Einstein developed a set of field equations in order to determine the metric coefficients in what would be otherwise the hyperbolic metric of special relativity. At least one metric, so derived, called the Schwarzschild metric can be written as a Hamilton-Jacobi equation (Landau and Lifshitz, *Classical Theory of Fields*, p. 306 4th edition). The Hamilton-Jacobi equation is equivalent to the eikonal equation of geometrical optics. The paths followed by the light rays, which can be interpreted as paths of particles, obey Fermat's principle of least time, or at least extremals. The 4-divergence of the energy-stress tensor in Einstein's equations ensure that the paths of corpuscles also follow the geodesic equation. These conditions provide the conservation of laws of energy and momenta.

So, from the point of view of the metric, we are studying the propagation of electromagnetic waves on curved space-times. Where do gravitational waves come in? A linearization of Einstein's equations readily show that the Ricci tensor, $R_{ik}$ can be written as the d'Alembertian, $\square$ of the perturbed metric, $h_{ik}$,

$$R_{ik}=\frac{1}{2}\square h_{ik},$$ which has now been made a function of time. From this, the conclusion that "gravitational fields, like electromagnetic fields, propagate in vacuum with the velocity of light." (Op. cit.) But this is very different from the metric that was derived on the basis of Einstein's equations, where the propagation speed in the metric is less than the speed of light on account of the presence of masses. In fact, if we choose the $x^i=x$ axis as the direction of propagation, the perturbed metric coefficients $h^k_i$ will satisfy

$$\left(\frac{\partial^2}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)h_i^k=0,$$

which a solution which is any function of $t\pm x/c$. The metric,

$$-c^2dt^2+dx^2,$$

is precisely that of special relativity. So it appears that we have made a complete circuit: Introduced variable metric coefficients, $g_{ik}$, linearized Einstein's equations by introducing a "small" term to $g_{ik}=g_{ik}^(0)+h_ik$, where $g_{ik}^{(0)}$ are the "unperturbed", flat space-time constant coefficients, and unsurprisingly obtained the wave equation!

When discussing the physics of gravitational waves, most, if not all, expositions begin with the quadrupole formula for the inspiraling phase of the merge of two black holes or a neutron star. Avoiding the thorny question that there is no metric which describes such an event, the calculation is handled entirely classically as that of the energy lost by a spinning dumbbell. The closest thing to an energy variable in general relativity is the energy-momentum pseudo-tensor $t^{ik}$, which, as the name implies, is not a tensor at all. That is, it contains at most first derivatives of the metric coefficients and can be made to vanish by a mere coordinate transformation. This raises the whole question as to whether general relativity allows energy to be localized.

The components of the pseudo-tensor are represented by quadratic terms in the perturbation since when $h=0$, the determinant of $g$ differs from that of the unperturbed metric, $g^{(0)}=-1$ by terms of second-order in the perturbation. Hence, the pseudo-tensor is represented to this order as

$$t^{ik}=\frac{c^4}{32\pi G}t^{\alpha,i}_{\beta}t^{\beta,k}_{\alpha},$$

where the comma denotes differentiation with respect to that coordinate. So, general relativity swaps real energy, and energy losses, for a pseudo-tensor that defies the localization of energy in the gravitational field, which, by matter of fact, doesn't exist!

Rather, if we start with the HJ equations

$$g^{ij}\frac{\partial S}{\partial x^i}\frac{\partial S}{\partial x^j}=n^2(r),$$

where $S$ is the action, or phase of the wave font. The fact that the equation is first-order means that the rays follow geodesics. Gravitational effects enter through the metric coefficients, $g^{ij}$ and the index of refraction, $n$, which in a spherical symmetric system can only depend on the radial coordinate, $r$. Landau and Lifshitz give the index of refraction as $(mc)^2+\mathcal{E}^2/[c^2(1-\mathcal{R}/r)]$ for the Schwarzschild metric in the relativistic limit, where $\mathcal{E}$ is the total, constant, energy and $\mathcal{R}$ is the Schwarzschild radius. Yet, they apply it to the advance of the perihelion of Mercury, where the total energy becomes non-relativistic and the necessity of introducing the rest momentum, $mc$, of the peripheral particle is to account for the gravitational potential---of all things.

The fact that gravitational waves and electromagnetic waves propagate at the speed of light in vacuum means that there is a certain similarity between them. The "vacuum" applies to flat-space time for gravitational waves, whereas electromagnetic waves will propagate at a speed, $c/n$ due to the effect of gravity. Without gravity, $n=1$, and that should correspond to the same "vacuum" as for gravitational waves. My question is: What allows general relativity to make reference to gravitational waves when all along it was describing the propagation of electromagnetic waves on curved space-time? For, in fact, the geodesic equations which they expound must refer to one and the same thing is there is any connection between the two.

I would even go further and speculate that any metic which cannot be cast as an HJ equation must be excluded; there is no physical reason for excluding the short-wavelength limit. And the transition between the optical geometrical limit and the finite wavelength limit cannot involve two different types of waves. What is even more discerning is the change in the background metric so that gravitational waves on a flat background must propagate at the speed of light. All traces of the gravitational potential have disappeared so that the claim that it is the propagation of gravitational because the wave equation has been derived from a linearization of the Einstein equations is fanciful.* *

There is an old adage that in the realm of the blind an one-eye man is king. Is this what LIGO is referring to when it says "we are preparing to see the universe with a new set of eyes that do not depend on light"?