# There is No Relation Between GR and EM

The analogy between General Relativity (GR) and Electromagnetism (EM) was too good to pass up. Being both field theories it seemed like a natural to try and cast GR as EM---at least its linearized version. From this saw the "birth" of Gravitational Waves (GW). However, it was a stillbirth.

Maxwell's equations are

i) Coulomb's law, or Laplace's equation

ii) Faraday induction whereby the curl of the electric field causes a rate of change in the magnetic field;

iii) there are no "magnetic charges" or the divergence of the magnetic field vanishes;

iv) Ampere's relation augmented by the Maxwell displacement current, a form of induction,

Because the tidal force is derived from a scalar potential, so that is second cross derivatives are equal, there is no Faraday induction. Mass in motion is not charge in motion. And since Ampere's law is a geodesic equation, its generalization with the addition of the Maxwell displacement current is not something the GR can handle.

The situation is even worse for GR. Consider a perfect fluid like in the FLRW metric. The radius of the universe, r(t), is a sole function of time. The GR field equations are given in terms of the components of the Ricci tensor. The time component is

R_{tt}=-3(d^2r/dt^2)/r, (1)

and the radial component is

R_{rr}=r(d^2r/dt^2)+2(dr/dt)^2. (2)

If we introduce Newton's second law into (1), it becomes an expression for the tidal force, but in which plane does it act? It has been claimed that (2) is Ampere's law for a stationary current [Costa & Hereiro, "Gravitoelectromagnetic analog based on tidal tensors") The authors were correct to point out that there is no gravitational analog to Faraday's law of induction, but missed the boat when the claimed that (2) was Ampere's law.

Weber wrote Ampere's law as the square of the velocity, V,

V^2_A=2r(d^2r/dt^2)-(dr/dt)^2>0, (3)

in his force equation

(F_W-e^2/r^2)/e^2/r^2)= 1-V^2/c^2,

where c became known as Weber's constant. According to Ampere', the kinetic energy term increases the force, F_W, while work (acceleration X displacement) of the charges opposed it. The factor of 2 in Ampere's equation (2) represents the ratio between longitudinal current elements to parallel current elements. What this has to do with gravitation is another story.

Nevertheless, a comparison between (2) and (3) shows that "work", r(d^2/dr^2/dt^2), and kinetic energy, (dr/dr)^2) hand in hand to increase the current. This cannot be correct. Velocity and acceleration always work against each other. This is clear in the Newtonian current,

V^2_N=2(dr/dt)^2-r(d^2r/dr^2)>0. (4)

The saving grace of the Ampere current as a radial geodesic [cf. *Seeing Gravity*, Sec 5.3.3] That the spatial component of the Ricci tensor, (2), does not have this opposing tendency between velocity and acceleration. Had the spatial component of the Ricci tensor been correct, the Ampere velocity would have been given by V^a=T^a_b u^b, where T is the energy stress tensor, and u is a 4-velocity. But this identification is superfluous since we "know" what the Ampere current is!

Without any recourse to the symmetry of the electric component of the Weyl tensor, we know that if tidal forces are not to lose their identity, there can be no induction like that in EM. If the spatial component of the Ricci tensor would have been correct, there would be no condition on the relative magnitude of the acceleration and velocity. Moreover, the fact that the tidal force is the time component of the Ricci tensor, does not tell us how the tidal force components act in the spatial planes so that to establish tidal equation. Oddly enough, this is contained in Einstein condition of emptiness that the sum of the Ricci components in any given spatial direction vanish [cf. *Seeing Gravity*, Sec. 2.6] But, in that case, the Ricci components are slowly vanish functions of the spatial coordinates for weak fields.