Black hole matter accumulation

[keepit]

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?

 

[Passionflower]

Whether something is a black hole not has nothing to do with the mass. The mass of a grain of salt could be a black hole as well.

Actually the amount of matter is not relevant, it is the ratio between area and mass that matters.

A non-spinning object becomes (or is already) a black hole if the ratio between the area it occupies and the area that represents its mass is smaller than 4. As soon as this happens the object's occupied area will shrink to zero.

E.g.:

<br /> {A_{occupied} \over A_{mass} } &lt; 4<br />

 

[Dale]

Passionflower, you may want to clearly define "area occupied" since most people will think in terms of "volume occupied".

 

[Passionflower]

Passionflower, you may want to clearly define "area occupied" since most people will think in terms of "volume occupied".

Actually I think it is simpler to express it in terms of area as volume is not Euclidean in curved spacetime and it avoids the usage of the r coordinate which is not a measure of distance in Schwarszschild coordinates.

For instance the volume occupied from the EH down to the singularity is actually infinite so that is not very helpful.

 

[Dale]

Actually I think it is simpler to express it in terms of area as volume is not Euclidean in curved spacetime and it avoids the usage of the r coordinate which is not a measure of distance in Schwarszschild coordinates.

I agree it is simpler, but you still should explain what it means for a mass to occupy an area just for clarity. Most new posters will not know what you mean.

 

[Rishavutkarsh]

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?

black hole and time stopping are far off . time stops relative to things outside it. And center of a black hole is a point of infinite density and zero volume . mass has to do with the size of black hole . it doesn't acquire mass it shrinks to zero volume making everything inescapable.

 

[WannabeNewton]

black hole and time stopping are far off . time stops relative to things outside it. And center of a black hole is a point of infinite density and zero volume . mass has to do with the size of black hole . it doesn't acquire mass it shrinks to zero volume making everything inescapable.

Kerr black holes have ring singularities which do not have zero volume.

 

[Dale]

Kerr black holes have ring singularities which do not have zero volume.

Are you sure about that? I thought that ring singularities were 1 dimensional. So they would have a circumference, but no volume. But I admit that I have not studied the Kerr solution in as much detail as I should have.

 

[WannabeNewton]

Are you sure about that? I thought that ring singularities were 1 dimensional. So they would have a circumference, but no volume. But I admit that I have not studied the Kerr solution in as much detail as I should have.

Yeah you're right I confused the two. Its basically the outer equator of a torus, so no volume.

 

[ZealScience]

But some black holes have less mass than those huge stars, you might have some misunderstanding.

And what do you mean by "time stop down for that mass"?

 

[keepit]

Zeal,
I didn't state my question very well. Rather than say "When a mass gets big enough" i should have said "when a mass qualifies as a black hole". sorry.
Anyhow i was just trying to understand better what is going on with the mass as it goes from not being a black hole to being a black hole. I guess time slows down gradually for the mass as it accumulates. Correct me if I'm wrong.
Thank you.

 

[Mordred]

The time slowing/stopping is from the perspective of the outside observer. For matter flowing into the singularlity time flows from its perspective as normal.

 

[ZealScience]

Zeal,
I didn't state my question very well. Rather than say "When a mass gets big enough" i should have said "when a mass qualifies as a black hole". sorry.
Anyhow i was just trying to understand better what is going on with the mass as it goes from not being a black hole to being a black hole. I guess time slows down gradually for the mass as it accumulates. Correct me if I'm wrong.
Thank you.

I think that going from normal stars to black holes is usually very rapid, which involves super nova explosion or collapse of giant stars. Using the Schwarzschild black hole as the simplest model, you can see that it is the radius that matters. To my perspective, if the radius of the event horizon is smaller than that of the star, it would not be a black hole. But usually collapse of stars and super nova explosion cause these rapid decrease of radius.

 

[zonde]

Gravitation potential becomes lower as we go toward center of mass. Because of that event horizon should first form at the center of gravitating body that is going to turn into the black hole and then move outward. I believe that is the way how birth of BH is modeled.

But at the center of gravitating body mass is not falling anywhere. So it should be extremely time dilated right before it turns into the black hole. So from where this "seed" black hole appears at the center of the body?

 

[stevebd1]

Zeal,
I didn't state my question very well. Rather than say "When a mass gets big enough" i should have said "when a mass qualifies as a black hole". sorry.
Anyhow i was just trying to understand better what is going on with the mass as it goes from not being a black hole to being a black hole. I guess time slows down gradually for the mass as it accumulates. Correct me if I'm wrong.
Thank you.

You might find the following post (and the thread) from https://www.physicsforums.com/showthread.php?t=473047" of interest (Note: M=Gm/c²)-

I think some insight can be gained from looking at the Schwarzschild interior metric. If we consider the time component only-

d\tau=\left( \frac{3}{2}\sqrt{1-\frac{2M}{r_0}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{r_0^{3}}}\right)dt

where r₀ is the radius of the spherical mass. If we consider a neutron star that has exceeded the TOV limit at 3 sol mass through accretion, r₀ will gradually reduce*. Considering M being a constant and r₀ reducing, when r₀ reaches 9M/4, d\tau at r=0 (i.e. the centre of the neutron star) becomes zero. As r₀ becomes less than 9M/4, a radius where d\tau=0 begins to move outwards from the centre, any volume of space within this radius is spacelike and because there are no stable radii in spacelike spacetime, this will induce the collapse of matter within to a form of singularity. This radius where d\tau=0 will continue to move outwards as r₀ reduces until they both meet at 2M and the interior solution becomes the vacuum solution.

*It's worth noting that in GR, pressure contributes to the stress energy tensor so the more compact a neutron star becomes, the greater the gravity.

 

[zonde]

If we consider a neutron star that has exceeded the TOV limit at 3 sol mass through accretion, r₀ will gradually reduce*.
*It's worth noting that in GR, pressure contributes to the stress energy tensor so the more compact a neutron star becomes, the greater the gravity.

And why exactly r₀ will be gradually reduced?
To me it seems the other way around - that r₀ should gradually increase. First, additional matter occupies some place and second, kinetic energy of accreted matter is converted into heat that would tend to expand the body.

 

[stevebd1]

And why exactly r₀ will be gradually reduced?
To me it seems the other way around - that r₀ should gradually increase. First, additional matter occupies some place and second, kinetic energy of accreted matter is converted into heat that would tend to expand the body.

The kinetic energy would convert into pressure (sometimes, pressure is described as confined KE) and Einstein's algebraic equation for gravity is g=\rho+3P where \rho is density and P is pressure and so the gravity would increase, pulling the star in on itself.

 

[zonde]

The kinetic energy would convert into pressure (sometimes, pressure is described as confined KE) and Einstein's algebraic equation for gravity is g=\rho+3P where \rho is density and P is pressure and so the gravity would increase, pulling the star in on itself.

Fine, confined KE will contribute to gravity by tiny bit so that expansion will be reduced by tiny bit but summary effect will be expansion not contraction.

 

[stevebd1]

Fine, confined KE will contribute to gravity by tiny bit so that expansion will be reduced by tiny bit but summary effect will be expansion not contraction.

Can you provide equations or proof that demonstrates this. The common census is that as a neutron star increases in mass, it's radius reduces.

 

[PeterDonis]

Fine, confined KE will contribute to gravity by tiny bit so that expansion will be reduced by tiny bit but summary effect will be expansion not contraction.

I'm having a discussion with zonde about a similar point in another thread; I posted this earlier today:

https://www.physicsforums.com/showpost.php?p=3497951&postcount=26

in which I reminded myself and zonde that a self-gravitating body has a negative heat capacity. That means that if you add energy to it, it expands and cools (i.e., its temperature goes *down*, not up), and if you take energy away from it, it contracts and heats up (i.e., its temperature goes *up*, not down). In other words, increasing the temperature of such a body does *not* create a tendency to expand against gravity.

Similar reasoning applies to the case under discussion here; the only difference is that, for an object above the maximum possible mass for a neutron star (the TOV limit), it will collapse even if it is not radiating away any energy to the outside universe. But that doesn't change the fact that the increasing temperature as it contracts does *not* compensate for the increased gravity that is pulling it in.

 

[PeterDonis]

Gravitation potential becomes lower as we go toward center of mass. Because of that event horizon should first form at the center of gravitating body that is going to turn into the black hole and then move outward. I believe that is the way how birth of BH is modeled.

This is correct as long as by "horizon" you mean "absolute horizon", i.e., the boundary of the region from which light signals cannot escape to infinity. However, this is *not* the same as an "apparent horizon", which is a "trapped surface" at which outgoing light signals no longer move outward. The distinction is important; see next comment.

But at the center of gravitating body mass is not falling anywhere. So it should be extremely time dilated right before it turns into the black hole. So from where this "seed" black hole appears at the center of the body?

When the absolute horizon first forms at the center of the gravitating body, there is no apparent horizon; outgoing light rays are still moving outward. That also means that there is no extreme time dilation, since the extreme time dilation near a black hole horizon is due to the presence of an apparent horizon (trapped surface) there, where outgoing light rays don't move outward. At the time when the absolute horizon forms, the time dilation at the center of the star (r = 0) doesn't change discontinuously at all. As the star collapses and the density at r = 0 increases, the "time dilation" there will gradually increase as well, but even that statement is subject to qualifications; see the end of this post.

Here's the way I picture the absolute horizon. Suppose that time t = 0, in the exterior Schwarzschild time coordinate that applies outside of the collapsing body, is the time at which the surface of the body is just passing inward through the Schwarzschild radius r = 2M. That is the time at which the apparent horizon forms, and the extreme time dilation occurs. Now consider a light ray emitted outward from r = 0 at some time prior to t = 0, such that that outgoing light ray just reaches r = 2M at t = 0. That light ray will then remain at r = 2M for all times greater than t = 0, since there is now a trapped surface there and outgoing light rays can no longer move outward. In fact, that light ray lies on the absolute horizon. But when the light ray first starts out at r = 0, it is moving outward, and observers inside the star who see it moving outward would not notice any unusual time dilation. The fact that that light ray will actually never get beyond r = 2M depends on the global structure of the entire spacetime; it's not something you can observe locally.

Regarding time dilation, it's also important to remember that time dilation is relative. An observer hovering near a black hole horizon doesn't notice any unusual time dilation locally; only by comparing his clock with that of someone far away (for example, by exchanging light signals) can he tell that much more time has passed in the faraway universe than has elapsed by his local clock.

Similarly, locally, observers inside the star after the absolute horizon has formed and has passed outward beyond their radius still would not notice anything unusual. Only by trying and failing to send light signals outward beyond the absolute horizon could they in principle tell that they were behind it. Eventually, of course, they would end up at the singularity at r = 0; either when the collapsing star passed inward through their radius and carried them with it, or, if they were able to escape the star's surface as it collapsed, eventually they would still hit the singularity since they are inside the horizon.

 

[zonde]

Can you provide equations or proof that demonstrates this. The common census is that as a neutron star increases in mass, it's radius reduces.

Do you agree that classical physics and in particular http://en.wikipedia.org/wiki/Virial_theorem" (thanks to PeterDonis for pointing that out) says that gravitating body has to expand as you add energy to the body?

It says that for gravitating body U_pot=-2U_kin. As a result adding more energy to the body increases it' s average potential energy (body expands) and decreases kinetic energy (it cools down on average).
U_{tot}+\Delta U=(U_{pot}+2\Delta U)+(U_{kin}-\Delta U)

If you agree with that then your statement becomes rather non trivial and I think that the burden of proof is entirely on you. It could mean that corrections to classical result under certain conditions would have to grow so large (in right direction) that they actually invert classical result.

 

[zonde]

This is correct as long as by "horizon" you mean "absolute horizon", i.e., the boundary of the region from which light signals cannot escape to infinity. However, this is *not* the same as an "apparent horizon", which is a "trapped surface" at which outgoing light signals no longer move outward. The distinction is important; see next comment.

When the absolute horizon first forms at the center of the gravitating body, there is no apparent horizon; outgoing light rays are still moving outward. That also means that there is no extreme time dilation, since the extreme time dilation near a black hole horizon is due to the presence of an apparent horizon (trapped surface) there, where outgoing light rays don't move outward. At the time when the absolute horizon forms, the time dilation at the center of the star (r = 0) doesn't change discontinuously at all. As the star collapses and the density at r = 0 increases, the "time dilation" there will gradually increase as well, but even that statement is subject to qualifications; see the end of this post.

Here's the way I picture the absolute horizon. Suppose that time t = 0, in the exterior Schwarzschild time coordinate that applies outside of the collapsing body, is the time at which the surface of the body is just passing inward through the Schwarzschild radius r = 2M. That is the time at which the apparent horizon forms, and the extreme time dilation occurs. Now consider a light ray emitted outward from r = 0 at some time prior to t = 0, such that that outgoing light ray just reaches r = 2M at t = 0. That light ray will then remain at r = 2M for all times greater than t = 0, since there is now a trapped surface there and outgoing light rays can no longer move outward. In fact, that light ray lies on the absolute horizon. But when the light ray first starts out at r = 0, it is moving outward, and observers inside the star who see it moving outward would not notice any unusual time dilation. The fact that that light ray will actually never get beyond r = 2M depends on the global structure of the entire spacetime; it's not something you can observe locally.

I am not sure I understood the difference between apparent and absolute horizon. In one case light that is going outwards can't reach infinity in other case light that is going outwards is not going outwards (obviously it can't reach infinity just the same).

Regarding time dilation, it's also important to remember that time dilation is relative. An observer hovering near a black hole horizon doesn't notice any unusual time dilation locally; only by comparing his clock with that of someone far away (for example, by exchanging light signals) can he tell that much more time has passed in the faraway universe than has elapsed by his local clock.

Yes of course time dilation is relative. But to find out what is time dilation at the center of gravitating body (let' s imagine that) we can observe light that is coming from the rest of the universe. From it' s blueshift we can determine our time dilation. So if we speak about time dilation at the center of gravitating body that is going to become black hole we can speak about blueshift of incoming radiation. So let me restate the question - would there be extreme blueshift of incoming radiation?

 

[PeterDonis]

I am not sure I understood the difference between apparent and absolute horizon. In one case light that is going outwards can't reach infinity in other case light that is going outwards is not going outwards (obviously it can't reach infinity just the same).

The apparent horizon is a local phenomenon; you just look at outgoing light beams in a small local region of spacetime and see if they are moving outward (increasing in radius).

The absolute horizon is global; you have to know the entire future path of light beams to know where the absolute horizon is. For example, you could be inside the absolute horizon and still not see an apparent horizon (as with the observers inside the star in my example, after the absolute horizon has passed them as it expands outward, before the star has collapsed inside r = 2M). That is, you could see outgoing light beams moving outward in your vicinity, but not be able to tell, at that time, that further in the future, those light beams will stop moving outward and be trapped when an apparent horizon forms in the future.

Yes of course time dilation is relative. But to find out what is time dilation at the center of gravitating body (let' s imagine that) we can observe light that is coming from the rest of the universe. From it' s blueshift we can determine our time dilation. So if we speak about time dilation at the center of gravitating body that is going to become black hole we can speak about blueshift of incoming radiation. So let me restate the question - would there be extreme blueshift of incoming radiation?

At the instant when the absolute horizon forms at r = 0 and begins expanding outward, no, the blueshift would not be "extreme"; more precisely, it would be the same at that point as it was before in the center of the star. As the star collapses, the blueshift at r = 0 becomes greater because the density at r = 0 increases, but the change is continuous and does not show any different behavior at the instant when the absolute horizon forms.

 

[PeterDonis]

Do you agree that classical physics and in particular http://en.wikipedia.org/wiki/Virial_theorem" (thanks to PeterDonis for pointing that out) says that gravitating body has to expand as you add energy to the body?

It says that for gravitating body U_pot=-2U_kin. As a result adding more energy to the body increases it' s average potential energy (body expands) and decreases kinetic energy (it cools down on average).
U_{tot}+\Delta U=(U_{pot}+2\Delta U)+(U_{kin}-\Delta U)

As I noted in the other thread we're having on gravitational collapse, this form of the virial theorem is limited, as I understand it, to the case of zero pressure, where all bodies are moving on geodesics (i.e., solely under the influence of gravity). I believe it gets more complicated when pressure is included, though I believe the general conclusion, that bodies in hydrostatic equilibrium expand (and cool) when you add energy and contract (and warm up) when you take away energy, still holds. However, see next comment.

If you agree with that then your statement becomes rather non trivial and I think that the burden of proof is entirely on you. It could mean that corrections to classical result under certain conditions would have to grow so large (in right direction) that they actually invert classical result.

All of the logic above about what happens to a gravitating body when it gains or loses energy only applies if the body is in a stable equilibrium, or more precisely if it is gaining or losing mass (energy) slowly enough that it can make a quasi-stable transition between stable equilibrium configurations. But the whole point of the TOV limit for neutron stars (or the Chandrasekhar limit for white dwarfs) is that for a body with total mass (energy) above the limit there is *no* stable equilibrium! (The reason is that the central pressure would have to be infinite for a stable equilibrium to exist.) What stevebd1 has been saying about pressure increasing total gravitational pull is an explanation of *why* there is no stable equilibrium; it's because as the body increases its pressure to hold itself up against gravity, that increase in pressure itself increases the gravity it has to hold itself up against. This then requires more pressure to hold the body up, which increases the total gravity further, etc...until you reach the limit beyond which even an infinite pressure can't hold the body up.

 

[stevebd1]

If you agree with that then your statement becomes rather non trivial and I think that the burden of proof is entirely on you. It could mean that corrections to classical result under certain conditions would have to grow so large (in right direction) that they actually invert classical result.

Not hard evidence but here's an extract from the 'curious about astronomy' website, http://curious.astro.cornell.edu/question.php?number=263"

..for objects less massive than Jupiter, adding mass increases their size. For objects more massive than Jupiter, adding even more mass decreases their size due to increased gravity and pressure. Since WDs and NSs are much more massive than Jupiter, their sizes decrease with increasing mass.

The link provides a bit more information.


On more mathmatical level, Fig. 3 on page 5 of this paper, http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.2708v2.pdf" shows the radius of neutron stars reducing as the mass increases.


It appears that it is generally accepted that neutron stars reduce in size as they grow in mass.

 

[PeterDonis]

..for objects less massive than Jupiter, adding mass increases their size. For objects more massive than Jupiter, adding even more mass decreases their size due to increased gravity and pressure. Since WDs and NSs are much more massive than Jupiter, their sizes decrease with increasing mass.

This link reminded me that the heat capacity issue is more complex than I had implied in my previous posts. The basic argument that I gave in previous posts for a self-gravitating body having a negative heat capacity depends, as I understand it, on the body having either zero pressure (the simple case where the virial theorem works as zonde's post indicated), or being in hydrostatic equilibrium with a non-zero pressure that is kinetic, i.e., it increases as temperature increases.

Let me give another short quote from the same web page, since the one stevebd1 gave, from later in the article, actually misstates things a bit (or else the article writer was pressed for time):

So let's consider a small giant planet, like Neptune. That planet is supported entirely by gas and degeneracy pressure. If we were to slowly add mass to Neptune, the planet would begin to grow in radius. The gravity and pressure would increase as well, of course, but not enough to offset the increase in volume. This will keep occurring until our planet is a few tens or hundreds the size of Jupiter.

As you can see, the actual "cutoff", where the change in radius on adding mass becomes negative instead of positive, is "a few tens or hundreds the size of Jupiter". That's about a tenth of the mass of the Sun. I believe the reason for the change in sign at around that size is that, up until then, degeneracy pressure is small compared to kinetic pressure (what the quote calls "gas pressure"). As degeneracy pressure gets larger and larger, the dependence of the total pressure on temperature gets weaker and weaker; or, put another way, the fraction of the total pressure that contributes to gravity (since all pressure does) but does *not* contribute to the expansion of the body (because it's not kinetic) gets larger and larger. I believe this is what causes the body, past a certain point, to get smaller instead of larger as it continues to gain mass (energy).

How does this affect heat capacity? I believe (but I haven't confirmed by calculation) that once a body is in the regime where degeneracy pressure is dominant, and it gets smaller when mass is added, its heat capacity becomes positive again. This is because, if adding energy causes the body to contract, the material in the body is basically falling a bit in its own gravitational field, which should cause it to gain kinetic energy and heat up. So I think it *is* possible for a white dwarf or neutron star to exist in stable thermal equilibrium with the outside universe, provided it is below the appropriate mass limit. Above the mass limit, as I said before, there is no possible stable equilibrium configuration, and the object must collapse.

 

[zonde]

Not hard evidence but here's an extract from the 'curious about astronomy' website, http://curious.astro.cornell.edu/question.php?number=263"

..for objects less massive than Jupiter, adding mass increases their size. For objects more massive than Jupiter, adding even more mass decreases their size due to increased gravity and pressure. Since WDs and NSs are much more massive than Jupiter, their sizes decrease with increasing mass.

Let's look at this reasoning for hidden assumptions.
What does it means to add mass? You can't simply import mass at the surface of the body.
So if we want more physical example we can take example where we move mass from more distant place toward body until it reaches surface. Now let's consider joined system of this further away matter and the body. In order to lower gravitational potential energy of this additional mass total system has to lower it's total energy i.e. it has to get rid of the energy.

So it turns out that hidden assumption in this reasoning from the article is that the body actually can emit heat.

If you do not make that assumption then another possible scenario for matter falling on surface of the body is conversion of additional kinetic energy along with part of body's own kinetic energy into potential energy and consequent expanding and cooling of the body. And at extreme case body can simply fall apart forming cloud of debris.

It appears that it is generally accepted that neutron stars reduce in size as they grow in mass.

I don't question that. But consensus viewpoint is not a proof of an argument.
For example this relationship can be valid but interpretation of physical mechanism behind this relationship can be incorrect.

 

[stevebd1]

Here is the paper that was the source for mass/radius chart (fig. 3) from the paper mentioned in post 26-

'Strange Quark Matter and Compact Stars' by Fridolin Weber
http://arxiv.org/abs/astro-ph/0407155

The chart is on page 29, Fig. 15. This paper appears to have more information/equations and also compares the various models to actual observed neutron stars.

 

[PeterDonis]

In order to lower gravitational potential energy of this additional mass total system has to lower it's total energy i.e. it has to get rid of the energy.

Not if the mass just falls from far away until it hits the body. Total energy is constant in that scenario; the body's gravitational potential energy when it's far away becomes kinetic energy when it hits the gravitating body. Assume (which is an idealization, yes) that none of the falling mass gets thrown back upward by the impact, it all becomes part of the gravitating body, along with all the kinetic energy it's carrying (which all becomes internal heat energy in the new, more massive gravitating body). That means the new combined body (original gravitating body, plus falling mass, plus all the kinetic energy gained by the mass as it fell) must reach a new static equilibrium configuration with a higher total energy than before. And if the original gravitating body was more massive than the threshold given earlier (ten to a hundred times the mass of Jupiter), the new static equilibrium configuration I just described will have a *smaller* radius than the old one did.

If you do not make that assumption then another possible scenario for matter falling on surface of the body is conversion of additional kinetic energy along with part of body's own kinetic energy into potential energy and consequent expanding and cooling of the body. And at extreme case body can simply fall apart forming cloud of debris.

If the body falls apart, it does not reach a new static equilibrium configuration. Such a case is not covered by the quotes stevebd1 gave. Those articles are only discussing stable, static equilibrium configurations. Obviously there is no guarantee, physically, that the system will end up in a static equilibrium after every possible change, but restricting discussion to static equilibrium seems reasonable in this thread given the question asked in the OP.

The other possibility you raise cannot result in a new static equilibrium; that is the point of the quotes stevebd1 gave. If the body were to start expanding and cooling as you say, due to some type of perturbation during the impact of the infalling mass, gravity would pull it back, and it would oscillate in radius until it settled down to a new static equilibrium, which, if the total energy was larger than before, would be *smaller* in radius than before.

 

[zonde]

Here is the paper that was the source for mass/radius chart (fig. 3) from the paper mentioned in post 26-

'Strange Quark Matter and Compact Stars' by Fridolin Weber
http://arxiv.org/abs/astro-ph/0407155

The chart is on page 29, Fig. 15. This paper appears to have more information/equations and also compares the various models to actual observed neutron stars.

This paper is more about what a neutron star might like once it has formed with given mass and radius.
But our discussion stared with the question about the process how it can get there.

Your point was that neutron star reduces in radius as it acquires matter because pressure contributes to gravity.

My point is that neutron star reduces in radius if (because) it radiates away "pressure" energy.

I don't see how the paper you gave can contribute to our discussion.

 

[zonde]

Not if the mass just falls from far away until it hits the body. Total energy is constant in that scenario; the body's gravitational potential energy when it's far away becomes kinetic energy when it hits the gravitating body. Assume (which is an idealization, yes) that none of the falling mass gets thrown back upward by the impact, it all becomes part of the gravitating body, along with all the kinetic energy it's carrying (which all becomes internal heat energy in the new, more massive gravitating body). That means the new combined body (original gravitating body, plus falling mass, plus all the kinetic energy gained by the mass as it fell) must reach a new static equilibrium configuration with a higher total energy than before. And if the original gravitating body was more massive than the threshold given earlier (ten to a hundred times the mass of Jupiter), the new static equilibrium configuration I just described will have a *smaller* radius than the old one did.



If the body falls apart, it does not reach a new static equilibrium configuration. Such a case is not covered by the quotes stevebd1 gave. Those articles are only discussing stable, static equilibrium configurations. Obviously there is no guarantee, physically, that the system will end up in a static equilibrium after every possible change, but restricting discussion to static equilibrium seems reasonable in this thread given the question asked in the OP.

The other possibility you raise cannot result in a new static equilibrium; that is the point of the quotes stevebd1 gave. If the body were to start expanding and cooling as you say, due to some type of perturbation during the impact of the infalling mass, gravity would pull it back, and it would oscillate in radius until it settled down to a new static equilibrium, which, if the total energy was larger than before, would be *smaller* in radius than before.

I can not agree with underlined statement. This is exactly opposite to conclusions that follow from virial theorem.
Please provide arguments.

 

[PeterDonis]

I can not agree with underlined statement. This is exactly opposite to conclusions that follow from virial theorem.
Please provide arguments.

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

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.

 

[mheslep]

Whether something is a black hole not has nothing to do with the mass. The mass of a grain of salt could be a black hole as well.

Actually the amount of matter is not relevant, it is the ratio between area and mass that matters.

A non-spinning object becomes (or is already) a black hole if the ratio between the area it occupies and the area that represents its mass is smaller than 4. As soon as this happens the object's occupied area will shrink to zero.

E.g.:

<br /> {A_{occupied} \over A_{mass} } &lt; 4<br />

Should there be an allowance for coulomb repulsion, or is that assumed in that inequality?

 

[stevebd1]

An extract from Wikipedia's entry for http://en.wikipedia.org/wiki/Degenerate_matter#Concept"-

Unlike a classical ideal gas, whose pressure is proportional to its temperature (P=nkT/V, where P is pressure, V is the volume, n is the number of particles—typically atoms or molecules—k is Boltzmann's constant, and T is temperature), the pressure exerted by degenerate matter depends only weakly on its temperature. In particular, the pressure remains nonzero even at absolute zero temperature. At relatively low densities*, the pressure of a fully degenerate gas is given by P=K(n/V)^5/3, where K depends on the properties of the particles making up the gas. At very high densities, where most of the particles are forced into quantum states with relativistic energies, the pressure is given by P=K'(n/V)^4/3, where K' again depends on the properties of the particles making up the gas.

Some idea of what K and K' represent are shown on page 16 of this paper where a slightly different interpretation of P has been used (P=K\rho_0^\Gamma where \small\Gamma represents either 5/3 or 4/3)-

'Neutron Stars' by Kostas Kokkotas
http://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/NS.BH.GW_files/GTR_course4.pdf

*'relatively low densities' are considered to be <<6e+018 kg/m³ for neutron stars.

 

[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}<br /> 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.