According to two sources on the Internet, the Ricci scalar is
R = 8π (ϱ - 3p)
in the Schwarzschild interior solution. There p is the positive pressure and ϱ is the total mass-energy density, including heat and other microscopic - not macroscopically translational - movement in the material.
The Einstein-Hilbert action is essentially a volume integral on
[1/2 R + ϱ] * redshift
= [3/2 ϱ - 3/2 p] * redshift
where the "redshift factor" is 1 in outer space and < 1 close to the mass. We use Planck units, so that G and c are 1.
The Komar mass of the system is a volume integral on
[ϱ + 3p] * redshift.
These formulas are very different. Can we reconcile them, so that the Einstein-Hilbert action would work in a sensible way?
One is always allowed to multiply a lagrangian by a constant, and that does not change the physics.
If we keep the mass M of the sphere constant and increase the radius r, then the pressure goes as r^-4 and the mass density goes as r^-3. Making the fluid very lightweight, we can make p essentially zero. Then the relevant formulas are
ϱ
and
3/2 ϱ.
We see that with a scale factor of 2/3 we can make the Einstein-Hilbert action to agree with the Komar mass for very lightweight material.
But that does not help with pressure. The Komar mass has +3p while Einstein-Hilbert has -p.
The Komar mass calculates the total energy of the system.
In our earlier blog post we raised the possibility that the Einstein-Hilbert lagrangian might contain kinetic energy T, even for a static solution. Then its value
V - T
would differ from the total energy of the system V + T. But if we have a macroscopically static system whose particles have random or other non-translational motion, then their kinetic energy must be counted in the potential energy V of the system. The differentiation of energy into V and T is used when we study the collective motion of the system. Then T is its translational kinetic energy,
T = 1/2 V v^2
at slow speeds v.
We do not know the microscopic origin of mass. It could be totally kinetic energy. However, we count such energy in V and not in T in a lagrangian. Likewise, if the energy in spacetime deformation does not have definite translational movement, its possible microscopic kinetic energy must be counted in V.
It seems to be impossible to reconcile the Komar mass and the Einstein-Hilbert action. Such action cannot calculate correctly the motion of a system under an external force?
We are assuming here that special relativity is correct and energy and momentum are conserved.
It is possible that even if the Einstein-Hilbert action miscalculates the total (potential) energy V of the sphere, it may get the inertial mass equal to the Komar mass. Then there would be no contradiction.
Writing a lagrangian in curved spacetime
Let us study forces and accelerations in a gravitational potential well.
We have a problem: in the ordinary lagrangian L_M in the action, how do we define the velocity of a particle, so that we can write its kinetic energy? In the Minkowski space, we can use global coordinates, but such coordinates are not available in general relativity.
Suppose that we have a 100 kg thin cloud of mass. Let us compare it to a 100 kg dense sphere where time runs roughly 10% slower than in outer space.
An observer in the thin cloud can use global Minkowski coordinates to determine his speed.
What does an observer within the dense 100 kg mass do? Maybe he is carrying an accelerometer. He assumes that he is at first static, and measures his speed by integrating what the accelerometer says.
Or, maybe there is a very strong rope which hangs from a global observer in the Minkowski space. The local observer can determine his speed by holding the rope tense and measuring his speed against the rope.
Static equilibrium
If we just study static systems, then we may be able to skip the velocity measurement problem. We may define the velocity as zero in a static system with a static metric.
The lagrangian or hamiltonian should then calculate only the amount of potential energy.
Suppose that we have an area in space where our lagrangian underestimates the energy which is contained therein. Is it possible that the lagrangian still can calculate, in some sense, "right" the equilibrium of the system?
Even if the lagrangian has wrong values for the energy, for example, wrong Hooke's constants for springs, it will keep the calculated energy constant. We have to analyze what the variation of the action really calculates.
A thought experiment: lower a new shell of fluid very slowly on the sphere
Let us have an incompressible fluid sphere and a rigid spherical shell around it. We attach pulley systems to the shell, so that we can lower a new shell of fluid on the surface of the fluid sphere.
We design the pulleys in a way that they allow a very slow lowering process, and the pulleys harvest all the energy which gravity releases in the process.
Let us model the process with the Einstein-Hilbert action. Most of the time the system is in an equilibrium. We can calculate the forces on different parts and they cancel each out.
We assume that the action was scaled, as explained above, so that it agrees with the Komar integral on ϱ over the fluid sphere.
We move the configuration an infinitesimal distance ds at a time. The energy needed is proportional to ds^2 and can be made arbitrarily small.
Let the pulley system lower the fluid and after that we pull the ropes and other gear up to the starting position.
Since the system was essentially static all the time and we used the Einstein-Hilbert action to develop its state, the integral of the Einstein-Hilbert action has to be the same in the starting configuration and in the end configuration.
What are the changes in the Komar masses in the process? There is a difference in how the Komar integral and the action integral handle pressure.
How large is the pressure in the rigid sphere and the pulleys at the start of the process and at the end of the process?
If we assume euclidean geometry, it turns out that the process moves an amount P of positive pressure from the rigid shell to the fluid sphere. The Komar mass stays constant and Birkhoff's theorem is satisfied.
The difference in the Komar mass integral and the action integral is how pressure contributes. The action integral is almost agnostic to pressure while it has a significant contribution to the Komar mass.
In this thought experiment, both the action integral and the Komar mass probably stay constant. There is no contradiction. The distribution of the action integral between the rigid shell and the fluid sphere is different from the Komar mass, but in the end that does not cause any contradiction.
The pressurized vessel
Ehlers, Ozsvath, Schücking, and Shang (2005) have calculated the result of enclosing a sphere of incompressible liquid inside a membrane with surface tension. Their conclusion is that general relativity and the Komar mass are consistent.
The pressurized vessel experiment is also called Tolman's paradox. If we do not take into account the negative pressure in the vessel wall, then Birkhoff's theorem is broken in the experiment - that is the "paradox".
Let us check their calculations.
The authors write that ϱ, r_0, and σ determine M, and M does not depend on the pressure in the membrane.
But can the pressure vary? The definition of ϱ depends on R, and R depends on r_0 and M. If ϱ and r_0 are fixed, then M is fixed, by definition.
There is an odd consequence of the interior Schwarzschild metric. The spatial metric apparently does not vary if we add more fluid and grow the radius.
Let us consider a sphere of fluid at the center. Let its radius in global coordinates be a fixed s. If we add more fluid on top of s, the pressure grows within s, but the spatial metric stays the same. This is counter-intuitive. We cannot force more fluid within s by increasing pressure. The metric of time does vary when pressure is increased. Like the pressure would be diverted to deform the time and not deform the space.
We have a problem in using the results of the Ehlers et al. paper: the membrane is of an infinitesimal thickness and there is no clear formula for the metric and the Ricci scalar inside such a wall. How to calculate the Einstein-Hilbert action?
C.W. Misner and P. Putnam, Phys. Rev. vol. 116, 1045 (1959).
Misner and Putnam have written another resolution of Tolman's paradox. Let us check their paper.
The authors write that ϱ, r_0, and σ determine M, and M does not depend on the pressure in the membrane.
But can the pressure vary? The definition of ϱ depends on R, and R depends on r_0 and M. If ϱ and r_0 are fixed, then M is fixed, by definition.
There is an odd consequence of the interior Schwarzschild metric. The spatial metric apparently does not vary if we add more fluid and grow the radius.
Let us consider a sphere of fluid at the center. Let its radius in global coordinates be a fixed s. If we add more fluid on top of s, the pressure grows within s, but the spatial metric stays the same. This is counter-intuitive. We cannot force more fluid within s by increasing pressure. The metric of time does vary when pressure is increased. Like the pressure would be diverted to deform the time and not deform the space.
We have a problem in using the results of the Ehlers et al. paper: the membrane is of an infinitesimal thickness and there is no clear formula for the metric and the Ricci scalar inside such a wall. How to calculate the Einstein-Hilbert action?
C.W. Misner and P. Putnam, Phys. Rev. vol. 116, 1045 (1959).
Misner and Putnam have written another resolution of Tolman's paradox. Let us check their paper.
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