Black hole radiation: energy at infinity of radiated particle

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)

energy per unit mass of a particle

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.

3 thoughts on “Black hole radiation: energy at infinity of radiated particle

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