On November 6, 2023 we tentatively proved that the Einstein field equations in most cases do not have a solution at all if there are accelerating masses. It might be that general relativity does not determine the field of a rotating body at all.
In literature, it is claimed that the Kerr metric can be derived by studying the Schwarzschild metric in a rotating frame. Let us check what we get if we try to determine the magnetic field of a ball of electric charge by switching to a rotating frame. This is a simple calculation, and we can test experimentally what the magnetic field is like.
(Picture Wikipedia)
The correct magnetic moment of a spinning uniform ball of charge
The magnetic dipole moment of a solid, uniformly charged sphere of uniform density is
m = Q L / (2 m)
= Q / m * 1/2 I ω
where Q is the charge, L is the angular momentum, and m is the mass of the sphere. The moment of inertia of the sphere is I and ω is its angular velocity. If we define
Lₑ = Σ q v r
as the "charge angular momentum", then
m = 1/2 Lₑ
= 1/2 Q I / m * ω
= 1/2 Q * 2/5 R² ω
= 1/5 Q R² ω,
where R is the radius of the ball.
In the equatorial plane, the magnetic field is
B = μ₀ / (4 π) * m / r³
= μ₀ / (4 π) * 1/5 Q R² ω / r³,
where r is the distance from the center. Very close to the surface of the ball the field is
B = μ₀ / (4 π) * 1/5 Q ω / R
= 1 / (4 π ε₀) * 1 / c² * 1/5 Q ω / R.
A very naive – and wrong – calculation
Let us compare this to a very naive calculation where we assume that the electric field of the ball "moves" at a velocity
v = ω R
at the surface of the ball. At the surface,
E = 1 / (4 π ε₀) * Q / R².
If v is slow, the magnetic field is
B = 1 / c² * v × E
= 1 / (4 π ε₀) * 1 / c² * Q ω / R.
The very naive method gives a magnetic field which is 5X the correct value.
Another naive calculation
<-- ω
___
/ \ 1/4 Q
\____/
• observer
We may approximate the rotating ball with a current loop. Let us assume that 1/4 of Q flows in a circular loop whose radius is R.
Our observer is at
r = R + r'
from the center. We ignore the far side of the loop and assume that the near side is straight. The magnetic field of a straight wire is
B = μ₀ / (4 π) * 2 I / r',
where I is the electric current and r' the distance from the wire. The current in our case is
I = 1/4 Q / (2 π R) * R ω.
We obtain
B = 1 / (4 π ε₀) * 1 / c²
* Q / (4 π) * ω / r'.
A reasonable value for r' might be R / 4. We get
B = 1 / (4 π ε₀) * 1 / c² * 1 / π * ω / R.
The estimate is 1.6X the correct value.
Yet another naive calculation
The magnetic field primarily rises from the charge moving very close to the observer. Let us guess that such a charge is Q / 10, and it moves at a velocity
v = ω R
at a distance R / 2 from the observer. The magnetic field is then
B = 1 / (4 π ε₀) * Q / 10 * 4 / R²
* 1 / c² * ω R
= 1 / (4 π ε₀) * 1 / c² * 2/5 * ω / R.
The value is 2X the correct value.
Estimating the gravitomagnetic moment of a rotating sphere: a very naive calculation
Let us first use a very naive method. We assume that we can switch to the rotating frame of the sphere, and in that frame the metric is Schwarzschild.
___
/ \ M
\____/ R = radius
^ V
\
• m
The test mass m approaches M, but goes a little bit sideways because m is not spinning along with M. Let us assume that m gathers some extra inertia as it descends, and that inertia is moving along M.
We assume that m is very close to the surface of M, the radius of M is R, and it is rotating at an angular velocity of ω.
If the gravity of M does the work W on m, then m acquires
W / c²
of extra inertia. That inertia is moving sideways at a velocity
v = ω R.
Let m descend down for a time t. Then
W = t V m G M / R²,
and m gets a sideways velocity
v = t V G / c² * M / R² * ω R,
which corresponds to a sideways acceleration
a = V G / c² * 5/2 * 1 / R³ * 2/5 M R² ω
= V * 5/2 G / c² * 1 / R³ * L,
where L is the angular momentum of the sphere. According to our August 10, 2023 definition, the gravitomagnetic moment is then
mg = 5/2 L.
The moment is 5X the analogous magnetic moment, just like we obtained with a similar very naive method in electromagnetism.
Gravitomgnetic moment: yet another naive calculation
Let us assume that M / 10 is moving at a distance R / 2 from the observer, at a velocity ω R. We can calculate like in the previous section:
W = t V m G M / 10 * 4 / R².
We obtain
mg = L,
that is, 2X the analogous electromagnetic moment.
Conclusions
The very naive method, where the entire electric field of a spherical charge is assumed to rotate "fixed" to the rotating sphere, yields a magnetic moment which is 5X too large.
Similarly, using the very naive method to the Schwarzschild metric gives a gravitomagnetic moment which is 5X the analogous electromagnetic moment.
In this blog we have been claiming that the Kerr metric overestimates the gravitomagnetic moment 4X or more. Maybe people doing the Kerr calculations believe that the gravity field rotates "fixed" to the rotating mass?
Our complicated calculations in August and September 2023 brought varying results. A key question is can we add the effects of rotating mass elements linearly?
Our October 18, 2023 blog post suggested that close to the mass M, one cannot obtain the correct Schwarzschild spatial metric by summing the Schwarzschild metric perturbations for each mass element dm. One has to consider the field of M as one "whole".
But that does not mean that if M rotates, then the "whole" field should rotate along with it. The gravitomagnetic effect calculated above depends only on the metric of time, because we calculate the work W done by gravity, and that work comes exclusively from the metric of time. One does obtain the metric of time by linearly adding the perturbations by each mass element.
However, we believe that the 2X extra inertia of a test mass in a radial motion relative to a mass element dm does affect the gravitomagnetic effect, and we did not include that in the calculations above.
On September 4, 2023 we sketched a "unified field theory", where electromagnetism is equivalent to gravity if we ignore the gravity of kinetic energy, of pressure, changes in the spatial metric, and so on. If that hypothesis is correct, then the gravitomagnetic field of a slowly rotating sphere should be exactly analogous to the corresponding electromagnetic field. But we have to check if the correct way to calculate the magnetic field of a rotating electric charge really is the classical one. Our October 9, 2023 blog post suggests that the classical calculation ignores some effects of inertia inside an electric field.
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