Deriving the Schwarzschild solution

From Wikipedia, the free encyclopedia

The Schwarzschild solution is one of the simplest and useful solutions of the Einstein field equations (see general relativity). It is worthwhile deriving this metric in some detail; the following is a reasonably rigorous derivation that is not always seen in the textbooks.

1 Assumptions and notation
2 Diagonalising the metric
3 Simplifying the components
4 Using the field equations to find A(r) and B(r)
5 Using the Weak-Field Approximation to find K and S
6 Dispensing with the static assumption - Birkhoff's theorem
7 See also

[edit] Assumptions and notation

Working in a coordinate chart with coordinates $\left(r, \theta, \phi, t \right)$ labelled 1 to 4 respectively, we begin with the metric in its most general form (10 independent components, each of which is an arbitrary function of 4 variables). The solution is assumed to be spherically symmetric, static and vacuum. For the purposes of this article, these assumptions may be stated as follows (see the relevant links for precise definitions):

(1) A spherically symmetric spacetime is one in which all metric components are unchanged under any rotation-reversal $\theta \rightarrow - \theta$ or $\phi \rightarrow - \phi$ .

(2) A static spacetime is one in which all metric components are independent of the time coordinate $t$ (so that $\frac {\part g_{\mu \nu}}{\part t}=0$ ) and hence the geometry of the spacetime is unchanged under a time-reversal $t \rightarrow -t$ .

(3) A vacuum solution is one which satisfies the equation $T a b = 0$ . From the Einstein field equations (with zero cosmological constant), this implies that $R a b = 0$ (after contracting $R_{ab}-\frac{R}{2} g_{ab}=0$ and putting $R = 0$ ).

[edit] Diagonalising the metric

The first simplification to be made is to diagonalise the metric. Under the coordinate transformation, $(r, \theta, \phi, t) \rightarrow (r, \theta, \phi, -t)$ , all metric components should remain the same. The metric components $g μ4$ ( $\mu \ne 4$ ) change under this transformation as:

$g_{\mu 4}'=\frac{\part x^{\alpha}}{\part x^{'\mu}} \frac{\part x^{\beta}}{\part x^{'4}} g_{\alpha \beta}= -g_{\mu 4}$ ( $\mu \ne 4$ )

But, as we expect $g μ4' = g μ4$ , this means that:

$g_{\mu 4}=\, 0$ ( $\mu \ne 4$ )

Similarly, the coordinate transformations $(r, \theta, \phi, t) \rightarrow (r, \theta, -\phi, t)$ and $(r, \theta, \phi, t) \rightarrow (r, -\theta, \phi, t)$ respectively give:

$g_{\mu 3}=\, 0$ ( $\mu \ne 3$ )

$g_{\mu 2}=\, 0$ ( $\mu \ne 2$ )

Putting all these together gives:

$g_{\mu \nu }=\, 0$ ( $\mu \ne \nu$ )

and hence the metric must be of the form:

$ds^2=\, g_{11}d r^2 + g_{22} d \theta ^2 + g_{33} d \phi ^2 + g_{44} dt ^2$

where the four metric components are independent of the time coordinate $t$ (by the static assumption).

[edit] Simplifying the components

On each hypersurface of constant $t$ , constant $θ$ and constant $φ$ (i.e., on each radial line), $g 11$ should only depend on $r$ (by spherical symmetry). Hence $g 11$ is a function of a single variable:

$g_{11}=A\left(r\right)$

A similar argument applied to $g 44$ shows that:

$g_{44}=B\left(r\right)$

On the hypersurfaces of constant $t$ and constant $r$ , it is required that the metric be that of a 2-sphere:

$dl^2=r_{0}^2 (d \theta^2 + \sin^2 \theta d \phi^2)$

Choosing one of these hypersurfaces (the one with radius $r 0$ , say), the metric components restricted to this hypersurface (which we denote by $\tilde{g}_{22}$ and $\tilde{g}_{33}$ ) should be unchanged under rotations through $θ$ and $φ$ (again, by spherical symmetry). Comparing the forms of the metric on this hypersurface gives:

$\tilde{g}_{22}\left(d \theta^2 + \frac{\tilde{g}_{33}}{\tilde{g}_{22}} d \phi^2 \right) = r_{0}^2 (d \theta^2 + \sin^2 \theta d \phi^2)$

which immediately yields:

$\tilde{g}_{22}=r_{0}^2$ and $\tilde{g}_{33}=r_{0}^2 \sin ^2 \theta$

But this is required to hold on each hypersurface; hence,

$g_{22}=\, r^2$ and $g_{33}=\, r^2 \sin^2 \theta$

Thus, the metric can be put in the form:

$ds^2=A\left(r\right)dr^2+r^2d \theta^2+r^2 \sin^2 \theta d \phi^2 + B\left(r\right) dt^2$

with $A$ and $B$ as yet undetermined functions of $r$ . Note that $A$ and $B$ must both be nowhere zero (otherwise the metric would be singular at those points where at least one of them is zero).

[edit] Using the field equations to find $A (r)$ and $B (r)$

To determine $A$ and $B$ , the vacuum field equations are employed:

$R_{ab}=\, 0$

Only four of these equations are nontrivial and upon simplification become:

$4 \dot{A} B^2 - 2 r \ddot{B} AB + r \dot{A} \dot{B}B + r \dot{B} ^2 A=0$

$r \dot{A}B + 2 A^2 B - 2AB - r \dot{B} A=0$

$- 2 r \ddot{B} AB + r \dot{A} \dot{B}B + r \dot{B} ^2 A - 4\dot{B} AB=0$

(The fourth equation is just $sin 2 θ$ times the second equation)

Subtracting the first and third equations produces:

$\dot{A}B +A \dot{B}=0 \Rightarrow A(r)B(r) =K$

where $K$ is a non-zero real constant. Substituting $A(r)B(r) \, =K$ into the second equation and tidying up gives:

$r \dot{A} =A(1-A)$

which has general solution:

$A(r)=\left(1+\frac{1}{Sr}\right)^{-1}$

for some non-zero real constant $S$ . Hence, the metric for a static, spherically symmetric vacuum solution is now of the form:

$ds^2=\left(1+\frac{1}{S r}\right)^{-1}dr^2+r^2(d \theta^2 + \sin^2 \theta d \phi^2)+K \left(1+\frac{1}{S r}\right)dt^2$

Note that the spacetime represented by the above metric is asymptotically flat, i.e. as $r \rightarrow \infty$ , the metric approaches that of the Minkowski metric and the spacetime manifold resembles that of Minkowski space.

[edit] Using the Weak-Field Approximation to find $K$ and $S$

To identify the constants $K$ and $S$ , use is made of the weak-field approximation; using this, the $g 44$ components can be compared:

$g_{44}=K\left(1 +\frac{1}{Sr}\right) \approx -c^2+\frac{2Gm}{r} = -c^2 \left(1-\frac{2Gm}{c^2 r} \right)$

where $G$ is the gravitational constant, $m$ is the mass of the gravitational source and $c$ is the speed of light. It is found that:

$K=\, -c^2$ and $\frac{1}{S}=-\frac{2Gm}{c^2}$

Hence:

$A(r)=\left(1-\frac{2Gm}{c^2 r}\right)^{-1}$ and $B(r)=-c^2 \left(1-\frac{2Gm}{c^2 r}\right)$

So, the Schwarzschild metric may finally be written in the form:

$ds^2=\left(1-\frac{2Gm}{c^2 r}\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(1-\frac{2Gm}{c^2 r}\right)dt^2$

[edit] Dispensing with the static assumption - Birkhoff's theorem

In deriving the Schwarzschild metric, it was assumed that the metric was vacuum, spherically symmetric and static. In fact, the static assumption is stronger than required, as Birkhoff's theorem states that any spherically symmetric vacuum solution of Einstein's field equations is stationary; then one obtains the Schwarzschild solution. Birkhoff's theorem has the consequence that any pulsating star which remains spherically symmetric cannot generate gravity waves (as the region exterior to the star must remain static).

Category: Exact solutions in general relativity

Deriving the Schwarzschild solution

From Wikipedia, the free encyclopedia

Contents

[edit] Assumptions and notation

[edit] Diagonalising the metric

[edit] Simplifying the components

[edit] Using the field equations to find $A (r)$ and $B (r)$

[edit] Using the Weak-Field Approximation to find $K$ and $S$

[edit] Dispensing with the static assumption - Birkhoff's theorem

[edit] See also

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Deriving the Schwarzschild solution

From Wikipedia, the free encyclopedia

Contents

[edit] Assumptions and notation

[edit] Diagonalising the metric

[edit] Simplifying the components

[edit] Using the field equations to find A(r) and B(r)

[edit] Using the Weak-Field Approximation to find K and S

[edit] Dispensing with the static assumption - Birkhoff's theorem

[edit] See also

Views

Navigation

interaction

Search

[edit] Using the field equations to find $A (r)$ and $B (r)$

[edit] Using the Weak-Field Approximation to find $K$ and $S$