The key problem is that a rotationally symmetric metric g has the (normalized) Ricci curvature always the same in the r (coordinate 1) direction and in the φ (coordinate 2) direction. By normalization we mean that we divide the Ricci curvature in the φ direction by gφφ, in order to make the unit of length in φ "standardized".
We do a similar operation to the Ricci curvature to the r direction.
The normalization operation is the same as the index raising operation for Rii. The normalized curvature is Rⁱⁱ.
An example: the metric of the surface of a sphere is rotation symmetric, and the normalized Ricci curvature is the same to all directions on a sphere.
A very simple example where the Einstein equations fail: only have the dimensions r and φ and put a radial pressure
We use polar coordinates with the metric signature (+ +). The metric g is approximately the flat metric
ηij =
1 0
0 r².
That is, g is a perturbation of the flat metric. The metric g is rotationally symmetric:
gij =
g₁₁ 0
0 r².
The inverse metric (= the inverse matrix of gij) is
g^(ij) =
1 / g₁₁ 0
0 1 / r².
The stress-energy tensor for a radial pressure p is
Tij =
p 0
0 0.
The trace is
Tr = g¹¹ * p
= p / g₁₁.
Trace reverse T to obtain the Ricci curvature tensor:
Rij ≈ Tij - 1/2 Tr ηij
=
1/2 p 0
0 -1/2 p r².
The normalized Ricci curvature has a different sign to the r direction from the φ direction.
But our calculations on April 2, 2024 showed that the normalized Ricci curvature to the r direction is the same as to the φ direction, if the metric is rotation symmetric. We have a contradiction now, because in R the Ricci curvatures have a different sign.
Thus, there exists no metric which satisfies the Einstein field equations in this simple case.
We have to figure out why Albert Einstein put trace reversing into his equations. On the Internet, someone explained that it has to do with conservation laws.
The problem which we uncovered on November 5, 2023 is that the "pressure charge" in general relativity can change, which breaks field equations. The trace reversing problem is probably something different.
Christoffel symbols
Let us calculate once again the Christoffel symbols. This helps us to check our April 2, 2024 calculations.
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.
Γ¹₁₁ = 1/2 g¹¹ * dg₁₁ / dr
= 1/2 g₁₁' / g₁₁,
Γ¹₂₁ = 0,
Γ²₁₁ = 0,
Γ¹₂₂ = 1/2 g¹¹ * -dg₂₂ / dr
= -r / g₁₁,
Γ²₂₁ = 1/2 g²² * dg₂₂ / dr
= 1 / r,
Γ²₂₂ = 0.
R₁₁
R₁₁ = dΓ¹₁₁ / dr - dΓ¹₁₁ / dr
- dΓ²₁₂ / dr
+ (Γ¹₁₁ + Γ²₂₁) Γ¹₁₁
- Γ¹₁₁ Γ¹₁₁ - Γ²₁₂ Γ²₂₁
= 1 / r²
+ 1 / r * 1/2 g₁₁' / g₁₁
- 1 / r * 1 / r
= 1/2 * 1 / r * g₁₁' / g₁₁.
R₂₂
R₂₂ = dΓ¹₂₂ / dr
+ (Γ¹₁₁ + Γ²₂₁) Γ¹₂₂
- Γ²₂₁ Γ¹₂₂ - Γ¹₂₂ Γ²₁₂
= dΓ¹₂₂ / dr
+ (Γ¹₁₁ - Γ²₂₁) Γ¹₂₂
= -1 / g₁₁ + r g₁₁' / (g₁₁)²
+ 1/2 g₁₁' / g₁₁ * -r / g₁₁
- 1 / r * -r / g₁₁
= 1/2 r g₁₁' / (g₁₁)².
The normalized Ricci curvature is the same to the r and the φ directions
Let us check the normalized Ricci curvatures:
R¹¹ = R₁₁ / g₁₁ = 1/2 * 1 / r * g₁₁' / (g₁₁)²,
R²² = R₂₂ / g₂₂ = 1/2 * 1 / r * g₁₁' / (g₁₁)².
The normalized curvatures are equal to the r and φ directions.
An example is the surface of a sphere. We can express its metric with a certain choice of g₁₁. The normalized curvature is the same to both directions, as we would expect.
We realize that this property is not unique for the surface of a sphere, but holds for all rotationally symmetric orthogonal metrics. For example, it is not possible that such a metric has a positive curvature to one direction and a negative curvature to another direction.
Only the zero stress-energy tensor has solutions – and every metric g₁₁ is a solution for it
What kind of stress-energy tensors can an arbitrary g₁₁ satisfy? Let us now calculate exactly. We do not assume that the metric is close to flat.
The Ricci scalar R is
R = Tr = 1 / g₁₁ * R₁₁ + 1 / g₂₂ * R₂₂
= 1/2 * 1 / r * g₁₁' / (g₁₁)²
+ 1/2 * 1 / r * g₁₁' / (g₁₁)²
= 1 / r * g₁₁' / (g₁₁)².
The Einstein tensor is
Rij - 1/2 R * gij = Tij
=
1/2 * 1 / r * g₁₁' / g₁₁ 0
- 1/2 1 / r * g₁₁' / (g₁₁)² * g₁₁
1/2 r g₁₁' / (g₁₁)²
0 - 1/2 1 / r * g₁₁' / (g₁₁)² * r²
=
0 0
0 0.
We obtain a strange result. Every metric g₁₁ is a solution for the zero stress-energy tensor Tij. No other stress-energy tensor has a solution.
Conclusions
We proved that a rotationally symmetric spatial metric in the plane has the same standardized Ricci curvature to the r and φ directions.
The Einstein equations do not produce meaningful results in this simple case of 0 + 2 dimensions.
We do not yet have any insight if the Einstein equations could be repaired by modifying the trace reversing operation.
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