Once we found the correct (?) approximate equations for the Ricci curvatures, it is surprisingly easy to find approximate solutions of g for any mass distribution in cylindrical coordinates.
Our equations do not contain any "second order" terms where there is a product of two metric perturbations. Finding exact solutions for the Einstein equations is notoriously difficult. Apparently, the second order terms make it difficult.
Nonuniform radial pressure and shear
The equations can be solved easily if the radial pressure is p / r, where p is a constant. What about more complicated pressure fields?
| p₁ > p₂
|
v
s shear
^ (F₁ + F₂) / 2
| <------
______
F₁ ^ | | | F₂ < F₁
| |______| v
------>
(F₁ + F₂) / 2
^
| p₂
------> s shear
^ r
|
-------> z
The radial pressure is initially
P / r,
where P is a constant. If we use a shear force F, we can manipulate the radial pressure. Let us set the shear s on the r, φ plane to the direction r to
s(r, z) = K - S(r) z,
where K is a (largish) constant > 0 and S(r) > 0.
Then the shear reduces the pressure p as we go closer to the center of the cylinder. Let in the diagram, the box be a cube 1 × 1 × 1. The net force |F₁| - |F₂| which lifts the box up is
S(r).
The pressure is then:
p(r) = P / r + Q(r),
where P is a constant, and Q'(r) = S(r).
We can assume an orthogonal metric
We assume that the metric is cylindrically symmetric. Can we assume that the metric is orthogonal?
If we start from the flat metric in cylindrical coordinates, then the metric is orthogonal, of course. But if we perturb the metric in a cylindrically symmetric way, then the coordinate lines of r = constant and z = constant can start to "bulge", when viewed in the perturbed metric. The coordinate lines are no longer orthogonal, when measured in the perturbed metric.
Can we define new coordinates in such a way that the coordinate lines become orthogonal again? It is obvious (?) that we can draw almost horizontal Z = constant lines in such a way that they intersect the r = constant lines at right angles. The new coordinates r, Z have an orthogonal metric.
Thus, we can assume that the metric g has no off-diagonal components.
The stress-energy tensor
The stress-energy tensor is
T =
ρ 0 0 0
0 p 0 s
0 0 0 0
0 s 0 0,
and its trace
Tr = g⁰⁰ ρ + g¹¹ p + 2 g¹³ s
= -ρ + p - 2 g₁₃ s
= -ρ + p.
The approximate Ricci tensor
R = T - 1/2 Tr g
=
1/2 ρ + 1/2 p 0 0 0
0 1/2 ρ + 1/2 p 0 s
0 0 1/2 ρ r² - 1/2 p r² 0
0 s 0 1/2 ρ - 1/2 p
Ricci curvatures
2 R₁₃ = dg' / dz + 1 / r * dg₁₁ / dz
= 2 K - 2 S z,
2 R₀₀ = -g'' - g' / r - d²g / dz²
= ρ + P / r + Q,
2 R₃₃ = -g₃₃'' - g₃₃' / r + d²g / dz²
- d²g₁₁ / dz²
+ 2 dg₁₃' / dz + 2 / r * dg₁₃ / dz
= ρ - P / r - Q
2 R₁₁ = g'' - g₃₃'' + g₁₁' / r - d²g₁₁ / dz²
+ 2 dg₁₃' / dz = ρ + P / r + Q,
2 R₂₂ / r² = g' / r - g₃₃' / r + g₁₁' / r
+ 2 / r * dg₁₃ / dz = ρ - P / r - Q.
The first equation gives:
dg₁₁ / dz = -r dg' / dz + 2 K r - 2 S z r
=>
d²g₁₁ / dz² = -r d²g' / dz² - 2 S r
= -r d²g' / dz² - 2 Q' r.
Sum the two last and use the two previous:
-2 d²g / dz² + 2 g₁₁' / r = 2 ρ + 2 P / r + 2 Q
<=>
g₁₁' / r = d²g / dz² + ρ + P / r + Q.
g₃₃' = g₁₁' + g' - ρ r + P + Q r
= r d²g / dz² + ρ r + P + Q r
+ g' - ρ r + P + Q r
= r d²g / dz² + g' + 2 P + 2 Q r
=>
g₃₃'' = d²g / dz² + r d²g' / dz² + g''
+ 2 Q + 2 Q' r.
g₃₃' / r + g₃₃'' = d²g / dz² + g' / r
+ 2 P / r + 2 Q
+ d²g / dz²
+ r d²g' / dz²
+ g'' + 2 Q + 2 Q' r
= d²g / dz²
+ r d²g' / dz² + 2 Q' r
+ 2 P / r + 4 Q
+ g'' + g' / r + d²g / dz²
= d²g / dz² - d²g₁₁ / dz²
+ 2 P / r + 4 Q
- ρ - P / r - Q
= d²g / dz² - d²g₁₁ / dz²
- ρ + P / r + 3 Q.
g₃₃'' + g₃₃' / r = d²g / dz² - d²g₁₁ / dz²
- ρ + P / r + Q.
We obtain a contradiction: 3 Q = Q.
Working with shear stresses
We are not familiar with working with shear stresses. The calculation above may be erroneous.
Is it possible to have shear stresses with such simple functions:
T₁₃ = T₃₁ = K - S(r) z ?
The torques of the shear stresses T₁₃ and T₃₁ cancel each other out. Our box in the above diagram will not start to spin. The shear causes the radial pressure p to be larger at larger r.
The shear does not produce a pressure in the z direction because the forces to the left and to the right are equal in the diagram.
Conclusions
We were able to derive the exact same contradiction in our April 24, 2024 post where we manipulate the radial pressure by putting a tangential pressure on the coordinate φ.
As if general relativity cannot handle pressure which changes spatially. Shear forces are then involved.
Recall that we on November 5, 2023 tentatively proved that general relativity cannot handle a pressure which changes in time.
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