UPDATE March 31, 2024: After adding the missing 1 / g₁₁ factor, the sign is correct. Wikipedia was right.
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UPDATE March 31, 2024: We added a missing 1 / g₁₁ factor to Γ¹₁₁ and Γ¹₂₂ in the calculation about the sphere. The sign error persists, but after correcting it, we get the right curvature 2 / R² for a sphere.
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The Riemann tensor defines curvature through a "parallel transport" of a vector around a loop.
(Picture by Hellerick
https://commons.m.wikimedia.org/wiki/File:Division_of_the_Earth_into_Gauss-Krueger_zones_-_Globe.svg )
Let a man walk from the equator in Africa to the North Pole, back to the equator in Asia, and back to the starting point in Africa.
The man is carrying a vector which initially points north. He tries to keep the direction of the vector "same", as he makes turns. The man will end up in Africa with the vector pointing east. The vector points to a different direction now. If the man would walk on a flat surface, the vector would still point to the original direction.
Vector transport
In the link there is a derivation of the formula:
Let the coordinate μ above be the angle φ in polar coordinates (coordinate 2) and ν be the radial coordinate r (coordinate 1).
The term Γ¹₁₁ Γ¹₂₂ above comes from the parallel transport of the vector Aλ in the diagram first up along the line 3 and then to the right along 4.
Since it is polar coordinates, the basis vectors eφ and er turn counterclockwise if we walk up along the line 3 in the diagram. The basis vector eφ turns to the negative direction of the r coordinate. By "turning" we mean that in the "natural coordinates" determined by proper distances, the basis vector turns. This explains why
Γ¹₂₂ = -r
is negative.
The sign error in the formula is at this place. The basis vector turns to the negative r direction, but the vector Aλ which we are transporting turns to the positive direction of r relative to the basis vectors. We really are interested in the behavior of the vector we are transporting, not in the basis vectors. The sign should be positive.
Suppose that the basis vectors turned some small positive angle α counterclockwise as we walked up. The vector which we are transporting gained a small positive r component,
dr = sin(α) |Aλ|.
dr dr
---> =======> ------->
transport
along 4
After walking up in the diagram along 3, we walk to the r direction along 4. The component dr "grows" because the metric of r is stretched as r becomes larger:
dgrr / dr > 0.
This explains why we multiply by Γ¹₁₁, which tells us how fast the metric of r grows.
If we transport the vector first along the line 1 and then along the line 2, there is a similar turn α counterclockwise, but no "growth" of the component dr. We conclude that the route 3, 4 turned the transported vector clockwise relative to the route 1, 2. This indicates positive curvature.
Thus, there is a sign error in the Riemann tensor formula in the case of
Γⁱii Γⁱjj,
but is the sign correct in other cases?
A correct sign case
The term
Γ²₂₁ Γ¹₁₁
in our March 24, 2024 blog post had a correct sign. Let us analyze it. The Christoffel symbol
Γ²₂₁
means that we move to the direction of the coordinate 1, or r, and watch how much the basis vector eφ grows to the direction of the φ coordinate.
There is no sign error here because we do not measure the turning of the basis vector eφ, only the growth of its length.
We obtain a tentative sign rule for the cross terms Γ * Γ in the Riemann or Ricci tensor.
Sign rule for the Γ * Γ terms. If in
Γⁱjk
we have i ≠ j, then the sign has to be flipped.
Covariant derivative
The sign error comes from the covariant derivative. Is the sign convention in the covariant derivative logical?
The covariant derivative ∇ⱼ u of a vector field u with respect to the coordinate j is defined as the usual derivative plus correction terms from the Christoffel symbols, in the case where the coordinates are not cartesian. The typical example is polar coordinates.
Is the Schwarzschild metric correct?
The sign error in the definition of Ricci curvature might mean that the familiar Schwarzschild metric from 1916 is incorrect. We have to check it.
We looked at one derivation of the Schwarzschild metric on the Internet, and there the sign of
Γ¹₁₁ Γ¹₂₂
in the definition of Riemann curvature was negative, that is, the correct one. Thus, the derivation most probably is correct. Any terms which would have the wrong sign, are zero.
The original Karl Schwarzschild (1916) paper, translated to English, is here:
Schwarzschild has the sign flipped in the definition of the Christoffel symbol.
Riemann curvature formula in old papers
Die Grundlage der allgemeinen Relativitätstheorie by Albert Einstein (1916) contains the following definition for the Riemann(–Christoffel) tensor:
Albert Einstein uses the metric signature
(- - - +),
where the last coordinate is time. We see that the Wikipedia factor g^kl, where the indices are raised, is missing from the definition of the Christoffel symbol by Einstein.
Gregorio Ricci-Curbastro and Tullio Levi-Civita in their 1900 paper attribute the Riemann tensor formula to Bernhard Riemann, from his 1861 paper Commentatio mathematica.
The Latin language Commentatio can be found in the link above.
The Wikipedia scalar curvature formula for the surface of a sphere
On March 24, 2024 we calculated for the surface of a sphere (or any radially symmetric metric) in polar coordinates that
R₁₁ = 1/2 g₁₁' / g₁₁ * 1 / r,
R₂₂ = 1/2 g₁₁' / (g₁₁)² * r,
where we have corrected the sign error of Wikipedia for R₂₂.
Wikipedia defines scalar curvature by
f(r)
^
|
R | ----___
| \
| |
-----------------------> r
R
The formula for a spherical surface of a radius R is:
R² = f(r)² + r²,
f(r) = sqrt(R² - r²),
f'(r) = 1/2 * 1 / sqrt(R² - r²) * -2 r
= -r / sqrt(R² - r²),
f'(r)² = r² / (R² - r²),
g₁₁(r) = 1 + f'(r)²
= 1 + r² / (R² - r²)
= R² / (R² - r²),
g₁₁'(r) = R² / (R² - r²)² * 2 r.
We obtain
R₁₁ = 1/2 g₁₁' / g₁₁ * 1 / r = 1 / (R² - r²),
R₂₂ = 1/2 g₁₁' / (g₁₁)² * r = 1 / R² * r².
Recall that g₂₂(r) = r². Then
g¹¹ R₁₁ = 1 / R²,
g²² R₂₂ = 1 / R²,
Scal = 2 / R².
The result is correct. The scalar curvature should be a constant 2 / R².
Analysis
The value of a tensor cannot change arbitrarily between different coordinate systems. If we have a static configuration, the conversion is trivial: for example, the mass density is easy to convert from kilograms per a cubic meter to kilograms per a cubic feet.
It might be that the Wikipedia formulae work correctly in spherical coordinates for spherically symmetric systems. Then the Schwarzschild solution is correct. That is, the Ricci curvature truly is zero outside the spherical mass. We calculated on March 11, 2024 that the Ricci curvature is zero in cartesian coordinates for a lightweight spherical mass. The cross terms Γ * Γ can be ignored in such a case.
If the system is not spherically symmetric, then the formulae may yield incorrect results. For example, the numerical simulations made by the LIGO team about merging black holes are highly suspect.
Our calculations in the March 20, 2024 blog post used cartesian coordinates and weak fields. The calculations may get Ricci curvature right. If those calculations are correct, then general relativity is fatally flawed. The geometric interpretation of gravity does not work, except in the special, symmetric case of the Schwarzschild metric. We have to abandon the notion of Ricci curvature in gravity.
Conclusions
The formula for the Riemann curvature tensor in literature seems to contain a sign error. The error has persisted ever since Bernhard Riemann introduced the formula in 1861.
^ vector attached to belt,
| initially pointing
| to north
|
o
| ---->
/\ man starts here
-------------------
| |
| | pseudo-square
| |
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We will write a new blog post where we try to determine the correct formula. We will start from the definition of the vector transport. In the case of a sphere, a man walks along great circles ("straight" geodesic lines) with a vector attached to his belt.
When the man turns at corners, he adjusts the vector accordingly, so that if the man would walk on a straight plane, the vector would always point to the same direction.
(Picture Wikipedia)
If the man in the diagram walks on a sphere, the angles of the pseudo-square at the corners are slightly larger than 90 degrees, let us say, 91 degrees. At each corner, the man only turns 89 degrees clockwise. He adjusts the vector 89 degrees counterclockwise relative to his body.
After the man has completed the loop, his body has turned 360 degrees clockwise. He has adjusted the vector 356 degrees counterclockwise relative to his body. The end result is that the vector turned 4 degrees clockwise! The clockwise rotation of the vector proves that there is positive curvature at the location.
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