General RelativityPhysics

Derivation of Schwarzschild Solution

Recall the definitions of the basic entities used in the equations.

Einstein Field Equations ( without cosmological constant ):

(1)   \begin{equation*} \displaystyle{R_{\mu \nu }=-\frac{8\pi G}{c^4}(T_{\mu \nu }-\frac{1}{2}Tg_{\mu \nu })} \end{equation*}

Ricci tensor :

(2)   \begin{equation*} \displaystyle{R_{\mu \nu }=R{^{\rho }}_{\mu \rho \nu }=\frac{\partial \Gamma _{\mu \rho }^{\rho }}{\partial x^{\nu }}-\frac{\partial \Gamma _{\mu \nu }^{\rho }}{\partial x^\rho }+\Gamma _{\mu \rho }^{\sigma }\Gamma _{\sigma \nu }^{\rho }-\Gamma _{\mu \nu }^{\sigma }\Gamma _{\sigma \rho }^{\rho }} \end{equation*}

Christoffel symbols :

(3)   \begin{equation*} \displaystyle{\Gamma _{\lambda \mu }^{\sigma }=\frac{1}{2}g^{\sigma \nu }(\frac{\partial g_{\mu \nu }}{\partial x^\lambda }+\frac{\partial g_{\lambda \nu }}{\partial x^\mu }-\frac{\partial g_{\mu \lambda }}{\partial x^\nu })} \end{equation*}

Contracted Christoffel symbols :

(4)   \begin{equation*} \displaystyle{(\Gamma _{\mu \rho }^{\rho })=(\frac{\partial ln\sqrt{g}}{\partial x^{\mu }})} \end{equation*}

Every solution of the field equations requires an ansatz; in this thread we will look at the simplest possible solution of the equations, which is the vacuum solution of a spherically symmetric gravitational field for a static mass. The solution is called the Schwarzschild Metric. The spherical symmetry and the condition that mass and resulting field are static leads to the following simple ansatz :

(5)   \begin{equation*} \displaystyle{ds^2=B(r)c^2dt^2-A(r)dr^2-r^2(d\theta ^2+sin^2\theta d\phi ^2)} \end{equation*}

with two as yet unspecified functions A(r) and B(r). Our task will be to find these two functions from the field equations.

In a vacuum ( T_{\mu \nu }=0 ) the Einstein Field Equations (1) reduce to

(6)   \begin{equation*} \displaystyle{R_{\mu \nu }=0} \end{equation*}

which is a set of partial differential equations for the unknown functions A(r) and B(r).


The non-vanishing elements of the Ricci tensor, as obtained from the Christoffel symbols (3), are thus :

(7)   \begin{equation*} \displaystyle{R_{00}=-\frac{B''}{2A}+\frac{B'}{4A}(\frac{A'}{A}+\frac{B'}{B})-\frac{B'}{rA} } \end{equation*}

(8)   \begin{equation*} \displaystyle{R_{11}=\frac{B''}{2B}-\frac{B'}{4B}(\frac{A'}{A}+\frac{B'}{B})-\frac{A'}{rA} } \end{equation*}

(9)   \begin{equation*} \displaystyle{R_{22}=-1-\frac{r}{2A}(\frac{A'}{A}-\frac{B'}{B})+\frac{1}{A}} \end{equation*}

(10)   \begin{equation*} \displaystyle{R_{33}=R_{22}sin^2\theta } \end{equation*}

From the above we obtain the system of equations

(11)   \begin{equation*} \displaystyle{R_{\mu \nu}=0} \end{equation*}


We now write

(12)   \begin{equation*} \displaystyle{\frac{R_{11}}{A}+\frac{R_{00}}{B}=-\frac{1}{rA}(\frac{A'}{A}+\frac{B'}{B})=0} \end{equation*}

and, doing some algebra, we obtain from this

(13)   \begin{equation*} \displaystyle{A(r)B(r)=const.} \end{equation*}

We also know that the gravitational field vanishes at infinity, i.e for \displaystyle{r \to \infty } we obtain

(14)   \begin{equation*} \displaystyle{A(r)\xrightarrow[]{r \to \infty }1} \end{equation*}

(15)   \begin{equation*} \displaystyle{B(r)\xrightarrow[]{r \to \infty }1} \end{equation*}

and therefore

(16)   \begin{equation*} \displaystyle{A(r)=\frac{1}{B(r)}} \end{equation*}

Now we can insert this into the remaining equations :

(17)   \begin{equation*} \displaystyle{R_{22}=-1+rB'+B=0} \end{equation*}

(18)   \begin{equation*} \displaystyle{R_{11}=\frac{B''}{2B}+\frac{B'}{rB}=\frac{1}{2rB}\frac{\mathrm{d} R_{22}}{\mathrm{d} r}=0} \end{equation*}

One can easily verify that these two differential equations are solved by

(19)   \begin{equation*} \displaystyle{B(r)=1-\frac{2a}{r}} \end{equation*}

(20)   \begin{equation*} \displaystyle{A(r)=\frac{1}{1-2a/r}} \end{equation*}

with an integration constant a. This constant is determined by the condition that the solution of the field equation must reduce the usual Newton’s law at infinity; therefore

(21)   \begin{equation*} \displaystyle{a=\frac{GM}{c^2}} \end{equation*}

Putting all this back into the ansatz (5) gives us the solution of the Einstein field equation we were looking for :

(22)   \begin{equation*} \displaystyle{ds^2=(1-\frac{2a}{r})c^2dt^2-\frac{dr^2}{1-\frac{2a}{r}}-r^2(d\theta ^2+sin^2\theta d\phi ^2)} \end{equation*}

This is called the Exterior Schwarzschild Metric, and its form is the simplest possible vacuum solution to the original field equations without cosmological constant.

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