Riemann curvature tensor formula is correct

UPDATE March 31, 2024: There is no sign error. It was our own calculation error.

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In the previous blog post we described the vector transport procedure which defines the Riemann curvature tensor. For simplicity, let us assume that there are only two coordinates, x and y, or 1 and 2.

 (x, y + dy)                   (x + dx, y + dy)

              --------------------

             |                      |

             |                      |  

             |                      |

              --------------------

 (x, y)                            (x + dx, y)

Above, we have drawn a small rectangle according to the coordinates.

(Ruler and protractor photo pxhere.com)

The metric g determines the "proper" distances between coordinate points. By proper we mean things that someone living on the surface measures with a ruler and a protractor. If we measure proper distances around (x, y), we can draw cartesian coordinates X and Y, for which the metric G is almost the euclidian metric:

            1     0

            0     1.

The above rectangle in the coordinate plane X, Y looks something like this:

             _______----______

             \                          / 

               \                      /

                 \                  /

                    -----___----

The rectangle is distorted and its sides are not straight lines.

This Wikipedia picture shows what happens to a longitude, latitude coordinate rectangle when we draw it in "more proper" coordinates X and Y. The lines are bulging outward.

In the definition of Riemann curvature, the man does not walk along coordinate lines, but along geodesics which are the "straightest" possible paths in the metric. Geodesics minimize the distance between the endpoints of the line. For a sphere, the geodesics are the great circles.

We measure the proper angles at each corner of the path. If the angles add up to 360 degrees, it is a flat plane. If the angles add up to > 360 degrees, the curvature is positive. Why proper angles? Because that is the angle which the man sees at the corner. He is not interested in the coordinates which we happened to choose. The man in interested in things which he can measure.

We have to find the formula which determines the proper angles at the corners. That will be the correct Riemann tensor formula.

Geodesics and coordinate lines

        ^ y

        |                     /    f(x) geodesic

        |                  /

        |            /

              •---------------------->  x coordinate line

Let us have a geodesic f which starts from a point P initially to the direction of the coordinate line x. The geodesic f diverts from the coordinate line.

The geodesic f differs from the coordinate line in two ways:

1.   fy(x) tells us how much the geodesic has diverted to the y direction from the x coordinate line;

2.   fx(x) tells us how much the proper distance measured along the geodesic differs from the proper distance measured along the x coordinate line.

The geodesic minimizes the proper distance between any two of its points. The variation where we slightly displace a part of the geodesic must always increase the proper distance measured along the geodesic.

Maybe we get a unique formula for f from the variational constraint?

https://en.wikipedia.org/wiki/Christoffel_symbols

If x and y are orthogonal in the metric g, then our diagram is concerned with

       Γ²₁₁  =  1/2  *  1 / g₂₂  *  -∂ g₁₁ / dy

and

       Γ¹₁₁  =  1/2  *  1 / g₁₁  *  ∂ g₁₁ / dx.

Does the Christoffel symbol reveal us the correct geodesic? People say that the above formula tells how much the basis vector of k turns or grows to the i direction if we move a short distance to the direction of l.

The Christoffel symbols consist of first derivatives. They cannot describe the gradual turning up of the geodesic f in the diagram. But they can describe the change in the metric g₁₁ as we walk along the x coordinate line. In the case of polar coordinates on a plane, Γ²₁₁ =  0 and Γ¹₁₁ = 0, which are correct descriptions of what would happen in the diagram in that case.

The formula of the covariant derivative

https://en.wikipedia.org/wiki/Covariant_derivative

The function x maps the cartesian coordinate plane ℝ² to a curved surface. The basis vectors are defined:

Let

be a mapping from ℝ² to vectors of the tangent plane at the image of the point in ℝ². Let

be another mapping. The covariant derivative of u along v is defined:

When calculating the "difference" of u at a location P and at a slightly displaced location Q, we need a way to transport the old value of u in the tangent plane of P to the tangent plane of Q. The Levi-Civita connection is supposed to do this transport in a "natural" way.

This reminds us of our vector transport problem. How to transport a vector in a "natural" way from P to Q?

The Levi-Civita connection

https://en.wikipedia.org/wiki/Affine_connection

We say that a vector field X is a parallel transport of the vector ξ along the curve γ, iff

The connection ∇ above measures how much the vector X "changes" when we walk along the curve γ. The vector X is not allowed to change at all.

https://en.wikipedia.org/wiki/Levi-Civita_connection

The connection which we use is the Levi-Civita connection. In the coordinate form,

       ∇X Y

is defined as:

Let X be the tangent vectors of a path γ. If the vector field Y describes the parallel transport of a vector along the path, then ∇X Y should be zero everywhere on the path. The numerical value is calculated from the metric using the Christoffel symbol formula. According to Wikipedia, the definition is essentially due to Elwin Bruno Christoffel in the year 1869.

A method to determine the rotation of a transported vector

Now we realize that there might be a relatively simple way to determine the amount of rotation in a loop.

 orthogonal coordinate lines

           

                ---/--/--|--\--\---        coordinate line

If orthogonal coordinate lines "fan out" like in the diagram above, then the coordinate line is not a geodesic. A transported vector will turn according to how much the lines fan out?

The calculation may be complicated by the fact that the coordinate line may be curvy to many directions?

                     ^

                   /

                 /

                 -------->

Also, the coordinates may be skewed.

Transport of a radial vector on a sphere along a coordinate line loop

We use on the spherical surface the metric and the r and φ coordinates of the previous blog post.

Let Y be a vector which everywhere points along the r axis, to the origin. The length of Y is constant in the r coordinate, but not in the proper length.

We choose the path γ as a small loop along coordinate lines of r and φ. On Earth that would mean a coordinate rectangle along meridians and latitude lines. The vector Y points everywhere along the meridian lines to the north.

(Picture by Hellerick

https://commons.m.wikimedia.org/wiki/File:Division_of_the_Earth_into_Gauss-Krueger_zones_-_Globe.svg )

 

                          walk around

                                  ---->

                          meridian line

                             -------------

                            |              |   latitude line

                            |              | 

                             -------------   γ path (= X)

                                 -----> Y

  ^  φ  (coordinate 2)            

  |                            

   ------> r (coordinate 1)

The vector Y is not a parallel transported around γ, because the latitude coordinate lines are not great circles. The meridians are great circles, and along them, Y is parallel transported.

Using the idea of the preceding section, the meridian coordinate lines "fan out" when we walk along a latitude line. The fanning is stronger close to the North Pole than close to the equator. 

In the formula

the derivative ∂j is zero because Y is constant. The terms with the Christoffel symbols Γ should tell us how much Y differs from a parallel transport. The difference tells us how much a parallel transported vector would have turned relative to Y after completing the loop. The X in the formula is the tangents of the path: γ'(t).

The r coordinate lines (latitude lines) in the rectangle are not great circles. The formula should show that the constant vector Y turns relative to a parallel transport.

The Christoffel symbol terms should recognize that we are not walking around a great circle, and calculate how much we divert from it.

The proper angle at each corner of the rectangle is 90 degrees. If the latitude lines were geodesics ("straight lines"), we would obtain a result that the rectangle is in a flat geometry: the sum of corner proper angles is 360 degrees. But it is on a surface of a sphere. The latitude lines are closer to the center of the rectangle than the corresponding great circle lines would be. That is, the latitude lines "curve inward".

Let us then analyze what the connection formula calculates when we walk along a latitude line.

1.   X² differs from zero and X¹ is zero.

2.   Y¹ differs from zero and Y² is zero.

3.   In the formula, j = 2 and k = 1.

4.   What is Γ¹₂₁ ∂₁ ? The coordinate vector ∂₁ points to the r direction (along a meridian). We have

       Γ¹₂₁  =  1/2 * 1 / g₁₁ * ∂₂ g₁₁  =  0.

5.   What is Γ²₂₂ ∂₂ ?

       Γ²₂₂  =  1/2 * 1 / g₂₂ * ∂₂ g₂₂  =  0.

The connection should be zero when we walk along r (meridian). But why it is zero along φ (latitude)?

Conclusions

The sign error was our own calculation error. Wikipedia was right.

Sign is not flipped in the Wikipedia Ricci tensor formula

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 Ricci tensor formula in Wikipedia comes from a similar formula for the Riemann curvature tensor. Let us study how the Riemann formula is derived.

https://en.wikipedia.org/wiki/Riemann_curvature_tensor

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

https://profoundphysics.com/the-ricci-tensor/

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?

https://en.wikipedia.org/wiki/Covariant_derivative

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:

https://arxiv.org/abs/physics/9905030

Schwarzschild has the sign flipped in the definition of the Christoffel symbol.

Riemann curvature formula in old papers

https://einsteinpapers.press.princeton.edu/vol6-doc/343

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.

https://books.google.fi/books/about/Ricci_and_Levi_Civita_s_Tensor_Analysis.html

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.

https://www.emis.de/classics/Riemann/

The Latin language Commentatio can be found in the link above.

The Wikipedia scalar curvature formula for the surface of a sphere

https://en.wikipedia.org/wiki/Scalar_curvature

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

                |                    |

                 -------------------

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.

Ricci curvature formula in Wikipedia is correct!

UPDATE March 31, 2024: After correcting the 1 / g₁₁ factor there no longer is a sign error!

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UPDATE March 31, 2024: We added a missing 1 / g₁₁ factor to Γ¹₁₁ and Γ¹₂₂.

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UPDATE March 26, 2024: The error seems to be in the sign of the cross term Γ¹₁₁ Γ¹₂₂ in Wikipedia. The Wikipedia formula is present already in Albert Einstein's 1915 papers. Is the sign convention different there?

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Let us analyze why the Schwarzschild internal solution gives an appropriate metric for a uniform ball, but we cannot find any metric for a uniform cylinder. We ignore pressure in this blog post.

(Pictures: Wikipedia)

An obvious difference between a ball and a cylinder is in the z coordinate. It does not affect anything for a cylinder, but has a prominent effect in the case of a ball.

In our previous blog posts we have used cartesian coordinates. Let us now use polar or cylindrical coordinates, in order to check the calculations that we have made in the past two weeks.

Stretched radial metric in a 2D plane

We treat the radius r as the first coordinate and the angle φ as the second coordinate. The metric in polar coordinates is

       ds²  =  g₁₁(r) dr²  +  r² dφ²,

that is, g₂₂(r) = r². We assume that g₁₁(r) is very close to 1.

We denote the derivative with respect to r by the prime '. The cross terms Γ * Γ are now important because g₂₂' is very large. In cartesian coordinates, the derivatives of each gij would be very small for weak fields.

https://en.wikipedia.org/wiki/Ricci_curvature

       Γ¹₁₁  =  1/2 * 1 / g₁₁ * g₁₁',

       Γ²₁₁  =  1/2 * 1 / r² * -dg₁₁ / dφ

                =  0,

       Γ¹₂₁  =  Γ¹₁₂  =  0,

       Γ²₁₂  =  Γ²₂₁  

                =  1/2 * 1 / r² * g₂₂'

                =  1 / r, 

       Γ¹₂₂  =  -1/2 * 1 / g₁₁ * g₂₂'

                =  -r / g₁₁,

       Γ²₂₂  =  0,

       R₁₁   =  dΓ²₁₁ / dφ  -  dΓ²₁₂ / dr

                   + Γ¹₁₁ Γ¹₁₁ +  Γ²₂₁ Γ¹₁₁

                   - Γ¹₁₁ Γ¹₁₁  -  Γ²₁₂ Γ²₂₁

               =  0  +  1 / r²  

 

                   + 1 / r * 1 / 2 * g₁₁' / g₁₁

                   - 1 / r²

               =  1/2 g₁₁' / g₁₁ * 1 / r.

The value agrees with our March 20, 2024 blog post.

       R₂₂  =  dΓ¹₂₂ / dr

                  + Γ¹₁₁ Γ¹₂₂  +  Γ²₂₁ Γ¹₂₂ 

                  - Γ²₂₁ Γ¹₂₂  -  Γ¹₂₂ Γ²₂₁

               =  -1 / g₁₁ + r / (g₁₁)² * g₁₁'

                   + 1/2 g₁₁' / g₁₁ * -r / g₁₁ 

                   + r / g₁₁ * 1 / r

               =  1/2 g₁₁' / (g₁₁)² * r.

The sign is not flipped relative to our March 20, 2024 post.

The formulae for the Christoffel symbols and the Ricci curvatures are incorrect above?

In the above calculations, if we replace the angle φ with a new variable θ:

       φ  =  C θ,

then the constant C is absorbed into the value of the metric g₂₂(r): g₂₂ gets larger by a factor C².

In the above calculations, Γ¹₂₂ and R₂₂ grow by the factor C². The value of R₂₂ is coordinate-dependent!

What is going on? Are the formulae incorrect?

https://en.wikipedia.org/wiki/Christoffel_symbols

Our earlier Ricci curvature calculations in the past two weeks concerned perturbations of the metric: raising of indexes might not be too relevant.

Let us analyze.

The crucial term above in R₂₂ is

        Γ¹₁₁ Γ¹₂₂,

especially

         Γ¹₂₂ =  1/2 g¹¹ * -∂₁ g₂₂

                 = -1/2 * 1 / g₁₁ *  g₂₂'.

The derivative is with respect to r, whose metric g₁₁ is close to 1. Raising the index on the infitesimal dx¹ only changes the value a little. Thus, the problem is not about raising indices in the derivatives.

The problem is that changing the variable from φ to θ inflates g₂₂ by a factor C². This inflation is nowhere compensated for.

We should "normalize" g₂₂ by dividing:

        g₂₂(x) / g₂₂(x₀)

if we want to study Ricci curvature at a point x₀? This may correspond to the basis vector

        ei 

in the formula above? The metric is defined:

        g₂₂  =  e₂ • e₂.

Suppose that we do the change of the coordinate variable φ = C θ. Then the basis vector e₂ becomes C times longer. Thus, the basis vector is not a unit vector in terms of the proper length, as measured by the metric. Rather, the basis vector depends on the coordinates which we choose.

This is strange. The natural way to define basis vectors would be to make their proper length 1.

Christoffel symbols are not tensors. Their value depends on the choice of the coordinates.

The Wikipedia article cites Bishop and Goldberg (1968): Tensor analysis on manifolds when it gives the formula:

Let us check the proofs that the Ricci tensor is coordinate-independent, even though Christoffel symbols are heavily coordinate-dependent.

Solution of the mystery. Tensors are not coordinate-independent! They transform with simple rules into new coordinate systems. The stress-energy tensor T in the Einstein field equations contains, e.g., the mass density relative to the coordinates, not relative to proper distances.

In cylindrical coordinates, a uniform cylinder does not have a uniform density in those coordinates! We should not have R₁₁ = R₂₂ when using polar coordinates, even though the equality holds in cartesian coordinates.

Set the metric of the plane to the surface of a sphere: this conclusively proves that the result above is wrong

The metric is

       ds²  =  g₁₁(r) dr² + r² dφ².

Let us set g₁₁(r) such that this describes the "polar cap" of a spherical surface:

      g₁₁(r)  =  1 / cos( arc sin(r / R) ),

where R is the radius of the sphere. As r grows, so does arc sin(r / R), and the cosine decreases. Thus, g₁₁'(r) > 0, if r > 0.

The Ricci curvature to every direction is positive on a spherical surface. But the calculation above claims that

       R₂₂  =  -1/2 g₁₁' / (g₁₁)²  *  r 

              <  0 !

Analysis of R₁₁ and R₂₂ on a spherical surface

Let us analyze how the formulae "detect" that the metric g is that of a polar cap. The detection should lead them to compute the curvature to be positive to every direction.

(Picture by Hellerick

https://commons.m.wikimedia.org/wiki/File:Division_of_the_Earth_into_Gauss-Krueger_zones_-_Globe.svg )

The polar cap is projected into a plane. Tangential distances around the origin are just like for a plane:

       r dφ.

The meridians in the picture of the globe are the coordinate lines of φ. The latitude lines are some coordinate lines of r, but note that in the projection in the plane, the lines are not evenly spaced. They are evenly spaced in the proper distance, but not in the coordinate distance. The variable r is the coordinate in the plane.

The formulae cannot tell apart the polar cap from the plane looking at g₂₂(r) only. The difference is in g₁₁(r). The radial metric is stretched ever more when we move away from the origin:

       dg₁₁(r) / dr  >  0.

By looking at this derivative should the formulae realize that we are on a sphere.

Let us now analyze the formulae in detail.

1.   Γ¹₁₁ = 1/2 * g₁₁' / g₁₁.

This symbol is the crucial one. It tells us how much the basis vector e to the positive direction of r becomes longer when we move along r to the positive direction. For a polar cap, this is positive. The surface of the sphere becomes ever steeper as we walk away from the North Pole.

2.   Γ²₁₂  =  Γ²₂₁  =  1/2 * 1 / r² * g₂₂'  =  1 / r.

This symbol is only about polar coordinates, not about the spherical surface. It measures the "speed" at which meridian lines move away from each other as r grows. The "speed" is a relative speed: how many percents per a unit of r. The meridian lines projected to the plane are the usual coordinate lines of φ in polar coordinates.

3.   Γ¹₂₂ = -1/2 * 1 / g₁₁ * g₂₂'  =   -r.

This symbol measures the "speed" in a different way: the sign is flipped and there is no "scaling" factor 1 / r². This is again only about polar coordinates, not about the spherical surface at all.

4.   R₁₁ =  dΓ²₁₁ / dφ  -  dΓ²₁₂ / dr

                   + Γ¹₁₁ Γ¹₁₁  +  Γ²₂₁ Γ¹₁₁

                   - Γ¹₁₁ Γ¹₁₁   -  Γ²₁₂ Γ²₂₁.

The symbols which are only associated with polar coordinates must yield a zero sum: a plane has zero curvature. The crucial non-zero term is

       Γ²₂₁ Γ¹₁₁  =  1 / r * 1 / 2 * g₁₁' / g₁₁.

Qualitatively, the term says that the coordinate lines associated with the variable 2, or φ, become more distant from each other as r grows, and that the radial metric is stretched as r grows.

              \        /   coordinate lines of φ

                \    /    

                  \/

            North Pole

stretch r when r is large  =>

            |            |

             |          |

               \     /    

                  \/

            North Pole

We see that the lines start to bend toward each other. This is just what happens when two people walk down from the North Pole along meridians. At the equator, they will walk in parallel.

We see that stretching the metric of r far away from the origin adds positive Ricci curvature. The formula for R₁₁ makes sense.

5.   R₂₂  =  dΓ¹₂₂ / dr

                  + Γ¹₁₁ Γ¹₂₂  +  Γ²₂₁ Γ¹₂₂ 

                  - Γ²₂₁ Γ¹₂₂    -  Γ¹₂₂ Γ²₂₁.

The only term associated with the spherical surface is

       Γ¹₁₁ Γ¹₂₂  = 1/2 g₁₁' / (g₁₁)²  *  -r.

The term is not logical. The symbol Γ¹₂₂ tells us that the meridian lines become more distant when r grows, and Γ¹₁₁ says that the metric of r is stretched when r becomes larger. The term should add positive curvature. But because of the minus sign in front of r, it adds negative curvature!

Why is there a minus sign in Γ¹₂₂?

Recall that Γ¹₂₂ measures the "speed" how fast coordinate lines of φ distance themselves from each other.

The term

       ∂₁ Γ¹₂₂

in Rij checks if the coordinate lines bend toward each other as r grows. If the speed slows down, that means positive curvature, as we explained above. The minus sign is required to flag this as positive curvature.

Conclusions

The formula for Ricci curvature in Wikipedia seems to be in error. The same formula is found in many sources. Maybe sign conventions save it there?

The problem is that in Rij, the derivative requires a different sign for Γ¹₂₂ than the cross term Γ¹₁₁ Γ¹₂₂ requires.

We are not sure about how to correct the formula.

Our own method of calculating perturbations of metric in cartesian coordinates is immune to the error, because we can ignore the cross terms Γ * Γ.

We will check if other methods of calculating the Christoffel symbols and Ricci curvature are correct. People on the Internet have calculated Ricci curvature for a polar cap using various methods – and did get a positive curvature to all directions. Thus, these various methods probably work correctly.

The metric inside a cylinder

UPDATE April 2, 2024: We forgot the factor g²¹ in the formulae. Since the coordinates are only slightly skewed, that factor makes Γ²₁₁ enormous, spoiling our attempt to calculate perturbations only.

After fixing a sign error, we no longer have a contradiction. However, we have to check carefully both the pressure case and uniform mass case.

----

UPDATE April 1, 2024: From Gauss's law we get the newtonian gravity force inside a cylinder. The mass within a radius r from the center is

       ~ r²,

and the area of the cylinder at the radius r is

       ~ r.

The density of lines of force at r is

       ~ r² / r  = r.

The force is

       F ~ r,

and the potential

       V ~ r².

Our formula

       g₀₀(x) = -p / 4 * x² - C

agrees with the newtonian potential.

We also added a note about Picard-Lindelöf and the uniqueness of g₀₀.

----

UPDATE March 31, 2024: We checked that factors g¹¹, etc. in the Christoffel symbols have a negligible effect in this calculation, as g is almost Minkowski, and only perturbed slightly.

----

Let us try to figure out what kinds of metrics there could exist inside pressurized matter. Karl Schwarzschild (1916) was able to find such a metric for incompressible fluid, which is pressurized by its own gravity.

We use the metric signature (- + + +).

Let us first consider an unphysical configuration where a ball of weightless matter has uniform pressure throughout.

https://en.wikipedia.org/wiki/Einstein_field_equations

The stress-energy tensor T within the ball looks something like

       T  =

               0          0          0          0

               0          p          0          0

               0          0          p          0

               0          0          0          p,

where p is the pressure. For now, we assume that the metric g is very close to the Minkowski metric η.

We can obtain the associated Ricci tensor R by "trace reversing":

       Tr  =  g⁰⁰ T₀₀ + g¹¹ T₁₁ + g²² T₂₂ + g³³ T₃₃

             =  -T₀₀ + T₁₁ + T₂₂ + T₃₃

             =  3 p.

Then the Ricci tensor

       R  =   T  -  1/2 Tr * g

            =   T  -  3/2 p  * g

            = 

                3/2 p     0            0             0

                0           -p / 2      0             0

                0            0           -p / 2       0

                0            0            0      -p / 2.

                •      •      •    static test masses

                •      •      •

                •      •      •

          => time passes

                  •    •    •

                  •    •    •

                  •    •    •

The Ricci curvature to the time direction, R₀₀ is positive. This means that if we let an initially static cubic constellation of test masses fall freely, the volume of the cube starts to shrink. This is just like with a positive mass density.

Let us then calculate the Ricci tensor for a positive mass density ρ and a zero pressure.

       TM  =  

                   ρ    0    0    0

                   0    0    0    0

                   0    0    0    0

                   0    0    0    0,

       RM =

                   ρ / 2   0        0          0

                   0        ρ / 2   0          0

                   0        0        ρ / 2     0

                   0        0        0     ρ / 2.

The Ricci curvature is positive to the time direction as well as to all spatial directions.

https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric

In the picture we have the Schwarzschild spatial metric for a plane passing through the center of the ball. The spatial metric in the interior Schwarzschild solution is stretched in the radial direction.

But for the pure pressure case, the Ricci curvature is negative to all spatial directions. The metric of time might be roughly the same in both cases. In the pressure case, we have to add "defocusing" to every direction. We have to shrink the radial metric inside the cap, making the spatial geometry "saddle-shaped"?

Then we face a dilemma. Outside the central ball, R₀₀ has to be zero. That requires that the gravity potential is ~ 1 / r, where r is the distance from the center.

Thus, the metric of time must be like in the Schwarzschild solution in vacuum. To make the Ricci tensor R there zero, also the spatial metric must be like in the Schwarzschild exterior solution. That is, the radial metric is stretched. We cannot fit the radial metric smoothly at the surface of the ball because the metric is shrunk inside and stretched outside.

There may be no (static) solution for pure pressure inside a ball.

Iterative solution of the Einstein equations does not succeed because the residual stress-energy tensor T contains pure pressure?

Suppose that we try to use an iterative method to solve the Einstein field equations. We have found an approximate metric g. We calculate the residual T in vacuum from the formula:

If T would be zero, then g would be the solution.

Let us assume that the residual T is small. If we were able to find a small metric perturbation h, such that

       η + h

on the left side of the equation would reproduce the residual T on the right side, then

        g - h

might be a better approximate solution, assuming that the equation is roughly linear.

For example, if T contains simply "mass" density in its T₀₀ component, we often can find a very good metric h: something like the Schwarzschild solution.

But if the residual T contains pure pressure, and there is no approximate solution of the equations for pure pressure, then the iteration stops. We cannot continue.

The problem may be the "rigidity" of the requirement that Ricci curvature is zero in vacuum. It is like the paper bending exercise of our March 8, 2024 blog post. The rigidity of paper severely restricts the shapes it can take. Locally, the shape has to be a cone or a cylinder.

If we could find a good approximate metric for any small T, then we would probably have a very efficient iterative solution procedure for the Einstein equations. But no one has found such a method in 109 years. Is it so that in most cases we cannot find any metric at all for a small residual T? Even if T would mimic realistic matter?

Pressure in a cylinder to the z direction

               negative pressure

               =========

              |                  |

               =========    positive pressure

              |                  |

               =========

               negative pressure

        ------> z

       |

       v   x

     y points out

     of the screen                  cylinder

      ● ------> z                 ============== 

     |                              positive pressure p

     v   x                                     <--->

We can build static structures where a cylinder has essentially a constant pressure in the z direction and no pressure in other directions. In the diagram we have three cylinders. The outer cylinders want to contract and have a negative pressure. They put a positive pressure into the central cylinder.

We assume that the mass of the cylinder is zero and the pressure p small.

Can we find an approximate metric for such a device?

Let us assume that the cylinder is relatively long. We try to find a cylindrically symmetric metric. We have a point (x, y, z) inside the cylinder, where |y| << |x|.

The stress-energy tensor T looks like this:

       T  = 

                0    0    0    0

                0    0    0    0

                0    0    0    0

                0    0    0    p

and the Ricci tensor:

       R  =

                p / 2    0         0            0

                0        -p / 2    0            0

                 

                0         0        -p / 2       0

                0         0         0       p / 2

We can make use of our March 11, 2024 calculations of the Schwarzschild metric.

We assume that r points approximately to the x direction. The metric is

       ds²  =  g₀₀ dt²  +  grr dr²  +  dn²  +  g₃₃ dz²,

where r is the radial vector normal to the z axis, and n is the normal vector to r in the x,y plane.

We assume that r is scaled in such a way, that the metric on n is 1.

In cylindrical coordinates:

       dr   =  (1 - 1/2 * y² / x²) dx

     

                  + y / x * dy,

       dn  =  (1 - 1/2 * y² / x²) dy

                  - y / x * dx,

We drop very small terms and write the spatial metric in terms of cartesian coordinates:

        ds²  =  grr * dx²

                    + (1 - grr) * y² / x² * dx²

                    + 2 (grr - 1) * y / x * dx dy

                    + dy²

                    + (grr - 1) y² / x² * dy²

                    + g₃₃ dz².

Curvatures due to the metric of time, spatial metric set flat

https://en.wikipedia.org/wiki/Ricci_curvature

We assume that all deviations from the flat Minkowski metric η are very small and that the partial derivatives of the components of the metric tensor, gij, are very close to zero. Then all the Christoffel symbols Γ are almost zero, and we can ignore the cross terms Γ * Γ in the formula of Rjk.

That is, we assume that the Rjk are linear in each gij.

Consequently, we can calculate the curvatures in parts: first the curvatures for the distorted metric g₀₀ of time, keeping the spatial metric flat, and then vice versa.

Let us calculate the curvatures which are due to the perturbed metric of time g₀₀:

       Γ¹₀₀  =  g¹¹ * 1/2 * -dg₀₀ / dx

                =  Γ⁰₁₀  =  Γ⁰₀₁,

       Γ²₀₀  =  g²² * 1/2 * -dg₀₀ / dx  *  y / x,

       R₀₀   =  dΓ¹₀₀ / dx  +  dΓ²₀₀ / dy,

               =  -1/2 g₀₀''  +  -1/2 g₀₀' / x,

where the prime ' denotes the derivative with respect to x. Can we treat g¹¹ and g²² as a constant 1 because their derivatives are very small and would multiply other very small derivatives? If the size of the system is one unit, then gij - 1, gij', and gij'' all have the same order of magnitude, and all have a very small absolute value. Any product of two of them is negligible. We can treat them as a constant 1.

       R₁₁   =  dΓ⁰₁₀ / dx

                =  1/2 * g₀₀'',

       Γ⁰₂₀  =  1/2 * -1 * dg₀₀ / dx  *  y / x,

       R₂₂   =  -Γ⁰₂₀ / dy

               =  1/2 g₀₀' / x,

       R₃₃  =  0.

Curvatures due to the spatial metric, the metric of time set to -1

Let us study the perturbation of the spatial metric.

       R₀₀        =  0,

                       The factor g²¹ is missing!

                        |

                        v

       Γ²₁₁ g₂₂ =     (dg₁₂ / dx  -  1/2 dg₁₁ / dy)

                     =  d((grr - 1) * y / x) / dx

                          - 1/2 d(grr + (1 - grr) * y² / x²) / dy

                     =  g₁₁' * y / x  -  (g₁₁ - 1) * y / x²

                            - 1/2 g₁₁' * y / x

                            - (1 - g₁₁) * y / x²,

       Γ²₁₂ g₂₂ =  1/2 dg₂₂ / dx

                     =  0,

       Γ³₁₃ g₃₃  =  1/2 dg₃₃ / dx.

Can we ignore the factors g₂₂, etc. after the Γ? Yes. Their derivative would multiply very small numbers, like (1 - g₁₁), and the product would be negligible.

       R₁₁   =  dΓ²₁₁ / dy  +  dΓ³₁₁ / dz

                   - dΓ²₁₂ / dx  -  dΓ³₁₃ / dx

               =  g₁₁' / x  -  g₁₁ / x²  + 1 / x²

                   - 1/2 g₁₁' / x 

                   - 1 / x²  +  g₁₁ / x²

                   - 1/2 g₃₃''

              =  1/2 g₁₁' / x  -  1/2 g₃₃''.

Let us calculate R₂₂. We can ignore the factors g¹¹, etc. in the Christoffel symbols, because their derivatives would multiply very small numbers.

       Γ¹₂₂  =  dg₂₁ / dy  -  1/2 dg₂₂ / dx

               =  d((grr - 1) * y / x) / dy

                   - 1/2 d(1 + (grr - 1) * y² / x²) / dx

   

               =  (g₁₁ -  1) / x,

       Γ³₂₂  =  dg₂₃ / dy  -  1/2 dg₂₂ / dz

                =  0,

       Γ⁰₂₀  =  -1/2 dg₀₀ / dx

                =  0,

       Γ¹₂₁  =  1/2 dg₁₁ / dy

                =  1/2 d(grr  +  (1 - grr) * y² / x² ) / dy

                =  1/2 g₁₁' * y / x  +  (1 - g₁₁) * y / x²,

       Γ³₂₃  =  1/2 dg₃₃ / dy

                =  1/2 dg₃₃ / dx * y / x,

       R₂₂   =  dΓ¹₂₂ / dx  +  dΓ³₂₂ / dz 

                   - dΓ⁰₂₀ / dy  -  dΓ¹₂₁ / dy  -  dΓ³₂₃ / dy

               =  g₁₁' / x  -  g₁₁ / x²  +  1 / x²

                   - 1/2 g₁₁' / x  -  1 / x²  +  g₁₁ / x²

                   - 1/2 g₃₃' / x

              =  1/2 g₁₁' / x

                  - 1/2 g₃₃' / x.

Let us calculate R₃₃:

       Γ¹₃₃  =  -1/2 dg₃₃ / dx,

       Γ²₃₃  =  -1/2 dg₃₃ / dx * y / x,

       R₃₃   =  dΓ¹₃₃ / dx  +  dΓ²₃₃ / dy 

                   - dΓ⁰₃₀ / dz  -  dΓ¹₃₁ / dz  -  dΓ²₃₂ / dz

                =  -1/2 * g₃₃''  -  1/2 * g₃₃' / x,

Analysis of R₀₀ inside the cylinder

The Ricci curvature to the direction of time, R₀₀, is positive inside the pressurized cylinder. That is, a cubical configuration of test masses tends to contract. We must have

       R₀₀  =  p / 2  =  -1/2 g₀₀''  +  -1/2 g₀₀' / x

The symmetry implies that

       g₀₀'(0)  =  0.

The graph of g₀₀(x) must look something like this:

                    0

       -----------------------> x

             __-------__

           /                 \

  

and the formula is

       g₀₀(x)  =  -p / 4 * x²  -  C,

where C is a positive constant. Recall that g₀₀ in empty space is -1.

http://mathonline.wikidot.com/the-existence-uniqueness-of-solutions-to-second-order-linear

The Picard-Lindelöf theorem guarantees that we found the unique solution to the second order differential equation, if we know g₀₀(0) and g₀₀'(0).

Analysis of R₃₃ inside the cylinder

       R₃₃ =  p / 2  =  -1/2 * g₃₃''  -  1/2 * g₃₃' / x.

The formula is

       g₃₃(x)  =  -p / 4 * x²  +  C',

where C' is a positive constant. Recall that g₃₃ in empty space is 1.

Analysis of R₁₁ and R₂₂ inside the cylinder

We sum the Ricci curvature from the perturbation of the metric of time to the curvature from the spatial metric perturbation:

       R₁₁ =  1/2 * g₀₀''  +  1/2 g₁₁' / x  -  1/2 g₃₃'',

       R₂₂ =  1/2 g₀₀' / x  + 1/2 g₁₁' / x  - 1/2 g₃₃' / x.

Above,

       g₃₃''       =  -p / 2,

       g₃₃' / x  =  -p / 2,

       g₀₀''       =  -p / 2,

       g₀₀' / x   =  -p / 2.

We have no contradiction!

A cylinder with no pressure

In the derivation of the contradiction above, we never used the fact that it is a pressure p inside the cylinder. We could as well assume that the pressure is zero and there is a mass density ρ. We have a contradiction also in that case.

In the literature we cannot find any discussion about the metric inside a cylinder. As if researchers would have overlooked the question altogether.

We have to check our calculations very carefully. Maybe there is an error?

Generally, since there are no existence theorems about a metric inside matter, it might be that such metrics only exist for very special configurations. The Schwarzschild interior solution is the well known example.

Our paper bending model on March 8, 2024 already suggested that solutions might exist only for beautifully symmetric configurations.

Conclusions

We have to check if a small departure from the cylindrical symmetry of the metric allows the Einstein equations to have a solution inside a cylinder. That is not likely, since our argument above already uses approximate values, and the contradiction in the values of R₁₁ and R₂₂ happens with a very large margin.

Also, we have to check the calculations carefully.

If the calculations are correct, this is a fatal blow to the Einstein field equations. The field equations have to be really badly wrong.

Our own Minkowski & newtonian gravity model has no problems handling a cylinder, with pressure, or without it.