black hole


by keepit
Tags: black, hole
?????
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#37
Sep18-11, 05:10 PM
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Quote Quote by Mordred View Post
The time slowing/stopping is from the perspective of the outside observer.
An insightful comment. But the question is: How does an outside observer know what rate time is proceeding at on a black hole (in the immediate vicinity of a black hole, well inside the Schwarzschild Radius)?
PeterDonis
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Sep18-11, 08:16 PM
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Quote Quote by zonde View Post
Why not? There just appears another energy in equations. Of course this additional energy will change energy/radius relationship.
The virial theorem, at least in the form you quoted it, requires all the particles to be moving on geodesics. That's not true in a fluid with non-zero pressure in hydrostatic equilibrium. The additional energy doesn't just change the energy/radius relationship; it changes the kinetic/potential energy relationship.

Quote Quote by zonde View Post
Increase in gravity just changes amount of energy that is gained from gravitational collapse.
Or, conversely, the amount of energy that must be expended to expand against gravity. Also, by "energy" here I believe you mean specifically "energy that is converted from gravitational potential energy into some other form" (or vice versa for the case of expansion instead of contraction), as distinct from "energy that comes into the system from outside".

Quote Quote by zonde View Post
So the question is how much of that energy can be stored in other forms (ordinary and degeneracy pressure) for a given radius. If less energy can be stored in pressure then we have energy excess if it is the other way around then we have energy deficit. So to get runaway process of collapse we should be able to store in "pressure" energy exactly the same amount that we get by going lower in gravitational potential.
I'm not sure I quite follow your reasoning here. I have several questions/issues:

First, we're not necessarily talking about a runaway collapse, just that when the body reaches a new static equilibrium after some energy has been added from an external source, the body's radius, if its mass is high enough (ten to a hundred times the mass of Jupiter, or more), will be smaller, not larger. So we're comparing static equilibrium configurations with slightly different total energy.

Second, to know whether a given configuration is in static equilibrium, we need the equation of state of the matter in the body. My point was that the equation of state is different for degeneracy pressure than for normal kinetic pressure; the dependence of pressure on temperature is much weaker if the pressure is degeneracy pressure than if it is kinetic pressure. That means that raising the temperature doesn't increase the pressure much, if at all, when the pressure is degeneracy pressure; but it also means that degeneracy pressure can increase a lot without increasing the temperature much, if at all.

Third, I'm not sure what you mean by "storing energy in pressure". Pressure has the units of energy density, but that doesn't mean it *is* energy density. Pressure happens to be proportional to energy density in an ideal gas, as we saw earlier, but that doesn't mean it "counts" as energy density. In terms of the stress-energy tensor, pressure is the diagonal space-space component, while energy density is the time-time component; they are physically distinct. And of course, the ideal gas equation of state is a very special case; for other equations of state, pressure may not even be proportional to energy density. If the pressure is degeneracy pressure, as I said above, the pressure will be only weakly, if at all, dependent on the temperature (hence energy density).
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#39
Sep19-11, 05:36 PM
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Quote Quote by keepit View Post
a black hole is first large object until it has enough mass to make it inescapable. Once the black hole accumulates the amount of mass to make it inescapable, doesn't time stop down for that mass that keeps accumulating after that point?
Such a simple question, such complicated answers. keepit - some people have hijacked your thread to engage in an apparent ongoing spat. But, I think your question is quite inspired. That is why it is so hard to answer. May I offer my own version of your question, and you tell me if I'm wrong. What you are really asking is "What does it mean to say that time stops?"
We are all familiar with the experiments that verify that time slows down. But the logical extreme of that is "time stops". If there is a region of space where time stops, what is the implication of that? And what does it mean if more mass packs on top of mass where time has stopped? If time has stopped on one particular chunk of mass, how can other mass pack on top of it? Packing means that time is moving, not stopped. A philosophical conundrum, yes?

Of course, I could be wrong. Is this what you are trying to get at?
PeterDonis
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Sep19-11, 09:58 PM
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Quote Quote by ????? View Post
If there is a region of space where time stops, what is the implication of that?
A black hole is not "a region of space where time stops". To observers far away from the hole, it appears that time is slowed down for objects close to the hole (observers outside the hole can't see anything inside it). But that apparent slowdown of time is an illusion. Objects falling into the hole do not see time stop; as Mordred pointed out in post #12, such objects see time flowing normally. After a finite time by their own clocks such objects will hit the singularity at r = 0, and classical GR predicts that they will be destroyed by infinite spacetime curvature at that point. But until then they experience normal time flow.
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Sep20-11, 05:50 AM
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Quote Quote by PeterDonis View Post
The virial theorem, at least in the form you quoted it, requires all the particles to be moving on geodesics. That's not true in a fluid with non-zero pressure in hydrostatic equilibrium. The additional energy doesn't just change the energy/radius relationship; it changes the kinetic/potential energy relationship.
I am afraid I don't follow you. Viral theorem does not describe zero pressure situation. As long as there is kinetic energy there is non-zero pressure.

Quote Quote by PeterDonis View Post
Or, conversely, the amount of energy that must be expended to expand against gravity. Also, by "energy" here I believe you mean specifically "energy that is converted from gravitational potential energy into some other form" (or vice versa for the case of expansion instead of contraction), as distinct from "energy that comes into the system from outside".
Yes

Quote Quote by PeterDonis View Post
I'm not sure I quite follow your reasoning here. I have several questions/issues:

First, we're not necessarily talking about a runaway collapse, just that when the body reaches a new static equilibrium after some energy has been added from an external source, the body's radius, if its mass is high enough (ten to a hundred times the mass of Jupiter, or more), will be smaller, not larger. So we're comparing static equilibrium configurations with slightly different total energy.
Well, runaway collapse probably came in because of my confusion. You can ignore it.

But argument about reduction in radius was different. It was about change in mass of object not about change in total energy of system.
As I understand that argument it assumes that system is at 0 temperature and there is only degeneracy pressure present and no kinetic pressure. So there is certain level of total energy when equilibrium is reached (all extra energy is radiated away) and consequently certain radius for that state.

Quote Quote by PeterDonis View Post
Second, to know whether a given configuration is in static equilibrium, we need the equation of state of the matter in the body. My point was that the equation of state is different for degeneracy pressure than for normal kinetic pressure; the dependence of pressure on temperature is much weaker if the pressure is degeneracy pressure than if it is kinetic pressure. That means that raising the temperature doesn't increase the pressure much, if at all, when the pressure is degeneracy pressure; but it also means that degeneracy pressure can increase a lot without increasing the temperature much, if at all.
Yes with one comment.
Degeneracy pressure does not depend from temperature at all. If you increase temperature of degenerate matter there appears non-zero kinetic pressure but it's contribution to summary pressure is rather small. That's the reason why pressure of degenerate matter depends very little from temperature (not because degeneracy pressure depends weakly from temperature).

Quote Quote by PeterDonis View Post
Third, I'm not sure what you mean by "storing energy in pressure". Pressure has the units of energy density, but that doesn't mean it *is* energy density. Pressure happens to be proportional to energy density in an ideal gas, as we saw earlier, but that doesn't mean it "counts" as energy density. In terms of the stress-energy tensor, pressure is the diagonal space-space component, while energy density is the time-time component; they are physically distinct. And of course, the ideal gas equation of state is a very special case; for other equations of state, pressure may not even be proportional to energy density. If the pressure is degeneracy pressure, as I said above, the pressure will be only weakly, if at all, dependent on the temperature (hence energy density).
When you compress body against pressure you perform work on the system and that is described as energy transfer to the system you are compressing.
My formulation probably was not very clear. Maybe "storing energy in compression" is more correct.
?????
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#42
Sep20-11, 07:01 AM
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Quote Quote by PeterDonis View Post
A black hole is not "a region of space where time stops". To observers far away from the hole, it appears that time is slowed down for objects close to the hole (observers outside the hole can't see anything inside it). But that apparent slowdown of time is an illusion. Objects falling into the hole do not see time stop; as Mordred pointed out in post #12, such objects see time flowing normally. After a finite time by their own clocks such objects will hit the singularity at r = 0, and classical GR predicts that they will be destroyed by infinite spacetime curvature at that point. But until then they experience normal time flow.
Good comment. Isn't that what I said?
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#43
Sep20-11, 01:31 PM
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Quote Quote by zonde View Post
I am afraid I don't follow you. Viral theorem does not describe zero pressure situation. As long as there is kinetic energy there is non-zero pressure.
Not necessarily. This may be more a matter of terminology than physics, but it is perfectly possible to have a system of particles that have non-zero kinetic energy, but which do not interact except that they all move in the potential of their combined gravitational field. This is the kind of system to which the formulation of the virial theorem that you gave (where the kinetic energy is minus one-half the potential energy) applies; the key is, as I said before, that if the only force acting is gravity, all the particles move on geodesics. But a fluid with a non-zero pressure is not this type of system; there are other forces than gravity present (the internal forces that cause the non-zero pressure), and they change the relationship between kinetic and potential energy.

For example, consider an "average" particle in the Earth's atmosphere, compared with a small test object in a free-fall orbit about the Earth at the same altitude. (We'll ignore the fact that an object in orbit inside the atmosphere would experience drag and would not stay in orbit; if you like, we can consider the second particle to be in orbit about a "twin Earth" that has no atmosphere but is otherwise identical to Earth.) The object in the free-fall orbit obeys the virial theorem in the simple form you stated it: its kinetic energy is minus one-half its potential energy. But the particle in the atmosphere does not; its kinetic energy is much *less* than minus one-half its potential energy, because its potential energy is the same (the altitude is the same), but its kinetic energy is just the temperature of the atmosphere in energy units, which is much smaller than the equivalent "temperature" of a particle in orbit. Put another way, the average velocity of a particle in the atmosphere is much *less* than the orbital velocity at the same altitude. So the virial theorem in the simple form you gave does not apply to a fluid with non-zero pressure.

Quote Quote by zonde View Post
But argument about reduction in radius was different. It was about change in mass of object not about change in total energy of system.
The mass of the object *is* the total energy of the system; at least, it is if mass is defined in a consistent way (as the externally measured mass, obtained by putting objects in orbit about the body at a large distance, measuring the distance and the orbital period, and applying Kepler's Third Law).

Quote Quote by zonde View Post
As I understand that argument it assumes that system is at 0 temperature and there is only degeneracy pressure present and no kinetic pressure. So there is certain level of total energy when equilibrium is reached (all extra energy is radiated away) and consequently certain radius for that state.
As I understand it, the intent of the calculations is to study the static equilibrium states as a function of total energy (which is the same as the externally measured mass, see above), without specifying exactly *how* the system goes from one static equilibrium state at a given total energy, to another at a slightly different total energy. The key is that the total energy is specified "at infinity" (which in practice means "as measured far away", using the procedure I described above). I don't think any specific assumptions are made about how the total energy of the system is partitioned internally--how much of it is rest mass of the parts, how much is internal kinetic energy, i.e., temperature, and so on.

Quote Quote by zonde View Post
Degeneracy pressure does not depend from temperature at all. If you increase temperature of degenerate matter there appears non-zero kinetic pressure but it's contribution to summary pressure is rather small. That's the reason why pressure of degenerate matter depends very little from temperature (not because degeneracy pressure depends weakly from temperature).
I think I agree with this, but I would have to do more digging into the details of how degeneracy pressure works to be sure. For the purposes of this discussion I'm fine with taking the above as correct.

Quote Quote by zonde View Post
When you compress body against pressure you perform work on the system and that is described as energy transfer to the system you are compressing.
As I noted above, I don't think the static equilibrium models go into this level of detail about how the system's total energy is partitioned internally. I understand what you're saying, but remember that, in relativity, energy and mass are different forms of the same thing; the energy that is added by compression, when the object is composed of degenerate matter, could just as well be taken up by nuclear reactions inside the object that effectively "store" that energy as the rest mass of newly created particles. That's basically what happens with neutron stars: the protons and electrons from atoms are forced by pressure to combine into neutrons, and the rest mass of a neutron is *more* than the rest mass of a proton and electron combined, so effectively the energy from the compression is being stored as rest mass. All this can happen without changing the temperature at all. But, as I said, I don't think the models that are predicting the overall equilibrium states go to this level of detail; they just adopt an overall equation of state for the matter making up the object.
PeterDonis
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#44
Sep20-11, 01:32 PM
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Quote Quote by ????? View Post
Good comment. Isn't that what I said?
I'm not sure; it seemed like you were saying the opposite, that time *is* stopped inside a black hole, or at least saying it was possible.
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Sep20-11, 01:42 PM
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Quote Quote by PeterDonis View Post
I'm not sure; it seemed like you were saying the opposite, that time *is* stopped inside a black hole, or at least saying it was possible.
Actually time does stop at the singularity as all geodesics simply end there.
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Sep20-11, 02:22 PM
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Quote Quote by Passionflower View Post
Actually time does stop at the singularity as all geodesics simply end there.
Yes, that's true. But "at the singularity" is not at all the same as "anywhere inside the black hole", which is what I read the OP and the commenter I was responding to to be claiming.
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#47
Sep20-11, 03:48 PM
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Quote Quote by PeterDonis View Post
Yes, that's true. But "at the singularity" is not at all the same as "anywhere inside the black hole", which is what I read the OP and the commenter I was responding to to be claiming.
Yes, passed the event horizon time for an observer goes on just fine, it is only at the singularity where it stops.
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#48
Sep20-11, 05:01 PM
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Quote Quote by PeterDonis View Post
I'm not sure; it seemed like you were saying the opposite, that time *is* stopped inside a black hole, or at least saying it was possible.
I asked "What does it mean to say that time has stopped?" I was asking keepit if this is what he was asking, since he was the original poster. Your response "classical GR predicts that they will be destroyed by infinite spacetime curvature at that point." I took this to me that you agreed that the notion that "time stops" somewhere in the vicinity of a black hole is an enigmatic term. I could have been more precise by saying that mass packing onto a black hole must first past through the Schwarzschild Radius, where presumably time stops. I was thinking of the case where the SR is in the vicinity of the surface on the mass. This is not the only possible case. I am still curious about the concept of time stopping and what that means.

I'm starting to feel like I may be hijacking this thread. In any case, keepit seems to have lost interest.
A-wal
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#49
Sep30-11, 07:33 AM
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Quote Quote by PeterDonis View Post
I'm not sure; it seemed like you were saying the opposite, that time *is* stopped inside a black hole, or at least saying it was possible.
Quote Quote by Passionflower View Post
Actually time does stop at the singularity as all geodesics simply end there.
Quote Quote by PeterDonis View Post
Yes, that's true. But "at the singularity" is not at all the same as "anywhere inside the black hole", which is what I read the OP and the commenter I was responding to to be claiming.
Quote Quote by Passionflower View Post
Yes, passed the event horizon time for an observer goes on just fine, it is only at the singularity where it stops.
What are you basing that on? It seems to me that if to any external observer a free-falling object will never ever reach an event horizon then it is in fact completely impossible for a free-faller to ever become anything other than an external observer. What makes you think the free-fallers experience of the event could be any different to from the more distant external observers perception of the same event when the event in question is whether or not one of them can move away from an object or not? I didnít think GR dealt with alternate realities? What would happen if you were to move the singularity in the equation so that time stops and length infinitely contracts at the event horizon instead? Wouldnít it be interesting if it showed that gravity and acceleration really are the same thing and you could look at either as curved space-time or as energy in flat space-time. It would mean the universe doesnít have to stop making sense at a singularity and that gravitons work in relativity. It would also mean something else. Quite profound I think. I forget. Iím still writing the reply for the other thread. Could take a while, Iím doing it as and when I feel like it. If you think I still donít get something I wish someone would tell me what it is because I havenít had a real answer yet. How can the free-faller both cross and at the same time not cross the horizon? I still donít think it makes any sense. Did you miss me? :)
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Sep30-11, 08:04 AM
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Quote Quote by A-wal View Post
What are you basing that on? It seems to me that if to any external observer a free-falling object will never ever reach an event horizon then it is in fact completely impossible for a free-faller to ever become anything other than an external observer.
An external observer seeing observers "frozen" at the event horizon is a coordinate based result. Singularities at the event horizon are purely due to the bad coordinate chart. Map to a coordinate chart non - singular at r = 2M. Also, you can easily very for yourself that for an observer falling in radially starting from rest at r = infinity, [itex]\frac{d\tau }{dr} = (\frac{r}{2M})^{1/2}[/itex] so [tex]\Delta \tau = -\int_{r_{i}}^{2M}(\frac{r}{2M})^{1/2}dr = \frac{4M}{3}[(\frac{r}{2M})^{3/2}]^{r_{i}}_{2M}
[/tex] which is clearly finite. You can also verify this for observers who don't start from infinity and still find that it takes finite proper time to cross the event horizon.
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#51
Sep30-11, 10:48 AM
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Quote Quote by WannabeNewton View Post
An external observer seeing observers "frozen" at the event horizon is a coordinate based result.
It is a physical truth, it has nothing to do with coordinates.
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Sep30-11, 10:57 AM
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Quote Quote by Passionflower View Post
It is a physical truth, it has nothing to do with coordinates.
So you're telling me that [itex]\frac{dt}{d\tau } = e(1 - \frac{2M}{r})^{-1}[/itex] being infinite at r = 2M is NOT a problem dependent on the coordinate chart when clearly [itex]U^{t} = \frac{dt}{d\tau }[/itex] is a coordinate dependent quantity and the singularity at r = 2M can be transformed away by a coordinate transformation. What correspond to physically meaningful are coordinate invariant quantities such as proper time.
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Sep30-11, 11:09 AM
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Quote Quote by WannabeNewton View Post
So you're telling me that [itex]\frac{dt}{d\tau } = e(1 - \frac{2M}{r})^{-1}[/itex] being infinite at r = 2M is NOT a problem dependent on the coordinate chart when clearly [itex]U^{t} = \frac{dt}{d\tau }[/itex] is a coordinate dependent quantity and the singularity at r = 2M can be transformed away by a coordinate transformation. What correspond to physically meaningful are coordinate invariant quantities such as proper time.
All that passionflower is (correctly) saying is that both of the following are physical facts, not coordinate features:

1) A distant observer never sees (gets any indication from light or any source) anything pass the event horizon. They also see time freeze for matter approaching the event horizon.

2) A free falling body will cross the event horizon in finite proper time, reaching the singularity in finite proper time.
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Sep30-11, 11:20 AM
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Quote Quote by PAllen View Post
All that passionflower is (correctly) saying is that both of the following are physical facts, not coordinate features:

1) A distant observer never sees (gets any indication from light or any source) anything pass the event horizon. They also see time freeze for matter approaching the event horizon.
I do not deny anything you said but I don't see how one can conclude that the statement in the second sentence is physical; we are talking about coordinate velocity here.
2) A free falling body will cross the event horizon in finite proper time, reaching the singularity in finite proper time.
Yes of course that is unequivocal so no argument there.


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