Deriving the Schwarzschild solution

Deriving the Schwarzschild solution

The Schwarzschild solution is one of the simplest and most useful solutions of the Einstein field equations (see general relativity). It describes spacetime in the vicinity of a non-rotating massive spherically-symmetric object. It is worthwhile deriving this metric in some detail; the following is a reasonably rigorous derivation that is not always seen in the textbooks.

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 the geometry of the spacetime is unchanged under a time-reversal $t \rightarrow -t$.

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

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 gμ4 ($\mu \ne 4$) 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 = gμ4 (metric components remain the same), 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 (line element) 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).

Simplifying the components

On each hypersurface of constant t, constant θ and constant ϕ (i.e., on each radial line), g11 should only depend on r (by spherical symmetry). Hence g11 is a function of a single variable:

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

A similar argument applied to g44 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 r0, 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:

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

with A and B as yet undetermined functions of r. Note that if A or B is equal to zero at some point, the metric would be singular at that point.

Calculating the Christoffel symbols

Another well-known notation for the metric tensor is

$g_{ik} = \begin{bmatrix} -f^2 & 0 & 0 & 0\\ 0 & h^2 & 0 & 0\\ 0&0&r^2 & 0\\ 0&0&0& r^2 \sin^2 \theta \end{bmatrix}$

From this form of the metric tensor one can calculate the Christoffel symbols

$\Gamma^0_{ik} = \begin{bmatrix} 0 & f'/f & 0 & 0\\ f'/f & 0 & 0 & 0\\ 0&0&0 & 0\\ 0&0&0& 0\end{bmatrix}$

Here comma means the r derivative of the functions.

$\Gamma^1_{ik} = \begin{bmatrix} ff'/h^2 & 0 & 0 & 0\\ 0 & h'/h & 0 & 0\\ 0&0& -r/h^2 & 0\\ 0&0&0& -r \sin^2 \theta /h^2 \end{bmatrix}$

$\Gamma^2_{ik} = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 1/r & 0\\ 0&1/r&0 & 0 \\ 0&0&0& -\sin \theta \cos \theta \end{bmatrix}$

$\Gamma^3_{ik} = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1/r\\ 0&0&0 & \cot \theta \\ 0&1/r&\cot \theta & 0\end{bmatrix}$

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

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

$\rm{R_{ab}=\, 0}$

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

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

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

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

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

Here dot means the r derivative of the functions. Subtracting the first and third equations produces:

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

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

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

which has general solution:

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

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

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

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.

Using the Weak-Field Approximation to find K and S

The geodesics of the metric (obtained where ds is extremised) must, in some limit (e.g., toward infinite speed of light), agree with the solutions of Newtonian motion (e.g., obtained by Lagrange equations). (The metric must also limit to Minkowski space when the mass it represents vanishes.)

$0=\delta\int\frac{ds}{dt}dt=\delta\int(KE+PE_g)dt$

(where E and Eg are _____?) The constants K and S are fully determined by some variant of this approach; from the weak-field approximation one arrives at the result:

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

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:

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

Alternative form in isotropic coordinates

The original formulation of the metric uses anisotropic coordinates in which the velocity of light is not the same in the radial and transverse directions. A S Eddington[1] gave alternative forms in isotropic coordinates. For isotropic spherical coordinates r1, θ, ϕ , coordinates θ and ϕ are unchanged, and then (provided r >= 2Gm/c2 [2])

$r = r_1 \left(1+\frac{Gm}{2c^2 r_1}\right)^{2}$ . . ., $dr = dr_1 \left(1-\frac{(Gm)^2}{4c^4 r_1^2}\right)$ . . ., and

$\left(1-\frac{2Gm}{c^2 r}\right) = \left(1-\frac{Gm}{2c^2 r_1}\right)^{2}/\left(1+\frac{Gm}{2c^2 r_1}\right)^{2}$ . . .

Then for isotropic rectangular coordinates x, y, z,

$x = r_1\, \sin(\theta)\, \cos(\phi) \dots,$ $y = r_1\, \sin(\theta)\, \sin(\phi) \dots,$ $z = r_1\, \cos(\theta) \dots$

The metric then becomes, in isotropic rectangular coordinates:

$ds^2= \left(1+\frac{Gm}{2c^2 r_1}\right)^{4}(dx^2+dy^2+dz^2) -c^2 dt^2 \left(1-\frac{Gm}{2c^2 r_1}\right)^{2}/\left(1+\frac{Gm}{2c^2 r_1}\right)^{2}$ . . .

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 gravitational waves (as the region exterior to the star must remain static).

References

1. ^ A S Eddington, "Mathematical Theory of Relativity", Cambridge UP 1922 (2nd ed.1924, repr.1960), at page 85 and page 93. Symbol usage in the Eddington source for interval s and time-like coordinate t has been converted for compatibility with the usage in the derivation above.
2. ^ H. A. Buchdahl, "Isotropic coordinates and Schwarzschild metric", International Journal of Theoretical Physics, Vol.24 (1985) pp.731-739.

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