# Question regarding GR wrt potential energy

1. Jan 11, 2016

### Buzz Bloom

The motivation for this post comes from a discussion on another thread
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
Rt > R0. ​
Using Newtonian gravity, the orbital velocity is
v = √(N m G / Rt)​
GR gives an approximate modified velocity
v ~= √(N m G / (Rt - Rs))​
where Rs is the Swartzchild radius
Rs = 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 Mg.

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 Mg, 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,
Mg = G N m / (2 c2 R0).
Some corrections may be needed.

First, the Mg 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 Mg.

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

Regards,
Buzz

Last edited: Jan 11, 2016
2. Jan 11, 2016

### pervect

Staff Emeritus
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.

3. Jan 12, 2016

### Buzz Bloom

Hi @pervect:

I am wondering since you did not comment on my derivation of Mg 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

4. Jan 12, 2016

### Staff: Mentor

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.

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.)

5. Jan 12, 2016

### Staff: Mentor

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.

6. Jan 13, 2016

### Buzz Bloom

Hi Peter:

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

7. Jan 13, 2016

### 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.

8. Jan 13, 2016

### Staff: Mentor

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.

9. Jan 13, 2016

### pervect

Staff Emeritus
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".

10. Jan 16, 2016

### Buzz Bloom

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-Rs))
where R is the radius of the test particle orbit, and Rs is the Shwartzchild radius corresponding to the total effective mass Ms of the shell
Rs = 2 G Ms / c2.
The effective mass M is calculated by
Ms = MN - Mg
where MN is the Newtonian mass
MN = N m
and
Mg is the mass equivalent of the gravitational binding energy of the shell
Mg = 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

11. Jan 16, 2016

### 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.

12. Jan 16, 2016

### 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

13. Jan 16, 2016

### Staff: Mentor

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 $M = 0$.

14. Jan 16, 2016

### Buzz Bloom

Hi Peter:

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

15. Jan 16, 2016

### 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.

16. Jan 16, 2016

### Staff: Mentor

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.

17. Jan 16, 2016

### Buzz Bloom

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

Regards,
Buzz

18. Jan 17, 2016

### Jonathan Scott

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.

19. Jan 17, 2016

### Jonathan Scott

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.

20. Jan 17, 2016

### Staff: Mentor

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.

21. Jan 17, 2016

### Jonathan Scott

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).

22. Jan 17, 2016

### 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?

Time for light to travel down to black hole event horizon: finite
Time for light to travel up from black hole event horizon: infinite

Last edited: Jan 17, 2016
23. Jan 17, 2016

### Staff: Mentor

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.

24. Jan 19, 2016

### Buzz Bloom

Hi Jonathan and Peter:

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

25. Jan 19, 2016

### Staff: Mentor

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.