The Einstein-Hilbert action is
S = ∫ over the whole spacetime
[1 / (2κ) R + L_M] sqrt(-g) (dx)^4.
There R is the Ricci scalar curvature and L_M is the lagrangian density of matter fields. If the metric tensor has all the off-diagonal elements zero, then g is the product of the diagonal elements.
The metric tensor reveals the mapping of our global coordinates (x_0, ..., x_3) to nature. If we have a map of Earth in the Mercator projection, the metric tensor would reveal how many kilometers in nature corresponds to a centimeter on the paper map to the x or y direction. Everybody knows that the metric varies a lot throughout the Mercator map. The map seriously exaggerates the area of polar countries.
The term sqrt(-g) (dx)^4 is the "volume element". It is the 4-volume of the 4-cube (dx)^4 in our global coordinates when it is mapped into nature. The unit of the volume is a cubic meter times a second.
The integral seems to be infinite. We need to normalize it in a suitable way to treat practical cases.
The energy interpretation of the Einstein-Hilbert action
The history of a physical system should be a local minimum of its action. It is the principle of least action.
Suppose that the physical system ends up in a static configuration. Then the action integral is simply the integral over the 3 spatial dimensions times the time elapsed. The action has to be a local minimum, which means that the 3D integral is at its local minimum. Any small change in the configuration of the system would raise the integral over the 3D space.
If we interpret the 3D action integral as potential energy, then we see that the system wants to settle in a minimum potential. For example, if we have a system of attached springs, it will be static if and only if the potential energy of string tensions is at a local minimum.
If we have a static gravitating system, we can interpret the 3D spatial integral of the Einstein-Hilbert action as the "potential energy" of the system.
The Schwarzschild solution
The Schwarzschild interior solution is a static spacetime with a sphere of incompressible uniform fluid in the middle. The pressure is zero at the surface and grows toward the center.
The system is clearly a static physical system. The external Schwarzschild metric is static, too.
If a curved metric can slow down the flow of time in the mass, then the integral in the Einstein-Hilbert action is reduced for a spacetime slice T_0 < t < T_1 where t is the global time of the static Minkowski observer far away from the mass. A curved metric will increase the integral over
1 / (2κ) R,
where R is the Ricci scalar curvature.
Since the spacetime does become curved in the Schwarzschild solution, the curved geometry must reduce the integral over the mass more than it increases the integral over R.
In the rubber sheet model, curvature requires energy, but more energy is released from the potential energy of a metal ball on the rubber sheet. We see that the analogy between the rubber sheet and general relativity works in this respect.
In the rubber sheet model, it is the whole deformed area which contains positive deformation energy. In the Einstein-Hilbert action, the deformation energy is calculated from areas where R != 0. In the Schwarzschild solution, R != only inside the mass in the center. The deformation of spacetime outside the mass does not contribute to the integral because R = 0 there. Does this mean that the R term inside the mass accounts also for the deformation energy outside the mass?
One may imagine that for the rubber sheet, the total deformation energy can be determined from the deformation in the area where the sheet is pushed down by a metal ball. But is it so? We need to check if someone has studied rubber sheets.
For the rubber sheet around a metal ball, on a radial line, the second derivative of the elevation of rubber is negative. On a tangential line, the second derivative is positive. For a patch of rubber to stay static, the forces that result from non-zero second derivatives must balance each other. This is similar to R being zero in the Schwarzschild solution around a sphere. R is the average curvature. If R = 0, curvatures on various axes must add to zero.
The analogy between a rubber sheet and general relativity looks strong.
Let us draw x and y axes on a rubber sheet A metal ball resting on the sheet causes both x and y axes to stretch close to the ball. That is, an ant living on the surface of the rubber sheet will measure a longer distance between any points P and P' after the rubber has stretched. The ant uses a rigid ruler.
The volume of rubber stays the same: a z axis embedded inside the rubber will contract as x and y stretch.
https://en.wikipedia.org/wiki/Poisson%27s_ratio
Poisson's ratio for 3-dimensional rubber is almost exactly 0.5: if we stretch 1 % in the x direction, the y and z directions will contract by 0.5 %. The volume stays the same.
In the Schwarzschild solution, a local observer outside the mass measures a temporal distance (t, t') shorter than the global observer, while he measures a radial distance (r, r') longer than the global observer. In the rubber model, the rubber has stretched along the axis r and contracted along the axis t. The rubber has not stretched at directions normal to r.
Suppose that the Schwarzschild mass is a thin shell. Let us extend the external Schwarzschild solution inside the shell.
In the interior Schwarzschild metric, at the center of the mass, the metric of spatial coordinates is the Minkowski metric. The time element is
4/9 (1 - r_s / r_g) dt^2,
where r_g is the radius of the sphere in global coordinates.
The volume of the rubber inside the shell has decreased: distances in time are less but they are not offset by larger distances along r. Here we have a discrepancy from a realistic rubber model.
An almost rigid body with a low mass close to a Schwarzschild sphere
We need a method to estimate how much the rigid body affects the minimal value of the Einstein-Hilbert action. Then we can calculate the force of anti-gravity on the rigid object.
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