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

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  7. 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 ?

    Reply

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