UPDATE November 3, 2023: If we add an infinitesimal mass dM into empty Minkowski space, the change of the metric g is infinitesimal at any one place, but the volume integral of the change is infinite over the whole space!
Thus, adding an infinitesimal mass dM is not equivalent to a "small" change of the metric. We are not allowed to use it in the variational calculus.
What if we add dM, but restrict the metric perturbation within some radius R from dM, and keep the flat Minkowski metric unchanged elsewhere? That is equivalent to adding a shell of negative mass -dM to a distance R from dM. The chance in the Ricci scalar over the entire space is probably zero in that case, because dM - dM = 0.
This may explain why dR / dg⁰⁰ = R⁰⁰, etc. That is, a flat space is a stationary point of the Einstein-Hilbert action, after all.
If we already have a mass M in the space, and add dM at the same location as M, then the redshift from M to R allows us to add a negative mass -dM' whose absolute value is less than dM, at a distance R. The total mass then grows by dM - dM', which probably means that R integrated over the whole space grows.
This reminds us of the problem with Birkhoff's theorem.
If we move a mass dM for a short distance, the change in the metric over the whole 3D space volume is how much?
It is very roughly proportional to the integral of
1 / r - 1 / (r + 1)
~ 1 / r²
integrated over r² dr. The integral is infinite. We may ask what are the infinitesimal variations which we are allowed to use in the variational calculus. Moving a small mass over a short distance corresponds to a "large" variation? Or do we have a reason to label it as a "small" variation?
----
1. a positive pressure seems to attract a test mass m in general relativity, but
2. Birkhoff's theorem seems to prohibit any attraction by pressure outside a spherically symmetric mass M.
In electromagnetism, the source of the field, that is, the electric charge Q, is conserved. In general relativity, the mass-energy T⁰⁰, and the mass flow T⁰¹, T⁰², T⁰³ are conserved.
The pressure T¹¹, T²², T³³ is another source of the gravity field. Is pressure conserved? It would be very strange. But if pressure is not conserved, we end up in a conflict with Birkhoff's theorem.
The Einstein-Hilbert action does not imply a zero Ricci scalar for a vacuum area?
The stationary action principle says that the variation of the action S is zero for small variations of the metric g.
Let us calculate a variation in two dimensions. If we are allowed to stretch the x and y distances at most by a factor 1.01 within a square whose side is 1, we can make a "bulge" in the plane, such that its height is roughly 0.07, and the radius of curvature is very roughly r = 14.
The scalar curvature is
2 / r².
We are able to add very roughly 0.01 units of scalar curvature into the square by allowing the metric to be stretched by 0.01.
What about shrinking the metric at most by a factor 0.99?
A positive Ricci scalar corresponds to a metric where the circumference of a circle is surprisingly short, compared to the radius. If we are allowed to shrink distances within a square, we obtain a corresponding circle whose circumference is surprisingly long, compared to the radius. We probably are able to construct a metric which adds roughly -0.01 of scalar curvature to the square.
We showed that a flat metric is not a stationary point of the Einstein-Hilbert action. This means that the Einstein field equations do not describe a lagrangian model of mechanics.
In the Wikipedia derivation of the Einstein field equations, the variation of R is calculated against a variation of single components of the metric, e.g., g⁰⁰. We vary two components of the metric simultaneously. Which is the correct procedure?
In a rubber model of gravity, any deviation from the flat metric increases the elastic energy of rubber. A rubber model is able to be a genuine lagrangian mechanical model.
Varying just a single component of a metric
_____
/ \
| |
\______/
^ y
|
------> x
Let us analyze the x, y plane. Suppose that we are allowed to stretch the y metric slightly in a certain area, but not the x metric. Does the ratio
circumference / radius
of a circle shrink for some circle? And does it shrink linearly relative to the perturbation of the y metric?
Conclusions
A possible way to fix the problem is to put |R| instead of R in the Einstein-Hilbert action. Though the derivative of |R| with respect to g is then not defined at g = η, which may spoil the variational calculus.
We have to check what literature says about stationary points of the Einstein-Hilbert action.
If it turns out that the Einstein field equations do not capture the action, that might explain the mystery around Birkhoff's theorem.
No comments:
Post a Comment