electromagnetism - Is it theoretically possible to shield gravitational fields or waves?

Electromagnetic waves can be shielded by a perfect conductor. What about gravitational fields or waves?

Answer

In a consistent theory of gravity, there can't exist any objects that can shield the gravitational field in the same way as conductors shield the electric field. It follows from the positive-energy theorems and/or energy conditions (roughly saying that the energy density cannot be negative).

To see why, just use the conductor to shield an ordinary electric field - which is what your problem reduces to temporarily for very low frequencies of the electromagnetic waves. The basic courses of electromagnetism allow one to calculate the electric field of a point-like charged source and a planar conductor: the electric field is identical to the original charge plus a "mirror charge" on the opposite side from the conductor's boundary. Importantly, the mirror charge has the opposite sign. In this way, one may guarantee that the electric field is transverse to the plane of the conductor. This fact makes the electromagnetic waves bounce off the mirror if you consider time-dependent fields.

If you wanted to do an "analogous" thing for gravity, you would first have to decide what boundary conditions you want to be imposed for the gravitational field by the mysterious new object - what is your gravitational analogy of "$\mathbf E$ is orthogonal to the conductor". The metric field has many more components. Most of them will require you to consider a negative "mirror mass" - but the mass in a region can't be negative, otherwise the vacuum would be unstable (one could produce regions of negative and positive energy in pairs out of vacuum, without violating any conservation laws).

Microscopically, one may also see why there can't be any counterpart of the conductor. The conductor allows to change the electric field discontinuously because it can support charges distributed over the boundary. The profile of the charge density goes like $\sigma\delta(z)$ if the conductor boundary sits at $z=0$.

However, you would need a similar singular mass distribution to construct a gravitational counterpart. If the distribution failed to be positively definite, it would violate the positive-energy theorem or energy conditions, if you wish. If it were positively definite, it would create a huge gravitational field. Locally, the boundary of the gravitational "conductor" would have to look like an event horizon. But we know that the event horizons cannot shield the interior from the gravitational waves - the waves as well as everything else falls inside the black hole. Moreover, the price for such "non-shielding" is that you have to be killed by the singularity in a finite proper time.

One may probably write some formal solutions that have some properties but they can't really work when all the behavior of gravity and the related consistency conditions are taken into account. One can't fundamentally construct a "gravitational conductor" because gravity means that the space itself remains dynamical and this fact can't be undone. In particular, in a consistent theory of gravity - in string/M-theory - you won't find any objects generalizing conductors to gravity.

By the way, there is one object that comes close to a "gravitational conductor" in M-theory - the Hořava-Witten domain wall, a possible boundary of the 11-dimensional spacetime of M-theory. It can be placed at $x_{10}=0$ and about $1/2$ of the components of the metric tensor become unphysical near this domain wall (boundary of the world that carries the $E_8$ gauge multiplet) because the domain wall also acts as an orientifold plane. But such an orientifold plane is not just some object (like a conductor) that is "inserted" into a pre-existing space that doesn't change. Quite on the contrary, the character of the underlying spacetime is changed qualitatively: the world behind the orientifold plane is literally just a mirror copy of the world in front of the plane.

So there is a lot of fundamentally incorrect thinking about gravity in all the answers that try to say "Yes". Gravity is not another force that is inserted into a pre-existing geometrical space; gravity is the curvature and dynamics of the spacetime itself. Once we say that the space is dynamical, we can't find any objects that would "undo" this fundamental assumption of general relativity.

There are also some more detailed confusions in the other solutions. First, a white hole corresponds to a time-reversed black hole with the same (positive) mass but it is an unphysical time reversal because it violates the second law of thermodynamics: entropy has to increase which means that the (large entropy) black hole can be formed, but it cannot be "unformed". But a white hole, which is forbidden thermodynamically (and microscopically, it corresponds to the same microstates as black hole microstates, they just never behave in any "white hole" way), is still something else than a negative-mass black hole.

A negative mass black hole isn't related to a positive-mass black hole by any "reflection". It is a solution without horizons, with a naked singularity, and can't occur in a consistent theory of gravity because it would cause instability of the vacuum. (Also, naked singularities can't be "produced" by any generic evolution in 3+1 dimensions because of Penrose's Cosmic Censorship Conjecture.) So white holes and negative-mass black holes are forbidden for different reasons. However, even if you considered them, it would still fail to be enough to create a "gravitational conductor" which is nothing else than the denial of the fact that the geometry of the spacetime is dynamical, and one can't freely construct infinite, delta-function-like mass densities without completely changing the shape of spacetime.