Tuesday, March 19, 2024

Searching for metrics with Ricci curvature zero

We know two exact static solutions of the Einstein field equations where Ricci curvature is zero in the vacuum area. The Schwarzschild solution and the Levi-Civita metric.

The Levi-Civita metric is not physically realistic though, because it assumes an infinitely long cylinder.

Is there any static solution for a finite cylinder? Or for two spherical masses held apart by a very lightweight rod?

Let us investigate the case of two lightweight masses. We assume that a "magic" mechanism keeps them static in space, so that they do not crash together.


                ●                     ●
               M₁                   M₂


Each M alone would have the Schwarzschikd metric around it. The field is weak. We may write the metric

       η + h,

where η is the Minkowski metric, and h is a perturbation.

Our first guess for two masses M is, of course, the sum of the perturbations

       η + h₁+ h₂.

Each h has Ricci curvature zero in vacuum.









The Cristoffel symbols Γ are linear in the metric g. If the directional Ricci curvatures Rⱼₖ were linear in the Christoffel symbols, then the sum of the perturbations h₁ + h₂ would have Ricci curvature zero. But we have the annoying cross terms:







Since the fields are weak, the Christoffel symbols have small absolute values, and the cross terms have very small absolute values. Since the equations are nonlinear, this does not guarantee that η + h₁+ h₂ is close to a solution. Maybe the solution does not exist at all?

This eerily reminds us of our paper bending experiments on March 8, 2024. If we were allowed to stretch the paper a little bit, then we could find a solution for many types of "masses". But we are not allowed to do that, and the cone is the only beautiful solution that we know.

On March 11, 2024 we calculated Ricci curvatures for various components of the Schwarzschild metric. Let us try to determine what the cross terms look like if we sum two Schwarzschild metrics.


                           M₁                           M₂
     ----> x            ●                              ●
    |                                  •  test mass
    v  y
      

In R₀₀, we have, for example, the product

       Γ¹₁₁  Γ¹₀₀.

Let us take the first Γ from h₁ and the second Γ from h₂. The cross term has an approximate value

       -1/2 * 1 / x₁²  *  1/2 * 1 / x₂²,

where x₁ is the x distance from M₁ and x₂ is the x distance from M₂. Recall that we set

       2 G M / c²  =  1

on March 11, 2024. Thus, they are actually very strong fields.


Iterative methods do not work?








Let us have the lightweight masses M₁ and M₂ above. Our first try to solve the Einstein field equations is the metric

       g  =  η + h₁+ h₂

above.

The metric does not precisely satisfy the Einstein field equations, because of the tiny cross terms between the perturbations h₁ and h₂:







The right side of the Einstein equations obtains some residual value T', where T' is a tensor close to zero.

Suppose further that only the 00 component of T' differs from zero. It is like a mass distribution M(r) in space, where r is the position vector. The mass M(r) is very small compared to M₁ and M₂.

Let us assume that we have a Schwarzschild-like solution

       η + hM

for the mass distribution M(r) in otherwise empty space. The perturbation hM is much closer to zero than h₁ and h₂ are. We obtain a much better approximate solution for the original equation in the metric:

       g - hM.

"Better" here means that the new residual value on the right, T'', is much closer to zero than T'.

If all the above assumptions were true, then we might be able to prove that the iteration quickly converges to a solution. We would have a powerful existence theorem. But such an existence theorem is not known. If our simple method would work, someone would have spotted it soon after the year 1915.

A problem is that the residual tensor T' has also other non-zero components besides the 00 component. T' does not describe a pure mass distribution.

Is there a "Schwarzschild" solution for an arbitrary stress-energy tensor component? For example, if we assume an isolated pressure component T₁₁ which occupies a spherical volume V in space?

For a moment, we do not worry if such isolated pressure can exist in the real world.


A Schwarzschild-like solution for pressure?


There is probably no Schwarzschild-like solution for a component of pressure. A central feature in the Schwarzschild solution is that it is spherically symmetric. Pressure is directed.

            
                    P pressure

                -------      T₁₁ non-zero
              -----------
                -------
                                           F' anomalous
                               F   ^    ^ 
                                       \  |  
                                           •  m test mass
     ^ y
     |
      ------> x


The test mass m stretches the radial metric around it. We expect the potential of m to be lower if m is to the x direction from the pressure volume P. The force F does not pull m directly toward P, but there is also an "anomalous" component F' which pulls m up so that m would be located to the x direction from P.

We have to check from the Einstein-Hilbert action if our reasoning is correct.

The anomalous force F' probably has the divergence

       ∇² F'

non-zero in vacuum. But is the anomalous force known to the metric g around P, or is F' a more complicated consequence of the Einstein-Hilbert action?

The metric in general relativity certainly understands some effects of pressure, but does the metric understand the force F'?


Conclusions


If a simple iterative method would work, then we would have a wealth of exact solutions in general relativity. We would probably have existence theorems, too.

We conclude that no one has found a good iterative method for solving the Einstein equations.

We will next investigate what general relativity understands about the pressure area P.

The Einstein approximation formula (1916) below suggests a very simple metric around the pressure area P. Does the formula give a good approximate metric?















Saturday, March 16, 2024

Degrees of freedom in general relativity

In this blog we have never studied degrees of freedom, and our understanding of the matter is not very good yet.

It is said that general relativity has 2 "degrees of freedom". There are 10 independent parameters in the 4 × 4 metric tensor g. The coordinate transformation freedom removes 4 + 4 of these, so that 2 remain. Also, in electromagnetism, we have two degrees of freedom.


Suppose that we have a spherically symmetric static electric charge. The electric Coulomb field around it has a "freedom" of only one real number: the charge.

Similarly, in general relativity, the field around a spherical mass is the Schwarzschild metric, and the "freedom" is the mass.

In our 2-dimensional surface model of March 8, 2024, the solution is a cone made of paper around a circular "mass". The freedom is the angle of the cone.

In all these cases, all "freedom" is eventually lost because of a symmetry, and because the solution must match the "charge" at the center.

Is there a paradox in this? A gravity field, or a wave, seems to be much more "complex" than an electromagnetic field. How can it have the same two degrees of freedom?


Controlling the metric around matter


Suppose that we have a configuration of light masses sitting in space. We can make the metric of time to match the newtonian gravity potential. That fixes one degree of freedom. After that, we can try to find a spatial metric which makes the Ricci tensor zero outside matter. In the Schwarzschild case, we can find such a spatial metric. Is there any guarantee that we can find it in the general case?

We can use the mass distribution to control the temporal metric g₀₀. If we can dynamically create pressure inside matter, we can presumably control also some spatial components of g. Does this remove too much freedom from the field, so that we cannot make Ricci curvature zero there?

On November 5, 2023 we tentatively proved that a change in pressure "breaks" the Einstein field equations. We cannot make the Ricci tensor zero in such a case.

Degrees of freedom may clarify our result. If the Ricci tensor R is zero around a spherically symmetric system, then the only possible freedom is the amount of mass at the center. But if we manipulate the pressure at the center, the system cannot be described by a single parameter.

In electromagnetism, the only relevant parameter is the amount of charge. Pressure plays no role.

How about a general system of masses? Can we "fix" the metric of time and some components of the spatial metric around it, so that the remaining freedom makes it impossible to make the Ricci tensor R zero?


Static masses: is there enough freedom to make R = 0 outside matter?


If we have two masses floating freely in space, it is not a static system. Gravity accelerates those masses.

Let us consider a subcase where the matter sits static, because of pressures which counteract gravity. Schwarzschild was able to handle a spherically symmetric ball of incompressible liquid. What about more complex shapes?


            ----------------------------------
           |                                        |
           |     distribution of          |
           |     mass and pressure |
           |                                        |
            -----------------------------------
                   block of matter


The block contains both mass and stress forces: pressure and shear forces.

The metric in the vacuum around the block must be able to match the mass distribution and the stresses inside the block. Is it plausible that such a metric can be found?

In the case of electromagnetism, the Coulomb field in the vacuum can easily match the charge distribution. Simply linearly sum the fields of elementary charges.

In the case of gravity, we can control g₀₀ around the block by tuning the mass distribution. We can probably control several other elements in the metric tensor by tuning the pressure and the shear stresses. How many degrees of freedom are there in elastic stresses of a material? The sum of forces on a cube of matter must be zero, to each direction x, y, z.

The pressure in the cube can have independent values to each direction x, y, z. The stresses, apparently, form a 3 × 3 stress tensor.

Such a tensor has 6 degrees of freedom? Is there a gauge invariance which would reduce them?

At first sight, the two degrees of freedom in general relativity are too few to match 6 degrees in the stress tensor and one degree in the mass distribution. This would imply that the Einstein field equations have no solutions in almost all cases which involve pressure or shear stresses.

Pressure at the surface of the block, obviously, cannot have a component which is normal to the surface. Also, shear stresses have to be 2-dimensional at the surface.

We are dealing with a 2 × 2 stress tensor, which has three independent components?

If the vacuum metric around the block seriously restricts the allowed mass density / stress distributions in the block, then we obtain strange forces which guide us to make the distributions into the allowed ones! This would be very strange, and an indication that general relativity is an incorrect theory.

Question. Did we heuristically prove that the Einstein field equations only have solutions for very special pressure distributions?


The question if the Einstein equations have a solution for an arbitrary mass distribution, is an open one. Above we employ the freedom in stresses. Do these definitely show that general relativity lacks the required degrees of freedom?

In our 2D surface model on March 8, 2024, we struggled to define the metric for "mass" distributions. We obviously in that model do not have any means of handling pressure distributions. The paper which we use is rigid, and cannot communicate the pressure in the "matter" which we embed.


A spherical mass with a tense membrane around it


General relativity says that the metric around a spherically symmetric system must be Schwarzschild, and the metric cannot change.

Let us put a rubber membrane around the sphere, and vary the tension in the rubber. The metric inside the membrane must change because the "source" of the gravity field changes. Could it be that the metric at the surface of the membrane stays constant?


Jürgen Ehlers et al. (2005) study this configuration and conclude that the metric stays the same outside the sphere, regardless of the pressure.

However, if the body is not spherical, what happens?


Conclusions


If we have a block of rubber, it can be strained with a stress tensor. The stretching of the block can be described with a "metric tensor". There intuitively exists a solution. Counting degrees of freedom does not bring anything more to this analysis.

For now, we will not analyze the degrees of freedom further.

We will look at zero Ricci curvature as a problem in optics. A bundle of parallel rays of light must not be focused or defocused if Ricci curvature is zero. How to design such an optical device? The Schwarzschild metric is one. But if we perturb that metric, does it necessarily follow that the resulting optics focus or defocus certain beams of light? If that is the case, there is no solution for the Einstein field equations, except for artificial, symmetric configurations.

Our analogy with 2D surfaces and bending a paper into a cone suggests that the Schwarzschild metric might be the only solution where the Ricci tensor R = 0, just like a cone (or a cylinder) is the only way to bend a paper if we want to keep its metric straight.

Friday, March 15, 2024

Metric around a cylinder: the Levi-Civita metric

UPDATE March 15, 2024: The Einstein approximation formula (see the post on August 15, 2023) produces roughly the Levi-Civita metric.

----

Tullio Levi-Civita (1919) was able to find a cylindrical metric where Ricci curvature is zero.











In the formula above, D and σ are constants. Let us set D = 1.

In newtonian gravity,

       dg₀₀ / dr  ~  -1 / r

around a cylinder. In the Levi-Civita metric,

       dg₀₀ / dr  ~  -4 σ / r  *  r^(4 σ).

If we set σ to a small positive constant, then the Levi-Civita metric of time mimics quite accurately the newtonian solution. The Levi-Civita metric stretches the radial metric and the metric to the z direction equally.

Let us try to calculate Ricci curvature around, and close to, a lightweight long cylinder. The signature is (- + + +).


    y axis points
    out of screen              cylinder
               ●--------> z   ==============
               |
               |                               r
               |
               |                               • m
               v  x
                 

Let us set

       g₀₀  =  -1  -  ln(r),

       grr   =   1  -  ln(r),

       gnn  =   1  -  ln(r),

       g₃₃  =   1  -  ln(r),

where gnn is the metric normal to the radius r and the z axis. We assume that the value of r is very close to 1.

We assume that r points approximately to the x direction, that is, |y| / |x| is small. Let us write

       ds²  =  grr dr²  +  gnn dn²  +  g₃₃ dz²,

               = (1 - ln(r))  *  (dx² + dy² + dz²)
     











The metric of time distorted, spatial metric flat


       Γ⁰₀₀  =  0,

       Γ¹₀₀  =  1/2 * -dg₀₀ / dx

                =  1/2 * 1 / x

                =  -Γ⁰₁₀  =  -Γ⁰₀₁,

       Γ²₀₀  =  1/2 * -dg₀₀ / dy

                =  1/2 * y / x²,

       R₀₀   =  dΓ¹₀₀ / dx  +  dΓ²₀₀ / dy,

               =  -1/2 * 1 / x²  +  1/2 * 1 / x²

               =  0,

       R₁₁   =  -dΓ⁰₁₀ / dx

                =  1/2 * 1 / x²,

       Γ⁰₂₀  =  1/2 * -1 * dg₀₀ / dy

                =  1/2 * y / x * 1 / x

                =  1/2 * y / x²,

       R₂₂   =  -Γ⁰₂₀ / dy

               =  -1/2 * 1 / x²,

       R₃₃   =  0.


The metric of time -1, the spatial metric stretched


       R₀₀  =  dΓ¹₀₀ / dx  +  dΓ²₀₀ / dy  +  dΓ³₀₀ / dz

               =  0,

       Γ²₁₁  =  -1/2 dg₁₁ / dy

                =  -1/2 d(1 - ln(r)) / dr * dr / dy

                =  1/2 * 1 / r * y / x

                =  1/2 * y / x²,

       Γ³₁₁  =  dg₁₃ / dx  -  1/2 dg₁₁ / dz

                =  0,

       Γ²₁₂  =  1/2 dg₂₂ / dx

                =  1/2 * -1 / x,

       Γ³₁₃  =  1/2 dg₃₃ / dx

                =  1/2 * -1 / x,

       R₁₁  =  dΓ²₁₁ / dy  +  dΓ³₁₁ / dz

                  - dΓ²₁₂ / dx  -  dΓ³₁₃ / dx

              =  1/2 * 1 / x²  +  0

                  - 1/2 * 1 / x²  - 1/2 * 1 / x²

              =  -1/2 * 1 / x².

Let us calculate R₂₂:

       Γ¹₂₂  =  -1/2 dg₂₂ / dx

                =  1/2 * 1 / x,

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

                =  0,

       Γ⁰₂₀  =  -1/2 dg₀₀ / dx

                =  0,

       Γ¹₂₁  =  1/2 dg₁₁ / dy

                =  -1/2 * y / x²,

       Γ³₂₃  =  1/2 dg₃₃ / dy

                =  -1/2 * y / x²

       R₂₂  =  dΓ¹₂₂ / dx  +  dΓ³₂₂ / dz 

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

               =  -1/2 * 1 / x²  +  0

                    - 0  +  1/2 * 1 / x²  +  1/2 * 1 / x²

               =  1/2 * 1 / x².

Then R₃₃:

       Γ¹₃₃  =  -1/2 dg₃₃ / dx

                =  1/2 * 1 / x,

       Γ²₃₃  = -1/2 dg₃₃ / dy

                =  1/2 * y / x²,

       R₃₃  =  dΓ¹₃₃ / dx  +  dΓ²₃₃ / dy 

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

              =  -1/2 * 1 / x²  +  1/2 * 1 / x²

                  - 0  -  0  -  0

              =  0.

Conclusions


The Levi-Civita metric is very simple, if we ignore the exponent 8 σ² of r in the first term. The spatial metric is stretched uniformly to x, y, and z directions, where the stretching depends on r.

We were able to confirm that the Ricci tensor is approximately 0.

The notion of uniform stretching of the spatial metric is strange. If our physical system consists of point particles moving at various speeds and bumping into each other, we can either interpret that local time has slowed down, or that all the distances have become longer.

Maybe we can measure "true" proper local distances through force fields, e.g., through the Coulomb field? If we double the distance, then the force is 1/4, while slowing down local time by a factor 1/2 only reduces the impulse given by the force by a factor 1/2. But this does not reveal a difference if we assume that the inertia of a charged particle doubled.

We here have a new relativity principle: relativity of time versus distances. In the cylinder case, if we interpret that the spatial metric is flat, and that local time has slowed down and the inertia of particles has increased, them the Ricci tensor is not zero.

Thursday, March 14, 2024

The Einstein equations are horribly nonlinear

UPDATE March 15, 2024: The Einstein approximation formula (see the post on August 15, 2023) produces a much more reasonable metric around the spherical shell, if we add perturbations. However, the "bulging coordinates" problem, which we studied in August 2023, may make the produced metric somewhat wrong.

----

We are trying to gain an intuitive sense about how to construct metrics such that their Ricci curvature is zero. It turns out that relying on (approximate) linearity leads to profoundly wrong results.

The Einstein equations say that in vacuum, the Ricci curvature is zero.









Let us have metric perturbations h₁ and h₂, such that the Ricci curvature of

        η + hi 

is zero. Here η is the Minkowski metric.

The derivatives in the Christoffel symbols Γ are assumed to have a very small absolute value. In the formula of the Ricci tensor, the cross terms Γ * Γ' then have extremely small absolute values. We are tempted to drop off the cross terms altogether. Then the Ricci tensor would be linear on perturbations.

Is the sum

       η + h₁ + h₂

a good guess for a metric which would have the Ricci tensor zero?

No. Let us construct a metric around a spherical shell of matter by adding the metric perturbations from each small element of the shell. We will show that the linearly summed metric is totally wrong. The correct metric, of course, is the Schwarzschild metric.
















A small element of the shell, dm, when alone in space, carries a Schwarzschild metric around it.

We assume that the radius of the shell is and mass / area of the shell is 1.

Let us then estimate the radial and the tangential metric around the shell, by adding the spatial metric perturbations. Let us denote the angle between OB and OC by α.

Let the radius of the sphere be 1. The element dm is at B in the diagram. We 

The element dm stretches the spatial metric at C by

       dm / (2 sin(α / 2)).

The radial stretch component is

       dm / (2 sin(α / 2))  *  sin(α / 2)

       = dm / 2,

and the tangential stretch component is

       ~ dm / (2 sin(α /2))  *  cos(α / 2).

The area of a C-centric narrow circular strip through B on the shell is

       2 π sin(α) dα.

The contribution to the stretching of the tangential metric is

                  π
       t₁  =   ∫   1 / (2 sin(α / 2)) * cos(α / 2)
                0
                      * 2 π sin(α) dα

            = 9.9,

where we used the numerical integrator at:




















The contribution to the radial metric is

                  π
       r₁  =   ∫   1 / 2 * 2 π sin(α) dα
                0

            = 2 π

            = 6.3.

We see that the sum of metric perturbations stretches the tangential metric at C more than the radial metric. Let us draw two circles around the sphere, such that the radius of one is 1, and the other is at a proper distance dr farther from 1. The length of the larger circle would be > 2 π dr larger than the smaller circle. The spatial metric would have a negative Ricci curvature around the sphere.

In the correct, Schwarzschild, metric, the spatial metric has a positive Ricci curvature around the sphere.


Conclusions


We showed that one cannot get the correct metric around a spherical shell by summing perturbations of the metric for each mass element in the shell. The sum of the perturbations is grossly wrong.

This implies that one cannot rely on sums of perturbations when working with general relativity. We can say that the Einstein equations are "horribly nonlinear".

In the next blog posts we will analyze this further. Suppose that we are working on some mass distribution and have found a metric g which is "almost" right: the Ricci tensor in the vacuum area only differs from zero by a tiny amount. But the correct metric may still be very different from g. Let us then calculate the Einstein tensor T for g. We are able to satisfy the Einstein equations if we place a small amount of matter at certain locations, so that the stress-energy tensor becomes T.

Is this physically reasonable? Placing a small amount of matter switches the metric radically. Nature does not seem to function like that.


Andrew Strominger et al. (2011) show that Navier-Stokes equations in p + 1 dimensions are equivalent to Einstein equations in a certain setup in p + 2 dimensions. We know that the Navier-Stokes equations behave horribly badly. There is turbulence, and the existence of solutions is a one of the Millennium Prize problems in mathematics:


The existence of molecules and atoms saves us from the potentially infinite complexity of the Navier-Stokes equations in the real world.

Monday, March 11, 2024

Ricci curvature in the Schwarzschild metric

UPDATE April 2, 2024: We forgot the factor g²¹ in the formulae.

We have to figure out how to use orthogonal coordinates.

----

We calculate separately the Ricci curvature due to the distorted metric of time, and the Ricci curvature due to the distorted radial metric. Since the Schwarzschild metric has Ricci curvature zero outside the central mass M, we expect that the sum of these separate curvatures is zero to every direction.

Let us have any configuration of masses. We only distort the metric of time according to the newtonian potential, that is, make clocks tick slower so that they explain the redshift. Generally, the metric of time makes Ricci curvature non-zero outside matter.

A difficult problem in general relativity is to find a spatial metric which resets Ricci curvature to zero. We hope that understanding the Schwarzschild metric would help us.


Schwarzschild metric


          y
             ^
             |               • (x, y, z)
             |        r
             |
             ●------------------> x
             Axis z points out
             of the screen


The mass M is at the origin. We assume weak fields. The metric signature is (- + + +).

Newtonian gravity satisfies

      dg₀₀ / dr  =  -2 G M / c²  *  1 / r²,

and the radial spatial metric

       grr  =  1  +  2 G M / c²  * 1 / r,

       dgrr / dr  =  -2 G M / c²  * 1 / r².

We set the coefficient 2 G M / c² to 1.

We assume that r points approximately to the x direction. Let us write

       ds²  =  grr dr²  +  dn₂²  +  dn₃²,

where dn₂ is normal to r, points approximately to the y direction and changes the "longitude" of r; dn₃ points approximately to the z direction and changes the "latitude" of r.

There

       dr   =  (1 - 1/2 * (y² + z²) / x²) dx
     
                  + y / x * dy  

                  + z / x * dz,

       dn₂  =  -y / x * dx 

                  + (1 - 1/2 * y² / x²) dy,

       dn₃  =  -z / x * dx

                   + (1 - 1/2 * z² / x²) dz,

                   - y / x * dy * z / x.

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

        ds²  =  grr * dx²

                    + (1 - grr) * (y² + z²) / x² * dx²

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

                    + 2 (grr - 1) * z / x * dx dz

                    + 2 (grr - 1) * y z / x² * dy dz

                    + dy²

                    + (grr - 1) y² / x² * dy²

                    + dz²

                    + (grr - 1) z² / x² * dz²,

where grr - 1 = 1 / r.


Spatial metric set flat: Ricci curvatures due to the metric of time










Let us first set the spatial metric flat. We calculate the Ricci curvatures which are due to the metric of time.

       Γ⁰₀₀  =  0,

       Γ¹₀₀  =  1/2 * -dg₀₀ / dx

                =  1/2 * 1 / x²

                =  Γ⁰₁₀  =  Γ⁰₀₁,

       Γ²₀₀  =  1/2 * -dg₀₀ / dy

                =  1/2 * y / x³,

       Γ³₀₀  =  1/2 * z / x³,

       R₀₀   =  dΓ¹₀₀ / dx  +  dΓ²₀₀ / dy  +  dΓ³₀₀ / dz

               =  -1 / x³  +  1/2 * 1 / x³  + 1/2 * 1 / x³

               =  0,

       R₁₁   =  -dΓ⁰₁₀ / dx

                =  -1 / x³,

       Γ⁰₂₀  =  1/2 * -1 * dg₀₀ / dy

                =  1/2 * y / x * 1 / x²

                =  1/2 * y / x³,

       R₂₂   =  -Γ⁰₂₀ / dy

               =  -1/2 * 1 / x³

               =  R₃₃.


The metric of time set to -1: Ricci curvatures due to the spatial metric


Let us then set the metric of time -1. We want to calculate the Ricci curvatures which are due to the spatial Schwarzschild metric.

We can drop terms which are contain a product where y or z occur three or more times.

       R₀₀  =  dΓ¹₀₀ / dx  +  dΓ²₀₀ / dy  +  dΓ³₀₀ / dz

               =  0,

       Γ²₁₁  =  1/2 * 1 / g₂₁ * dg₁₁ / dx 
   
                    + dg₁₂ / dx

                    -  1/2 dg₁₁ / dy

                =  1/2 * x / ((grr - 1) y) * -1 / x²

                   + d(1 / r  * y / x) / dx 

                   - 1/2 d(grr  -  1 / r * (y² + z²) / x²) / dy

                =  -1/2 * 1 / y

                    - 2 y / x³

                    + 1/2 * y / x³

                    - 1/2 * -2 * y / x³

                =  -1/2 * y / x³,

       Γ²₁₂  =  1/2 dg₂₂ / dx

                =  -3/2 * y² / x⁴.

The values for the third coordinate z are symmetric with y.

       R₁₁  =  dΓ²₁₁ / dy  +  dΓ³₁₁ / dz

                  - dΓ²₁₂ / dx  -  dΓ³₁₃ / dx

              =  -1 / x³.

Let us calculate R₂₂:

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

                =   d(1 / r  *  y / x) / dy

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

                =   1 / x²

                      - 1/2 * -3 y² / x⁴,

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

               =  d(1 / r * y z / x²) / dy

                   - 1/2 d(1 + 1 / r * y² / x²) / dz

               =  z / x³,

       Γ⁰₂₀  =  -1/2 dg₀₀ / dx

                =  0,

       Γ¹₂₁  =  1/2 dg₁₁ / dy

                =  1/2 d(grr  -  1 / r * (y² + z²) / x²) / dy

                =  -1/2 y / x³

                    + 1/2 * -2 y / x³

                =  -3/2 y / x³,

       Γ³₂₃  =  1/2 dg₃₃ / dy

                =  0,

       R₂₂  =  dΓ¹₂₂ / dx  +  dΓ³₂₂ / dz 

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

              =  -2 / x³  +  1 / x³

                  + 3/2 * 1 / x³

              =  1/2 * 1 / x³

              =  R₃₃.

We showed that the spatial metric, indeed, resets the Ricci curvature of the temporal metric to zero.


Conclusions


We can use the calculations above to get an intuitive understanding of what "happens" in a metric, and how to make Ricci curvature zero.

We are particularly interested if we can repeat the trick for the field of a long lightweight cylinder. Close to the cylinder, the newtonian potential is something like

        m c² (ε ln(r)  -  C),

where m is a test mass, ε > 0 is small, r is the distance from the cylinder, and C is a small positive constant. The potential corresponds to the following metric of time:

       g₀₀  =  -1  -  2 ε ln(r) + 2 C 

Can we make Ricci curvature zero by stretching the radial metric by a factor

       1  -  ε ln(r) + C,

that is,

       grr  =  1  - 2 ε ln(r) + 2 C ?

Friday, March 8, 2024

Ricci curvature in 2 dimensions: two circular masses make an ugly solution?

Let us have a two-dimensional surface embedded in three spatial dimensions. The surface then has locally the Ricci curvature zero, if we can form the surface by bending a rigid sheet of paper. That is, the curvature can only be zero if the surface is a part of a cone or a "cylinder" locally.














Let us then embed a point "mass" M into the surface. We demand that Ricci curvature is positive at the point M. The obvious solution is a cone whose tip is M.

Note the analogue to Birkhoff's theorem. The theorem states in 1 + 3 dimensions that the only static metric around M is the Schwarzschild metric. Analogously, in two dimensions, the only possible surface around M is a cone.

Let us embed a circularly shaped mass into the 2D surface. The obvious solution is that we cut the head off the cone. The hole is circularly shaped.


    form a half-cone
              ____   circular mass
            /        \   
            \____/
          |            |
          |            |  bend
          |            |
              ____   
            /        \ X singularity?
            \____/   
                       circular mass
    form a half-cone


Let us then have two circular masses on a flat sheet of paper. We begin by bending the paper down along the two vertical lines.


                          /\
                        /    \
                      /        \

                      wedge
  

After that, we can form two half-cones at the top end and bottom end in the diagram by cutting off infinitely many infinitesimally wide wedges whose tip is on either circle.

But the solution is ugly. There is a transition from a cone shape to a flat shape along the egde of a circle. If the transition is sharp, we get a singularity-like point, marked with X in the diagram.

What if the transition is not sharp? Is Ricci curvature still zero outside the circle? Maybe not. At least, the solution is ugly.


Fitting an oversized carpet into a room


             wrinkle
       _______/\________ carpet


Our efforts to make Ricci curvature zero outside matter bring to our mind the carpet analogy from our blog post on November 9, 2023. One can make a part of the carpet flat on the floor, but since the carpet is oversized, there will always be a wrinkle. A wrinkle is a "singularity".


A rubber sheet model of gravity is flexible


The requirement that Ricci curvature is zero outside matter leads us to use a rigid sheet of paper above. It is conceivable that many configurations do not have a solution at all: an attempt to solve the problem inevitably leads to wrinkles in the paper, or to other types of singularities.

A rubber sheet model would be more flexible. Intuitively, we can always find a solution because a rubber sheet can be stretched. Thus, we know, or can guess, that a rubber sheet model always works. This is different from the Einstein field equations, where it is not clear if solutions exist at all for nontrivial cases.

In our blog we have been claiming for two years that the Einstein field equations may be too "rigid", so that they do not have solutions.


Conclusions


Requiring that Ricci curvature is zero outside matter can lead to singularities or ugly solutions. Maybe we have to give up the assumption that Ricci curvature is zero outside matter – also in the case of static configurations? For dynamic changes of pressure we already know (tentatively) that the Einstein field equations break.

Thursday, March 7, 2024

Hypothesis of simplicity: "pathologically" nonlinear differential equations cannot occur in nature

The Einstein field equations are nonlinear differential equations. In mathematics, we can prove very little about the existence and uniqueness of solutions to such equations. We have a few examples where the existence is proved by finding an exact solution to the equations. In general relativity, the prime example of a solution is the Schwarzschild metric around a spherically symmetric mass.

But is it just a lucky coincidence that a simple and beautiful metric exists for a spherically symmetric mass?

Our own Minkowski & newtonian gravity model reproduces the Schwarzschild metric in the case of a spherically symmetric mass M, if we tune the "shipping" of energy to be the radial distance r of the test mass m to the center of the mass M. This is probably just a coincidence. There is no obvious reason why our model would make Ricci curvature zero outside more complex mass distributions.

In the fall of 2023 we tentatively proved that changes in pressure "break" the Einstein field equations. The equations in that case do not have a solution, even though our own Minkowski & newtonian model does have a simple solution also in that case.


Nonlinear differential equations in physics


Question. Do the Einstein field equations have a solution for any complex mass distribution at all? For example, for a finitely long uniform cylinder?


If there is no solution to certain nonlinear equations, the symptom often is that an attempt to solve the equations leads a singularity. We know that general relativity is rife with singularities.

In physics, differential equations which describe an interacting system, are almost always nonlinear. However, the problem of nonlinearity often is not as bad as in general relativity:

1.   There may be efficient iterative methods which find an approximate solution easily.

2.   The system consists of atoms or molecules, and the system is microscopically totally different from its high-level differential equations. An example is the Navier-Stokes equations: they ignore molecules.

3.   The system may be quantum, and we can only make a single measurement from it. The existence of a continuous solution which fills Minkowski space, is not too relevant if we cannot really probe that solution efficiently.


In general relativity, the metric is assumed to be a continuous function which fills the entire universe. A singularity in that metric would often have drastic physical consequences.


General relativity claims things about mathematics


Let us assume that the Einstein equations are a correct physical theory. That assumption implies that the equations do have solutions for all realistic mass distributions.

On the other hand, it is difficult or impossible to prove mathematically that such solutions do exist. It may even be more probable that solutions in most cases do not exist.

General relativity tries to be a mathematical theory. This does not sound reasonable!

Another question is if nature itself is able to solve difficult nonlinear differential equations. Do we expect too much of the mathematical capabilities of nature?


Hypothesis of simplicity in nature: there cannot exist "pathologically" nonlinear differential equations in nature


The hypothesis in the title claims that nature does not possess "magical" capabilities in solving equations. For every nonlinear equation occurring in nature, there has to exist an "efficient" method to generate approximate solutions.


Atomism


Our simplicity hypothesis is related to atomism.


Atomism claims that there is no truly "continuous" matter. There were conceptual difficulties in imagining how continuous substances could have chemical reactions and form chemical compounds. Greek philosophers Leucippus and Democritus, in the 5th century BC, solved the problem by assuming the existence of atoms.

Their solution turned out to be correct. John Dalton in his book in 1808 noted that chemical elements react in ratios p / q, where p and q are often small natural numbers. In 1808 there already was strong evidence that matter is divided into atoms. Quantum physics confirmed the existence of atoms in the early 20th century. 

In the case of general relativity, the metric of spacetime is assumed to be a continuous "substance" which has astounding mathematical capabilities. It can magically solve nonlinear differential equations – maybe even in cases where there does not exist any solution!

Atomism wants to remove the inherent complexity of continuous substances. Our simplicity hypothesis has a similar goal.


Conclusions


Our simplicity hypothesis has ramifications throughout physics. In quantum field theory, the series of Feynman diagrams seems to diverge, but intermediate results produce astoundingly accurate predictions – up to 14 significant decimals. We will investigate the convergence problem in the future.

In the fall of 2023 we tentatively proved that the Einstein equations break in a pressure change. It might be that the equations do not produce meaningful results for any complex mass distribution. In that case, the correct "metric" of spacetime might have the Ricci curvature tensor non-zero even in vacuum locations.

We will next try to determine an approximate metric for a long cylinder and check if the Einstein equations can produce a reasonable metric.


The Levi-Civita metric around an infinitely long cylinder is very strange. The force of gravity is not ~ 1 / r around the cylinder, where r is the distance from the center of the cylinder.