The Schwarzschild Solution

Abstract

So far we have only used the linearized Einstein equations, and considered geometric configurations typical of the weak-field approximation. In this chapter we will apply for the first time the full Einstein equations without approximations, and we will obtain a particular exact solution for a static, spherically symmetric gravitational field.

Appendices

Exercises Chap. 10

10.1

Killing Vectors and Static Gravitational Fields A space-time manifold is characterized by a time-like killing vector $\xi ^\mu$. Show that the geometry is static (i.e. that there is a chart where the metric tensor satisfies the conditions $\partial _0 g_{\mu \nu }=0$ and $g_{i0}=0$) if and only if

$$\begin{aligned} \xi _{[\alpha } \nabla _\mu \xi _{\nu ]}=0. \end{aligned}$$

(10.80)

10.2

Squared Riemann Tensor for the Schwarschild Metric Compute the curvature invariant $R_{\mu \nu \alpha \beta } R^{\mu \nu \alpha \beta }$ for the Schwarzschild metric (10.19).

10.3

Geodesic Motion in the Rindler Manifold Consider the two-dimensional Rindler manifold of Exercise 6.1, described by the metric

$$\begin{aligned} ds^2= \xi ^2 d\eta ^2- d \xi ^2, \end{aligned}$$

(10.81)

and show that a test particle moving geodesically from $\xi _0$ towards the origin reaches the point $\xi =0$ (on the border of the Rindler manifold) in a finite proper time interval. The geodesics trajectory cannot be extended beyond that point, and this shows that the Rindler chart (associated to the metric (10.81)) does not represent a maximally analytically extended system of coordinates for the Minkowski space-time.

Solutions

10.1

Solution Let us choose a chart where the time axis is oriented along the $\xi ^\mu$ direction, namely a chart where $\xi ^\mu = \delta ^\mu _0$. In this case $\xi _\mu = g_{\mu 0}$ and $\xi ^\mu \xi _\mu =g_{00}>0$. The Killing condition $\delta _\xi g_{\mu \nu }=0$, written explicitly according to Eq. (3.53), reduces to

$$\begin{aligned} \partial _0 g_{\mu \nu }=0, \end{aligned}$$

(10.82)

so that the metric is not time dependent. We also obtain, in this chart,

$$\begin{aligned} \nabla _\mu \xi _\nu = \nabla _\mu g_{\nu \alpha } \xi ^\alpha = g_{\nu \alpha }\nabla _\mu \xi ^\alpha = g_{\nu \alpha } \varGamma _{\mu 0}{}^\alpha ={1\over 2} \left( \partial _\mu g_{0\nu }- \partial _\nu g_{\mu 0}\right) = \partial _{[\mu } g_{\nu ]0}. \end{aligned}$$

(10.83)

If the metric is static it satisfies the condition $g_{i0}=0$, so that

$$\begin{aligned} \xi _{[\alpha } \nabla _\mu \xi _{\nu ]}=g_{0[\alpha } \partial _\mu g_{\nu ]0} \equiv 0 \end{aligned}$$

(10.84)

(because, in the above expression, the metric is non vanishing only for $\alpha =0$ and $\nu =0$).

Conversely, let us suppose that Eq. (10.80) is satisfied, and show that we can always find a chart where $g_{i0}=0$.

Let us keep for the moment the coordinates where $\xi ^\mu = \delta ^\mu _0$ and the metric is time independent. Writing explicitly Eq. (10.80), and contracting with $\xi ^\alpha$, we obtain

$$\begin{aligned}&\xi ^\alpha \big ( \xi _\alpha \nabla _\mu \xi _\nu + \xi _\mu \nabla _\nu \xi _\alpha + \xi _\nu \nabla _\alpha \xi _\mu - \xi _\alpha \nabla _\nu \xi _\mu - \xi _\mu \nabla _\alpha \xi _\nu - \xi _\nu \nabla _\mu \xi _\alpha \big )= \nonumber \\&~~~~~~~~~~~= \xi ^2 \nabla _\mu \xi _\nu +{1\over 2} \xi _\mu \nabla _\nu \xi ^2 +\xi _\nu \xi ^\alpha \nabla _\alpha \xi _\mu - \left\{ \mu \leftrightarrow \nu \right\} =0, \end{aligned}$$

(10.85)

where $\xi ^2 = \xi ^\mu \xi _\mu$. From Eq. (10.83) we have:

$$\begin{aligned} \xi ^\alpha \nabla _\alpha \xi ^\mu = \nabla _0 \xi _\mu = \partial _{[0}g_{\mu ]0}=-{1\over 2} \partial _\mu g_{00}= -{1\over 2} \nabla _\mu \xi ^2. \end{aligned}$$

(10.86)

Inserting this result into Eq. (10.85), and dividing by $\xi ^4$, we are led to the condition

$$\begin{aligned}&\xi ^{-2} \left( \nabla _\mu \xi _\nu - \nabla _\nu \xi _\mu \right) - \xi _\mu \nabla _\nu \xi ^{-2} + \xi _\nu \nabla _\mu \xi ^{-2} \nonumber \\ \equiv&\nabla _\mu \left( \xi ^{-2} \xi _\nu \right) - \nabla _\nu \left( \xi ^{-2} \xi _\mu \right) =0, \end{aligned}$$

(10.87)

which is solved by

$$\begin{aligned} \xi _\nu = \xi ^2 \partial _\nu \phi , \end{aligned}$$

(10.88)

where $\phi$ is an arbitrary scalar function. In the chart where we are working, on the other hand, we have $\xi _0= g_{00}= \xi ^2$, which implies $\partial _0\phi =1$, namely

$$\begin{aligned} \phi = x^0+f(x^i), \end{aligned}$$

(10.89)

where f is an arbitrary function of the spatial coordinates.

Let us now consider the coordinate transformation

$$\begin{aligned} x^0 \rightarrow x^{{\prime }0}= \phi = x^0+ f(x^i), ~~~~~~~~~~~ x^i \rightarrow x^{{\prime }i}=x^i. \end{aligned}$$

(10.90)

The components of our Killing vector are left unchanged,

$$\begin{aligned} \xi ^{{\prime }\mu }= {\partial x^{{\prime }\mu } \over \partial x^\nu } \xi ^\nu = {\partial x^{{\prime }\mu } \over \partial x^0 } = \delta ^\mu _0, \end{aligned}$$

(10.91)

and so is also the metric component $g_{00}$:

$$\begin{aligned} g'_{00}= {\partial x^{\alpha } \over \partial x^{{\prime }0} } {\partial x^{\beta } \over \partial x^{{\prime }0} } g_{\alpha \beta }= \delta ^\alpha _0\delta ^\beta _0 g_{\alpha \beta }= g_{00}. \end{aligned}$$

(10.92)

For the mixed components, instead, we find

$$\begin{aligned} g'_{i0}= {\partial x^{\alpha } \over \partial x^{{\prime }i} } {\partial x^{\beta } \over \partial x^{{\prime }0} } g_{\alpha \beta }&=g_{\alpha \beta } \delta ^\beta _0 \left( \delta ^\alpha _j\delta ^j_i- \delta ^\alpha _0 \partial _if\right) \nonumber \\&=g_{i0}- g_{00} \partial _if \equiv 0. \end{aligned}$$

(10.93)

The result is zero because, in the old chart,

$$\begin{aligned} g_{i0}= \xi _i= \xi ^2 \partial _i \phi = g_{00} \partial _if. \end{aligned}$$

(10.94)

This shows that if the Killing vector satisfies the condition (10.80) it is always possible to find a chart where the metric components $g_{i0}$ are all vanishing, as appropriate to a geometry of static type.

10.2

Solution The Schwarzschild metric (10.19) has the same structure as that of the metric (6.92) introduced in Exercise 6.6, with

$$\begin{aligned} f(r)= g_{00}=-{1\over g_{11}}=1-{2m\over r}. \end{aligned}$$

(10.95)

By using the result (6.94) we then immediately obtain the following non-zero components of the Riemann tensor:

$$\begin{aligned}&R_{01}\,^{01}= -{1\over 2} f''={2m\over r^3}, ~~~~~~~ R_{23}\,^{23}=-{1\over r^2}(f-1)={2m\over r^3}, \nonumber \\&R_{02}\,^{02}=R_{03}\,^{03}=R_{12}\,^{12}=R_{13}\,^{13}= -{1\over 2r} f' =-{m\over r^3}. \end{aligned}$$

(10.96)

Hence:

$$\begin{aligned} R_{\mu \nu \alpha \beta } R^{\mu \nu \alpha \beta }= & {} R_{\mu \nu }\,^{\alpha \beta } R_{\alpha \beta }\,^{\mu \nu } \nonumber \\= & {} 4R^2_{01}\,^{01}+4R^2_{02}\,^{02}+4R^2_{03}\,^{03}+4R^2_{12}\,^{12}+4R^2_{13}\,^{13}+4R^2_{23}\,^{23} \nonumber \\= & {} {48 m^2 \over r^6}. \end{aligned}$$

(10.97)

10.3

Solution The connection for the metric (10.81) has been already computed in Exercise 6.1. The time-component of the geodesic equation is

$$\begin{aligned} \ddot{\eta }+ {2\over \xi } \dot{\eta }\dot{\xi }=0, \end{aligned}$$

(10.98)

where the dot denotes differentiation with respect to the proper time $\tau$. Its integration gives

$$\begin{aligned} \dot{\eta }= k \xi ^{-2}, \end{aligned}$$

(10.99)

where k is an integration constant. In addition, from the normalization of the four-velocity vector we have

$$\begin{aligned} \dot{x}^\mu \dot{x}_\mu = \xi ^2 \dot{\eta }^2 - \dot{\xi }^2 ={k^2\over \xi ^2}- \dot{\xi }^2= c^2. \end{aligned}$$

(10.100)

Separating the variables, and integrating, we then obtain the proper-time interval $\varDelta \tau$ needed to reach the origin $\xi =0$, starting from the point $\xi =\xi _0$:

$$\begin{aligned} \varDelta \tau = -\int _{\xi _0}^0 {\xi d \xi \over \sqrt{k^2-c^2 \xi ^2}}={1\over c^2} \left( k- \sqrt{k^2-c^2 \xi ^2_0}\right) . \end{aligned}$$

(10.101)

This integral is convergent, and the corresponding proper time interval is finite. Note that at $\xi =0$ the parametrization of the Minkowski space-time in terms of the Rindler coordinates is no longer defined, since the transformation (10.59) is singular.

The integration constant k can be expressed in terms of the velocity $\dot{\xi }_0$, evaluated at the initial time $\tau =0$. From Eq. (10.100) we have, in fact,

$$\begin{aligned} k^2 = \xi _0^2\left( c^2+\dot{\xi }_0^2\right) , \end{aligned}$$

(10.102)

and we can also rewrite the final result (10.101) in the equivalent form

$$\begin{aligned} \varDelta \tau = {\xi _0\over c^2} \left( \sqrt{c^2+ \dot{\xi }_0^2}- \dot{\xi }_0\right) . \end{aligned}$$

(10.103)