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Gravitational Waves Are Not Rotational Waves

From the generalized Helmholtz's representation theorem for tensors,

E + curl S = (grad g + g grad) + curl curl g, (1)

where E and S are symmetric 2-tensors, g, will either satisfy

d^2/dt^2 g - c_p^2 div grad g =0, (2)


d^2/dt^2 g + c_s^2 curl curl g =0 (3)

wave equations.

The propagation velocities are

c_p^2 =(lambda+mu)/p (4)

c_s^2 = mu/p. (5)

The subscripts are taken from seismic primary and secondary waves and lambda and mu are the Lame constants with p indicating the density.

It should be clear that gravitation favors (2), and not (3). In fact, the Einstein tensor (linearized) is

G_(ij) =(1/2) (curl curl h)_(ij) (6)

which is known as the incompatibility tensor in the theory of elastic deformations where h is the linearized metric in a Minkowskian (flat) background. In 4D, the incompatibility tensor in (6) is

(curl curl h)_(44) = e_(4klp)e_(4mnp) h_(ln,km)

where the indices in the Levi-Civita pseudo-tensor (k,l) and (m,n) indicate the spatial components, while p represents a dummy index that vanishes after summation.

The vanishing of (6) gives the compatibility conditions for the validity of expressing the elastic strain as a symmetrical gradient of the acceleration, g. When viewed in this light, the Einstein tensor has nothing whatsoever to do with a dilatative field, div g. And even if (2) is valid for wave propagation, the propagation speed c_p > c_s, which if the latter were the speed of light, would mean that we are talking about superluminal velocities for the propagation of gravitational waves.

Consequently, the only thing that makes sense in Einstein's general theory of relativity is his condition of "emptiness" that requires (6), or the Ricci tensor itself, to vanish. Instead of being a condition of "emptiness" are conditions of compatibility. This being the case, his field equations which equate

geometry = energy stresses

is completely meaningless.

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