UPDATE April 29, 2024: In this example, we can use symmetry to argue that dg₁₃ / dz, dg₁₃' / dz, and dg₁₃'' / dz are all almost zero at the midpoint of a long cylinder. We do not need to orthogonalize the metric.
Note that the dg₁₃ / dz are not zero at the midpoint. Symmetry only makes the "skew" g₁₃ zero at the midpoint, but not its z derivatives.
^ r barrel distortion
| ____------____
| / | \ coordinate
| | | | lines
--------> z =====
short cylinder
In the barrel optical distortion, the skew "leans" to the right on the left side of the cylinder and to the left on the right part.
----
We use the metric signature (- + + +). The metric g is assumed to be very close to the flat cylindrical metric:
η =
-1 0 0 0
0 1 0 0
0 0 r² 0
0 0 0 1.
Let the pressure in the tangential φ direction be q and the pressure in the radial direction p. We have
p(r) = P / r + Q(r),
where P is a constant. Here
Q'(r) = q / r.
We assume that r and φ are orthogonal in the metric g. There might be a shear stress s between r and φ?
The stress-energy tensor is
Tμν =
ρ 0 0 0
0 p s 0
0 s q r² 0
0 0 0 0
To determine the stress-energy tensor T, we used the formulae of the Karl Schwarzschild 1916 paper of an interior solution. There,
T²₂ = -q
denotes the "proper" positive pressure q which an observer would measure (though the minus sign in -q remains unexplained). The component
T₂₂ = T²₂ g₂₂ = q r².
Karl Schwarzschild uses the metric signature (- - - +), where time is the 4th component.
The trace of the stress-energy tensor T is
Tr = g⁰⁰ ρ + g¹¹ p + g²² q r² + 2 g¹² s
≈ -ρ + p + q.
The Ricci tensor is
Rμν ≈ T - 1/2 Tr g
=
1/2 ρ + 1/2 p 0 0 0
+ 1/2 q
0 1/2 ρ + 1/2 p s 0
- 1/2 q
0 s 1/2 ρ r² - 1/2 p r² 0
+ 1/2 q r²
0 0 0 1/2 ρ - 1/2 p
- 1/2 q
Ricci curvatures
2 R₁₃ = dg' / dz + 1 / r * dg₁₁ / dz = 0,
2 R₀₀ = -g'' - g' / r - d²g / dz² = ρ + p + q,
2 R₃₃ = -g₃₃'' - g₃₃' / r + d²g / dz²
- d²g₁₁ / dz²
+ 2 dg₁₃' / dz
+ 2 / r * dg₁₃ / dz = ρ - p - q,
2 R₁₁ = g'' - g₃₃'' + g₁₁' / r - d²g₁₁ / dz²
+ 2 dg₁₃' / dz = ρ + p - q,
2 R₂₂ / r²
= g' / r - g₃₃' / r + g₁₁' / r
+ 2 / r * dg₁₃ / dz = ρ - p + q.
The first equation gives:
dg₁₁ / dz = -r dg' / dz,
=>
d²g₁₁ / dz² = -r d²g' / dz².
Sum the two last and use the two previous:
-2 d²g / dz² + 2 g₁₁' / r = 2 ρ + 2 p + 2 q
<=>
g₁₁' / r = d²g / dz² + ρ + p + q.
At z = 0, we get from the equation for R₂₂:
g₃₃' = g₁₁' + g' - ρ r + p r - q r
= r d²g / dz² + ρ r + p r + q r
+ g' - ρ r + p r - q r
= r d²g / dz² + g' + 2 p r
=>
g₃₃'' = d²g / dz² + r d²g' / dz² + g''
+ 2 p + 2 p' r.
Do g₃₃' and g₃₃'' satisfy the equation for R₃₃?
Calculate g₃₃' / r + g₃₃'' at z = 0:
g₃₃' / r + g₃₃'' = d²g / dz² + g' / r
+ 2 p
+ d²g / dz²
+ r d²g' / dz²
+ g''
+ 2 p + 2 p' r
= d²g / dz² - d²g₁₁ / dz²
- ρ - p - q
+ 4 p + 2 p' r
= d²g / dz² - d²g₁₁ / dz²
- ρ + 3 (P / r + Q)
+ 2 (-P / r² + q / r) * r
- q
= d²g / dz² - d²g₁₁ / dz²
- ρ + P / r + 3 Q + q.
The equation for R₃₃ says:
g₃₃'' + g₃₃' / r = d²g / dz² - d²g₁₁ / dz²
- ρ + P / r + Q + q.
A contradiction: 3 Q = Q.
The contradiction is the same as in our April 22, 2024 post about an r, z shear inside a cylinder.
What about the equation for R₁₁ at z = 0? It says:
g'' - g₃₃'' + g₁₁' / r - d²g₁₁ / dz² = ρ + p - q
<=>
g₃₃'' = g'' + d²g / dz² + ρ + p + q
+ r d²g' / dz² - ρ - p + q.
=>
p + p' r = q
<=>
P / r + Q - P / r² * r + Q' r = q
<=>
P / r + Q - P / r + q = q
<=>
Q = 0.
The contradiction is the same as for R₃₃.
An analysis of the shear
|
| p₁
|
v
_____
| |
|_____|
^
| p₂
^ r
|
--------> φ
Let the box be 1 × 1 × 1 in the coordinates r, φ, and z. Since the pressure p₁ from up is larger than the pressure p₂ from down, there have to exist shear forces which support the box.
If we would draw the shear forces into the diagram, they would look like the ones which we drew on April 22, 2024.
There is a paradox, though: if the shear s decreases when we go to the right in the diagram, what happens when we have gone a full circle 2 π?
Conclusions
This example is simpler than our April 22, 2024 post where the cylinder had shear between the r and z coordinates. We obtain the exact same contradiction as on April 22.
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