# Does a gravitating body have to undergo gravitational collapse?

1. Aug 30, 2011

### zonde

Imagine body that has mass and volume such that it is just a bit short of turning into the black hole. Now let's say that the body has been around for some time and has reached present state in some smooth process so that it can be in equilibrium state if such an equilibrium exists.
I assume that this equilibrium state is such that density of mass increases proportionally with the radius so that event horizon does not form anywhere inside the body. Now as the pressure increases so does temperature. Now if the heat will escape from inside of the body toward outside then inside will collapse.
But can it really escape? Photons going in outward direction are redshifted but photons going inward are blueshifted. So it seems that inside of the body can not really cool down.
Of course photons will escape from surface of the body but then there are photons coming from the rest of the universe that would be highly blushifted.

So does this reasoning seems valid? Can a static body exist without undergoing gravitational collapse even without considering degeneracy pressure?

2. Aug 30, 2011

### Staff: Mentor

This is a non-trivial constraint. I believe there is a lower limit to the radius of such a static equilibrium state; I think it's 9/8 of the Schwarzschild radius. I'll see if I can find an online reference (I believe the limit was first derived in a paper by Einstein some time in the 1930's). So any object with a radius less than that *must* collapse to a black hole; it can't exist in a static equilibrium state at all.

It's worth expanding a little bit on how this happens. Heat escaping from the body means the body itself is losing energy; losing energy means it has to contract, i.e., its radius has to decrease. (This is assuming that there is no other source of energy inside the body, for example, it can't undergo nuclear reactions as a star does.) Eventually, its radius will decrease enough that it is smaller than the limit I referred to above, and at that point, it will collapse.

(Of course this is not the *only* way an object could collapse to a black hole. But we're not talking about something like a supernova explosion; we're talking about a slow, "smooth" process, as you put it, where everything changes very gradually.)

No, this is not correct. The blueshift/redshift is really a red herring; in order to measure the net energy loss of the body, you have to pick a single radius at which to do it. Normally this is done "at infinity", i.e., at a very, very large radius, far away from the body, where spacetime is very, very close to flat. At that large radius, you balance outgoing radiation against incoming radiation, and the blueshift/redshift doesn't enter into it. If (as is highly likely with a radiating body in our universe), there is more outgoing radiation than incoming radiation, then the body is losing energy.

You could also, of course, do the energy balance at a much smaller radius, say just outside the surface of the body. There, the incoming radiation would indeed be blueshifted, but the outgoing radiation would *not* be redshifted (because it hasn't yet "climbed out" of the body's gravity well). So the balance would still work out the same. (Basically, since you have to do the comparison at a constant radius, changing the radius changes both the outgoing and incoming radiation in concert, so the balance between them doesn't change.)

Bottom line, under the conditions as described above, no; it would continually lose energy by radiation until its radius was below the limit I described above, and then it would collapse.

Of course, any real object under such conditions would be subject to degeneracy pressure, since that will occur whenever you squeeze matter to a high enough density. So an object like a white dwarf or a neutron star can indeed exist indefinitely in a stable equilibrium after it has radiated away all the energy it can (provided its radius is larger than the limit I gave above--that limit applies even in the presence of degeneracy pressure).

3. Aug 31, 2011

### zonde

Well, yes balance between incoming and outgoing radiation doesn't change with radius.
But I still think that blueshift/redshift is relevant.

Let's look at such a setup. We have two bodies held at different heights above the surface of planet. Bodies are emitting only black body radiation. They emit this radiation only toward each other as they are surrounded by ideal mirrors from other sides.

Let me illustrate this schematically
Code (Text):

b1
|            |
|<R<<<<<<<<B<|
|            |         (planet)
|>R>>>>>>>>B>|
|            |
b2

"b1" and "b2" are two surfaces of bodies emitting black body radiation and ">R>" and ">B>" are redshifted or blueshifted photons of black body radiation.

As you can see net balance across two interfaces for incoming and outgoing radiation is zero and yet two bodies will have different temperature after reaching equilibrium.

4. Aug 31, 2011

### Staff: Mentor

This is true in the idealized situation you've described, but it's not relevant to your original question because you've changed the situation. Your original question was whether a gravitating body in static equilibrium could lose energy by radiation. There were no ideally reflecting mirrors involved; obviously the presence of ideally reflecting mirrors changes the physics of the situation. In the presence of ideally reflecting mirrors as you describe, obviously the body cannot lose any net energy to the universe outside the upper mirror.

The key to whether the body can lose energy, in the original scenario, is that, as you appear to agree, the net energy balance across a surface at a given radius is independent of the radius. The actual equilibrium temperature, given the net energy balance, may be dependent on radius, but what matters for whether the body can lose energy by radiation is the net energy balance, not the equilibrium temperature. As long as the net energy balance is negative (the body radiates away more energy than it takes in, as evaluated across a surface at a given radius), the body will lose energy.

5. Sep 3, 2011

### zonde

I think that I am more interested about temperature gradient that does not contribute to conduction. There is some ground level temperature of universe so if we have this gradient steep enough energy loss can go to zero even with high temperature at the middle of gravitating body.

And after a bit of thinking it seems to me that this question is more about classical physics.
If we model temperature as motion of particles then this motion slows down as we go against gravity and speeds up as we go along. Basically if we don't have convection in material (it's solid) then equilibrium temperature gradient implied by this classical model can be much more steeper than from gravitational redshift.

6. Sep 3, 2011

### Staff: Mentor

Do you really mean just "conduction" or do you mean "heat transfer" in general? Conduction is only possible through a material, not through vacuum, so you can easily have two objects at widely different temperatures separated by vacuum that can't exchange heat through conduction (or convection, for that matter). But they can still do so through radiation. If you are trying to say there can be a temperature gradient that does not contribute to heat transfer in any form, I'm not sure that's possible. Any object at a non-zero temperature radiates.

If this is just another way of describing the conversion between kinetic energy and gravitational potential energy of a given particle, then OK. But that in no way rules out heat transfer between particles.

No, you would still have conduction (and radiation; individual atoms in the solid can still exchange radiation with each other).

7. Sep 4, 2011

### zonde

Hmm, I thought that my example clearly demonstrated that this is possible.
Heat transfer is difference between incoming and outgoing radiation. If outer layer of gravitating object is at temperature around 2.7K it will be in equilibrium with CMB radiation. So as we go closer to the center of gravitating object and redshift/blueshift increases we will have increase in temperature that can not be radiated away.

So basically you are saying that temperature and conduction can not be modeled as motion of individual particles, right?

8. Sep 4, 2011

### Staff: Mentor

Your example shows this *only* if the rate of temperature increase is *exactly* the same as the rate of gravitational blueshift. But in your example, that is only true because the two mirrors are separated by vacuum, and only exchange heat by means of radiation that blueshifts/redshifts with radius at exactly the "gravitational" rate. Your example does not account for heat transfer *within* the gravitating body inside the inner mirror (we'll assume that the outer mirror is exactly at the CMBR temperature so it is in equilibrium with the rest of the universe). Inside the gravitating body, there is not a vacuum, there is matter present; and in the presence of matter (a) the rate of "gravitational blueshift" is different than in vacuum, and (b) the rate of temperature increase as you move inward is governed by factors other than gravity; in the simplest case, the matter is a perfect fluid and is in hydrostatic equilibrium. Unless you can show that in the presence of matter, such as a fluid in hydrostatic equilibrium, the temperature rises as you go inward at *exactly* the same rate as the gravitational blueshift, you haven't supported your case. See next comment.

I'm not saying that at all. What I *am* saying is that the motion of the individual particles, in any real substance, is affected by factors other than gravity. Even if we leave out degeneracy pressure, what about ordinary fluid pressure? (Which is due to the fact that when individual particles in the fluid get close enough, they repel each other--in ordinary fluids the repulsion is due to electromagnetic interactions between the electron shells of the fluid atoms/molecules.)

For an actual example, consider the Earth, which behaves fairly close to a perfect fluid (the Earth does have deviations from perfect fluidity because portions of it are solid, but we'll leave out any such complications here, since they only cause an error of a few percent in the calculation). The temperature at Earth's surface is about 300 K; the temperature at the center is about 5,700 K, or about 19 times that at the surface. However, the gravitational blueshift from the surface to the center of the Earth is only about one part in a billion. So temperature rises inside the Earth a *lot* faster than gravitational blueshift does, because hydrostatic equilibrium causes the ordinary pressure to rise very quickly (the pressure at the center is about 3 *million* times that at the surface), which forces the temperature to rise by ordinary thermodynamics.

Reference for temperature and pressure at center of Earth:

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

Thread where formula for time dilation at center of Earth is given (the blueshift is given by the same formula):

So for any non-vacuum substance to behave as you're suggesting, it would have to have no ordinary pressure, not just no degeneracy pressure. But for any real substance to be in a stable equilibrium at a given radius, it *has* to have ordinary pressure; otherwise there would be nothing opposing its gravity and it would just collapse. But in any substance with ordinary pressure included, as we've just seen, temperature rises a *lot* faster than gravitational blueshift.

It is true that the Earth has internal sources of heat (radioactivity) and receives heat from the Sun, which is how it stays in equilibrium at a hotter temperature than the CMBR. But even if we consider a hypothetical Earth in a stable equilibrium with its surface at the CMBR temperature, the temperature inside would still rise a lot faster than the gravitational blueshift, because of hydrostatic equilibrium.

Last edited: Sep 4, 2011
9. Sep 5, 2011

### zonde

Not sure I understand you.
Let's say that temperature gradient from gravitational blueshift/redshift is "1".
And so you are saying that if temperature gradient is exactly 1 then it will not contribute to heat transfer but if temperature gradient is 1+x then heat transfer will be proportional to 1+x rather than x. Seems weird.

Ordinary fluid pressure is momentum density i.e. motion of the individual particles.
When individual particles are so close that they repel each other then this is called degeneracy pressure.
I believe that electromagnetic interactions are what we can call "collisions".

I will try to respond to the rest of your post later.

10. Sep 6, 2011

### Staff: Mentor

No, I'm saying that you can only attribute the temperature gradient to gravitational blueshift if the rate of increase in temperature is exactly the same as the rate of increase in blueshift. But the rate of increase in temperature is much *larger* than the rate of increase in blueshift, at least in the case I described (the Earth). So the temperature gradient must have some other cause.

Not exactly. Ordinary fluid pressure is caused by momentum *transfer* between the individual particles (or between the particles and external objects like the walls of a container of gas). The momentum transfer occurs when particles get close enough to collide. See further comment below.

No, this is not correct. Degeneracy pressure is a quantum phenomenon; it is caused by the Pauli exclusion principle. It has nothing to do with the ordinary electromagnetic repulsion between the individual particles; it happens even if the particles are electrically neutral and have no repulsive force between them, for example, in neutron stars.

Yes, although I would put it the other way around; what we normally call "collisions" between individual fluid particles, which allow them to exchange momentum, are really repulsive electromagnetic interactions between the particles. Those interactions actually happen at some distance, but for many purposes we can idealize them as happening only on "contact" between the particles.

11. Sep 6, 2011

### zonde

Yes, there are other causes for temperature gradient in case of Earth. But the point I was trying to make was about heat transfer that is caused or not caused by particular temperature gradient and not about causes of temperature gradient.

You said:
I considered my example as an argument in support that there can be such "temperature gradient that does not contribute to heat transfer".

How exactly your argument that "fluid pressure is caused by momentum *transfer*" is correcting my argument that "fluid pressure is momentum density"?
Let me rephrase my statement: Ordinary fluid pressure is proportional to momentum density.
Have you any corrections for that statement?

Degeneracy pressure is not caused by Pauli exclusion principle. Degeneracy pressure is explained by Pauli exclusion principle.
And there is no "ordinary electromagnetic repulsion between the individual particles" that would contribute to pressure if we speak about neutral fluid.

Collisions can happen even between electrically neutral particles like neutrons. I am not sure that I am buying your "repulsive electromagnetic interactions" model.

12. Sep 6, 2011

### zonde

I have to say that I changed my mind. It seems that for fluids pressure is not proportional to momentum density. This hold only for ideal gas.

13. Sep 6, 2011

### Staff: Mentor

I understood your argument slightly differently. I understood your argument to be that there can be a temperature gradient *solely due to the difference in gravitational potential* that does not contribute to heat transfer. My point is that that can only be the case if the temperature gradient is the same as the potential gradient. The only example we've discussed so far where that is the case is your example with the two mirrors, and the only reason the gradients are the same there is that the mirrors are separated by vacuum. Inside the inner mirror (i.e., within the substance of the gravitating body whose surface is the inner mirror), the temperature gradient will *not* be the same as the gravitational potential gradient.

I know you changed your mind about saying that fluid pressure *is* momentum density, but let me clarify the point I was making here anyway. My point is that fluid pressure requires momentum *exchange* between particles; just momentum density by itself isn't enough.

I'm not sure I see the distinction you are trying to make here. Maybe I wasn't clear enough about what I meant. When I said degeneracy pressure is "caused" by the Pauli exclusion principle, I meant that it is a quantum phenomenon that happens because the particles in question are fermions, so no two of them can be in the same state. When a fluid composed of such particles is compressed, it resists the compression because the more compressed it is, the more the wave functions of the individual particles overlap, meaning they are closer to being in the same state.

Ordinary fluid pressure, by contrast, is a classical phenomenon; it does not depend on any quantum properties of the individual fluid particles. A fluid composed of bosons still exhibits ordinary fluid pressure, even though bosons are not subject to the Pauli exclusion principle and there is nothing prohibiting every boson in the fluid from being in the same quantum state.

This is true, and I did not phrase this part of what I was saying very well. The key point is that "collisions" can happen between the fluid particles. See next comment.

You are correct that "collisions" between fluid particles do not *have* to be caused by electromagnetic repulsion. But they *can* be. In an ordinary fluid composed of atoms or molecules, the collisions are due to electromagnetic repulsion between the electrons in the outer shells of the atoms or molecules. And yes, this happens even though the atoms or molecules, as a whole, are electrically neutral. That is all I meant by my "repulsive ectromagnetic interactions model". Do you dispute that this happens in an ordinary fluid composed of atoms or molecules? If so, how do you think the collisions happen in such a fluid?

In a fluid composed of neutrons, such as the interior of a neutron star, you are correct that there are no "collisions" due to electromagnetic repulsion. However, neutrons still interact by means of the strong nuclear force, so they can still exchange momentum by "collisions" mediated by that force. The strong nuclear force is attractive at ranges comparable to the size of an atomic nucleus, so the interactions between neutrons at that range are not what we normally think of as "collisions". However, it is believed that the strong force becomes repulsive at shorter ranges, and "collisions" between neutrons mediated by this repulsion are believed to produce a measurable ordinary fluid pressure in neutron stars (although the majority of the observed pressure is still due to degeneracy pressure).

Last edited: Sep 6, 2011
14. Sep 6, 2011

### Staff: Mentor

Can you give an explicit demonstration of this? I can't get this from either form of the ideal gas law that I'm familiar with:

$$P = nkT$$

where n is the number density (number of particles per unit volume), k is Boltzmann's constant, and T is temperature; or:

$$P = \rho R T$$

where $\rho$ is the mass density, R is the gas constant per unit mass for the particular substance, and T is temperature. I can't see any way to get from either of these equations to the claim that fluid pressure is proportional to momentum density. I can see the proportionality to *energy* density (which is basically what both equations say), but that's not the same as *momentum* density.

15. Sep 6, 2011

### utesfan100

Might the temperature, roughly an estimate of the average particle energy, not be Lorentzian invariant?

This appears to be a solid argument for temperature measurements being hotter from frames deeper within a gravitational field, with the increase in temperature exactly that needed to balance the blue shifting of the black body radiation.

16. Sep 6, 2011

### Staff: Mentor

Energy by itself isn't Lorentz invariant, so one would not expect temperature to be either. For an individual point particle, the invariant object is the energy-momentum 4-vector. For a continuous substance like a fluid, the invariant object is the stress-energy tensor. In either case, energy (and hence temperature) is only one component of the full object, and so it will be different in different frames.

This is actually a separate question from the above. Asking whether a quantity is "Lorentz invariant" is asking whether its value at a single event in spacetime is the same for observers in different states of motion passing through that event. What you are talking about here is the change in the value of the quantity as you change locations--i.e., the difference in its value at different events. When you do that, you have to be very careful about specifying how you are going to compare observations made at different events.

For example, consider the "mirror" scenario that zonde described earlier in this thread. The temperature at the "lower" mirror, in that scenario, is "blueshifted" from the temperature at the "upper" mirror by exactly the right amount to balance the blueshift in radiation as it goes deeper into the gravitational field. However, that only holds because, as I've noted several times now, the space between the mirrors is vacuum, and the only heat exchange between the mirrors is radiation. Of course radiation traveling in vacuum in a gravitational field is going to blueshift/redshift by exactly the "right" amount to balance the change in gravitational potential; but that in itself does not show anything about what will happen when the intervening space is *not* vacuum, or when other mechanisms of heat transfer are available.

I suppose it might still be useful to compare, for a given situation, the "blueshifted" temperature at some radius with the *actual* temperature measured at that radius. For example, I quoted temperatures for the Earth in an earlier post: a surface temperature of 300 K, and a temperature at the center of about 5700K. One could also defined a "blueshifted" temperature at the center, by taking the 300 K at the surface and multiplying by a "blueshift factor" based on the change in gravitational potential from surface to center. Since that change is about one part in a billion, the "blueshifted" value of the surface temperature, at the center, would be about 300.000000003 K. Obviously this is a lot cooler than the *actual* temperature at the center, which was my point. But you could, I suppose, say that the "actual" change in temperature, "corrected for blueshift", was not 5700/300 (a factor of 19), but 5700/300.000000003 (a factor of a smidgen less than 19).

17. Sep 8, 2011

### zonde

And your idea is that within body after reaching equilibrium (no net transfer of heat) temperature gradient is 0, right?
If that's right what mechanism of heat transfer is responsible for that?

Does photon gas have pressure? Photons can collide with walls of container (mirrors) but there wouldn't be collisions between photons.

There are no boson fluids. The only real boson we can investigate more or less directly is photon and there are no photon fluids that we know of. We can speak only about something like photon gas.

Maybe you mean fluids consisting of "bosonic-composites" like Helium-4? They are still subject to Pauli exclusion principle.

Hmm, there of course is repulsion between particles of fluid. Do we have to agree exactly how we call this repulsion and how this repulsion should be "explained"? What do you say?

18. Sep 8, 2011

### zonde

Not sure that I can give you satisfactory answer.
Well it seems right to me. Pressure is caused by kinetic energy of particles (I am using kinetic theory). If you decrease volume of ideal gas pressure increases in inverse proportion. Density is some physical quantity per volume. So it fits.

19. Sep 9, 2011

### Staff: Mentor

All I've been saying so far is that the temperature gradient will not be the same as the gradient in gravitational blueshift for a non-vacuum body. The question of whether there can still be *any* temperature gradient in a body at equilibrium is more complicated, and I'll discuss it in a separate post after I've thought about it some more.

To first order, you are correct, there are no collisions directly between photons (there are indirect, higher-order photon interactions when quantum vacuum polarization is taken into account). However, a photon gas is a little complicated because you can't really model it as a typical non-relativistic fluid (because photons are by definition relativistic particles). To really model a photon gas, you have to use a relativistic stress-energy tensor, and the only ones I've ever seen for a "photon gas" (such as the one used to describe a radiation-dominated universe) all have nonzero pressure components.

A gas is a fluid. "Fluid" does not just mean liquid.

The individual fermions that are bound together to form the He-4 atom do, yes. But He-4 atoms themselves, as particles in a fluid, don't. If they did, He-4 would not become superfluid at low enough temperatures; a superfluid is a Bose-Einstein condensate.

As long as you agree that there is *some* kind of repulsion between particles in a fluid, I have no problem. I was only objecting because you seemed to be saying that the *only* possible type of repulsion was electromagnetic, or perhaps you thought that's what I was saying. Obviously there are other types of repulsion possible, and it now seems to me that you agree with that.

It's good that the equations I posted seem right to you, since their derivations are a fundamental part of the kinetic theory of gases.

20. Sep 9, 2011

### zonde

Thanks for correction. I somehow associated "fluid" exclusively with liquid.

Oh, I didn't mean equations. As I understood your question it was if using ideal gas equations I can demonstrate that pressure is proportional to momentum density.