Question regarding GR wrt potential energy

[Buzz Bloom]

The motivation for this post comes from a discussion on another thread

https://www.physicsforums.com/threads/questions-regarding-mass-volume-x-density.851172/​

It seems worthwhile to me to try a simplification in this post to hopefully avoid the complications raised in the cited thread.

Assume an otherwise empty universe with a very thin spherical shell of radius R₀ at time t = 0, with a uniform density of dust particles. There is zero radiation, zero pressure, and zero temperature.

If it is reasonable to do so, make additional assumptions:

each dust particle is a very tiny sphere of one milligram mass;
there are N>>1 of these particle;
they are approximately arranged uniformly on the surface of the sphere; and
the fact that it is impossible to make the arrangement completely uniform (and therefore the density is not exactly uniform) is not relevant to the reasoning presented later in this post.​

If it is not reasonable to make these assumptions, I would appreciate a post explaining why so that I can try a restatement in a different and acceptably reasonable way.

There is a test particle of negligible mass in a circular orbit around the spherical shell at a radius of

R_t > R₀.​

Using Newtonian gravity, the orbital velocity is

v = √(N m G / R_t)​

GR gives an approximate modified velocity

v ~= √(N m G / (R_t - R_s))​

where R_s is the Swartzchild radius

R_s = 2 G N m / c².​

The source of the non-exactness for v is that the total gravitational mass of the shell is not exactly = N m. I understand that the dust particles in the shell interact with each other gravitationally, and this interaction creates an additional gravitational mass-energy M_g.

I would much appreciate some help in understanding how to calculate this additional mass-energy (hopefully without having to do any tensor calculus). This is the main intended focus for starting this thread.

The following is my naive attempt to make this calculation. I expect that it is probably wrong, but I am hopeful that I can be educated to understand why it is wrong.

If we ignore the needed added mass M_g, then what effect will result from the fact that all the particles in the shell for time t > 0 will be falling towards the center of the sphere? As the particles fall they acquire kinetic energy. Will the added kinetic energy has a gravitational effect on the orbit of the test particle?

I am guessing that the falling of the particles with increasing kinetic energy will not actually effect the orbit. If I am mistaken I hope someone will correct this guess. My reasoning is that the sum of kinetic and potential energy is a constant as the particles fall. The stationary shell at t=0 has potential energy and the addition of kinetic energy equals the reduction of potential energy.
With Newtonian gravity, the shell as a whole exerts an attraction on each particle as if the shell were replaced by a mass at the center equal to N m / 2. Therefore,

M_g = G N m / (2 c² R₀).
​

Some corrections may be needed.

First, the M_g mass based on the potential energy will also have a gravitation effect on the particles, so there will be second order potential energy effects, etc., adding to the above M_g.

Second, The calculation of M_g due to the potential energy may require a GR correction to the Newtonian calculation.

All comments are most welcome.

Regards,
Buzz

 

[pervect]

Because of a lack of pressure, the thin spherical shell is going to collapse - this is true in Newtonian mechanics as well as in GR.

I don't think I've seen the thin spherical shell collapse in MTW, but a solid spherical dust-cloud collapse is certainly in there, complete with a metric. As I recall is a time-reversed cosmological solution. I don't think there will be any real difference between the hollow spherical collapse and the solid spherical collapse in the exterior region.

Basically, what should happen is that there will be absolutely no effect on the orbit of a test particle outside the collapsing shell as the shell collapses. Assuming no cosmological constant, the metric outside the shell will be pure Schwarzschild, which seems to be the gist of your question as I understand it. Additionally, because the Schwarzschild metric is asymptotically flat and has no gravitational waves, you can use either/both of the ADM or the Bondi masses, which will be constant in the asymptotically flat space-time.

 

[Buzz Bloom]

Hi @pervect:

Thanks for your post.

I am wondering since you did not comment on my derivation of M_g that you intended this to mean that my derivation is OK. If so, do you also agree that the correction I described as the "first" correction is also correct? And is there also a need for the "second" correction? If so, what would this "second" correction look like?

I am trying to understand as best I can at an intuitive level some subtle aspects of mass-energy in GR before tackling the complexities of the ADM and Bondi energies as presented in Wikipedia, and perhaps later in MTW.

Regards,
Buzz

 

[PeterDonis]

The source of the non-exactness for v is that the total gravitational mass of the shell is not exactly = N m.

If you are defining "m" as the mass of a single dust particle as measured locally, then yes, this is correct; the total gravitational mass of the shell, measured as you describe, by putting a test particle in orbit about the shell, will be less than N m.

I understand that the dust particles in the shell interact with each other gravitationally, and this interaction creates an additional gravitational mass-energy Mg.

Only if you interpret "additional" to mean "negative". As above, the total gravitational mass of the shell--i.e., the total mass that appears in your equation for the orbital velocity of the test particle--is less than N m. The difference is called "gravitational binding energy"; it is the energy that would have to be extracted from the system in order to take N dust particles, each separated by a very large distance from all the others, and bring them together into a spherical shell with a finite radius that was at rest at some instant of time. (Note that if you let the dust particles fall inward starting from infinite radius, the gravitational binding energy is zero and the total mass is equal to N m. The reason the total mass of the shell in your scenario is less than N m is that you specified that the shell was at rest at a finite radius at some instant of time.)

 

[PeterDonis]

because the Schwarzschild metric is asymptotically flat and has no gravitational waves, you can use either/both of the ADM or the Bondi masses

There are complications here. Suppose we have N dust particles, each of rest mass m as measured locally, each separated by a very large distance from all the others (so their gravitational interaction is negligible), and all at rest relative to each other. We stipulate that the total configuration is spherically symmetric. This system will have ADM mass N m. It will also have Bondi mass N m, if we evaluate the Bondi mass at a point on future null infinity that is close to spacelike infinity (so no radiation has escaped).

Now suppose we bring all the dust particles together into a spherical shell with a finite radius and at rest at some instant of time. We have to extract energy from the dust particles in order to do that (for example, we could have them emit radiation that escapes to infinity). However, the ADM mass of the total system is still N m; the process of forming the shell can't change it. The Bondi mass, however, will change (provided we evaluate it at a point on future null infinity after the radiation has escaped); it will now be the value I described in my previous post (N m less the gravitational binding energy).

Note that, in doing all this, we are implicitly using a metric that is not the Schwarzschild metric everywhere; the spacetime has to have at least one region where the metric is something like the outgoing Vaidya metric (the region where the radiation is escaping). So if we put our test particle orbit far enough away (so that it is outside all of the dust particles both before and after they are brought together into the spherical shell), its orbit will change--it will see radiation passing by as the dust particles are brought together, and its orbital parameters will change to correspond to a smaller mass, a reduction from mass N m to mass N m minus the gravitational binding energy.

Of course, we could ignore all these complications and just stipulate that the system starts out with the dust particles at a finite radius at some instant of time, no radiation anywhere, and metric Schwarzschild outside that radius for all time. But then (even ignoring the issue of how the dust particles got to that point--they will have had to be in an expanding spherical shell before that instant of time, and we will end up having a spacetime with a white hole as well as a black hole, with all the attendant issues), we have no way of giving any physical meaning to the statement "the gravitational mass of the system is less than N m", because there was never any physical system with mass N m to compare it to.

 

[Buzz Bloom]

Hi Peter:
Thanks for your helpful posts.

I am wondering about the relationship between the concepts "binding energy" and "potential energy". It seems to me to be the same thing, at least in Newtonian physics. It may well be that physics vocabulary has changed since I studied undergraduate physics in the 1950s.

In Newtonian physics the energy E needed for a test particle at distance R from a point mass M to no longer be gravitationally bound to M is

E = (1/2) m v²​

where v is the escape velocity for the test particle. Therefore

E = (1/2) m (2 G M / R) = G M m / R,​

which is the potential energy (ignoring the sign).

Regards,
Buzz

 

[Jonathan Scott]

"Binding energy" is the potential energy within a single object (or system) due to its own components.

This particular example has an interesting complication in GR because there is no pressure involved, and because it is dynamic.

If one looks at the total energy of a static system in GR, then for each pair of particles which makes it up, their energy is effectively lowered by the time dilation due to the other, both ways, which means their total energy is decreased by twice their potential energy relative to each other. However, if the system is static, then the force between them times the distance between them is equal to their potential energy, but with a positive sign. This quantity is equal to the integral of the pressure due to the force between those two particles over the line between those particles. (Any non-gravitational forces must balance the gravitational forces over a complete plane for the system to remain static). If this pressure integral is treated as a contribution to the total energy (as in the "Komar Mass" expression) then the total energy matches the energy of the components minus the binding energy, as for Newtonian theory.

I personally find it very difficult to see how treating the pressure integral like energy could be valid, as the pressure can be temporarily changed, and can of course be zero temporarily in a situation like a non-interacting dust cloud. This certainly doesn't work in the Newtonian equivalent. This is the subject of a "Tolman Paradox" (not the only one, the "antitelephone" being more famous), where Tolman pointed out that it seems that GR predicted that changes in the internal structure of a star could abruptly affect its gravitational field.

An alternative view which does match up with the Newtonian view but seems to look at things very differently from the normal GR view is to assume that there is effectively some positive energy in the gravitational field, given in Newtonian terms by $(1/8\pi G) g^2$ for field $g$, in an analogous way to the energy density in an electrostatic field, as described for example in the article "Gravitational field energy density for spheres and black holes" by D Lynden-Bell and J Katz: http://adsabs.harvard.edu/full/1985MNRAS.213P..21L

I think that this "energy density of the field" is closely related to Einstein's original "gravitational energy" pseudotensor. It describes the effective energy as seen from a particular point of view, and is useful for comparison with the Newtonian viewpoint.

With this alternative view, the energy of each source mass is reduced by the time dilation of the potential, again making the total energy decrease by twice the Newtonian potential energy, but the energy of the field then adds back the same amount as the potential energy (which can be shown by integrating it over all space in a similar way to the standard electromagnetic energy calculation), making the overall energy decrease correctly match the Newtonian potential energy.

 

[PeterDonis]

I personally find it very difficult to see how treating the pressure integral like energy could be valid, as the pressure can be temporarily changed

But if it changes, then that will cause changes in other stress-energy tensor components, because of the local conservation law $\nabla_\mu T^{\mu \nu} = 0$. Also, you have to include in the analysis anything that is confining the matter under pressure (for example, if you have a gas inside a container, you have to include the stresses in the container walls). As you note, in a static system, including these items in the integral ensures that everything balances correctly.

The different feature of the OP's example is that the system is not static. That means the Komar mass can't even be defined, since it requires a timelike Killing vector field, which doesn't exist in the OP's example. It also means that concepts like "gravitational binding energy" are problematic in general; however, the OP's assumption of spherical symmetry avoids many of those problems, since we can describe his scenario using a combination of particular known solutions (Schwarzschild and outgoing Vaidya, as I posted previously) whose properties are simple enough to allow a thought experiment like the one I described--taking a bunch of very widely separated dust particles and bringing them together into a spherical shell at rest at an instant at a finite radius--to yield reasonable results.

 

[pervect]

I won't go into the details, due to a lack of time, and a feeling that this has been talked about before without much success, but I do want to point out that according to the textbooks (specifically MTW's Gravitation", localizing the energy of the gravitational field is, unfortunately not possible. See section $20.4, pg 466-467 of MTW's "Gravitation".

If one doesn't have the textbook, it's possible to use Google search to find the relative section. Searching for following rather snappy quote (use quotation marks to indicate to the search engine that it's an exact quote) should find the relevant section.

""No, the question is wrong. The motivation is wrong. The result is wrong. The idea is wrong"

It should come back as "Gravitation - Part 3 - Page 467 - Google Books Result".

 

[Buzz Bloom]

making the overall energy decrease correctly match the Newtonian potential energy.

Hi @Jonathan Scott:

If I interpret the above correctly, you are agreeing that the orbital velocity squared of the test particle would be
v² = (G M /(R-R_s))
where R is the radius of the test particle orbit, and R_s is the Shwartzchild radius corresponding to the total effective mass M_s of the shell
R_s = 2 G M_s / c².
The effective mass M is calculated by
M_s = M_N - M_g
where M_N is the Newtonian mass
M_N = N m
and
M_g is the mass equivalent of the gravitational binding energy of the shell
M_g = G N m / (R₀ c²)
where R₀ = the radius of the shell at time t = 0.

Therefore

Please let me know if I have misunderstood you.

Regards,
Buzz

 

[Jonathan Scott]

I have done no specific calculation, and I suspect that mixing the Newtonian approximation for kinetic energy with the Schwarzschild solution will produce inaccurate results.

What I was pointing out is that for a semi-Newtonian approximation for GR involving multiple masses, one can preserve Newtonian conservation of energy by assuming that the rest energy of each object is reduced by the local time dilation factor due to other masses (which reduces both sides by the potential energy of the relevant pair of masses), but that there is positive energy in the gravitational field (or more usefully in the difference between the squares of the fields of the individual masses and the square of their combined field) which balances it all out. With that scheme, any change in (relativistic) kinetic energy is equal and opposite to the change in potential energy to maintain a constant total.

 

[Buzz Bloom]

Hi @Jonathan Scott, @PeterDonis, and @pervect:

Jonathan has pointed out to me that my derivation of a formula for the orbital velocity of a test particle, given the assumptions for the thought experiment described in my post #1, is likely to be inaccurate. I am assuming that it should be possible to derive such an accurate formula, or if not, an equation from which a numerical calculation can be made with the accuracy limited only by the numerical methods used. I also gather from previous posts that none of you know any specific references that have explored this specific problem. I would much appreciate any specific help on how to go about developing such a formula or a numerically solvable equation.

Regards,
Buzz

 

[PeterDonis]

my derivation of a formula for the orbital velocity of a test particle, given the assumptions for the thought experiment described in my post #1, is likely to be inaccurate

For a test object in a circular orbit around a mass $M$, the correct relativistic formula for orbital velocity is $v^2 = GM / \left( R - R_\text{s} \right)$. If you check, you will see that this gives $v = c$ for $R = 3GM / c^2$, or $R = (3/2) R_\text{s}$. So there are no circular orbits possible inside that radius. (Also, it turns out that the circular orbits with $R < 6 GM / c^2$, or $R < 3 R_\text{s}$, are unstable against small perturbations). But I don't see that that creates any issue in your scenario.

So your formula for orbital velocity in terms of $M$ is correct; what you need to focus on is how $M$, the mass of the shell, is determined. It will be less than $N m$, as has already been discussed, and the difference can be interpreted as "gravitational binding energy". You might want to start by looking at the case of a spherical shell of matter being slowly lowered onto a static spherical mass like a planet. If the planet starts with mass $M$ and has radius $R$ ("radius" here means "circumference divided by $2 \pi$), and the shell starts at infinity with mass $m$ and is slowly lowered until it is at radius $R$, how much energy is extracted by the slow lowering process? If we call that energy $E$, then the final mass of the planet + shell should be $M + m - E$, so $E$ can be thought of as the "gravitational binding energy" of the shell. Then consider how the same process would work if the "planet" wasn't actually there, so $M = 0$.

 

[Buzz Bloom]

Hi Peter:

Thanks very much for your helpful post.

I intend to try to work out a formula for v₂, and If that fails, I will try to develop a spreadsheet to perform a numerical calculation. I will post the result when I finish and ask for someone to check it for correctness.

Regards,
Buzz

 

[Jonathan Scott]

Apologies for my not having read the calculation carefully enough when I suggested it might involve a Newtonian approximation, and my thanks to PeterDonis for picking that up.

I think you can probably start with a massless shell (which will not have any binding energy and will not cause any potential) then integrate adding mass to it. That will tell you what the total mass should be with the correct binding energy taken into account. However, that will not be equal to the normal GR way of adding up the total energy of a static object in that case because of the absence of pressure (and the dynamic situation), which is what I was trying to say before. I think that the GR way of integrating the total energy would be equivalent to multiplying the rest energy of the material which was used to create the shell by the time dilation factor corresponding to the potential at the surface of the shell, and that would be less than the original rest energy of the material by twice the potential energy. So it's not clear to me whether the M used in the Schwarzschild solution to describe what happens should be less than the original rest energy of the material by once or twice the potential energy, and if anyone can clarify that I'd be very interested to know.

 

[PeterDonis]

it's not clear to me whether the M used in the Schwarzschild solution to describe what happens should be less than the original rest energy of the material by once or twice the potential energy, and if anyone can clarify that I'd be very interested to know.

I know what I think the answer is, but I'd prefer to wait until Buzz has posted his calculations before giving it away. But the key idea, as I said before, is to imagine the shell being slowly lowered from infinity in a spherically symmetric manner, and ask how much energy is extracted during the slow lowering process. To fully match the scenario posed in the OP, we would imagine that, once the shell has been slowly lowered to radius $R_0$, it is then released into free fall, so at the instant of release, all of the dust particles in the shell are at rest relative to each other.

 

[Buzz Bloom]

I think that the GR way of integrating the total energy would be equivalent to multiplying the rest energy of the material which was used to create the shell by the time dilation factor corresponding to the potential at the surface of the shell, and that would be less than the original rest energy of the material by twice the potential energy.

Please explain "the time dilation factor corresponding to the potential at the surface of the shell". What is the math for this?

Regards,
Buzz

 

[Jonathan Scott]

To fully match the scenario posed in the OP, we would imagine that, once the shell has been slowly lowered to radius $R_0$, it is then released into free fall, so at the instant of release, all of the dust particles in the shell are at rest relative to each other.

But immediately before release, the integral of the pressure is equal to the binding energy and immediately after release (even before anything gets moving), the integral of the pressure is zero. That's what bothers me about this case.

 

[Jonathan Scott]

Please explain "the time dilation factor corresponding to the potential at the surface of the shell". What is the math for this?

Regards,
Buzz

That's simply the time factor in the Schwarzschild metric (proper time per coordinate time).

But I may be confused about that too; I remember an exercise in MTW about the binding energy which never seemed to make sense.

 

[PeterDonis]

immediately before release, the integral of the pressure is equal to the binding energy

Not if the shell has zero pressure. Imagine each of the dust particles in the shell being suspended by a string; the strings are used to slowly lower each dust particle (and extract energy from each particle while doing so); then, once all the particles are stationary at radius $r_0$, the particles are all released at the same instant. Then the shell has zero pressure throughout; the force keeping each particle from free-falling during the slow lowering process is the tension in each of the strings. When each particle is released, the tension in its string goes to zero at the release point; a wavelike disturbance then propagates up each string as the tension in each infinitesimal piece of the string goes to zero.

 

[Jonathan Scott]

Not if the shell has zero pressure. Imagine each of the dust particles in the shell being suspended by a string; the strings are used to slowly lower each dust particle (and extract energy from each particle while doing so); then, once all the particles are stationary at radius $r_0$, the particles are all released at the same instant. Then the shell has zero pressure throughout; the force keeping each particle from free-falling during the slow lowering process is the tension in each of the strings. When each particle is released, the tension in its string goes to zero at the release point; a wavelike disturbance then propagates up each string as the tension in each infinitesimal piece of the string goes to zero.

If you describe any self-contained piece of apparatus which performs that suspension, then you will find that the volume integral of the pressure over that apparatus is still equal to the binding energy, and it drops instantly to zero (well, at the speed of propagation of tension in the medium) when the shell is released.

The simplest self-contained model is that the dust is initially in the form of a rigid shell when then crumbles isotropically. Again, you'll find that if you integrate the pressure over the volume of the shell the result is equal to the binding energy. (The pressure may locally include additional forces for purpose of rigidity, but all the forces which are not counteracted only by gravity must cancel out over any given plane otherwise the initial state will not be static).

I think your suggestion is equivalent to physically removing the apparatus containing the pressure term at the time of release. I presume that would mean that by Birkhoff's theorem, when the apparatus passes out of the the sphere containing the relevant orbit, the orbit would change accordingly.

However, for my suggestion of a crumbling shell, I don't know where the pressure term goes. If one considers the relationship of tension and elasticity in the material, it is true that some energy can be stored in the tension, but that energy is less the stiffer the material, and there is no reason to assume it can store anything which adds up to the binding energy. (You can in fact model the suspending material by a spring to compare the energy stored to the total potential energy; it turns out that because the maximum force is fixed as the force of attraction, you could only store enough energy in it if the initial length stretched to infinity).

 

[jartsa]

I just want to say something about things under stress in gravity field.

There exists an anisotropy of the coordinate-speed of light in a gravity field. So also coordinate speed of sound is anisotropic in any object in gravity field. So inside a static compressed rod in a gravity field there are more phonons moving upwards than downwards, which means the rod has momentum in the upwards direction in the frame where the rod is static.

So have you guys taken this into account?I'm talking about this anisotropy:
Time for light to travel down to black hole event horizon: finite
Time for light to travel up from black hole event horizon: infinite

 

[PeterDonis]

inside a static compressed rod in a gravity field there are more phonons moving upwards than downwards, which means the rod has momentum in the upwards direction in the frame where the rod is static.

No, this is not correct. Anisotropy in the speed of light (or sound, assuming you are correct about that--I don't think you can assume it, and you would need a detailed material model to derive any such result) does not mean more photons are moving up than down. Coordinate speeds are just coordinate speeds; they don't have any physical meaning in and of themselves.

 

[Buzz Bloom]

Hi Jonathan and Peter:

Thanks for your posts.

Unfortunately I am still in the dark about the mathematical form for the binding energy that goes with the scenario I am investigating. Can either of you help me with this?

I also don't understand why calculating the escape velocity of a single particle is the shell, and multiplying by Nm/c², fails to correctly calculate the binding energy of the shell at t=0. The equation I used to make this calculation may be incorrect because I failed to take into account the effect of the non-flat space on escape velocity. I was not able to find any GR based equation for escape velocity, although Wikipedia did include the equation I used for circular orbit velocity with a Schwartzschild metric correction.

Regards,
Buzz

 

[PeterDonis]

I think your suggestion is equivalent to physically removing the apparatus containing the pressure term at the time of release. I presume that would mean that by Birkhoff's theorem, when the apparatus passes out of the the sphere containing the relevant orbit, the orbit would change accordingly.

I think this is correct, yes. Basically, during the slow lowering process, the energy being removed from the dust is being stored in the apparatus. Then, when the dust is released, the stored energy is propagated outward, ultimately to infinity; and when it passes the finite radius of the orbit of the test object, that orbit will change to reflect the (reduced) mass of the dust shell alone, rather than dust + apparatus.

 

[Jonathan Scott]

I think this is correct, yes. Basically, during the slow lowering process, the energy being removed from the dust is being stored in the apparatus. Then, when the dust is released, the stored energy is propagated outward, ultimately to infinity; and when it passes the finite radius of the orbit of the test object, that orbit will change to reflect the (reduced) mass of the dust shell alone, rather than dust + apparatus.

But surely my model of a crumbling shell should give the same result but doesn't have a mechanism to remove energy?
I must admit I haven't looked into this in detail for some time, and I'm quite rusty on it. There must be some explanation in GR, but I don't know what it is.

 

[PeterDonis]

surely my model of a crumbling shell should give the same result but doesn't have a mechanism to remove energy?

First, how did the shell get to be static at that radius in the first place? Was it slowly lowered? Did it collapse from some larger radius, and then radiate away the kinetic energy from the collapse? If it radiated energy away before, it could do so again.

Second, any "crumbling shell" model has to explain where the pressure goes, locally, when it vanishes. It can't just disappear because that would violate $\nabla_\mu T^{\mu \nu} = 0$. It has to be transferred to some other stress-energy component locally, before it can be "removed" to infinity. In a typical collapse scenario--for example, a star that stops generating heat--it gets transferred to the inward momentum of the collapsing matter; then, as the collapsing matter particles collide with each other, it gets transferred to radiation that escapes to infinity.

 

[Jonathan Scott]

Second, any "crumbling shell" model has to explain where the pressure goes, locally, when it vanishes. It can't just disappear because that would violate $\nabla_\mu T^{\mu \nu} = 0$. It has to be transferred to some other stress-energy component locally, before it can be "removed" to infinity. In a typical collapse scenario--for example, a star that stops generating heat--it gets transferred to the inward momentum of the collapsing matter; then, as the collapsing matter particles collide with each other, it gets transferred to radiation that escapes to infinity.

Remember that the zero divergence means that energy and momentum are conserved (and continuous). It does not mean that pressure or stress is conserved, although there are obviously temporary transients in those if the pressure suddenly changes.
Perhaps the solution lies in the fact that the "pressure" in the tensor is actually simply a two-way flow of momentum, so if dust particles start moving past each other instead of pushing against each other that would still count as pressure on average, but that definitely doesn't change the total energy.

 

[PeterDonis]

Remember that the zero divergence means that energy and momentum are conserved (and continuous). It does not mean that pressure or stress is conserved, although there are obviously temporary transients in those if the pressure suddenly changes.

I didn't say pressure was "conserved"; I said its behavior is constrained by the zero divergence condition--that is what tells you "where the pressure goes" when it changes. The zero divergence condition applies to all stress-energy tensor components, not just energy and momentum. Pressure and stress are stress-energy tensor components. Try writing down an individual component of $\nabla_\mu T^{\mu \nu}$ where $\nu$ is a spatial index, not the time index (the latter is what gives what is normally viewed as energy-momentum conservation).

 

[Buzz Bloom]

I also don't understand why calculating the escape velocity of a single particle is the shell, and multiplying by Nm/c2, fails to correctly calculate the binding energy of the shell at t=0.

I just noticed that I had carelessly made two errors in this sentence. Here is what I should have said:

I also don't understand why calculating the escape velocity of a single particle in the shell, and multiplying its square by Nm/c2, fails to correctly calculate the binding energy of the shell at t=0.
​

Regards,
Buzz

 

[Jonathan Scott]

I didn't say pressure was "conserved"; I said its behavior is constrained by the zero divergence condition--that is what tells you "where the pressure goes" when it changes. The zero divergence condition applies to all stress-energy tensor components, not just energy and momentum. Pressure and stress are stress-energy tensor components. Try writing down an individual component of $\nabla_\mu T^{\mu \nu}$ where $\nu$ is a spatial index, not the time index (the latter is what gives what is normally viewed as energy-momentum conservation).

That sounds like wishful thinking to me; I'd like to see a worked example, but I think I satisfied myself that nothing magic happens some time ago after we were discussing the Komar mass expression.
The divergence condition applies separately to each of the four components, energy and each component of momentum, and effectively says that the amount moving through the three spacelike walls of the infinitesimal box at each point is equal and opposite to the rate of change with respect to time within that box. It doesn't impose any condition that's not already there in Newtonian mechanics (except of course that the covariant aspect means that it is conserved relative to local space rather than the coordinate system).
Although there are transient effects of a wave of unbalanced pressure passing through the system, I'm not aware of anything which prevents the initial state having a certain amount of energy-momentum plus pressure and the final state having the same amount of energy-momentum but no pressure.
It may be possible to calculate the GR stuff in this case from first principles, but it seems to be beyond my time and skills at present.

 

[Jonathan Scott]

Although there are transient effects of a wave of unbalanced pressure passing through the system, I'm not aware of anything which prevents the initial state having a certain amount of energy-momentum plus pressure and the final state having the same amount of energy-momentum but no pressure.

I should add that of course there is a significant difference between those situations, in that the initial state is static, and in the final state the lack of pressure means that the situation is dynamic and acceleration must be present, although if the pressure change is sudden, the energy and momentum will not have had time to change significantly from the initial state.

 

[PeterDonis]

The divergence condition applies separately to each of the four components, energy and each component of momentum, and effectively says that the amount moving through the three spacelike walls of the infinitesimal box at each point is equal and opposite to the rate of change with respect to time within that box.

Exactly--it relates the quantity of one thing to the rate of change of something else. See below.

I'm not aware of anything which prevents the initial state having a certain amount of energy-momentum plus pressure and the final state having the same amount of energy-momentum but no pressure.

There's nothing which prevents that, as long as the change in pressure is properly related to the rate of change of something else. See below.

in the final state the lack of pressure means that the situation is dynamic and acceleration must be present, although if the pressure change is sudden, the energy and momentum will not have had time to change significantly from the initial state

The energy and momentum won't have had time to change significantly, but their rate of change will have to. That's what the covariant divergence equation says.

(Note that the description you gave implicitly uses a global, non-inertial coordinate chart, in which the energy/momentum is not changing in the initial state but is in the final state. In a local inertial frame, in the initial state, the pressure that is present is related to the energy flux--which is there because an infinitesimal piece of the matter, being under pressure, has nonzero proper acceleration and therefore is changing its energy/momentum as viewed in a local inertial frame. In the final state, the pressure is zero and the infinitesimal piece of matter is now in free fall, with zero change in energy/momentum as viewed in a local inertial frame. The covariant divergence equation requires the two changes--difference in pressure vs. difference in rate of change of energy/momentum--to be equal.)

 

[Jonathan Scott]

Now I'm confused. Thanks for playing, but I think you just moved the goalposts off the pitch, to use a soccer analogy.

The conventional assertion is that in GR, the diagonal pressure terms in the tensor effectively contribute to the curvature of space in the same way as energy, and that is what I find hard to believe, because the pressure can drop to zero (almost instantaneously) without any change in the energy-momentum. If Birkhoff's theorem covers this case, the external field should be unchanged by any internal spherically symmetrical changes, but the pressure can abruptly change while the other terms remain unchanged. I don't see how any rates of change relate to this effect at all.

A very similar point is the subject of one of Tolman's paradoxes, and in some other thread recently I saw a reference to a paper which attempts to resolve that and concludes, roughly, after expanding the GR tensors, that the pressure term doesn't actually affect the external field but something else does, having an equivalent effect. I tried to understand the paper at the time, but it seemed to be looking at things in a way which I didn't trust, so I couldn't trust the conclusion either.

I don't understand this well enough to find the answer at the moment, and I'm not even sure I'd recognize it if I saw it. However, I'll add it to my list of things to think about if I ever get the time before my brain totally wears out, and if I make any interesting progress I'll start a new thread on it.

 

[Buzz Bloom]

I would like to return the discussion to the questions I asked in post #1. The discussion has suggested that the equation I wrote (repeated below) for the self potential energy of a spherical dust shell with zero velocity, temperature and pressure might be wrong, but there have been no specific statements that it is definitely wrong.

M_g = G N m / (2 c² R₀).

Rather, much of the discussion has been about how the assumed t=0 configuration of the shell got to be that way during t<0.

Why is the history of the shell during t<0 relevant to a calculation of its potential energy at t=0? Why isn't the formula for mass equivalent M_g of the potential energy at t=0 correct?

I have searched the internet for any discussion of potential energy with respect to GR, or the Schwartzschild metric, and failed to find anything. The conclusion that seems plausible to me is that there is no GR change to the Newtonian calculation for potential energy with respect to a spherically symmetric mass distribution. If someone definitely know this to be wrong, I would much appreciate a correction.

Regards,
Buzz