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Suppose you were held by a very strong rope at constant Schwarzchild coordinate $r = 2M (1 + \epsilon)$ just above the event horizon of a Schwarzchild black hole of mass $M$. You would feel a proper acceleration of magnitude $$a_\text{gravity}(r) = \frac{M}{r^2 \sqrt{1-\frac{2M}{r}}} \sim \frac{1}{4M \sqrt{\epsilon}}$$ from the rope (in units where $G = c = \hbar = 1$). According to eqs. (1.3) and (3.1) of this paper, you would also observe Hawking radiation with effective temperature $$ T(r) = \frac{T_H}{\sqrt{1 - \frac{2 M}{r}}} \sim \frac{T_H}{\sqrt{\epsilon}}$$ coming out of the black hole, where $T_H := 1/(8 \pi M)$ is the Hawking temperature. The temperature and the acceleration would both diverge as you approach the horizon, but at a constant ratio $$ \frac{a_\text{gravity}(r)}{T(r)} = \frac{8 \pi M^2}{r^2} \sim 2 \pi.$$

By the Stefan-Boltzmann law, you would observe a total emitted power per unit area (or equivalently radiation pressure, in units where $c=1$) of $$ P = \frac{\pi^2}{60} T^4 = \frac{1}{245760 \pi^2 M^4 \epsilon^2}.$$

The total surface area of a black hole is a subtle concept, due to the spacetime curvature, so let's restrict ourselves to considering a small region just outside the horizon (i.e. a region with diameter much less than the Schwarzchild radius, which sets the curvature scale), which we can locally approximate by Minkowski spacetime.

Suppose you were to then unfurl a solar sail with mass surface density $\sigma$ (including the contribution from your own mass). The radiation pressure would accelerate you and the sail at $$ a_\text{Hawking} = \frac{P}{\sigma} = \frac{1}{245760 \pi^2 \sigma M^4 \epsilon^2}$$ away from the hole.

We see that $a_\text{gravity}$ diverges much more slowly than $a_\text{Hawking}$ at small $\epsilon$. Indeed, if $$\epsilon < \frac{1}{256 M^2 (15 \pi^2 \sigma)^{2/3}},$$ then the acceleration from the Hawking radiation wins, and would seem to blow you away from the hole! As a sanity check, as $M$ grows larger (cooler black hole) or $\sigma$ grows larger (denser and less efficient sail), the Hawking radiation becomes less effective at pushing you away.

Obviously this would be a ludicrous setup for a real black hole, but in principle, would it be possible to use such a solar sail to ride the Hawking radiation out and escape the hole? (Note that this idea is closely related to that of a black hole starship.) And if you did, how would a nearby free-falling observer passing you into the black hole describe the process? After all, according to the linked paper, to her the black hole would only be radiating at a quite gentle temperature of $2 T_H$ and would only provide a bounded radiation pressure.

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  • $\begingroup$ Some radiation's direction is so vertical that it never falls back to the black hole, this is the reason why the two sides of the solar sail do not receive exactly the same amount of radiation. Far from horizon almost any radiation does not fall back. $\endgroup$ – stuffu Feb 14 '17 at 14:13
  • $\begingroup$ Maybe there aren't any answers yet because it sounds like you've answered your own question. Asking "are there any holes in this argument?" might not get answers quickly. $\endgroup$ – DanielSank Feb 15 '17 at 9:08
  • $\begingroup$ @DanielSank My question lies in the last two sentences of my post - if the infalling observer sees negligible Hawking radiation, then from her perspective, what allows the (apparently unpowered) solar-sail user to escape the black hole's enormous gravity? $\endgroup$ – tparker Feb 15 '17 at 9:53
  • $\begingroup$ I think you have incorrectly calculated the power - you need to multiply by the area of the BH $\endgroup$ – Akoben Feb 18 '17 at 17:04
  • $\begingroup$ @Akoben I never calculated the power - the fourth equation gives the power per unit area of the emitting body. $\endgroup$ – tparker Feb 18 '17 at 20:26
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Hopefully I have understood your situation correctly, if not then please let me know and i'll delete this answer.

For a radially in-falling observer to hover at $r = 2M + \epsilon$, they would need to provide an opposing acceleration of $a \sim \frac{1}{4M\sqrt{\epsilon}}$, as you stated.

Any amount more than this critical amount will cause the observer to move radially outwards, away from the black hole. If the observer had a rocket strapped to her back, then she would indeed be propelled outwards by the additional acceleration provided by the hawking radiation, and any infalling observer would see nothing special: just a radially boosted observer sailing past.

The rope in your problem makes this tricky however. If the rope is tied to some fixed point far from the black hole, then you can not ride the hawking radiation from the black hole. This is because the moment you move radially outwards from your position, the rope will stop providing any force, as it presumably goes slack. Thus, gravity will immediately grab a hold of you again and drag you back to $r = 2M + \epsilon$.

If your rope is tied to an accelerating rocket or you happen to have luckily grabbed on to the tentacle of a giant galactic space squid desperately trying to save you, then you will again be in the situation where it is as if you have a rocket attached to your back and an in-falling observer will again see nothing weird.

With regards to whether or not you can escape a black hole eventually powered only by Hawking radiation, you might want to take a look at the calculation that has been done in this paper (see also this paper).

Conceptually, a hovering observer will measure hawking radiation and this will give an acceleration to the observer, potentially allowing them to escape the black hole. However, with regards to turning off the acceleration that keeps you static, you would need to wait until the amount of hawking radiation was large enough to sustain your motion without the tension of the rope.

This may in principle be possible, but you would have to wait for an incredibly long time, since black holes evaporate slowly and give out very little hawking radiation until they have very small mass. At this point, quantum gravity becomes important so who knows.

Another point is about the sail. By calculating the acceleration at $r = 2M(1+\epsilon)$, you are only getting information about what the acceleration would be there at some fixed $\sigma$, which would die out as you got further away. In order to calculate this more effectively. To maintain the same acceleration, you would need to increase the area of your sail as you moved out.

The power is given by $$ P \propto A_{BH}T(r)^4 \propto \frac{1}{M^2(1-\frac{2M}{r})^2} $$ While the acceleration by $$ a_{hawking} = \frac{P}{\sigma} = \frac{A_{sail}}{mM^2(1-\frac{2M}{r})^2} $$ Where $\sigma = m/A_{sail}$.

This needs to be bigger than the gravitational acceleration felt by the observer, meaning our area needs to be: $$ A_{sail} > \frac{mM^3(1-\frac{2M}{r})^{3/2}}{r^2} $$

For $r = 2M(1+\epsilon)$, this means that $A_{sail} > mM\epsilon^{3/2}$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Apr 7 '17 at 16:54
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Suppose you are at being held by an external force at a fixed radius $R$ with $$ \frac{R}{2M} -1 = \epsilon < \epsilon^* = \frac{1}{256 M^2 (15 \pi^2 \sigma)^{2/3}}.$$

From your perspective, since the net effect of gravity and Hawking radiation is to push outwards on you, the external force must be pushing you towards the black hole! So it seems to reason that if you then turn off that external force, you will start moving away from the event horizon.

But that reasoning is wrong, which can be argued either from the material in the linked paper above or from the perspective of a different observer.

An observer at infinity (i.e. an observer fixed at some other radius $R'$ very very far away) sees that the Hawking radiation at the constant temperature $T_H$ (independent of \epsilon) is exerting a relatively tiny amount of pressure, nowhere near enough to balance out gravity. From this perspective, the external force is pulling you out of the black hole, and if it is turned off you start falling into the black hole... as usual.

The logical loophole is that the very act of turning off the external force will change the temperature of the Hawking radiation you experience. This is in referenced in the paper as the "FFAR observer", and it is shown in the paper that this observer again experiences a Hawking temperature near $T_H$ that does not blow up at the horizon.

To sum up, if you turn off the external forces and try to ride the Hawking radiation away from the black hole, you will realize that the external forces were holding you out of the black hole all along and start falling in.

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  • $\begingroup$ Not sure I totally buy this. If you're being accelerated, then you can feel which direction you're being accelerated in. If the external observer sees the rope as holding you out of the hole, then how could you feel the rope as pushing you into the hole? The rope would go slack. I could buy the idea that the Hawking radiation pressure you feel (in excess of the gentle $2 T_H$) isn't a "real" force, but just a "reflection" of the force of the rope. What is the form of the Hawking radiation's contribution to the stress-energy tensor near the horizon? That should be frame-independent. $\endgroup$ – tparker Feb 17 '17 at 1:43
  • $\begingroup$ Suppose instead of suddenly opening the sail, you very slowly unfurl it so its area is $A(\tau)$, and in a synchronized way, loosen the tension $F$ in the rope in such a way that the total outward force $$F(\tau) + \frac{A(\tau)}{245760 \pi^2 M^4 \epsilon^2} = \frac{m}{4 M \sqrt{\epsilon}}$$ stays constant, and strong enough to keep you at contant $r$. Then once you hit the critical sail area $A$ that makes $\epsilon^*(A) = \epsilon$, shouldn't the Hawking radiation pressure take over from the rope tension at that point, so that $F \to 0$ but you remain stationary? $\endgroup$ – tparker Feb 17 '17 at 1:52
  • $\begingroup$ In response to tparker's comment: "If you're being accelerated, then you can feel which direction you're being accelerated in." Suppose you are currently being held at fixed radius. You have two net forces acting on you and their contributions to your acceleration must add to your total proper acceleration: $$ a_{Hawking} + a_{external} = a_{gravity} = \frac{1}{4 M \sqrt{\epsilon}}$$. Clearly different observers must agree on the value of $a_{gravity}$, and according to your source they DISagree on the value of $a_{Hawking}$. By subtraction, they must disagree as well on $a_{external}.$ $\endgroup$ – user2694879 Feb 21 '17 at 21:29
  • $\begingroup$ An observer from far away sees a very small value of $a_{Hawking}$ (that does not diverge as $\epsilon \to 0$ and thus sees $a_{external} > 0$, while you see $a_{external} < 0$. It seems like the two observers disagree on which way you will start moving when you turn off $a_{external}$. But your own source says that without $a_{external}$, you become a different "FFAR" observer and see a very small value of $a_{Hawking}$ that cannot overcome $a_{gravity}$. The perspectives of the FFAR observer and the far away one agree that you cannot ride the Hawking radiation away from the black hole. $\endgroup$ – user2694879 Feb 21 '17 at 21:42
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In addressing this question people are fixating a little bit too much on the rope. The rope is merely a conceptual tool to put the test object into an initial state that is stationary with respect to the black hole. It could be a rope, it could be a rocket, anything that puts the test object into an initial state that is stationary is fine.

From that initial state — regardless of how it got there — the test object deploys the reflective sail. Once the sail is deployed, it is the Hawking radiation, not the rope (or rocket or whatever) supplying the outward acceleration.

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Being outside of EH, this is a mechanics question. If the acceleration can beat the gravity, you sail out, if it does not, you don't. Rest are calculations.

Also, Can we rely on the rate of Hawking radiation being uniform?

What about the radiation/stuff falling into BH which will hit the sail from other side thereby cancelling/overriding the momentum due to Hawking radiation?

Will the rope continuously wind, or it will go slack once you move a little? If it winds up and maintains same force, then you just need to move a little and rope will do the rest.

You have to wait for a moment when the Hawking radiation overwhelms the gravity plus momentum of external stuff falling on the sail. Then the "constant force" "winding" rope will pull you faster and faster. Like a big bang moment! Till then, the rope will hold you just above EH.

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Can you ride Hawking radiation away from a black hole?

No. Let's go through this carefully.

Suppose you were held by a very strong rope at constant Schwarzchild coordinate $r = 2M (1 + \epsilon)$ just above the event horizon of a Schwarzchild black hole of mass $M$.

OK, we're used to such strong ropes, no problemo!

You would feel a proper acceleration of magnitude $a_\text{gravity}(r) = \frac{M}{r^2 \sqrt{1-\frac{2M}{r}}} \sim \frac{1}{4M \sqrt{\epsilon}}$ from the rope (in units where $G = c = \hbar = 1$).

Whoa. Check out what Einstein said about a gravitational field: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". That c=1 is an issue. If it really was a c=1 your rope would be slack. But let's move on, because we're confident that your rope wouldn't be slack, that it would be digging into your waist a tad, and that you'd be subject to a time-dilation factor $t_{0}=t_{f}{\sqrt {1-{\frac {r_{s}}{r}}}}$.

According to eqs. (1.3) and (3.1) of this paper, you would also observe Hawking radiation with effective temperature $ T(r) = \frac{T_H}{\sqrt{1 - \frac{2 M}{r}}} \sim \frac{T_H}{\sqrt{\epsilon}}$ coming out of the black hole, where $T_H := 1/(8 \pi M)$ is the Hawking temperature.

But would you? I thought that was a black hole? I can understand black holes by paying attention to "Einstein and the evidence". I note what Einstein said about the speed of light above, I note that optical clocks go slower when they're lower, I note that at the event horizon of a black hole the coordinate speed of light is zero, and I note Clifford Will's confrontation between general relativity and experiment. General relativity is one of the best-tested theories we've got. The initial test came in 1919, only three years after publication, with a war on. However Hawking radiation has been around for forty years, and there's no actual evidence for it. So if there's some conflict between the predictions of the two theories, I know which side I'm leaning to. General relativity says the coordinate speed of light at the event horizon is zero, so it sounds to me as if that's the speed of the Hawking radiation too.

The temperature and the acceleration would both diverge as you approach the horizon, but at a constant ratio $ \frac{a_\text{gravity}(r)}{T(r)} = \frac{8 \pi M^2}{r^2} \sim 2 \pi.$

Are you sure? The force of gravity at some location relates to the local gradient in the coordinate speed of light, which goes to zero at the event horizon. Which suggests the temperature at the event horizon is zero too. N'est pas?

By the Stefan-Boltzmann law, you would observe a total emitted power per unit area (or equivalently radiation pressure, in units where $c=1$) of $ P = \frac{\pi^2}{60} T^4 = \frac{1}{245760 \pi^2 M^4 \epsilon^2}.$

There you go with that c=1 again. It isn't a c=1, it's a c=0.

The total surface area of a black hole is a subtle concept, due to the spacetime curvature, so let's restrict ourselves to considering a small region just outside the horizon (i.e. a region with diameter much less than the Schwarzchild radius, which sets the curvature scale), which we can locally approximate by Minkowski spacetime.

I'm afraid that's a contradiction in terms. See section 20 of Relativity: the Special and General Theory where Einstein said this:

“We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes”.

You can't transform away a real gravitational field. If you could your rope would go slack. See this again and note how it says SR is nowhere precisely realized in the real world. Your small region is an infinitesimal region. It's a region of zero size. That's no region at all. The principle of equivalence was "Einstein's happiest thought", but IMHO it's important not to take it too far.

Suppose you were to then unfurl a solar sail with mass surface density $\sigma$ (including the contribution from your own mass). The radiation pressure would accelerate you and the sail at $ a_\text{Hawking} = \frac{P}{\sigma} = \frac{1}{245760 \pi^2 \sigma M^4 \epsilon^2}$ away from the hole.

What radiation pressure? Would that be the radiation pressure of light leaving the black hole at a speed c=0?

We see that $a_\text{gravity}$ diverges much more slowly than $a_\text{Hawking}$ at small $\epsilon$. Indeed, if $\epsilon < \frac{1}{256 M^2 (15 \pi^2 \sigma)^{2/3}},$ then the acceleration from the Hawking radiation wins, and would seem to blow you away from the hole!

If it did, it would blow away other material too, and black holes wouldn't get any bigger. That doesn't sound right to me.

As a sanity check, as $M$ grows larger (cooler black hole) or $\sigma$ grows larger (denser and less efficient sail), the Hawking radiation becomes less effective at pushing you away.

I thought M wasn't going to get any bigger?

Obviously this would be a ludicrous setup for a real black hole

Agreed.

but in principle would it be possible to use such a solar sail to ride the Hawking radiation out and escape the hole?

If the setup is ludicrous, then no.

(Note that this idea is closely related to that of a black hole starship.)

IMHO the problem with that is that it doesn't pay sufficient attention to general relativity. There's a bit of an issue wherein the falling body falls faster and faster but the coordinate speed of light is getting lower and lower. Friedwardt Winterberg wrote a paper about this in 2001.

And if you did, how would a nearby free-falling observer passing you into the black hole describe the process? After all, according to the linked paper, to her the black hole would only be radiating at a quite gentle temperature of $2 T_H$ and would only provide a bounded radiation pressure.

Pass. I think you may have stumbled on a contradiction there Mr Parker!

Edit 19/02/2017 score -7 : perhaps I can clarify by saying the issue with Hawking radiation is that it totally ignores gravitational time dilation. Since gravitational time dilation occurs because "the speed of light is spatially variable", and since "the curvature of light rays occurs only in spaces where the speed of light is spatially variable" Hawking radiation ignores the very reason the gravitational field is there in the first place. No wonder it leads to contradiction. On top of that the given explanation relies on particles popping into existence, and negative-energy particles to boot. See this answer where I gave some details of that. Virtual particles are virtual, they don't pop in and out of existence like magic, there are no negative-energy particles, and there's been no evidence for Hawking radiation for 43 years now.

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    $\begingroup$ Your edit on Hawking radiation shows that you have never actually seen or done the calculations involved in understanding Hawking radiation: At no point are "virtual particles" needed. In fact, that's just a (seriously flawed) simplified picture invented for laymen who cannot actually go through the required calculations. That you feel confident in dismissing Hawking radiation based on this "lie for children" (as you like to call such things) shows not only your ignorance of the physics but also your inability to judge your own knowledge of this topic. $\endgroup$ – Danu Feb 22 '17 at 9:29
  • $\begingroup$ The speed of light is not zero at the horizon - it's c, for an observer at the horizon falling in. $\endgroup$ – Akoben Feb 28 '17 at 20:08

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