Furthermore, M should attract a test mass m more than the corresponding electric charge Q attracts q, when m moves relative to the entire M. The gravity charge of m should appear to be equal to its inertia γ m. This differs from electromagnetism, where the electric charge of q stays the same if it moves relative to Q, while the inertia grows by a factor γ.
The Coulomb force seems to be Lorentz covariant once we correct the error in E_x' = E_x (August 9, 2023).
Gravity forces depend on γ m
Hypothesis of gravity forces on m. We define that the gravitoelectric force on a test mass m is
F_e = γ m E_g,
and the gravitomagnetic force is
F_m = γ m B_g,
where
γ = 1 / sqrt(1 - |v|² / c²)
and v is the velocity of m relative to the laboratory frame.
These both forces have a "magnetic" character because they depend on the velocity v of m.
Lorentz transformations of E_g and B_g. In the comoving frame of M, M is static. It only has a gravitoelectric field
E_g' = G M / r'²,
where the vector E_g points toward M. Suppose that M moves to the positive direction of the x axis in the laboratory frame. Then
E_g_y = γ E_g_y',
E_g_z = γ E_g_z',
E_g_x = γ³ E_g_x'
are the Lorentz transformations of the electric field x, y, z components. Also, in the laboratory frame there is a gravitomagnetic field
B_g = 1 / c² v × E.
Sanity checks
● M
• m
|
v v
Let us first check how the model works if the test mass m moves radially away from M at a velocity v. The inertia grows as γ in v, and so does the attractive force. The acceleration stays the same. This makes sense.
The behavior is the same, whatever the direction of v.
Let us then consider another case:
M
● ---> v
• m
The gravitoelectric field E_g of M increases by a factor γ relative to the case when M is static. This is because E_g' in the comoving frame of M is Lorentz transformed to the laboratory frame. The result is reasonable. The kinetic energy of M gravitates.
^ v
|
● M
r = distance (m, M)
• m
We defined that the gravitoelectric field E_g of a static M has to be Lorentz transformed through the formula
E_g = γ³ E_g'
to the laboratory frame. In the comoving frame of M the field
E_g' = G M / r'²,
where r' is the distance (m, M). Then
E_g = G γ M / (r' / γ)².
This is reasonable. The distance (m, M) is
r = r' / γ
in the laboratory frame. The mass M possesses kinetic energy, so that its gravitating mass is γ M.
^ v
|
● M
• m
\
v V
If m is static, we already studied that configuration. If m moves at a velocity V, it feels the second order gravitomagnetic force. Its inertia and gravity charge match. The behavior is reasonable.
Lorentz covariance of the fields
Let us first check a simple case.
M
● -----> v
r = distance (m, M)
• m
In the comoving frame of M, we have
E_g' = G M / r'².
We Lorentz transform E_g' to the laboratory frame:
E_g = G γ M / r².
The acceleration of m in the laboratory frame is, calculated from E_g:
a = G γ M / r².
Let us switch back to the comoving frame of M. There the acceleration of m is
a' = G M / r².
The Lorentz transformation of a' to the laboratory frame is
a = γ² a'.
Our model does not work! The results for the acceleration do not match!
We probably have to add the pincushion effect of the previous blog post, to fix the problem. If we would define the Lorentz transformation of E_g' to the laboratory frame to be γ² E_g', then the results would match, but then the kinetic energy of M would cause a 2X gravity. We do not want that.
Let us check the second simple case:
^ v
|
● M
r = distance (m, M)
• m
In the comoving frame of M, the gravitoelectric field is
E_g' = G M / r'²
= G M / (γ r)²,
and the acceleration is the same:
a' = G M / (γ r)².
We Lorentz transform the field E_g' and the acceleration a' to the laboratory frame:
E_g = γ M / r²,
a = γ M / r².
The results match! Moreover, it is reasonable that the gravity charge of M in the laboratory frame is M plus its kinetic energy divided by c². The model works in this case.
How to correct the problem?
We have to combine the formulae above to the pincushion trick. Our goal is that if a part M' moves back and forth inside M, the gravity attraction of M' is γ M'. But if the test mass flies entirely by M, then the gravity attraction of M is γ² M.
A distorted spatial metric, or some kind of a magnetic effect is required to do the trick.
● M
• m
-----> x
Let us have a light mass M static in the laboratory frame. We believe that the acceleration of m is almost exactly
a = G M / r²
regardless of the velocity of the test mass m We believe that a Lorentz transformation is the correct way to switch to a moving frame. This implies that
a' = G γ² M / r²
is the correct acceleration in a frame which moves along the x axis relative to the laboratory.
So far there is no problem. We can define the transformations of E_g and B_g to satisfy this.
The problem is the fact that this transformation gives too much gravity attraction to the kinetic energy of a part M' which is moving inside M.
M'
<-- ● --> v
• m
The part M' cannot move a long distance because it must stay inside M. Thus, a solution is to implement the gravity attraction in a way which increases the attraction of the kinetic energy in M' less if M' moves back and forth.
The "bounce" on M', when M' changes the course must cancel about a half of the gravity attraction which M' got from kinetic energy. Spatial metric or a "magnetic" field can probably do the trick.
But can this save us? What happens if the test mass m moves rapidly back and forth and M stays static?
● M
<-- • --> v
m
Our rules at the start of this blog post say that the attractive force on m grows by the factor γ. Also, the inertia of m grows by the same factor γ. The acceleration of m toward M is not affected. Thus, the spatial metric trick or a magnetic force trick might do the job.
If we use the spatial metric trick, then this gives us an "explanation" why the spatial metric has to be manipulated in some way in gravity: to prevent kinetic energy from having double gravity.
A "pincushion" force to fix the problem
M
● ---> v
r = distance(m, M)
• m
^ y
|
-------> x
Hypothesis. In the configuration above, the total force on m is
F = m G γ² M / r²,
and it is toward M. A subpart of the force F has the following formula:
F₂ = 1 / (2 c²) * d²r / dt² * m G M / r
= 1/2 v² / c² * m G M / r²,
where F₂ points toward M.
Moreover, the force F₂ behaves like a pseudoforce on m which comes from pincushion coordinates. If M turns back, so that v flips sign, then the momentum p_y which m received from F₂ flips its sign, too. The test mass m "bounces" back, retracing its earlier route.
Imagine a test mass m bouncing back and forth horizontally, along a straight line, in the above pincushion coordinates in Minkowski space. A pseudoforce acts on m, pulling it toward the center, if we calculate in the pincushion coordinates. When m bounces back, its momentum p, calculated in the pincushion coordinates, flips sign.
Can we find an intuitive explanation for the pincushion force from the shared inertia between M and m?
Another observation: in our analysis, the test mass m obeys the Schwarzschild solution for M as long as M moves at a constant velocity. The crucial thing in our analysis is what happens when M turns back. This question is not something what the Schwarzschild solution tells us. In that sense, we are not "correcting" the Schwarzschild solution, but analyzing it in more detail, so that we can deduce what happens when M turns back.
Analyzing the effect of the spatial metric and coordinates on the orbit of the test mass m: we can "undo" effects of spatial metric
● M
r = distance (m, M)
^ a
|
• ----> v
m
^ y
|
------> x
In the Schwarzschild metric, the the radial metric is stretched by a factor
1 + 1/2 r_s / r.
Could this explain a part of the acceleration of the test mass m in the above configuration?
No, not in the Schwarzschild coordinates. The derivative of the cartesian coordinate x metric is
d g_xx / dy = 0.
The coordinate acceleration
d²y / dt² = G M / r²
comes entirely from the temporal metric.
Let us then use the Einstein approximate metric and its coordinates from August 28, 2023. The metric perturbation of the Minkowski signature (- + +) metric η in cartesian coordinates is h:
h =
4 G / c² *
1/2 M / r 0 0
0 1/2 M / r 0
0 0 1/2 M / r
In these coordinates, the coordinate acceleration is
a = G M / r² + G M v² / c² * 1 / r².
The second term can be viewed as a spurious acceleration from the spatial metric. It is as if M would have double the kinetic energy 1/2 M v².
Suppose that the test mass "turns back", in some sense, in its movement. If we view the distortion of the spatial metric as simply an artefact of squeezed or stretched rulers, then in the pincushion diagram, the test mass really traces its original route back. The effect of the spatial metric is undone!
If m is initially static and M moves back and forth, then this model might be the correct one: the effect of the spatial metric around M is undone.
The pincushion force probably comes from inertia
Our pincushion force is intended to restore Lorentz covariance to gravity. If the force is ad hoc, why should we trust the model? It probably is not ad hoc, but we can derive it from the inertia of m in the field of M.
● M
r = Minkowski
distance (m, M)
^ v
|
• m
In our own Minkowski & newtonian model of gravity, the stretching of the radial metric in the Schwarzschild solution is caused by the test mass m having more inertia in the radial direction than in the tangential direction. The distance r above is the coordinate distance in the underlying Minkowski metric. Rulers may contract, but we only talk about coordinate distances which are the "true" spatial distances of things.
When m approaches M at a velocity v, energy E is shipped from M to m. This energy E increases the kinetic energy of m. Shipping energy over the distance r involves inertia: this is why the inertia of m in the radial direction is larger.
M
-v <---- ● ----> v
r
• m
^ y
|
------> x
Let M then move back and forth like in the diagram. Let M move to the right at the moment. The distance r grows when M moves to the right. The inertia of m associated with m is moving to the right. It would pull m up like a small weight hanging from a rope.
The inertia of m held by M is something like
I = m G M / r * 1 / c²,
that is, the negative potential energy of m in the field of M.
If m does not want to move to the y direction, then it must keep the inertia at the distance r. We obtain a hypothetical centrifugal force which pulls m up in the diagram:
F_c = I v² / r
= v² / c² * m G M / r².
Maybe m and the inertia reach a compromise, and both give in? The force is then
F₂ = 1/2 v² / c² * m G M / r².
We found an explanation for the force F₂!
Hypothesis. The test mass m resists changes in the radial velocity dr / dt of the inertia I associated with M (the inertia I is "located" at M). Because of the resistance, r only varies 1/2 of what it would vary otherwise.
As M moves back and forth, then r has a positive acceleration d²r / dt² when M is moving to the right at the velocity v, or to the left at the velocity -v. When M turns back, r has a positive acceleration. Since r always returns to the same value after one cycle, accelerations cancel over a cycle. The cycle does not impose any force on m, on the average. We obtained the desired result.
Why does m want to keep r constant? An explanation from the energy shipping in the field of m and M
● M
r
• ----> v
m
^ y
|
------> x
Let us consider the configuration above. Suppose that m want to continue on the straight line in the direction of the x axis. If it does so, then energy is shipped from m to the combined field of M and m. The average shipping distance is r.
When m ships the energy E a distance r to the positive y direction, the position of m and M must move to the negative y direction, to conserve the center of mass.
If m has moved a distance x to the right from the x coordinate position of M, then the distance (m, M) has grown by
sqrt(r² + x²) - r = 1/2 x² / r,
if x is small. The test mass m had to ship
E = m G M / r² * 1/2 x² / r
of energy over the distance r. To balance this, m itself had to move to the negative y direction, and, presumably, also M moves to the negative y direction. If m and M each respond to 1/2 of the shipping, then m must move this distance:
-1/2 E / c² * r / m
= -1/2 m G M / r² * 1/2 x² / r
* r / c² * 1 / m
= -1/2 * 1 / c² * G M / r² * 1/2 (v t)²,
where t is the time that it took m to move the distance x. Let us denote by a the resulting acceleration of m to the positive y direction:
1/2 a t² = -1/2 * 1 / c² * G M / r²
* 1/2 v² t².
Then
a = -1/2 v² / c² * G M / r²,
and
m a = -F₂.
The force F₂ has the wrong sign!
How to fix this?
The error probably is in the assumption that m and M both respond 1/2 to the shipping. If M takes essentially all the responsibility, then our calculation yields 0. We were not able to explain F₂.
Conclusions
We were wrong to trust the Schwarzschild solution in the radial acceleration?
The calculation in the previous section probably has wrong assumptions. If it were true, the following would hold:
● M
r
• ----> v
m
In the above configuration, the acceleration of m is not
G M / r²,
like Schwarzschild claims, but it is
a = 1 / γ * G M / r²,
where
γ = 1 / sqrt(1 - v² / c²).
If we Lorentz transform this to the comoving frame of m, the acceleration is
a' = γ² a
= γ G M / r².
In the comoving frame of m, the gravitating mass is M + 1/2 M v² / c². This model only counts the kinetic energy once in the gravitating mass, not twice. We are saved from the complicated "undoable" gravity force.
However, above we ignored the metric perturbation which M causes. It is of the order
sqrt(1 - r_s / r).
If the test mass m is very fast, it spends a very short time close to M. Then the metric perturbation may dominate over the γ calculations in determining the deflection of m. The path of a photon is determined by the coordinate speed of light at different positions. Our γ calculation might not be relevant at all for a photon.
We will write a new blog post about this.
The geodesic equation seems to be wrong. The Schwarzschild solution would recognize F₂ if the geodesic equation were corrected. The geodesic equation does not understand that if tangential momentum is converted to radial, the coordinate velocity of m slows down.
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