Imagine an observer is very far from the black hole so that their local space-time is Minkowskian. In this case, we can apply special relativity to the derivation of the entropy of the black hole.
We begin with the equation for the entropy of a black hole as derived out by Hawking:
$$
S=\frac{c^3 k}{4 \hbar G}A
$$
Where $A$ is the surface area of the black hole.
The relationship between energy $E$, temperature $T$ and the entropy $S$ is:
$$
\partial E=T \partial S
$$
Using Einstein's famous relation, we obtain:
$$
\partial E = \gamma c^2 \partial M
$$
Thus,
$$
T=\gamma c^2 \left(\frac{\partial S}{\partial M}\right) ^{-1} = \frac{4 \gamma \hbar G}{c k} \left(\frac{\partial A}{\partial M} \right)
$$
Now with the temperature and Area of the black hole, the luminosity is given as:
$$
L=A\sigma T^4= \frac{32 \pi }{15 c^6} \gamma ^4 h G^4 A \left(\frac{\partial A}{\partial M}\right)^{-4}
$$
and since $\frac{- \partial E}{\partial t} = L$, we obtain:
$$
\frac{-\partial M}{\partial t} = \frac{32 \pi }{15 c^8} \gamma ^3 h G^4 A \left(\frac{\partial A}{\partial M}\right)^{-4}
$$
and finally,
$$
-\partial t = \frac{15 c^8} {32 \pi \gamma ^3 h G^4 A }\left(\frac{\partial A}{\partial M}\right)^{4} \partial M
$$
There is a problem here however. For two observers, Alice who is boosted and Bob who is at rest, unless the black hole's surface area transforms by a factor of $\gamma^{2/3}$, which seems a very odd transformation factor, even after taking time dilation into account, they will disagree on how long the black hole took to evaporate.
Does this derivation make sense? I have been trying many different possible surface area transformations on the black hole, and under this premise, they all seem to imply that either $G$ or $h$ is not Lorentz invariant (or both).
———–A little more lost in the weeds, not related———–
For instance, Naïvely using the static Schwarzschild radius (even though I know it is derived without a preference to direction and a few other simplifications that makes it not a good candidate here) implies $G^2/h \propto \gamma^2$, and assuming relativistic energy has an effect on the surface area implies $G^2/h \propto \frac{1}{\gamma^4}$ or more probably $G \propto \gamma^{-2}$… which I assume is false, but does happen to mean that a photon can never be doppler shifted into appearing as a black hole, and same for any other mass that was not already a black hole in a rest frame, which is quite nice.
Anyways. I am a bit confused by this and would love some verification on this derivation being completely false or not.
Thank you!
Best Answer
The answer to the title question is negative: Hawking radiation or more precisely Unruh vacuum, a state of quantum field that could be used to describe Hawking radiation, is not Lorentz invariant.
OP's calcultations took the wrong turn almost from the start, since the first law of thermodynamics written for a moving body must necessarily include additional term with the differential of black hole's linear momentum $\mathcal{P}$, treated as a thermodynamic variable: $$dE=TdS+\eta d \mathcal{P},$$ where $\eta$ is a variable thermodynamically conjugate to linear momentum, the velocity. Consequently, results of such calculations if done properly should just reproduce usual expressions of relativistic mechanics rather than being nontrivial statements about Hawking radiation.
Note, that there are no general and unambiguous rules about the behavior of thermodynamic variables under Lorentz transformations, though there has been several proposals over the years. For a relatively recent review of the problem aimed at undergraduate-level students of thermodynamics (without references to GR or black hole physics) see:
For a discussion of thermodynamics for a black hole moving with constant velocity see e.g. this paper: