The answer is that the electro-kinetic momentum, aka vector potential, is what generates gravitational waves. Gravitational waves are described by a linear(ized) wave equation, which has no nonlinear generalization.

In analogy with with elasticity, the general displacement field is now the acceleration, a_j, and we ask under what circumstance does a second-order symmetric tensor g_(ij) happen to be a strain tensor for the displacement field, a_i; that is,

g_(ij) = d_i a_j +d_j a_i= g_(ji). (1)

The answer to this question was found by Saint-Venant in 1864 in which he found that (1) applies when

(curl curl g)_(ij) = e_(iab) e_(jcd) d_a d_c g_(bd) = 0, (2)

where e_(ijk) is the Levi-Civita completely asymmetric tensor.

Equation (2) is a generalization of the condition for that a vector field be a gradient field, i.e., the curl of the field must vanish. And that is what is precisely done in general relativity. For small perturbations from a flat metric

g_(ij) = delta_(ij) + e h_(ij),

where delta is the Kronecker delta and e is a small parameter such that square terms and higher can be neglected, the Einstein tensor in 3D can be written as

G_(ij) = R_(ij) - (1/2) R,

=(1/2)e e_(ikl) e_(jmn) d_k d_m h_(ln), (3)

which, upon comparison with (2), is the Saint-Venant operator operating on the linearized metric. It's vanishing is precisely the condition for h_(ij) to be a strain tensor. What changes in 4D?

As we mentioned, specialization to the case where h_(ij) is time independent. This isolates the terms h_(00) and h_(0j), and the derivatives of these terms are with respect to the space coordinates only. What is done [See Ohanian & Ruffini, Gravitation and Spacetime] is to construct the analog of the Faraday tensor,

f_(ij) = (k/2){d_j h_(0i) - d_i h_(0j)}, (4)

where k=[16 pi G]^(1/2), the Einstein gravitational coefficient in units where c=1. Now comparing (1) and (4), we have introduced an asymmetric (pseudo) tensor, and cannot, therefore expect, that the "electric" field will be a gradient field.

In terms of the Faraday tensor (4), the equation of motion reads

dv_k/dt =- f_(0k) +2 f_(kj)v^j, (5)

for a 3-velocity, v_k. We can write (5) in vector notation,

dv/dt = E + v X B, (6)

where the E and B are the gravitational, and gravitomagnetic fields, respectively,

E_k =(k/2) d_k h_(00), (7)

and

B_k = k { d_k h_(0i) - d_i h_(0k)}v^i. (8)

Now, it is clear from (7) that E is a gradient field so that Faraday's law

curl E = - (1/2) dB/dt (9)

must vanish. This is equation (3.103) of Ohanian and Ruffini. It is apparent that they did not connect this to the fact that their gravitational acceleration, -(k/2) grad h_(00), is defined following their equation (3.104).

Furthermore, they write the second circuital equation (their equation (3.103)),

curl B = -16 pi G P, (10)

as if it were the definition of the Poynting vector, which they associate with the 0k component of the energy-stress tensor. Even if (9) and (10) were valid, they would not lead to a wave equation when either E or B is eliminated, since the Maxwell displacement current, dE/dt, is absent in (10). So the Poynting vector, P, is the momentum-energy flux of "nothing".

Since (7) corresponds to (1), the game is over. The description in terms of a scalar Newtonian potential of the tidal forces implies that (8)---the so-called gravitomagnetic field---vanish. Hence---just as in 3D---the vanishing of the Einstein tensor is the condition for the existence of a symmetric strain tensor. It is symmetry and traceless, which upon diagonalization represents the tidal force components in the 3 orthogonal directions. Without the time derivative of the vector potential, A, appearing in (7), there can be no Faraday induction (7) where B= curl A.