UPDATE April 16, 2024: We did not prove anything yet on April 8, 2024. More research is needed. The rubber block squeezing model is our best bet.
| squeeze
v
2 R ________
| / \
| | | semisphere
| \_________/
cylinder
^
| squeeze
We want to squeeze an elongated rubber block, with no shear. Is it so that the outside shape of the block has to stay constant? If yes, then the pressure inside the rubber in the semispherical end parts will be larger because there the volume is ~ R³, while in the cylindrical middle part, the volume is ~ R². It will not work.
The Schwarzschild metric would correspond to the semispheres, and the Levi-Civita metric to the cylindrical part.
local
| squeeze
v
----------------------------- --> negative pressure
rubber <-- positive pressure
-----------------------------
^
| local
squeeze
In the diagram we have a long cylindrical rubber block. The surface of the block probably is horizontally stretched near the point where we apply the squeeze: there is a negative horizontal pressure in the surface. But at the center of the block there probably is a positive horizontal pressure which prevents the rubber from flowing away from the point which we squeeze.
There is "shear" in the configuration. Shear in the meaning that in a "natural" coordinate system there is shear. The Cauchy stress tensor can always be diagonalized, and the shear disappears.
----
The difficulty in calculating Christoffel symbols and Ricci curvatures lies in the three indices which are carried around. The formulae are better suited for a computer than a human. We may in the future repeat the calculations in Mathematica or some other symbolic calculator.
Another difficulty is that it is not easy to see if the results are reasonable. The human intuition about metrics in three or four dimensions is not too well developed.
The coordinates
The coordinates are the standard cylindrical coordinates t, r, φ, and z, which are given numbers 0, 1, 2, 3, in this order. The metric signature is (- + + +).
^ r
|
| L length
0 =========== cylinder
--------------|-------------> z
0
We have z = 0 at the midpoint of the cylinder. The radial coordinate is r = 0 at the center of the cylinder
We denote the derivative with respect to r by the prime '.
The assumptions
1. The metric g is a small perturbation of the flat metric η.
2. The metric g is static, i.e., it has no time dependence.
3. The metric g is rotationally symmetric around the axis of the cylinder.
4. The metric g is mirror symmetric around the midpoint of the cylinder.
5. The cylinder is uniform, rotationally symmetric, and of a finite length L. There is no significant pressure.
6. The newtonian gravity potential gives a fairly good approximation of the metric of time, g₀₀.
7. The metric of the φ coordinate is the simple g₂₂ = r².
8. We allow the off-diagonal component g₁₃ to be non-zero (r and z metrics are skewed). Other off-diagonal components are zero.
If assumption 6 would not be true, then we would observe strange behavior in gravitational lenses in the sky. Also, the Cavendish torsion balance experiment would give odd results for a cylindrical weight. Orbits of planets might differ from the ones calculated with post-newtonian methods.
Assumption 6 could also be taken as an axiom: a theory of gravity must be close to newtonian gravity for weak fields.
The same grounds apply for assumption 1. If a small mass would create large perturbations of the flat metric, then we would see odd phenomena in gravitational lenses.
Assumption 6 can be made true by defining r = 1 / (2 π) * the proper length of the circle around the axis of the cylinder.
The inverse metric g⁻¹
The formulae of Christoffel symbols require us to know the inverse matrix of g, because there are the factors like g¹¹ in the symbols. Let the metric be
g =
g₀₀ 0 0 0
0 g₁₁ 0 g₁₃
0 0 r² 0
0 g₁₃ 0 g₃₃,
where g₀₀ is very close to -1, g₁₁ and g₃₃ are very close to 1, and g₁₃ is very close to 0. Then the inverse matrix is approximately
g⁻¹ ≈
1 / g₀₀ 0 0 0
0 1 / g₁₁ 0 -g₁₃
0 0 1 / r² 0
0 -g₁₃ 0 1 / g₃₃,
The elements of g⁻¹ are denoted g¹¹, and so on.
Christoffel symbols
We ignore all terms where there is the product of two "perturbations". That is, any product of two derivatives of g₀₀, g₁₁, or g₃₃, and any product of g₁₃ with a derivative of those.
Note that one can permute the subscripts of a Christoffel symbol, and the value does not change:
Γ¹₁₃ = Γ¹₃₁.
All the components of the metric g depend on r and z (1 and 3). No component depends on t or φ (0 and 2).
A non-zero Christoffel symbol must involve a derivative with respect to 1 or 3. This means that 1 or 3 has to be among k, i, j in the formula above.
Christoffel symbols involving 0 and 1 are of the form Γ¹₀₉, Γ⁰₁₉, or Γ⁹₀₁, where 9 denotes any coordinate.
Γ¹₀₀ = 1/2 g¹¹ * -dg₀₀ / dr,
Γ⁰₁₀ = 1/2 g⁰⁰ * dg₀₀ / dr ≈ -1/2 g₀₀'.
All others are zero. Next we check Γ³₀₉, Γ⁰₃₉, and Γ⁹₀₃.
Γ³₀₀ = 1/2 g³³ * -dg₀₀ / dz,
Γ⁰₃₀ = 1/2 g⁰⁰ * dg₀₀ / dz ≈ -1/2 dg₀₀ / dz.
The Christoffel symbol involving three times 1 is:
Γ¹₁₁ = 1/2 g¹¹ * dg₁₁ / dr.
The Christoffel symbols involving 1 twice are of the form Γ¹₁₉ or Γ⁹₁₁.
Γ¹₁₃ = 1/2 g¹¹ * dg₁₁ / dz,
Γ³₁₁ = g³³ * dg₁₃ / dr
- 1/2 g³³ * dg₁₁ / dz.
The symbols involving 1 and 2 are of the form Γ¹₂₉, Γ²₁₉, or Γ⁹₁₂.
Γ¹₂₂ = 1/2 g¹¹ * -dg₂₂ / dr
= -g¹¹ * r,
Γ²₁₂ = 1/2 g²² * dg₂₂ / dr
= 1 / r.
The symbols involving 1 and 3 are of the form Γ¹₃₉, Γ³₁₉, or Γ⁹₁₃.
Γ¹₃₃ = g¹¹ * dg₁₃ / dz
+ 1/2 g¹¹ * -dg₃₃ / dr
Γ³₁₃ = 1/2 g³³ * dg₃₃ / dr.
The symbol involving three times 2 is zero. The symbols involving two 2 and one 3 are Γ²₂₃ and Γ³₂₂.
Γ³₂₂ = 1/2 g³¹ * -dg₂₂ / dr
= -g³¹ * r.
The symbols involving two 3 and one 2 are Γ³₂₃ and Γ²₃₃, which are zero.
Γ³₃₃ = 1/2 g³³ * dg₃₃ / dz.
The Christoffel symbols calculated above are identical to the ones on April 8, 2024.
The off-diagonal curvature R₁₃
In vacuum, the stress-energy tensor T is zero. The Einstein field equations say that all components of the Ricci curvature tensor must be zero. In particular, the off-diagonal component R₁₃ must be 0.
A product Γ * Γ can only be significant if it contains Γ¹₂₂ or Γ²₁₂.
R₁₃ = dΓ¹₁₃ / dr + dΓ³₁₃ / dz
- dΓ⁰₃₀ / dr - dΓ¹₁₃ / dr
- dΓ³₃₃ / dr
+ Γ²₂₁ Γ¹₁₃
= 1/2 g³³ * d²g₃₃ / (dr dz)
- 1/2 g⁰⁰ * d²g₀₀ / (dz dr)
- 1/2 g³³ * d²g₃₃ / (dz dr)
+ 1 / r * 1/2 g¹¹ * dg₁₁ / dz
= -1/2 g⁰⁰ * d(g₀₀') / dz
+ 1 / r * 1/2 g¹¹ * dg₁₁ / dz.
≈ 1/2 ( d(g₀₀') / dz + 1 / r * dg₁₁ / dz ).
R₀₀
R₀₀ = dΓ¹₀₀ / dr + dΓ³₀₀ / dz
+ Γ²₂₁ Γ¹₀₀
≈ 1/2 g¹¹ * -d²g₀₀ / dr²
+ 1/2 g³³ * -d²g₀₀ / dz²
+ 1 / r * 1/2 g¹¹ * -dg₀₀ / dr
≈ 1/2 ( -g₀₀'' - g₀₀' / r - d²g₀₀ / dz² ).
R₁₁
R₁₁ = dΓ¹₁₁ / dr + dΓ³₁₁ / dz
- dΓ⁰₁₀ / dr - dΓ¹₁₁ / dr
- dΓ²₁₂ / dr - dΓ³₁₃ / dr
+ Γ²₂₁ Γ¹₁₁ - Γ²₁₂ Γ²₂₁
≈ dg₁₃' / dz - 1/2 d²g₁₁ / dz²
+ 1/2 g₀₀''
+ 1 / r² - 1/2 g₃₃''
+ 1 / r * 1/2 g₁₁' - 1 / r²
= 1/2 ( 2 dg₁₃' / dz - d²g₁₁ / dz²
g₀₀'' - g₃₃'' + g₁₁' / r ).
R₂₂
R₂₂ = dΓ¹₂₂ / dr + dΓ³₂₂ / dz
+ (Γ⁰₀₁ + Γ¹₁₁ + Γ²₂₁ + Γ³₃₁) Γ¹₂₂
- Γ²₁₂ Γ¹₂₂ - Γ¹₂₂ Γ²₁₂
= dΓ¹₂₂ / dr + dΓ³₂₂ / dz
+ (Γ⁰₀₁ + Γ¹₁₁ - Γ²₂₁ + Γ³₃₁) Γ¹₂₂
= d(-g¹¹ * r) / dr
+ d(-g³¹ * r) / dz
+ ( 1/2 g⁰⁰ * dg₀₀ / dr
+ 1/2 g¹¹ * dg₁₁ / dr
- 1 / r
+ 1/2 g³³ * dg₃₃ / dr )
* -g¹¹ * r.
We have:
d(-g¹¹ * r) / dr = -d(1 / g₁₁ * r) / dr
= 1 / (g₁₁)² * g₁₁' * r
- 1 / g₁₁,
d(-g³¹ * r) / dz = r dg₃₁ / dz,
R₂₂ = g₁₁' / (g₁₁)² * r - 1 / g₁₁ + r dg₃₁ / dz
- r / g₁₁ *
( -1/2 g₀₀'
+ 1/2 g₁₁' / g₁₁
- 1 / r
+ 1/2 * 1 / g₃₃ * g₃₃')
= 1/2 r g₁₁' / (g₁₁)²
+ 1/2 r g₀₀' / g₁₁
- 1/2 r g₃₃' / (g₁₁ g₃₃)
+ r dg₃₁ / dz.
R₃₃
R₃₃ = dΓ¹₃₃ / dr + dΓ³₃₃ / dz
- dΓ⁰₃₀ / dz - dΓ¹₁₃ / dz - dΓ³₃₃ / dz
+ Γ²₁₂ Γ¹₃₃
= g¹¹ * d²g₁₃ / (dz dr)
+ 1/2 g¹¹ * -d²g₃₃ / dr²
- 1/2 g⁰⁰ * d²g₀₀ / dz²
- 1/2 g¹¹ * d²g₁₁ / dz²
+ 1 / r * (g¹¹ * dg₁₃ / dz
+ 1/2 g¹¹ * -dg₃₃ / dr)
≈ dg₁₃' / dz + 1 / r * dg₁₃ / dz
- 1/2 g₃₃''
+ 1/2 d²g₀₀ / dz²
- 1/2 d²g₁₁ / dz²
- 1/2 g₃₃' / r.
The equations
2 R₁₃ = d(g₀₀') / dz + 1 / r * dg₁₁ / dz = 0,
2 R₀₀ = -g₀₀'' - g₀₀' / r - d²g₀₀ / dz² = 0,
2 R₃₃ = -g₃₃'' - g₃₃' / r + d²g₀₀ / dz²
- d²g₁₁ / dz²
+ 2 dg₁₃' / dz + 2 / r * dg₁₃ / dz = 0,
2 R₁₁ = g₀₀'' - g₃₃'' + g₁₁' / r
- d²g₁₁ / dz²
+ 2 dg₁₃' / dz = 0,
2 R₂₂ / r² = g₀₀' / r - g₃₃' / r + g₁₁' / r
+ 2 / r * dg₁₃ / dz = 0.
Our April 8, 2024 calculations contained a significant error in R₃₃.
Conclusions
The calculations above were written carefully and might be correct.
We need to understand the role of shear and the R₁₃ curvature if we wish to derive a contradiction. We will next study the relationship between the Einstein field equations and rubber models.
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