Chapter 18: Problem 26

In the Schwarzschild solution show the only possible closed photon path is a circular orbit at \(r=3 m\), and show that it is unstable.

Short Answer

Expert verified

The only closed photon path in the Schwarzschild solution is a circular orbit at \(r=3m\), and the orbit is unstable.

Step by step solution

Setting up the equation

In the Schwarzschild metric, the motion of a photon is determined by the equation \( \frac{1}{b^{2}} = \frac{1}{r^{2}} - \frac{1}{r^{3}} \). The letter 'b' is the impact parameter related to the angular momentum per unit energy, 'r' is the radial distance from the mass 'm', and 'm' is the mass of the object causing the gravitational field. We are looking for solutions where the photon moves in a circle of fixed radius. For this to happen, the photon needs to be at a local extremum of the effective potential. To derive this potential with respect to the radius 'r', set the derivative of the expression on the right-hand side of the equation to zero.

Deriving the closed path

The derivative of \( \frac{1}{r^{2}} - \frac{1}{r^{3}} \) with respect to 'r' is \(- \frac{2}{r^{3}} + \frac{3}{r^{4}} \). Setting this equal to zero gives the equation: \(- \frac{2}{r^{3}} + \frac{3}{r^{4}} = 0 \), which simplifies to \(- 2r + 3 = 0\). Solving for 'r' gives \(r=3m\). So, the only circular path is for \(r=3m\).

Proving instability

To show that the orbit at \(r=3m\) is unstable, we have to show that a small deviation from \(r=3m\) leads to progressively larger deviations, i.e., the potential must have a local maximum at \(r=3m\). Taking the second derivative of \( \frac{1}{r^{2}} - \frac{1}{r^{3}} \) gives \(\frac{6}{r^{4}} - \frac{12}{r^{5}}\). Replacing \(r\) with \(3m\) gives a positive value, indicating the potential has a local maximum and the orbit is indeed unstable.