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The Hawking temperature of a Schwarzschild black hole is given in SI units as $$T_{H}=\frac{\hbar c^3}{8 \pi G k_{B} M},$$ where $k_{B}$ is the Boltzmann constant. I would like to know how $\hbar$ and $k_{B}$ show up in the temperature. I mean where in the original derivation by Hawking do these constants show up?

I have looked into the original paper by Hawking, "Particle Creation by Black Holes". There he begins the calculation by writing down the massless scalar wave equation in curved background $$ \frac{1}{\sqrt{-g}}\partial_{\mu} \left(\sqrt{-g}\,g^{\mu\nu} \partial_{\nu} \phi \right)=0.$$

Now as far as I can understand Hawking temperature shows up in the exponent when the modes $\sim\phi$ are traced from the surface of collapsing body to the future infinity. Alternative calculations without invoking the collapse geometry suggest modes tunnel through the horizon. So at first I thought $\hbar$ naturally shows because from quantum mechanics we have $\phi \sim e^{\frac{i}{\hbar}S}$. But what is bothering me here is that above wave equation does not have any $\hbar$ in it. In fact such wave equation comes from a Lagrangian of the form $$ I\left[\phi\right] = \int \left[\frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi \right]\sqrt{-g} d^4x,$$ where $\hbar$ does not show up. And I am not sure whether such Lagrangian should come with a $\hbar$ based on dimensional analysis.

My second source of confusion is related to the Boltzmann constant. Again, I have no idea how and where $k_{B}$ emerges in the derivation. Without the notion of temperature, $k_{B}$ seems unrelated to such calculations which involve wave equations, Bogoliubov transformations etc...

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  • $\begingroup$ Are you familiar with how $k_B$ normally shows up in statistical mechanics, starting from the general definition of temperature? And are you familiar with how $\hbar$ normally shows up in the commutation relations for the field operators? $\endgroup$ May 25, 2020 at 13:28
  • $\begingroup$ As far as I understand both show up on dimensional grounds. $k_{B}$ shows up when writing the thermal average of say molecules : $< m v > = k_{B} T$. And $\hbar$ appears in the geometric quantization, when normalizing the phase space volume: $ 1/\hbar \oint p dx $. What I don't understand in the context of Hawking temperature is whether $\hbar$ and $k_{B}$ (and specifically $\hbar$) were there to begin with or they were put by hand to mimick the usual thermal ensemble coming from thermodynamics. $\endgroup$
    – user91411
    May 25, 2020 at 13:41
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    $\begingroup$ $\hbar$ was there to begin with. I don't remember if Hawking used units with $\hbar=k_B=1$, but he did use the fact that the field operators don't commute with each other (that was a crucial ingredient), and $\hbar$ is in that commutation relation (if we don't hide it by using units $\hbar=1$). Regarding $k_B$: Hawking's calculation showed that the state really is a thermal state, so it really does have a true temperature in the usual stat-mech sense. Then $k_B$ shows up as usual, as a "units conversion factor" to express temperature in Kelvin instead of in units of energy. $\endgroup$ May 25, 2020 at 18:51

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You are correct that the action he used does not contain $\hbar$ in it. After all, it is a classical action which describes simply a wave equation in curved spacetime. There is no problem in considering these sorts of phenomena classically and, in principle, there is no reason for $\hbar$ appear anywhere in computations involving the wave equation.

Nevertheless, Hawking uses the wave equation as the classical equation of motion for a quantum field, which means he does not treat the field as a real function, but rather as an operator. This operator must satisfy commutation relations which do involve $\hbar$, even though (as far as I remember) Hawking uses units with $\hbar = 1$. The fact that the calculation uses operators everywhere (for example, in the very notion of a Bogoliubov transformation) already necessitates $\hbar$ to be present. Of course, Hawking's units hide this fact.

As for the Bolztmann context, you said

Without the notion of temperature, $k_B$ seems unrelated to such calculations [...]

This statement is correct and actually applies to every expression in Physics that involves $k_B$, not only the Hawking temperature. The sole physical meaning of $k_B$ is to serve as a conversion factor between energy scales and more conventional temperature scales. The reason $k_B$ appears in the expression—actually, in any expression—is just so you can read the temperature in Kelvin. Temperature could just as well be expressed, e.g., in Joules or in your favorite unit of energy. The Wikipedia page for the Boltzmann constant seems to mention a bit about this. Notice also that in every physical equation written in the SI in which temperature appears it is accompanied by the Boltzmann constant (sometimes, hidden in another constant, such as the molar gas constant, which is just the Boltzmann constant times Avogadro's number).

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