Does a gravitating body have to undergo gravitational collapse?

[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?

 

[PeterDonis]

Interesting question. A few comments:

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.

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.

Now if the heat will escape from inside of the body toward outside then inside will collapse.

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.)

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.

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.)

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

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).

 

[zonde]

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.)

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

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.

 

[PeterDonis]

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.

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.

 

[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.

 

[PeterDonis]

I think that I am more interested about temperature gradient that does not contribute to conduction.

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.

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.

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.

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.

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

 

[zonde]

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.

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.

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).

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

 

[PeterDonis]

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.

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.

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

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):

https://www.physicsforums.com/showthread.php?t=40391

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.

 

[zonde]

Your example shows this *only* if the rate of temperature increase is *exactly* the same as the rate of gravitational blueshift.

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.

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.)

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.

 

[PeterDonis]

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.

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.

Ordinary fluid pressure is momentum density i.e. motion of the individual particles.

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.

When individual particles are so close that they repel each other then this is called degeneracy pressure.

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.

I believe that electromagnetic interactions are what we can call "collisions".

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.

 

[zonde]

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.

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:

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.

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

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.

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?

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.

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.

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.

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

 

[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.

 

[PeterDonis]

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.

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

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.

How exactly your argument that "fluid pressure is caused by momentum *transfer*" is correcting my argument that "fluid pressure is momentum density"?

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.

Degeneracy pressure is not caused by Pauli exclusion principle. Degeneracy pressure is explained by Pauli exclusion principle.

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.

And there is no "ordinary electromagnetic repulsion between the individual particles" that would contribute to pressure if we speak about neutral fluid.

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.

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

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).

 

[PeterDonis]

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.

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:

[tex]P = nkT[/tex]

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

[tex]P = \rho R T[/tex]

where [itex]\rho[/itex] 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.

 

[utesfan100]

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.

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.

 

[PeterDonis]

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

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 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.

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).

 

[zonde]

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.

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?

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.

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

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.

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.

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).

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?

 

[zonde]

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:

[tex]P = nkT[/tex]

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

[tex]P = \rho R T[/tex]

where [itex]\rho[/itex] 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.

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.

 

[PeterDonis]

And your idea is that within body after reaching equilibrium (no net transfer of heat) temperature gradient is 0, right?

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.

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

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.

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.

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

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

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.

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?

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.

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.

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.

 

[zonde]

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

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

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.

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.

And I have to correct myself that I meant this proportionality does not hold for liquids (instead of fluids) or for real gases with high density once we talk about this. Err, maybe your question was about this part?

 

[PeterDonis]

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.

And I have to correct myself that I meant this proportionality does not hold for liquids (instead of fluids) or for real gases with high density once we talk about this. Err, maybe your question was about this part?

No, my point was that the ideal gas equation says that pressure is proportional to *energy* density, not *momentum* density. The two are not the same.

 

[zonde]

No, my point was that the ideal gas equation says that pressure is proportional to *energy* density, not *momentum* density. The two are not the same.

I got it now. And I think I have no objections to your statement.

 

[PeterDonis]

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.

I'm still thinking, but I have at least a basic heuristic picture that I want to go ahead and post.

Consider the following situation: we have two surfaces, similar to what zonde proposed in his "mirror" scenario, but without the mirrors.

Surface R is the surface of a gravitating body. We don't care what it's made of except that it is in some kind of static equilibrium. It is at some radius r.

Surface S is a surface above the gravitating body, at which the radiation temperature is in equilibrium with the CMBR; in other words, we are assuming there is basically a "heat bath" from this surface outward that is at a constant uniform temperature of 2.7 K (the CMBR temperature). Surface S is at some radius s which is larger than r, enough so that there is a measurable gravitational redshift/blueshift between the two surfaces.

Note that we will assume that surface S does not move; its radius s is constant. However, we will see that we need to allow surface R to move; its radius r may change.

Let's suppose that surface R starts out at the *same* temperature as surface S, 2.7 K. What will happen? Well, since the radiation coming into R from S is blueshifted, it will be at some temperature higher than 2.7 K. The outgoing radiation from R to S will be similarly redshifted, but there is a heat bath at S so S's temperature won't change (some extra heat from the heat bath is simply taken to compensate for the reduced heat coming from R). So the temperature at R will rise. This means that the total energy contained in the gravitating body whose surface is R will rise.

Can the body simply radiate this energy away and return to its original equilibrium, at 2.7 K at radius r? I don't see how, because the blueshifted radiation from S will keep on coming in and increasing its temperature again. So the body has to reach some new static equilibrium configuration that includes the extra energy.

How will R change its configuration to accommodate this increased energy? There are only three possibilities:

(1) It could stay at the same radius r. This would require the density and pressure in its interior to increase, until its temperature at the surface (R at the original radius r) was equal to the blueshifted temperature of the incoming radiation from S.

(2) It could *decrease* its radius, but compensate by increasing density and pressure even more than in #1 above, so that its surface temperature at the new smaller radius would be equal to the (higher) blueshifted temperature of radiation from S at that smaller radius.

(3) It could increase in radius until it came to a new equilibrium with a higher surface temperature that was just equal to the blueshifted temperature of incoming radiation from S at the larger radius (i.e., higher than 2.7 K, but not as high as in #1 above). In this case, it would be possible in principle for the central density and pressure of the body to be unchanged, by making the gradients of both gentler than they were before.

I believe that the only one of the above possibilities that would actually be realized is #3. This is because #1 and #2 both require the pressure in the body's interior to increase, and the natural outcome of that is for the body to expand. So while I could see #1 (or less likely #2) happening as part of a transient condition, I can't see them as a new stable equilibrium. Only #3 looks to me like a possible new stable equilibrium. (There may well be some slick argument based on the equations of hydrostatic equilibrium that makes this obvious.)

So we have two conclusions. First, at the surface of a gravitating body, where the only heat exchange with the rest of the universe is by radiation, the equilibrium temperature will be the "blueshifted" temperature of the background radiation of the universe as a whole. Thermal equilibrium maintained by radiation through vacuum *does* have to take into account the gravitational redshift/blueshift. Second, the body as a whole equilibrates itself to the background radiation temperature, properly blueshifted, by adjusting its radius.

What I'm wondering is whether the above analysis still applies when we look at the interior of the body, its temperature profile and heat transfer within it. I'm not sure it does, because the interior of the body is not vacuum. That adds two factors that may make a difference: first, there is nonzero pressure inside the body, whereas pressure in vacuum is zero; and the interior of the body can move heat around in other ways besides radiation. I'm still thinking about how that affects the analysis.

 

[zonde]

What I'm wondering is whether the above analysis still applies when we look at the interior of the body, its temperature profile and heat transfer within it. I'm not sure it does, because the interior of the body is not vacuum. That adds two factors that may make a difference: first, there is nonzero pressure inside the body, whereas pressure in vacuum is zero; and the interior of the body can move heat around in other ways besides radiation. I'm still thinking about how that affects the analysis.

If we use kinetic theory to model heat inside body then heat energy is average kinetic energy of individual particles. But for particle that is moving across different gravitation potentials kinetic energy changes.
So it seems to me that applying kinetic theory to body in gravitation field there should appear temperature gradient.

Higher pressure can result from higher temperature or from higher density but if we model conduction as collisions between particles then for the same density particles moving at higher speed will participate in more collisions then slower particle so to balance number of collisions between faster moving particle and slower moving particle we have to increase density of slower moving particles. So we can't compensate temperature gradient using different densities because density gradient would tend to be exactly opposite as needed for that compensation.

 

[PeterDonis]

If we use kinetic theory to model heat inside body then heat energy is average kinetic energy of individual particles. But for particle that is moving across different gravitation potentials kinetic energy changes.

For an individual particle, yes. But in kinetic theory we have to look at statistical averages of large numbers of particles, in equilibrium, not just at the dynamics of one individual particle. It could be that, averaged over time and over many particles, the kinetic energy loss when a single particle rises in the gravitational field is compensated by something else. Or it could be that some additional factors *add* to the kinetic energy change with height in the time average over many particles, so the actual temperature gradient is *greater* than the gradient in gravitational potential. (That's certainly the case with the Earth, for example.)

Higher pressure can result from higher temperature or from higher density but if we model conduction as collisions between particles then for the same density particles moving at higher speed will participate in more collisions then slower particle so to balance number of collisions between faster moving particle and slower moving particle we have to increase density of slower moving particles. So we can't compensate temperature gradient using different densities because density gradient would tend to be exactly opposite as needed for that compensation.

At a fixed pressure, you are correct: temperature and density would be expected to be inversely related. (In the second form of the ideal gas law I gave, the inverse relationship at fixed pressure is obvious.) However, for a body in a gravitational field, pressure and density are related by hydrostatic equilibrium. In the non-relativistic case, assuming spherical symmetry, the equation is:

[tex]\frac{dP}{dr} = - \rho g[/tex]

where g is the "acceleration due to gravity" at radius r. (When you include relativistic effects, the equation becomes a lot more complicated: it becomes the Tolman-Oppenheimer-Volkoff equation. But even the above non-relativistic version is enough to show the basic form of the pressure-density relationship.)

What this means is that, as you go deeper into the interior of the body, pressure has to *rise* in order to maintain hydrostatic equilibrium. As pressure rises, both density *and* temperature will rise also. The question is how much each will rise, relative to the other. If the temperature gradient is to be equal to the gravitational potential gradient, and no more, then the temperature can't rise very much; almost *all* of the rise in pressure has to be compensated by a rise in density. But your argument above would seem to suggest the opposite: that more of the rise in pressure should be compensated by a rise in temperature, and less by a rise in density.

I should also note that I think I left out a key factor in my analysis even of the "vacuum" case in my previous post. I'll do a separate follow-up post about that.

 

[PeterDonis]

I have remembered that in my analysis in post #23 I left out a key factor: that a self-gravitating body (meaning a body whose structure is determined only by its own gravity), interacting with the rest of the universe only via absorbing and emitting radiation, has a negative heat capacity. If you add energy to it (it absorbs more radiation than it emits), it expands, but as it expands, it cools (because it has to expand against its own gravity); conversely, if it loses energy (by emitting more radiation than it absorbs), it contracts and heats up (because it "falls" in its own gravity). See the Wikipedia page here:

http://en.wikipedia.org/wiki/Heat_capacity#Negative_heat_capacity_.28stars.29

I *believe* that the above applies in the case of a body in hydrostatic equilibrium, provided that the equation of state is such that the pressure depends on the temperature (in other words, it is "kinetic pressure" instead of some non-temperature source such as degeneracy pressure). If that is correct, then I was wrong in post #23 when I said that the gravitating body would rise in temperature if it were at 2.7K and absorbed blueshifted CMBR radiation that was at greater than 2.7K. The body would indeed gain energy, and expand, but as it expanded it would *cool*. Conversely, if the body were at *greater* than the blueshifted CMBR temperature at its radius, it would indeed lose energy and contract, but as it contracted it would *heat up*.

This means that it is basically impossible for a self-gravitating body without degeneracy pressure to be in true thermal equilibrium via radiation with the surrounding universe. But it also means that, without some other means of maintaining its radius, *any* such gravitating body which is at a temperature greater than the surrounding radiation temperature *must* eventually collapse. Stars avoid this fate by generating energy within themselves by nuclear reactions to compensate for the energy lost by radiation. White dwarfs and neutron stars, of course, are supported by degeneracy pressure and will not collapse.

 

[zonde]

For an individual particle, yes. But in kinetic theory we have to look at statistical averages of large numbers of particles, in equilibrium, not just at the dynamics of one individual particle. It could be that, averaged over time and over many particles, the kinetic energy loss when a single particle rises in the gravitational field is compensated by something else. Or it could be that some additional factors *add* to the kinetic energy change with height in the time average over many particles, so the actual temperature gradient is *greater* than the gradient in gravitational potential. (That's certainly the case with the Earth, for example.)

If we take into account that there is some repulsion between particles then exchange of kinetic energy happens at some distance between mass centers of particles and that should lower temperature gradient.

But another point is that temperature gradient is determined by other rules here than in case of radiation. The thing is that kinetic energy is converted to potential energy not proportionally to the kinetic energy but proportionally to the mass:
[tex]U_{pot}=-\frac{MGm}{r}[/tex]
So you can reach zero temperature (all kinetic energy converted into potential energy) at finite radius.

Therefore it's not clear at all that your connection with Earth case is correct.

I have remembered that in my analysis in post #23 I left out a key factor: that a self-gravitating body (meaning a body whose structure is determined only by its own gravity), interacting with the rest of the universe only via absorbing and emitting radiation, has a negative heat capacity. If you add energy to it (it absorbs more radiation than it emits), it expands, but as it expands, it cools (because it has to expand against its own gravity); conversely, if it loses energy (by emitting more radiation than it absorbs), it contracts and heats up (because it "falls" in its own gravity). See the Wikipedia page here:

http://en.wikipedia.org/wiki/Heat_capacity#Negative_heat_capacity_.28stars.29

I *believe* that the above applies in the case of a body in hydrostatic equilibrium, provided that the equation of state is such that the pressure depends on the temperature (in other words, it is "kinetic pressure" instead of some non-temperature source such as degeneracy pressure). If that is correct, then I was wrong in post #23 when I said that the gravitating body would rise in temperature if it were at 2.7K and absorbed blueshifted CMBR radiation that was at greater than 2.7K. The body would indeed gain energy, and expand, but as it expanded it would *cool*. Conversely, if the body were at *greater* than the blueshifted CMBR temperature at its radius, it would indeed lose energy and contract, but as it contracted it would *heat up*.

This means that it is basically impossible for a self-gravitating body without degeneracy pressure to be in true thermal equilibrium via radiation with the surrounding universe. But it also means that, without some other means of maintaining its radius, *any* such gravitating body which is at a temperature greater than the surrounding radiation temperature *must* eventually collapse. Stars avoid this fate by generating energy within themselves by nuclear reactions to compensate for the energy lost by radiation. White dwarfs and neutron stars, of course, are supported by degeneracy pressure and will not collapse.

Thanks for bringing up virial theorem. It certainly gives additional ground for discussion.

But I think you are letting out one crucial point when you are saying that there is no stable thermal equilibrium when ignoring other factors. Body is loosing it's heat from it's surface. You are assuming that loss of heat is proportional to the average temperature of the body and that surface temperature of the body is proportional to the average temperature.

But equilibrium temperature gradient within the body can make difference if it plays the role of additional variable. That's actually what I wanted to find out when I started this discussion.

 

[PeterDonis]

But another point is that temperature gradient is determined by other rules here than in case of radiation. The thing is that kinetic energy is converted to potential energy not proportionally to the kinetic energy but proportionally to the mass:
[tex]U_{pot}=-\frac{MGm}{r}[/tex]
So you can reach zero temperature (all kinetic energy converted into potential energy) at finite radius.

For a free "test body" moving only under the influence of gravity, yes, this is correct. So it would also be true for "dust", a collection of particles that do not interact with each other and so have zero pressure. However, I don't think it's true for a collection of interacting particles with non-zero pressure.

Body is loosing it's heat from it's surface. You are assuming that loss of heat is proportional to the average temperature of the body and that surface temperature of the body is proportional to the average temperature.

Actually, the assumption is a bit weaker than that. The assumption is that the surface temperature changes in the same direction as the average temperature; it is not required that the constant of proportionality between the two be independent of temperature or energy, only that it doesn't change sign.

But equilibrium temperature gradient within the body can make difference if it plays the role of additional variable. That's actually what I wanted to find out when I started this discussion.

I don't think the specific form of the equilibrium temperature gradient makes a difference as long as it remains monotonically increasing with decreasing radius. However, I haven't confirmed that mathematically.

 

[PeterDonis]

Thanks for bringing up virial theorem. It certainly gives additional ground for discussion.

One thing to bear in mind is that the virial theorem gets more complicated when you are dealing with a non-zero pressure. The simple form in which you state it only applies to the "zero pressure" case, when all particles are moving solely under the influence of gravity (meaning they all move on geodesics). I'll most a more detailed comment along these lines in the other thread where we are discussing the role of pressure in gravitational collapse.

 

[zonde]

For a free "test body" moving only under the influence of gravity, yes, this is correct. So it would also be true for "dust", a collection of particles that do not interact with each other and so have zero pressure. However, I don't think it's true for a collection of interacting particles with non-zero pressure.

This simply does not make sense. You can't have zero pressure if you have non-zero kinetic energy (I mean "internal" KE, not KE of bulk motion).

Maybe you mean pressure from repulsion like degeneracy pressure?

 

[Agerhell]

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.

I find your temperature discussion interesting. If we assume that we have an extremely dense planet such as that light originating from the surface of that body is redshifted by a factor of two (twice as long wavelenghts) as compared to light originating from a place were the gravitational effects are miniscule. What actually would happen is that people living on the surface of that planet will think that the temperature of the universe is not 2.7 Kelvin but 5.4 Kelvin. This as all electromagnetic waves from space will appear blueshifted by a factor of two which as Wien's displacement law shows will make the people on the planet think the temperature is twice as high. Do note that even if the light appears blueshifted on the surface of the planet it will not magically gain any energy from somewhere traveling down to the planet.

So the people on the planet will measure their temperature as 5.4 Kelvin, in thermal balance with the cosmic background radiation, but as the thermal radiation from the planet reaches a distant observer the observer will think that the planets surface temperature is only 2.7 Kelvin, in balance with the cosmic background radiation, because of the redshift. So both the distant observer and the planets inhibitants will think that the planet has the same temperature as the cosmic backgrund radiation, however they will disagree on the temperature of the cosmic background. Was this what you were wondering?

There are some problems with this explanation. For instance Stefan-Boltsmanns law states that the radiated power should go as the temperature to the power of four... That means that the people at the surface of the planet, that measures their temperature as 5.4 Kelvin would expect them to radiate 16 times as much energy then if the temperature had been 2.7 Kelvin. If the Stefan-Boltzmanns law holds for the people on the planet then the distant observer should be surprised that the planet radiates 8 times as much energy as a black-bodyradiator at 2.7 Kelvin should... Hmmm now I am somewhat confused...

 

[PeterDonis]

This simply does not make sense. You can't have zero pressure if you have non-zero kinetic energy (I mean "internal" KE, not KE of bulk motion).

Maybe you mean pressure from repulsion like degeneracy pressure?

No, I was talking about ordinary "kinetic" pressure. The key thing is that, in the simple form of the virial theorem that you stated, all the particles have to move on geodesics; that is, they must all be in free fall. That's what I meant by "moving solely under the influence of gravity". But if there is pressure present, then at least some of the particles can't be moving on geodesics; they can't be in free fall. The particles that make up the Earth, for example, are not in free fall (except, possibly, at the very center).

Conversely, if we consider, say, the Solar System, with the Sun and each of the planets, moons, asteroids, comets, etc. as a "particle", then to the extent we can view the system as a whole as a "fluid" at all, it is a fluid with zero pressure, because all of the "particles" are in free fall. None of them are "pushing on" any others the way individual pieces of the Earth push on each other. Perhaps it is better to say, in cases like the Solar System, that the model of the system as a "fluid" breaks down, rather than that it is a "fluid" with zero pressure. The reason the latter term is used is that the stress-energy tensor of the system as a whole, as used for example in matter-dominated FRW models in cosmology, is the same as the stress-energy tensor of a fluid with zero pressure.

 

[PeterDonis]

Stefan-Boltsmanns law states that the radiated power should go as the temperature to the power of four... That means that the people at the surface of the planet, that measures their temperature as 5.4 Kelvin would expect them to radiate 16 times as much energy then if the temperature had been 2.7 Kelvin. If the Stefan-Boltzmanns law holds for the people on the planet then the distant observer should be surprised that the planet radiates 8 times as much energy as a black-bodyradiator at 2.7 Kelvin should... Hmmm now I am somewhat confused...

The gravitational redshift/blueshift factor goes as the inverse square root of the radius. The radiated power goes as the inverse area of the sphere as a function of radius, i.e., as the inverse square of the radius. So the radiated power, measured at any given radius, goes as the fourth power of the temperature at that radius, adjusted for redshift/blueshift. Everything fits together fine.

 

[Agerhell]

The gravitational redshift/blueshift factor goes as the inverse square root of the radius. The radiated power goes as the inverse area of the sphere as a function of radius, i.e., as the inverse square of the radius. So the radiated power, measured at any given radius, goes as the fourth power of the temperature at that radius, adjusted for redshift/blueshift. Everything fits together fine.

I missed the time dilation. As the distant observer and the planets inhabitants have different measures of time as well as of frequency, a certain amount of energy radiated out from the planet per time unit in this particular example will be perceived as only a fourth of that amount of energy per time unit for the distant observer.

Does the planets inhabitants and the distant oberserver somehow disagree on the planets surface area, is that what you are saying? If you have some "gravitational length contraction" which is the same factor as the redshift and the time-dilation then the puzzle would be solved...

 

[PeterDonis]

I missed the time dilation. As the distant observer and the planets inhabitants have different measures of time as well as of frequency, a certain amount of energy radiated out from the planet per time unit in this particular example will be perceived as only a fourth of that amount of energy per time unit for the distant observer.

Does the planets inhabitants and the distant oberserver somehow disagree on the planets surface area, is that what you are saying? If you have some "gravitational length contraction" which is the same factor as the redshift and the time-dilation then the puzzle would be solved...

No, the surface area of the planet looks the same to the inhabitants and the distant observers. There is no puzzle; you agreed to the resolution yourself in the first paragraph of the above quote: "a certain amount of energy radiated out from the planet per time unit in this particular example will be perceived as only a fourth of that amount of energy per time unit for the distant observer." I.e., the observed energy radiated matches the fourth power of the observed temperature: one-fourth the energy for twice the distance for radiated power; 1/sqrt(2) the temperature for twice the distance due to gravitational redshift. One-fourth is the fourth power of 1/sqrt(2).