Flat space limit of the Schwarzschild metric and Hawking temperature

Question

The Schwarzschild metric reduces to the Minkowski metric in the limit of vanishing $M$, but the Hawking temperature which is proportional to $1/M$ diverges in the same limit. This would imply that flat spacetime has infinite rather than zero temperature. What am I missing?

EDIT: This question is back up on the front page because I've posted an answer of my own - let me know if I'm thinking about this right.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

Answer 1

There's a topology change in the spacetime when you complete the $M\rightarrow 0$ limit--The Minkowski spacetime is non-singular, while the Schwarzschild spacetime contains a spacetime singularity for any (positive) value of $M$.

You could also look at it in terms of the total energy available for a small mass BH is going to be miniscule, and you would expect the evaporation time to be rapid (especially since you're going to be triggering more decay modes at high temperature), with very hot evaporation remnants likely to scatter a large distance quickly, so its effect on the spacetime will be to produce a gas of high-temperature particles that quickly diffuse to an essentially zero temperature gas.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

Comments

Is the topology change relevant? Because presumably, in the complete quantum gravity theory of a black hole, there will be no distinction between the black hole scenario and the post-evaporation scenario in terms of singularities, since singularities don't exist in the complete theory. The second explanation makes sense to me though. +1 for that :)

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

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@dbrane: But, there is no complete theory, so we're stuck with semiclassical results. And in a complete theory, you would still expect a qualitative difference between a state with a classical black hole limit and one that is without one, and that this would mimic the topological difference you see classically.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

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When I said "in the complete quantum gravity theory", I meant the complete QG theory that exists out there Platonically but which we don't yet know :) It's clear to me that a topology change would make a huge difference but not so much for something that simply mimics it. In a complete QG theory, what should still survive is the fact there is a horizon in the black hole case and none in the M→0 limit.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

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@dbrane: depends on how you feel about cosmic censorship, for one thing. (though I actually have no idea how the Hawking calculation works out for something like a $a>M$ spacetime). And a spacetime with an evaporating black hole wont' have an event horizon anyway.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

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"a spacetime with an evaporating black hole wont' have an event horizon" Why not? How can you even have evaporation with out an event horizon?

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

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@dbrane: The evaporation happens due to an apparent horizon. If the black hole goes away at late times, then the final state is a horizonless, singularity-free spacetime--any geodesic inside the horizon will eventually be able to exit it (at least in the case of a timelike singularity, which is the case for a hole with even an infinitesimal charge or angular momentum). Another way of phrasing this is that a black hole with a decreasing radius forms a timelike surface in the whole spacetime, and there is a formal result that shows that timelike surfaces cannot be trapping horizons.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

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Wow, I should have realized that. I hadn't heard of the distinction between an apparent horizon and an event horizon before (one of the relativity courses I took defined the apparent horizon as the radius of the smallest stable photon orbit - bah, misleading). I read your answer at physics.stackexchange.com/questions/970/… and that really helped a lot. Thanks!

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

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@dbrane: that's not really a bad definition of an apparent horizon, the formal one basically just formalizes this idea--the definition of the event horizon,however, is the one that's a bit trickier--it depends on the end state of the spacetime, and you take all of the light rays that hit infinity, and then you take their topological boundary.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

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Hang on, the innermost-stable-photon-orbit (ISPO) definition can't possibly be the same as the way you defined it in the link, which I'm going to paraphrase as "the surface that traps you if the BH were to stop evaporating once you crossed this surface". Because assuming no evaporation, you could have a photon that crosses the ISPO and still escapes to infinity whereas you couldn't have a photon that can cross the surface as defined in your link and then escapes to infinity (assuming no evaporation). Right?

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

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@dbrane: sorry, I was getting a little cavalier about what I meant by orbit. I meant the photon path that tries to escape the black hole by pointing radially outward, but neither falls into the black hole nor escapes. Formally, an apparent horizon is a closed two-surface with two perpendicular null vectors $k$ and $\ell$ that satisfy $k^{a}\ell_{a}=-1$ and $\left(g^{ab}+k^{a}\ell^{b}+\ell^{a}k^{b}\right)\nabla_{a}\ell_{b}=0$. What this means is that $\ell_{a}$ has no component of its gadient lying on the AH, and therefore, at some infinitesimally later time, will still be at the 'same place'

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

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You are right to say that it is most assuredly not the $r=3M$ surface of photon orbit in the Schwarzschild solution.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Jerry Schirmer

Answer 2

The lower the black hole mass, the higher the temperature of the Hawking radiation. This prediction is based on approximations that seem warranted as long as the black hole has mass $M >> M_\text{Planck}$. No-one knows what precise phenomena occur when the black hole mass approaches $M_\text{Planck}$. Drawing conclusions on the $M \to 0$ limit based on GR is certainly not justified.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Johannes

Comments

Thanks dmckee. Gives me the opportunity to learn how to add formulas. (I cannot edit texts of others, so no clue how equations are implemented.)

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Johannes

Answer 3

I'm not particularly satisfied with the answers that cite topology change or breakdown of semi-classicality. Particularly the latter doesn't explain why a metric which appears more and more flat seems to produce hotter and hotter Hawking radiation well before the Planckian regime is approached.

But I've thought about this for a while and finally have found a way to be at peace with this behaviour (save for the incompatibility of asymptotic flatness and a temperature at infinity that QGR pointed out, which I've posed as a separate question).

Here's how I'm thinking about it - do leave a comment if you disagree:

Close to the horizon, a stationary observer sees the Rindler metric and the Unruh temperature here is equal to the Hawking temperature - the observer has to fire their engines with a proper acceleration equal to the appropriate proper acceleration in the Unruh case. This also means that a free-falling observer sees no radiation at all (this is true at least close to the horizon, which I think is enough).

So, as I send the limit $ M \rightarrow 0 $, the appropriate observer that I'm mapping on to as I approach flat spacetime is not the inertial observer in flat spacetime, but the eternally accelerating observer with proper acceleration $ 1/M $. (Unruh proper acceleration is identified with the surface gravity)

So of course I should expect to see a thermal state of infinite temperature in this limit to flat spacetime since I'm not being mapped to an inertial observer in Minkowski, but to an eternally accelerating Rindler observer with infinite acceleration.

If I were always considering geodesic (free-falling/inertial) observers, at least close to the horizon, this limit makes perfect sense because there is no radiation (thermal or otherwise) during and after taking the limit.

PS: Note that after asking the question, I confused this limit taking process with black hole evaporation - which is not really what I was asking - I just wanted to know what the temperature of the spacetime looks for flatter and flatter spacetimes - and not the complicated stuff that happens when a black hole actually evaporates. Or in other words, I wanted to know what the $M=0$ element in the set of eternal black holes looks like, and what it looks like is the spacetime seen by a Rindler observer with infinite acceleration in flat spacetime.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

Answer 4

Jerry’s answer is pretty good, and I would say that topology, or $H^2({\cal M})$ is given by the evaluation of a curvature two-form $\Omega_{ab}~=~R_{abcd}\omega^c\wedge\omega^d$ on a two surface, and where $ R_{abcd}$ is the Riemann curvature tensor. The result is a “charge,” which is a measure of the topology. A very small black hole has a huge curvature near the horizon, which is different from the Minkowski spacetime of zero curvature.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Lawrence B. Crowell

Answer 5

This problem is analogous to the "ultraviolet catastrophe" in the pre-quantum understanding of black-body radiation. There, because one did not take into account the discrete character of radiation, the total energy emitted by a black-body was divergent. Planck came along and remedied the situation by noting that photons were emitted in discrete packets rather than the previous assumption that all frequencies occurred in the radiation.

Similarly, as @Johannes noted in his answer, the calculations which lead to the prediction that a black hole radiates at a temperature $T \propto 1/M$ are valid only for macroscopic black holes. As and when, we understand how the discrete character of spacetime modifies this dependence, will we be able to answer this question. The simplest thing to do is to guess that such corrections will modify the temperature-mass dependence to be of the form:

$$ T \propto \frac{1}{M + M_O} $$

where $M_O$ is the minimal size a black-hole can attain, leading to a finite maximum temperature that a system containing a decaying black hole can attain $T_{max} \sim 1/M_O$. The conjecture would be that a black-hole decays until we are left with a flat background populated by a gas of these black-hole quanta of mass $M_O$ and temperature $T_{max}$.

Of course, this is all very hand-wavy.

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Edit: A generalization of the above approach is to postulate that the exact form of the dependence of $T_{max}$ on $M$ is given by some function $f(M/M_O)$ which tends to $ 1/M $ in the limit $M/M_O \rightarrow \infty$.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user user346

Answer 6

Classically the story is as follows. Flat space has no horizon and thus no Hawking temperature. A tiny black hole is very different from flat space, no matter how small its mass, since it does have a horizon. Thus, the limit $M\to 0$ differs essentially from the limiting point $M=0$, and you cannot conclude anything about the latter by studying the former.

Another way to see that the conclusion that flat space has infinite Hawking temperature obviously is incorrect is to consider different limits of black hole space-times to Minkowski space. There are black holes with positive specific heat (take e.g. BTZ in 3 dimensions or the exact string black hole or Schwarzschild-AdS above the Hawking-Page transition), and if you take any of these black holes and take appropriate limits of mass (and possibly other parameters) to zero then also Hawking temperature drops to zero.

Perhaps the issue that confuses you is the negative specific heat of semi-classical Schwarzschild black holes, which is responsible for the non-smoothness of the limit $M\to 0$. One may argue on general grounds (or with some simple models) that quantum effects should convert the negative specific heat into a positive one for tiny black holes. Then the limit to vanishing mass would again be smooth.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Daniel Grumiller

Answer 7

For asymptotically flat spaces, a black hole can never be in thermal equilibrium with the rest of spacetime. If it were, the rest of spacetime would be filled with radiation at the Hawking temperature, and the resulting nonzero stress-energy tensor everywhere is inconsistent with the assumption of asymptotic flatness.

So, assuming spacetime sufficiently far away from the black hole is vacuumlike (i.e. no cosmic microwave background radiation), we have a temperature gradient ranging from the Hawking temperature right outside the event horizon to zero very far away.

But once you accept a temperature gradient, the temperature of the black hole no longer tells us anything about the temperature of spacetime far away from it. The Hawking temperature may be around the Planck temperature, but far away, the temperature is still approaches zero.

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AdS black holes are an entirely different matter. I know this question isn't about AdS black holes, but they're instructive as we can have thermal equilibrium everywhere for AdS black holes. This is because we have a warp factor for clock rates which rescales the temperature as measured by a local observer compared to the temperature as measured by a distant observer. So, even at thermal equilibrium, the local temperature falls off exponentially beyond the AdS radius as we move away from the black hole. Any backreaction will remain finite because of this warp factor.

There are two cases to consider here:

• the black hole radius is smaller than the AdS radius
• the black hole radius is larger than the AdS radius

For the latter case, as the size of the black hole goes up, so does the temperature, and there's no upper limit to the temperature. This means the black hole has a positive heat capacity, and remains in thermal equilibrium with its environment.

For the former case, the temperature goes down as the size of the black hole goes up, leading to a negative heat capacity. So, even if we start off with an "equilibrium" state, it is unstable. If the black hole expands, it becomes cooler than its environment, and so, it will absorb a net amount of radiation and expand even further beyond the AdS radius after which it will start to cool off, eventually settling down to the thermal state in the latter case.

With a U-shaped temperature relation as a function of black hole size, we can see it's not possible to have black hole temperatures below the AdS scale, $k$. Instead, we have a phase transition to a thermal state with no black holes.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user QGR

Comments

I take your point about the incompatibility with asymptotic flatness, but I don't see how the temperature can be obtained to be zero at infinity. See en.wikipedia.org/wiki/Hawking_radiation#Emission_process and also a new question that I've posted.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user dbrane

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This answer is correct--- the Hawking vacuum is unphysical, it describes a black hole in equilibrium with a fixed temperature field theory with no back-reaction. It cannot exist stably by itself in flat space-time if you include backreaction.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user Ron Maimon

Answer 8

Assuming you're not interested in thermal equilibrium everywhere, then as the black hole mass approaches the Planck mass, the thermodynamic limit for black hole microstates breaks down, and we can no longer speak of its temperature meaningfully.

This post imported from StackExchange Physics at 2014-03-24 03:51 (UCT), posted by SE-user QGR