The slowing down of time in low gravitational potential may be explained by this kinetic energy (or, velocity) which a static observer on the surface has relative to an observer who is not orbiting. This would connect special and general relativity.
The positive energy content in the universe, the dark energy, causes negative pressure. It is logical that positive pressure would be associated with a negative energy content of space inside a body of mass.
We will continue studying the pressurized vessel thought experiment.
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If we believe in conservation of energy, the hamiltonian or the lagrangian of a theory has to conserve energy.
It has been known for a long time that the lagrangian in the Einstein-Hilbert action miscalculates the energy of spacetime deformation. Several pseudotensors were developed to determine the correct energy content of a system. The Landau-Lifshitz and Einstein pseudotensors are the best known ones.
Also, formulas like the ADM mass and Komar mass were introduced. They calculate the correct gravitational and inertial mass of a system in an asymptotic Minkowski space.
To determine the equilibrium of a system we need a hamiltonian which determines the forces correctly, and that requires that the hamiltonian must calculate the correct potential energy for each subsystem.
A typical lagrangian is T - V, where T is the kinetic energy and V is the potential energy in various subsystems. If there are no external forces, V contains the rest mass mc^2 and the energies in the deformation of springs, etc., in the system.
In a static system, there is no kinetic energy, and the potential energy formula of the lagrangian, with the sign flipped, is the hamiltonian. It calculates the total energy of the system.
We cannot find the correct equilibrium if the hamiltonian has an error in the energy of some subsystem. Just this is the case with the Einstein-Hilbert action.
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/ \______
| __###-----/\/\/\/\/\
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balloon piston spring
Let us imagine a system where we have a rubber balloon and a piston which pushes air to inflate the balloon. There is a spring which pushes the piston. We can use a hamiltonian to determine the equilibrium for the piston position. But if our hamiltonian has a grossly wrong value for the balloon deformation energy, the result will be incorrect.
In general relativity, we can convert mass-energy into spacetime deformation energy with the exact same system as in the diagram above. Pressure makes the space to curve more, and we can pump more air into the balloon than in newtonian mechanics.
Since the hamiltonian which we get from the Einstein-Hilbert action has an error in the deformation energy, it will give a wrong answer for the equilibrium.
The pressure at the center of a neutron star generates up to 1/4 of the gravitational field there, the mass only 3/4. The error in the Einstein-Hilbert action may show up when we calculate the interior solution for a neutron star.
+ electric force +
● <--------------------------> O
dense mass low-density mass
Let us study a dynamic system where a dense mass interacts with a low-density mass through an electric force.
If we use the lagrangian in the Einstein-Hilbert action, it miscalculates the energies of the two masses and their acceleration. We get wrong paths for the masses, and conservation of momentum is broken.
How is it possible that this problem with the Einstein-Hilbert action has been ignored since the year 1916?
If we would be constructing a bridge and in its strength calculations we would use grossly erroneus formulas for the deformation energy, engineers would immediately recognize that there is a serious problem.
It looks like people have been concentrating on the Einstein field equation and have forgotten about the hamiltonian or the lagrangian. The Einstein field equation is derived by variating the Einstein-Hilbert action. If the action is incorrect, then the field equation is, too.
However, the metric outside the gravitating body may be right when derived from the erroneous Einstein-Hilbert action. Currently, we have no means for measuring the metric inside a large gravitating body. An error in the metric there would go undetected.
The Ricci tensor has to be zero outside a gravitating body?
The Ricci tensor is identically zero outside a gravitating body in Einstein's theory. If we shoot a small spherical dust cloud from a point in spacetime, its volume will very precisely be the same in the Minkowski space as in a laboratory falling freely close to a gravitating body. That means that the Ricci tensor has all the components zero. The form of the cloud will stretch, however, because of tidal forces.
For the corrected Einstein-Hilbert action, the Ricci tensor should probably be zero, too.
In newtonian gravity, the dust cloud volume behaves like in an empty Minkowski space. We thus know that the Ricci tensor outside the mass for the corrected general relativity must be zero or almost zero if the mass is not a neutron star or a black hole. It is a reasonable hypothesis that it is exactly zero.
At this point, we have not yet calculated the Ricci tensor for various rubber models of gravity. We did show in an earlier blog post that a rubber model has the spatial metric close to the Schwarzschild one. We did not calculate the temporal metric yet.
How to calculate the correct spacetime deformation energy?
For a rubber sheet model we can write a lagrangian where the energies are right. A big difference to the Einstein-Hilbert action is that variation of one component g_ij of the metric will cause a huge or infinite deformation energy unless we vary also other components of the metric. If you want to increase the distance of two pencil-marked points on the x axis of a rubber sheet, you may press a pit in the rubber between the points. But that pit stretches also distances in the y direction.
In the Einstein-Hilbert action, the deformation energy is negative - it is the integral on R, and R is usually positive. In the lagrangian, mc^2 and other potential energies appear negative.
In the Einstein-Hilbert action we probably can vary an individual component g_ij of the metric without causing a huge decrease in the lagrangian value.
It remains to be seen if we can write a lagrangian where the variation is as simple as in the Einstein-Hilbert action.
The Einstein-Hilbert action may estimate the energy of the system approximately right if the pressure inside the mass M is negligible. The integral E on R may in this case be the deformation energy. The binding energy is -E and an energy 2E was "freed" from M through the redshift.
A correct Einstein-Hilbert lagrangian might be something like
-V(curvature) + T_curv + L_M
where V(curvature) is the spacetime deformation energy and T_curv is the kinetic energy of spacetime curvature. L_M is the usual lagrangian T - V for particles and fields, except the gravitational field. We assume "minimal couplings" between the curvature and other fields.
Clocks slow down within the mass and make the integral of the above lagrangian over spacetime smaller, and effectively release mass-energy from the mass. It is the redshift factor.
In Einstein's gravity, for a static solution, the spacetime outside the mass has a zero Ricci tensor and a zero stress-energy tensor. In rubber sheet models, stretched rubber contains energy throughout the asymptotic Minkowski space. Should this deformation energy gravitate? Can we mine the deformation energy far away from a mass and convert it to conventional mass-energy? If the rubber sheet has assumed its energy minimum, mining the deformation energy would require us to cut the rubber. It might be that the energy does not produce more gravitation.
A correct Einstein-Hilbert lagrangian might be something like
-V(curvature) + T_curv + L_M
where V(curvature) is the spacetime deformation energy and T_curv is the kinetic energy of spacetime curvature. L_M is the usual lagrangian T - V for particles and fields, except the gravitational field. We assume "minimal couplings" between the curvature and other fields.
Clocks slow down within the mass and make the integral of the above lagrangian over spacetime smaller, and effectively release mass-energy from the mass. It is the redshift factor.
In Einstein's gravity, for a static solution, the spacetime outside the mass has a zero Ricci tensor and a zero stress-energy tensor. In rubber sheet models, stretched rubber contains energy throughout the asymptotic Minkowski space. Should this deformation energy gravitate? Can we mine the deformation energy far away from a mass and convert it to conventional mass-energy? If the rubber sheet has assumed its energy minimum, mining the deformation energy would require us to cut the rubber. It might be that the energy does not produce more gravitation.
The role of Birkhoff's theorem
Birkhoff's theorem states that the outside metric of a spherically symmetric system is always a static Schwarzschild geometry. This holds even if we are converting mass-energy to pressure and spacetime deformation inside the spherical system.
The theorem is a principle of energy conservation. An outside observer measures the gravitational mass of the system constant even though the form of energy within the system may change.
If the Einstein-Hilbert action miscalculates the total energy of the system, why does energy conservation still hold in Birkhoff's theorem?
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