# Schwarzschild metric: four-momentum of a photon

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapters 9; Problem 9.6.

This is a first example of the use of the time component of the Schwarzschild metric. This metric is, for a spherical mass ${M}$:

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

Suppose we have an observer at a Schwarzschild radius ${R}$ from the centre of a star of mass ${M}$, and this observer watches a photon move radially outward. The observer measures the energy of the photon to be ${E}$. We can use this to calculate the four-momentum of the photon.

In special relativity, for an observer at rest the observer’s four-velocity is ${u^{i}=\left[1,0,0,0\right]}$ so the scalar product of the observer’s four-velocity with another object’s momentum (as measured by the observer) is

 $\displaystyle \mathbf{p}\cdot\mathbf{u}_{obs}$ $\displaystyle =$ $\displaystyle g_{ij}p^{i}u^{j}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p^{t}u^{t}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p^{t}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -E \ \ \ \ \ (5)$

since the time component of an object’s four-momentum is its energy. Since this is a tensor equation, it should be true in curved space-time as well. In the Schwarzschild metric, an observer at rest has

$\displaystyle u^{t}=\left[\left(1-\frac{2GM}{R}\right)^{-1/2},0,0,0\right] \ \ \ \ \ (6)$

Therefore, we get

 $\displaystyle \mathbf{p}\cdot\mathbf{u}_{obs}$ $\displaystyle =$ $\displaystyle g_{ij}p^{i}u^{j}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{R}\right)p^{t}\left(1-\frac{2GM}{R}\right)^{-1/2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p^{t}\left(1-\frac{2GM}{R}\right)^{1/2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -E\ \ \ \ \ (10)$ $\displaystyle p^{t}$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{R}\right)^{-1/2} \ \ \ \ \ (11)$

For a photon, ${\mathbf{p}\cdot\mathbf{p}=0}$, and for a photon moving in the radial direction ${p^{\theta}=p^{\phi}=0}$ so

 $\displaystyle \mathbf{p}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle g_{ij}p^{i}p^{j}\ \ \ \ \ (12)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{R}\right)E^{2}\left(1-\frac{2GM}{R}\right)^{-1}+\left(1-\frac{2GM}{R}\right)^{-1}\left(p^{r}\right)^{2}\ \ \ \ \ (13)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -E^{2}+\left(1-\frac{2GM}{R}\right)^{-1}\left(p^{r}\right)^{2}\ \ \ \ \ (14)$ $\displaystyle p^{r}$ $\displaystyle =$ $\displaystyle E\sqrt{1-\frac{2GM}{R}} \ \ \ \ \ (15)$

Thus the photon’s four-momentum in the Schwarzschild basis is

$\displaystyle \mathbf{p}=\left[E\left(1-\frac{2GM}{R}\right)^{-1/2},E\sqrt{1-\frac{2GM}{R}},0,0\right] \ \ \ \ \ (16)$

## 7 thoughts on “Schwarzschild metric: four-momentum of a photon”

1. Liron

If the observer was at the Schwarzschild radius then they would be right on the event horizon and 1-2GM/R = 0 so pt of the photon would have a divide by zero error. What about if they were at another radius r ?