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I want to calculate the Hawking temperature of the metric

$$\mathrm ds^{2}=-\left(1-\frac{r^{2}}{l^{2}}\right)\mathrm dt^{2}\ +\frac{\mathrm dr^{2}}{\left(1-\frac{r^{2}}{l^{2}}\right)}+\ r^{2}\mathrm dx^{2}.$$

Since the metric is diagonal, I guess the Hawking formula for the temperature of this metric isn't complicated. Do you know how one can calculate the temperature?

Thank you!

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    $\begingroup$ Whether it is hard or easy to calculate this depends on where you are starting from. There is a 1-line argument where you Wick rotate to the Euclidean and then note that there's a periodicity in the time coordinate that gives you the temperature. But this argument is only meaningful if you are familiar with thermal instantons. $\endgroup$
    – Andrew
    Nov 9, 2021 at 5:42

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The Hawking effect is an effect associated to black hole spacetimes. Since every stationary black hole solution to the Einstein–Maxwell equations is essentially a rotating, charged black hole due to the no-hair theorems and the metric you provided is not that of a rotating, charged black hole, it does not characterize a black hole within the usual notion of General Relativity. Hence, I'll assume your question refers to computing the temperature associated with the Unruh effect, which is also an effect concerning temperature due to quantum effects in non-inertial frames of reference. This temperature is often referred to as the Hawking temperature. Furthermore, the main difference between the effects for the purposes of this question is that in the Hawking effect the "particles" are seem as coming from the black hole region, while in the Unruh effect they come from everywhere. When one is close to the black hole, the Hawking effect can be approximated by the Unruh effect.

The Unruh effect occurs in spacetimes possessing what is known as a bifurcate Killing horizon. In short terms, a Killing horizon is a null surface orthogonal to a Killing field (such as the event horizon of the Schwarzschild black hole), and a bifurcate Killing horizon is a pair of such surfaces crossing one over the other (such as the future and past event horizons of the Schwarzschild black hole). In spacetimes with such a structure, observers following the orbits of the Killing field which generates the bifurcate Killing horizon perceive the vacuum (namely, the unique Killing-invariant, nonsingular state of the quantum field) as being a thermal state with temperature given up to a Doppler shift factor by the Hawking temperature $$T = \frac{\kappa}{2\pi},$$ where $\kappa$ is the so-called surface gravity of the Killing horizon. Hence, all you need to do to compute the Hawking temperature is to compute the surface gravity.

This subject is covered in many textbooks in relativity. The details of the Unruh effect which I mentioned can be found on Chap. 5 of Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics and the equation I wrote above corresponds to Wald's Eq. (5.3.2). More on the surface gravity can be found on Wald's book on QFTCS and BH Thermodynamics or, for further examples, on Poisson's A Relativist's Toolkit or on Chap. 12 of Wald's General Relativity.

Assuming this spacetime does have a bifurcate Killing horizon (I did not attempt to prove or disprove it), I'll sketch how to perform the computation of the surface gravity through Wald's Eq. (12.5.18) (GR book), but I won't write all the details since that is also against the site's policy. The relevant expressions are $$V \equiv \sqrt{- \chi^a \chi_a},$$ $$a^c = \frac{\chi^b \nabla_b \chi^c}{- \chi^a \chi_a}, \tag{12.5.17}$$ $$\kappa = \lim_{r \to l} (a V), \tag{12.5.18}$$ where I've specified (12.5.18) to our particular situation and one defines $a = \sqrt{a^c a_c}$. The static Killing field is $\chi^a = \left(\frac{\partial}{\partial t}\right)^a$, and hence the metric provided leads to $$V = \sqrt{1 - \frac{r^2}{l^2}}.$$

An easy way to compute $a^c$ is noticing that Killing's equation allows us to write $$a^c = \frac{\nabla^c V^2}{2 V^2},$$ from which it follows that $$a_c = - \frac{r}{V^2 l^2} (\textrm{d}r)_c,$$ and hence one can compute $$a_c a^c = \frac{r^2}{V^2 l^4}.$$

Bringing everything together, we find $$\kappa = \lim_{r \to l} \frac{r}{l^2} = \frac{1}{l}.$$

This agreed, up to a sign, with Poisson's Eq. (5.39). I believe the discrepancy is because Poisson's formula assumes the Killing field to be timelike at infinity, while for this particular metric it is spacelike at infinity.

For completeness, in this notation, the temperature measured by any local observer will be (Wald's QFTCS Eq. (5.3.3)) $$T = \frac{\kappa}{2\pi V}.$$

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  • $\begingroup$ Thank you! I found this paper: aip.scitation.org/doi/abs/10.1063/1.4943704 is it useful? $\endgroup$
    – Star21
    Nov 8, 2021 at 21:04
  • $\begingroup$ @Star21 You're welcome! There is more than one way of arriving at the same result, and I believe their treatment is different from the one I proposed, though likely will lead to the same results. I do have a feeling their approach might be more involved for your particular case: since your metric is static (which is not true for the Vaidya spacetime), I believe Killing horizon techniques might be an easier approach. Nevertheless, of course, both approaches should lead to the same answer (as a disclaimer, I did not read the paper you mentioned in detail, only its abstract) $\endgroup$ Nov 8, 2021 at 21:09
  • $\begingroup$ I see those books, but I didn't understand how can find k from the metric. $\endgroup$
    – Star21
    Nov 8, 2021 at 22:25
  • $\begingroup$ @Star21 My favorite method is usually to use Wald's GR book Eq. (12.5.18), on p. 332. The Killing field for your metric is $\chi^a = \left(\frac{\partial}{\partial t}\right)^a$, since it is a static metric. You then compute the redshift factor $V = \sqrt{- \chi^a \chi_a}$ using your metric and the acceleration $a$ using Eq. (12.5.17). Then you just have to plug in the values on Eq. (12.5.18). Also, Poisson's Eq. (5.39) is an explicit expression that only requires you to plug in the metric you have. $\endgroup$ Nov 8, 2021 at 22:42
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    $\begingroup$ -1. The first paragraph of your answer is wrong. The no-hair theorems apply to solutions of EFE's. A “given metric” does not have to be such solution, so making inferences based on this theorem is a logical fallacy. $\endgroup$
    – A.V.S.
    Nov 9, 2021 at 9:50

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