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# The Absence of Induction in Einstein Field Equations Means No Gravitational Waves

Updated: Jul 17, 2022

There has been much ado concerning the putative correspondence between Einstein's field equations, at least in their linearized form with electrodynamics. Unfortunately, such a correspondence does not exist.

Based on the geodesic equation for a charged particle in an electromagnetic field,

m[d^2x^(c) /dt^2+ G^(c)_(ab) dx^(a)/dt dx^(b)/dt]= (e/c)F^(c)_(a)dx^(a)/dt, (1)

where the latin indices run over 0,1,2,3, G^(c)_(ab) are the connection (Christoffel) coefficients, and F^(c)_(a) is the Faraday tensor, a completely anti-symmetric second-order tensor.

In the weak field limit, slowly moving limit v<<c,

G^(c)_(ab) dx^(a)/dt dx^(b)/dt -> G^(c)_(00) c^2 + 2G^(c)_{0a) c v^(a).

Now, one wants to capitalize on the possible correspondence between the Christoffel symbols and the Faraday tensor. Lowering indices so that the second becomes the first Christoffel symbols, there results

G_(a0b)=(1/2)(h_(0b),a -h_(0a),b + h_(a,b),0) (2)

where comma dentos differentiation, In the static case, (2) is entirely antisymmetric, and so behaves like the Faraday tensor---as far as symmtry goes. In the time-dependent case, the anti-symmetric component,

f_(ab)= (1/2)(h_(0b),a - h_(0a,b))/2

is extracted out, and claimed that it satisfies a Bianchi identity,

f_(ab),c + f_(bc)_a + f_(ca),b=0,

which gives two of the pseudo-Maxwell equations

div H=0 (3)

and

curl g=-(1/c)dH/dt, (4)

where g is analogous to the electric field E, and H is called the gravitomagnetic vector, in analogy to B.

But note that G is not a tensor so that the transformation properties are quite different than Maxwell's equations. But the seeming logic facade allows the Lorentz force to be generalized to

d^2x/dt^2= (e/mc){E+ v X B} = (1/c){g + v X H},

which is completely fallacious, as we shall now show.

To add to the problems, let us introduce Einstein's field equations,

R_(ab) = -(8 pi G/c^4) S_(ab)

where S_(ab) is the modified energy-stress tensor. The 00 and 01 components are

S_(00)= p c^2/2, and S_(0i)=p c v_i, where p is the density. The Ricci tensor, in the linear approximation (which neglects the gravitational energy that is represented by a pseudo-tensor, being a bilinear product of Christoffel symbols) is

R_(ab)= G^(c)_(ac),b - G^(c)_(ab),c.

Now, on the one hand, for a=0, this gives 4 equations,

div g = - 4pi G p, (5)

while for b=0

curl H = -16 pi/c G p v (6)

Equation (5) corresponds to the Gauss (Coulomb) equation, and (6) is the fourth Maxwell equation. Whereas (4) corresponds to Faraday's law of induction, (6) DOES NOT contain the displacement current. In order to get a circuit equation you would need the Hodge dual, or star, of f_(ab), which doesn't exist because f_(ab) is not a tensor! The Hodge dual would essentially swap 2g for H and H for -2g. In so doing Faraday's law becomes the law satisfied by Ampere's law together with Maxwell's displacement current. Without the possibility of such a swap, the circuit equations have been broken, or, more precisely, can't be formed, and, consequently, there can be no periodic traveling waves--not even in the linear approximation!

The same problem occurred to Einstein when he associated gravitational energy with a bilinear product of Christoffel symbols. The Christoffel symbol is a connection, not a tensor. And anything derived from it like f_(ab) is not a tensor. The fact that its Hodge dual doesn't exist is an affirmation that gravitational waves don't exist.

It is, therefore, somewhat surprising that linearization of the Ricci tensor gives

d^c d_c h_(ab) =(d_0^2 - div.grad)h_(ab) =-16 pi G/c^4 S_(ab) (7)

the wave equation together with the source terms. Now since h_(00)= phi, the Newtonian gravitational potential, and h_(01), the corresponding vector potential, (7) can be written as

d^c d_c phi = (d_0^2-div.grad)phi = -8pi G p = 2 div g (8)

and

d^c d_c A=(d_0^2-div.grad)A = 16 pi G/c pv= - curl H. (9)

The factor of 2 shouldn't worry us because it is the spin of the gravitor, or should it? Multiplying (8) by -2 and taking the time derivative, and adding it to the divergence of (9) give the equation of continuity if we look at the second equalities. If we look at the equality generated by the first set, we get

-2 d_0 phi + div A= 0, (10)

which has the wrong sign when comparing it to the correct gauge condition. Finally, the sum resulting from the third equalities leads to a g which is time-independent. More correctly it is nonsense since the divergence of a curl is zero in the third equation and this requires the diverence of the flux p v to vanish as well as div A. Finally, to this list we add the "colpo di grazie" by taking the curl of (9) to get

where the second term vanishes since (3) holds. Even if H were time-independent, the signs don't correspond in (11).

The "convenient" gauge condition,

d_a h^a_b -d_b h^c_c/2 =0, (12)

which looks a little suspicious when we set indices a=b. These 4 gauges conditions give (10) and d_0 A=0. But, plugging this condition back into (10) gives d_0^2 phi=0, or at least it is second order in v/c [ E G Harris, "Analogy between general relativity and electromagnetism for slowly moving particles in weak gravitational fields"] So what is this second-order term doing in the wave equation (8)???

The main take-away from all this, apartment from the glaring incongruencies, is that since Einstein's field equations do not contain inductive terms, corresponding to Faraday's law and Maxwell displacement current, they do not form circuit equations, and, therefore cannot support traveling waves. The linearization of the Ricci tensor to obtain a wave equation is, at most, a gimic since Poisson's equation is not comparable to a wave equation. In other words, there is no justification for replacing the Laplacian by the d'Alembertian in that equation.