NOTE: the calculations in this post are complicated and have not been checked. There are probably errors.
----
Let us calculate the orbital speed of an equatorial circular prograde Kerr orbit and check how it compares to newtonian gravity. We will treat the case where a test mass orbits at a relatively large distance r, and the rotation of the central mass (black hole) is very slow.
The magnetic effect on the orbit of an electric charge
q • -->
Q ● --------->
If we have a large electric charge Q moving fast, and a test charge q of the opposite sign moving slowly to the same direction, then the magnetic field of Q repulses q.
Let us determine the magnetic field of a rotating shell or a rotating ball of electric charge.
Marvin Zanke in the link calculates the magnetic moment of a spherical shell or a ball of charge.
For a spinning circular loop of charge the moment is
m = I π R²,
where I is the electric current and R is the radius of the circle. The circle rotates around its center. Using the angular velocity ω we get
m = Q / (2 π R) * ω R π R²
= 1/2 Q ω R²,
where Q is the electric charge of the loop. For a spherical uniform shell, the formula is
m = 1/3 Q ω R²,
and for a uniform ball
m = 1/5 Q ω R².
Let us compare these values to the "charge angular momentum" J of the configurations. For the circular loop,
m = 1/2 J.
In the link we have the moment of inertia for a shell and a ball. In both cases, we get m = 1/2 J.
The equatorial magnetic field strength for a magnetic dipole m is
B = μ₀ / (4 π) * m / r³,
where r is the distance from the center of the rotation.
Let us calculate how the spinning of a central charge affects an equatorial circular orbit of a test charge around it.
Let us set Q = c = μ₀ / (4 π) = 4 π ε₀ = 1.
We assume that the test charge q has a mass which is equal to q. The velocity is v, which can vary from 0 to 1.
We set the fictitious centrifugal force equal to the Coulomb force minus the magnetic force:
q v² / r = q / r² - 1/2 J / r³ * q v
<=>
ω² r³ = 1 - 1/2 J ω
If |J| << 1 and r > 1, then ω is approximately 1 / r^1.5.
ω² = 1 / r³ - 1/2 J / r^4.5
= 1 / r³ * (1 - 1/2 J / r^1.5)
ω = 1 / r^1.5 * (1 - 1/4 J / r^1.5)
= 1 / r^1.5 * 1 / (1 + 1/4 J / r^1.5)
= 1 / (r^1.5 + J / 4).
In a linear motion, the magnetic force is really a Lorentz transformation of the electric force
Let us have a charge Q moving at a constant velocity v to the x direction, and a test charge q close to it. The natural frame to determine the path of q is the frame fixed to Q. In that frame, Q does not have a magnetic field, and we can solve the path of q using just Coulomb's force.
Let q move initially along with Q with the same velocity vector v = 0.1 c, above Q (above means to the z direction). The electric field E of Q to the z direction in the laboratory frame is 0.5% stronger than in the comoving frame.
The mass-energy of the test charge, and its inertia, is 0.5% larger in the moving frame.
When q moves closer to Q, have to add 0.5% to the energy gained from E, because the energy "travels" with q at the speed 0.1 c and has 0.5% more mass-energy in the laboratory frame.
All this would make q to accelerate toward Q 0.5% faster in the moving frame.
But the magnetic field to the y direction exerts a force on q such that it cancels 1% of the electric force. The end result is that the acceleration of q in the z direction is 1% less, measured in the laboratory frame.
The proper time of Q runs 0.5% slower than in the laboratory frame. In this case, the role of the magnetic field is actually to calculate the effect of time dilation.
Note that on a circular orbit around Q, the test charge q does not see the electric field of Q strengthened by 0.5% because the field is not "squeezed". There is no need to transform the electric field. This explains why we could use q / r² as the Coulomb force in the calculation of the previous section.
Magnetic newtonian gravity
Electric charges of the same sign repulse each other. Gravity charges of the same sign attract each other.
Thus, we expect magnetic newtonian gravity to produce a repulsive force in the case above.
test mass
• -->
repulsion
___
/ \
\____/
<----
rotation
For a prograde circular equatorial orbit around a spinning central mass whose total mass-energy is M, magnetic gravity repulses a test mass m. We expect a test mass m to orbit the central mass M slower than for a static mass with the same mass-energy M.
Circular equatorial Kerr orbits: the correction is fourfold compared to the electromagnetic effect
Bardeen et al. (1972) give the angular velocity as
where G = c = M = 1, and J = a.
The r in the formula is the Boyer-Lindquist r coordinate. It is more realistic to use the cartesian coordinate. In cartesian coordinates, r corresponds to a radius
R = sqrt(r² + a²).
Then
Ω = 1 / ( (R² - a²)^0.75 + a ).
If |a| << 1, then approximately
(R² - a²)^0.75
= R^1.5 * (1 - a² / R²)^0.75
= R^1.5 - R^1.5 * 0.75 * a² / R²
= R^1.5 - 0.75 a² / sqrt(R).
For |a| << 1, we can treat a² = 0, and the formula is
Ω = 1 / (R^1.5 + a).
We see that for a small angular momentum (a or J), the gravitomagnetic effect is 4 times as large as for the electric analogue.
"Apparent kinetic energy" of the central mass explains 1/2 of the Kerr correction
a = J / M
v' <- ● M / 2
R
x center
R
M / 2 ● -> v'
m • ---> v = ω r
Let a test mass orbit a central mass M at a velocity v. If M is not rotating, then test mass m "sees" the near side and the far side of M to move at a speed v relative to it.
Let us then assume that the central mass M rotates and a = J / M.
The test mass m sees less kinetic energy in the near side of M and more kinetic energy in the far side. Let us calculate how this affects the pull of gravity.
Let us set G = M = c = 1. We assume that |a| is very small. Let the radius of the central mass be 1. The moment of inertia for a uniform ball is I = 2/5.
We have
J = a = 2/5 Ω,
where Ω is the angular velocity of the central mass.
We calculate with two symmetric "elements" of the central mass. The distance of the elements from the center is R. Let the mass of each element be M / 2 = 1/2.
The velocity of an element is
v' = Ω R = 5/2 a R.
The kinetic energy of the near element in the frame of the test mass m is
1/4 (v - 5/2 a R)² = 1/4 v² - 5/4 a R v,
and the far side
1/4 (v + 5/2 a R)² = 1/4 v² + 5/4 a R v,
where we omitted terms containing a².
The distance to the near element is r - R and to the far element r + R. If a is different from zero, then the test mass "sees" less kinetic energy in the two elements:
-5/4 a R v / (r - R)² + 5/4 a R v / (r + R)²
= -5/4 a R v * 4 R r / (r² - R²)²
= -5 a R² v r / (r² - R²)².
Let us approximate this for a large r.
The above calculation holds if the elements are as in the diagram. If the elements are turned by an angle α, then v' is multiplied by a factor cos(α) and the R in 4 R r / (r² - R²)² is also multiplied by cos(α). Since cos²(α) is on the average 1/2, we must multiply the formula above with 1/2.
The moment of inertia of a ball of a radius 1 and mass 1 is
I = ∫ R² dM = 2/5,
where we in the integral sum for all particles dM in the ball.
A "typical" R² thus is 2/5. We get the following correction to gravity if r is large:
-1/2 * 2/5 * 5 a v / r³
= -a v / r³.
The acceleration of gravity is (remember that we set G = M = 1):
1 / r².
Let us determine the velocity of the test particle on a circular orbit:
v² / r = 1 / r² - a v / r³.
Since v is approximately 1 / sqrt(r), we have
ω² r = 1 / r² - a / r^3.5
<=>
ω² =1 / r³ * (1 - a / r^1.5).
If a is small, then
ω = 1 / r^1.5 * 1 / (1 + 1/2 a / r^1.5)
= 1 / (r^1.5 + 1/2 a).
The correction is exactly a half of the one in the Kerr metric. This correction comes from the fact that the "gravity charge" of the sphere depends on the speed of the orbiting test particle. This correction does not exist if the charges are electric. This is not really a "magnetic" correction.
For gravity, we probably have to sum the magnetic correction and the gravity charge correction. We can now explain 3/4 of the Kerr correction.
The correction from frame dragging is already included in the magnetic effect
The remaining part of the Kerr correction has to be something associated with gravity, but not with an electric charge.
Let us have m orbiting at a distance r. How much is the frame dragged there?
The formula gives the frame dragging angular velocity as a function of r for the Kerr metric. In the formula,
α = J / (M c).
Conclusions
We were able to explain 1/4 of the Kerr correction from the analogous magnetic effect for a rotating electric charge.
We were able to explain 1/2 of the correction from the "apparent kinetic energy".
We were not able to explain 1/4.
We face a mystery: how can frame dragging in gravity be 4-fold relative to the electromagnetic analogue? We have to investigate this.
No comments:
Post a Comment