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A Pressure and Newtonian potential energy

  1. Oct 25, 2017 #1

    Jonathan Scott

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    Recent discussions on pressure as a source of gravity and the related Tolman paradox have reminded me that few people seem to be aware that even in Newtonian gravity the pressure is related to the potential energy, which I've mentioned a few times on these forums before.

    This is easy to show, but a bit messy to describe. I'd like to know whether this is a known named result, and whether someone can give me a simpler way of describing it in Newtonian terms (preferably without requiring the use of GR-style tensor notation as in the Komar mass expression).

    Specifically, if you have an isolated static system of masses in free space internally supported in equilibrium by any means (rods, strings, springs, balloons or whatever), then if you integrate the normal pressure component over planes perpendicular to three perpendicular axes through the whole system, the total of the three integrals is a positive quantity equal and opposite to the gravitational potential energy of the system.

    To show this, you can divide up the system conceptually into point masses where each one is joined to all other point masses by lines representing their gravitational interaction. For the system to be static, the total force (including gravitational) across any plane must be zero. The gravitational force along the line between masses ##m_i## and ##m_j## is then ##-Gm_i m_j/r_{ij}^2##, so this is the contribution to the overall force across any plane cutting the line between those masses, which must be equal and opposite to the integral over that plane of the pressure. The force is the same at all points on the line between the masses. If the pressure is integrated along that line, that simply multiplies the result by ##r_{ij}## giving a contribution ##Gm_i m_j / r_{ij}## for that pair of masses, equal and opposite to the potential energy between those masses.

    More generally, if the displacement vector ##(x,y,z)## between two masses is not perpendicular to the plane, the normal pressure is reduced by the cosine of the angle between the displacement line and the perpendicular to each plane, and so is the perpendicular component of the length of the line, so the contributions for three planes add up as ##x^2/r^2##, ##y^2/r^2## and ##z^2/r^2##, adding up to the same total as the original case. As these quantities add linearly for all contributing point masses, the total for the system is equal and opposite to the total potential energy of the system.

    This means for example that if the system just consists of a perfect fluid held together by gravity, the gravitational potential energy is equal and opposite to the volume integral of three times the pressure.

    Note that in the Newtonian case it is clear that although this pressure integral is equal and opposite to the potential energy for a static system, it cannot be identified with the potential energy, because for example if a support is removed and the system starts to collapse, the Newtonian potential energy is initially unaffected, but terms in the pressure integral can vanish immediately, at least to orders of magnitude smaller than their original values.

    (For this reason, I feel that Tolman was on to something and there is something wrong either with the conventional interpretation that pressure is a source of gravity in GR or with the field equations themselves, but that's a different topic).
     
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  3. Oct 25, 2017 #2

    PeterDonis

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    Not to my knowledge, but if it isn't it is somewhat surprising, since as you say it's easy to show, and it does seem to be something that's worth having a named result for. :wink:

    Intuitively, I would say that, in hydrostatic equilibrium, the force due to pressure at any point must just balance the gravitational force. The gravitational force is the gradient of the potential, so integrating it over the volume of the object, heuristically, gives you the gravitational potential energy itself, and therefore you would expect that integral to be proportional to the volume integral of the pressure. The factor of 3 would come from the fact that there are 3 spatial dimensions.

    As you say, we've had this discussion before. :wink:

    One thing I would mention is that the idea of "removing a support instantly" doesn't work as it stands in GR, because of the automatic conservation of the source--the Bianchi identities. There is nothing corresponding to those in Newtonian gravity, so, for example, in Newtonian gravity it's perfectly possible to construct a model where at time ##t = 0## a supporting strut suddenly just disappears, or the pressure inside an object in hydrostatic equilibrium suddenly just vanishes. In GR it's not possible to construct such a model, because it would violate the Bianchi identities, which means there is no solution of the Einstein Field Equation that describes such things.

    I realize that "suddenly disappears" and "suddenly vanishes" are idealizations, and a more rigorous model would replace them with something more like "changes, not instantaneously, continuously, but really fast compared to all other timescales in the problem". But I still think that it's very important to pay attention to the constraints on the model from the Bianchi identities, since those constraints are quite a bit more restrictive than just "the change needs to be continuous".
     
  4. Oct 25, 2017 #3

    Jonathan Scott

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    Thanks for another way of looking at it.

    I'm wondering whether there is a simple way to write down the integral which I've mostly described in words, without needing to express it as a sum of three separate integrals along arbitrary axes.

    In this case, there's no problem with "instantly" being a little slower. Removing a support doesn't mean removing matter or energy; it could just be taking a very thin rod which has a cut through it between frictionless faces and pushing the two parts a little so they slip past each other. Although this would temporarily cause a wave to propagate within the parts of the rode at up to the speed of sound in the material, the kinetic and potential energy in that wave would depend on the characteristics of the rod material and would typically be orders of magnitude too small for the pressure to continue to match the potential energy. Basically, there could be a little energy stored when the force in the rod causes it to be compressed elastically by a small distance, which can be arbitrarily small for a stiff rod, but since the pressure integral term is equal to exactly the same force multiplied by the full length of the rod, the energy from compression would be tiny in comparison.

    (And as I've mentioned before in the GR-related threads, what is conserved is energy and momentum, not pressure or stress, so the conservation laws do not prevent a fairly abrupt change of pressure).
     
  5. Oct 25, 2017 #4

    PeterDonis

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    For the spherically symmetric case, I think it can be written as an integral in one variable, since everything is a function of one variable (the radial coordinate).

    As I said, we've had this discussion before. :wink: I don't see the point of rehashing it here since we are just talking about heuristic, intuitive speculations; one would need to actually develop a specific mathematical model based on the EFE and see what it says.
     
  6. Oct 25, 2017 #5

    PeterDonis

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    No, what is conserved is the stress-energy tensor, in the sense that its covariant divergence is zero. That tensor includes pressure and stress as well as energy and momentum.
     
  7. Oct 25, 2017 #6

    Jonathan Scott

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    I'm still looking for a neat way to describe the integral in the general Newtonian case. I guess it will probably have to involve some sort of tensor notation, but just 3D tensors in flat space.

    For purposes of this thread, I was talking about the Newtonian case. It is very clear in the Newtonian case that the pressure integral is exactly equal to the potential energy when static, but nowhere near equal to the potential energy when released into a dynamic situation, for example by a rod being split as previously described, even though the Newtonian potential energy of the system is initially unchanged (although part of it will soon be converted to kinetic energy if a support has been removed).
     
  8. Oct 25, 2017 #7

    Jonathan Scott

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    The stress-energy tensor represents the flow of the four components of energy-momentum density in the four directions of space-time, where for example the flow of x-momentum density in the x-direction is pressure. In flat space, each term of the ordinary divergence is effectively a continuity equation for one of those four components of the energy-momentum, giving global conservation of each component separately. As I understand it, the covariant divergence also takes into account the effect of curved space, so what is conserved is effectively the flow of energy-momentum density relative to a local free fall frame. So what is conserved is not the stress-energy tensor itself, but the energy and momentum whose flow is represented by the stress-energy tensor.

    I thought this had been clarified before. I don't really want to discuss it again in this thread, which is supposed to be about the Newtonian pressure integral. It obviously still places constraints on what can happen with changing pressure, but pressure is definitely not "conserved".
     
  9. Oct 25, 2017 #8

    PeterDonis

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    For the general case, I don't think there's any way to avoid either tensors or something equally forbidding. :wink:

    Sure, but, as you note, if you insist on using this language, "energy-momentum density" includes pressure and stresses. So your statement that pressure is somehow "not included" is not correct.

    I thought so too, but the way you are stating it here does not correspond to what I thought had been clarified. See above.

    In any case, I think it's highly preferable to express all this in math, not ordinary language, since, as we have just shown, ordinary language is vague.
     
  10. Oct 27, 2017 #9

    Jonathan Scott

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    I think the pressure integral I've been describing can be simplified to simply taking the sum of the pressure in any three perpendicular directions at each point then taking the volume integral of that sum. That's like integrating the trace of the ordinary 3D stress tensor. Does that make it any more familiar to anyone?
     
  11. Oct 27, 2017 #10

    PeterDonis

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    Not in the Newtonian case, but in the GR case this works out (once you include the correction factor for spacetime curvature and the 00 component of the stress-energy tensor) to be the Komar mass integral, for which it is well known that the contributions of pressure and gravitational potential energy exactly cancel (since the Komar mass only applies to stationary spacetimes, the concept of "gravitational potential energy" is well-defined). So I would say what you propose is certainly reasonable for the Newtonian case.
     
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