Black hole matter accumulation

[zonde]

As I said before, the virial theorem does not apply to a fluid in hydrostatic equilibrium with a non-zero pressure.

Why not? There just appears another energy in equations. Of course this additional energy will change energy/radius relationship.

Also, as far as arguments go, your intuition that a body should expand when it gains energy depends on the body's pressure being kinetic (i.e., dependent on temperature). As I noted in a previous post, as degeneracy pressure becomes a larger and larger fraction of total pressure, the dependence of pressure on temperature gets weaker and weaker. Once degeneracy pressure is high enough, the increase in gravity caused by an increase in total energy (including the extra gravity caused by the increased pressure) compresses the body more than the increase in pressure and temperature expands it, so on net it contracts when it gains energy.

Increase in gravity just changes amount of energy that is gained from gravitational collapse. 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.

 

[?????]

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]

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.

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".

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).

 

[?????]

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]

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.

 

[zonde]

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.

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

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.

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).

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.

 

[?????]

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?

 

[PeterDonis]

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.

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).

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.

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.

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]

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.

 

[Passionflower]

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.

 

[PeterDonis]

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.

 

[Passionflower]

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.

 

[?????]

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]

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.

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.

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 didn’t 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? Wouldn’t 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 doesn’t 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. I’m still writing the reply for the other thread. Could take a while, I’m doing it as and when I feel like it. If you think I still don’t get something I wish someone would tell me what it is because I haven’t had a real answer yet. How can the free-faller both cross and at the same time not cross the horizon? I still don’t think it makes any sense. Did you miss me? :)

 

[WannabeNewton]

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, $\frac{d\tau }{dr} = (\frac{r}{2M})^{1/2}$ so $$\Delta \tau = -\int_{r_{i}}^{2M}(\frac{r}{2M})^{1/2}dr = \frac{4M}{3}[(\frac{r}{2M})^{3/2}]^{r_{i}}_{2M}$$ 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.

 

[Passionflower]

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.

 

[WannabeNewton]

It is a physical truth, it has nothing to do with coordinates.

So you're telling me that $\frac{dt}{d\tau } = e(1 - \frac{2M}{r})^{-1}$ being infinite at r = 2M is NOT a problem dependent on the coordinate chart when clearly $U^{t} = \frac{dt}{d\tau }$ 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.

 

[PAllen]

So you're telling me that $\frac{dt}{d\tau } = e(1 - \frac{2M}{r})^{-1}$ being infinite at r = 2M is NOT a problem dependent on the coordinate chart when clearly $U^{t} = \frac{dt}{d\tau }$ 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.

 

[WannabeNewton]

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.

 

[PAllen]

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.

Come again? I'm not talking about velocity here at all (coordinate or otherwise), but for example, emission of gamma rays from some radioactive infalling matter. The gamma rays get red shifted, and the rate of decay for the isotope goes down, approaching infinite redshift and no decay. This is physical observation by the outside observer. It has nothing to do with coordinates. That's what I thought was obvious by 'time freezing'. It is also true that motion will be observed to freeze (to the extent it can be observed with ever fainter, redder, signal), but I wasn't referring to that. Anyway, an observer taking pictures, measuring radiation, etc. - these are physical observations not coordinate effects. No matter what coordinates you use, you will predict the same observations by the distant observer.

This is a distinction between 'instruments at different places and different states of motion will make different measurements' and 'interpreting measurements (of e.g. light) from distant objects will require conventions'. The former describes invariant physical facts; the latter describes coordinate effects (conventions).

 

[Passionflower]

What correspond to physically meaningful are coordinate invariant quantities such as proper time.

Let me ask you these questions:

Do you think it is physical meaningful to measure radar pulses from another source?
What do you think would an observer far away from a BH measure about an observer who is approaching the EH while sending out radar pulses?
Do you think such a physical measurement is coordinate dependent and if so why?

 

[WannabeNewton]

I am not saying that the null rays won't take longer and longer to travel outwards. I am saying that $U^{t} = \frac{dt}{d\tau } \to \infty$ as $r\rightarrow 2M$ is coordinate based.

 

[PeterDonis]

I asked "What does it mean to say that time has stopped?"

Ok, that clarifies things somewhat. The answer to your question as you stated it is that it can mean a variety of things: just saying "time has stopped" by itself doesn't pin down what you're talking about sufficiently to give a precise answer. For one thing, "time" by itself is not a precise term in relativity (special or general); you have to specify whose time (i.e., which observer's proper time), or what coordinate system's time. See below.

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.

It is, but that isn't what I was saying when I said there is infinite spacetime curvature at the singularity. The singularity is *inside* the event horizon, not "somewhere in the vicinity of the black hole", by which you seem to mean "near the event horizon".

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.

No, it doesn't, and that is *not* what I was saying. See below.

This is not the only possible case. I am still curious about the concept of time stopping and what that means.

Okay, let's try to disentangle some things by describing everything in terms of actual physical observables:

(1) Suppose you are hovering at some constant radius r far away from a black hole. You watch someone else free-fall past you, towards the hole, and watch as they send light signals back to you while they fall. If they send you light signals at constant intervals of time by their own clock, you will see the light signals arrive farther and farther apart by your clock; if we imagine them sending out a last light signal at the instant they cross the hole's horizon, that signal will never reach you at all.

(2) Suppose now that you are the free-faller; you fall past someone hovering far away from the hole, and send light signals back towards them. Everything around you seems normal: your clock ticks away just as it always has, as far as you can tell, and the intervals between each light signal you send out remain the same to you. In fact, from local observations in your vicinity you can't even tell when you cross the horizon; you have to calculate that if you want to know when, by your clock, you can stop sending light signals (because they will never reach the hoverer once you cross the horizon). However, in a finite time by your clock after you cross the horizon, you will see the tidal forces in your vicinity rise rapidly, ultimately diverging to infinity as you reach the singularity, still in a finite time by your clock.

Some people describe what I described in (1) above by saying that, from your viewpoint far away from the hole, "time slows down" for the person free-falling as they get close to the hole, and that when they reach the horizon, "time stops" for them. However, as you can see from (2), that description is, at the very least, misleading, since it doesn't convey the fact that the free-faller sees his own time flowing normally, and it doesn't cover at all what the free-faller experiences after he crosses the horizon. Another way of saying this is that the "time" coordinate that is natural to the hoverer, far away from the hole, only covers a portion of the spacetime; it assigns a "time" value of "plus infinity" to the black hole's horizon, which means it simply can't deal with the region of spacetime inside the horizon.

One could also describe what I described in (2) above by saying that "time stops" for the free-faller when he hits the singularity. However, a better way to describe that would be to say that "spacetime stops"; the singularity is an "edge" of the spacetime, and worldlines that hit it simply stop there. Physically, this isn't really reasonable (another sign of this is that, as I noted above, spacetime curvature diverges to infinity at the singularity), but standard general relativity can't give us any help in fixing that: we need some other theory, such as a quantum gravity theory, to take over and tell us what actually happens when standard GR tells us that "spacetime curvature diverges to infinity".

 

[PeterDonis]

Hi, A-wal! Yes, I've missed you.

Btw, for everyone else, the other thread A-wal refers to is here:

https://www.physicsforums.com/showthread.php?t=337236&page=32

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.

I understand that it seems this way to you, but can you *prove* it? Show me an actual argument, starting from premises we all accept, that shows that this *has* to be the case. (Either here or in the other thread is fine with me, although since you are the OP in the other thread, it might be better there.)

How can the free-faller both cross and at the same time not cross the horizon?

Again, all you have shown is that the external observer does not *see* the free-faller cross the horizon. You have given no justification for claiming that that implies that the free-faller does not cross the horizon, period.

 

[DaveC426913]

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 didn’t think GR dealt with alternate realities?

It doesn't; it deals with the relative simultaneity of events. Two observers see the same event happen at different times.

To one observer (internal), crossing the event horizon happens in a split second.
To another (external) observer it happens after the universe has grown old and died.Pretty standard relativity, applied in an unexpected way.

 

[Passionflower]

How can the free-faller both cross and at the same time not cross the horizon? I still don’t think it makes any sense. Did you miss me? :)

A free falling observer physically crosses the event horizon however no information can reach an outside observer about this. Outsiders simply cannot see and measure it.

 

[Dale]

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.

I think it depends on what you mean by "see".

Sometimes you simply mean what is happening in a particular coordinate system. In this case the frozen-ness is a coordinate dependent phenomenon, but it is observer independent. I.e. all observers will agree that the Schwarzschild coordinate time goes to infinity as the falling object approaches the horizon. I think this is the "see" meant by WannabeNewton.

Other times you mean literally what you visually see, in particular the redshift. In this case the frozen-ness is coordinate independent, but it is observer dependent. I.e. regardless of what coordinate system you use hovering observers will detect an infinite redshift while free-falling observers will not. I think this is the "see" meant by Passionflower.

I think the disagreement is semantic not substantive as I believe that both of you know enough to understand correctly.

 

[Dale]

I haven’t had a real answer yet

Not understanding an answer is not the same as not receiving one.

 

[Dale]

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, $\frac{d\tau }{dr} = (\frac{r}{2M})^{1/2}$ so $$\Delta \tau = -\int_{r_{i}}^{2M}(\frac{r}{2M})^{1/2}dr = \frac{4M}{3}[(\frac{r}{2M})^{3/2}]^{r_{i}}_{2M}$$ 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.

A-wal is anti-math. He won't even try to understand this.

 

[WannabeNewton]

Sometimes you simply mean what is happening in a particular coordinate system. In this case the frozen-ness is a coordinate dependent phenomenon, but it is observer independent. I.e. all observers will agree that the Schwarzschild coordinate time goes to infinity as the falling object approaches the horizon. I think this is the "see" meant by WannabeNewton.

Yeah I should have been more clear with my words. I was just talking about the mathematics in that the coordinate time going to infinity was a coordinate related quantity. I didn't mean to talk about what physically was being seen or not seen.

A-wal is anti-math. He won't even try to understand this.

My apologies hehe.

 

[Dale]

My apologies hehe.

No need to apologize, I thought the explanation was spot-on. Just don't expect A-wal to respond reasonably.

 

[A-wal]

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, $\frac{d\tau }{dr} = (\frac{r}{2M})^{1/2}$ so $$\Delta \tau = -\int_{r_{i}}^{2M}(\frac{r}{2M})^{1/2}dr = \frac{4M}{3}[(\frac{r}{2M})^{3/2}]^{r_{i}}_{2M}$$ 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.

They never cross if they can always escape.

I understand that it seems this way to you, but can you *prove* it? Show me an actual argument, starting from premises we all accept, that shows that this *has* to be the case. (Either here or in the other thread is fine with me, although since you are the OP in the other thread, it might be better there.)

Yes, that’s for the other thread. Here I want to concentrate strictly on how an external observers and internal observers experience of the same event are incompatible with each other in standard GR. At least they are as far as I can see.

Again, all you have shown is that the external observer does not *see* the free-faller cross the horizon. You have given no justification for claiming that that implies that the free-faller does not cross the horizon, period.

If they can always escape then they can never cross it.

It doesn't; it deals with the relative simultaneity of events. Two observers see the same event happen at different times.

To one observer (internal), crossing the event horizon happens in a split second.
To another (external) observer it happens after the universe has grown old and died.

This isn’t just relative simultaneity. This is a direct contradiction.

Pretty standard relativity, applied in an unexpected way.

Hold on, you’re saying nothing can reach an event horizon within the lifetime of the universe. I agree!

A free falling observer physically crosses the event horizon however no information can reach an outside observer about this. Outsiders simply cannot see and measure it.

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.

Which is it? Everything is to do with coordinates. The problem here is different coordinate systems that simply aren’t compatible with each other because they say different things.

Not understanding an answer is not the same as not receiving one.

Fine. I haven't had an answer I understand yet. Personally I think that's because I haven't been given an answer that makes sense yet.

A-wal is anti-math. He won't even try to understand this.

Because to me this is an art, not a science. It’s not a choice, it’s just that I visualise everything when I do this. That’s how I understand any aspect of it and I’m finding this aspect impossible to visualise. We got sidetracked in the other thread with all the different aspects of relativity, which is great. It helped me to get a much sharper broader picture, but I lost track of this part and this is the part that I find deeply wrong with standard GR. That doesn’t mean it is wrong, but it is right then I will understand why, and how it’s even possible.

No need to apologize, I thought the explanation was spot-on. Just don't expect A-wal to respond reasonably.

 

[Dale]

That’s how I understand any aspect of it and I’m finding this aspect impossible to visualise.

Largely because you refuse to even attempt to learn the math which expresses all of these concepts in a single coherent, logical, and rigorous framework. Your anti-math prejudice is preventing you from learning, so there is little anyone else is going to be able to do for you here.

WannabeNewton's explanation in Schwarzschild coordinates goes along with my earlier explanation to you for the same in Rindler coordinates. It is all there. But you won't even make an effort. Each time you receive an answer you close your eyes and mind and look away and go back to claiming that you have not received an answer.

Answer me this. How can you reasonably expect anyone to be able to present a logical picture to you if you refuse to look at the math which expresses the logic when it is presented to you?

 

[DaveC426913]

It doesn't; it deals with the relative simultaneity of events. Two observers see the same event happen at different times.

To one observer (internal), crossing the event horizon happens in a split second.
To another (external) observer it happens after the universe has grown old and died.

This isn’t just relative simultaneity. This is a direct contradiction.

Where exactly is the contradiction?

The same event happens, it just happens at two different times depending on the observer's FoR.

It's identical to a scenario in SR where one observer is traveling at, for all intents and purposes, c. While a split second passes for him before his dropped spoon hits the deck of his rocketship, to an external observer, the entire universe ages and dies before the spoon hits the deck.

Same event two different timescales.Explicitly, where is the contradiction?

Hold on, you’re saying nothing can reach an event horizon within the lifetime of the universe. I agree!

From one frame of reference, sure. From another it takes split second. Why is this so difficult?

 

[PeterDonis]

They never cross if they can always escape.

And can you prove that they can always escape, without *assuming* that they can? Every argument you've offered has had a hidden assumption that's equivalent to your conclusion.

Which is it? Everything is to do with coordinates. The problem here is different coordinate systems that simply aren’t compatible with each other because they say different things.

They say different things because they're *about* different things. One is about how a hovering observer far away from the hole assigns space and time coordinates to events. The other is about how an observer free-falling into the hole assigns space and time coordinates to events. Neither of them is saying anything that contradicts the other. The first simply can't assign coordinates to events inside the horizon, while the second can. That's not a contradiction, just a limitation of the first observer's coordinates.

I’m finding this aspect impossible to visualise.

Have you looked at a Kruskal chart or a Penrose diagram of Schwarzschild spacetime? Those are easy ways to visualize exactly what we've been saying. In fact, I've referred to them repeatedly in the other thread.