\grad^c\grad_c$''.
The last term on the right can be simplified further:
\begin{eqnarray*}
83
\end{eqnarray*}
Equation (83\ref{83}) gives some components of the Ricci tensor in terms of
$\lambda$.  We proceed (as always in such situations) to eliminate
the Ricci tensor in favor of the matter variables.  For a perfect fluid,
Einstein's equation becomes:
\begin{eqnarray}
84
\end{eqnarray}



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