The metric g is assumed to be very close to the flat cylindrical metric:
η =
-1 0 0 0
0 1 0 0
0 0 r² 0
0 0 0 1.
Thin disk
^ r
| thin disk, radius R, thickness L
| |
------|-----------------> z
|
z = 0
We denote the mass density of the thin disk by ρ(r). The disk has a radius R and its center is located at z = 0. The thickness of the disk is L.
The stress-energy tensor is
T =
ρ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0,
and the Ricci tensor
R =
1/2 ρ 0 0 0
0 1/2 ρ 0 0
0 0 1/2 ρ r² 0
0 0 0 1/2 ρ.
The formulae are:
2 R₁₃ = dg' / dz + 1 / r * dg₁₁ / dz = 0,
2 R₀₀ = -g'' - g' / r - d²g / dz² = ρ,
2 R₃₃ = -g₃₃'' - g₃₃' / r + d²g / dz²
- d²g₁₁ / dz²
+ 2 dg₁₃' / dz + 2 / r * dg₁₃ / dz = ρ,
2 R₁₁ = g'' - g₃₃'' + g₁₁' / r - d²g₁₁ / dz²
+ 2 dg₁₃' / dz = ρ,
2 R₂₂ / r² = g' / r - g₃₃' / r + g₁₁' / r
+ 2 / r * dg₁₃ / dz = ρ,
where the prime ' denotes the r derivative, and the plain letter g denotes the metric component g₀₀.
Let us sum the equations for R₀₀, R₁₁, R₂₂, and subtract the equation for R₃₃:
2 g₁₁' / r - 2 d²g / dz² = 2 ρ
<=>
g₁₁' = r d²g / dz² + r ρ.
The equation for R₀₀ yields:
g₁₁' = -r g'' - g'.
The integral function is:
g₁₁ = -r g' + 1,
which has the correct value 1 at large r.
The equation for R₂₂ gives:
g₃₃' = g₁₁' + g' + 2 dg₁₃ / dz - r ρ
= -r g'' - r ρ + 2 dg₁₃ / dz.
Let us set the skew g₁₃ identically zero. Then
g₃₃' = -r g'' - r ρ.
The integral function is:
g₃₃ = 2 + g - r g' + Int(r, r ρ),
where
∞
Int(r, r ρ) = ∫ r ρ(r) dr
r
denotes the definite integral of r ρ from r to infinity. Then g₃₃ has the correct value 1 at large r.
Analysis of g₃₃
Let us compare the value of g₃₃ at r = 0 inside the thin disk and just outside the surface of the disk.
There is a major difference in the value: inside, g₃₃ is larger by Int(r, r ρ). The g₃₃ metric inside the disk is much more stretched than just outside.
Could this just be an artefact from the choice of coordinates?
We can stretch the proper distance of the coordinates lines z = constant outside the disk, in order to get the metric component 33 to match inside and outside. But is this redefinition "natural"? What does it look like from the viewpoint of geodesics which run to the r direction?
| geodesic to r direction
|
|
-----------------> z
The geodesics are determined by the g₁₁ metric component. The geodesics turn to the side where g₁₁ is larger. Above we derived:
g₁₁ = -r g' + 1.
Let us keep the total mass M of the thin disk constant. We increase its radius R and make the disk thinner (thickness L), keeping the density ρ constant.
The newtonian gravity field is almost homogeneous close to the the surface of the disk, close to the center of the disk. By enlarging the disk with the procedure above, we can make
|g'(0) - g'(L)|
as small as we like. The radial metric g₁₁ is essentially the same inside and outside the disk. The geodesics to the r direction essentially run at z = constant.
On the other hand, the integral
Int(r, r ρ),
stays very much different inside the disk and outside the disk.
If geodesics determine the "proper" coordinate lines, then the g₃₃ metric, indeed, is very different inside the disk and outside it.
Also,
g₃₃ = 2 + g - r g' + Int(r, r ρ),
the value of g₃₃ can be quite large inside the disk, near the center. The integral Int(r, r ρ) can get a large value if the disk has a large radius R, even though the newtonian gravity potential is very close to zero, and g = g₀₀ is very close to -1.
A mass ring causes a major g₃₃ metric change
We have
∞
Int(r, r ρ) = ∫ r ρ(r) dr.
r
■ ring
R
---------|-----------------> z
0
■ ring
Let us put a mass M as a ring of a width L and a radial thickness K at a radius R around z = 0. If we increase R, then we make K smaller:
K ~ 1 / R
to keep the mass of the ring constant M. The integral Int(0, r ρ) stays constant as we increase R.
Thus, the "enclosed volume" of the ring, the volume with
-L / 2 < z < L / 2,
r < R,
has the g₃₃ metric stretched, and the stretching is independent of R.
When R is very large, then g₀₀ is very close to -1 everywhere. But we still have a substantial change in the spatial metric in a large volume. Is this reasonable?
We may also let ρ to approach zero as R grows, but we keep the total mass M constant. Still, the metric in the enclosed volume remains significantly stretched. The value of the integral is not much affected.
What about rubber sheet models of gravity? If we put a total mass M on a tense rubber sheet, and let the area density ρ of M to approach zero, then, obviously, the rubber sheet approaches a flat configuration.
The predictions of general relativity do not look reasonable. Also, we found here a major difference between general relativity and rubber sheet models of gravity.
The metric inside a spherical shell of mass
Let the radius of a thin spherical shell be R. Its center is at r = 0 and z = 0. Let us use our April 13, 2024 formulae to determine the metric inside a thin, uniform spherical shell of mass. Above we derived:
g₁₁ = -r g' + 1,
g₃₃ = 2 + g - r g' + Int(r, r ρ).
The formulae look different. Inside the sphere, the derivative g' is zero. What is the value of Int(r, rρ)? It is a constant because for a "latitude" α on the sphere, we have
r ~ sin(α)
and the projection of the area density of the shell on the equatorial plane is
~ 1 / sin(α).
Also, g is constant. The metric inside the shell is flat, because both g₁₁ and g₃₃ are constant.
The metric component g₃₃ inside the shell is constant, but it is not 1. Note that we have made the skew g₁₃ identically zero in the solution. These are not the standard Schwarzschild coordinates, since in them the skew would not be zero. We have g₃₃ > 1 at z = 0 and r > R, while in the standard Schwarzschild coordinates, the corresponding tangential metric is 1.
The value of
g - r g' + Int(r, r ρ)
must probably stay constant as we dive through a thin shell?
Otherwise, g₃₃ would differ inside and outside the thin shell.
But the product of
r ≈ R sin(α)
and
g' ~ sin(α)
is different for different α. How can it be? We had the skew zero outside the shell. Is it so that one cannot make it zero inside the shell?
The coordinate lines of the Schwarzschild metric in the standard Schwarzschild metric outside the shell have the "barrel" distortion.
In order to orthogonalize, we have to change the coordinate lines of z to look like a "hourglass".
But inside the shell, the metric is flat. The z coordinate lines must be straight, in order for them to be orthogonal to the r coordinate lines.
The "abrupt" turn of the coordinate lines at the shell probably explains why g₃₃ seemingly changes at the shell. The component g₃₃ is not discontinuous. The shell has a finite thickness > 0.
Let us determine the internal circumference of the shell and compare to the external.
In the Schwarzschild interior and exterior solutions in spherical coordinates, the tangential metric is set to 1. The circumference is the same there. We have already checked that the Schwarzschild exterior solution, written in cylindrical coordinates, satisfies the April 13, 2024 formulae.
The Schwarzschild solution explains why g₃₃ changes when we go through the shell. The radial metric in spherical coordinates is stretched, the tangential metric is not. When the latitude α is not 90 degrees, then some of the radial stretching is projected to g₃₃. Inside the shell there is no radial stretching. Thus, we expect the g₃₃ metric to change as we go through the shell. That is exactly what happens.
Conclusions
General relativity gives a strange prediction for the g₃₃ metric inside a thin disk, and close to the surface of the disk. Our Minkowski & newtonian gravity model predicts a much smaller change in the g₃₃ metric.
In the future, observations of disk-shaped gravity lenses in the sky may decide which gravity model is correct.
