Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Box 16.2.

In radiation from a black hole, we found an estimate for the energy of a radiated particle as measured in the frame of an observer who is momentarily at rest at the point at which the particle pair is created.

$\displaystyle E=\frac{\hbar}{2\sqrt{2GM\epsilon}} \ \ \ \ \ (1)$

To work out the energy at infinity in the Schwarzschild (S) frame, we can use the particle’s four-momentum in the form

$\displaystyle E=-\mathbf{o}_{t}\cdot\mathbf{p} \ \ \ \ \ (2)$

where ${\mathbf{o}_{t}}$ is the time basis vector and ${\mathbf{p}}$ is the four-momentum of the particle in the observer’s frame. In this frame, ${\mathbf{o}_{t}=\left[1,0,0,0\right]}$ and ${E}$ is the time component of ${\mathbf{p}}$ so 2 follows.

If we work out this equation in S coordinates, we have

$\displaystyle \mathbf{o}_{t}=\left[\left(1-\frac{2GM}{r}\right)^{-1/2},0,0,0\right] \ \ \ \ \ (3)$

so using the S metric, we have

 $\displaystyle E$ $\displaystyle =$ $\displaystyle -g_{ij}o_{t}^{i}p^{j}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\left(1-\frac{2GM}{r}\right)^{-1/2}p^{t}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{1-\frac{2GM}{r}}m\frac{dt}{d\tau} \ \ \ \ \ (6)$

Since the energy per unit mass of a particle is given by

$\displaystyle e=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau} \ \ \ \ \ (7)$

we get

$\displaystyle E=\frac{me}{\sqrt{1-2GM/r}} \ \ \ \ \ (8)$

Since ${e}$ is the energy per unit mass at infinity, ${me=E_{\infty}}$ is the total energy of the particle at infinity, so

$\displaystyle E_{\infty}=\sqrt{1-\frac{2GM}{r}}E \ \ \ \ \ (9)$

For a particle that is created at ${r=2GM+\epsilon}$ with energy given by 1, the energy at infinity is

$\displaystyle E_{\infty}=\sqrt{1-\frac{2GM}{2GM+\epsilon}}\frac{\hbar}{2\sqrt{2GM\epsilon}} \ \ \ \ \ (10)$

This can be expanded in a series around ${\epsilon=0}$ to get

$\displaystyle E_{\infty}=\frac{\hbar}{4GM}-\frac{\hbar}{\left(4GM\right)^{2}}\epsilon+\frac{3}{2}\frac{\hbar}{\left(4GM\right)^{3}}\epsilon^{2}+\ldots \ \ \ \ \ (11)$

Thus for very small distances from the event horizon, the energy the particle has at infinity tends to a constant.