UPDATE April 14, 2024: We added a missing term dg₃₁' / dz to R₁₁. R₀₀ got a new contribution.
There is a serious sign error in g₀₀' and g₀₀'' !
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UPDATE April 13, 2024: The long calculations below certainly contain errors. But if the end result is correct, then there is no vacuum solution for the Einstein equations around a finitely long cylinder. We will write a new blog post where we try to verify the result.
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On April 2, 2024 we tentatively proved that inside most cylindrically symmetric configurations we cannot find
any metric g which satisfies the Einstein field equations.
We used two major assumptions:
1. weak fields: the metric g is very close to the flat Minkowski metric η;
2. the metric g is cylindrically symmetric.
Let us investigate if we can drop the assumption that g has no z dependence.
Empirical observations of the metric near or inside a cylinder: gravity lenses and the Cavendish balance
(Photo ESA/Hubble & NASA Acknowledgement: Judy Schmidt)
Fortunately, gravity lens observations in the sky give us some constraints on how metrics can behave in the real world.
Suppose that we have an elongated object in the sky, like several galaxies in a row. If the gravity lens effect would substantially differ from the one which we obtain by summing perturbations of the metric linearly, we would have observed it.
This proves empirically that g₀₀, g₁₁, and g₃₃ near or inside a cylinder cannot obtain surprisingly large or small values, and their derivatives must behave in a reasonably smooth way.
Another way to measure g₀₀ empirically is to use a Cavendish torsion balance. We can measure the gravity field of various bodies fairly accurately, up to several decimals. If the metric of time g₀₀ would substantially differ from newtonian gravity near a cylinder, experimenters would probably have noticed it.
Yet another way to measure g₀₀ is to observe orbits of planets.
Empirical observations all suggest that the metric does not wildly vary from the one which we get by summing Schwarzschild metric perturbations linearly.
A finite cylinder
^ r
| cylinder
----> z ==============
L <--- F
If the cylinder is of a finite length, then the newtonian gravity potential V is not exactly cylindrically symmetric. There is also weak force component F which points toward the middle point of the cylinder. The force component is ~ 1 / L², where L is the length of the cylinder. We can make the component as small as we like.
The mismatches which we obtained on April 2, 2024 are large. It would be surprising if a small deviation of the metric could remove the mismatch. But let us investigate.
Question. If the cylinder has a uniform mass density and is of a finite length, does that spoil the approximate solution for the metric which we found on April 2, 2024?
We suspect that the Einstein equations only have a solution for an infinitely long uniform cylinder. The Levi-Civita metric is a vacuum solution for an infinitely long cylinder.
The curvature in the time direction R₀₀
On April 2, 2024 we calculated for a cylindrically symmetric orthogonal metric that
R₀₀ = -1/2 g₀₀'' / g₁₁ + 1/2 g₀₀' g₁₁' / (g₁₁)²
- 1/2 g₀₀' / g₁₁ *
(-1/2 g₀₀' / g₀₀ + 1/2 g₁₁' / g₁₁
+ 1 / r + 1/2 g₃₃' / g₃₃)
≈ -1/2 g₀₀'' - 1/2 g₀₀' / r
The solution is
g₀₀(r) = -1/4 ρ r² - C,
if the cylinder has a uniform mass density ρ.
Let us have a cubic constellation of initially static test masses. The curvature R₀₀ tells us how fast does the volume of the cube shrink if the test masses fall freely. The term g₀₀' describes how fast the sides of the cube approach each other, and the term g₀₀'' tells how fast the bottom and the top of the cube approach each other. Intuitively, it would require very large perturbations of the spatial metric to affect the shrinking process appreciably.
We dropped the term
-1/2 g₀₀' / g₁₁ * 1/2 g₃₃' / g₃₃
in the approximation above. The term "competes" with the terms -1/2 g₀₀'' and 1/2 g₀₀' / r.
In order for the term to compete significantly, the value of g₃₃' should be of the order 1. The distortion of the spatial metric caused by a steel sphere of a diameter 10 cm is of the order
rs / r,
where
rs = 2 G M / c²
= 5 * 10⁻²⁷ m
is the Schwarzschild radius of a 4 kg sphere of steel. If the spatial metric distortion inside a cylinder of steel would be as large as ~ 1, it would be easy to observe in practice! We know empirically that the spatial metric inside a cylinder is very close to flat.
Is it possible to affect R₀₀ significantly with a small distortion of the spatial metric? What about skewed coordinates?
The curvature in the z direction R₃₃: allow z dependence and non-orthogonality
The curvature in the t direction cannot be changed without huge changes in the metric of space. The curvature in the z direction is entirely analogous to the time direction. The curvature for a cylindrically symmetric metric is:
R₃₃ = dΓ¹₃₃ / dr
+ (Γ⁰₀₁ + Γ¹₁₁ + Γ²₂₁ + Γ³₃₁) Γ¹₃₃
- 2 Γ³₁₃ Γ¹₃₃
= -1/2 g₃₃'' / g₁₁ + 1/2 g₃₃' g₁₁' / (g₁₁)²
- 1/2 g₃₃' / g₁₁ *
(1/2 g₀₀' / g₀₀ + 1/2 g₁₁' / g₁₁
+ 1 / r - 1/2 g₃₃' / g₃₃)
≈ -1/2 g₃₃'' - 1/2 g₃₃' / r.
Can we change R₃₃ substantially by distorting either g₁₁ or g₃₃ or g₁₃?
Non-orthogonality
If the metric g is slightly skewed, then off-diagonal components g₁₃, etc., have values which slightly differ from zero. The inverse matrix g⁻¹, similarly, has components g¹³, etc., which slightly differ from zero, but have the opposite sign.
The Christoffel symbol terms with the factor
1/2 g¹³
in front of them obtain very small absolute values if all the derivatives in the brackets are small. We can (probably) ignore such terms. The only large derivative is
dg₂₂ / dr.
The symbols
Γ³₂₂ = 1/2 g³¹ * -dg₂₂ / dr = -g³¹ r
and
Γ³₁₁ = g³³ dg₁₃ / dr = g³³ g₃₁'
get new values which substantially differ from zero. The symbol Γ¹₃₃ gets a significant new contribution from the skew:
ΔΓ¹₃₃ = g¹¹ dg₃₁ / dr.
The derivative dΔΓ¹₃₃ / dr contributes to R₃₃:
g¹¹ g₃₁''.
In the products Γ * Γ, the new values contribute significantly if
Γ¹₂₂ = 1/2 g¹¹ * -dg₂₂ / dr = -r / g₁₁
or
Γ²₁₂ = 1/2 g²² * dg₂₂ / dr = 1 / r
is involved:
Γ²₂₁ Γ¹₃₃ = 1 / r * g¹¹ g₃₁'.
Since g¹¹ is very close to 1, we conclude that the skew adds
g₃₁'' + g₃₁' / r
to R₃₃. The terms look a lot like the ones which R₃₃ already has.
On March 10, 2024 we tried to calculate (and made errors) the Schwarzschild metric in cartesian coordinates. There the nondiagonal component of the metric for the x and y coordinates (g₁₂) was very substantial. In the case of a long cylinder we expect the skew to be much less, probably insignificant.
We do not yet make g₀₀ to depend on z
We assume that newtonian gravity describes g₀₀ very accurately. Then, by making the cylinder very long and lightweight, we can make g₀₀ to be essentially independent of z close to the midpoint of the cylinder. This helps us greatly in the calculations.
Make g₃₃ to depend on z
The only significant change is a non-zero value:
Γ³₃₃ = 1/2 g³³ dg₃₃ / dz.
It does not contribute anything significant to R₃₃.
Make g₁₁ to depend on z
ΔΓ³₁₁ = 1/2 g³³ * -dg₁₁ / dz,
Γ¹₁₃ = 1/2 g¹¹ * dg₁₁ / dz.
The derivative
-dΓ¹₁₃ / dz = -1/2 g¹¹ d²g₁₁ / dz²
contributes to R₃₃.
Contributions to R₃₃
The new formula is
R₃₃ ≈ -1/2 g₃₃'' - 1/2 g₃₃' / r
+ g₃₁'' + g₃₁' / r
- 1/2 d²g₁₁ / dz².
We must investigate if this can help us to find a metric which matches a nonuniform cylinder.
Contributions to R₁₁
dΓ³₁₁ / dz = -1/2 g³³ d²g₁₁ / dz²
+ d(g³³ g₃₁') / dz,
R₁₁ ≈ 1/2 (g₀₀'' + g₁₁' / r - g₃₃'')
- 1/2 d²g₁₁ / dz².
+ dg₃₁' / dz.
Contributions to R₂₂
dΓ³₂₂ / dz = -dg³¹ / dz.
R₂₂ ≈ 1/2 r (g₀₀' + g₁₁' - g₃₃')
+ r dg³¹ / dz.
Off-diagonal curvature R₁₃
Γ¹₃₃ = 1/2 g¹¹ * -dg₃₃ / dr = -1/2 g₃₃' / g₁₁,
Γ³₁₃ = 1/2 g³³ * dg₃₃ / dr = 1/2 g₃₃' / g₃₃,
R₁₃ = dΓ¹₁₃ / dr + dΓ³₁₃ / dz
- dΓ¹₃₁ / dr
- dΓ³₃₃ / dr
+ Γ²₂₁ Γ¹₁₃
= d(1/2 g₃₃' / g₃₃) / dz
- d(1/2 g³³ dg₃₃ / dz) / dr
+ Γ²₂₁ Γ¹₁₃
≈ 1 / r * 1/2 g¹¹ * dg₁₁ / dz.
If the stress-energy tensor contains no shear stress, then R₁₃ = 0, and g₁₁ cannot depend on z.
But if we present the Schwarzschild metric in cylindrical coordinates, there g₁₁ does depend on z. We assumed that g₀₀ does not depend on z. Could that explain? Yes.
Updated Ricci curvatures where a skew and variation on z is allowed: gravity lenses in the sky
If we calculated right (probably not), we obtain new equations for a nonuniform finite cylinder.
2 R₀₀ = -g₀₀'' - g₀₀' / r
= ρ(r),
2 R₃₃ = -g₃₃'' - g₃₃' / r
+ g₃₁'' + g₃₁' / r
= ρ(r),
2 R₁₁ = g₀₀'' + g₁₁' / r - g₃₃''
- d²g₁₁ / dz² + dg₃₁' / dz.
= ρ(r),
2 R₂₂ / r² = g₀₀' / r + g₁₁' / r - g₃₃' / r
+ r dg₃₁ / dz
= ρ(r).
At the midpoint of the cylinder we can assume that dg₃₁ / dz = 0 and dg₃₁' / dz = 0. The skew g₃₁ at the midpoint is zero, and so are its derivatives with respect to r.
Is it possible to have an "elongated" metric with no "shear"?
Question. If there is no pressure, then the Ricci curvature must have all nondiagonal components zero. But can that be true if we rotate the coordinate system slightly?
Suppose that the Ricci curvature is non-zero in the x direction and zero in the y direction. What happens in a slight coordinate rotation? It cannot be zero in the y direction?
An analogous problem: suppose that we have a pressure in the x direction but not in the y direction. If we rotate the coordinates, we do not get a shear force. We only have pressure. Thus, a coordinate rotation cannot create a shear force.
Question. If we have an "elongated" system, is it possible that the metric has the Ricci curvature tensor purely diagonal? Does elongation necessarily introduce a "shear" in the Ricci curvature?
If we have a cubic configuration of test masses falling freely in newtonian gravity, the tidal forces probably can squeeze or stretch the cube, but cannot distort it in a "lopsided" way: there is no "shear" which would distort the cube.
But if we add the necessary distortion of the spatial metric, do we get a shear when we move test particles along geodesics to a certain spatial coordinate direction, say, z?We do not have any shear in the Schwarzschild metric or the Levi-Civita metric. But what about a cylinder of a finite length? Such a cylinder is an elongated system.
(Photo amazon.com)
Let us think about a block of rubber. We can certainly squeeze it in a spherically symmetric way without introducing shear. We can also squeeze it in a cylindrically symmetric way with no shear. But can we make the squeezing elongated, like around a cylinder of a finite length, without introducing shear?
We may require that the displacement of the rubber diminishes as we go farther from the squeezed place, and that the displacement very far away is almost spherically symmetric.
Let us have the displacement vector fields D₁ and D₂ for two spherically symmetric squeezes of rubber, with different centers P₁ and P₂.
no pressure
push ---> • • <--- push
P₁ P₂
The displacement field D₁ + D₂ is not sensible because between P₁ and P₂, there would be no pressure to counteract the pressure which pushes from the left and the right. One can easily sum the "metric perturbations" which the squeezing introduces, but the sum is not physically reasonable.
If we replace the rubber block with a compressible liquid, then there must exist a solution where there is no shear force. But the density variations of the liquid may be such that there is no way to realize them in rubber without shear.
The Einstein equations in vacuum around a cylinder
What about the metric in the vacuum around the cylinder? Is it possible to find a metric which is "elongated" and has no off-diagonal components?
We now make g₀₀ to depend on z.
Γ⁰₃₀ = 1/2 g⁰⁰ dg₀₀ / dz,
Γ³₀₀ = 1/2 g³³ * -dg₀₀ / dz.
The off-diagonal curvature is then
R₁₃ = 1 / r * 1/2 g¹¹ * dg₁₁ / dz
- dΓ⁰₃₀ / dr
= 1 / r * 1/2 g¹¹ * dg₁₁ / dz
- d(1/2 g⁰⁰ dg₀₀ / dz) / dr
≈ 1 / r * 1/2 g¹¹ * dg₁₁ / dz
- 1/2 g⁰⁰ d(g₀₀') / dz
≈ 1/2 (1 / r * dg₁₁ / dz + d(g₀₀') / dz).
R₃₃ gets a new contribution:
-dΓ⁰₃₀ / dz = -1/2 g⁰⁰ d²g₀₀ / dz².
R₀₀ gets a new contribution:
dΓ³₀₀ / dz = 1/2 g³³ * -d²g₀₀ / dz².
Suppose that we (miraculously) calculated right in this blog post. What do the equations say about vacuum solutions?
2 R₁₃ = 1 / r * dg₁₁ / dz
+ d(g₀₀') / dz = 0,
2 R₀₀ = -g₀₀'' - g₀₀' / r
- d²g₀₀ / dz² = 0,
2 R₃₃ = -g₃₃'' - g₃₃' / r + d²g₀₀ / dz²
+ g₃₁'' + g₃₁' / r = 0,
2 R₁₁ = g₀₀'' + g₁₁' / r - g₃₃''
- 1/2 d²g₁₁ / dz²
+ dg₃₁' / dz = 0,
2 R₂₂ / r² = g₀₀' / r + g₁₁' / r - g₃₃' / r
+ r dg₃₁ / dz = 0.
Suppose that we have a long cylinder of a finite length L. Can we prove that the equations above cannot be satisfied?
The formula for g₀₀ we get from newtonian gravity. If the metric of time would not be close to the newtonian approximation, we would have noticed that in gravity lenses in the sky.
V gravity potential
• m test mass
|
v F
^ r
|
----> z ==============
L length
= ------------------------> move a segment
Δz
Initially, m is at the midpoint of the cylinder. Let us move a short segment from one end of the cylinder to the other. The length of the segment is Δz. The segment moves from a distance L / 2 - 1/2 Δz to a distance L / 2 + 1/2 Δz. How does the potential V of m change?
The change in V is
~ 2 Δz / (L - Δz) - 2 Δz / (L + Δz)
= 4 Δz² / (L² - Δz²)
~ Δz² / L².
The potential V is something like
V ~ ln(r) + C z² * (1 - 1/2 r² / L²)
~ -g₀₀ + 1,
if we define the midpoint of the cylinder as z = 0. There C is a positive constant. Then
g₀₀' = -1 / r - C z² r / L².
The equation about R₁₃ tells us that
dg₁₁ / dz = -r * d(g₀₀') / dz = 0
=>
d²g₁₁ / dz² = 2 C z r² / L².
We work at the midpoint z = 0. Summing the equations for R₁₁ and R₂₂, and summing the equation for R₀₀ and subtracting the equation for R₃₃ yields:
2 g₁₁' / r = 1/2 d²g₁₁ / dz² + 2 d²g₀₀ / dz²
= C z r² / L² - 4 C
=>
g₁₁' / r = 1/2 C z r² / L² - 2 C.
The equation for R₂₂ gives:
g₀₀' - g₃₃' = -1/2 C z r³ / L² + 2 C r.
=>
g₀₀'' - g₃₃'' = -3/2 C z r² / L² + 2 C.
The equation for R₁₁ gives:
g₀₀'' - g₃₃'' = ...
Maybe we have to modify the cylinder, so that the potential gets a more complicated formula where r and z have a complex interaction? In the Schwarzschild metric,
V ~ -1 / sqrt(r² + z²).
The Rij equations for an arbitrary newtonian potential V
We know that the newtonian gravity potential satisfies ∇²V = 0 in vacuum. Can we prove that all such V satisfy the equations?
Probably not. The potential V has also other features besides having the laplacian zero.
0 = ∇²V = d²V / dx² + d²V / dz²,
where x is a cartesian coordinate to the r direction. The "flow" of lines of force, ∇V, must be zero.
r
^
| ____
| \___/
In cylindrical coordinates, it is
0 = 1 / r * dV / dr
+ d²V / dr² + d²V / dz².
which is equivalent to
1 / r * dg₀₀ / dr
+ d²g₀₀ / dr² + d²g₀₀ / dz²
= g₀₀' / r + g₀₀'' + d²g₀₀ / dz²
= 0
=>
d²g₀₀ / dz² = -g₀₀'' - g₀₀' / r.
R₁₃ = 1/2 (1 / r * dg₁₁ / dz + d(g₀₀') / dz)
= 0.
The equation for R₁₃ says:
dg₁₁ / dz = -r * d(g₀₀') / dz
=>
d²g₁₁ / dz² = -r * d( d²(g₀₀) / dz² ) / dr
= r * g₀₀'''.
At the midpoint (z = 0) we have:
2 R₀₀ = -g₀₀'' - g₀₀' / r
+ g₀₀'' + g₀₀' / r = 0,
2 R₃₃ = -g₃₃'' - g₃₃' / r
- g₀₀'' - g₀₀' / r = 0,
2 R₁₁ = g₀₀'' + g₁₁' / r - g₃₃''
- 1/2 r g₀₀''' = 0,
2 R₂₂ / r² = g₀₀' / r + g₁₁' / r - g₃₃' / r = 0.
The equation about R₀₀ is identically true. It is the well-known fact that the volume of a free-falling cubical configuration of test masses does not change.
We have three equations, for R₁₁, R₂₂, R₃₃, and three unknowns, g₀₀, g₁₁, g₃₃. Is there a solution? There is an additional requirement that g₀₀ must arise from the potential V of some mass distribution. If we fix the mass distribution, then we have three equations for g₁₁ and g₃₃.
We could try the newtonian potential V of two symmetrically placed point masses M on the z axis, at locations -1 and 1.
^ r
|
| M M
-------●------|------●-------> z
-1 0 1
-V(r, 0) = 1 / sqrt(1 + r²)
= g₀₀(r).
We sum the two last equations and eliminate g₃₃ using the equation for R₃₃:
2 g₀₀'' + g₁₁' / r - 1/2 r g₀₀'''
+ 2 g₀₀' / r + g₁₁' / r = 0.
We obtain an equation for g₁₁ which we can solve. A trivial solution for the equation of R₃₃ is to set g₃₃ = -g₀₀.
We found a solution, and did not need to assume anything about the mass distribution, except that it is symmetric relative to z = 0.
If there are any problems in solving the equations, they come from the values z ≠ 0, where the off-diagonal component g₁₃ is non-zero. This is reasonable. In our rubber block squeezing model, we certainly can satisfy the shearlessness condition at the midpoint of the rubber block. If there are any problems with shear, they will arise at the ends of the block.
Actually, this is a way to derive the Schwarzschild metric. Because of the spherical symmetry, the solution at z = 0 must describe the entire metric.
Setting g₀₀ = 1 / r yields:
4 / r³ + 3 / r³ - 2 / r³ = -2 g₁₁' / r
=>
g₁₁' = -5/2 * 1 / r².
We do not understand the origin of the 5/2, but otherwise, the result is as expected.
Test an "elongated" metric
The Schwarzschild metric in vacuum has the "shear" zero. What happens if we try to "squeeze" the metric near a cylindrical source? Far away, it will still be the Schwarzschild metric.
The full equations are listed, once again, below. We can assume that g₀₀ is very close to the newtonian potential V. Then the R₀₀ equation is identically true.
2 R₁₃ = 1 / r * dg₁₁ / dz
+ d(g₀₀') / dz = 0,
2 R₀₀ = -g₀₀'' - g₀₀' / r
- d²g₀₀ / dz² = 0,
2 R₃₃ = -g₃₃'' - g₃₃' / r + d²g₀₀ / dz²
+ g₃₁'' + g₃₁' / r = 0,
2 R₁₁ = g₀₀'' + g₁₁' / r - g₃₃''
- 1/2 d²g₁₁ / dz²
+ dg₃₁' / dz = 0,
2 R₂₂ / r² = g₀₀' / r + g₁₁' / r - g₃₃' / r
+ r dg₃₁ / dz =
Conclusions
Let us close this blog post. This was plagued by calculation errors. In our April 13, 2024 we try to get the formulae right.
It is certain that we cannot get a contradiction by looking at the midpoint z = 0 alone. The Schwarzschild metric is a solution. We have to study the whole range of z values.
