UPDATE February 23, 2024: The method below assumes that we can sum the (Schwarzschild) metric perturbations caused by each mass element dm at the test mass m. This is probably wrong. The summing is not linear at all. An example: a spherical shell of mass. The metric around it is the familiar Schwarzschild. But if we sum the metric perturbations for each element dm in the shell, we get a very different result, where the tangential metric is stretched a lot and the radial metric only a little.
In the case of the long cylinder, the stretching of the metric in the x direction (below) is probably very little and the stretching to the y direction is more than what we calculate below.
----
We will also calculate the metrics using the Einstein approximation formula and compare to our results.
Summing Scwarzschild metrics using Schwarzschild coordinates
cylinder length 2 L
dm ρ = dm / dx
======================•===
β₀ = angle to end
of cylinder
R = distance (m,
cylinder)
| / r = distance(m, dm)
| /
| / β = angle y axis vs. (m, dm)
•
m test mass
^ y
|
------> x
The test mass m is directly under the center of the cylinder. To obtain the metric at m, we sum the metric perturbation generated by each element dm of the cylinder. We use the Schwarzschild metric and coordinates for dm. The Schwarzschild radius:
r_s = 2 G dm / c².
The mass dm has two effects at the test mass m:
1. it slows down clocks;
2. it squeezes rulers in the radial direction (m, dm).
Let us sum these effects for all parts dm of the cylinder, to obtain the metric at m.
The slowdown of time caused by dm is a perturbation to the Minkowski metric η:
-1 0 0
0 1 0
0 0 1
The perturbation is
dg_tt = r_s / r
= 2 G dm / c² * 1 / (R / cos(β))
= 2 G / c² * 1 / R * cos(β) dm.
The stretching of the spatial metric by dm involves off-diagonal elements in the metric since a square stretched from an oblique angle is no longer a rectangle. However, these off-diagonal elements must cancel at the test mass m, because of the symmetry.
The perturbation to the diagonal elements of the metric is
dg_xx = 2 G / c² * 1 / R * cos(β) sin(β) dm,
dg_yy = 2 G / c² * 1 / R * cos(β) cos(β) dm.
We have to integrate over the whole cylinder. The mass element is
dm = ρ R / cos²(β) * dβ.
We have:
Δg_tt =
β₀
2 * ∫ 2 G / c² * ρ / cos(β) * dβ
0
= 4 G / c² * ρ
* ln( 1 / cos(β₀) + tan(β₀) ).
If β₀ is almost π / 2 then the argument of ln is roughly 2 L / R. The derivative in that case is
d(Δg_tt) / dy = 4 G / c² * ρ * 1 / R.
This corresponds to a gravity acceleration
2 G ρ / R
toward the cylinder, which is the familiar newtonian formula.
The perturbations to the spatial metric:
Δg_xx = 4 G / c² * ρ
β₀
* ∫ cos³(β) sin(β) dβ
0
= 4 G / c² * ρ
* -1/4 (cos⁴(β₀) - 1)
= G / c² * ρ (1 - 1/4 R⁴ / L⁴),
Δg_yy = 4 G / c² * ρ
β₀
* ∫ cos⁴(β) dβ
0
= 4 G / c² * ρ
* (3/8 β₀
+ 1/4 sin(2 β₀) + 1/32 sin(4 β₀)
= 4 G / c² * ρ
* (3/16 π - 3/8 R / L
+ 1/4 * 2 R / L + 1/32 * 4 R / L)
= G / c² * ρ * (3/4 π + 1/4 R / L).
We now have the metric perturbation near the center of a static cylinder:
h =
G / c² * ρ *
4 ln(2 L / R) 0 0 0
0 1 - 1/4 R⁴ / L⁴ 0 0
0 0 3/4 π + 1/4 R / L 0
0 0 0 0
where we assumed that R / L is small.
What about the metric in the z direction in our diagram of the cylinder?
A short movement up from the screen to the z direction is tangential movement for all parts dm of the cylinder. The spatial metric is not stretched in the tangential direction. The perturbation in g_zz is zero. We add a fourth row and column of zeros to the matrix of h.
Orbits of the test mass m
If general relativity is Lorentz covariant, it does not matter if we calculate the orbit of a moving test mass against a static metric, or if we have a static test mass in the field of a moving cylinder.
If the test mass is initially static, then the above metric means that its y acceleration is the newtonian one.
What if the test mass m is moving to the positive x direction at a velocity v? The g_xx metric above is almost exactly constant close to the cylinder. The correcting term is only
~ R⁴.
Thus, the y acceleration for small R is essentially the newtonian one.
===================
^ V
|
|
• --> v
m
^ y
|
------> x
What about m moving to the positive x direction at a velocity v and toward the cylinder at a velocity V? We know from previous blog posts that the test mass gains more inertia as it approaches the cylinder. That is why the x velocity slows down. In the geodesic equation this is reflected by the proper time of m, τ slowing down. Let us check this using the geodesic equation and the metric which we calculated above. We will calculate
d²x / dt².
We have μ = x. A relevant Christoffel symbol is
Γ_tty =
1/2 (dg_tt / dy + dg_ty / dt - dg_ty / dt)
= 1/2 dg_tt / dy.
It contributes to the x acceleration
1/2 dg_tt / dy * V v.
The same contribution
1/2 dg_tt / dy * V v
comes from Γ_tyt. All the others are probably zero because of the symmetry and the diagonality of the metric.
Raising the time index in
Γ^t_ty
flips the sign of the Christoffel symbol, since it is multiplied with 1 / g_tt = -1. We can assume that g_tt is almost exactly -1 if we make the mass of the cylinder small.
Suppose that the test mass moves closer to the cylinder Δy in a time Δt and gains the energy W. The test mass gains inertia worth 2 W / c². Let us calculate the x acceleration with our own methods.
The x velocity v slows down by a fraction
2 W / c² / m.
The acceleration in the x direction is
a_x = -2 W / (m c²) * v / Δt.
If dg_tt / dy = D, then
W = 1/2 D Δy m c².
We get
a_x = -D Δy m c² / (m c²) * v / Δt
= -dg_tt / dy * V v.
The acceleration agrees with the geodesic equation. Our own method is much easier to calculate in your head than the route through the geodesic equation.
Lorentz transformation to get the metric for a moving cylinder
We start from the static metric η + h and switch to a frame which moves at a velocity v to the negative direction of the x axis. The transformation Λ to the coordinates (marked with the prime ') in the moving frame is
t' = γ (t + v x),
x' = γ (x + v t),
y' = y,
where
γ = 1 / sqrt(1 - v² / c²).
The inverse transformation is:
t = γ (t' - v x'),
x = γ (x' - v t'),
y = y'.
We have
R = sqrt(x² + y²),
if we put the origin at the center of the cylinder. Remember that our metric only hold close to the center of the cylinder.
The relevant part of the metric perturbation is:
h =
G / c² * ρ *
4 ln(2 L / sqrt(x² + y²)) [ = T] 0
0 [X =] 1 - 1/4 (x² + y²)² / L⁴
or
dτ² =
-(1 + ρG/c² * 4 ln(2 L / sqrt(x² + y²)) dt²
+ (1 + ρG/c² * (1 - 1/4 (x² + y²)² / L⁴) dx².
Let us denote the tt element of h by T and the xx element by X. Using the formulae:
dt = γ (dt' - v dx'),
dx = γ (dx' - v dt')
we obtain the transformation:
T γ² (dt' - v dx')² + X γ² (dx' - v dt')²
= γ² * ( T dt'² - 2 T v dt' dx' + T v² dx'²
+ X dx'² - 2 X v dx' dt' + X v² dt'² ).
The relevant part of the perturbation matrix in the moving frame becomes:
h_v =
γ² *
T + X v² -(T + X) v
-(T + X) v X + T v²
Very close to the cylinder this is:
h_v =
γ² G / c² * ρ *
4 ln(2 L / R) + v² - (4 ln(2 L / R) + 1) v
- (4 ln(2 L / R) + 1) v 1 + 4 ln(2 L / R) v²
============== --> v
^
| V
|
• m
Let us use the geodesic equation to determine the x acceleration of a test mass m, which in the moving frame has v_y = V and v_x = 0. The relevant Christoffel symbols should be about g_xt and g_tx.
Since dx/dt = 0, the second term in the geodesic equation is zero. The relevant Christoffel symbols are:
Γ_xty = 1/2 dg_xt / dy,
Γ_xyt = 1/2 dg_xt / dy.
The derivative
dg_xt / dy = -(dT + dX) v / dy
= -v dT / dy,
because X is essentially a constant. The x acceleration from the geodesic equation is:
a_x = 1 / g_xx * v dT / dy * dy / dt * dt / dt
= dg_tt / dy * V v
where we assumed that the mass of the cylinder is small, so that g_xx is essentially 1. Our result agrees with the calculation which we made in a frame where the cylinder is static. The flip of sign in a_y comes from the fact that in the last calculation x moves to the left relative to the cylinder.
General relativity should be Lorentz covariant. Then it does not matter if one lets the cylinder move or the test mass m move. The result should be the same. In our example this was true.
Close to the cylinder, and when v is small, the acceleration is
a_x = V v
* d( γ² G / c² * ρ * (4 ln(2 L / R) + v²) ) / dy
= 4 V v G / c² * ρ R * 1 / R²
= V * 4 G / c² * v ρ / R.
This agrees with our calculation on August 14, 2023.
The Einstein approximation formula and the cylinder
We suspect that the Einstein approximation formula (1916) is incorrect. Let us use it to calculate the metric around the cylinder, and let us compare to our own, Schwarzschild-based results above.
For a static cylinder, the "mapping" of the stress-energy tensor to the location of the test mass m gives:
β₀
γ'_tt = 2 * 4 G / c² ∫ ρ / cos(β) * dβ
0
= 8 G / c² * ρ
* ln( 1 / cos(β₀) + tan(β₀) ).
The metric perturbation is obtained by "trace reversing":
h' =
G / c² * ρ *
4 ln(2 L / R) 0 0
0 4 ln(2 L / R) 0
0 0 4 ln(2 L / R)
where we assumed that R / L is small. The tt component is the same as we calculated above from Schwarzschild metrics. But the xx and yy components are completely different.
For a moving cylinder (velocity v), the mass density per length ρ is multiplied by γ² because there is kinetic energy, and the cylinder is length contracted. The components tx and xt get a contribution from the movement of the mass:
h_v' =
γ² G / c² * ρ *
4 ln(2 L / R) 4 ln(2 L / R) v 0
4 ln(2 L / R) v 4 ln(2 L / R) 0
0 0 4 ln(2 L / R)
This is very different from h_v.
Conclusions
Our own Schwarzschild-based method gives sensible and Lorentz covariant results for the cylinder. The Einstein approximation formula produces very different results, and the results are not Lorentz covariant. We believe that the Einstein approximation formula is erroneous. Gravitoelectromagnetism relies on the Einstein formula. This is the reason why results from gravitoelectromagnetism are not Lorentz covariant.
We will next proceed to calculate the approximate metric around a rotating ball, using the Schwarzschild-based method. We will check if orbits of the test mass agree with what we calculated on August 10, 2023.
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