Pressure, acceleration and potential energy

  • #1
Jonathan Scott
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I've now found a different way to look at potential energy and its location in the Newtonian approximation to relativistic gravity which may perhaps shed some light on how the corresponding terms work in GR. This is related to a previous thread of mine on "Pressure and Newtonian potential energy" but I've now made some progress on the details and a more secure mathematical basis.

Consider a flat space stress-energy tensor describing some self-contained system of masses, not necessarily static, in Cartesian (x,y,z) coordinates with gravity being treated as an acceleration field ##\mathbf{g}##, with energy density ##\rho c^2## and local directional pressure ##P_{xx}## in the ##x## direction (the negative of the normal stress ##\sigma_{xx})##, with no transverse stresses. The continuity equation for the ##x## component of momentum is then as follows:
$$\frac{\partial(\rho v_x)}{\partial t} + \frac{\partial(P_{xx})}{\partial x} = \rho g_x$$
This is equivalent to the ##x## component of the covariant divergence in the weak approximation in GR.

We can take the dot product of both sides with the vector ##-\mathbf{x}## relative to some arbitrary origin. The ##x## component of the result is as follows:
$$-x \cdot \frac{\partial(\rho v_x)}{\partial t} - x \cdot \frac{\partial(P_{xx})}{\partial x} = -x \cdot \rho g_x$$
From my previous thread, the RHS of this integrated over all three spatial dimensions is equal to the Newtonian potential energy, because if one considers the gravitational force between every pair of masses making up the collection, taking the dot product with the position means that one has overall taken the dot product of the two equal forces, of opposite sign, with the vector for the displacement between them, which is equal to the potential energy of that pair but positive. Note that the RHS is not affected by whether the object is static or accelerating. In the case where there is no pressure, the mass is in free fall and the rate of change of momentum is equal to the gravitational force. In the static case, the rate of change of momentum and the term containing the pressure integrates to the potential energy.

The term containing the pressure is not simply the pressure, but its integral is equal to the integral of the pressure over the whole object, which we can confirm as follows (with force ##F_{xx}## being the area integral of the pressure):
$$\iiint P_{xx} + x \cdot \frac{\partial(P_{xx})}{\partial x} \, dx \, dy \, dz = \iiint \frac{\partial}{\partial x} \left ( x \cdot P_{xx} \right ) \, dx \, dy \, dz = \int \frac{\partial}{\partial x} \left ( x \cdot F_{xx} \right ) \, dx = \left [x \cdot F_{xx} \right ]$$
That is, the pressure plus that term with its sign switched to positive integrates to the difference in ##x\cdot F_{xx}## between the limits of integration, which is zero when the integration extends to beyond the edge of the system of masses, so the integral of the pressure minus the integral of that term is zero over the whole system. However, the pressure is not necessarily locally equal to that term; the equality only holds for the integral over the whole system.

So the integral of the directional pressure terms (negative normal stresses) in all three dimensions over a system is equal to the integral of ##-x \cdot \partial(P_{xx})/\partial x##, and the integral of that term plus the force density term (which only appears when acceleration is present) adds up to the Newtonian potential energy of the system.

Note that the expression involving ##-x \cdot \partial(P_{xx})/\partial x## is a local equation, which says that there is a local quantity which can be integrated to give the potential energy, and that quantity is proportional to the local energy density, so it is associated with the existing mass density, not with support structures or stresses, and it cannot change abruptly (although the total potential energy can change gradually as it is converted to or from kinetic energy).

The effective energy of an object of local total energy ##E## in a Newtonian potential ##\phi = -Gm/r## is expected to be given in General Relativity by ##E \, e^{\phi/c^2}##, applying the time dilation associated with the potential. For multiple objects this counts the binding energy twice, but there is an alternative expression for the effective source strength based on the above equations which gives the correct effective binding energy (where ##\mathbf{g} = -\nabla \phi##):
$$\nabla \left ( \mathbf{x} \, E \, e^{\phi/c^2} \right ) = E \, e^{\phi/c^2} \left ( 1 - \frac{\mathbf{x}\cdot\mathbf{g}}{c^2} \right ) $$
If ##\mathbf{x}## is relative to an arbitrary location then this still does not give a specific value for the potential energy of each part (which is similar to the Newtonian case) but if it is however taken to be relative to the centre of mass of the system (taking into account the potential) then that is equivalent to the following constraint on the origin to use for ##\mathbf{x}##, which involves an expression that just appeared in the previous equation, summed for all components of the system:
$$\sum \left ( \mathbf{x} \, E \, e^{\phi/c^2} \right ) = 0$$

So far this hasn't helped me very much with my attempts to understand in what way the binding energy is reflected in the SET and in the EFE in the general dynamic case, but it does make the important point that even if the source strength in the static case is effectively increased by an amount equal to the pressure, that doesn't mean that the pressure is the source. The above shows that there is another quantity which provides the same correction to the effective source strength and which has the same integral as the pressure in the static case but is also valid in the dynamic case. Perhaps something like that quantity is hiding somewhere in some form of the EFE or a time derivative and gives rise to the pressure in static cases and some other term in other cases. Any constructive suggestions for how to progress from here would be gratefully received.

In my previous thread I thought that perhaps the gravitational induction linear frame-dragging effect of the accelerating matter could explain the dynamic case, but I have now ruled that out. On calculating the effect more carefully I found some additional terms which cancelled out those induction terms, and it also would not have worked for transverse accelerations, which could cancel by symmetry.
 

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  • #2
PeterDonis
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Perhaps something like that quantity is hiding somewhere in some form of the EFE or a time derivative and gives rise to the pressure in static cases and some other term in other cases.
We've had this discussion before. The pressure is what appears in the EFE. It's what appears in the stress-energy tensor. And the stress-energy tensor is the source. I don't see why you keep insisting on ignoring that and looking for something else.
 
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  • #3
PeterDonis
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but is also valid in the dynamic case
In the dynamic case, the spacetime is not stationary, so there is no well-defined potential ##\phi## and your whole method breaks down.
 
  • #4
Jonathan Scott
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We've had this discussion before. The pressure is what appears in the EFE. It's what appears in the stress-energy tensor. And the stress-energy tensor is the source. I don't see why you keep insisting on ignoring that and looking for something else.
But pressure is definitely not conserved; from the calculations above, the integral of the pressure is equal to the potential energy minus the acceleration term, and in a spherical collapse situation the acceleration term is definitely non-zero.
 
  • #5
PeterDonis
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pressure is definitely not conserved
No, it's not. What is conserved is the SET as a whole, in the sense that its covariant divergence is zero. No other conservation law holds for all spacetimes. We've had this discussion before as well.
 
  • #6
Jonathan Scott
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In the dynamic case, the spacetime is not stationary, so there is no well-defined potential ##\phi## and your whole method breaks down.
This is for situations involving non-relativistic speeds and weak fields. Newtonian mechanics is easily close enough. Anyway, what I'm showing mathematically is true for the Newtonian approximation; how it maps to GR is what puzzles me.
 
  • #7
Jonathan Scott
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No, it's not. What is conserved is the SET as a whole, in the sense that its covariant divergence is zero. No other conservation law holds for all spacetimes. We've had this discussion before as well.
I have used the fact that the covariant divergence is zero in my calculation. It does not place any restriction on the pressure changing without any change to the energy and momentum. That would involve a condition on d/dt of the pressure, which doesn't show up anywhere.
 
  • #8
Dale
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it does make the important point that even if the source strength in the static case is effectively increased by an amount equal to the pressure, that doesn't mean that the pressure is the source.
The EFE makes it clear. The pressure is part of the source.

But pressure is definitely not conserved; from the calculations above
You don’t need the calculations above. Pressure is obviously not conserved. That is irrelevant. There is no requirement that sources must be conserved.

We have had this discussion before too. In EM neither the charge density nor the current density is conserved, charge is conserved. But the charge density and the current density are the sources.

Similarly in GR. None of the components of the SET are conserved, energy and momentum are conserved. But the SET is the source of gravity.
 
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  • #9
Dale
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What is conserved is the SET as a whole, in the sense that its covariant divergence is zero.
If you have a quantity, A, which is the density and flux of some other quantity, B, then A having 0 covariant derivative means that B is locally conserved. So here the stress energy tensor is the density and flux of energy and momentum. So the conserved quantity is the four momentum.

Edit: of course, sometimes the quantity A, with 0 covariant divergence, is called a conserved current. Particularly in regards to Noether’s theorem.
 
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  • #10
Jonathan Scott
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The EFE makes it clear. The pressure is part of the source.
The pressure terms clearly contribute to the shape of space in static case via the SET and EFE. The Newtonian calculation however confirms that the non-zero pressure integral is effectively caused by the gravitational field, so in that sense pressure is a side-effect of the existing source distribution.

The Newtonian calculation also shows that there is another physical quantity whose integral is equal to the pressure term in the static case but which also has the same integral in dynamic cases. This term is present even when the pressure is zero, which can occur very suddenly without changing the energy or momentum distribution on the same time scale. I'd like to know what that term would correspond to in the SET, and whether it could explain how the distant field is unaffected in the Birkhoff's theorem case. As the SET in the EFE involves the metric as well as the local stress-energy, could it be related to a derivative of the metric?
 
  • #11
Dale
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I'd like to know what that term would correspond to in the SET,
Your term is such a hodgepodge of tensor components and three vectors that I doubt it has any clear covariant interpretation. Certainly none seems obvious to me.

Can you confirm explicitly that you now understand and agree that pressure is part of the source of gravity in GR?

We can take the dot product of both sides with the vector −x−x-\mathbf{x} relative to some arbitrary origin.
This is a bizarre quantity. Do you have any reference that takes this quantity and uses it for anything? I don’t see any meaning for it.
 
  • #12
Jonathan Scott
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My calculations are in flat space treating a gravitational field as generating a force proportional to the energy.

Taking the dot product of both sides of an equation with the same thing gives an equation which is still true by linearity. The resulting quantity is the one which I mentioned in the previous thread whose integral gives the potential energy.
 
  • #13
Dale
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Taking the dot product of both sides of an equation with the same thing gives an equation which is still true by linearity.
Sure, I understand the mathematical manipulation. It is just a bizarre quantity. It is completely obscure to me, and I don’t see how computations using it are supposed to help clear up any confusion about pressure.

Can you confirm explicitly that you now understand and agree that pressure is part of the source of gravity in GR?
 
  • #14
Jonathan Scott
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The pressure terms in the SET obviously contribute as a gravitational source in the static case, as is well known. As the pressure terms and in particular their volume integral are determined by the internal gravitational forces, there is of course no choice about them; they are not an independent quantity which can be added or removed.
 
  • #15
Dale
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The pressure terms in the SET obviously contribute as a gravitational source in the static case,
They contribute in all cases. The EFE is valid for dynamic spacetimes too.

As the pressure terms and in particular their volume integral are determined by the internal gravitational forces, there is of course no choice about them; they are not an independent quantity which can be added or removed.
I can think of many many counterexamples.
 
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  • #16
PeterDonis
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Thread closed for moderation, to determine whether the discussion is within PF rules in the absence of any references.
 
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  • #17
PeterDonis
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The moderators have reviewed the thread, and it will remain closed.
 
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