3
$\begingroup$

I recently read Gibbons and Hawking's paper Action integrals and partition functions in quantum gravity, Phys. Rev. D 15 (1977) 2752. I am interested in repeating their calculations.

It is fairly simple to calculate the Euclidean action for the Schwartzchild spacetime. When I tried to do it for the Reissner-Nordstrom spacetime, I came across a problem. As suggested in the paper, when calculating the Euclidean action for the electromagnetic field, we can use the following relation:

$-\frac{1}{16\pi}\int F_{ab}F^{ab}\sqrt{-g}d^4x=-\frac{1}{8\pi}\int F^{ab}A_ad\Sigma_b$

by Stoke's theorem. Then, they chose a so-called regular gauge, such that the 4 potential takes the following form,

$A_a=(Q/r-Q/r_+)\nabla_at$

where $r_+$ is the radius of the outer horizon.

Now, I have three questions:

  1. I do not perticularly understand what "regular gauge" means. It sounds to me, as long as the potential vanishes at the horizon, the potential is in "regular gauge".

  2. As we know, the action is independent of gauge. Why did they choose a gauge? In fact, after my calculation, in order to get their result, you can do the above integral on the right hand side at the outer horizon, and at the same time, do another integral $\frac{1}{8\pi}\int [K]d\Sigma$ also at the outer horizon, and add them together. But this does not sound reasonable. In a spacetime with a horizon, the boundary consists of the one at infinity and the other at the horizon. We should do both integrals at the outer horizon and the infinity, but this will not give you the correct answer.

  3. You can also carry out the calculation in a different way. You first evaluate $\frac{1}{8\pi}\int [K]d\Sigma$ at the infinity, which turns out to be one part of Gibbons and Hawking's result. Then, you do electromagnetic integral, without using their "regular gauge" potential, but $Q/r\nabla_a t$, at both the outer horizon and the infinity, you get the other part of their result. Adding them together, you get the correct answer. There is one unsatisfactory thing about this method: you only integral $\frac{1}{8\pi}\int [K]d\Sigma$ at the infinity...

Which one is the correct one? Or neither of them (method 2 and 3) is correct?

Improve this question
$\endgroup$
2
  • $\begingroup$ Did you find the answer to this question? $\endgroup$
    – apt45
    Commented Jan 20, 2021 at 4:03
  • $\begingroup$ I was reading the same paper and I had your same question 2. Why do they choose a gauge if the action is gauge invariant? Also, I don't understand why they integrate only for $r\geq 2m$ and not for smaller values of $r$. Here's my question physics.stackexchange.com/questions/608911/… $\endgroup$
    – apt45
    Commented Jan 20, 2021 at 4:23

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.