The Schwarzschild Metric & Black Holes
The Schwarzschild Metric & Black Holes
The Schwarzschild solution is the unique spherically symmetric vacuum solution to the EFE (Birkhoff's theorem). It describes spacetime outside any spherically symmetric mass and predicts the existence of black holes — regions bounded by an event horizon from which nothing, not even light, can escape.
Schwarzschild Metric
\(ds^2 = -\!\left(1-\frac{r_s}{r}\right)c^2 dt^2 + \left(1-\frac{r_s}{r}\right)^{-1}\!dr^2 + r^2 d\Omega^2\), where \(r_s = 2GM/c^2\) is the Schwarzschild radius and \(d\Omega^2 = d\theta^2 + \sin^2\theta\,d\phi^2\). The metric is singular at \(r = r_s\) (coordinate singularity) and \(r=0\) (true curvature singularity).
Birkhoff's Theorem & Event Horizon
The unique spherically symmetric vacuum solution is Schwarzschild, and it is static regardless of radial dynamics of the source. The surface \(r = r_s\) is a null hypersurface (event horizon), a coordinate singularity removable by Kruskal-Szekeres coordinates. The Kretschner scalar \(K = 48G^2M^2/(c^4 r^6)\) diverges only at \(r=0\).
Example 1: Schwarzschild Radius of the Sun
Compute \(r_s\) for \(M_\odot = 1.99\times 10^{30}\,\text{kg}\).
Solution: \(r_s = 2GM_\odot/c^2 = 2(6.67\times10^{-11})(1.99\times10^{30})/(9\times10^{16}) \approx 2.95\,\text{km}\). The Sun's actual radius \(\approx 7\times10^5\,\text{km}\) is far larger, so the Sun is not a black hole.
Example 2: Gravitational Redshift
Find the redshift of a photon emitted at \(r_e\) received at \(r_r \gg r_s\).
Solution: \(1+z = \nu_e/\nu_r = \left(1-r_s/r_e\right)^{-1/2}\left(1-r_s/r_r\right)^{1/2} \approx \left(1-r_s/r_e\right)^{-1/2}\). As \(r_e\to r_s\), \(z\to\infty\): infinite redshift at the horizon.
Practice
- Verify the Schwarzschild metric satisfies \(R_{\mu\nu}=0\) by computing \(R_{tt}\).
- Find the innermost stable circular orbit (ISCO) at \(r = 3r_s\).
- Compute the proper time for a freely falling observer to reach \(r=0\) from \(r=r_s\).
- Describe the Kruskal extension and the meaning of the white hole region.
Show Answer Key
1. Schwarzschild: $ds^2 = -(1-r_s/r)c^2dt^2+(1-r_s/r)^{-1}dr^2+r^2d\Omega^2$. Static, spherically symmetric vacuum: compute $R_{\mu\nu}$ from the Christoffel symbols. $R_{tt}$ involves $\Phi''+\Phi'^2+2\Phi'/r$ (where $e^{2\Phi}=1-r_s/r$). Setting $R_{tt}=0$ (and all other components) is satisfied by $\Phi = \frac{1}{2}\ln(1-r_s/r)$. ✓
2. Effective potential for circular orbits: $V_{\text{eff}} = (1-r_s/r)(1+L^2/(r^2c^2))$. Circular: $V'_{\text{eff}}=0$. Stable: $V''_{\text{eff}}>0$. ISCO: $V''_{\text{eff}}=0$ simultaneously. Solving: $r_{\text{ISCO}} = 3r_s = 6GM/c^2$. Below this, no stable circular orbits exist — particles spiral into the black hole.
3. Radial free fall from $r=r_s$: proper time $\Delta\tau = \int_0^{r_s}\frac{dr}{c\sqrt{r_s/r-1}}$... For fall from $r_0$: $\Delta\tau = \frac{\pi r_s}{2c}\sqrt{r_0/r_s}(r_0/r_s)$. From $r_s$ to $r=0$: $\Delta\tau = \frac{\pi r_s}{2c} = \frac{\pi GM}{c^3}$. For $M = M_\odot$: $\Delta\tau \approx 16\,\mu$s. The observer reaches the singularity in finite proper time despite the coordinate singularity at $r=r_s$.
4. Kruskal coordinates $(T,X)$: the Schwarzschild coordinate singularity at $r=r_s$ becomes regular. The maximal extension reveals four regions: I (exterior), II (black hole interior, $r