UPDATE April 14, 2024: We corrected a sign error in g₀₀' and g₀₀''. Correcting the error seems to remove contradictions.
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On March 20, 2024 we tried to solve the metric inside a uniform cylinder, using cartesian coordinates and a metric perturbation. We found out that there is no solution to the Einstein field equations. However, our calculation contained errors because we forgot the factor g²¹ of skewed coordinates x and y. Also, there were sign errors.
Let us repeat the task, this time using cylindrical coordinates. They are orthogonal.
We are looking for a static, cylindrically symmetric solution which does not depend on time or on the z coordinate.
The stress-energy tensor and the metric
The coordinates are denoted t (coordinate 0), r (1), φ (2), and z (3). The stress-energy tensor is:
T =
ρ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0.
We assume that the metric g of spacetime is very close to the flat Minkowski metric η in cylindrical coordinates:
η =
-1 0 0 0
0 1 0 0
0 0 r² 0
0 0 0 1.
We use the metric signature (- + + +).
We can then calculate an approximate value of the Ricci curvature tensor R by "trace reversing".
Tr = g⁰⁰ T₀₀ + g¹¹ T₁₁ + g¹¹ T₂₂ + g³³ T₃₃
= -T₀₀ + T₁₁ + 1 / r² * T₂₂ + T₃₃
= -ρ,
where we assumed that g₀₀ is almost exactly -1.
R = T - 1/2 Tr * η
= T + 1/2 ρ * η
=
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 Ricci curvature to the φ direction appears to be r²-fold, compared to other directions. But that is just an artefact from the fact that the unit of length to that direction is r-fold compared to other directions. The Ricci curvature of a spherical surface is ~ 1 / R² in cartesian coordinates, where R is the radius. If we switch to a unit of length which is r-fold, then R shrinks by ~ 1 / r, and the curvature increases ~ r².
Let us check that we obtained the right Ricci tensor. The Ricci scalar R is
R = g⁰⁰ R₀₀ + g¹¹ R₁₁ + g²² R₂₂ + g³³ R₃₃
= -1/2 ρ + 1/2 ρ + 1/2 ρ + 1/2 ρ
= ρ.
The Einstein tensor is:
R - 1/2 R g =
ρ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
= T.
The line element in cylindrical coordinates is
ds² = g₀₀ dt² + g₁₁ dr² + g₂₂ dφ² + g₃₃ dz²,
where we choose g₂₂ to be exactly:
g₂₂ = r².
Is this choice allowed? We can define the value of the coordinate r by
r = proper circumference / (2 π).
Thus, it should be allowed that we fix the metric of φ to r².
The metric g which we will calculate will not exactly be the flat Minkowski metric η in cylindrical coordinates. Consequently, the Einstein tensor R - 1/2 R g will slightly differ from the stress-energy tensor T, because g is used twice in the calculation of the term R g.
Christoffel symbols
The metric does not depend on t, φ, or z. The derivatives with respect to them are zero. We denote the derivative with respect to r by the prime '.
A non-zero Christoffel symbol must involve the coordinate 1 (r) because gkl can only be non-zero if k = l, and only a derivative with respect to r can be non-zero.
The non-zero Christoffel symbols involving the coordinate 0 (t) are of the form Γ¹₀₉, Γ⁰₁₉, or Γ⁹₀₁, where 9 denotes any coordinate, and we have used the fact that one is allowed to permute the subscripts of Γ.
Γ¹₀₀ = 1/2 g¹¹ * -dg₀₀ / dr = -1/2 g₀₀' / g₁₁
Γ⁰₁₀ = 1/2 g⁰⁰ * dg₀₀ / dr = 1/2 g₀₀' / g₀₀.
The Christoffel symbol with only 1 (r) is
Γ¹₁₁ = 1/2 g¹¹ dg₁₁ / dr = 1/2 g₁₁' / g₁₁.
The non-zero Christoffel symbols involving the coordinate 2 (φ) are of the form Γ¹₂₉, Γ²₁₉, or Γ⁹₁₂:
Γ¹₂₂ = 1/2 g¹¹ * -dg₂₂ / dr = -1/2 * 2 r / g₁₁
= -r / g₁₁,
Γ²₁₂ = 1/2 g²² * dg₂₂ / dr = 1 / r.
The non-zero Christoffel symbols involving the coordinate 3 (z) are of the form Γ¹₃₉, Γ³₁₉, or Γ⁹₁₃:
Γ¹₃₃ = 1/2 g¹¹ * -dg₃₃ / dr = -1/2 g₃₃' / g₁₁,
Γ³₁₃ = 1/2 g³³ * dg₃₃ / dr = 1/2 g₃₃' / g₃₃
R₀₀
Only a derivative against the coordinate 1 (r) can differ from zero.
R₀₀ = dΓ¹₀₀ / dr
+ (Γ⁰₀₁ + Γ¹₁₁ + Γ²₂₁ + Γ³₃₁) Γ¹₀₀
- 2 Γ⁰₀₁ Γ¹₀₀
= -1/2 g₀₀'' / g₁₁ + 1/2 g₀₀' g₁₁' / (g₁₁)²
- 1/2 g₀₀' / g₁₁ *
(-1/2 g₀₀' / g₀₀ + 1/2 g₁₁' / g₁₁
+ 1 / r + 1/2 g₃₃' / g₃₃)
≈ -1/2 g₀₀'' - 1/2 g₀₀' / r.
Here we used the assumption that g₀₀', etc., are very small. We also assumed that g₁₁ is very close to 1.
We have
R₀₀ = -1/2 g₀₀'' - 1/2 g₀₀' / r
= 1/2 ρ.
We require that g₀₀'(0) = 0. The derivative cannot be non-zero at the center.
The solution is
g₀₀(r) = -1/4 ρ r² - C.
The Picard-Lindelöf theorem implies that it is the unique solution. The solution agrees with the newtonian gravity potential.
R₁₁
R₁₁ = -dΓ⁰₁₀ / dr - dΓ²₁₂ / dr - dΓ³₁₃ / dr
+ (Γ⁰₀₁ + Γ²₂₁ + Γ³₃₁) Γ¹₁₁
- (Γ⁰₁₀)² - (Γ²₁₂)² - (Γ³₁₃)²
= -1/2 g₀₀'' / g₀₀ + (1/2 g₀₀')² / (g₀₀)²
+ 1 / r²
- 1/2 g₃₃'' / g₃₃ + (1/2 g₃₃')² / (g₃₃)²
+ 1/2 g₁₁' / g₁₁ *
(1/2 g₀₀' / g₀₀ + 1 / r + 1/2 g₃₃' / g₃₃)
- (1/2 g₀₀' / g₀₀)²
- 1 / r²
- (1/2 g₃₃' / g₃₃)²
≈ 1/2 (g₀₀'' + g₁₁' / r - g₃₃'')
= 1/2 ρ.
R₂₂
R₂₂ = dΓ¹₂₂ / dr
+ (Γ⁰₀₁ + Γ¹₁₁ + Γ²₂₁ + Γ³₃₁) Γ¹₂₂
- 2 Γ²₂₁ Γ¹₂₂
= -1 / g₁₁ - r * -1 / (g₁₁)² * g₁₁'
- r / g₁₁ *
(1/2 g₀₀' / g₀₀ + 1/2 g₁₁' / g₁₁
- 1 / r + 1/2 g₃₃' / g₃₃)
= -1 / g₁₁ + 1 / g₁₁ + r g₁₁' / (g₁₁)²
- r / g₁₁ *
(1/2 g₀₀' / g₀₀ + 1/2 g₁₁' / g₁₁
+ 1/2 g₃₃' / g₃₃)
≈ 1/2 r (g₀₀' + g₁₁' - g₃₃')
= 1/2 r² ρ.
R₃₃
R₃₃ = dΓ¹₃₃ / dr
+ (Γ⁰₀₁ + Γ¹₁₁ + Γ²₂₁ + Γ³₃₁) Γ¹₃₃
- 2 Γ³₁₃ Γ¹₃₃
= -1/2 g₃₃'' / g₁₁ + 1/2 g₃₃' g₁₁' / (g₁₁)²
- 1/2 g₃₃' / g₁₁ *
(1/2 g₀₀' / g₀₀ + 1/2 g₁₁' / g₁₁
+ 1 / r - 1/2 g₃₃' / g₃₃)
≈ -1/2 g₃₃'' - 1/2 g₃₃' / r
= 1/2 ρ.
The solution is
g₃₃(r) = -1/4 ρ r² - C'.
The Picard-Lindelöf theorem implies that it is the unique solution.
The metric g₁₁(r)
The equations for R₁₁ and R₂₂ are:
1/2 (g₀₀'' + g₁₁' / r - g₃₃'') = 1/2 ρ,
1/2 r (g₀₀' + g₁₁' - g₃₃') = 1/2 r² ρ
<=>
1/2 (g₀₀' / r + g₁₁' / r - g₃₃' / r) = 1/2 ρ.
The equations are equivalent, since g₀₀'' = g₀₀' / r, and similarly, g₃₃'' = g₃₃' / r. We have
ρ = g₀₀'' + g₁₁' / r - g₃₃''
= -1/2 ρ + g₁₁' / r + 1/2 ρ
= g₁₁' / r
=>
g₁₁(r) = ρ r² / 2 + constant.
A cylinder where ρ(r) is not constant
Above we found a solution for a cylinder where the mass density is a constant ρ. But we had to use the results that g₀₀'' = g₀₀' / r, and similarly, g₃₃'' = g₃₃' / r, which are not true for a general cylindrically symmetric density distribution ρ(r).
Let us repeat the calculations in the general case.
The equations for Rii are:
2 R₀₀ = -g₀₀'' - g₀₀' / r
= ρ(r),
2 R₃₃ = -g₃₃'' - g₃₃' / r
= ρ(r),
2 R₁₁ = g₀₀'' + g₁₁' / r - g₃₃''
= ρ(r),
2 R₂₂ / r² = g₀₀' / r + g₁₁' / r - g₃₃' / r
= ρ(r).
If we sum the two last equations, sum the first and subtract the second, we obtain:
g₁₁' / r = ρ(r).
Radial pressure
Suppose that we put a tight membrane around the cylinder and squeeze it. There is a negative pressure inside the membrane. It has to be counteracted by a positive radial pressure in the cylinder.
If the counteracting pressure is solely radial, then it must have a formula
p / r,
where p is a constant. This in order to cancel the force F which squeezes the cylinder.
______
/ \
| O | <------ tight membrane
\_______/
^
| strong tube O at the center
The pressure would be infinite at r = 0. To prevent that, we drill a round hole into the cylinder at the center, and put a strong round tube there to counteract the radial force.
We assume that the metric g is almost exactly the Minkowski metric η with cylindrical coordinates. The stress-energy tensor is
T =
0 0 0 0
0 p / r 0 0
0 0 0 0
0 0 0 0,
Tr = p / r * 1 / g₁₁
≈ p / r.
We trace-reverse T to obtain the approximate Ricci tensor.
R ≈ T - 1/2 Tr η
=
1/2 p / r 0 0 0
0 1/2 p / r 0 0
0 0 -1/2 p r 0
0 0 0 -1/2 p / r.
The equations for Rii are:
2 R₀₀ = -g₀₀'' - g₀₀' / r
= p / r,
2 R₃₃ = -g₃₃'' - g₃₃' / r
= -p / r,
2 R₁₁ = g₀₀'' + g₁₁' / r - g₃₃''
= p / r,
2 R₂₂ / r² = g₀₀' / r + g₁₁' / r - g₃₃' / r
= -p / r.
Summing the two last equations, summing the first and subtracting the second yields:
p / r = g₁₁' / r.
Then
g₀₀'' - g₃₃'' = 0
and
g₀₀' / r - g₃₃' / r = -2 p / r.
Conclusions
If the calculations are correct, then we are only able to find an approximate metric for a uniformly dense cylinder.
UPDATE: They were not correct!
On November 5, 2023 we tentatively proved that a changing pressure breaks the Einstein field equations.
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