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Suppose that we have a horizontal stressed large thin rubber sheet, such that the tension is uniform throughout its area. We paint polar coordinates on its surface, and also paint inside the thin sheet vertical coordinates.
The painted coordinates will be the global coordinates of a global observer.
The horizontal coordinates correspond to spatial coordinates in the Schwarzschild solution, and the vertical coordinate is the time.
Note that these are not the global coordinates of the standard Schwarzschild solution. Besides r, also the coordinate normal to r will stretch in the rubber. In the standard solution, r is the physical radial distance seen by an outside observer. In the rubber global coordinates it will not be the physical distance.
The rubber will stretch in such a way that its volume stays constant.
Suppose that the rubber sheet is very slippery. We put many small light circular steel disks on the rubber. The disks slide together and cause a depression in the rubber.
Is the solution for the rubber sheet the Schwarzschild solution?
Negative pressure and the Weak Energy Condition
We can model mass and the pressure caused by its gravitation simply by putting small steel disks on the rubber.
We can model positive pressure: take a steel ring and try to fit many small steel disks within. The rubber will bulge so that more disks fit in the area within the ring.
A shear stress on a square of rubber is a negative pressure along the diagonal which wants to grow and a positive pressure along the other diagonal.
A shear stress on a square of rubber is a negative pressure along the diagonal which wants to grow and a positive pressure along the other diagonal.
How can we model negative pressure? In a static setup, a negative pressure is usually accompanied by positive pressure nearby.
The Weak Energy Condition seems to require that the positive mass density always "wins" a negative pressure. In the rubber model, that might mean that if we have a spring pulling some objects together, the mass of the spring will cause the rubber sheet to bulge downward, which means a positive Ricci scalar curvature.
Our anti-gravity device works by causing a large negative pressure in the grid close to a Schwarzschild mass. The Weak Energy Condition may require that the mass of the device must be so big that weight of the mass wins the anti-gravity generated by the negative pressure.
However, there is no proof that the Weak Energy Condition must hold. It is a conjecture.
Positive pressure tends to bulge the rubber sheet and make it thinner. A negative pressure should make it thicker. A thicker sheet means that time flows faster than in the Minkowski space far away.
If we have a steel ring and negative pressure within it, could we use springs to pull more of the rubber sheet inside the steel ring, to make rubber thicker there?
A positive mass stretches the rubber and makes it thinner. A positive pressure works much in the same way. A negative mass and a negative pressure should have the opposite effect.
We have not detected negative mass in nature, but negative pressure certainly exists.
Our anti-gravity device works by causing a large negative pressure in the grid close to a Schwarzschild mass. The Weak Energy Condition may require that the mass of the device must be so big that weight of the mass wins the anti-gravity generated by the negative pressure.
However, there is no proof that the Weak Energy Condition must hold. It is a conjecture.
Positive pressure tends to bulge the rubber sheet and make it thinner. A negative pressure should make it thicker. A thicker sheet means that time flows faster than in the Minkowski space far away.
If we have a steel ring and negative pressure within it, could we use springs to pull more of the rubber sheet inside the steel ring, to make rubber thicker there?
A positive mass stretches the rubber and makes it thinner. A positive pressure works much in the same way. A negative mass and a negative pressure should have the opposite effect.
We have not detected negative mass in nature, but negative pressure certainly exists.
A 2-dimensional rubber sheet is not rigid enough?
Suppose that we do not pre-stress the sheet but attach it to a large circular frame without tension.
Imagine a long equilateral triangle made of the rubber sheet. Assume that we pull with a force F from the sharp tip of the triangle. The negative pressure is then proportional to 1 / x, where x is the distance from the tip. The triangle will stretch an integral of 1 / x from, say 1, to infinity. The integral function is ln(x) and the integral is infinite.
If we pull a 3D rubber cone from the tip, the negative pressure is F / x^2, and the integral function is -1/2 F / x. The integral is finite for an infinitely long cone. The stretching is linearly proportional to F.
Problems with the thin 2-dimensional rubber sheet
If we use a pre-stressed sheet which is attached to a circular frame, then the sheet will probably behave like a harmonic spring if we press it with a finger at the center. Let the pressing small force be F.
Then the vertical depression z of the rubber is
F = k z.
Let the rubber now form an angle α(r) relative to horizontal, where r is the radial distance from the center. The increase in the radius, measured along the new rubber surface is
∫ from 0 to r [1 - cos(α(r))] dr
= ∫ from 0 to r [α(r)^2 / 2] dr
= ∫ from 0 to r [α(r)^2 / 2] dr
Suppose that we double the depression z at the center. The solution z(r) probably is such that the angle α(r) doubles and α(r)^2 is 4-fold. We see that the stretching of the original rubber metric is 4-fold for a 2-fold load F. But in the Schwarzschild solution, the metric stretches linearly with M.
If we do not pre-stress the rubber before attaching it to the frame, then for a very flexible rubber sheet and a strong force F,
F = k z^3,
A circular elastic rubber membrane without pre-stressing, under a load from a weight in the middle
__
(__( dN ________
dR R
What happens with a very light load F is a harder question. Let us have a roughly square piece of rubber of a size dR × dN at some distance R from the center. More precisely, the piece is a sector of an annular ring.
The rubber is probably curved vertically downward along R, but curved upward along the normal N of R.
Let P be a point on the rubber, originally at a distance r from the center vertical axis. Let R(r) be the new distance measured straight from the center vertical axis, and S(r) the distance measured along the rubber surface. Let z be the vertical displacement of the stretched membrane.
Let N be a local normal axis to R at P. Let (R, N) define a local cartesian coordinate system. There is tension radially as well as normal to it. To keep the infinitesimal square of rubber vertically stationary, the force pulling the rubber down must be equal to the force pulling it up. From the partial second derivatives of z we get:
(1) F_R d^2 z(R, N) / dR^2
= -F_N d^2 z(R, N) / dN^2,
where F_R is the radial negative pressure on the square, and F_N is the negative pressure along N. We have assumed that the thickness of the rubber is the same on each side of the square.
The negative pressure F_R is proportional to the radial stretching (strain) of the rubber:
F_R = k (dS(r) / dr - 1),
and the negative pressure F_N is from the stretching of the circle of an original radius r:
F_N = k (R(r) - r) / r.
Let us look from the Internet if someone has solved the equations. A brief search does not return anything. If there is a uniform pressure on the membrane, then this is the well-known soap bubble film problem, and the solution is a part of a spherical surface.
We can write
dS^2 = dR^2 + dz^2.
If dz / dR is small, then
dS = dR (1 + 1/2 (dz / dR)^2).
We still need to calculate what is d^2 z(R, N)/ dN^2:
(R + ΔR)^2 = R^2 + N^2,
2 R ΔR + ΔR^2 = N^2
ΔR = N^2 / (2R),
if N is small. We have
z(R, N) = z(R + N^2 / (2R))
= z(g(N)).
We use the formula for the second derivative of a function composition:
z''(g(N)) * (g'(N))^2 + z'(g(N)) * g''(N)
= z''(g(N)) * (N / R)^2 + z'(g(N)) * 1 / R.
When N = 0, we have
d^2 z(R, N) / dN^2 = dz(R) / dR * 1 / R.
The differential equation (1) becomes:
(R' - 1) (1 + 1/2 (z')^2) z'' = (1 / R - 1 / r) z'.
If z' is small, then we have
(2) (R' - 1) z'' = (1 / R - 1 / r) z'.
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We still need another equation to solve R as a function of r. Our differential equation (1) above was about vertical forces. What about the radial force? The radial negative pressure at the square side whose distance is R, is
(R'(r) - 1) [1 + 1/2 z'(R)^2].
The radial negative pressure at the square side at R + ΔR is
(R'(r + Δr) - 1) [1 + 1/2 z'(R + ΔR)^2].
In this calculation we cannot assume that the thickness of the rubber is the same at R and R + ΔR. Also, the rubber "square" must be treated as a sector of an annular ring. If there is no stress, then the area of the sector at R is
ΔN T,
where T is the rubber thickness. The area at R + ΔR is
(1 + ΔR / R ) ΔN T.
How does pressure affect these? Let us think of a practical example. We have a rubber disk whose radius is 100. There is a weight M in the middle, its radius is 1. The stretching of the rubber is 1% close to M. If cos(α) = 0.99, then α = 0.14. That would correspond to a mild slant close to M.
Let us consider a sector at R = 2. Let ΔR be 0.2. The stretching may make the rubber thickness T to be 1% lower at R = 2, and the thickness T is roughly 0.9% lower at R = 2.2. On the other hand, ΔR / R is 10%. This example shows that we can neglect the changes in rubber thickness within the sector. The dominant thing is that the area of the sector is smaller closer to the center.
In our example, z'(R) may be 0.14, and 1/2 z'(R)^2 = 0.01. Thus:
1 + 1/2 z'(R)^2 = 1 + 1/2 z'(R + ΔR)^2
almost exactly. We then have that the radial force at R is proportional to
R'(r) - 1
and at R + ΔR it is
[1 + ΔR / R(r)] (R'(r + Δr) - 1).
We have ΔR = R'(r) Δr. Thus
R'(r) - 1 = [1 + R'(r) / R(r) Δr]
* (R'(r) + R''(r) Δr - 1)
= R'(r)
+ R''(r) Δr
- 1
+ R'(r)^2 / R(r) Δr
+ R'(r) R''(r) / R(r) Δr^2
- R'(r) / R(r) Δr
<=>
(3) 0 = R''(r) R(r) + R'(r)^2 - R'(r)
<=>
R''(r) = R'(r) [1 - R'(r)] / R(r),
where we let Δr approach zero. If r_f is the radius of the frame where we attached the rubber, then
R(r_f) = r_f.
If r_w is the radius of the weight, we have
R(r_w) > r_w,
because the rubber has stretched horizontally under the weight. If there is no weight, then R(r) = r should be the only solution.
Let R(r) = r + f(r). The equation (3) becomes
0 = f''(r) (r + f(r)) + (f'(r) + 1)^2 - f'(r) - 1
= f''(r) (r + f(r)) + f'(r)^2 + f'(r).
If f(r) is very small and also f'(r)^2 is very small, then we have:
0 = r f''(r) + f'(r),
a linear differential equation. Let us denote f'(r) = y. Then:
0 = r y' + y.
Let us try y = C / r:
0 = -C r / r^2 + C / r.
It holds for any C. Then
f(r) = C ln(r) + C_1
<=>
R(r) = r + C ln(r) + C_1
If there is no weight in the middle, then, for example, R(1) = 1 and R(100) = 100. Then C_1 = 0 and C = 0. The solution is reasonable.
From R(r_f) = r_f we get
0 = C ln(r_f) + C_1.
If R(r_w) = r_w', then
r_w' - r_w = C ln(r_w) + C_1.
We can solve C, C_1:
r_w' - r_w = C (ln(r_w) - ln(r_f))
<=>
C = (r_w' - r_w) / (ln(r_w) - ln(r_f))
C_1 = -(r_w' - r_w) * ln(r_f) / (ln(r_w) - ln(r_f)).
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Let us calculate an example. We set r_w = 1 and r_f = 100. Then
C = -(r_w' - 1) / ln(100)
C_1 = r_w' - 1
R(r) = r + (1 - r_w') ln(r) / ln(100) + r_w' - 1.
If r_w' = 1.1, then
R(r) = r - 0.022 ln(r) + 0.1.
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If we have drawn coordinates on r in an unstretched rubber, then
R'(r) = 1 - C / r
and the new local radial metric is:
ds^2 = (1 - C / r)^2 dr^2.
If C / r is small, the local radial metric is
ds^2 = (1 - 2C / r) dr^2.
In the Schwarzschild solution, the local radial metric is
ds^2 = (1 - r_s / r) dr^2,
but the Schwarzschild global coordinate r is not the same as our rubber r, but rather it is our R(r).
If R(r) is very close to r, then the Schwarzschild metric, indeed, agrees with our rubber metric, if 2C = r_s.
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The equation (2) can be used to solve z(R):
(R' - 1) z'' = (1 / R - 1 / r) z'
<=>
C z'' / r = (r - R) / (R r) z'
= -(C ln(r) + C_1) / r^2 * z',
where we assumed R is approximately the same as r. Let us denote y = z'. Then
C / r y' = -(C ln(r) + C_1) / r^2 * y
<=>
0 = (C ln(r) + C_1) / r * y + C y'
<=>
0 = (ln(r) + C_1 / C) / r * y + y'
<=>
y' = - (ln(r) + K) / r * y,
where K is a constant.
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