Why the black hole density is called the infinite density.

[sshai45]

There is no *matter* in the hole (once the object that collapses to form it has vanished into the singularity), but there is "energy" and "mass" there. One way to tell is, as I said in my previous post, to put an object in orbit about the hole. Another, as you note, is to observe that a real black hole can radiate energy.

As far as "where the energy comes from", the best simple answer is probably that it comes from the curvature of spacetime; as the hole radiates, its mass slowly decreases (for example, if you were in orbit about the hole, you would see your orbital parameters slowly changing to reflect a slowly decreasing mass), so the spacetime around the hole slowly becomes less curved. However, there is a *lot* more lurking here, enough for multiple threads (and there have been plenty on these forums already).

Yet then where is this energy, anyways, when you just said the equations describing the situation are the "vacuum" Einstein Field Equations and the "stress energy tensor" is "zero" everywhere?

 

[yuiop]

Careful, the t-coordinate does not represent time (time on a clock at infinity) over the whole spacetime.

What does it represent then?

<EDIT> Having given it yet more thought, the t coordinate in the interior solution is the same as the t coordinate in the exterior solution (ie the reference coordinate clock at infinity) because the 2 solutions join smoothly at the boundary.

For example, when the dust sphere has a radius of 9/8 Rs and the test clock is at the surface of the dust sphere so that r = R = 9/8 Rs the proper time rate according to the interior solution is:

\frac{d\tau }{dt} = \frac{3}{2}\sqrt{1-\frac{8}{9}}-\frac{1}{2}\sqrt{1-\frac{ (9/8)^2}{(9/8)^{3}}} = \frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{8}{9}} = \frac{1}{3}

while the proper time rate according to the exterior solution is:

\frac{d\tau }{dt} = \sqrt{1-\frac{8}{9}} = \frac{1}{3}

The solutions are in agreement, so the proper times are relative to the reference Schwarzschild coordinate clock at infinity in both cases. Perhaps I misunderstand what you are getting at?

 

[Passionflower]

What does it represent then?

Inside it represents space and the r coordinate represents time.

That does not, as some often do, imply, that space becomes time and vice versa.
It is only the coordinates that change their mapping.

 

[PeterDonis]

Yet then where is this energy, anyways, when you just said the equations describing the situation are the "vacuum" Einstein Field Equations and the "stress energy tensor" is "zero" everywhere?

That's part of the "a lot more lurking here". But the short answer is that a black hole that is radiating is *not* a solution of the vacuum Einstein Field Equations. Those equations are classical and don't include quantum effects. Hawking radiation is a quantum effect. That's why I was careful to state that the particular vacuum solution that describes a "static" black hole only applies if we ignore quantum effects. If you want a longer answer, you should probably start a separate thread since this is really a separate question from the question in the OP.

 

[Austin0]

Consider a sphere of dust of even density and radius R. When the sphere collapses to a radius of 9/8 Rs where Rs is the Schwarzschild radius equal to 2M/c^2, a clock at the centre of the mass stops ticking relative to Schwarzschild coordinate time.

No. Gravity decreases inside the Earth because more and more of the Earth's mass is above you instead of below you. Once the collapsing object has formed a singularity at the center, r = 0, there is no place anywhere where any of the mass of that object is above you. It's all been crushed into the singularity and ceased to exist.

Would this mean that the collapsing dust cloud [above] as it was approaching the Schwarzschild radius would be collapsing more rapidly on the surface with an increasing density and dilation gradient from there to a minimum at the center?

Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface??
Thanks

 

[PeterDonis]

Would this mean that the collapsing dust cloud [above] as it was approaching the Schwarzschild radius would be collapsing more rapidly on the surface with an increasing density and dilation gradient from there to a minimum at the center?

First, please note my later post in response to yuiop; his specific formulas only apply (I believe) to the case of a static equilibrium, *not* to the case of a collapsing object.

That said, your general intuition here is correct. As the object collapses, the density at its center increases, and the "time dilation" at its center becomes more extreme, relative to an observer far away (another way to look at it would be to say that the "potential well" inside the object becomes deeper). See next comment.

Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface??

Because there is a "potential well" there, and it gets deeper as the object collapses. Please note that the "potential", or "time dilation" factor, which is what we're discussing here, is different from the "acceleration due to gravity", which is what gets smaller as you go deeper inside a gravitating body (like the Earth) because more of it is above you, as I said in that previous post you quoted.

Also note that, strictly speaking, "acceleration due to gravity" only applies to a *static* observer, one who is "hovering" at a constant radial coordinate r. Once the object has collapsed inside its Schwarzschild radius, there are no such observers in the interior vacuum region inside the event horizon (i.e., outside the surface of the collapsing object but still below the horizon). (I'm not sure if there can still be static observers inside the collapsing object once it has collapsed inside the horizon; I think not, but I have not looked at the math in detail.)

 

[yuiop]

Would this mean that the collapsing dust cloud [above] as it was approaching the Schwarzschild radius would be collapsing more rapidly on the surface with an increasing density and dilation gradient from there to a minimum at the center?

That seems a reasonable assumption. However it more likely that the dust cloud would be denser nearer the centre during the initial stages before the time dilation effects become significant and start evening out or even reversing the density gradient. The equation I used assumes an even density gradient which is unlikely in practice. However, it can be shown that if the density increases towards the centre inversely proportional to radius of the dust sphere, that the results are the same. Also, as Peter pointed out the interior solution is a static solution and in reality a dust sphere so close to the Schwarzschild density is almost certainly going to be collapsing rapidly and not static. It is a pity we do not have a solution for the dynamic case.

Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface??

If you had a very deep well that went to the centre of the Earth, the acceleration of gravity reduces as you get nearer the centre, but the gravitational time dilation increases. Gravitational time dilation is a function of gravitational potential which can be expressed as -GM/r. As r gets smaller, -GM/r gets larger (and more negative).

NOTE: I am assuming an idealised non rotating Earth or a well dug from a pole. Rotation of the Earth will cause increased time dilation nearer the surface due to velocity induced time dilation of SR.

 

[PeterDonis]

It is a pity we do not have a solution for the dynamic case.

If you mean an analytical solution, we do for the idealized case of a perfectly spherically symmetric collapse. It's the Oppenheimer-Snyder solution, from their 1939 paper. The abstract is here:

http://prola.aps.org/abstract/PR/v56/i5/p455_1

Unfortunately the paper itself is behind a paywall.

The interior of that solution is basically the time reverse of the Big Bang: a contracting FRW metric which is matched at the boundary to the exterior vacuum Schwarzschild metric. The density inside the collapsing object, in this model, increases as the object collapses, ultimately going to infinity as the radius of the object goes to zero (i.e., as it forms the singularity at r = 0).

There have been plenty of numerical simulations of less idealized collapses; unfortunately I don't have any good general references handy.

 

[yuiop]

Unfortunately the paper itself is behind a paywall.

Any equations? I would be interested in how they avoid negative or imaginary proper time during the collapse.

 

[PeterDonis]

Any equations? I would be interested in how they avoid negative or imaginary proper time during the collapse.

The OH solution is treated in most GR textbooks; in MTW it's in section 32.4. I can post some more details from there if desired, but I don't have time at the moment. But the fact that the proper time behaves itself in the interior should be obvious just from the fact that it's a contracting FRW metric; proper time, which is just FRW coordinate time, obviously increases monotonically during the collapse, just from looking at the FRW line element.

The only technical point is that the most natural time coordinate in the FRW interior, the one in the standard FRW metric, doesn't match up with the most natural time coordinate in the Schwarzschild exterior, which is the Schwarzschild one; but as the exercises in MTW show, one can "fix" this, if desired, by finding a single time coordinate that covers both regions.

 

[Austin0]

That said, your general intuition here is correct. As the object collapses, the density at its center increases, and the "time dilation" at its center becomes more extreme, relative to an observer far away (another way to look at it would be to say that the "potential well" inside the object becomes deeper). See next comment.
Because there is a "potential well" there, and it gets deeper as the object collapses. Please note that the "potential", or "time dilation" factor, which is what we're discussing here, is different from the "acceleration due to gravity", which is what gets smaller as you go deeper inside a gravitating body (like the Earth) because more of it is above you, as I said in that previous post you quoted.

That seems a reasonable assumption. However it more likely that the dust cloud would be denser nearer the centre during the initial stages before the time dilation effects become significant and start evening out or even reversing the density gradient. The equation I used assumes an even density gradient which is unlikely in practice. However, it can be shown that if the density increases towards the centre inversely proportional to radius of the dust sphere, that the results are the same.
If you had a very deep well that went to the centre of the Earth, the acceleration of gravity reduces as you get nearer the centre, but the gravitational time dilation increases. Gravitational time dilation is a function of gravitational potential which can be expressed as -GM/r. As r gets smaller, -GM/r gets larger (and more negative).

SO consider a homogeneous cloud well outside R_s. R=4 R_s
Would, in this case, the contraction acceleration and density occur with a differential decreasing towards the center?

Thank you both for clearing up my confusion regarding potential and acceleration factor.
This actually answered a question I have posted before without answer ; What is the dilation factor mid point between two equally massive bodies compared to on the surface of one. From the above I would now assume it would be greater?

PS IS the time running backwards at the center after collapse derived from Rindler or is it intrinsic in the Schwarzschild equations? Thanks

 

[sshai45]

The interior of that solution is basically the time reverse of the Big Bang: a contracting FRW metric which is matched at the boundary to the exterior vacuum Schwarzschild metric. The density inside the collapsing object, in this model, increases as the object collapses, ultimately going to infinity as the radius of the object goes to zero (i.e., as it forms the singularity at r = 0).

So does this mean that if you could stand inside the star as it collapses to a BH, you'd find that the matter and what not around you was changing in a way that looked like a little mini "Big Crunch"?

 

[PeterDonis]

SO consider a homogeneous cloud well outside R_s. R=4 R_s
Would, in this case, the contraction acceleration and density occur with a differential decreasing towards the center?

For the case of a perfectly *homogeneous* dust cloud, the density would be the same everywhere in the cloud (at least if you are using the standard "comoving" coordinates); that's the scenario that Oppenheimer and Snyder analyzed (because it's the only one they could treat analytically). The "contraction acceleration" is not really a good term to use for this scenario, because there is no single point within the cloud that counts as the "center" towards which everything else is collapsing; *every* point within the cloud is the same as every other (just as in our expanding universe, there is no single "center" from which everything is expanding; *every* point in the universe is the same as every other).

What is the dilation factor mid point between two equally massive bodies compared to on the surface of one. From the above I would now assume it would be greater?

There is no analytical solution for the case of two massive bodies; GR is not linear so you can't just superpose two one-body solutions to get a solution for two bodies. However, for the case where the distance between the bodies is large compared to their masses, I would expect the "potential" at the midpoint between them to be greater (i.e., closer to the "potential" at infinity) than the "potential" at the surface of either body.

PS IS the time running backwards at the center after collapse derived from Rindler or is it intrinsic in the Schwarzschild equations? Thanks

Time does not run backwards at the center after collapse. The statement yuiop made to that effect was not correct; his formulas only apply to a static equilibrium, not a collapsing object, and there is no static equilibrium with a radius less than 9/8 the Schwarzschild radius, which was the condition he needed to make it look like time was "running backwards" at the center.

 

[PeterDonis]

So does this mean that if you could stand inside the star as it collapses to a BH, you'd find that the matter and what not around you was changing in a way that looked like a little mini "Big Crunch"?

Yes.

 

[Passionflower]

the most natural time coordinate in the Schwarzschild exterior, which is the Schwarzschild one;

Could you perhaps give a reason why you think there is a natural time coordinate and why you think it is the time coordinate in Schwarzschild coordinates.

 

[Whovian]

Our mathematics probably doesn't work there.

Sincerest apologies for attacking wording, but our mathematics works fine. It's our physics that doesn't work.

 

[PeterDonis]

Could you perhaps give a reason why you think there is a natural time coordinate and why you think it is the time coordinate in Schwarzschild coordinates.

"Natural" because it matches up with the time translation symmetry of the spacetime. But as you have pointed out, the Schwarzschild time coordinate is no longer timelike inside the horizon; so really it is only a natural *time* coordinate outside the horizon; inside the horizon it's still "natural" (because d/dt is still a Killing vector), but it's not a "time" coordinate any more because it's spacelike.

 

[Passionflower]

I think the whole idea which coordinates are more natural a little arbitrary but if I would be forced to pick one it would not be Schwarzschild coordinates.

 

[PeterDonis]

I think the whole idea which coordinates are more natural a little arbitrary

I could have been more precise by saying "coordinates which match up best with a particular symmetry of the spacetime", but that would have been a mouthful.

but if I would be forced to pick one it would not be Schwarzschild coordinates.

Which one would you pick?

 

[Passionflower]

Which one would you pick?

If forced I would go with Gullstrand–Painlevé coordinates.

 

[alexg]

No. Gravity decreases inside the Earth because more and more of the Earth's mass is above you instead of below you.

Peter, a minor point, almost a diversion, but doesn't the Shell Theorem state that the gravity inside the Earth remains the same until you reach the center? While there is more mass above you, you are closer to the rest of the mass on the other side, and it balances out.

 

[yuiop]

Peter, a minor point, almost a diversion, but doesn't the Shell Theorem state that the gravity inside the Earth remains the same until you reach the center? While there is more mass above you, you are closer to the rest of the mass on the other side, and it balances out.

The shell theorem (that you are thinking of) only applies when you are in the cavity of a hollow shell and in that case it states there is no net gravitational acceleration anywhere inside the cavity due to the cancelling effects. (There is however a lower gravitational potential inside the cavity than outside.)

 

[PeterDonis]

Peter, a minor point, almost a diversion, but doesn't the Shell Theorem state that the gravity inside the Earth remains the same until you reach the center? While there is more mass above you, you are closer to the rest of the mass on the other side, and it balances out.

No. I assume you mean the Shell Theorem as stated, for example, on this Wiki page:

http://en.wikipedia.org/wiki/Shell_theorem

The theorem basically says two things:

(1) If you're on the outside of (i.e., above, further from the center than) a spherically symmetric mass distribution, you feel the gravity of the entire mass as if it were concentrated at the center of the sphere.

(2) If you're on the inside of (i.e., below, closer to the center than) a spherically symmetric mass distribution, you feel *no* net gravity from it, because the contributions from all different parts of it cancel out.

If you're in the interior of an idealized spherically symmetric body (like an ideal non-rotating spherical Earth), both (1) and (2) above apply. (2) applies to the part of the body that's above you, and says that that part contributes nothing to the gravity you feel. (1) applies to the part of the body that's below you, and says that you feel the gravity of that part normally, i.e., just as if the mass that's below you were concentrated at the center.

As you descend through the body, more and more of its mass is above you (and so doesn't contribute to the gravity you feel), and less and less is below you. So the gravity you feel gets less and less, reaching zero just as you reach the center.

 

[Austin0]

No. I assume you mean the Shell Theorem as stated, for example, on this Wiki page:

http://en.wikipedia.org/wiki/Shell_theorem

The theorem basically says two things:

(1) If you're on the outside of (i.e., above, further from the center than) a spherically symmetric mass distribution, you feel the gravity of the entire mass as if it were concentrated at the center of the sphere.

(2) If you're on the inside of (i.e., below, closer to the center than) a spherically symmetric mass distribution, you feel *no* net gravity from it, because the contributions from all different parts of it cancel out.

If you're in the interior of an idealized spherically symmetric body (like an ideal non-rotating spherical Earth), both (1) and (2) above apply. (2) applies to the part of the body that's above you, and says that that part contributes nothing to the gravity you feel. (1) applies to the part of the body that's below you, and says that you feel the gravity of that part normally, i.e., just as if the mass that's below you were concentrated at the center.

As you descend through the body, more and more of its mass is above you (and so doesn't contribute to the gravity you feel), and less and less is below you. So the gravity you feel gets less and less, reaching zero just as you reach the center.

Does density have a significant effect?
What is the difference between a solid spherical body and a uniformly distributed spherical particle cloud ,which of course can have a fairly dense distribution if we want?

 

[PeterDonis]

Does density have a significant effect?
What is the difference between a solid spherical body and a uniformly distributed spherical particle cloud ,which of course can have a fairly dense distribution if we want?

Density doesn't matter, nor does the exact state of matter in the body (solid, liquid, gas, dust, etc.), as long as the mass distribution is exactly spherically symmetric. That means any variable, such as density, can only be a function of radius (i.e., distance from the center). That's the key condition that leads to the conclusions of the theorem.

 

[yuiop]

Does density have a significant effect?
What is the difference between a solid spherical body and a uniformly distributed spherical particle cloud ,which of course can have a fairly dense distribution if we want?

If we have a low density body and a high density body with equal masses, then the effects of gravity on an external test particle are identical (in Newtonian terms). Both behave as if they are a point particle containing all the mass. The lower density body will have a larger radius. For two bodies with equal radius and differing densities, the lower density body will of course have a lower lower mass and therefore less of a gravitational effect.

In General Relativity, if we have a solid body in equilibrium such that there are stresses and pressures within the body, then its external gravitational effects will be different from a lower density particle cloud that is not in equilibrium even if it has the same total mass. The stresses and pressures add to the effective total mass of the solid body and the motion of the particles in the particle cloud will also have an effect.

 

[yuiop]

Well even that question is not simple.

The class of observers that are stationary in a Schwarzschild spacetime can certainly chart the event horizon as a sphere with a certain distance removed from them but not a sphere with a distance removed from a center, because there is no center.

Why do we need a centre? We could have a spherical shell around a vacuum with no clearly defined centre and still define the sphere in terms of its diameter or its circumference divided by 2pi or in terms of its surface area (A) as r = sqrt(A/4pi) or in terms of its volume (V) as r = (3V/(4pi))^(1/3) etc.

Lets say we added mass to the Sun until it collapsed into a black hole and continued to add mass until it engulfed Mercury, Earth, Mars etc until only Pluto remained orbiting it. Would you agree that the black hole occupied a volume of space? Would you agree if occupies a volume then we can in some sense say it has a radius? Would you agree that to say it has no radius is to suggest that all points on the event horizon are at the same spatial point? Would you agree that the finite proper time taken by free falling observer to fall from the event horizon to the singularity contradicts the idea of their being volume or radius? Would you agree that no radius implies that the singularity and the event horizon are at the same spatial location?

 

[PeterDonis]

In General Relativity, if we have a solid body in equilibrium such that there are stresses and pressures within the body, then its external gravitational effects will be different from a lower density particle cloud that is not in equilibrium even if it has the same total mass.

Careful! This is not correct as you state it. The "total mass" is one of the "external gravitational effects", so it *is* affected by the things you mention. I think you realize this, since you next say:

The stresses and pressures add to the effective total mass of the solid body and the motion of the particles in the particle cloud will also have an effect.

"Effective total mass" and "total mass" are really the same thing. I think what you are getting at is that the same "amount of stuff" can have a different "effective total mass", but you have to be careful how you characterize "amount of stuff" if you want to make it distinct from "total mass". For example, you could say "the same total number of atoms can have a different total mass, depending on its configuration"--"total number of atoms" is a more precise way of defining "amount of stuff". (MTW uses "total number of baryons" in the same way.)

I should note that there is a lot more lurking in the details of how "total mass" is calculated, as shown, for example, in this thread:

https://www.physicsforums.com/showthread.php?t=585547