UPDATE September 3, 2023: The assumption that the acceleration of the parts of disk does not affect the acceleration of the test mass, is incorrect. If we let the parts "fly loose", that significantly changes the system. See our post today.
----
UPDATE August 18, 2023: We had omitted the Schwarzschild correction of the radial velocity on the radial acceleration. We added it below, and now the frame dragging is to the positive direction of the rotation.
We will continue checking the proofs. It would be great if we would find the complete equations of motion for the Schwarzschild metric calculated somewhere.
----
<- rotation
● rotating ball
^ V
|
• --> acceleration
test mass
Definition of frame dragging. Let us have a test mass radially approaching a rotating, spherically symmetric ball at a slowish velocity V. We define frame dragging as the sideways acceleration of the test mass. The acceleration is measured by a faraway observer.
The analogy between electromagnetism and gravitoelectromagnetism
We have the following definition different from Wikipedia: we define the gravitomagnetic field B_g by the formula on the force on a test mass m moving at a velocity V:
F = m V × B_g,
while in Wikipedia the definition is
F = 4 m V × B_g.
In gravitoelectromagnetism, we have the analogies:
G ~ 1 / (4 π ε₀) = Coulomb's constant;
G / c² ~ μ₀ / (4 π).
A magnetic moment m produces a magnetic field:
We define the gravitomagnetic moment of a rotating mass through the analogy: a rotating mass has a gravitomagnetic moment m_g if it in its equatorial plane produces a gravitomagnetic field
B_g(r) = G / c² * m_g / r³,
or, equivalently, a force
F = m V × G / c² * 1 / r³ * m_g.
Magnetic moment versus "charge angular momentum"
Let us have a circular loop where a uniformly distributed charge Q is rotating at a velocity v. The radius of the loop is r.
<-- v
_____
/ \ r = radius
| |
\______/
Q
We may define the charge angular momentum as the vector
J_e = Q v r,
which points straight out of the screen in the above diagram, for a positive charge Q. For a negative charge Q, the vector points inside the screen.
The magnetic moment of the loop is the vector
m = I S,
where I is the electric current and S is the area of the loop.
We have
I = Q / (2 π r) * v,
and
m = Q / (2 π r) * v * π r²
= 1/2 Q v r
= 1/2 J_e.
We see that the magnetic moment m is 1/2 of the charge angular momentum J_e. Since a rotating, uniformly charged disk consists of many circular loops, its magnetic moment m, similarly, is 1/2 J_e.
Lorentz transformations
Frame dragging by a lightweight, slowly rotating uniform ball
We assume that the ball is not heavy, it has a uniform density, and its density is low. Then the gravity field is weak and we can use the linearity of solutions conjecture. We work in a frame where the center of the ball is static.
<-- ω
_____
/ \ R₀ = radius
| | M
\______/
^ V
|
|
m • --> a acceleration
We divide the ball into small parts and calculate the frame dragging individually for each part. The total frame dragging is the sum over all parts. We will concentrate on a thin equatorial disk in the plane of the test mass and the center of the ball.
center of ball
X
^ v
/
dm' • • dm
\
v' v
^ V
|
|
• m
^ x
|
---------> y axes of the laboratory frame
Above we have the general case of the test mass m and two parts of the ball, dm and dm' in a plane which passes through m and the center of the ball X. The part dm' is the mirror image of dm against the line (m, X). The distance (X, dm) is R and the velocity
|v| = ω R.
We assume that both V and v are very small and
v << V.
The sum of newtonian accelerations calculated for dm and dm' points directly from m to X in the laboratory frame. Our task is to calculate corrections to the newtonian gravity of dm and dm'. In newtonian gravity we ignore kinetic energy. The corrections are of two types:
1. Schwarzschild corrections because of the metric, and
2. Lorentz corrections as we transform accelerations from a comoving frame of dm to the laboratory frame.
Switch to a comoving frame of dm
We switch to a comoving frame of dm. We can then use the Schwarzschild metric around dm.
Assumptions about the metric around an accelerating mass. The Schwarzschild metric is negligibly distorted by the acceleration of dm. We will consider a very short time interval. We assume that we can then ignore the acceleration of dm and treat the comoving frame as an inertial frame.
We want to find out the y acceleration of m in the laboratory frame. It comes from relativistic corrections.
R_m = distance (m, X)
X center of ball
| \
| \ R
| \
| \
α ● dm is static
^ V' = V - v
|
| β angle (m, X) vs. (m, dm)
| /
|/
•
m
Above we have a comoving frame of dm. The distance (m, X) is R_m. The distance of dm from the center of the ball X is R and α is the angle between (X, m) and (X, dm). The angle (m, X) versus (m, dm) is β.
We have to be careful with Lorentz transformations. Time dilation and length contraction are ~ v² / c², and we can ignore them, just like any square term of v.
The Lorentz correction to V - v is roughly
~V * v V / c².
We can ignore it because its ratio to v is
~ V² / c²,
and by choosing V small, the ratio is extremely small. However, we have to check below that a small error like that cannot become magnified by some mechanism. We must make sure that it stays insignificant relative to the ~ v V correction terms which we calculate. The same holds for length contraction and time dilation: we must check that they do not get magnified.
The velocity of m relative to dm,
V' = V - v,
is the important vector.
Let us decompose V' into a radial velocity vector which points toward dm:
v_r = V cos(β) - v sin(α + β),
and the tangential velocity orthogonal to it:
v_ t = -V sin(β) - v cos(α + β).
Conservation of angular momentum - the Schwarzschild correction to the tangential velocity and tangential acceleration
We use conservation of angular momentum in the Schwarzschild metric (see our August 5, 2023 post).
The specific angular momentum h stays constant for the test mass. L denotes the angular momentum measured by a faraway observer, μ = m is the reduced mass of the test mass, dτ is a short proper time measured by the test mass, and dφ is the change in the angle φ in that time in the polar Schwarzschild coordinates.
● dm r = distance(m, dm)
^ v_r
|
|
v_t <-- • m
Let us study the acceleration of m in the above diagram. We have
dτ = dt * (1 - 1/2 r_s / r)
* (1 - 1/2 V''² / c²),
where r_s = 2 G dm / c² and V'' is the velocity of the test mass measured by a local observer who is static relative to dm. We assume that r_s and V'' are small.
If dt is short, then the line (m, dm) only turns a little, and the radial acceleration contributes negligibly to the tangential acceleration.
In newtonian gravity, there is no tangential acceleration at all. The formula above for dφ / dτ tells us that the Schwarzschild orbit differs from newtonian in tangential movement.
Let m move a short radial distance Δr in a short time Δt. If we let the time Δt be very short, then Δr is essentially the same for newtonian movement. The locally measured velocity increases by ΔV''.
In newtonian movement,
r² dφ / dt
is a constant. But in the Schwarzschild metric we have to use dτ instead of dt, and correct the newtonian movement:
dτ = dt * (1 - 1/2 r_s / (r - Δr))
* (1 - 1/2 (V'' + ΔV'')² / c²).
The Schwarzschild and newtonian tangential velocities v_t agreed at the start of the interval Δt, but at the end they differ. For Schwarzschild, dτ is then less than at the start: dφ has to be less, too. "Frame dragging" by the mass dm slows down tangential movement of m if m approaches dm.
The relative difference in dτ is
(1 - Δr * 1/2 r_s / r²)
* (1 - ΔE_kin / (m c²)),
when Δr is small.
There ΔE_kin = m c² * Δr * 1/2 r_s / r², because potential energy of m is converted to kinetic energy.
We see that the tangential velocity v_t changes by
-v_t * 2 Δr * 1/2 r_s / r²
in the time Δt.
Schwarzschild tangential acceleration correction
Since Δr / Δt = v_r, this amounts to a tangential acceleration correction of
-v_r v_t * r_s / r²
in the Schwarzschild coordinates of dm.
We will study below the four terms in -v_r v_t r_s / r²:
V² cos(β) sin(β) r_s / r²
+ V cos(β) v cos(α + β) r_s / r²
- V sin(β) v sin(α + β) r_s / r²
- v sin(α + β) v cos(α + β) r_s / r².
Corrections to the radial acceleration toward dm
There is a Schwarzschild correction which comes from a radial movement, and a Lorentz correction when we transform the newtonian gravity acceleration a' to the laboratory frame.
The gravity field of dm is very weak and v and V are slow. The acceleration is almost the newtonian one:
a' = G dm / r².
However, we have to correct the radial acceleration in the Schwarzschild metric for the same reason as we corrected the tangential acceleration above. Let us have a test mass approaching dm radially.
As the test mass m approaches dm, it gains more inertia and its original velocity component v_r (before the journey) slows down. Let us consider a very short journey.
The proper time τ of the test mass slows down relative to t. Then dr in the above formula must be less than in the corresponding newtonian orbit. The correction is -v_r² r_s / r². This is
-v_r² r_s / r² * sin(β)
in y acceleration. The corrections for dm' and dm almost cancel out each other, but leave for dm
da_y = 2 V cos(β) v sin(α + β) r_s / r² * sin(β)
= 4 G / c² * dm / r² * v sin(α) V sin(β),
where we assumed that β is small. We do not Lorentz transform da_y, because the transformation would change it only a little.
Above we have the Lorentz correction from a' when we move to the laboratory frame. The comoving frame is marked with the prime'.
We can ignore the middle term, since it is ~ v². The Lorentz correction from the first term is
-a' (2 v • V' / c²),
and the third term
-V' (a' • v / c²).
We can replace V' by V because the value of the terms only changes by ~ v².
We assume that R_m is large. Then we can treat a' and V as parallel vectors and we can easily sum the two terms:
-3 a' * v • V / c².
The correction for the y acceleration is
-3 G / c² * dm / r² * v sin(α) V * sin(β).
We add this to the Schwarzschild correction da_y above, and obtain
G / c² * dm / r² * v sin(α) V * sin(β).
If β is small and R_m is large relative to R, then sin(β) = sin(α) R / R_m and we can approximate 1 / r² = 1 / R_m². Recall that v = ω R.
The acceleration is
ω V G / c² * 1 / R_m³ * sin²(α) R² dm.
where
dm = ρ dα R dR.
We integrate over a thin equatorial disk of the rotating ball. Above, ρ is the mass per area of that disk.
We integrate over 0 < α < π and 0 < R < R₀. If we keep R constant, then the integral over sin²(α) is π / 2. The integral over R³ is R₀⁴ / 4.
The integral over the half disk where β > 0 is the same as for β < 0.
The integral over the entire disk is
1/2 ω V G / c² * 1 / R_m³ * ρ π R₀² * R₀² / 2.
We have
ρ π R₀² * R₀² / 2
= M R₀² / 2
= I,
where I is the moment of inertia of the disk. Since ω I = J, we have as the acceleration:
V G / c² * 1 / R_m³ * 1/2 J,
or force:
F = m V * G / c² * 1 / R_m³ * 1/2 J.
The gravitomagnetic moment m_g from this term is 1/2 J.
Lorentz transformations to the laboratory frame of the four terms in the Schwarzschild tangential acceleration correction
The first term V² cos(β) sin(β) r_s / r² would dominate because V >> v, but we eliminate it by summing to the corresponding term for the mirror image dm': sin(β) has the opposite sign for dm'. It would cancel entirely without Lorentz corrections.
Concerning the accuracy of V, note that an error ~ V² v in its value only results in an error ~ V³ v in the first term, and is negligible relative to other corrections. We do not need to worry that we did not Lorentz transform V to the comoving frame.
Looking at the Wikipedia acceleration formulae above, the Lorentz correction to a tangential acceleration a' is approximately the factor
-a' (3 v • V / c²),
The Lorentz correction differs for dm and dm'. The sum of the corrected first term for dm and dm' is then at most
6 |v| |V| / c² * V² cos(β) sin(β) r_s / r².
If we choose v and V very small, the sum is negligible relative to the 2nd and the 3rd terms.
• dm''
X center of the ball
• dm
^ x
|
---------> y
The second term V cos(β) v cos(α + β) r_s / r² would dominate, but we reduce it by summing it to the corresponding term for dm'', where dm'' is the mirror image of dm as in the diagram above. Then cos(α) has the opposite sign for dm'', but the factor 1 / r² causes a difference.
We have to check if the Lorentz transformation affects the sum much. The sum (= acceleration) is something like
V v cos(α) * 2 R r_s / r³
without the Lorentz transformation. The Lorentz correction to the acceleration is at most
6 |v| |V| / c² * V cos(β) v cos(α + β) r_s / r³.
By making v and V small we can make the correction negligible.
The third term -V sin(β) v sin(α + β) r_s / r² does not change sign for 0 < α < π. The effect of the Lorentz correction can be made negligible by choosing v and V small.
The fourth term v sin(α + β) v cos(α + β) r_s / r² is negligible if we choose v very small.
We conclude that the 2nd and the 3rd term contribute to the tangential acceleration of the test mass m. We have to multiply them by cos(β) to obtain the y acceleration in the laboratory frame:
ω V * 2 G / c² * dm R / r²
* ( cos(β) cos(β) cos(α + β)
- cos(β) sin(β) sin(α + β) ).
Above, R, r, α, β vary according to the part dm. We have used the formula
r_s = 2 G / c² * dm.
Integrating the 3rd term ("dm' approaches faster")
The origin of the 3rd term is the following: the velocity vector V points to the left of dm. If the test mass dives fast into the field of dm, then that leftward motion decelerates, that is, the test mass accelerates to the right. It is a mirror image for dm'. Since the test mass approaches dm slower than it approaches dm', the sum is an acceleration to the left, negative frame dragging.
If β is small and R_m is large relative to R, then sin(β) = sin(α) R / R_m and we can approximate 1 / r² = 1 / R_m². The acceleration is:
-ω V * 2 G / c² * dm R / R_m²
* sin(α) R / R_m * sin(α)
= -2 ω V G / c² * 1 / R_m³ * sin²(α) R² dm.
The integral is -2 times the integral for the radial acceleration above. The associated gravitomagnetic moment is -J.
Integrating the 2nd term ("dm moves to the right")
The origin of the 2nd term is simple: when the test mass dives deeper into the field of dm, it is pulled to the right along with dm. The pull of gravity of dm'' is smaller: it cannot cancel the effect. This is positive frame dragging.
The 2nd tangential acceleration term is
ω V * 2 G / c² * dm R / r²
* cos²(β) cos(α + β).
We can drop cos²(β) from the formula, because it differs from 1 at most approximately
1/2 R₀² / R_m²
and the 1 / r² factor makes the resulting error from dropping that factor
~ 1 / R_m⁴,
while the integral over the term is ~ 1 / R_m³.
We should integrate the following over the equatorial disk to get the force F on the test mass:
F = 2 m V * G / c² * dm ω R cos(α + β) / r².
If we map the formula to electromagnetism, it is
F = 2 * q V * μ₀ / (4 π) * dq ω R cos(α + β) / r².
The integral actually is 2 times q V times the magnetic field B at the test mass/charge. We can calculate B from the magnetic moment m of the rotating disk:
B = μ₀ / (4 π) * 1 / r³ * 1/2 J_e.
Mapping back to gravity, we have
F = 2 * m V * G / c² * 1 / r³ * 1/2 J
= m V * G / c² * 1 / r³ * J.
The 2nd term adds J to the gravitomagnetic moment m_g.
What corrections did we make?
Let us recapitulate the corrections.
1. Schwarzschild correction from the radial velocity of m toward dm, on the radial acceleration of m.
2. Lorentz correction of the newtonian gravity acceleration of m toward dm.
3. Schwarzschild correction from the radial velocity of m toward dm, on the tangential acceleration of m.
The sum of corrections
If the test mass m is far away, we assume that the frame dragging to m is the same if it is slightly off the plane of the disk. One can then slice a rotating ball into many horizontal disks and calculate the sum to get the frame dragging by the ball. The gravitomagnetic moment formulae are the same for a disk and a ball.
The radial correction amounted to a gravitomagnetic moment 1/2 J.
The third term of the tangential acceleration contributed a gravitomagnetic moment of -J.
The second term contributed J.
The total is 1/2 J. This differs from the literature where everyone is claiming that the gravitomagnetic moment is 2 J! (Using the Wikipedia definition of gravitoelectromagnetism 2 J corresponds to 1/2 J, but we use our own definition.)
A half of the second term is analogous to the magnetic field of electromagnetism. The third term comes from the fact that the the test mass dives faster into the field of the left side of the ball than the right side. The radial term comes from the fact that the test mass approaches the left side faster than the right side.
The gravitomagnetic moment 1/2 J is like the magnetic moment of the analogous rotating charged ball, 1/2 J_e. We do not know if this is a coincidence.
Eliminating the accelerated motion of dm by letting the ball parts "fly loose"
The parts dm rotate around the center of the ball, and are thus in an accelerated motion. Let us make the ball very lightweight, so that the parts are only kept together by a mechanical structure.
Let us then release the structure and let the parts fly free. The parts now fly along straight paths and we do not need to worry about their acceleration. Immediately after the release, the parts still move approximately like they did inside the ball. This suggests that the acceleration did not change the solution much.
Frame dragging by a rotating charge
In the calculation above, did we use anything which differs for a mass versus an electric charge?
In the radial and tangential calculations we used the Schwarzschild metric to determine the slowdown of a radial or tangential velocity as the test mass dives deeper into the field of dm. If we believe Calculation A of our August 5, 2023 post, then there is an exactly analogous process with an electric charge.
Otherwise, we use special relativity, which works in the same way for weak gravity fields and electric fields.
An atomic nucleus is a charged body which might rotate at a fast pace. We have to check what is known about frame dragging in such a case.
Conclusions
The fundamental idea in the calculation is the linearity of solutions conjecture for weak gravity fields. We reduce the problem into summing many Schwarzschild metrics, one for each part of the system. We have to check the literature: has anyone found a flaw from this approach?
The method works also for electric charges. There we do not face the nonlinearity problem of general relativity.
How well do we know that the parameter a in the Kerr solution really is J / (Mc) for the rotating system?
Roy P. Kerr (1963) himself refers to a third order Einstein-Infeld-Hoffmann (1938?) approximation for a spinning particle. We have to check that.
Apparently, only a small number of people have worked with the gravity field of a spinning mass, since the equations are so very complicated. It is not clear if the Kerr vacuum solution corresponds to any source of gravity. We have to check how Hartle and Thorne associated a certain angular momentum J with their approximate metric.
The radial acceleration effect in our calculation comes from special relativity and the newtonian attractive force. We may imagine that we divide the rotating equatorial disk into a very small number of parts, say, 16 parts. Could the 15 other parts affect the metric around dm enough, so that it renders our calculation incorrect?
Gravitomagnetic effects are very weak. A small error in the approximation technique can spoil the effects. This is true for our method, and true for linearized Einstein equations, too.
Our method is bound to work for the gravitomagnetic effect of a moving single mass, since we use the true-and-tested Schwarzschild solution for it. However, our analysis of conservation of angular momentum has to be checked carefully.
No comments:
Post a Comment