On May 29, 2024 we used a rogue metric variation to prove that the Einstein field equations do not have any solution for a rotating mass.
The Einstein field equations have no solution for a rotating mass, but we can still try to build a metric g which is "in the spirit" of general relativity. We can then compare it to the Kerr metric.
Let us recapitulate our argument against the existence.
A rogue metric variation δg proves that the Einstein field equations have no solution for a rotating mass
Let g be the solution of the Einstein field equations for a rotating uniform mass sphere M. M is kept from collapsing by pressure and the centrifugal "force".
rotation
----->
M
• •
• • • •
• • • • m ----> lift higher
• • • •
• • ---> pressure
---> centrifugal
"force"
<-----
rotation
^ y
|
------> x
The action integral S of LM contains:
( kinetic energy of m
- energy of pressure on m
- m c² )
* sqrt(-det(g)).
Let us assume that the spatial coordinates are cartesian. Using a rogue metric variation δg, we move the coordinate lines around a particle m a little bit to the right, relative to the spacetime geometry. The integral of R does not change.
The y metric does not change at m. Let the absolute value of the metric of time |g₀₀| ≈ 1 grow by a small amount ε at m, which means that the proper time at m runs by a factor
1 + ε / 2
faster.
The coordinate velocity of m does not change. Since proper time now runs faster, an observer standing on m measures that the proper kinetic energy of m decreased by a factor
1 - ε.
This is partially compensated by |g₀₀| at m growing by the factor
1 + ε / 2.
Thus, the product
(kinetic energy of m) * sqrt(-det(g))
decreases by a factor 1 - ε / 2.
The product
(energy of pressure on m) * sqrt(-det(g))
is the "pressure potential" of m. The product
m c² * sqrt(-det(g))
tells us the newtonian gravity potential of m. If we move m to the right, the newtonian gravity potential increases more than the pressure potential decreases (remember the centrifugal "force" which supports m).
We conclude that the integral of LM decreases if we move m to the right. This contradicts the assumption that g is a solution of the Einstein field equations.
Stretched radial metric grr in cartesian coordinates
Let us have a mass element dm at x, y coordinates (0, 0). We have an observer at coordinates (x, -y), where |y| >> |x|. What is the metric in cartesian coordinates?
We have:
r² = x² + y²
=>
r dr = x dx + y dy
=>
dr = x / r * dx + y / r * dy.
Let dn be the normal to dr:
dr² + dn² = dx² + dy².
The metric:
ds² = (1 + hrr) dr² + dn²
= hrr dr² + dr² + dn²
= hrr dr² + dx² + dy².
There:
hrr dr² = hrr *
(x² / r² * dx²
+ 2 x y / r² * dx dy
+ y² / r² * dy²).
Lorentz transforming a Schwarzschild metric perturbation
We want to calculate the metric produced by a moving mass element m. The metric perturbation h created by m is approximately Schwarzschild, and in many cases we can obtain an approximate solution of the Einstein equations by summing metric perturbations.
^ y
|
|
m ●----> vx
|\
| \
| \
vy v v
R
• observer (x, -y)
The mass element m is at x, y coordinates (0, 0). The observer is at (x, -y), where
|x| << |y|.
We denote x² + y² = R².
The components of the velocity v are vx and vy.
Let η be the flat metric. We want to transform, at the observer,
η + h
to coordinates moving at a velocity v. We will transform twice, for each component of v. Let us calculate the transformation for vx.
The metric at the observer is:
c² dτ² = -(1 - rs / r) c² dt²
+ (1 + rs / r * x² / r²) * dx²
+ (1 + rs / r * y² / r²) * dy²
+ 2 rs / r * x y / r² * dx dy
+ dz²,
where rs = 2 G m / c² and r = -y. Lorentz says:
dt = dt' + vx / c² * dx',
dx = dx' + vx dt',
dy = dy',
dz = dz',
c² dτ² ≈ -(1 - vx² / c²) (1 - rs / r) c² dt'²
+ (1 + rs / r * x'² / r²) * dx'²
+ (1 + rs / r * y'² / r²) * dy'²
+ 2 rs / r * x' y' / r² * dx' dy'
+ dz'²
+ (2 vx - 2 vx + rs / r * 2 vx) dt' dx',
where we used the facts that x' ≈ x, y' = y, and that the product rs vx * |x|/ r is very small. The value of r is approximately the same in the new coordinates.
The difference to the static metric is
vx² dt'²
and the cross term
2 vx rs / r * dt' dx'.
The velocity vx makes the metric perturbation of time t to "spill over" to the cross term dt' dx'.
The velocity vy, obviously, has a similar effect. A difference is that also the perturbation of the y metric spills over to the cross term.
Summing the effects of vx for a rotating circle
We must sum the term for each mass element dm:
rs / r * vx * dt' dx'.
Let
r' = distance (observer, center of circle).
• dm'
\
\
\
×
| \ ^ ω
| \ R /
| \
| • dm M
α
r' = distance(×, observer)
• observer
Let us integrate it for a circle of a radius R, which turns at an angular velocity ω. The mass of the circle is M. The distance of dm from the observer is:
r = sqrt( (r' - R cos(α))² + R² sin²(α) )
≈ r' (1 - R / r' * cos(α)).
Since vx has opposite values for dm and its mirror image dm' relative to the center of the circle, we can sum their contribution. We have:
rs = 2 G / c² * M / (2 π) * dα,
vx = R ω cos(α),
π / 2
h₀₁ = ∫
-π / 2
2 G / c² * M / (2 π)
* R ω cos(α)
* 1 / r' * (1 + R / r' * cos(α)
- 1 + R / r' * cos(α))
* dα.
The integral
π / 2
∫ 2 cos²(α) dα = π.
-π / 2
We get:
h₀₁ = G / c² * M ω R² / r'²
= G / c² * J / r'²,
where J is the angular momentum of the circle.
The part of the metric perturbation due to the rotation of the disk or the sphere is:
0 G / c² * J / r'² 0 0
G / c² * J / r'² 0 0 0
0 0 0 0
0 0 0 0.
Summing the effects of vy for a rotating circle
We studied this extensively in August 2023.
The symmetry of the disk makes the cross term of t' y' in the metric zero if the observer is at (0, -y) in the diagram.
^ y'
|
● -----> x'
M (0, 0)
r'
• observer (0, -y')
<----> cross term of t' x'
When the observer moves to the x direction, the cross term will have a non-zero value. But is the value negligible?
Let the observer move to a position (x₀', -y₀'), where |x'| is small. The "radial" coordinates there are rotated through an angle x₀' / y₀' = β.
x'' = x' + β y',
y'' = y' - β x'.
Let us have a cross term in the metric at the new position:
C t' x'' = C t' x' + C t' β y'.
The cross term t' y' in the old coordinates is
C x₀' / y₀' * t' y'.
We have
C = 2 rs / r * vx.
If we make y₀' large, then the cross term of t' and y' is negligible relative to the cross term of t' and x'.
It looks like that vy would not contribute at all. The reason is that each dm has a centripetal acceleration.
We are not sure if this is the correct "approximate" solution in general relativity. If the mass M is lightweight, then a negative pressure keeps the parts dm in the circular orbit. If we let a rogue variation to move a part dm farther from the center of the circle, then the action integral of the matter lagrangian LM decreases. That shows that the Einstein field equations have no solution.
The fact that the t' y' cross term is negligible is due to the same reason why the Einstein equations have no solution! It does not sound like a proper approximate solution!
Test mass m acceleration due to the rotation of M
Now we can calculate the gravitomagnetic effects on m.
J
| ● M
v ω
r'
^ V
|
• m
^ y
|
-------> x
The x coordinate acceleration is
d²x / dτ² = -Γ¹₀₀ * dt / dτ * dt / dτ
- 2 Γ¹₀₂ * dt / dτ * dy / dτ
- Γ¹₂₂ * dy / dτ * dy / dτ
= -2 * 1/2 g¹¹ dg₀₁ / dy * 1 * V
- 2 * 1/2 g⁰¹ dg₀₀ / dy * 1 * V
≈ -V dg₀₁ / dy
= V * G / c² * 2 J / r'³.
The gravitomagnetic moment, as defined by our August 10, 2023 blog posts, is 2 J. It is 4 times the analogous magnetic moment of a rotating electric charge. This is probably the origin of the strange 4-fold gravitomagnetism in literature.
Tangential movement of a test mass
Above we calculated the effect of vx if the test mass m approaches M. What about m moving tangentially?
● M
• -----> V
m
d²y / dτ² = -Γ²₀₀ * dt / dτ * dt / dτ
- 2 Γ²₀₁ * dt / dτ * dx / dτ,
Γ²₀₀ = 1/2 g²² * -dg₀₀ / dy
= 1/2 * (1 + rs / r) * rs / r²
= 1/2 rs / r² + 1/2 rs² / r³,
Γ²₀₁ = 1/2 g²² * -dg₀₁ / dy
≈ 1/2 * -G / c² * 2 J / r'³.
= -G / c² * J / r'³.
Since both sides in the equation for d²y / dτ² contain ... / dτ², we can instead write .../ dt².
The y coordinate acceleration due to vx and V is
d²y / dt² = V * G / c² * 2 J / r'³.
This agrees with the previous section. The gravitomagnetic moment is 2 J.
On top of that we have the regular
1/2 rs / r² = G M / c² * c² * 1 / r²
= G M / r²
gravity acceleration.
But is the y coordinate "natural" here? If we make coordinate lines to "bulge", we can obtain spurious coordinate accelerations, which depend on V.
A single moving mass element dm creates a time-dependent metric. We have forgotten that in the analysis above!
The gravitomagnetic field of a single moving mass
Let us calculate a very simple example.
M ● ----> v
r
m • -----> v
^ y
|
• ----> x
observer
The mass M is moving at a velocity v to the x direction, and so is the test mass m.
If both were static in our coordinates, then a static observer would measure a y acceleration of
d²y / dt² = G M / r²,
where r is the Schwarzschild radial coordinate of m.
But proper time in the system M, m is slower than for the observer. He measures an acceleration of
d²y / dt² = (1 - v² / c²) G M / r².
The term
v * v / c² * G M / r²
can be interpreted as a "magnetic" effect. It corresponds to a gravitomagnetic field of
v / c² * G M / r²,
which is analogous to a magnetic field
B = v / c² * E.
There is no factor 4.
Let us use the geodesic equation to calculate the acceleration. The metric in the laboratory coordinates is:
c² dτ² ≈ -(1 - v² / c²) (1 - rs / r) c² dt²
+ (1 - v² / c²) * dx²
+ (1 + rs / r) * dy²
+ dz²
+ rs / r * 2 v dt dx,
where
rs = 2 G M / c².
The acceleration:
d²y / dt² = -Γ²₀₀ * dt / dt * dt / dt
-2 Γ²₀₁ * dt / dt * dx / dt
= (1 + v² / c²) G M / r²
- 2 * 1/2 g²² * -dg₀₁ / dy * v
≈ (1 + v² / c²) G M / r²
- 1 * 1 * rs / r² * v * v
= G M / r²
- v² G M / c².
M
● --> v
^ V
|
• m
^ y
|
-----> x
Let us calculate the x acceleration.
d²x / dτ² = -2 Γ¹₀₂ * dt / dτ * dy / dτ,
= -2 * 1/2 g¹¹ dg₀₁ / dy * 1 * V
- 2 * 1/2 g⁰¹ dg₀₀ / dy * 1 * V
= -V d(v rs / r) / dy
= V v * 2 G M / c² * 1 / r².
What is wrong in the above calculation?
An obvious problem is that dm is accelerating. Can we sum the metric perturbations as if the mass elements dm were moving at a constant velocity?
We wrote at length about this problem in August 2023.
Now we know that the Einstein equations do not have a solution at all for a rotating system, because it is "dynamic".
Conclusions
If we want to preserve Gauss's law for newtonian gravity for a gravitational wave, we need magnetic gravity which is exactly analogous to magnetism. The condition R₀₀ = 0 states Gauss's law for gravity.
In the above calculations, we were able to track the origin of the strange factor 4 in literature, of the gravitomagnetic effect of a rotating sphere.
If gravitomagnetism for gravitational waves is exactly analogous to electromagnetism, then it would be surprising if "static" gravitomagnetic fields differ from electromagnetism by the factor 4. How does nature know to make the effect 4-fold in certain configurations?
We will investigate this in the next blog post.
No comments:
Post a Comment