Why wouldn't black hole singularity evaporate before it can form?

[PAllen]

Since you can make the Hawking radiation rate arbitrarily small by having a sufficiently large M it seems clear to me that "pre-evaporation" cannot be a general case.

I lean toward the majority view on this, but I don't think this argument is enough to settle the question. For a big black hole, suppose you are seeing an incredibly low rated of pre-hawking radiation for 10^^(10^^(10^^...)) years before the horizon has formed. Then, without comparing rates and a model of horizon formation subject to decreasing mass, you still don't know which wins. There exists an actual computation claimed to show evaporation always wins (reference 17, I believe) in the paper I cite. The paper I referenced aims to refute that, but neither side claims the question is subject to trivial argument.

 

[Dale]

For a big black hole, suppose you are seeing an incredibly low rated of pre-hawking radiation for 10^^(10^^(10^^...)) years before the horizon has formed.

As long as the BH is sufficiently massive that it is colder than the CMB I don't think that matters.

 

[PAllen]

As long as the BH is sufficiently massive that it is colder than the CMB I don't think that matters.

The CMB gets colder over time, eventually below the Hawking temperature of any size BH. Also, note that the question of whether a horizon forms is a global, invariant question. Thus, analysis in any coordinate system is valid. In particular, analysis using and SC type time foliation is valid. This gives infinite time before horizon formation for evaporation to succeed. So you have to show how horizon formation wins the race even in these coordinates.

 

[Dale]

The CMB gets colder over time, eventually below the Hawking temperature of any size BH.

I don't see how future cooling is relevant to the collapse. As long as the BH is colder today than the CMB today then it has been colder than the CMB for the past "10^^(10^^(10^^...)) years" too. So what if that changes in the future, it doesn't impact the collapse.

 

[PeterDonis]

is it possible to define the energy with in a sphere defined a radius from the singularity, from the point of view of an in falling observer, inside the horizon.

I don't think so, because inside the horizon there are no static objects, and that includes spheres of constant radius. So there's no way to add up the "energy inside a sphere of constant radius" at an instant of time inside the horizon.

 

[stevendaryl]

Lo and behold, see what I found in my notes and had completely forgotten:

from Steve Carlip...
http://www.physics.ucdavis.edu/Text/Carlip.html#Hawkrad



thoughts?? My amateur conclusion is that Carlip doesn't think any Hawking radiation exists inside a BH horizon.

I don't think that intuitive description of black hole radiation is original to Carlip; I think that Hawking himself may have come up with it.

However, as I understand it, physicists have not convincingly proved that the intuitive description in terms of virtual pairs actually connects with the more careful analysis, having to do with the different definitions of "vacuum" and "particle number" used by stationary and infalling observers. In other words, I don't think that the intuitive description has been shown to be correct.

 

[PAllen]

I don't see how future cooling is relevant to the collapse. As long as the BH is colder today than the CMB today then it has been colder than the CMB for the past "10^^(10^^(10^^...)) years" too. So what if that changes in the future, it doesn't impact the collapse.

I thought we were discussing collapse, not eternal black holes. The debate on this concerns collapse models. Thus, eternal past is irrelevant. Thus you have a collapse, with infalling matter and also a tiny bit of CMB. The question remains: for a distant observer, the collapse process 'never' completes, the evaporation does (even accounting for CMB). At least that is the argument of Krauss et. al. I think this is most likely wrong, but reading over the respective papers, I don't think the debate can be trivialized. I think you have to model both horizon formation and evaporation as quantum processes, and there won't be a final answer until there is an accepted theory of quantum gravity.

 

[lukesfn]

Neither of the papers you linked makes any claim that there is any Hawking radiation observed inside the horizon. You can't apply formulas developed outside the horizon, for what is accepted to be a horizon based phenomenon, inside the horizon just because you want to.

I can't think of a reason why the formulas wouldn't extend beyond the event horizon. I don't know what the difference would be there. They predict a non zero hawking radiation observed by a free faller at the horizon. If it changes in a non-continuous way at the horizon, it would seem to me to break the equivalence principal.

My intuition (and I don't have the mathematical skills to back this up) is that the situations outside the event horizon and inside the event horizon are sufficiently different that you can't extrapolate from one region to the other. From what I've read, Hawking radiation is associated with a horizon, and for an infalling observer, there is no horizon once you've passed the Schwarzschild radius. Inside the Schwarzschild radius, the coordinate r becomes time-like and so the horizon, which is at a fixed value of r, becomes something in the past, rather than something that persists.

Obviously I am not an expert, but there is no reason obvious to me why the math should suddenly change at the EH, however, the hawking radiation observed inside the BH comes from a non-static falling "Horizon" inside the static EH that the in-faller overtook in it's past. Non of the observed radiation can escape the BH, everything is in-falling. It all seems to fit together perfectly and make perfect sense to me, however, the in-faller will beat the radiation in the race to the singularity. Which implies it overtakes some of the energy in that space, which implies that not all of the energy of a BH is concentrated at the singularity, however, it seems that GR has trouble with locating energy in this situation anyway, so perhaps this is all ok.

And I am left wanting a more convincing proof that there is a problem.

Fair enough. Even if my calculations of hawking radiation are correct, I can't be sure they are a problem if I don't have a way to calculate the energy of the region they come from.

We have clear solutions showing collapse and aggregations in a finite amount of time as measured by the collapsing/aggregating matter, and Hawking radiation becomes arbitrarily small as M gets large. So to me what seems unconvincing is the implication that Hawking radiation could prevent collapse or aggregation.

But I assume the hawking radiation you are talking about here is as observed at infinity, not as observed by the collapsing matter at it's location inside the EH, which increases as the radius gets smaller, there for the second half of that argument is not relevant to preventing a singularity, even if there is a EH.

I don't think so, because inside the horizon there are no static objects, and that includes spheres of constant radius. So there's no way to add up the "energy inside a sphere of constant radius" at an instant of time inside the horizon.

I just starting to realize this could be a problem my self. I'm doing some reading. It is surprising how many different definitions of energy there are in GR.

Ironically, perhaps, the observed Hawking radiation gives an estimate of the energy of the region it came from. Interestingly, even if it doesn't make sense to talk about the energy of such a region the radiation wouldn't cause that region to loose any energy, because it is radiating into its self.

I was just trying to imagine a region of space, moving out at the speed of light, starting from just inside the EH, and falling to the singularity, and calculating how much energy it looses, however, I think I would just run into the same problem because the energy loss needs to be calculated by an observer at infinity, but the radiation won't get there.

 

[PAllen]

I can't think of a reason why the formulas wouldn't extend beyond the event horizon. I don't know what the difference would be there. They predict a non zero hawking radiation observed by a free faller at the horizon. If it changes in a non-continuous way at the horizon, it would seem to me to break the equivalence principal.

The point is that the derivation of the Hawking radiation assumes it is observed outside the horizon, and is caused by the horizon. In the realm of quantum theory (unlike classical) the horizon is not necessarily locally undetectable. Look up black hole firewalls for the current hot (pun intended) debate on the quantum nature of horizons.

More importantly, I asked you for references to anyone proposing Hawking radiation is detectable inside the horizon. You gave links to two papers that say no such thing. That leaves you with nothing but personal opinion in disagreement with all derivations and discussions of Hawking radiation.

 

[Dale]

I thought we were discussing collapse, not eternal black holes.

Yes, me too.

 

[lukesfn]

The point is that the derivation of the Hawking radiation assumes it is observed outside the horizon, and is caused by the horizon. In the realm of quantum theory (unlike classical) the horizon is not necessarily locally undetectable. Look up black hole firewalls for the current hot (pun intended) debate on the quantum nature of horizons.

More importantly, I asked you for references to anyone proposing Hawking radiation is detectable inside the horizon. You gave links to two papers that say no such thing. That leaves you with nothing but personal opinion in disagreement with all derivations and discussions of Hawking radiation.

Well, unfortunately, that is probably the best I can do right now, if you find it easier to wave your hands and believe that an observer instantly stops seeing radiation at the moment they cross the horizon while abandoning the equivalence principal rather then then simply assume there is no reason for the maths to disappear at the EH, I doubt I can stop you

I am aware of the firewall which brings the equivalence principal into question.

However, I really don't think that is relevant. I think that the point of confusion here is that hawking radiation is traditionally derived from the frame of an observer at infinity. Any radiation observed inside the horizon will not contribute to the radiation observed at infinity, so in a sense, it definitely isn't hawking radiation as traditionally derived, and it won't contribute to the evaporation of the BH.

But that doesn't mean that radiation wouldn't be observed, but it is not like you cross the EH, and you can suddenly see all the way to the singularity. Visually, there is still some kind of "horizon" which could be observed, which should have observable radiation observed coming from it. This was my premiss at the start of the thread. Later I found the maths to back it up.

It seems like such a simply obvious concept to me, and I assume everybody here knows much more about this kind of thing to me, I'm quite surprised by the resistance. Weather or not anything useful can be learned from such radiation is a fair question, however, I don't think handwaving away a simple continuous mathematical curve will lead to any kind of understanding.

 

[atyy]

I can't think of a reason why the formulas wouldn't extend beyond the event horizon. I don't know what the difference would be there. They predict a non zero hawking radiation observed by a free faller at the horizon. If it changes in a non-continuous way at the horizon, it would seem to me to break the equivalence principal.

One speculative idea about why the infalling observer doesn't see Hawking radiation is called "black hole complementarity". "According to this principle, the matter which has fallen past the event horizon and the Hawking radiation are not different objects. They are complementary descriptions of a single system, viewed from very different reference frames which are related by an enormous Lorentz boost." http://arxiv.org/abs/hep-th/9506138

Another similar idea is Unruh radiation. In flat spacetime, the non-accelerating observer sees no radiation, but the accelerating observer sees radiation from the Rindler horizon.

 

[PeterDonis]

it is not like you cross the EH, and you can suddenly see all the way to the singularity.

It's true that you can't see "all the way to the singularity" when you cross the EH, but that's because the singularity is in your future. The singularity is not a place in space; it's a moment of time.

Visually, there is still some kind of "horizon" which could be observed

Andrew Hamilton has some good animations of what it might look like if you fell into a black hole:

http://casa.colorado.edu/~ajsh/schw.shtml

They basically bear out the idea that thinking of spacetime inside the horizon as being similar to our ordinary space doesn't work very well, particularly when you include the angular coordinates. Typical representations of a black hole interior suppress the angular coordinates because of the spherical symmetry, and just look at the $t - r$ plane.

One counterintuitive property is that the horizon, meaning the one at $r = 2M$, always appears to be ahead of you as you fall, even after you've crossed it; this is because, when you include the angular coordinates, it turns out that ingoing light from the horizon curves around and comes at you from the front.

which should have observable radiation observed coming from it.

Possibly, but I'm not sure that the links you have provided actually say that, as PAllen has pointed out. See further comments below.

It seems like such a simply obvious concept to me, and I assume everybody here knows much more about this kind of thing to me, I'm quite surprised by the resistance.

I would agree that there is one thing that is simple and obvious: if there is a quantum field present in the spacetime (which there must be for Hawking radiation to be observed from the exterior), then that quantum field will be present inside the horizon as well as outside. But that's quite different from claiming that the presence of the field on the inside *must* result in observable radiation; that depends on what state the field is in, as well as the state of motion of the observer.

As I understand the standard argument for Hawking radiation, it relies on the fact that particle number is observer-dependent; more precisely, the particle number operator for an accelerated observer is different from the particle number operator for an inertial observer. That means that one and the same quantum field state can appear to be vacuum to an inertial observer, but appear to contain particles to an accelerated observer. The standard model that is used to derive Hawking radiation assumes that the field is in such a state--one that looks like vacuum to an observer free-falling into the hole, but looks like it contains particles to an observer accelerating so as to stay at a constant altitude above the hole's horizon.

(I know there are other arguments in the literature for why inertial observers should observe radiation too; as I understand it, these arguments attribute some different state to the quantum field, one that does *not* necessarily look like a vacuum to an inertial observer. More on that in a moment.)

Now if the quantum field is in a state such as I described above, that looks like vacuum to an inertial observer, that should hold inside the horizon as well as outside, so an observer free-falling into the hole should not observe radiation at any point; to him, the quantum field is always in a vacuum state. So any argument that claims to show that such a free-falling observer *would* observe radiation would, it seems to me, have to argue that the quantum field is in some other state, not the state that looks like vacuum to an inertial observer. And *that*, as I understand it, changes the whole basis for the Hawking radiation calculation. I'm not sure how such a model would look, but I think it's fair to say that it's not "simple and obvious" at this point.

 

[lukesfn]

One counterintuitive property is that the horizon, meaning the one at $r = 2M$, always appears to be ahead of you as you fall, even after you've crossed it;

I don't think I have much trouble visualising these cases, but I think you've made a small error. Light from the horizon at r=2M moving directly outward, will remain at the horizon. Light moving directly inward will appear to as if it came from behind you. (Assuming you are inside the EH in its path)

The visual horizon we are talking about is not at r = 2M, it is further in, and you can catch up to falling light that came from it, that would hit you in the front, even though it's was originated from a position behind you. It doesn't have to pass around you though, because at the time it originated you where in front of it. You don't have to consider any angular effects to get this. Everything is in the same straight line away from or towards the singularity.

Anyway, I have said repeatedly that any observed radiation inside the horizon would be in falling and have originated at a position further out.

Now if the quantum field is in a state such as I described above, that looks like vacuum to an inertial observer, that should hold inside the horizon as well as outside, so an observer free-falling into the hole should not observe radiation at any point; to him, the quantum field is always in a vacuum state. So any argument that claims to show that such a free-falling observer *would* observe radiation would, it seems to me, have to argue that the quantum field is in some other state, not the state that looks like vacuum to an inertial observer. And *that*, as I understand it, changes the whole basis for the Hawking radiation calculation. I'm not sure how such a model would look, but I think it's fair to say that it's not "simple and obvious" at this point.

If read some about this, and I think it may have been addressed inside the papers I linked. Obviously they put a case for observed Hawking radiation during free fall. Unfortunately, my level of understanding is way too insufficient to get much understanding out of most of those kind of discussions, I just assumed they are correct because it fits with my intuition. Furthermore, It just seems straight forward to me that if there arguments hold outside the EH, they would also hold just inside, unless you bring in new physics at the EH.

 

[atyy]

More importantly, I asked you for references to anyone proposing Hawking radiation is detectable inside the horizon. You gave links to two papers that say no such thing. That leaves you with nothing but personal opinion in disagreement with all derivations and discussions of Hawking radiation.

Barbado, Barcelo and Garay's paper http://arxiv.org/abs/1101.4382 that lukesfn linked to does say that some free falling observers will see a non-zero temperature at the horizon. And if one assumes that there is no discontinuity at the horizon for a big black hole, then presumably these observers will also see a non-zero temperature after crossing the horizon. (The authors do say that there is a free-falling observer for whom the temperature will be zero at the horizon.)

Their introduction says "In order to have Hawking radiation at infinity, it is necessary to set the quantum field in a particular quantum state: the Unruh vacuum state [8, 9]. This state is commonly described as being a vacuum state for observers freely falling at the event horizon of the black hole." Most of the paper does not use the Unruh state, but at the end their section 6 says "One might worry that some of the results shown in this paper related with the late-time behaviour of the system, are peculiarities of an unusual choice of the vacuum state. However, this is not the case. In this section, we will briefly show that at late times this state is really equivalent to the Unruh vacuum state. Thus, all the properties found in the previous sections that applies for late times, are immediately present also in the Unruh state."

Their suggestion does seem different from Lowe et al's http://arxiv.org/abs/hep-th/9506138 (unless the free falling observer here is specifically the non-generic free-falling observer that Barbado et al say will see zero temperature at the horizon). "According to the principle of black hole complementarity [1], an observer who remains outside the horizon of a black hole can describe the black hole as a very hot membrane, the stretched horizon, which lies just above the mathematical event horizon, and absorbs any matter, energy, and information which fall onto it. The information which is absorbed by the stretched horizon is eventually re-emitted in the Hawking radiation, albeit in a very scrambled form. An observer who falls freely into the black hole sees things very differently: no membrane, no high temperature, no irregularities of any kind as the observer crosses the event horizon."

 

[PeterDonis]

Light from the horizon at r=2M moving directly outward, will remain at the horizon.

Yes.

Light moving directly inward will appear to as if it came from behind you. (Assuming you are inside the EH in its path)

*If* it's moving radially only. (Hamilton's animations don't include this possibility because they represent what things look like in front of you, not behind you, as you fall.) That's why I mentioned including the angular coordinates. Other light rays that are ingoing from the horizon, but *not* moving purely radially, will curve around and come at you from the front as you are falling. (Obviously this light will have been emitted from the horizon earlier than the light that moves purely radially and catches up with you from behind.)

The visual horizon we are talking about is not at r = 2M, it is further in

Then I don't understand what you're referring to. Can you give a reference?

you can catch up to falling light that came from it, that would hit you in the front, even though it's was originated from a position behind you. It doesn't have to pass around you though, because at the time it originated you where in front of it.

This doesn't make sense to me; I think you need to clarify exactly what you mean by "position" here.

Anyway, I have said repeatedly that any observed radiation inside the horizon would be in falling and have originated at a position further out.

Not necessarily. Light that is radially outgoing can come at you from the front, even though it is infalling, because it is infalling more slowly than you are; it will be emitted from a smaller radial coordinate than you, but because its radial coordinate is decreasing more slowly than yours, you can catch up to it and pass it, and you will see it as coming from in front of you.

I just assumed they are correct because it fits with my intuition.

I don't think this is a good heuristic to use in subjects like these.

It just seems straight forward to me that if there arguments hold outside the EH, they would also hold just inside

The region inside the horizon is not static, whereas the region outside is; that is one major difference between the two that does not require any "new physics", it's already part of the classical black hole spacetime. So any argument that, directly or indirectly, implies that the region of spacetime it applies to is static, would only apply outside the horizon. I don't know for sure that the various Hawking radiation arguments fall into this category, but I think it's important to realize that "it just seems straightforward to me" is, again, not a good heuristic, because you're ignoring an obvious difference between the two regions.

 

[PAllen]

Barbado, Barcelo and Garay's paper http://arxiv.org/abs/1101.4382 that lukesfn linked to does say that some free falling observers will see a non-zero temperature at the horizon. And if one assumes that there is no discontinuity at the horizon for a big black hole, then presumably these observers will also see a non-zero temperature after crossing the horizon. (The authors do say that there is a free-falling observer for whom the temperature will be zero at the horizon.)

First, while classically, a horizon is locally undetectable by a free faller, this is not necessarily so quantum mechanically. Passing a membrane emitting light from one side, thus seeing light stop when you pass it, is not any sort of prohibited discontinuity.

I reviewed the paper again and still see no statements implying Hawking radiation inside the horizon, nor any part of the derivation that implies it for me. I see the opposite - they talk about how everything ends up explainable in terms the infaller's speed, start time of free fall, and what static (Unruh) observers see at different times (early, late). Since there are no Unruh observers inside the horizon, I see no implication of Hawking radiation inside. Perhaps, an implication is the possibility of ordinary Unruh radiation seen by an inside observer with proper acceleration (not a free faller), but they don't explore this.

Further, since they highlight as surprising that there is temperature perceived by most free fallers when crossing the horizon, I would think they would even more strongly highlight a claim that radiation continues inside. Instead, they make no statements of such a claim.

 

[atyy]

First, while classically, a horizon is locally undetectable by a free faller, this is not necessarily so quantum mechanically. Passing a membrane emitting light from one side, thus seeing light stop when you pass it, is not any sort of prohibited discontinuity.

I reviewed the paper against and still see no statements implying Hawking radiation inside the horizon, nor any part of the derivation that implies it for me. I see the opposite - they talk about how everything ends up explainable in terms the infaller's speed, start time of free fall, and what static (Unruh) observers see at different times (early, late). Since there are no Unruh observers inside the horizon, I see no implication of Hawking radiation inside. Perhaps, an implication is the possibility of ordinary Unruh radiation seen by an inside observer with proper acceleration (not a free faller), but they don't explore this.

Further, since they highlight as surprising that there is temperature perceived by most free fallers when crossing the horizon, I would think they would even more strongly highlight a claim that radiation continues inside. Instead, they make no statements of such a claim.

Because they say "crossing", and because the geometry is smooth at the horizon in their case, it makes me think that they mean the non-zero temperature will be present on both sides of the horizon. As I understand, the temperature is a proxy for excitations of the vacuum that are nearly thermal.

 

[PAllen]

Because they say "crossing", and because the geometry is smooth at the horizon in their case, it makes me think that they mean the non-zero temperature will be present on both sides of the horizon. As I understand, the temperature is a proxy for radiation whose spectrum is nearly thermal.

Perhaps. Suppose you interpret it this way. The the most reasonable conclusion from this, noting that their derivation of specific crossing temperature involved the limiting behavior of blue shift approaching infinite relative to Unruh observers, while unruh temperature approaches zero towards the horizon. The limiting Unruh observer at the horizon is said to have zero temperature. Then, just past the horizon, you now have infinite redshift relative to this limiting Unruh observer. Thus, instead of 0*∞ you have 0/∞. Thus, again one is led to conclude zero Hawking radiation inside.

 

[lukesfn]

(Hamilton's animations don't include this possibility because they represent what things look like in front of you, not behind you, as you fall.)

Actually, there is a Hamilton animation of a radial dive with 180 degree fish eye on the diagonal, some of what you are seeing in the animation is coming from behind. I don't think the camera is pointed in your direction of motion. It isn't 100% clear if you can see directly behind you or not in that animation.

It isn't quite so easy for me to tell how much light coming in from an angled fall can appear to be in front of you, but I don't think it would be directly in front of you, I think it would it be close to hitting you directly in the side.

I think the animations show this. Hamilton discribes that without the fish eye distortion the singularity would appear as a plane through your middle.

Then I don't understand what you're referring to. Can you give a reference?

I will try to explain differently.

The horizon that appears to be in front is not the horizon at r=2m, since as I described, light from that location can be observed coming from behind directly you, but never directly infront. There is a "horizon" with a smaller radius, from where no light is able to reach you until you fall further. This is the horizon observed in front of you as you fall inside the EH, or, to be clear the surface that looks black. (Except for any hawking like radiation that might be observed coming from it)

For lack of a better word, I sometimes call it the optical horizon. This is where it would make sense for any observed hawking like radiation to originate from.

Not necessarily. Light that is radially outgoing can come at you from the front, even though it is infalling, because it is infalling more slowly than you are; it will be emitted from a smaller radial coordinate than you, but because its radial coordinate is decreasing more slowly than yours, you can catch up to it and pass it, and you will see it as coming from in front of you.

That depends on your coordinate system.

Any light moving radial that hits you in the front will have originated from a radial distance greater then your radial distance where it is observed.

When I say it will have come from be hind, I mean, a greater radial distance from where it was observed.

I don't think this is a good heuristic to use in subjects like these.

Intuition can certainly lead you the wrong way sometimes, however, I am not convinced I am guilty of anything you are not in this case.

The region inside the horizon is not static, whereas the region outside is;

I just can't think of any reason why that would be relevant case, just like it isn't relevant to the equivalence principal.

Because they say "crossing", and because the geometry is smooth at the horizon in their case, it makes me think that they mean the non-zero temperature will be present on both sides of the horizon. As I understand, the temperature is a proxy for excitations of the vacuum that are nearly thermal.

Thanks a lot for reminding me of the important relevant bits of that paper in your other post, it was a long time ago since I actually read it.

I don't pretend to read the mind of the authors. It was my conjecture that their results imply that an in-faller would continue to observe the radiation described after crossing the EV. I've never tried to pretend other wise. I've never intentionally claimed that they directly talking about hawking radiation anywhere inside the EH.

Like PAllen, I wonder why they didn't mention the implication, because it could be an interesting and straightforward implication to me. (However, I did read the paper looking for such implications, so I obviously had some bias)

Perhaps. Suppose you interpret it this way. The the most reasonable conclusion from this, noting that their derivation of specific crossing temperature involved the limiting behavior of blue shift approaching infinite relative to Unruh observers, while unruh temperature approaches zero towards the horizon. The limiting Unruh observer at the horizon is said to have zero temperature. Then, just past the horizon, you now have infinite redshift relative to this limiting Unruh observer. Thus, instead of 0*∞ you have 0/∞. Thus, again one is led to conclude zero Hawking radiation inside.

That would seem to contradict the maths I already provided in this thread. I've forgotten exactly what happens here, but I think the blue shift to red shift change simply changes the infinity from positive to negative in the math. I think your intuition about swapping from a multiply to a divide probably is wrong.

 

[PAllen]

That would seem to contradict the maths I already provided in this thread. I've forgotten exactly what happens here, but I think the blue shift to red shift change simply changes the infinity from positive to negative in the math. I think your intuition about swapping from a multiply to a divide probably is wrong.

I'll turn it around. You are misapplying math without understanding the physics. Blueshift is multiply by Doppler factor; red shift is divide by it. The paper you posted clearly stated the points I made atyy: that Unruh observers can be used to explain quantitative features of Hawking radiation. There are no Unruh observers inside the horizon because there can't be a static world line, and the metric no longer has a timelike killing vector. The exterior Unruh observer's radiation is redshifted. Even if you assume (and I am not sure it is valid) that Hawking radiation detected away from the horizon is isotropic, such ingoing radiation will be signficantly redshifted.

Especially, any concept of Hawking radiation originating from the singularity or anywhere inside the horizon is completely at odds with the paper.

 

[lukesfn]

I'll turn it around. You are misapplying math without understanding the physics. Blueshift is multiply by Doppler factor; red shift is divide by it. The paper you posted clearly stated the points I made atyy:

Perhaps we better actually go check the maths I did and see if we can find a flaw.

I think these are the most relevant posts, I recommend you take a look.
Post 22
Post 30

And the most relevant part of those posts below:

I will return to the formula I gave earlier. Under certain assumptions, this gives observed hawking radiation for a free faller where m is mass of the BH, r is the observers radius, and /r₀ is the radius the object was dropped from. (Before being dropped, the object was hovering at a constant radius)
1/(4m) (1 + 2m/r) (1 - 2m/r)^1/2
×{(1 - 2m/r₀)^1/2 + (2m/r×(1-r/r₀))^1/2 }^1/2
×{(1 - 2m/r₀)^1/2 - (2m/r×(1-r/r₀))^1/2 }^-1/2

Now if we simplify things by taking the limit /r₀→∞
1/(4m) (1 + 2m/r) (1 - 2m/r)^1/2
×{1 + (2m/r)^1/2 }^1/2
×{1 - (2m/r)^1/2 }^-1/2

In the approximation in the paper it came from, this formula now gives a temperature of hawking radiation for an observer free falling from ∞. This formula can be continued all the way to the singularity. (Note that I don't think that the paper ever tried to extend this formula inside the horizon, I am doing that on my on initiative)

My comment about the positive minus infinity thing may have been off the mark, but possibly the multiply divide thing is not relevant to the maths I used. I'd have to get my self back in the right head space to remember what is going on. It was quite a few months back I did it.

You are right that there are no real Unrah observers in the horizon. In line with this some value in the math becomes imaginary. However the complex numbers all nicely cancel out and we are always left with a real value and a very nice smooth curve that passes r=2m. Unfortunately, it has been a long time since I looked at this, and I have forgotten the physical meaning of each term and why it all made sense.

 

[PAllen]

Perhaps we better actually go check the maths I did and see if we can find a flaw.

I think these are the most relevant posts, I recommend you take a look.
Post 22
Post 30

And the most relevant part of those posts below:


My comment about the positive minus infinity thing may have been off the mark, but possibly the multiply divide thing is not relevant to the maths I used. I'd have to get my self back in the right head space to remember what is going on. It was quite a few months back I did it.

You are right that there are no real Unrah observers in the horizon. In line with this some value in the math becomes imaginary. However the complex numbers all nicely cancel out and we are always left with a real value and a very nice smooth curve that passes r=2m. Unfortunately, it has been a long time since I looked at this, and I have forgotten the physical meaning of each term and why it all made sense.

Your 'extension' is ignoring the basis on which the approximations are derived and is inconsistent with the physics of the paper (Hawking radiation originates from static observers with proper acceleration). I encourage pursuit of such ideas - but not here. Personal theories are not actually allowed here.

 

[lukesfn]

Your 'extension' is ignoring the basis on which the approximations are derived and is inconsistent with the physics of the paper (Hawking radiation originates from static observers with proper acceleration). I encourage pursuit of such ideas - but not here. Personal theories are not actually allowed here.

I disagree, it see it as a straight forward implication of the math, and the only reasons put forward why it shouldn't be are speculative, or based on misconceptions.

(Hawking radiation originates from static observers with proper acceleration).

You are again talking about hawking radiation that could be observed at infinity, I am not, so I am not against the premiss of the paper at all, and it is all perfectly consistent with it, and the maths.

Most of the discussion on this thread has been with moderators who could have shut me down a long time ago, in fact, I ask one to please let me know if I was venturing too far into dangerous water, but once found the papers and did the maths, they seemed ok with discussing it's implications.

To deny it, you have to speculate that the equivalence principal doesn't hold, and that the smooth mathematical curve suddenly becomes divergent and jumps to zero from, which I would have thought would be a much bigger stretch then simply accepting it.

Anyway, I am wasting way too much time here, I have what I want, I have already been pointed in a new direction of enquiry by those who took the time to listen which will send me towards more reading else where. I'm very sorry to say, I feel you are more interested in putting me in my place then being helpful, and I really expect more from a Patron and Sci Adviser. I understand how ever that it is easy to miss read somebodies tone in this kind of correspondence, and that there is a lot of abuse of these forums and so I thank you for taking the time to engage at all.

 

[PAllen]

To deny it, you have to speculate that the equivalence principal doesn't hold, and that the smooth mathematical curve suddenly becomes divergent and jumps to zero from, which I would have thought would be a much bigger stretch then simply accepting it.

If you'll bear with me a little, I don't see the applicability of the principle of equivalence at all. First, it is a local, approximate, classical principle. More importantly, it makes a completely different prediction from what the paper you reference establishes.

Specifically, the principle of equivalence would imply that (correctly, for most purposes) that a BH horizon is locally equivalent to a Rindler horizon. However, for the latter, Unruh radiation ceases immediately on commencing free fall. In fact, many authors have relied on this to argue that there is no Hawking radiation observed at all for a free falling observer. The paper you reference makes a plausible case that this is not so - that is, that the principle of equivalence is false for the combination of a global invariant feature (the existence of static observers with proper acceleration) and QFT, which is non-local in nature.

So what is, exactly, the principle of equivalence argument as you see it?

 

[PeterDonis]

The horizon that appears to be in front is not the horizon at r=2m, since as I described, light from that location can be observed coming from behind directly you, but never directly infront. There is a "horizon" with a smaller radius, from where no light is able to reach you until you fall further. This is the horizon observed in front of you as you fall inside the EH, or, to be clear the surface that looks black. (Except for any hawking like radiation that might be observed coming from it)

This still doesn't make it clear. Once again, do you have a reference? What you're saying doesn't match up with the math as I understand it, so I either need a reference for where you are getting this from, or you'll need to show me the math yourself.

That depends on your coordinate system.

What does? Even though $r$ is often called a "radial coordinate", it can be given an invariant definition (as $\sqrt{A / 4 \pi}$, where $A$ is the area of a 2-sphere at $r$).

Any light moving radial that hits you in the front will have originated from a radial distance greater then your radial distance where it is observed.

Yes, this is true, but the light still hits you from the front. See below.

When I say it will have come from be hind, I mean, a greater radial distance from where it was observed.

This is true, but the same could be said for you: you are "coming from a greater radial distance" just as much as the light is. And if the light is radially outgoing, then its radial coordinate changes more slowly than yours does, which doesn't make much sense if the light is "coming from behind" you.

I just can't think of any reason why that would be relevant case

Then you're not thinking very hard, or reading very carefully. It has been mentioned several times that the derivations of Hawking radiation in the literature appear to depend on the region of spacetime in question being static. If that's the case, then those derivations are not valid inside the horizon, since spacetime there is not static.

 

[lukesfn]

Specifically, the principle of equivalence would imply that (correctly, for most purposes) that a BH horizon is locally equivalent to a Rindler horizon. However, for the latter, Unruh radiation ceases immediately on commencing free fall. In fact, many authors have relied on this to argue that there is no Hawking radiation observed at all for a free falling observer. The paper you reference makes a plausible case that this is not so - that is, that the principle of equivalence is false for the combination of a global invariant feature (the existence of static observers with proper acceleration) and QFT, which is non-local in nature.

So what is, exactly, the principle of equivalence argument as you see it?

I really need to stop posting here... I am using time I don't have... but...

If the paper is correct, an in-falling observer sees hawking radiation at the EH, but not inside the EV, then the equivalence theorem would be in trouble since it implies that hawking radiation emanates from the Horizon from a local point of view. Therefor an equivalence principle lab could do an experiment to detect the location of the EV. Perhaps QFT allows this some how, but it would still leave the equivalence principal in trouble. It isn't clear that you can neglect the sudden change in radiation.

I think that the paper is correct and I believe the Rindler Horizon analysis for the hawking radiation as you described is flawed. The problem I see is that hawking radiation generally originates from a location out side the lab. I think it is a mistake to view it as a local effect.

If hawking radiation comes from EH, and you are at a great distance from it, then the equivalence principle can't say anything about that radiation, it is all very external.

In the case an equivalence principal lab falling through the EH:
a) Before the horizon the lab can't say anything about the hawking radiation emanating from the horizon because nothing happens internally.
b) Falling through the horizon, the lab falls through the hawking radiation already at the horizon, so it doesn't see anything happen internally.
c) Falling after inside the horizon, the lab catches up to falling radiation in front of it, but again, nothing happens internally.

The can only show that faller at the horizon will not see hawking radiation coming from the horizon.

If non of this is true, and all free fallers see no hawking radiation, I would love an explanation of the following case:
A distant radiationally stationary observer sees the Hawking radiation coming from the BH, then the observer is dropped, then after time of seeing no radiation, when the observer reaches a smaller radius, it is held stationary again and starts seeing radiation. I the time of free fall, does the BH loose mass due to hawking radiation? If it does, how come the observer didn't see any? Where did the energy go?

This still doesn't make it clear. Once again, do you have a reference? What you're saying doesn't match up with the math as I understand it, so I either need a reference for where you are getting this from, or you'll need to show me the math yourself.

No reference sorry. I guess I just sound confusing. After you fall through the EV horizon, there appears to be a big black thing in front of you. Now, imagine, hypethetically, some light came off an area just out side that black thing and hit you it the front. All motion being radial, and you facing towards the singularity. What was the location of that particle relative to you? What was the radial location of the surface just behind it.

The particle didn't come from 2M=r because if it did, it couldn't have hit you in the front.

This big black thing could be considered a horizon, but it is not the EH at 2M=r.

Yes, this is true, but the light still hits you from the front. See below.

Then we agree.

This is true, but the same could be said for you: you are "coming from a greater radial distance" just as much as the light is. And if the light is radially outgoing, then its radial coordinate changes more slowly than yours does, which doesn't make much sense if the light is "coming from behind" you.

Hense my clarification.

Then you're not thinking very hard, or reading very carefully. It has been mentioned several times that the derivations of Hawking radiation in the literature appear to depend on the region of spacetime in question being static. If that's the case, then those derivations are not valid inside the horizon, since spacetime there is not static.

More likely I am just too ignorant to see the importance, for me the maths still works, and just because it was derived in one coordinate system that requires a static spacetime, I don't know why it shouldn't extend into a dynamic region if it does so smoothly. All the math of it made sense to me when I did the analysis way back when.

 

[PeterDonis]

After you fall through the EV horizon, there appears to be a big black thing in front of you.

And why do you think this? Did you just come up with it on your own, or did you read it somewhere, or did you deduce it from something you read somewhere? This is what I'm looking for when I ask for a reference: what you're saying is not obvious to me, so I need to know where it comes from.

Now, imagine, hypethetically, some light came off an area just out side that black thing and hit you it the front. All motion being radial

If that's the case, how can light come off an area outside the black thing? Any point radially below you should be within the area of your visual field occupied by the black thing.

just because it was derived in one coordinate system that requires a static spacetime

I didn't say anything about coordinates. I said that, as I understand it, the argument requires a static region of spacetime. That's an invariant statement that has nothing to do with coordinates; Schwarzschild spacetime outside the horizon is static regardless of which coordinates you use to describe it.

 

[lukesfn]

And why do you think this? Did you just come up with it on your own, or did you read it somewhere, or did you deduce it from something you read somewhere? This is what I'm looking for when I ask for a reference: what you're saying is not obvious to me, so I need to know where it comes from.

Really? Sorry, I've lost my patience at this point.

If that's the case, how can light come off an area outside the black thing? Any point radially below you should be within the area of your visual field occupied by the black thing.

Are you serous? Sorry again, I think I will leave the puzzle to you work out the unnecessary assumptions you have made here.

I didn't say anything about coordinates. I said that, as I understand it, the argument requires a static region of spacetime. That's an invariant statement that has nothing to do with coordinates; Schwarzschild spacetime outside the horizon is static regardless of which coordinates you use to describe it.

I knew I would get picked up on this for my sloppy use of terminology on this one. I am not going to go and re check all the terminology definitions right now. However, the fact that an unrah observer can not be real inside the horizon, does not stop the maths from working beautifully. Infact, the symmetry of the imaginary numbers works our beautifully. I just don't see the relevance of these arguments the math. It doesn't stop the math from working and to me would seem ad hoc to change it to work the way you think it should.

However, I have bias since I am using it to try to show something that seemed obvious to me in the first place without the maths.

If you want to close this thread for being overly speculative, please do, I am wasting too much time, and have not enough patience left to explain minor points that I had incorrectly assumed would be very simply understood without needing to get pedantic.

Probably if we where face to face these things would be cleared up quickly... but...

If your curious about this topic, I am sure you can all explore it all much better your self without my in adequate explanations.

Thank you for everybody's time though. Even through my frustration here, I have been pushed to learn't much.

 

[PAllen]

If the paper is correct, an in-falling observer sees hawking radiation at the EH, but not inside the EV, then the equivalence theorem would be in trouble since it implies that hawking radiation emanates from the Horizon from a local point of view. Therefor an equivalence principle lab could do an experiment to detect the location of the EV. Perhaps QFT allows this some how, but it would still leave the equivalence principal in trouble. It isn't clear that you can neglect the sudden change in radiation.

I think that the paper is correct and I believe the Rindler Horizon analysis for the hawking radiation as you described is flawed. The problem I see is that hawking radiation generally originates from a location out side the lab. I think it is a mistake to view it as a local effect.

If hawking radiation comes from EH, and you are at a great distance from it, then the equivalence principle can't say anything about that radiation, it is all very external.

In the case an equivalence principal lab falling through the EH:
a) Before the horizon the lab can't say anything about the hawking radiation emanating from the horizon because nothing happens internally.
b) Falling through the horizon, the lab falls through the hawking radiation already at the horizon, so it doesn't see anything happen internally.
c) Falling after inside the horizon, the lab catches up to falling radiation in front of it, but again, nothing happens internally.

The can only show that faller at the horizon will not see hawking radiation coming from the horizon.

If non of this is true, and all free fallers see no hawking radiation, I would love an explanation of the following case:
A distant radiationally stationary observer sees the Hawking radiation coming from the BH, then the observer is dropped, then after time of seeing no radiation, when the observer reaches a smaller radius, it is held stationary again and starts seeing radiation. I the time of free fall, does the BH loose mass due to hawking radiation? If it does, how come the observer didn't see any? Where did the energy go?

First we need to agree on what the EP is. Here are two definitions:

1) (Modern). See, for example, http://relativity.livingreviews.org/Articles/lrr-2006-3/fulltext.html , sections 2.1 and 3.1.

2) Einstein's historic formulation: physics inside an accelerating rocket 'far from all matter' is locally identical to physics in a lab sitting (or maintaining static world line) on (near) a gravitating body (to the extent the tidal effects are minimal).

[There are arguments to show that these are essentially equivalent.]

I brought up a common way the EP is used to discuss physics near (either side of) a supermassive BH horizon - it should be locally identical to physics anywhere else; and (because of low curvature), the BH horizon should be (locally) equivalent to a Rindler Horizon. This is certainly true classically. I brought up that many authors have used this to motivate why Hawking radiation is not seen by free fall observers - that it should be locally equivalent to Unruh radiation. I did not mean to imply I thought this view was necessarily correct - just that it is common, and suggests (assuming the paper you linked is correct) that the EP is an unreliable guide to predictions about Hawking radiation.

Separate from this, I asked you how you were using the EP. It seems you think the EP prohibits local detectability of the event horizon. This does not follow from either variant I posted above. Once you admit that there is specific physics associated with the horizon (i.e. that it is not locally just like a Rindler horizon), the EP no longer tells you anything about what a specific observer sees. All it can do for you is say if one observer sees x, then we can predict what a nearby observer in some state of motion will see, by local lorentz invariance. If a specific theory makes the EH locally detectable (e.g. the firewall hypothesis), the EP is in no way in conflict with that, any more than the local detectability of the Earth's' surface is a problem for EP.

---

As to your last paragraph/question I've quote:

- I hope I have clarified that I don't necessarily believe the 'free faller sees no Hawking radiation' argument

- Even so, if I did insist the 'free faller sees no radiation', this argument is easy to answer. The EP is a local statement. You can't integrate over dispersed collection of free fallers and draw any valid conclusions - this is a severe misuse of the principle. This would be valid only if there were global inertial frames.