UPDATE November 6, 2021: The answer to Question 1 is incorrect if we keep the test mass outside the body of the spherical mass. The increase dV in the volume only happens if we lower the test mass inside the spherical mass. The whole process caused by pressure is very complicated. See the last section of our blog post today.
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| | | rigid grid
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● infinitesimal test mass m
There is some positive pressure in the grid because it has to resist contraction under its own gravity.
Let us then calculate the metric around the grid from the Einstein field equations.
Does the calculated metric describe right the orbit of an infinitesimal test mass? Probably not. When the test mass approaches the grid, the Schwarzschild metric around the test mass distorts the spatial metric in the grid. The grid would become deformed, and that requires energy. The test mass feels a repulsive force from the rigidity. On May 28, 2019 we wrote about this "antigravity device".
In the idealized case where the mass-energy of the grid is zero, there is no mass-energy nor any pressure in the grid. The Einstein field equations for the grid, ignoring the test mass, calculate that the metric around the grid is the flat Minkowski metric. That metric does not describe the real behavior of the test mass. The test mass feels a repulsive force.
The orbit of an infinitesimal test mass defines the "true" geometry of spacetime. If there is a complex interaction, and we ignore the test mass when solving the Einstein field equations, we obtain a different metric which is not the "true" geometry of spacetime.
The Einstein field equations are derived from the Einstein-Hilbert action using a simple variation of the system. That simple variation ignores complex interactions.
Thus, in the case of a rigid grid, the metric derived from the Einstein field equations does not describe right the orbit of an infinitesimal test mass.
The infinitesimal test mass causes a deformation of the flat metric. That, in turn, introduces negative and positive pressure in various parts of the grid.
Question 1. If we calculate the corrected metric for the grid, including this "backreaction" from the test mass, does the corrected metric describe the orbit of the test mass correctly?
Question 2. In which cases does the metric derived from the Einstein field equations describe right the orbit of an infinitesimal test mass?
We believe that the external Schwarzschild metric is approximately the true geometry of spacetime around a spherical mass, though we have not seen a proof.
In popular science books it is claimed that a ray of light obeys the metric which we calculate from the Einstein equations. The ray moves along a "straight" path. But in our grid example, a test mass does not move along a straight path of the calculated metric.
A thought experiment: a spherical mass supported by springs, and an infinitesimal test mass lowered close to it
● large mass supported with springs
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• infinitesimal test mass dm
We assume that the infinitesimal test mass is surrounded by a Schwarzschild metric. We assume that the deviations from the flat metric are small, so that we can, in some way, sum the deviations.
Initially the large mass is static in its lowest energy state.
We lower the test mass very slowly to the large mass. The spatial metric inside the large mass gets slightly stretched in the direction of the test mass. The pressure in the springs does some work as the spatial metric stretches. The person doing the lowering can harvest the work done by pressure. He interprets that pressure contributes to the gravitational attraction of the large mass.
The person can then use the energy which he harvested and pull the mass very slowly back up.
However, if the person pulls the test mass quickly up, then the large mass ends up being slightly displaced from its lowest energy state. The springs become compressed on the side where the test mass was.
The end result of a swift pull is that energy was transferred from the person into vibration of the central mass.
During the whole process, all the interaction of the test mass was through its gravity.
We may say that the gravity field around the central mass for swift movements of the infinitesimal test mass is not a conservative force field.
This is in contrast to Birkhoff's theorem which states that the gravity field around a spherical mass is the static Schwarzschild metric. Even an infinitesimal test mass perturbs the system in such a way that it no longer is spherically symmetric.
Our thought experiment suggests that if we temporarily increase the pressure inside the central mass, then the test mass will feel a stronger pull from the central mass. Increasing the pressure can temporarily increase the gravity of a spherically symmetric object. Birkhoff's theorem does not prevent this.
Tolman's paradox is that changing slow mass into radiation increases its gravity. Our thought experiment suggests that this indeed is the case. There is no paradox if one realizes that the Birkhoff theorem is not valid for an infinitesimal test mass.
Based on our new thought experiment, we can now answer Question 1 negatively:
Answer to Question 1. Correcting the metric with the backreaction to the test mass dm does not yield the right value for the gravitational pull. The following reasoning proves that:
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UPDATE November 7, 2021: See our blog post on November 6, 2021, the last section. The process is much more complicated, and the analysis below is incorrect.
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When we very slowly lower the infinitesimal test mass dm close to the spherical mass, it stretches the metric inside the spherical mass and increases the volume by some
dV ~ V dm,
where V is the volume of the spherical mass.
We can harvest the energy which is roughly
dE = p dV
~ p V dm,
where p is the pressure of the spherical mass.
When we swiftly pull the test mass out, a significant portion of dE becomes vibrational energy of the spherical mass.
The extra pressure that is created inside the spherical mass during the swift pull is
dp ~ p dV / V.
The gravitational pulling force F on the test mass from the pressure p inside the spherical body is
F ~ p V dm.
The added pressure adds something like
dF ~ dp V dm
~ p dV dm
~ p V dm²
to the gravitational pulling force of the pressure. The energy that we must spend to win the extra pull is
~ s p V dm²,
where s is the distance that we lift the test mass. If dm is small, this energy is much less than dE.
Birkhoff's theorem
Let us, instead of an infinitesimal test mass, use a spherical shell of infinitesimal mass which we lower close to the central mass. Then there is no increase dV in the volume of the central mass? How does pressure in that case create attraction?
The answer may be that the spherical shell lowers the potential of the central mass so much that it compensates the missing contribution of pressure.
Question 3. What about changing the pressure of the central mass temporarily? Can we find a solution where the external metric stays constant as the static Schwarzschild metric?
In a static configuration, the attraction of the pressure is visible in the Schwarzschild external metric. It may be so that in a dynamic configuration the contribution of the pressure change is not visible in the external metric. If this is true, then we avoid a contradiction with Birkhoff's theorem. If this is the right explanation, then the static external Schwarzschild metric is the correct solution for the Einstein-Hilbert action, but the orbit of an infinitesimal test mass does not obey that metric. Rather, the test mass feels the pull from the changing pressure.
Is the Ricci scalar R the right term in the Einstein-Hilbert action?
The metric which we calculate from the Einstein-Hilbert action does not predict correctly the orbit of an infinitesimal test mass. It only predicts the orbit approximately right under most circumstances.
The question arises if the R term in the action is correct. Why calculate the Ricci scalar for a metric which is not the "true" geometry of spacetime?
We do not yet know the answer to that question. Our own approach is to treat the newtonian gravity force as an ordinary force, and not to claim that there is a curved metric of spacetime at all. That is, in our own approach the metric, or its Ricci scalar, does not appear.
Our own approach seems to reproduce effectively the external Schwarzschild metric. Since the Einstein-Hilbert action gives the same metric, the term R must be approximately right.
The slowing down of time in the Schwarzschild metric has been tested empirically. There probably is no empirical data on how well general relativity fares with gravity of pressure.
Conclusions
The grid experiment shows that solving the Einstein field equations for a system alone does not necessarily produce a metric which would describe correctly the orbit of an infinitesimal test mass. The "backreaction" of the system to the test mass is important.
Our second thought experiment showed that even in the basic case of a spherical central mass supported by pressure, the static Schwarzschild external metric does not predict correctly the orbit for a swift movement of an infinitesimal test mass.
Also, the interpretation of the metric as the "geometry" of spacetime is not correct under a complex interaction. Even if we calculate a new metric from the backreaction of the system, it does not predict the orbit of an infinitesimal test mass correctly.
In our October 10, 2021 blog post we claimed that one cannot meaningfully define the "geometry of spacetime" if we have a complex lagrangian. We got more evidence which supports our claim.
The Schwarzschild interior solution assumes that the Einstein field equations calculate the correct metric for incompressible fluid. We need to check if that really is the case.
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