Inertia inside an electric field: the Lorentz transformation of E_x

Let us return to the problem of determining the acceleration of a test charge q near a moving charge Q. On September 9, 2023 we made a calculation error and claimed that the field Eₓ' is not the same as Eₓ.

But we did not then analyze the extra inertia of q inside the electric field of Q. Let us do it now.

A static Q and a moving frame

                 r = distance (q, Q)

                  ---> F

                • --> a                       ● 

               q                               Q            

               m

  

The test charge q is negative and the charge Q positive.

The acceleration a is from the familiar Coulomb force

       a  =  1 / m  *  1 / (4 π ε₀)  *  q Q / r²

           =  F / m.

Let us then have a test charge q moving to the right at the speed v.

                                   r

                      --> A

                   • ---> v                    ●

                  q                             Q

                  m

What is the acceleration A? Let us switch to comoving frame of q, which we denote with the prime'. The electric field in that frame at q is

       E'  =  E,

that is, the same as in the laboratory frame. Then

       a'  =  a,

and

       A  =  a / γ³,

where

       γ  =  1 / sqrt(1  -  v² / c²).

However, this ignores the deceleration which is due to the extra inertia of q close to Q. If the absolute value of potential energy of q inside the field of Q is

       W  >  0,

we have claimed that the extra inertia in a tangential motion relative to Q is

       W / c²,

and in a radial motion

       2 W / c².

Let us assume that v is not very large and W is neither. Let q move toward Q for a short time t. How large is the deceleration correction because of the increased inertia? The potential energy of q changes by

       W  =  v t F.

Since the inertia grows by 2 W / c², the velocity v decelarates by

       v * 2 W / (m c²)

       =  2 t v² / c²  *  F / m.

The correction to the acceleration in the laboratory frame is

       -2 v² / c²  *  a.

We have

       A  =  a (1  -  3/2 v² / c²).

The corrected A is

       Ac  =  a (1  -  7/2 v² / c²)

             =  a / γ⁷.

The corrected acceleration in the comoving frame of q is

       ac'  =  a / γ⁴,

quite different from the value a which we got from the Lorentz transformation of the electric field E!

                   

            v <--- •                   ●

                      q                   Q

What if q moves to the left at a speed v? Let us consider this in the comoving frame of q.

The Lorentz transformation of the electric field E gives the same result as in the previous case where v was to the right.

Since the inertia of q inside the electric field of Q is decreases to the left, q gets an additional acceleration to the left. The result for ac' is the same as for when v is to the right.

The case where both q and Q are negative

 

          A <-- • ---> v                          ●

                   q  (neg)                      Q (neg)

                   m 

Let W be the absolute value of the potential of q in the electric field of Q, and W is small. We have claimed that the inertia of q is diminished by

       W / c²

in a tangential motion relative to Q, and diminished by

      2 W / c²

in a radial motion.

As q moves closer to Q, the inertia of q decreases, which speeds up its velocity v. The acceleration A has to be corrected to have a smaller absolute value. We conclude that also in this case,

       ac'  =  a / γ⁴.

Transformation of the electric field Eₓ

Theorem. If we define the electric field E in a frame through the acceleration of an initially static test charge q in that frame, then the transformation of the electric field is

       Eₓ'  =  Eₓ / γ⁴,

where Eₓ is the electric field in the frame where the charge Q is static.

Lorentz transformations of the electric and magnetic fields

                        v

                      <---

                     \  |  /

                       \|/

                 ----- ● Q ---

                       /|\    

                     /  |  \  field lines

    ^ y'

    |

     -----> x

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

In the configuration above, the Lorentz transformation of Ey is

       Ey'  =  γ (Ey  -  v Bz).

In the formula, a non-zero value of Bz reveals that the frame is moving relative to Q.

We could define

       Ey'  =  γ Ey,

where Ey is the electric field in the frame where Q is static. This definition would be analogous to the transformation for Eₓ which we introduced in the previous section. In this sense, our transformation is kind of a Lorentz transformation for the electric field.

Conclusion

We have a hypothesis that a test charge q feels an increased or decreased inertia in the field of another charge Q. Though this hypothetical change is not present in quantum mechanical systems like the hydrogen atom.

If the inertia hypothesis is correct, it affects significantly the radial acceleration of q relative to Q. The "Lorentz transformation" of Eₓ becomes

       Eₓ'  =  Eₓ / γ⁴,

where

       γ = 1 / sqrt(1  -  v² / c²),

v is the radial velocity of q relative to Q, the x axis is parallel to (q, Q), and Eₓ is the electric field in the frame where Q is static.

We could define a "magnetic field" which would allow us to drop the dependency of the transformation formula to a frame where Q is static. The "magnetic field" would code the speed of Q in a frame.

In our September 9, 2023 blog post we aimed to reveal the effect of inertia on the Lorentz transformation of Eₓ, but we miscalculated and got confused.

Momentum of a box full of bouncing particles

We were wrong in our claim on August 25, 2023 that special relativity would break conservation of center of mass if rest masses exist. We assumed that the y velocity of the bouncing particle stays the same when the particle is pulled with a force F to the x direction. That assumption is wrong.

Let us analyze the momentum and inertia of the box full of bouncing particles.

                      --------

                         ^

                         |

                         |  vy 

                         |

                         •  m

                      --------

                        ● M  =  m / sqrt(1  -  vy² / c²)

      ^  y

      |

       -----> x

The particle m is bouncing up and down in the box. The particle M is static, with the same mass-energy as m.

Let us assume that a force F pulls M and m both to the positive x direction for a time t, giving an impulse of

       pₓ  =  F t.

What is the x velocity of each particle after the impulse?

We assume that for a particle m moving at a velocity v, the momentum p satisfies

       p²  =  v² m² / (1  -  v² / c²),

and the momentum vector p is parallel with the velocity vector v. Then

       p² -  p² v² / c²  =  v² m²

   <=>

       v²  =  p²  /  (m²  +  p² / c²).

For M, from the above formula we get the velocity

       Vₓ²  =  pₓ²  /  (m² / (1 - vy² / c²))  +  pₓ² / c²).

For the bouncing particle m, the square of the x component of the velocity is

       vₓ²  =  pₓ² / (pₓ² + py²)

                  * (pₓ² + py²)  /  (m²  +  (pₓ² + py²) / c²)

              =  pₓ²  /  (m² + (pₓ² + py²) / c²)).

Let us assume c = 1. To prove Vₓ = vₓ, we have to show

       m² / (1 - vy²)  +  pₓ²  =  m² + pₓ² + py².

   <=>

       m²  =  (m² + py²)  *  (1 - vy²).

The energy-momentum relation says

       E²  =  p²  +  m²,

and for the bouncing particle,

       E²  =  m² / (1 - vy²).

Our equation becomes

       m²  =  m² / (1 - vy²)  *  (1 - vy²),

which is true.

When a particle is x accelerated, its y momentum stays the same, but not the y velocity.

The force F pulls to the x direction. The y velocity slows down. Thus, the acceleration of the particle is not solely to the x direction, but also to the negative y direction. The acceleration is not to the direction of the force.

Conclusions

We did have a calculation error in our August 25, 2023 blog post. Calculating with velocities is error-prone in special relativity. It is better to use momenta.

On September 9, 2023 we claimed that the Lorentz transformation of the electric field Eₓ is wrong in the literature. Our claim was wrong. We erroneously assumed that the "inertia" of a particle m against a change in its velocity is γ m, like its mass-energy is.

Above we showed that the "inertia" really is γ m when the force F is normal to the velocity v of the particle. The bouncing particle behaves like a static particle whose mass M = γ m. But if the force F pulls to the direction of v, the behavior is very different. A simple example is this: we can increase the x velocity of a static particle M by any number < c, but that is not possible for a particle m which already moves to the x direction.

We will next look at the inertia of a test charge q inside the electric field of a charge Q. Does the inertia spoil the Lorentz transformation Eₓ' = Eₓ?

The standard Big Bang hypothesis is probably too bold an assumption: a tower of turtles

The Big Bang hypothesis was born when Alexander Friedmann in 1922 found a very nice and simple solution for the Einstein field equations. The solution assumes that the universe topologically differs from Minkowski space. The spatial extent of the universe is like the surface of a balloon which is inflated and expands.

https://en.wikipedia.org/wiki/Friedmann_equations

Our own work in the past two months and earlier makes the concept of curved spacetime suspect  –  it may be impossible to describe gravity through a "metric". We thus exposed a weak point in various Big Bang models: they assume that spacetime is curved, and in addition to that, that spacetime even differs topologically from the familiar flat Minkowski space that we know.

Let us analyze these assumptions.

An interaction is expected to bend light – that does not prove that spacetime itself is curved

Bending of light when it passes the Sun is not a surprise at all. We expect gravity to pull photons. More surprising is that the bending is double the newtonian expectation. It is due to the radial metric stretching.

We have in this blog presented arguments for that the inertia, which a gravity field imposes on a test mass, explains both the slowing down of the time and the stretching of the radial metric close to a mass M. The interaction field can carry energy. One would expect that the field imposes inertia on a test mass m. It is not a surprise at all.

Why would we then assume that spacetime is curved? That is a bold and unnecessary assumption for a very mundane process of an interaction.

We have argued that the electric field expresses phenomena which are similar to gravity. Why would these phenomena in gravity be fundamentally different, so that they show that "spacetime is curved"?

Claiming that the Big Bang is the starting point of the entire universe breaks the Copernican principle

In astronomy we have observed that we live inside an explosion cloud which is expanding. The entire observable universe seems to be inside this explosion cloud.

There are explosions of various sizes in the observable universe, for example, supernova explosions.

Why should we assume that our explosion cloud is the entire universe and that the universe started from that explosion? That is an unnecessary assumption. It breaks the Copernican principle that we should not assume that we are located in a special place and time in the universe. The standard Big Bang model claims that quite a short time, only 13.7 billion years has passed since the beginning of the entire universe. We are special in the standard Big Bang model.

https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric

The Friedmann (FLRW) Big Bang model is very similar to a newtonian expansion of a dust ball. This suggests that we actually live inside an ordinary newtonian explosion. Why should we assume a topologically strange model to explain a simple newtonian explosion?

If we would be free to design the birth of a universe, why should it be so similar to an ordinary newtonian explosion? We could have a fundamentally different mechanism which creates the universe. Why pick a newtonian explosion?

Why the observable universe seems to be spatially flat?

If we use the Friedmann model, we have to fine-tune the mass-energy density of the universe at the early stage, to ensure that the spatial metric is flat after 13.7 billion years. A slight change of the density can make the universe to collapse very quickly.

People devised the inflation model to explain this fine-tuning. The inflation model is very speculative. There should be very good evidence for it before we can accept it.

A simpler explanation for the flatness might be that we live in flat Minkowski space. However, we still have to tune the explosion to be such that it does not collapse back into a black hole quickly. Maybe there is a law of nature which says that the explosion has to disperse indefinitely and it cannot collapse back?

The Milne model

We have suggested in this blog that the total gravity charge of the observable universe is zero. Then the expansion would continue at a constant speed with no deceleration.

https://en.wikipedia.org/wiki/Milne_model

Does this solve the flatness problem?

https://en.wikipedia.org/wiki/Flatness_problem

Currently, the mass density of the observable universe seems to be about 30% of the critical density ρc of the standard Big Bang model.

In the Milne model, the Hubble "constant" is

       H = C / t,

where C is a constant and t is the time from the explosion.

How does the density of mass-energy ρ develop in the Milne model?

If a is the scale factor, the apparent energy density of photons declines as

       ~  1 / a⁴,

because the observer sees arriving photons redshifted.

The energy density of massive particles declines roughly as

       ~  1 / a³.

The problem becomes:

Why is the density ρ of the same order of magnitude as ρc now that 13.7 billion years has passed since the explosion? The value of ρ could be much larger or much less. It was much larger at a time t = 10 million years, and it will be much less at a time t = 1,000 billion years.

In the Milne model the spatial metric is always flat because the total gravity charge is zero. The fine-tuning is in the mass-energy density of ordinary and dark matter, relative to the critical density ρc.

Let us change the mass-energy density ρ somewhat, say, 50% from the current value. Let us then calculate backward the history of the universe. In the Milne model, the universe would have roughly the same history as we know it, but the mass-energy density would be 50% larger at every stage. The Milne model does not require a fine-tuning of the density to one part in 10⁶⁰ in the early stages, like the FLRW model does.

What evidence is there that the topology of the observable universe is as in the FLRW model?

The angular spectrum of temperature differences in the cosmic microwave background is rather "uniform", with some bumps and troughs. Supporters of the inflation hypothesis believe that inflation would create such a spectrum from "random quantum fluctuations" in a tiny primordial space.

Such random fluctuations could occur also in a newtonian explosion.

The troughs and bumps in the spectrum are explained by "baryonic oscillation" of matter soon after the Big Bang. We admit that this does support the FLRW model, though the same process might happen also in a newtonian explosion.

Conclusions

If we are able to show that a "metric" cannot describe the gravity field, then it is very unlikely that a curved spacetime like that in the FLRW model would exist. Far more probable is that the metric of spacetime is Minkowski, and the Big Bang is an ordinary, newtonian explosion. The Big Bang is not the start of the entire universe.

The Big Bang is very much like a newtonian explosion also in the FLRW model. Occam's razor implies that we should not assume the FLRW model at all, but stick to the simplest explanation: an ordinary newtonian explosion.

The Milne model removes the need for an extreme fine-tuning of the mass-energy density ρ in the early stages of the universe. Our observation is a serious blow to the Big Bang plus inflation hypothesis. The volatility of the Big Bang model is a result of the Einstein equations. The inflation hypothesis is highly speculative. We may interpret the history of modern cosmology since 1922 in this way:

1.   People tried to apply to cosmology erroneous Einstein assumptions about curved spacetime.

2.   They obtained the FLRW model which is extremely volatile for initial conditions. This was a sign that the model is wrong.

3.   Alan Guth and others in the 1980s tried to fix the volatility problem with a very speculative inflation hypothesis.

4.   It was found out that also the inflation hypothesis has to be fine-tuned to make inflation to work! Paul Steinhardt has criticized the inflation hypothesis on these grounds.

5.   The expansion is not slowing down as expected. People invented a speculative cosmological constant, or dark energy, to explain this. The Milne model probably does not need dark energy to explain the astronomical observations.

This history of cosmology reminds us of the tower of turtles cosmological model:

https://en.wikipedia.org/wiki/World_Turtle

https://en.wikipedia.org/wiki/Turtles_all_the_way_down

Gravitating mass, inertia, and centripetal acceleration

In general relativity, the gravity potential steepens at low radii r, relative to the newtonian potential.

                       ●  M

                       r

                   m •     

                       |   

                       |   rope

                       | 

                       |   ^

                       |   |  F force

                       |

                         \ o

                            |

                           /\    observer

An observer uses a rope to lower a test mass m very slowly close to a mass M. Let us calculate the potential very naively.

The remaining fraction of the mass-energy of a static test mass is

       sqrt( 1  -   (2 G M / c²) / r ).

The remaining mass-energy is zero at the Schwarzschild radius

       rₛ  =  2 G M / c².

In the newtonian potential, the remaining fraction of mass-energy is

       sqrt( 1  -  (G M / c²) / r ).

It is zero at

       rₛ / 2  =  G M / c².

The horizon radius in the newtonian potential is just 1/2 of the Schwarzschild radius rₛ.

What is the gravitating mass of m?

In the above equations we assumed that the gravitating mass-energy of a test mass m remains m when it is lowered close to M. That is a reasonable assumption if m falls freely toward M and the potential energy which m loses is converted to kinetic energy of m.

But we used a rope to lower the mass m slowly. It did not gain kinetic energy.

The paradox is solved by the fact that the field of M shortens the rope as we lower m downward. The observer can let the force F do work for a surprisingly long additional length of the rope.

The gravitating mass of m and the force F do shrink according to the fraction

       sqrt(...),

but that is compensated by the radial metric stretching according to

       1 / sqrt(...).

The gravity force F shrinks to zero when m is at the horizon.

A perpetuum mobile? No

Let us lower the test mass m in such a way that first it is allowed to gain kinetic energy. After a while, we let it fall at a constant radial coordinate velocity.

The gravitating mass of m is larger than with the slow lowering procedure. Gravity does more work for the whole trip down. Can we recover more energy this way?

No. If we let m fall freely to some distance r from M, its total mass-energy, measured from far away, is still m c². Lowering that mass-energy from the distance r to the horizon will give the energy m c² to the observer – no more.

Energy is conserved. In this context, the notion of "work done by gravity" is misleading.

If we lower the mass m slowly with a rope, we extract the energy from m at larger radii than in the partial free fall case. That is the difference between the two procedures.

The centripetal force and acceleration

To keep a test mass m on an orbit, gravity must give it enough acceleration. The inertia of the test mass m tries to make the test mass to fly along a straight line.

                          ● M

                                       ^ 

                                     /

                      •  ---------

                     m

Thus, the centripetal force is gravity. Crucial in this is what is the inertia of m in this configuration.

We have claimed in this blog that the inertia against a tangential acceleration relative to M is

       m / sqrt(1  -  rₛ / r)

and the inertia against a radial acceleration is

       m / (1  -  rₛ / r).

Which is the relevant inertia on a circular orbit?

Let us look at circular orbits in the Schwarzschild metric.

https://en.wikipedia.org/wiki/Schwarzschild_geodesics

There μ = m, because we assume that m is very small. The angular velocity ω in the Schwarzschild metric is the same as in newtonian gravity. Is this a coincidence?

If we believe the Schwarzschild metric, then the gravitating mass of m is its total mass-energy, and its inertia on a circular orbit has to be the same, because m stays on the newtonian orbit.

This is strange. The gravitating mass-energy of a static m is

       m sqrt(1  -  rₛ / r).

Can its inertia against the acceleration on a circular orbit be that low?

The acceleration on a circular orbit is neither tangential nor radial. The tangential and the radial velocities stay constant. Thus, the inertia might be that low.

The extra inertia in a tangential motion of m relative to M would be a separate "load" that m has to carry, but that load would "naturally" orbit M so that m does not need to exercise a force to keep the load on a circular orbit. The following mechanical model would explain this:

              <-- ω

              _____      ring of inertia

            /            \

           |      ● M  |  

            \______/

                   • --> v

                  m

When m comes close to M, m attaches itself to a "ring of inertia", which slows down the tangential motion of m. Once m is attached to the ring, the extra inertia does not affect a circular orbit of m. The ring rotates around M at an angular velocity ω without any extra effort.

The extra inertia is a property of the common field of m and M. The inertia is not an independent body which could fly around on its own. It is not strange at all that the extra inertia can stay on a circular orbit without any extra effort from m.

Similarly, the extra inertia does not increase the gravitating mass of m.

The circular orbit of a photon around M

Let us calculate the coordinate radius for a circular orbit of a photon around M.

The tangential velocity of a photon is

       v = c  sqrt(1  -  2 G M / c² * 1 / r).

We have

       G M / r²  =  v² / r

   <=>

       G M  =  v² r

                =  c² r  -  2 G M

   <=>

       r  =  3 G M / c²

  

           =  3/2 rₛ.

If we use a newtonian gravity potential, then

       v  =  c  (1  -  G M / c² * 1 / r),

       G M  =  v² r

                 =  c² r  -  2 G M  +  G² M² / c²  *  1 / r

   <=>

       0  =  c² r²  -  3 G M r  + G² M² / c²

   <=>

       r  =  3/2 G M / c² 

         +- sqrt(9 G² M²  -  4 c² G² M² / c²)  /  (2 c²)

           =  3/2 G M / c²

               +- sqrt(5) / 2 * G M / c².

The sensible value is

      r  =  (3/2 + sqrt(5) / 2)  G M / c²

          =  2.62 G M / c².

Conclusions

The gravitating mass of m close to a mass M is

       m  sqrt(1  -  rₛ / r),

where rₛ = 2 G M / c². The inertia of the gravitating mass alone is the same

       m  sqrt(1  -  rₛ / r).

Also, m feels extra inertia if its tangential velocity or its radial velocity changes. The total inertia of m for a tangential movement is

       m / sqrt(1  -  rₛ / r)

and for a radial movement

       m / (1  -  rₛ / r).

It is not a coincidence that circular orbits in the Schwarzschild metric obey the newtonian equation. The gravity force is newtonian, and the gravitating mass of m is the same as its inertial mass, for a circular motion.

Flaws in general relativity: a summary

UPDATE November 13, 2023: The question of Lorentz covariance of general relativity is still unclear. On November 7, 2023 we tentatively proved that the Einstein equations have no solution for any realistic physical system. The question of Lorentz covariance becomes partially moot.

We are undecided about linearity/nonlinearity of gravity.

----

Let us summarize our findings so far about what is wrong and right in general relativity.

What is wrong in general relativity

1.  The Einstein field equations are not Lorentz covariant. They calculate a wrong orbit for a test mass m in the case where the central mass M moves. The reason is that the field equations think that the kinetic energy of M gravitates like the rest mass of M, while the gravitational pull really is fourfold, if m passes past M.

2.  The steepening of the gravity potential in the Schwarzschild solution is highly suspect: it leads to very strange results. The potential probably has to be newtonian.

3.  General relativity does not satisfy a weak equivalence principle. Though this is not really wrong, since we do not think that gravity should satisfy it.

4.  The concept of a "metric" handles accelerating gravity sources in a wrong way. We may be forced to abandon the concept of a metric and treat the interactions between the test mass m and the parts of M "privately".

5.  If gravitational waves truly can shorten spatial distances between two events, they open a way for superluminal communication. That has to be prevented.

6.  The Gödel universe and the Kerr solution seem to contain timelike loops. That is unlikely to be a physically correct prediction.

7.  The claim that spacetime can be "curved" and that its topology can differ from the Minkowski space is a bold hypothesis, for which we have no evidence whatsoever. The hypothesis is almost certainly wrong.

What is correct in general relativity

1.  In the weak field, the Schwarzschild solution seems to be right. It agrees with our own Minkowski & newtonian gravity model.

2.  The geodesic equation calculates orbits correctly for a metric (if there exists a correct metric).

3.  Linearized Einstein equations calculate the energy content of gravitational waves correctly.

All "exotic" black hole physics is incorrect

1.  It is very unlikely that a one-way membrane  like the event horizon, can exist physically. The system probably "freezes" before such a strange object can form.

2.  A singularity does not make much sense. The freezing process probably prevents a singularity from forming.

3.  Black hole thermodynamics is entirely wrong. There is no need for a black hole to have large entropy.

4.  Hawking radiation does not exist. There is no black hole information paradox.

Conclusions

General relativity got right weak field gravity effects, like the bending of a ray of light as it passes the Sun, and gravitational waves. There has to be some truth to the Einstein field equations, even though they handle moving masses incorrectly. We have to analyze what exactly is right in the field equations. Equivalence principles might be the reason why the field equations get some things right.

In July 2023 our goal was to find out if we can disassemble a black hole by spinning it fast. We now know more about gravity, and can try to resolve the question.

Private interactions and the metric give contradictory results

In our previous blog post we claimed that "private interactions" between the test mass m and the parts of a mass flow can give a different acceleration for m than what we get if we sum the metric perturbations for each part and use the metric to determine the acceleration.

This is a surprising result because one would guess that the acceleration is linear in perturbations.

The metric of a mass flow

Let us determine the metric of the mass flow which we introduced in the previous blog post.

            1                                 3

                \                           /          ^

                  \                       /          /   v

                    \__________/

                             2

                             ^   V

                             |

                             •  m

We sum the metric perturbations for each part of the mass flow. We are interested in the metric close to the test mass m. The parts 1 and 3 point directly at m.

https://en.wikipedia.org/wiki/Lorentz_transformation

Let us first determine the metric perturbation which is caused by the part 1 of the mass flow.

                               

                      r' = distance (M', m)

                  M'

          -------•-----------                        • m

         R₂                   R₁     

         ρ = mass / length

              mass flow

     ------> x

Let us temporarily choose the part 1 as the x axis. Then it is easy to calculate the perturbation at m, which is also located on the x axis. Let the moving frame be comoving with the part 1. Let us have ρ dr as a mass element of the flow 1.

Then the line element for the metric around ρ dr is:

       ds²  =  (-1  +  rₛ / r')  dt'²

                + (1  +  rₛ / r')  dr'²

                + (r' dφ')².

We write

       dr' =  dx',

       r' dφ' = dy',

because the contribution to dr' from dy' close to m is essentially zero. Then

       ds²  =  (-1  +  rₛ / r')  dt'²

                + (1  +  rₛ / r')  dx'²

                + dy'².

The Lorentz transformation gives:

       dt'   =  γ dt  -  γ v / c²  *  dx,

       dx'  =  γ dx  -  γ v * dt,

       dy'  =  dy.

Let us set c = 1 to simplify the calculations. The metric in the laboratory frame is

       ds² =  γ²  (-1  +  rₛ / (γ r))

    

                         * (dt²  -  2 v dt dx  +  v² dx²)

                + γ²  (1  +  rₛ / (γ r))

                         * (dx²  -  2 v dt dx  +  v² dt²)

               + dy²

              =  γ²  (-1  +  rₛ / (γ r)  +  v²)  dt²

                 + γ²  (1  +  rₛ / (γ r)  -  v²)  dx²

                  - 2 γ v  rₛ / r  *  dt dx

   

                  - 2 γ v  rₛ / r  *  dx dt

                 + dy².

Note that the metric is at one moment of the laboratory coordinate time t. Since the part ρ dr is moving at a velocity v, the metric is time-dependent.

We dropped the terms ~ v² rₛ, because we assume that rₛ  <<  v  <<  1. The metric is almost orthogonal since v rₛ is very small.

To obtain the metric near m, we need to integrate over the radii R₁ + x to R₂ + x, where x is the displacement from the test mass m on the x axis. That is, we set x = 0 at m.

However, in the geodesic equation we are interested in the derivatives of the metric g. It is better to calculate the derivatives directly.

The perturbations of individual parts ρ dr are time-dependent. But the sum of all the perturbations is time-independent. The time derivatives of the integrated metric are zero. It is enough to calculate the derivatives with respect to x and y.

We have:

       dgₜₜ / dx  =  dgₓₓ / dx

                                       R₂ + x

                          =  2 γ      ∫      G ρ / r²  *  dr

                                     R₁ + x

                          =  2 γ G ρ

                              *  (1 / (R₂ + x)  -  1 / (R₁ + x))

                          = 2 A(x),

where A(x) is the newtonian gravity acceleration caused by the mass in the part 1.

We have

       dgₜₓ / dx   =  dgₓₜ / dx

                            =  -4 v A(x).

The acceleration from the geodesic equation

https://en.wikipedia.org/wiki/Geodesics_in_general_relativity

          1

              \  

                \      \      v

                  \      v   

                           β

                          \   ^    V

                            \ |

                              •  m

                                \

           y                     v   a

     __--->               

     \  

       \   

        v  x

Using the coordinates of the previous section, the velocity of m is

       Vₓ  =  -V cos(β),

       Vy  =   V sin(β).

We are interested in the acceleration of m to the x and y directions. The y acceleration relative to the proper time τ of the test mass m is

       d²y / dτ²  =  0,

that is, the y velocity Vy slows down as the proper time of m slows down. The Christoffel symbols relevant for the x acceleration are:

       Γₓₓₓ  =   A(x),

       Γₓₜₜ    =  -A(x),

       Γₓₜₓ   =  4 v A(x) - 4 v A(x)  =  0,

       Γₓₜₜ   =  4 v A(x) - 4 v A(x)  =  0.

The acceleration is

       d²x / dτ²  =  1 / gₓₓ(x)

                          * ( A(x) * (dx / dτ)²

                              + A(x) * (dt / dτ)² )

   <=>

       d²x / dt²  =  1 / gₓₓ(x)

                          * ( -A(x) V_x² + A(x) ).

The accelerations do not depend on the sign of v. If we calculate the accelerations due to the part 3 of the mass flow, they are symmetric relative to the the part 1. This means that the parts 1 and 3 do not cause any horizontal acceleration.

The part 2 does cause acceleration to the right.

If we calculate the accelerations using the private interaction model, the result agrees for the part 2, but does not agree for the parts 1 and 3. In the private model those parts cause acceleration to the right.

The accelerations of the mass flow in the turns of the flow between 1 and 2, and 2 and 3, contribute a net acceleration straight up.

Thus, the private interaction model gives a different result from the metric.

How can a metric perturbation be "nonlinear" in such a way that the sum gives a different acceleration from the sum of individual accelerations?

       1

           \

             \

               \________   2

                    ---> v    

                       ^   V

                       |

                       • m

The reason is that the metric does not understand what happens when a mass element ρ dr turns from the part 1 to the part 2. Thus it is the acceleration of the mass elements, after all, which spoils the calculation with the summed metric perturbation.

The acceleration of m is due to it diving deeper into the gravity field of the moving mass elements ρ dr in the part 1 of the mass flow. But when an element comes to the turn between 1 and 2, the element suddenly "disappears" from the integral. The potential of that element vanishes and m magically jumps up into a higher potential.

Can we somehow fix the metric so that it would not be confused by this case?

The "private interaction" model does handle this.

The metric should be augmented with a mechanism which tells us what happens to the test mass m when a mass element is accelerated.

A partial solution: calculate the metric letting the parts of M "fly loose"?

Since the metric cannot handle an acceleration of a mass element of M, let us remove all the accelerations for a short moment. All the parts of M fly freely and there is no acceleration in their path. Then we can calculate the sum of the metric perturbations.

Though this will not work if the acceleration of the mass elements happens to impose an "important" acceleration just at that moment.

Also, the solution only works for a short moment, as the free-flying parts of M soon are dispersed and do not describe the true state of M any more.

In the case of a rotating disk, letting the parts "fly loose" makes the configuration and the metric time-dependent. It is not a beautiful solution.

Conclusions

We were able to solve the mystery of why the calculations on August 10 and August 29, 2023 gave different values for the frame dragging by a rotating disk. Our August 10 calculation claimed that the gravitomagnetic moment is 1/2 J, while the August 29 calculation said that there is no frame dragging at all. The reason is that the concept of a "metric" does not understand a configuration where parts of the mass M are under an acceleration.

The value for the gravitomagnetic moment of a rotating disk is 2 J in the literature. We believe that value is incorrect.

It is not clear if we can mend the concept of a metric to handle mass elements which are under an acceleration. If not, then the "private interaction" model, which we have introduced, is the correct way to describe a many body system where the parts are accelerating.

Gravity of an accelerating mass M

Now that we have gained a lot of experience about inertia, metrics, and the geodesic equation in the past two months, it is time to revisit the mystery (solved?) on September 3, 2023. Can we get conflicting results for two methods:

1. we calculate the Schwarzschild acceleration of m for each part of M, and sum the accelerations;

2. we sum the metric perturbations for each part of M and then calculate the acceleration of m using the summed metric?

If M, for example, is a rotating disk, its parts are under a constant centripetal acceleration. We suspected on September 3, 2023 that the acceleration of the parts spoils the method 2 above.

We did get consistent result for a moving cylinder on August 28, 2023, using the methods 1 and 2 above. The cylinder moved at a constant speed, no part of it was under an acceleration.

An accelerating M and a test mass m by its side

                       M

                        ● -----> a_x

                        r

                        • --> a_x'

     ^  y            m

     |

      -------> x

Both M and m are initially static. The mass M starts to accelerate right. What happens to m?

Our own Minkowski & newtonian gravity model suggests that M pulls m along with it. The masses M and m share some inertia. The inertia starts to accelerate with M.

The mass m has gained the extra inertia

       m  *  1/2 r_s / r

in the field of M. This extra inertia resists movements of m in the field of M. If M starts to accelerate, then we can guess that the inertia tries to keep m moving along with M.

If m is very close to the horizon of M, then the speed of light there relative to M can be tiny, as observed by a faraway observer. The test mass m must move along with M  –  otherwise it would break the speed limit.

Does the metric of general relativity say anything about the problem?

Let us imagine that M has already gained a tiny velocity v, and m is somehow "thrown" into the metric of M. The test mass m "enters" the field of M at the opposite velocity -v.

If m enters the field of M from far away, the slow metric of time close to M slows down the movement of m, as shown by the constants of motion for the Schwarzschild metric: if the proper time of m slows down, so does its velocity.

We can guess that m attains an acceleration

       a_x'  =  a_x  *  (1  -  1 / sqrt(1 + r_s / r))

in the diagram above. If r is large, then the value is

       a_x'  =  a_x  *  1/2 r_s / r.

The test mass m behind M

               m

                • -->              r             ● -----> a_x

                a_x'                          M

       -----> x

In the radial direction, m has double the extra inertia of the tangential direction. Both m and M are initially static. Then M starts to accelerate to the right. What is the acceleration a_x' of m?

The formula might be

       a_x'  =   a_x  * 2 (1  -  1 / sqrt(1 + r_s / r))

                =   a_x  *  r_s / r,

for large r.

Can general relativity suggest the same formula? Yes. If we "throw" m to the metric of M, m loses coordinate velocity because its proper time τ slows down, and because the radial metric is stretched. We obtain a double effect relative to the "by the side" case.

Does this affect the field of a rotating disk?

                     <--  ω

                       ____

                    /          \

                    \_____/

                        ^   V

                        | 

                        • m

Every part of the disk accelerates toward the center of the disk. The parts close to m share more inertia with m. There might be an additional force which pulls m toward the disk if m is static? We are not sure.

Suppose then that m approaches the disk at a velocity V. Then m starts to share more inertia with the near part of the disk than with the far part. This is the traditional magnetic force which we have always included in our calculations.

Could there be a subtler mechanism through which the acceleration of the parts of the disk pulls m to the side?

We wrote on September 3, 2023:

"If we let the disk parts fly loose, so that there is no acceleration in their paths, then the parts on the left of the disk really start a collective movement toward the test mass, and our argument on August 10 that the left parts are "approaching" the test mass faster, is true.

But in a rotating disk, the collective field of the left side of the disk is time-independent. The collective field is not approaching the test mass faster on the left than on the right."

The discrepancy between the August 10, 2023 calculation of the sum of the accelerations, and the August 29, 2023 sum of the metric perturbations probably came from the fact that the acceleration calculation lets the parts of the disk "fly loose", without being accelerated toward the center of the disk. The situation is "dynamic".

The sum of the metrics, on the other hand, takes into account that the parts cannot fly away, they are stuck to their position in the disk. The metric becomes time-independent.

Which is the correct way to calculate? Probably neither one. They both ignore the acceleration of m when its extra inertia bound to a part of disk accelerates as the disk turns.

The magnetic effect of a simple mass flow

Let us analyze a simpler configuration:

                  1                            3

                      \                      /          ^

                        \                  /          /   v

                          \_______/      mass flow

                                2

                                ^  V        ^

                                |              |  F' force

                                 •  m

                                   \

                                     v    F force

                                  \   F''

                                   v 

                                      ^

                                        \  F'''

    ^ y

    |

     -------> x

We tune V and v so that the mass in the part 3 of the mass flow is almost static relative to the test mass m. The parts 1 and 3 point directly toward m.

When m moves toward the mass flow, it acquires inertia from the mass flowing at the part 1 of the mass flow. The acquired inertia pushes m down to the right with a force F.

The acquired inertia is "carried" by the mass flow from the part 1 to the part 3. The inertia accelerates upward when it takes the turns as it moves from 1 to 3. This exerts another force F' which pushes m up.

The force F' partially cancels the y component of F. The net force

       F + F'

pushes m to the right, and also down.

We conclude that the parts 1 and 2 do exert a "magnetic", horizontal force on m. If we let the parts 1 and 2 "fly loose", so that they do not turn at the bends of the mass flow above, then the effect is essentially the same as for the configuration where the mass flow takes the turns (= accelerates at the turns).

Let us then analyze the extra inertia which m acquires as it approaches the horizontal part 3. The extra inertia pushes m to the right and down with a force F'' but when the extra inertia turns to the part 3, it accelerates to the left and and pushes m up and to the left with a force F'''. Note that if we naively calculate the metric of the mass flow, it is not aware of the force F'''.

The sum

        F + F' + F'' + F'''

pushes m down and to the right.

Since the part 1 is length contracted, gravity pulls m more to the left. Does this cancel the entire magnetic effect of the mass flow? Probably not. If V is much larger than v, then we can ignore all effects ~ v².

Let us analyze under the assumption V >> v.

1.  F_x ~ v V,

2.  F_x' ~ v V,

3.  F_x'' ~  v V,

4.  F_x'''  ~  v V.

All the forces are relevant.

We have yet another force F'''' which is caused by the part 3 slowing down m as m approaches 3.

The force which the part 1 exerts on the test mass m is, to a larige extent, due to the slowing down of the time close to 1. If we imagine that the part 1 "flies loose", then m approaches the slow time zone of 1 at a velocity ~ V + v, while it only approaches the slow time zone of 3 at a velocity ~ V - v. The configuration is asymmetric horizontally.

However, the metric close to 1, 2, 3 is time-independent, and the time runs at about the same rate close to 1 and 3. The metric does not reveal the time asymmetry between 1 and 3.

Accelerations probably are not too relevant, after all

       

                  <--- ω

                   ____

                /          \       M

                \_____ /      

           

                     •  m

      ^ y

      |

       ------> x

Consider again the case of a rotating disk. The configuration is almost symmetric in the x direction. On the left side the parts are accelerating to the right. Some inertia of m accelerates right, causing a force F on m to the positive x direction.

But the force F is canceled by the corresponding effect on the right side.

The acceleration of the far side pushes m down and the acceleration of the near side pulls m up. The sum of forces pulls m upward, but there is no horizontal force.

Conclusions

We now have a very simple hypothetical formula which tells us how the acceleration of M affects a test mass m.

The discrepancy in our calculations of the magnetic effect of a rotating disk on August 10 and August 29, 2023 is not explained by the centripetal acceleration of the parts of the disk.

The discrepancy, probably, is due to the fact that the acceleration of m has to be calculated from its "private" interaction with each part of the disk. If we first sum the metric perturbations for each part, we lose information which is required in calculating the acceleration. This means that the concept of a "metric" is flawed, as we have suspected for the past two months. We will write a new blog post which examines in detail this question.