UPDATE April 2, 2024: We forgot the factor g²¹ in the formulae. Since the coordinates are only slightly skewed, that factor makes Γ²₁₁ enormous, spoiling our attempt to calculate perturbations only.
After fixing a sign error, we no longer have a contradiction. However, we have to check carefully both the pressure case and uniform mass case.
----
UPDATE April 1, 2024: From Gauss's law we get the newtonian gravity force inside a cylinder. The mass within a radius r from the center is
~ r²,
and the area of the cylinder at the radius r is
~ r.
The density of lines of force at r is
~ r² / r = r.
The force is
F ~ r,
and the potential
V ~ r².
Our formula
g₀₀(x) = -p / 4 * x² - C
agrees with the newtonian potential.
We also added a note about Picard-Lindelöf and the uniqueness of g₀₀.
----
UPDATE March 31, 2024: We checked that factors g¹¹, etc. in the Christoffel symbols have a negligible effect in this calculation, as g is almost Minkowski, and only perturbed slightly.
----
Let us try to figure out what kinds of metrics there could exist inside pressurized matter. Karl Schwarzschild (1916) was able to find such a metric for incompressible fluid, which is pressurized by its own gravity.
We use the metric signature (- + + +).
Let us first consider an unphysical configuration where a ball of weightless matter has uniform pressure throughout.
The stress-energy tensor T within the ball looks something like
T =
0 0 0 0
0 p 0 0
0 0 p 0
0 0 0 p,
where p is the pressure. For now, we assume that the metric g is very close to the Minkowski metric η.
We can obtain the associated Ricci tensor R by "trace reversing":
Tr = g⁰⁰ T₀₀ + g¹¹ T₁₁ + g²² T₂₂ + g³³ T₃₃
= -T₀₀ + T₁₁ + T₂₂ + T₃₃
= 3 p.
Then the Ricci tensor
R = T - 1/2 Tr * g
= T - 3/2 p * g
=
3/2 p 0 0 0
0 -p / 2 0 0
0 0 -p / 2 0
0 0 0 -p / 2.
• • • static test masses
• • •
• • •
=> time passes
• • •
• • •
• • •
The Ricci curvature to the time direction, R₀₀ is positive. This means that if we let an initially static cubic constellation of test masses fall freely, the volume of the cube starts to shrink. This is just like with a positive mass density.
Let us then calculate the Ricci tensor for a positive mass density ρ and a zero pressure.
TM =
ρ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0,
RM =
ρ / 2 0 0 0
0 ρ / 2 0 0
0 0 ρ / 2 0
0 0 0 ρ / 2.
The Ricci curvature is positive to the time direction as well as to all spatial directions.
In the picture we have the Schwarzschild spatial metric for a plane passing through the center of the ball. The spatial metric in the interior Schwarzschild solution is stretched in the radial direction.
But for the pure pressure case, the Ricci curvature is negative to all spatial directions. The metric of time might be roughly the same in both cases. In the pressure case, we have to add "defocusing" to every direction. We have to shrink the radial metric inside the cap, making the spatial geometry "saddle-shaped"?
Then we face a dilemma. Outside the central ball, R₀₀ has to be zero. That requires that the gravity potential is ~ 1 / r, where r is the distance from the center.
Thus, the metric of time must be like in the Schwarzschild solution in vacuum. To make the Ricci tensor R there zero, also the spatial metric must be like in the Schwarzschild exterior solution. That is, the radial metric is stretched. We cannot fit the radial metric smoothly at the surface of the ball because the metric is shrunk inside and stretched outside.
There may be no (static) solution for pure pressure inside a ball.
Iterative solution of the Einstein equations does not succeed because the residual stress-energy tensor T contains pure pressure?
Suppose that we try to use an iterative method to solve the Einstein field equations. We have found an approximate metric g. We calculate the residual T in vacuum from the formula:
If T would be zero, then g would be the solution.
Let us assume that the residual T is small. If we were able to find a small metric perturbation h, such that
η + h
on the left side of the equation would reproduce the residual T on the right side, then
g - h
might be a better approximate solution, assuming that the equation is roughly linear.
For example, if T contains simply "mass" density in its T₀₀ component, we often can find a very good metric h: something like the Schwarzschild solution.
But if the residual T contains pure pressure, and there is no approximate solution of the equations for pure pressure, then the iteration stops. We cannot continue.
The problem may be the "rigidity" of the requirement that Ricci curvature is zero in vacuum. It is like the paper bending exercise of our March 8, 2024 blog post. The rigidity of paper severely restricts the shapes it can take. Locally, the shape has to be a cone or a cylinder.
If we could find a good approximate metric for any small T, then we would probably have a very efficient iterative solution procedure for the Einstein equations. But no one has found such a method in 109 years. Is it so that in most cases we cannot find any metric at all for a small residual T? Even if T would mimic realistic matter?
Pressure in a cylinder to the z direction
negative pressure
=========
| |
========= positive pressure
| |
=========
negative pressure
------> z
|
v x
y points out
of the screen cylinder
● ------> z ==============
| positive pressure p
v x <--->
We can build static structures where a cylinder has essentially a constant pressure in the z direction and no pressure in other directions. In the diagram we have three cylinders. The outer cylinders want to contract and have a negative pressure. They put a positive pressure into the central cylinder.
We assume that the mass of the cylinder is zero and the pressure p small.
Can we find an approximate metric for such a device?
Let us assume that the cylinder is relatively long. We try to find a cylindrically symmetric metric. We have a point (x, y, z) inside the cylinder, where |y| << |x|.
The stress-energy tensor T looks like this:
T =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 p
and the Ricci tensor:
R =
p / 2 0 0 0
0 -p / 2 0 0
0 0 -p / 2 0
0 0 0 p / 2
We can make use of our March 11, 2024 calculations of the Schwarzschild metric.
We assume that r points approximately to the x direction. The metric is
ds² = g₀₀ dt² + grr dr² + dn² + g₃₃ dz²,
where r is the radial vector normal to the z axis, and n is the normal vector to r in the x,y plane.
We assume that r is scaled in such a way, that the metric on n is 1.
In cylindrical coordinates:
dr = (1 - 1/2 * y² / x²) dx
+ y / x * dy,
dn = (1 - 1/2 * y² / x²) dy
- y / x * dx,
We drop very small terms and write the spatial metric in terms of cartesian coordinates:
ds² = grr * dx²
+ (1 - grr) * y² / x² * dx²
+ 2 (grr - 1) * y / x * dx dy
+ dy²
+ (grr - 1) y² / x² * dy²
+ g₃₃ dz².
Curvatures due to the metric of time, spatial metric set flat
We assume that all deviations from the flat Minkowski metric η are very small and that the partial derivatives of the components of the metric tensor, gij, are very close to zero. Then all the Christoffel symbols Γ are almost zero, and we can ignore the cross terms Γ * Γ in the formula of Rjk.
That is, we assume that the Rjk are linear in each gij.
Consequently, we can calculate the curvatures in parts: first the curvatures for the distorted metric g₀₀ of time, keeping the spatial metric flat, and then vice versa.
Let us calculate the curvatures which are due to the perturbed metric of time g₀₀:
Γ¹₀₀ = g¹¹ * 1/2 * -dg₀₀ / dx
= Γ⁰₁₀ = Γ⁰₀₁,
Γ²₀₀ = g²² * 1/2 * -dg₀₀ / dx * y / x,
R₀₀ = dΓ¹₀₀ / dx + dΓ²₀₀ / dy,
= -1/2 g₀₀'' + -1/2 g₀₀' / x,
where the prime ' denotes the derivative with respect to x. Can we treat g¹¹ and g²² as a constant 1 because their derivatives are very small and would multiply other very small derivatives? If the size of the system is one unit, then gij - 1, gij', and gij'' all have the same order of magnitude, and all have a very small absolute value. Any product of two of them is negligible. We can treat them as a constant 1.
R₁₁ = dΓ⁰₁₀ / dx
= 1/2 * g₀₀'',
Γ⁰₂₀ = 1/2 * -1 * dg₀₀ / dx * y / x,
R₂₂ = -Γ⁰₂₀ / dy
= 1/2 g₀₀' / x,
R₃₃ = 0.
Curvatures due to the spatial metric, the metric of time set to -1
Let us study the perturbation of the spatial metric.
R₀₀ = 0,
The factor g²¹ is missing!
|
v
Γ²₁₁ g₂₂ = (dg₁₂ / dx - 1/2 dg₁₁ / dy)
= d((grr - 1) * y / x) / dx
- 1/2 d(grr + (1 - grr) * y² / x²) / dy
= g₁₁' * y / x - (g₁₁ - 1) * y / x²
- 1/2 g₁₁' * y / x
- (1 - g₁₁) * y / x²,
Γ²₁₂ g₂₂ = 1/2 dg₂₂ / dx
= 0,
Γ³₁₃ g₃₃ = 1/2 dg₃₃ / dx.
Can we ignore the factors g₂₂, etc. after the Γ? Yes. Their derivative would multiply very small numbers, like (1 - g₁₁), and the product would be negligible.
R₁₁ = dΓ²₁₁ / dy + dΓ³₁₁ / dz
- dΓ²₁₂ / dx - dΓ³₁₃ / dx
= g₁₁' / x - g₁₁ / x² + 1 / x²
- 1/2 g₁₁' / x
- 1 / x² + g₁₁ / x²
- 1/2 g₃₃''
= 1/2 g₁₁' / x - 1/2 g₃₃''.
Let us calculate R₂₂. We can ignore the factors g¹¹, etc. in the Christoffel symbols, because their derivatives would multiply very small numbers.
Γ¹₂₂ = dg₂₁ / dy - 1/2 dg₂₂ / dx
= d((grr - 1) * y / x) / dy
- 1/2 d(1 + (grr - 1) * y² / x²) / dx
= (g₁₁ - 1) / x,
Γ³₂₂ = dg₂₃ / dy - 1/2 dg₂₂ / dz
= 0,
Γ⁰₂₀ = -1/2 dg₀₀ / dx
= 0,
Γ¹₂₁ = 1/2 dg₁₁ / dy
= 1/2 d(grr + (1 - grr) * y² / x² ) / dy
= 1/2 g₁₁' * y / x + (1 - g₁₁) * y / x²,
Γ³₂₃ = 1/2 dg₃₃ / dy
= 1/2 dg₃₃ / dx * y / x,
R₂₂ = dΓ¹₂₂ / dx + dΓ³₂₂ / dz
- dΓ⁰₂₀ / dy - dΓ¹₂₁ / dy - dΓ³₂₃ / dy
= g₁₁' / x - g₁₁ / x² + 1 / x²
- 1/2 g₁₁' / x - 1 / x² + g₁₁ / x²
- 1/2 g₃₃' / x
= 1/2 g₁₁' / x
- 1/2 g₃₃' / x.
Let us calculate R₃₃:
Γ¹₃₃ = -1/2 dg₃₃ / dx,
Γ²₃₃ = -1/2 dg₃₃ / dx * y / x,
R₃₃ = dΓ¹₃₃ / dx + dΓ²₃₃ / dy
- dΓ⁰₃₀ / dz - dΓ¹₃₁ / dz - dΓ²₃₂ / dz
= -1/2 * g₃₃'' - 1/2 * g₃₃' / x,
Analysis of R₀₀ inside the cylinder
The Ricci curvature to the direction of time, R₀₀, is positive inside the pressurized cylinder. That is, a cubical configuration of test masses tends to contract. We must have
R₀₀ = p / 2 = -1/2 g₀₀'' + -1/2 g₀₀' / x
The symmetry implies that
g₀₀'(0) = 0.
The graph of g₀₀(x) must look something like this:
0
-----------------------> x
__-------__
/ \
and the formula is
g₀₀(x) = -p / 4 * x² - C,
where C is a positive constant. Recall that g₀₀ in empty space is -1.
The Picard-Lindelöf theorem guarantees that we found the unique solution to the second order differential equation, if we know g₀₀(0) and g₀₀'(0).
Analysis of R₃₃ inside the cylinder
R₃₃ = p / 2 = -1/2 * g₃₃'' - 1/2 * g₃₃' / x.
The formula is
g₃₃(x) = -p / 4 * x² + C',
where C' is a positive constant. Recall that g₃₃ in empty space is 1.
Analysis of R₁₁ and R₂₂ inside the cylinder
We sum the Ricci curvature from the perturbation of the metric of time to the curvature from the spatial metric perturbation:
R₁₁ = 1/2 * g₀₀'' + 1/2 g₁₁' / x - 1/2 g₃₃'',
R₂₂ = 1/2 g₀₀' / x + 1/2 g₁₁' / x - 1/2 g₃₃' / x.
Above,
g₃₃'' = -p / 2,
g₃₃' / x = -p / 2,
g₀₀'' = -p / 2,
g₀₀' / x = -p / 2.
We have no contradiction!
A cylinder with no pressure
In the derivation of the contradiction above, we never used the fact that it is a pressure p inside the cylinder. We could as well assume that the pressure is zero and there is a mass density ρ. We have a contradiction also in that case.
In the literature we cannot find any discussion about the metric inside a cylinder. As if researchers would have overlooked the question altogether.
We have to check our calculations very carefully. Maybe there is an error?
Generally, since there are no existence theorems about a metric inside matter, it might be that such metrics only exist for very special configurations. The Schwarzschild interior solution is the well known example.
Our paper bending model on March 8, 2024 already suggested that solutions might exist only for beautifully symmetric configurations.
Conclusions
We have to check if a small departure from the cylindrical symmetry of the metric allows the Einstein equations to have a solution inside a cylinder. That is not likely, since our argument above already uses approximate values, and the contradiction in the values of R₁₁ and R₂₂ happens with a very large margin.
Also, we have to check the calculations carefully.
If the calculations are correct, this is a fatal blow to the Einstein field equations. The field equations have to be really badly wrong.
Our own Minkowski & newtonian gravity model has no problems handling a cylinder, with pressure, or without it.
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