Stress and pressure as gravitational source

  • #51
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The Komar mass should be valid at either stationary spacetime, but I don't know why you would think that it would be valid in a non stationary spacetime that transitions between the two. The Komar mass simply doesn't exist in the spacetime that you are describing.

My solution: don't use the Komar mass for non-stationary spacetimes since they violate the basic assumptions required for the Komar mass.
 
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  • #52
Jonathan Scott
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Because you can go quickly from one equilibrium to another, and the Komar mass should be valid at both equilibria. If it has changed substantially, this would mean the orbit of a test body shifts due to internal changes in a massive body that do not result in any flow of radiation or mass to/from the body.
I should point out that I didn't actually have a direct problem with the equilibrium case. I had already worked out that for semi-Newtonian gravity (treating gravity as a force) the integral of the pressure for any equilibrium configuration is always equal and opposite to the potential energy of the configuration. (Is this a widely-known fact? It was something I discovered for myself only after looking for the Newtonian equivalent of the Komar mass pressure term). Assuming the weak field approximation, the effective energy as decreased by its own potential (time dilation) is equal to the local energy of the components minus twice the potential energy of the configuration, so the total of the two parts is exactly equivalent to the Newtonian total energy. What bothered me is what happened in the non-equilibrium case when the pressure was unbalanced.

However, the Ehlers paper suggests that it doesn't work the way I thought!
 
  • #53
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If we ignore the initial effect of the decompression wave, which can be arbitrarily small for an arbitrarily rigid rod, then as far as I can see, there is no change in any other term of the SET apart from the pressure. There is a step change in the rate of change of the momentum with respect to time which occurs at the same time as the step change in the pressure with respect to space passes through the relevant location. These two terms match to preserve continuity.
Yes. I agree. For simplicity if we consider a box of pressurized gas then at, say the -x face of the box there is a sudden change change in pressure as a function of x (dp/dx > 0). If the box is rigid then the continuity of the SET requires that there be a change in the shear stress in the wall of the box along y and z to compensate for the change in the pressure along x. If the box cannot support a shear stress, then there must be a temporal change in the x momentum density to compensate for the spatial change in pressure along x. This is what forces the gas to spew out until pressure is spatially equalized.

Of course, you do have to be careful. As a pressure changes in time, it generally disturbs the geometry in space also. Depending on the details a pressure with ##\partial_t## is likely be associated with alterations to ##\partial_x## also. But I am sure that some clever person could come up with a pressure that has a nonzero ##\partial_t## but a zero ##\partial_x## throughout the spacetime.

The question remains, given all that, what is the problem? Why should any of that alter the fact that the SET is the source of gravity in GR and pressure is a component of the SET?
 
  • #54
PAllen
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I should point out that I didn't actually have a direct problem with the equilibrium case. I had already worked out that for semi-Newtonian gravity (treating gravity as a force) the integral of the pressure for any equilibrium configuration is always equal and opposite to the potential energy of the configuration. (Is this a widely-known fact? It was something I discovered for myself only after looking for the Newtonian equivalent of the Komar mass pressure term). Assuming the weak field approximation, the effective energy as decreased by its own potential (time dilation) is equal to the local energy of the components minus twice the potential energy of the configuration, so the total of the two parts is exactly equivalent to the Newtonian total energy. What bothered me is what happened in the non-equilibrium case when the pressure was unbalanced.

However, the Ehlers paper suggests that it doesn't work the way I thought!
Can you describe what you mean in reference to the following scenario:

I have box of hydrogen and oxygen, at some pressure. 'Somehow' spontaneous combustion occurs throughout the box. The box is able to prevent any radiation from escaping, and itself has negligible mass and thickness (though it does need to have stress as needed for the stated behavior). Comparing the two equlibria, total energy density is the same in both, pressure is radically different (only water vaper pressure after). So, how is pressure equal and opposite potential energy in both cases? Do you include chemical potential energy?

Ehler's answer is that even though pressure is a source term in the SET, integrated for both equlibria, the pressure does not contribute externally measured gravitational mass in either case. I agree with Dalespam, that Komar mass simply doesn't apply in between. The Bondi mass formulation (which equals Komar mass for stationary cases) proves (non-locally - by integrating to null infinity in AF spacetime) that the mass doesn't change in between as well. Of course, for spherical case, Birkhoff applies, but we need not assume a sphere - I could have a cubical box. Conservation of Bondi mass when there is no radiation says if we meet the problem statement (no radiation) there will be no mass change, even in radically non-stationary evolution.

Lacking, so far in this discussion, is a quasi-local explanation of constancy in between, in terms of SET components.
 
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  • #55
Jonathan Scott
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For the Newtonian case, you simply need to include the (negative) pressure in the walls of the container; for equilibrium the overall force (including gravitational) must be zero through any plane. Note that changing the internal configuration can change the potential energy, but if there is no external change in energy, the total is still the same.
 
  • #56
Jonathan Scott
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I guess from the fact that anyone is asking, the fact that the volume integral of the pressure in the Newtonian equilibrium case is equal to the potential energy isn't that well known. Just consider the force between any two masses in the configuration and you get the usual ##-G m_1 m_2 / r^2##, so for equilibrium the integral of the perpendicular pressure over any plane slice of the system in between the masses balances that force (regardless of any local higher or lower pressure areas) and integrating that over the distance ##r## between them gives ##G m_1 m_2 / r##, the potential energy of that pair of masses (but positive, as this is the pressure resisting the force). The forces add up as vectors, and the overall integral is equal and opposite to the potential energy.
 
  • #57
Jonathan Scott
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I should probably have made it clear that I'm talking about the three separate pressure terms for each perpendicular plane (like those which appear in the SET).

Edit: That means for example that if the perpendicular pressure in a small volume is p in each direction, the result is 3p times the volume, as usual for the Komar mass.
 
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  • #58
PAllen
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I guess, for me, the upshot (for now) is as follows:

1) In equilibrium, you can use Komar mass, but as shown by Ehlers, the factors that produce equilibrium against pressure result that the 3p term of Komar mass does not actually add anything to the mass over the body as a whole.

2) Out of equilibrium, even instantly, you just can't use Komar mass formula at all. Bondi mass covers the whole evolution and is equal to Komar mass for equilbrium and conserved up to radiation (including GW) - but only defined for AF spacetimes. This guarantees that if you compute Komar at an equilibrium, and radiation is not significant, then Bondi mass equals that Komar mass during non-equlibrium. During non-equlibrium, you simply can't talk about the 3p term of Komar mass. If you want a uniform approach to computation over the evolution, you have to use Bondi mass throughout.

3) Integrating pressure by itself over a volume and expecting this integral to be a separately detectable component of gravitational mass at a distance is simply too simplistic an expectation.
 
  • #59
Jonathan Scott
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Earlier in this thread I explained what happens to SET terms when pressure drops locally to zero. However, I now realise there's a problem with the explanation, in that the gravitational force is treated as external which means that momentum is not conserved, which means that the divergence of the SET can't be zero. I know it's the "covariant" divergence, but I need to understand what that implies in terms of gravitational effects. I'll think about it for a bit and if I can't resolve it I'll start a new thread.
 
  • #60
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Earlier in this thread I explained what happens to SET terms when pressure drops locally to zero. However, I now realise there's a problem with the explanation, in that the gravitational force is treated as external which means that momentum is not conserved, which means that the divergence of the SET can't be zero. I know it's the "covariant" divergence, but I need to understand what that implies in terms of gravitational effects.
For me the easiest way is to look locally at a momentarily comoving free-fall frame.

So, for example, consider a non-rotating spherical fluid planet in static equilibrium. It is a non-flowing fluid, so the shear stress components are always 0. There is no pressure gradient in ##\theta## or ##\phi##, but there is a pressure gradient in ##r##. Given the no-shear condition, the negative pressure gradient in r would imply that there is an increase in momentum density wrt time, per the continuity. Since it is static, that seems to be a contradiction.

However, as you say, it is a covariant derivative, not an ordinary one. So, looking at a locally free-falling frame we see that the fluid is indeed accelerating in the positive r direction and therefore there is, in fact, an increase in momentum density in the r direction which leads to the 0 covariant divergence.
 
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