Gravitational Waves: From Schwarzschild to Michelson

We know very well how a static gravitational field influences the propagation of light from Schwarzschild's solution of Einstein's field equations. But, do we know how light behaves in a passing gravitational wave? It is argued that gravitational waves are "ripples in spacetime", and their effect is to stretch and contract it so that spacetime is a new aether and not a vacuum in which light (and gravitational waves) propagate. This is why an interferometer can be used to detect them by the effect that gravitational waves have on light propagating in an arm of a Michelson interferometer. Forget about the effect that gravitational waves have on the arms of the interferometer and the apparatus that measure the shift in the fringes of the reflected light since one of the mirrors has been jolted by the passing gravitational wave. If we do forget all the other effects then we can proceed---and proceed they do!

A passing gravitational wave can affect masses in free fall, and for simplicity the LIGO team uses a transverse-traceless gauge. The metric is conceived by considering propagation along the $z$-axis with the wave having its polarization along the $x$ and $y$ axes. If the wave amplitude is $h(t)$, a sole function of time, the metric is


which can be considered as a slice in the $z$ direction. It is also assumed--not proven--that gravitational waves propagate at the speed of light so that $ds=0$. Transforming to polar coordinates $(r,\varphi)$, and using a dot to indicate the derivative with respect to $t$, the metric reduces to

$$c^2=(1+h\cos 2\varphi)\dot{r}^2+(1-h\cos 2\varphi)r^2\dot{\varphi}^2-2hr\sin2\varphi\dot{r}\dot{\varphi}.$$

The lagrangian describing this geometry is

$$\mathcal{L}=\frac{1}{2}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}=(1+h\cos 2\varphi)\dot{r}^2+(1-h\cos 2\varphi)r^2\dot{\varphi}^2-2hr\sin 2\varphi\dot{r}\dot{\varphi}.$$

The Euler-Lagrange equations giving the geodesic equations of motion are

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}^{\mu}}-\frac{\partial\mathcal{L}}{\partial x^{\mu}}=0.$$

The first thing to notice is that the angular momentum is not conserved,

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\varphi}}\neq 0,$$

so that we cannot use it to derive the equation of the ray trajectory, determined by the vanishing of the Lagrangian

$$(1+h\cos 2\varphi)\left(\frac{dr}{d\varphi}\right)^2-2hr\sin 2\varphi\frac{dr}{d\varphi}+(1-h\cos 2\varphi)r^2=0,$$

where we have changed the differentiation variable to $\varphi$. 

The two roots of the equation are

$$\frac{dr}{d\varphi}=r\frac{h\sin 2\varphi\pm\sqrt{h^2-1}}{1+h\cos 2\varphi}.$$

But, the gravitational wave amplitude is supposedly very small which "causes distances between freely falling test masses (or mirrors suspended in an arm of an interferometer) separated only in the $x$ direction to change by a factor of


At the same time, an equal and opposite change occurs in distances between masses separated only in the $y$ direction." (P. Saulson, AJP 65 (1997) 501-505) Hence, the smallness of the wave amplitude makes the equation of a ray trajectory complex! If the trajectory of the wave doesn't exist, what are we analyzing?

To make matter even still worse, this simple model of a gravitational wave does not conserve angular momentum. The geodesic equation for the radial coordinate is coupled to that of the angular coordinate, and both involve the unknown rate of change of the gravitational wave amplitude in time. For instance, the geodesic equation for the radial coordinate is

$$(1+h\cos2\varphi)\ddot{r}-r\dot{\varphi}^2+\dot{h}\left(\dot{r}\cos 2\varphi-r\dot{\varphi}\sin 2\varphi\right)-2h\sin 2\varphi\dot{r}\dot{\varphi}-hr\cos 2\varphi\dot{\varphi}^2-hr\sin 2\varphi\ddot{\varphi}=0.$$

Indeed, there is no light wave propagating in a confined region of space with a metric as that supposed by the proponents of gravitational waves. It should be emphasized that what we see, or should see, is an electromagnetic wave affected by a passing gravitational wave, and not the gravitational wave itself. The cause for such a wave can be attributed to almost an infinite of other causes, (e.g., noise, earthquakes, etc.), but the beauty is that by their own model of the effect of such a wave, the ray trajectory of light doesn't exist!

And do you think that Paulson, or whoever, answered the question "Are gravitational waves observable by examining light in an interferometer?" Absolutely not! If Schwarzschild has taught us anything is that light in a gravitational field is identical (not analogous) to a medium of a different index of refraction.  In it, the speed of light is $c/n$, where $n$ is the index of refraction. In a gravitational field with spherical symmetry, the index of refraction is a function only of the radial coordinate. In the Schwarzschild field, the index of refraction is $(1-\mathcal{R}/r)^{-1}$.

Therefore, the statement that "[g]ravitational waves or no, light travels through the arm at the speed of light, $c$", is completely inaccurate. So "[t] physically observable meaning of the stretching of space [time?] is that the light light in it has to cover extra distance, and so will arrive late." Everything stretches or shrinks, even the beam splitter!