The metric loses information about magnetism for rotating circle

In our June 4, 2024 blog post we remarked that for a rotating circle, the contribution of the velocity component vy of a mass element dm to gravitomagnetism is asymptotically zero, because of the symmetry. How then can vy significantly affect the magnetic field B of a circular current loop?

We wrote about "private" interactions on September 23, 2023.

The metric g₀₂ is not aware of handedness and loses information

The problem seems to be that the metric of general relativity "loses information", which would be required to calculate the magnetic effect.

                         v

                      <-----  rotation

                   ----------- 

                /                 \

              |         ×           |   rotating circle M or

                \                 /     current loop, charge Q

                  • -----------•

                  |             dm' or dq'

                  v   vy 

 mass element

 dm or

 electron dq

                         ^  V

                         |

                         •  test mass m or

                            test charge q

                      ×   ×  B magnetic field

                      ×   ×

         ^ y

         |

          ------> x

When we calculate the magnetic field B at the test charge q, we use the vector (q, dq) in the calculation. In the diagram, the magnetic field dB caused by vy points into the screen (marked with ×), because dq is approaching, at it is to the left of q.

Also the electron dq' creates a magnetic field dB which points into the screen at q.

In the case of the metric g produced by the rotating circle M, the cross term g₀₂ of

         dt dy

is not aware if the contribution came from the left or from the right. The cross term does not know the "handedness". Since the contributions from each side cancel out each other in g₀₂, they do not produce any gravitomagnetic force on m.

The contribution of vy cancels a large part of the magnetic field B produced by vx. This explains, partially, why gravitomagnetism is stronger by a factor 4 than the analogous magnetism.

Should gravity be aware of handedness?

Is general relativity wrong in ignoring the handedness?

Gauss's law seems to require that for "free" fields, gravitomagnetism must be identical to the analogous magnetism. Can we allow that it differs for a rotating circle?

Could electromagnetism have a 4-fold magnetic field B for a closed current loop?

The magnetic field in electromagnetism is probably dictated by special relativity, and Gauss's law.

For a rotating current loop, there is no outgoing radiation. It might be that the Gauss's law could be satisfied with different strenghts of the magnetic field B. Also, it might be that special relativity would allow the same.

                  ω rotation

                <---

                     ___   Q charge

                  /

                |        × center

                  \____   

                      B 

                  •   •   •

                  |   |   |   E

Suppose that we have a charged half-circle rotating around the center. The magnetic field in this case has to make sure that Gauss's law is satisfied. The lines of of force of (partially induced by dB / dt) the electric field E must not break.

The half-circle gives out electromagnetic radiation.

Let us then combine the half-circle with another half-circle. We have a full circle now, and there no longer is electromagnetic radiation going out.

Could it be that B can change significantly for a full circle?

If that is the case, then nature must somehow know that the circle is now full, and it can strengthen B. We could imagine that if there is no radiation out, then B can somehow grow stronger. But this sounds far-fetched.

Empirically we know that magnetism works the same way for a half-circle and a full circle.

Gravitomagnetism has different fluxes inside the circle and outside it

In electromagnetism, vy reduces the magnetic field outside the current loop, because dB from vy points to the opposite direction to dB from vx.

But inside the loop, these dB point to the same direction.

The metric in gravity loses the information about vy outside the loop. The magnetic field is 4X because of that.

What about inside the loop?

                                                     ^  ω

                                                        \

             |                                  |

          |                   ^ v                |

          |                   |                    |

             |                •                 |

                              m

   ^ y

   |

    ----> x

The information about vy is lost also inside the loop.

In electromagnetism, the effect of vy is to strengthen the magnetic field B inside the loop. This means that the gravitomagnetic field is surprisingly weak inside the loop.

It looks like the hypothetical gravitomagnetic field lines do not close in gravitomagnetism in general relativity!

This may be yet another fatal flaw in general relativity.

If magnetic field lines break, then we have "gravitomagnetic monopoles" in empty space. Does this break the Einstein equations there?

Geodesic deviation equation

                           ^   ^ test masses turn up

                           |   |

                      Bg   

                 ... ----------

                          ×   ×   ×    test masses m

                 ... ----------

                          ×   ×   ×    going into the screen

                 ... ----------

                                ^

                                |

                   gravitomagnetic

                   field lines break here

     ^ z

     |

     •-----> x

     y points into the screen

Suppose that gravitomagnetic field lines break in empty space.

It looks ugly, but we are not able to prove that the Ricci tensor R then necessarily differs from zero. 

If "gravitoelectric" field lines would break, then a cube of initially static test masses will increase in volume, and R₀₀ is then non-zero. But breaking magnetic lines do not seem to cause the same effect.

Conclusions

For a rotating circle, disk, or sphere, the metric of general relativity loses information, compared to the magnetic field B of the analogous electric current loop.

This makes the gravitomagnetic field ugly: gravitomagnetic field lines break. We conclude that general relativity probably has a wrong approach to the problem.

We believe that the correct gravitomagnetic field of a rotating object is fully analogous to the magnetic field of the analogous electric current, but we are not yet able to prove that that is the case.

The metric of a rotating sphere

On May 29, 2024 we used a rogue metric variation to prove that the Einstein field equations do not have any solution for a rotating mass. 

The Einstein field equations have no solution for a rotating mass, but we can still try to build a metric g which is "in the spirit" of general relativity. We can then compare it to the Kerr metric.

Let us recapitulate our argument against the existence.

A rogue metric variation δg proves that the Einstein field equations have no solution for a rotating mass

https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action

Let g be the solution of the Einstein field equations for a rotating uniform mass sphere M. M is kept from collapsing by pressure and the centrifugal "force".

                       rotation

                          -----> 

                            M

                         •       •

                  •      •       •      •

                  •      •       •       • m ---->  lift higher

                  •      •       •      •

                         •       •           --->  pressure

                                              --->  centrifugal

                                                      "force"

                         <----- 

                      rotation

     ^ y

     |

      ------> x

The action integral S of LM contains:

       ( kinetic energy of m

        - energy of pressure on m

        - m c² )

       * sqrt(-det(g)).

Let us assume that the spatial coordinates are cartesian. Using a rogue metric variation δg, we move the coordinate lines around a particle m a little bit to the right, relative to the spacetime geometry. The integral of R does not change.

The y metric does not change at m. Let the absolute value of the metric of time |g₀₀| ≈ 1 grow by a small amount ε at m, which means that the proper time at m runs by a factor

       1  +  ε / 2

faster.

The coordinate velocity of m does not change. Since proper time now runs faster, an observer standing on m measures that the proper kinetic energy of m decreased by a factor

        1  -  ε.

This is partially compensated by |g₀₀| at m growing by the factor

       1  +  ε / 2.

Thus, the product

       (kinetic energy of m)  *  sqrt(-det(g))

decreases by a factor 1 - ε / 2.

The product

       (energy of pressure on m)  *  sqrt(-det(g))

is the "pressure potential" of m. The product

       m c²  *  sqrt(-det(g))

tells us the newtonian gravity potential of m. If we move m to the right, the newtonian gravity potential increases more than the pressure potential decreases (remember the centrifugal "force" which supports m).

We conclude that the integral of LM decreases if we move m to the right. This contradicts the assumption that g is a solution of the Einstein field equations.

Stretched radial metric grr in cartesian coordinates

Let us have a mass element dm at x, y coordinates (0, 0). We have an observer at coordinates (x, -y), where |y| >> |x|. What is the metric in cartesian coordinates?

We have:

       r²  =  x²  +  y²

  =>

       r dr  =  x dx  +  y dy

  =>

       dr  =  x / r  *  dx  +  y / r  *  dy.

Let dn be the normal to dr:

       dr²  +  dn²  =  dx²  +  dy².

The metric:

       ds²  =  (1  +  hrr) dr²  +  dn²

               =  hrr dr²  +  dr²  +  dn²

               =  hrr dr²  +  dx²  +  dy².

There:

       hrr dr²  =  hrr  * 

                        (x² / r²  *  dx²  

                         + 2 x y / r²  *  dx dy

                         + y² / r²  *  dy²).

Lorentz transforming a Schwarzschild metric perturbation

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

We want to calculate the metric produced by a moving mass element m. The metric perturbation h created by m is approximately Schwarzschild, and in many cases we can obtain an approximate solution of the Einstein equations by summing metric perturbations.

                       ^  y

                       |

                       |

                 m  ●----> vx  

                       |\     

                       |  \ 

                       |    \

                       vy    v   v

                            R

                                • observer  (x, -y)

The mass element m is at x, y coordinates (0, 0). The observer is at (x, -y), where

       |x|  <<  |y|.

We denote x² + y² = R².

The components of the velocity v are vx and vy.

Let η be the flat metric. We want to transform, at the observer,

       η  +  h

to coordinates moving at a velocity v. We will transform twice, for each component of v. Let us calculate the transformation for vx.

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

The metric at the observer is:

       c² dτ²  =  -(1  -  rs / r) c² dt²

                        + (1  +  rs / r  *  x² / r²)  *  dx²

                        + (1  +  rs / r  *  y² / r²)  *  dy²

                        + 2 rs / r  *  x y / r²  *  dx dy

                        + dz²,

where rs = 2 G m / c² and r = -y. Lorentz says:

       dt   =  dt'  +  vx / c²  *  dx',

       dx  =  dx'  +  vx dt',

       dy  =  dy',

       dz  =  dz',

       c² dτ²  ≈  -(1  -  vx² / c²) (1  -  rs / r) c² dt'²

                        + (1  +  rs / r  *  x'² / r²)  *  dx'²

                        + (1  +  rs / r  *  y'² / r²)  *  dy'²

                        + 2 rs / r  *  x' y' / r²  *  dx' dy'

 

                        + dz'²

    

                        + (2 vx  -  2 vx  +  rs / r * 2 vx) dt' dx',

where we used the facts that x' ≈ x, y' = y, and that the product rs vx * |x|/ r is very small. The value of r is approximately the same in the new coordinates.

The difference to the static metric is

       vx² dt'²

and the cross term

       2 vx rs / r  *  dt' dx'.

The velocity vx makes the metric perturbation of time t to "spill over" to the cross term dt' dx'.

The velocity vy, obviously, has a similar effect. A difference is that also the perturbation of the y metric spills over to the cross term.

Summing the effects of vx for a rotating circle

We must sum the term for each mass element dm:

       rs / r  *  vx  *  dt' dx'.

Let

       r'  =  distance (observer, center of circle).

            •  dm'

               \

                  \

                     \

                       ×

                       |  \                    ^   ω

                       |     \   R           /

                       |        \

                       |          • dm       M

                           α

                       r' = distance(×, observer)

          

                       •  observer

Let us integrate it for a circle of a radius R, which turns at an angular velocity ω. The mass of the circle is M. The distance of dm from the observer is:

       r  =  sqrt( (r'  -  R cos(α))² + R² sin²(α) )

           ≈  r' (1  -  R / r'  *  cos(α)).

Since vx has opposite values for dm and its mirror image dm' relative to the center of the circle, we can sum their contribution. We have:

       rs   =  2 G / c²  *  M / (2 π)  *  dα,

       vx  =  R ω cos(α),

                    π / 2

       h₀₁  =     ∫  

                  -π / 2

                     2 G / c²  *  M / (2 π) 

                     * R ω cos(α)

                     * 1 / r'  *  (1  +  R / r'  * cos(α)

                                       - 1  +  R / r'  *  cos(α))

                     * dα.

The integral

           π / 2

            ∫  2 cos²(α) dα  =  π.

       -π / 2

We get:

       h₀₁  =  G / c²  *  M ω R² / r'²

              =  G / c²  *  J  / r'²,

where J is the angular momentum of the circle.

The part of the metric perturbation due to the rotation of the disk or the sphere is:

      0                        G / c² * J / r'²    0      0

      G / c² * J / r'²    0                        0      0

      0                        0                        0      0

      0                        0                        0      0.

Summing the effects of vy for a rotating circle

We studied this extensively in August 2023.

The symmetry of the disk makes the cross term of t' y' in the metric zero if the observer is at (0, -y) in the diagram.

                       ^ y'

                       |

                      ● -----> x'

                     M  (0, 0)

                      

                      r'

              

                      • observer (0, -y')

                 <----> cross term of t' x'

When the observer moves to the x direction, the cross term will have a non-zero value. But is the value negligible?

Let the observer move to a position (x₀', -y₀'), where |x'| is small. The "radial" coordinates there are rotated through an angle x₀' / y₀' = β.

       x''  =  x'  +  β y',

       y''  =  y'   -  β x'.

Let us have a cross term in the metric at the new position:

       C t' x''  =  C t' x'  + C t' β y'.

The cross term t' y' in the old coordinates is

       C x₀' / y₀'  *  t' y'.

We have

       C  =  2 rs / r  *  vx.

If we make y₀' large, then the cross term of t' and y' is negligible relative to the cross term of t' and x'.

It looks like that vy would not contribute at all. The reason is that each dm has a centripetal acceleration.

We are not sure if this is the correct "approximate" solution in general relativity. If the mass M is lightweight, then a negative pressure keeps the parts dm in the circular orbit. If we let a rogue variation to move a part dm farther from the center of the circle, then the action integral of the matter lagrangian LM decreases. That shows that the Einstein field equations have no solution.

The fact that the t' y' cross term is negligible is due to the same reason why the Einstein equations have no solution! It does not sound like a proper approximate solution!

Test mass m acceleration due to the rotation of M

Now we can calculate the gravitomagnetic effects on m.

                     J

           |       ● M

           v ω

                     r'

                     ^  V

                     |

                     • m

     ^ y

     |

      -------> x

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

The x coordinate acceleration is

       d²x / dτ²  =  -Γ¹₀₀  *  dt / dτ  *  dt / dτ

                            - 2 Γ¹₀₂  *  dt / dτ  *  dy / dτ

                            - Γ¹₂₂  *  dy / dτ  *  dy / dτ

                        =  -2 * 1/2 g¹¹ dg₀₁ / dy  *  1  *  V

                            - 2 * 1/2 g⁰¹ dg₀₀ / dy  *  1  *  V

                        ≈  -V dg₀₁ / dy

                        =  V  *  G / c²  *  2 J  / r'³.

The gravitomagnetic moment, as defined by our August 10, 2023 blog posts, is 2 J. It is 4 times the analogous magnetic moment of a rotating electric charge. This is probably the origin of the strange 4-fold gravitomagnetism in literature.

Tangential movement of a test mass

Above we calculated the effect of vx if the test mass m approaches M. What about m moving tangentially?

                            ●  M

     

                             • -----> V

                            m

       d²y / dτ²  =  -Γ²₀₀  *  dt / dτ  *  dt / dτ

                            - 2 Γ²₀₁  *  dt / dτ  *  dx / dτ,

       Γ²₀₀  =  1/2 g²²  *  -dg₀₀ / dy

                =  1/2  *  (1  +  rs / r)  *  rs / r²

                =  1/2 rs / r²  +  1/2 rs² / r³,

       Γ²₀₁  =  1/2 g²²  *  -dg₀₁ / dy

                ≈  1/2  *  -G / c²  *  2 J  / r'³.

                =  -G / c²  *  J / r'³.

Since both sides in the equation for d²y / dτ² contain ... / dτ², we can instead write .../ dt².

The y coordinate acceleration due to vx and V is

       d²y / dt²  =  V  *  G / c²  *  2 J  / r'³.

This agrees with the previous section. The gravitomagnetic moment is 2 J.

On top of that we have the regular

       1/2 rs / r²  =  G M / c²  *  c²  *  1 / r²

                         =  G M / r²

gravity acceleration.

But is the y coordinate "natural" here? If we make coordinate lines to "bulge", we can obtain spurious coordinate accelerations, which depend on V.

A single moving mass element dm creates a time-dependent metric. We have forgotten that in the analysis above!

The gravitomagnetic field of a single moving mass

Let us calculate a very simple example.

              M ● ----> v

                   r

              m • -----> v

       ^ y

       |

        • ----> x

 observer

The mass M is moving at a velocity v to the x direction, and so is the test mass m.

If both were static in our coordinates, then a static observer would measure a y acceleration of

       d²y / dt²  =  G M / r²,

where r is the Schwarzschild radial coordinate of m.

But proper time in the system M, m is slower than for the observer. He measures an acceleration of

       d²y / dt²  =  (1  -  v² / c²) G M / r².

The term

       v  *  v / c²  *  G M / r²

can be interpreted as a "magnetic" effect. It corresponds to a gravitomagnetic field of

       v / c²  *  G M / r²,

which is analogous to a magnetic field

       B  =  v / c²  *  E.

There is no factor 4. 

Let us use the geodesic equation to calculate the acceleration. The metric in the laboratory coordinates is:

       c² dτ²  ≈  -(1  -  v² / c²) (1  -  rs / r) c² dt²

                        + (1  -  v² / c²)  *  dx²

                        + (1  +  rs / r)  *  dy²

 

                        + dz²

    

                        + rs / r  *  2 v dt dx,

where

       rs  =  2 G M / c².

The acceleration:

       d²y / dt²  =  -Γ²₀₀  *  dt / dt  *  dt / dt

                            -2 Γ²₀₁  *  dt / dt  *  dx / dt

                        =  (1  +  v² / c²)  G M / r²

                             - 2  *  1/2 g²²  *  -dg₀₁ / dy  *  v

                        ≈  (1  +  v² / c²) G M / r²

                            - 1  *  1  *  rs / r²  *  v  * v

                        =  G M / r²

                            - v² G M / c².

                  M

                   ● --> v

                   ^  V

                   |

                   •  m

   ^ y

   |

    -----> x

Let us calculate the x acceleration.

       d²x / dτ²  =  -2 Γ¹₀₂  *  dt / dτ  *  dy / dτ,

                        =  -2 * 1/2 g¹¹ dg₀₁ / dy  *  1  *  V

                            - 2 * 1/2 g⁰¹ dg₀₀ / dy  *  1  *  V

                        =  -V d(v rs / r) / dy

                        =  V v  *  2 G M / c²  * 1 / r².

What is wrong in the above calculation?

An obvious problem is that dm is accelerating. Can we sum the metric perturbations as if the mass elements dm were moving at a constant velocity?

We wrote at length about this problem in August 2023.

Now we know that the Einstein equations do not have a solution at all for a rotating system, because it is "dynamic".

Conclusions

If we want to preserve Gauss's law for newtonian gravity for a gravitational wave, we need magnetic gravity which is exactly analogous to magnetism. The condition R₀₀ = 0 states Gauss's law for gravity.

In the above calculations, we were able to track the origin of the strange factor 4 in literature, of the gravitomagnetic effect of a rotating sphere.

If gravitomagnetism for gravitational waves is exactly analogous to electromagnetism, then it would be surprising if "static" gravitomagnetic fields differ from electromagnetism by the factor 4. How does nature know to make the effect 4-fold in certain configurations?

We will investigate this in the next blog post.

Gravitomagnetism and Gauss's law

Edward M. Purcell calculated the energy of an electromagnetic wave from a retardation rule and the principle that Gauss's law must hold for the electric field. Gauss's law states that lines of force can never break.

https://physics.weber.edu/schroeder/mrr/MRRtalk.html

In the diagram we have the electric lines of force for a charge which was suddenly moved to the left. The lines are kept non-breaking by a magnetic field which varies with time.

Ricci curvature R₀₀ = 0 means that the lines of force must not break

Having the Ricci curvature R₀₀ = 0 (the curvature to the time direction) means that a cube of test masses m, which are initially static, must preserve its volume when the masses are let to float freely.

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

The movement of test masses is like a flow of incompressible liquid. The lines of force acting on the test masses must not break – otherwise the density of test masses would not stay constant. The vector field defined by the electric field E must be "sourceless", or solenoidal.

Is it so that gravitomagnetism has to be exactly analogous to ordinary magnetism? Yes, for single accelerating masses there can be no factor 4

Newtonian gravity and its lines of force look exactly like the Coulomb force. Let us assume that retardation of gravity is identical to the electric force. Let us also assume that weak gravity fields obey Gauss's law.

Then a good hypothesis is that gravitomagnetism acts on test masses exactly like magnetism acts on test charges. 

In our November 13, 2023 we presented a hypothesis that gravitomagnetism is exactly analogous to magnetism. We have here more evidence that supports that hypothesis.

However, we must note that the energy density of a gravitational wave is 16 times the analogous electromagnetic wave. Changes in the spatial metric seem to carry this large energy density. The gravitomagnetism which we consider in this blog post concerns the metric of time. The metric of time produces the newtonian gravity force.

https://inspirehep.net/files/9d435cdaeb6fef055827a8d77facaf12

A. Barros (2016) writes that for weak fields, the Kerr metric is exactly analogous to the electric and magnetic field of a spinning charge. The lagrangian is identical.

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

On August 7, 2023 we wrote that in the literature, the gravitomagnetic field of a spinning mass is claimed to be 4-fold compared to the analogous spinning electric charge. Is this possible? What happens in electromagnetism if we make the magnetic field 4-fold?

If there would just be retardation and no magnetic field would exist, then in the diagram, the electric lines of force would break at the circle.

https://en.wikipedia.org/wiki/Maxwell%27s_equations

A changing magnetic field is able to connect the ends of the lines of force through the formula:

If we make B 4-fold, then the electric field along the circle will be too strong. We end up with too many lines of force along the circle.

The gravitomagnetic field for a moving single mass cannot be 4-fold relative to the analogous magnetic field of a moving single charge.

In this example, the gravitomagnetic field is already "detached" from the point mass which generated it. The field of slowly moving masses may be quite different from the field of a rapidly accelerated mass.

What about "current loops"? A rotating mass is a "loop" of a mass current. Could it be that a loop somehow has a 4-fold gravitomagnetic field?

That would be strange. How would nature know which movement of an atom constitutes a mass current loop?

The spatial metric around a static spherical mass shell cannot be calculated by summing metric perturbations for each atom, as we have noted in this blog. Could it be that there is a similar nonlinear effect which radically changes the gravitomagnetic effect of a rotating mass shell?

Using the Einstein approximation formula to calculate the gravitomagnetic effect of a sphere

https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession

The strange factor 4 in the literature probably originates from calculations of Josef Lense and Hans Thirring in 1918. They defined a metric which is a weak field approximation of the Kerr metric.

On May 29, 2024 we showed that no rotating mass M can have a metric which solves the Einstein field equations. Therefore, we cannot define what is the "correct" metric around a rotating mass in general relativity. Anyway, we have to check how Lense and Thirring got the factor 4.

https://www.zarm.uni-bremen.de/uploads/tx_sibibtex/2001LaemmerzahlNeugebauer.pdf

Claus Lämmerzahl and Gernot Neugebauer (2001) give an elementary derivation of the Lense-Thirring metric. They seem to apply the Einstein approximation formula (see the blog post August 16, 2023) to the metric tensor, including the mass flow in the rotating body:

where the "frame-dragging" off-diagonal component is:

in the metric:

Let us analyze this. How does the factor 4 come about?

Let us have a mass circle which rotates around its center. Below we have drawn two elements dm of the circle. We calculate what the Einstein approximation formula says.

We assume weak fields.

                                                ^  v

                                           \       \ 

                          ×               dm

          \               center

         dm           r = radius

                          α = angle(center-dm, x axis)

  ^ y

  |

   -----> x

                        ^   V

                        |

                        •    m test mass (0, -y)

We assume that dm = dα. Then the mass of the entire circle is 2 π and its angular momentum L = 2 π v r. The coordinates of the circle center are (0, 0) and the test mass m (0, -y).

In the diagram, the contribution of the two elements dm to the t, x component of tilde-g₁ at the test mass m is κ c² times:

       dm sin(α) v / (y  -  r sin(α))

       - dm sin(α) v / (y  +  r sin(α))

       ≈  dα sin(α) v  *  1 / y  *  2 r / y  *  sin(α)

       =  2 v r / y²  *  sin²(α) dα.

The integral of sin²(α) from 0 to π is π / 2. The contribution of the full circle is

       κ c² π v r / y²

       = 8 π G / c²  *  π v r / y²

       = 4 π G / c²  *  L / y²

The formula we obtained is totally different from the one in the paper. Strange. Maybe the authors have set 8 π / c² = 1?

Anyway, let us continue our calculation. The Einstein approximation formula indicates that the metric g at the test mass m is something like

      g  =

   ≈ -1                             4 π G / c²  *  L / y²    0    0

   4 π G / c²  *  L / y²    ≈ 1                              0    0

   0                                 0                               ≈ 1  0

   0                                 0                                0  ≈1,

in cartesian coordinates t, x, y, z. What is the strength of magnetic gravity there?

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

Let us use the geodesic equation to determine the x acceleration of a test mass m which approaches the circle at a velocity V. In the formula above, s is the proper time τ of the test mass.

We want to determine

       d²x / dτ²  =  -Γ¹₀₀  *  dt / dτ  *  dt / dτ

                            - 2 Γ¹₀₂  *  dt / dτ  *  dy / dτ

                            - Γ¹₂₂  *  dy / dτ  *  dy / dτ

                        =  -Γ¹₀₀  -  2 Γ¹₀₂ V  -  Γ¹₂₂ V².

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

       Γ¹₀₀  =  1/2 g¹⁰ (0  +  0  -  0)

                    + 1/2 g¹¹ (0  +  0  -  0)                =  0,

       Γ¹₂₂  =  1/2 g¹⁰ (0  +  0  -  0)

                    + 1/2 g¹¹ (0  +  0  -  0)                =  0,

       Γ¹₀₂  =  1/2 g¹⁰ (0  +  dg₀₀ / dy  -  0)

                    + 1/2 g¹¹ (0  +  dg₀₁ / dy  -  0)

                ≈  1/2 dg₀₁ / dy.

On the last line we use the fact that both g¹⁰ and dg₀₀ / dy have very small absolute values and their product is negligible, and that g¹¹ ≈ 1. We then have:

      d²x / dτ²  =  -V  *  d (4 π G / c² * L / y²) / dy

                       =  V  *  8 π G / c²  *  L / y³.

The gravitomagnetic force on the test mass is

       Fgm  =  m V  *  8 π / c²  *  G  *  L / y³.

Let us compare this to the magnetic field of a electrically charged rotating sphere. The magnetic dipole moment m of a solid, uniformly charged sphere is

       m  =  1/2 Le,

where Le is the "charge angular momentum" of the sphere, defined analogously with the mass angular momentum:

       Le  =  Σ  qi v r.

                 i

In the equatorial plane, the magnetic field of the rotating sphere is:

       B  =  μ₀ / (4 π)  *  m / y³

            =  1 / (8 π ε₀ c²)  *  Le / y³.

The magnetic force on a test charge q is:

       Fem  =  q V  *  1 / 2  *  1 / c²

                   *  1 / (4 π ε₀)  *  Le / y³.

The gravitomagnetic force is 16 π times the analogous magnetic force. This does not make sense. The Einstein approximation formula produces wildly wrong results.

The gravitomagnetic effect of a moving cylinder in general relativity: the factor 4 comes up here

On April 26, 2024 were able to solve an approximate metric around a long uniform cylinder, from the Einstein field equations. Let us use our result to derive the path of a test mass which approaches the cylinder obliquely. If we imagine that the cylinder is moving and the test mass approaches it along a normal, then we also have calculated the gravitomagnetic effect in general relativity.

    z       ρ = mass / length

 --->      ===============  static long cylinder

|

v r                    Vr

                          ^     ^  V

                          |   /

                          | /

                          • ---> Vz

                         m

                         test mass

The cylindrical coordinates are t, r, φ, z.

We derived:

       dg₁₁ / dr  =  r d²g₀₀ / dz²

                       ≈  0.

If the cylinder is very long, then |d²g₀₀ / dz²| is very small, and we can assume that the metric component g₁₁ does not depend on r.

On April 28, 2024 we derived:

       dg₃₃ / dr  =  dg₀₀ / dr  +  r d²g / dz²

                        ≈  dg₀₀ / dr.

Let us assume that the test mass m is near the midpoint of the cylinder. Then off-diagonal components in the metric g are very small and can be neglected.

The metric of time g₀₀ (which is close to -1) is calculated from newtonian gravity:

       dg₀₀ / dr   =  -2 G m ρ / r  *  1 / (m c²)

                         =  -2 G / c²  *  ρ / r,

where ρ is the mass of the cylinder per length. Note that dτ²  =  g₀₀ dt², from which the factor 2.

We use the geodesic equation again:

If we assume that

       |V|  <<  c.

Then

       d²z / dτ²  =  -Γ³₀₀ dt / dτ * dt / dτ

                            - 2 Γ³₀₁ dt / dτ  *  dr / dτ

                            - Γ³₁₁ dr / dτ  *  dr / dτ

                            - 2 Γ³₀₃ dt / dτ  *  dz / dτ

                            - 2 Γ³₁₃ dr / dτ  *  dz / dτ

                            - Γ³₃₃ dz / dτ  *  dz / dτ.

The Christoffel symbols:

       Γ³₀₀  =  1/2 g³³  (0  +  0  -  0),

       Γ³₀₁  =  1/2 g³³  (0  +  0  -  0),

       Γ³₁₁  =  1/2 g³³  (0  +  0  -  0),

       Γ³₀₃  =  1/2 g³³  (0  +  0  -  0),

       Γ³₁₃  =  1/2 g³³  (dg₃₃ / dr  +  0  -  0),

       Γ³₃₃  =  1/2 g³³  (0  +  0  -  0).

We obtain:

       d²z / dτ²  =  -dg₃₃ / dr  *  -Vr  *  Vz 

                        =  -Vr Vz  *  2 G / c²  *  ρ / r.

The proper time τ of the test mass m slows down as it descends closer to the cylinder. Let us calculate the effect. The slowdown is due to two things:

1.   the metric of time |g₀₀| is slower close to the cylinder, and

2.   the velocity of the test mass grows close to the cylinder, which slows down its proper time.

Item 1:

Let m approach the cylinder for a unit time. The rate dτ / dt (≈ 1) slows down by

       Vr G / c²  *  ρ / r,

which translates to a z acceleration:

       -Vr Vz G / c²  *  ρ / r.

Item 2:

The effect is as large as for item 1:

       -Vr Vz G / c²  *  ρ / r.

The combined effect of all of the above is

      -Vr Vz  *  4 G / c²  *  ρ / r.

We finally found a possible origin of the strange factor 4 in the literature!

A half of the effect is due to the stretching of the z metric g₃₃ close to the cylinder. One fourth comes from the slowdown of time close to the cylinder. One fourth comes from the increased kinetic energy. Are these really "magnetic" effects?

The analogue between gravity and electromagnetism, in the case of a moving cylinder and a test charge approaching it

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

              ρ = charge / length

            ==================   --> Vz 

                       cylinder

                           ^

                           |

                           • q test charge

The "electric field":

Gravity:

       Eg  =  G ρ / r.

Electromagnetism:

       E  =  1 / (4 π ε₀)  *  ρ / r,

where ρ is the charge per length.

The "magnetic field":

Gravity: 

       Bg  =  4 Vz / c²  *  Eg.

Electromagnetism: 

       B  =  Vz / c²  *  E.

A test mass moving along a moving cylinder

Let us check if the analogue extends to other movements than a normal movement.

              ============  static cylinder

                        •---> Vz 

    

       d²r / dτ²  =  -Γ¹₀₀ dt / dτ * dt / dτ

                            - 2 Γ¹₀₁ dt / dτ  *  dr / dτ

                            - Γ¹₁₁ dr / dτ  *  dr / dτ

                            - 2 Γ¹₀₃ dt / dτ  *  dz / dτ

                            - 2 Γ¹₁₃ dr / dτ  *  dz / dτ

                            - Γ¹₃₃ dz / dτ  *  dz / dτ.

The three last terms are associated with the velocity

       Vz  =  dz / dτ,

and can create a "magnetic" force. We can discard the one with the dr / dτ factor.

       Γ¹₀₃  =  1/2 g¹¹ (0  +  0  -  0),

       Γ¹₃₃  =  1/2 g¹¹ (0  +  0  -  dg₃₃ / dr)

                =  -1/2 g¹¹ d(-2 G / c²  *  ρ / r) / dr

                ≈  -G / c²  *  ρ / r².

Then

       d²r / dτ²  =  Vz²  *  G / c²  *  ρ / r².

The associated force is proportional to the square of the velocity Vz. It is not a "magnetic" force.

Let us assume that the cylinder moves to the right at a very slow speed v. This simulates an electric current, which moves very slowly inside a wire.

            ==========  --> v

                    • ------> Vz 

A "magnetic" force arises:

        d²r / dτ²  =  2 Vz v  * G / c² * ρ / r².

But depends on 1 / r², not on the "electric" field which is ~ 1 / r.

We cannot find anything which would resemble electromagnetism. Thus, gravitomagnetism makes no sense for a movement parallel to the cylinder.

Conclusions

For a moving cylinder, there is not a good analogy with magnetism in general relativity.

If we look at a "free" gravity field, it would be nice to have a mechanism which imitayes the magnetic field, in order to preserve Gauss's law.

We will write another blog post about the gravitomagnetic field of a rotating mass. It is not clear to us what kind of a magnetic force is natural for general relativity. Is the Kerr metric natural?

In our own Minkowski & newtonian gravity model, gravitomagnetism is not yet clear, either.