Sparse contracting shell of masses breaks Gauss's law? No!

Our effort to break Gauss's law for the electric field on January 3, 2025 failed. A uniform charged shell satisfies Gauss's law because the accelerating charges that our test charge q "sees", create a correcting electric field E'.

The form of a line of force for an accelerated charge

Suppose that an electric charge q or a mass m with a newtonian-like gravity force is under a constant acceleration a. What form do lines of force take?

 

           -------___    ___-------  line of force

                          • q

                          |

                          v   a

If q is at a laboratory time t₀ at a location x₀, moving at a constant speed v, then the line of force at a later time t at a distance t c from x₀ points to 

       x₀ + v (t - t₀).

This is the regular retardation formula. But q is not moving at a constant speed. The line of force will be bent.

Let us calculate the form of the horizontal line of force in the diagram.

The form is a parabola. The line of force is as if q would have progressed farther that its real position.

We had in this blog been assuming that retardation shows q lagging behind. This means that our efforts to disprove Gauss's law cannot succeed in the way we presented below.

A sparse shell of masses, accelerating radially

Let us try to break the law using a sparse shell of masses m. The masses form a matrix on a spherical shell whose radius is R.

                                  • m

                                  |

                                  v  a

 

              • --> a          ×          a <-- •

             m                                     m

                                  ^  a

                                  |

                                  • m       R radius

We accelerate the masses inward for a time

       Δt

at an acceleration a, so that the radial velocity becomes a nonrelativistic

       v  <<  c.

The masses move the distance

       s  =  1/2 a Δt²

closer to the center.

Let the spacing between the masses be L.

   

                 m'                    m

                  •          L          •

                 |                      |   

                 v    v                v    v

Let

       t  =  L / c.

The mass m sees the velocity of m' to be

       v'  =  v  -  a t

and m sees the location m' to be a distance

      1/2 a t²

higher in the diagram than the location of m. 

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The simple retardation formula should work in this case? No! We have to derive a more precise formula. The line of force is bent by the acceleration.

In the first section we now derived the formula. It shows that our arguments below do not work.

####

The four m' closest to m will pull m up with a force

       F  ~  4 m / L²  *  sin(1/2 a t² / L)

           ≈  2 m / L³  *  a t²,

and the associated not gained energy in the push is

       Wm  =  F s

                ~  2 m / L³  *  a t²  *  s

                ~  m / L³  *  t²

This means that we gain strangely little energy from the pull of gravity when we contract the shell.

What happens if we halve the value of L? Then m is 1/4 of the old value and t is halved. The value of Wm is halved. But the number of masses m quadruples in the shell. The missing energy W in the entire shell is doubled!

Maybe the missing energy goes to gravitational waves which leave the system?

That is extremely unlikely. If we halve L, then the field of all the masses m on the shell will more closely mimic the field of a uniform massive shell. A uniform massive shell does not generate any gravitational wave at all. That is, the outgoing radiation will have less energy, while the missing energy W is doubled.

Let us calculate the energy which initially might reside in the gravitational waves. The energy of a wave generated by m is directly proportional to m. If we halve L, the sum of the masses m does not change. The initial energy of the gravitational waves does not change appreciably.

Conclusions

We did not succeed in breaking Maxwell's equations yet. We have to look at the Poynting vector and the action of classical electromagnetism. Are there weak points there?