Schwarzschild Solution: Derivation
Introduction
The Schwarzschild solution, discovered by Karl Schwarzschild in 1916 (just months after Einstein published general relativity), describes the spacetime geometry outside a spherically symmetric, non-rotating mass. It's the first exact solution to Einstein's field equations and describes black holes.
We'll derive this solution step by step from Einstein's field equations using spherical symmetry.
Step 1: Symmetry Assumptions
Key Assumptions
- Spherical symmetry: The metric depends only on radial coordinate r
- Static: The metric is time-independent (no time derivatives)
- Asymptotically flat: At large distances, spacetime approaches Minkowski space
- Vacuum solution: We solve Einstein's equations in vacuum (outside the mass)
The most general spherically symmetric, static metric in spherical coordinates (t, r, θ, φ) can be written as:
$$ds^2 = -e^{2\alpha(r)} c^2 dt^2 + e^{2\beta(r)} dr^2 + r^2 d\Omega^2$$
where $d\Omega^2 = d\theta^2 + \sin^2\theta \, d\phi^2$ is the metric on a unit 2-sphere, and α(r) and β(r) are unknown functions we must determine.
Step 2: Einstein Field Equations in Vacuum
In vacuum (no matter or energy), Einstein's field equations reduce to:
$$R_{\mu\nu} = 0$$
This says the Ricci tensor must vanish everywhere outside the mass. The Ricci tensor is computed from the Christoffel symbols, which are derived from the metric.
Step 3: Christoffel Symbols
The Christoffel symbols are given by:
$$\Gamma^\rho_{\mu\nu} = \frac{1}{2}g^{\rho\sigma}\left(\frac{\partial g_{\sigma\mu}}{\partial x^\nu} + \frac{\partial g_{\sigma\nu}}{\partial x^\mu} - \frac{\partial g_{\mu\nu}}{\partial x^\sigma}\right)$$
For our metric, the non-zero Christoffel symbols include:
$$\Gamma^t_{tr} = \alpha'(r), \quad \Gamma^r_{tt} = \alpha'(r) e^{2(\alpha - \beta)}, \quad \Gamma^r_{rr} = \beta'(r)$$
$$\Gamma^r_{\theta\theta} = -r e^{-2\beta}, \quad \Gamma^r_{\phi\phi} = -r\sin^2\theta \, e^{-2\beta}$$
$$\Gamma^\theta_{r\theta} = \Gamma^\phi_{r\phi} = \frac{1}{r}, \quad \Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$$
$$\Gamma^\phi_{\theta\phi} = \cot\theta$$
Here, primes denote derivatives with respect to r: α'(r) = dα/dr.
Step 4: Ricci Tensor Components
The Ricci tensor is computed from:
$$R_{\mu\nu} = \partial_\rho \Gamma^\rho_{\mu\nu} - \partial_\nu \Gamma^\rho_{\mu\rho} + \Gamma^\rho_{\mu\nu}\Gamma^\sigma_{\rho\sigma} - \Gamma^\rho_{\mu\sigma}\Gamma^\sigma_{\nu\rho}$$
After substantial calculation (which I'll spare you the details of!), the independent components are:
$$R_{tt} = e^{2(\alpha-\beta)}\left[\alpha'' + \alpha'^2 - \alpha'\beta' + \frac{2\alpha'}{r}\right]$$
$$R_{rr} = -\alpha'' - \alpha'^2 + \alpha'\beta' + \frac{2\beta'}{r}$$
$$R_{\theta\theta} = e^{-2\beta}\left[r(\beta' - \alpha') - 1\right] + 1$$
$$R_{\phi\phi} = \sin^2\theta \, R_{\theta\theta}$$
Step 5: Solving Rμν = 0
Setting each component to zero and solving the coupled differential equations:
From Rθθ = 0:
$$e^{-2\beta}\left[r(\beta' - \alpha') - 1\right] + 1 = 0$$
$$\Rightarrow \frac{d}{dr}\left(r e^{-2\beta}\right) = 1$$
$$\Rightarrow r e^{-2\beta} = r - r_s$$
$$\Rightarrow e^{2\beta} = \frac{1}{1 - r_s/r}$$
where rs is an integration constant (the Schwarzschild radius).
From Rtt + Rrr = 0:
$$\alpha' + \beta' = 0 \quad \Rightarrow \quad \alpha = -\beta + \text{const}$$
Choosing the constant so that the metric is asymptotically flat (approaches Minkowski as r → ∞):
$$e^{2\alpha} = 1 - \frac{r_s}{r}$$
Step 6: The Schwarzschild Metric
Substituting our solutions for α and β back into the metric:
Schwarzschild Metric
$$ds^2 = -\left(1 - \frac{r_s}{r}\right)c^2 dt^2 + \frac{dr^2}{1 - r_s/r} + r^2(d\theta^2 + \sin^2\theta \, d\phi^2)$$
where the Schwarzschild radius is:
$$r_s = \frac{2GM}{c^2}$$
This is determined by matching to the Newtonian limit at large distances.
Birkhoff's Theorem
Theorem:
The Schwarzschild solution is the unique spherically symmetric vacuum solution to Einstein's field equations.
Implications:
- Any spherically symmetric mass distribution has Schwarzschild geometry outside it
- A pulsating spherical star doesn't produce gravitational waves (spherical symmetry prevents it)
- The exterior geometry doesn't depend on the star's internal structure—only its total mass M
Physical Interpretation
For the Sun:
$$r_s = \frac{2GM_\odot}{c^2} \approx 3 \text{ km}$$
The Sun's actual radius is 696,000 km, so we're far from the Schwarzschild radius.
For Earth:
$$r_s = \frac{2GM_\oplus}{c^2} \approx 9 \text{ mm}$$
Earth would need to be compressed to 9 millimeters to become a black hole!
What is a Black Hole?
A black hole forms when mass M is compressed within its Schwarzschild radius rs. At r = rs, the metric component gtt vanishes and grr diverges—this is the event horizon, the point of no return.