Can one have gravitational mirror?

Gravitational lensing phenomena are due to light *deflection*, i.e. the change in direction of a light beam, analogous to refraction by ordinary materials but, for building a decent "gravitational mirror" you would need a different phenomena: reflection.

Reflection in an ordinary mirror happens because electrons in the atoms of a metal layer on the back surface of the mirror, absorb and re-emit the photons back. There is no analogous phenomenon due to gravity, but you can think nevertheless of the possibility of a very strong *deflection* that turns a light beam back to you.

The formula Eddington used during the famous solar eclipse of 1919 for testing GR was first derived by Einstein. He departed from the assumption of having a nearly flat metric, a slightly perturbed version of the flat metric $\eta_{\mu\nu}$ of special relativity:

$$g_{\mu\nu}\approx \eta_{\mu\nu}+h_{\mu\nu}$$
$$\lvert h_{\mu\nu} \rvert \ll 1$$

He obtained a formula for the deflection angle, in terms of the impact parameter $d$, or minimum distance to the deflecting point mass:

$$\alpha=\frac{4GM}{c^{2}d}$$

Now you can see that the only way to have a 180 degrees refraction, i.e. a quite big $\alpha$ value, is by means of sending a light beam with a very small impact parameter $d$, quite close to the deflecting point mass. That is a problem, not only because then the nearly-flat metric will no longer hold, but also because usual astrophysical objects (stars) are extensive. You need an object that is both very small and very dense, that is, a black hole.

In summary, there is the theoretical possibility of a extreme light deflection, quite close to a black hole, so that a light beam you send happens to come back to you after a 180 degrees deflection. You can name that "mirror" in some sense, but it would not be enough for having a decent, spatially extended image of you.

Can gravitational waves be reflected?

In General Relativity, gravitational waves are supposed to propagate along null geodesics, exactly as light does. That is, you can have *deflection* of gravitational waves, but again no reflection.

EDIT: You could theoretically send a light beam as close as you want to a sort of point-like black hole without an event horizon, a so-called *naked singularity*, see here

## Best Answer

Good question! Indeed, there is an analogous phenomenon in GR to the emission by a Rutherford atom, but your formula does not describe it: as written, it would apply to dipolar emission of gravitational radiation, but gravitational emission only starts at the quadrupole order of the perturbative expansion, which is suppressed by a factor $(v/c)^2$ compared to the dipole.

The reason for this, in Newtonian terms, is that when differentiating the gravitational dipole you find the total momentum of the emitting system, which is conserved, so the second derivative of the gravitational dipole vanishes (see here) --- this reflects the fact that while in electromagnetism we only have one conservation law for the charge, in GR the full four-momentum is conserved.

The formula describing the energy loss of a binary system due to quadrupolar emission can be found in this Wikipedia article; note the $c^5$ term in the denominator as opposed to your $c^3$. There is also a hidden factor of $\omega^6$ (as opposed to $\omega^4$) in the Wikipedia formula: an alternative way to express the radiated power (averaged over a period) is $$ - \left\langle \frac{\mathrm{d} E}{\mathrm{d} t} \right\rangle = \frac{32}{5} \frac{G \mu^2}{c^5} R^4 \omega^6\,, $$ where $\mu = (1/M_s + 1/M_e)^{-1}$ is the reduced mass of the Earth-Sun system. This reflects the higher power of $v$.

To give a sense of the difference, I've plugged some numbers in and your expression would predict a rate of energy loss of about $3 \times 10^{14} \,\mathrm{W}$ in the current Sun-Earth configuration, while the correct GR expression predicts $2 \times 10^{2} \,\mathrm{W}$.