Questions: The Schwarzschild Solution

Question 1

At the Schwarzschild radius r = 2GM/c², the metric component g_{rr} diverges. This means:

• A. There is a physical singularity — spacetime curvature becomes infinite at this radius
• B. The coordinate system breaks down, but spacetime geometry is perfectly regular — freely falling observers pass through without experiencing infinite tidal forces
• C. The metric is undefined and must be replaced by a different theory of gravity
• D. Time stops at this radius, so nothing can ever cross it

Answer: (B)

The divergence at r = 2GM/c² is a coordinate singularity, not a physical one. The curvature invariant R_{μνρσ}R^{μνρσ} is finite and well-behaved at r = r_s, confirming that spacetime is smooth there. Alternative coordinate systems (Eddington-Finkelstein, Kruskal-Szekeres) eliminate the singularity and show that freely falling observers cross the horizon in finite proper time. The apparent 'infinite time' in Schwarzschild coordinates is a coordinate artifact — it is the coordinate time t of a distant observer that diverges, not the proper time of the infalling observer.

Question 2

Birkhoff's theorem states that the Schwarzschild solution is the only spherically symmetric vacuum solution, even if the source is not static (e.g., a radially pulsating star).

• T. True
• F. False

Answer: True

Birkhoff's theorem proves that any spherically symmetric vacuum solution must be static and equal to the Schwarzschild metric. This means the exterior spacetime of a radially pulsating or collapsing spherically symmetric star is still Schwarzschild — the time-dependent internal dynamics do not radiate gravitational waves (spherical symmetry forbids it) and do not affect the external geometry. This is the GR analog of Newton's shell theorem (a spherically symmetric mass distribution creates the same external field as a point mass at the center).

Question 3

Derive the Schwarzschild radius for a mass equal to that of the Sun (M_☉ ≈ 2 × 10³⁰ kg) and explain its physical significance.

Model answer: r_s = 2GM/c² = 2(6.67 × 10⁻¹¹)(2 × 10³⁰)/(3 × 10⁸)² ≈ 2.95 km. This is the radius to which the Sun would need to be compressed for its surface to coincide with the event horizon of a black hole. For a normal star like the Sun (radius ~700,000 km), r_s is deep inside the star where the vacuum Schwarzschild solution does not apply. The Schwarzschild radius sets the scale at which general relativistic effects become extreme: gravitational time dilation becomes infinite, escape velocity reaches c, and an event horizon forms.

The remarkable smallness of r_s compared to the Sun's actual radius illustrates why Newtonian gravity is an excellent approximation for most stellar physics. The ratio r_s/R_☉ ≈ 4 × 10⁻⁶ is the dimensionless measure of how relativistic the Sun's gravitational field is at its surface.

Question 4

How does the Schwarzschild metric reduce to Newtonian gravity far from the central mass?

Model answer: For r >> 2GM/c², the metric component g_{00} = -(1 - 2GM/rc²) ≈ -(1 + 2Φ/c²), where Φ = -GM/r is the Newtonian gravitational potential. The geodesic equation for a slowly moving particle (v << c) in this weak-field metric reduces to d²r/dt² ≈ -GM/r², which is Newton's inverse-square law. The spatial metric deviation from flatness is also small (g_{rr} ≈ 1 + 2GM/rc²), contributing only post-Newtonian corrections. Thus the Schwarzschild metric smoothly connects to Newtonian gravity in the weak-field limit, as required for any viable theory of gravity.

The weak-field expansion of the Schwarzschild metric is the starting point for computing post-Newtonian corrections: the perihelion precession of Mercury, gravitational time dilation, and the Shapiro time delay are all first-order corrections beyond the Newtonian limit.