Well, the theories definitely make meaningful predictions. Most people just don't trust them.nismaratwork said:Let me be more precise: the math holds, but the theories fail to make meaningful predictions... better?
Chalnoth said:Well, the theories definitely make meaningful predictions. Most people just don't trust them.
Chalnoth said:Actually, that's not the case. There is no orbit just above the event horizon. The smallest possible orbits are unstable, and some distance from the event horizon. I forget the exact numbers, but I seem to remember that photons orbit a non-rotating black hole at ~4/3 the Schwarzschild radius or somewhere thereabouts. Obviously matter would have to be further out.
Well, yes, there is definitely a semantic issue here.nismaratwork said:I'm one of those people, and for a prediction to be meaningful it has to be testable. I think we may be arguing semantics here...
Chalnoth said:Well, yes, there is definitely a semantic issue here.
What I would suggest, first, is that it does not automatically follow that extending a theory outside of observable limits will necessarily be unreliable. We could, in principle, show that certain aspects of the theory outside of our observable limits, for instance, can be directly tied to things within our observable limits, making it very strange for things to vary just beyond. For a very simple case, we do not expect the assumption of homogeneity to break down close to the edge of our observable universe, but extend for quite some distance beyond it, because it would be difficult to conceive of a theory that broke that symmetry in some significant way that is also completely unobserved within the visible region. We can't expect the assumption of homogeneity to extend forever, but we can expect it to extend for some significant distance beyond the observable universe.
But there are independent reasons to not trust General Relativity immediately inside the event horizon. Here I pose three different points:
1. Hawking Radiation ensures that black holes always exist for a finite amount of time for an external observer.
2. An external observer will never see anything actually pass the event horizon of a black hole (the proper time coordinate of an infalling observer past the horizon is identified with times beyond positive infinity for an external observer). This may indicate that for a real infalling observer, the black hole will evaporate before the infalling observer ever enters the horizon (caveat: this is an unsolved problem in GR. We know of some special cases where this does not hold. For instance, an evaporating black hole that extends infinitely into the past certainly does have some infalling observers reaching the singularity. But we don't know, at present, what this means for real, astrophysical black holes.)
3. We know that information is conserved in the formation and destruction of a black hole, indicating that the information about what falls into a black hole is somehow encoded in the Hawking radiation that comes from the horizon.
These three points, to me, seem to indicate that something very strange is going on at the event horizon of a black hole that we just do not understand, and quantum gravity is likely to have quite a lot to say to the behavior of a black hole right at the event horizon. I do not think this behavior is in principle unknowable. Just that at present our knowledge is insufficient to be even reasonably confident of any inferences we might make about it, as we can be reasonably confident that the assumption of homogeneity extends some distance beyond our visible universe.
Well, matter always travels at lower than the speed of light, so clearly orbits for matter must be beyond the photon sphere. The photon sphere is, after all, just the orbit of matter in the high-energy limit, and it corresponds to unstable orbits. Any lower-energy orbit will be necessarily further away.ClamShell said:Why is that? Are you in fact referring to the unstable orbits that
slow(and stop) BH spin? I like the unstable orbits better than the
stable orbits if we are still referring to non-rotating BH's. Unstable
orbits carry away the energy that is spinning a BH...another reason
for non-rotating BH's to become more common. That stable orbits
for matter are further out than the photon sphere is not clear to me.
Chalnoth said:Well, matter always travels at lower than the speed of light, so clearly orbits for matter must be beyond the photon sphere. The photon sphere is, after all, just the orbit of matter in the high-energy limit, and it corresponds to unstable orbits. Any lower-energy orbit will be necessarily further away.
As for slowing down the BH spin, well, I wasn't referring to that issue at all. It was my understanding that the primary slowdown of BH spin comes from the generation of relativistic jets due to matter entering the ergosphere and being expelled out the poles.
ClamShell said:Anybody got concrete facts here.
Anyway, 212000 km/s may still be a property
of unstable orbits near(but outside) the event horizon...what we do "know"
is that V_escape >= c, for photons inside the event horizon trying to get out.
Or we don't...since they are behind the event horizon and cannot be observed...
and therefore cannot be described via maths. Remember, the whole point is
not to rule-out the wormhole via untested "pseudo-concepts".
Well, I think we'd need to delve into the math to be certain, which I haven't done in quite a while, but just bear in mind that any matter will gravitate like a photon as long as its kinetic energy is much greater than its mass energy. The orbital radius of a particle must therefore be a continuous function of velocity which limits to the photon sphere at v=c.ClamShell said:Photons are deflected more than matter(twice?) in a spherical g field...therefore
photons need to "orbit" further away or be spiraled into the BH. Does that make
sense? Or is my dyslexia acting up.
Bear in mind that this is the case with gravitational orbits as well, provided you include both potential and kinetic energy.ClamShell said:It was non-intuitive for Bohr to say that
electron orbits are lower in energy when closer to the nucleus, worked for him.
Since anything that escapes the black hole must first cross the event horizon, and since the escape velocity is the speed of light at the event horizon, nothing can escape a black hole, even without worrying about whatever exotic things might or might not be happening inside.ClamShell said:Or we don't...since they are behind the event horizon and cannot be observed...
and therefore cannot described via maths. Remember, the whole point is
not to rule-out the wormhole via untested "pseudo-concepts".
Chalnoth said:Well, I think we'd need to delve into the math to be certain, which I haven't done in quite a while, but just bear in mind that any matter will gravitate like a photon as long as its kinetic energy is much greater than its mass energy. The orbital radius of a particle must therefore be a continuous function of velocity which limits to the photon sphere at v=c.
Must this function be monotonic? I believe so, but I confess I'm not absolutely certain. And my Google-fu is failing me in finding a more authoritative source at the moment.
Bear in mind that this is the case with gravitational orbits as well, provided you include both potential and kinetic energy.
Since anything that escapes the black hole must first cross the event horizon, and since the escape velocity is the speed of light at the event horizon, nothing can escape a black hole, even without worrying about whatever exotic things might or might not be happening inside.
Unless you count Hawking radiation. But since Hawking radiation is completely thermalized, well, this implies that escape out of a black hole involves utter destruction.
Well, as I alluded to a bit earlier, the information about what went into the black hole is actually encoded in the Hawking radiation that leaves it, so in a sense, what goes in must come out. It may not literally be particle-for-particle, of course, but somehow the precise quantum-mechanical description of a black hole must allow at least the information regarding the matter that formed the black hole to leave as Hawking radiation.nismaratwork said:I would add that HR doesn't really "escape" the event horizon, but rather the portion that seems to escape was part of pair creation outside of the horizon. This is of course why HR isn't superluminal either, and its origin in a quantum process unrelated to the original infalling matter is precisely why it carries to information about that original matter with it. HR is essentially quantum static originating just beyond the EH, with a component that is left within the EH which we cannot observe.
Your statement that the EH is still an absolute limit for escape of matter, or anything at c still holds.
Chalnoth said:Well, as I alluded to a bit earlier, the information about what went into the black hole is actually encoded in the Hawking radiation that leaves it, so in a sense, what goes in must come out. It may not literally be particle-for-particle, of course, but somehow the precise quantum-mechanical description of a black hole must allow at least the information regarding the matter that formed the black hole to leave as Hawking radiation.
ClamShell said:Finally, somebody likes Entropy_in = Entropy_out...a BH, wormH, WH construct
satisfies this without a "pasta" machine model. I suggest that what goes in is
exactly what comes out. IE, if Alice goes in, Alice comes out...not Bob.
Well, that is a possibility, but I'm actually rather skeptical. We do know that the matter coming out of a black hole has a thermal spectrum, after all. Among other things, that means that you get a different distribution of particle types coming out than went in. So I don't think it's quite as simple as the same thing coming out as went in, but rather that what comes out is a unitary evolution of what went in (meaning that if one had a hypothetical perfect computer and new the underlying physics perfectly, one could calculate every bit of Hawking Radiation coming out of a black hole if one knows everything that falls into it).ClamShell said:Finally, somebody likes Entropy_in = Entropy_out...a BH, wormH, WH construct
satisfies this without a "pasta" machine model. I suggest that what goes in is
exactly what comes out. IE, if Alice goes in, Alice comes out...not Bob.
A white hole is just the time reverse of a black hole. This means that white holes are basically a contradiction in terms, because they indicate a system that decreases in entropy as time goes forward, which contradicts what we mean by time going forward.jarednjames said:On a more serious note, if a black hole gives out what it takes in (entropy_in = entropy_out), does this mean that white holes aren't require to 'balance' things?
jarednjames said:But if Jared goes in past the event horizon, Jared won't be coming out. I may reappear a bit later as a burst of radiation of some form. But I don't think I'd be picking up with my life any time soon.
On a more serious note, if a black hole gives out what it takes in (entropy_in = entropy_out), does this mean that white holes aren't require to 'balance' things?
ClamShell said:I'm suggesting that the BH horizon is the input node, the wormH is the communication
link, and the WH horizon is the output node. IE, the BH, wormH, WH is a single construct
of which we can only see the BH event horizon...that there is a stage behind the curtain.
ClamShell said:I'm suggesting that the BH horizon is the input node, the wormH is the communication
link, and the WH horizon is the output node. IE, the BH, wormH, WH is a single construct
of which we can only see the BH event horizon...that there is a stage behind the curtain.
skeptic2 said:Why do we assume that objects inside the event horizon do not travel faster than c?
That's a circular argument.ClamShell said:Perhaps the simple answer is that if they went faster than c,
they would(could) escape.
Photons always travel at c so there's no real argument about photons exceeding c. Is there any reason why massive particles cannot travel faster than c inside the EV?ClamShell said:Or perhaps it is closely tied to V_escape = root(GM/R)*root(2) that can result in V_escape > c, for R < R_eh, and the photons have
petered-out to c when they get to the event horizon. IE, a misconception that escape velocity > c implies anything more than a more difficult job for photons going c to get out.
skeptic2 said:Photons always travel at c so there's no real argument about photons exceeding c. Is there any reason why massive particles cannot travel faster than c inside the EV?
ClamShell said:OK, for particles of matter:
Or perhaps it is closely tied to V_escape = root(GM/R)*root(2) that can result in V_escape > c, for R < R_eh, a misconception that escape velocity > c implies anything more than a more difficult job for matter to get out.
My personal favorite is that the hypothetical structure on the other side
of the horizon is an "independent coordinate system" and obeys
the very same rules that we have here. Circular again?
Or would you prefer the concept that the Schwarzschild Metric
really represents two "independent coordinate systems" with
different rules?
nismaratwork said:Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?
nismaratwork said:Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?
nismaratwork said:Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?
ClamShell said:Dense mass without a horizon seems to be like a brick wall to falling matter.
Dense mass with a horizon seems to be like a curtained stage with the brick
wall there or not there or both or neither. IE, if we linger long enough just
above the horizon, by the time we finally cross the horizon, the brick wall
will have evaporated.
skeptic2 said:I understand that when matter reaches the singularity it may no longer be matter but it's no longer traveling either. Between the EV and the singularity, how fast can massive particles travel and why? Is there some prohibition against exceeding c in that region?
nismaratwork said:There's no wall, especially since you have to remember that everything falling into a black hole is ripped apart by gravitational tidal forces, and blasted by radiation. The EH has no physical existence, it's just the ever-changing (as long as there is infalling matter or HR) demarcation of the point of no return. Degenerate matter is dense, yes, but even that would be "sphagettified" as it fell into a BH. In a very real sense, anything on OUR side of the EH can never be observed by us to cross the EH, so there is the theoretical notion of a wall. Keep in mind that the infalling mass will not experience any such barrier, and crosses the EH without any resistance. I believe that your understanding of Einstein's view of gravity and spacetime is incomplete, and to grasp just what a black hole is, you need to understand that first.
nismaratwork said:Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation.
skeptic2 said:I understand that when matter reaches the singularity it may no longer be matter but it's no longer traveling either. Between the EV and the singularity, how fast can massive particles travel and why? Is there some prohibition against exceeding c in that region?
nismaratwork said:There's no wall, especially since you have to remember that everything falling into a black hole is ripped apart by gravitational tidal forces, and blasted by radiation.
These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd to0 conclude that it acts as physical barrier to the falling observer.
The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuum polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ...
ClamShell said:It is accepted by previous posters, that distant observers will never see a
a test mass cross the horizon. I take this to mean that when it does finally
happen(relative to the test mass), the stage and its contents will have evaporated.
Supposedly by Hawking radiation. And that a distant observer does not have a
long enough duration to observe this. But the test mass(by its own clock)
would experience nothing in particular because (after infinity by distant
observers clocks), the BH will have evaporated. A no show.
George Jones said:Consider two observers, observer A that falls across the the event horizon and observer B that hovers at a finite "distance" above the event horizon, and two types of (uncharged) spherical black holes, a classical black hole that doesn't emit Hawking radiation and a semi-classical black hole that does.
For the classical black hole case, B "sees" A on the event horizon at infinite future time, and B never sees the singularity.
For the semi-classical black hole case, at some *finite* time B simultaneously "sees": A on the event horizon; the singularity. In other words, the singularity becomes naked, and A winks out of existence at some finite time in the future for B.
In both cases, A crosses the event horizon, remains inside the event horizon, and hits the singularity. In both cases, B, does not see (even at infinite future time) A inside the event horizon, as this view is blocked by the singularity.