# 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?

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My favorite joke answer to this question is that they would be shielded by G-mu-nu metal! – nibot Jan 24 '11 at 23:44
Cavorite of course – docscience Feb 27 at 16:18

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.

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š: nice answer and welcome! Note that you can include TeX by using dollars in the usual way. I went ahead and added formatting to your answer; I hope you don't mind. – Marek Jan 13 '11 at 23:05
Dear Marek, on the contrary, thank you for your services - and nice words. I am already using LaTeX - the preview helped so that I could be sure that the simplest dollar solution worked and I didn't have to search for help for the right code to type TeX which could be more complicated, too. – Luboš Motl Jan 14 '11 at 11:05
Welcome to physics.SE! Also note that you can trigger a notification of your reply for the first person mentioned with "@" (first three letters suffice, actually) in your comments (the OP of the question/answer you're commenting on will automatically be notified, however) – Tobias Kienzler Jan 17 '11 at 12:07
Hi Lubos, youre response is correct of course under the usual textbook assumptions, but nevertheless there does exist the exotic possibilities that in fact an energy condition could be violated. Something as mundane as the QFT Casimir effect will give a very small negative energy density (for a very small delta t). Formally, there also exists exact solutions to the Einstein field equations that also possess unusual gravitational properties, typically involving massless scalars weakly coupled to the curvature. – Columbia Jan 17 '11 at 21:01
Thanks for your responses, folks. And yes, @Columbia - there exist possible loopholes in literature. However, I think it's a premature statement to say that these loopholes exist in the real world as well. In particular, the Casimir effect - or other advanced effects - may produce a negative energy density for a little while. That's no coincidence: the uncertainty principle allows energy to move by $\hbar/dt$ for a little while $dt$. However, one can't create a permanent shielding by anything of this kind. And massless scalars almost certainly don't exist in the reality, either... – Luboš Motl Jan 19 '11 at 10:42

One potential solution, related to a problem discussed by Kip Thorne, would be to construct an enormous array of resonant bar detectors that attenuate the amplitude of the GWs as they pass through. If it is big enough, and the frequency is correct, you could absorb all the energy.

A more active device would be like noise-cancelling headphones: an array of quadrupole oscillators that can be tuned to exactly phase-cancel the incoming GWs, effectively reflecting them back to the source.

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nice answer. Gravitational-wave cancelling, never heard of that :D There might be some minor technical problems ;) – Robert Filter Jan 13 '11 at 17:24
That was my thought (cancelling headphones like). However that requires foreknowledge of the incoming waves. Since the waves travel at the speed of light, how can you obtain knowledge of how to cancel them in time? – Omega Centauri Jan 13 '11 at 19:56
Just like headphones, perfect cancellation would be nearly impossible, but since the frequency changes very slowly in time, you can measure the phase and freq of the first part of the wave, then lock in the cancellation. – Jeremy Jan 14 '11 at 13:53

This is an interesting question. I don't know of any definite answer, but can point you to some (speculative) work done along these lines by Raymond Chiao. His arguments are quite fascinating and if they do end up working the implications would be breathtaking to say the least.

Here are links to two of his papers The Interface between Quantum Mechanics and General Relativity and Laboratory-Scale Superconducting Mirrors for Gravitational Microwaves.

His basic argument is that the equivalence principle is a statement about classical objects and must be modified when looking at the dynamics of quantum objects such as the cooper pairs in a BEC. He suggests, that the interaction between gravitational wave and a superconducting surface they are incident on can be amplified by a factor of $10^{42}$ due to this modification. Normally any such effects would be suppressed by the ratio of the strengths of the gravitational to the electromagnetic force ( $\sim 10^{-42}$). However, due to the anomalous behavior of the cooper pairs of a superconductor this factor can in effect become of $\mathcal{O}(1)$. I leave it up to you to read through his papers and determine if his arguments are physically sound.

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Nice answer and nice papers :) – Robert Filter Jan 13 '11 at 21:15

a possible realization for your "spacetime engineering question" could be white holes.

This hypothetical object can be described by the Schwarzschild metric $$g = -c^2\left( 1-\frac{2Gm}{c^2 r} \right)dt^2 + \left( 1-\frac{2Gm}{c^2 r} \right)^{-1}dr^2 +r^2 d\Omega^2$$

but in contrast to a black hole you interprete the time direction the other way round - nothing will ever be able to enter the object like nothing will ever leave a black hole (ok, maybe some Hawking radiation and virtual particles...).

I tried to construct another "ordinary spacetime" with Maclaurin spheroids and counterrotating black holes on an axis of symmetry (stationarity for a stable solution!) but my ability to visualize such a situation was not sufficient :)
Sincerely

Robert

### Further notes

For those who are interested in experimental realizations (in the sense of analogue gravity) of white-hole horizons permitting light to enter a certain region via the nonlinear Kerr-effect:

Prediction: Philbin et al Fiber-Optical Analog of the Event Horizon, Science 2008 (arXiv-link)

Realizations: Analogue gravity and ultrashort laser pulse filamentation (EPL) and Hawking Radiation from Ultrashort Laser Pulse Filaments (PRL).

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@Robert - that is just a plain old Schwarzschild metric. How does that correspond to a while hole? – user346 Jan 13 '11 at 17:25
@space_cadet: As I understood it, both spacetimes have the same metric. Since it is invariant under time-inversion, one can interprete all phenomena from black holes the other way round. If this interpretation is not correct, please let me know. Greets – Robert Filter Jan 13 '11 at 17:33
@Robert there is a thermal aspect to black (and white) holes which determines whether matter should fall inwards or come "falling" outwards. Even if the microscopic metric is the same, what is it that causes this change to occur at a macro level? This is analogous to the statement that the laws of mechanics are time-reversal invariant but the macroscopic laws of thermodynamics are not. Time-reversal invariance of the metric alone, in other words, is insufficient to construct a white hole. Also it is not clear to me how a "white hole" would help shield against gravitational waves. – user346 Jan 13 '11 at 17:43
space_cadet has this right. A white hole is either identical to a black hole, or an exponentially unlikely configuration that violates the second law of thermodynamics. Neither one is very interesting. – Matt Reece Jan 13 '11 at 20:47
@space_cadet: Seems like your next Nature... "Non-equilibrium white hole configuration as earth's meteorite shield" ;) – Robert Filter Jan 14 '11 at 8:06

Suppose you were in a spacecraft in the neighborhood of two mutually orbiting black holes. The system is decaying and producing gravity waves which you detect. When the two BHs finally coalesce in the near future you know that lots of gravitational radiation will result and the quadrupole oscillation will turn you and your craft into pieces. There is one way to shield your self from this, so to speak, which is to get out of Dodge fast! If you accelerate to a high gamma velocity relative to this system you will Doppler shift the oncoming gravitational radiation in your frame. If the radiation reaches you with a wavelength $\lambda~>>~L$, for $L$ the dimensions of your craft you will avoid being ripped apart.

Electric field shields result from there being $\pm e$ charges, and an electric field can be stopped by a conducting shell or Faraday cage. Similarly for magnetic shields, high $\mu$ metals have magnetic dipoles which conduct $B$ fields. You don’t have this luxury with gravity. With a gravity wave it should be possible to set up lots of bulk matter which oscillates and converts energy to heat. Feynman used an argument with beads on a rod to show that energy from a gravity wave could be detected. However, gravity waves couple to matter very weakly, so you need lots of material to attenuate a gravity wave.

One possible way to partially shield one’s self from a gravity wave would be to hide behind a black hole. This might be about jumping out of the pan into the fire though. However, the gravity wave will not pass through the black hole. What will of course happen is that the black hole will absorb some $\delta mc^2$ of energy and adjust its horizon size to a larger radius. This will then manifest itself in the reaction of spacetime near the black hole which will be in part in the form of gravity waves.

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I notice that some times words appear double, but when I try to edit that disappears. Odd!? – Lawrence B. Crowell Jan 19 '11 at 20:28

Yes, gravitational waves can be shielded. As gravitons are the particles which can interact with matter, the matter can absorb gravitons.

This would depend on the wavelength however. For uniform substance the rate of absorption would generally increase with the wavelength. The rate of absorption also can be increased by constructing mechanical oscillators with frequency in resonance with that of gravitational waves. Such oscillators would transform gravitational waves into mechanical energy.

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You are the only person talking about gravitons, which is the nature of the question I have about this topic. A particle can only be absorbed once, so shading / shielding must occur. Right? Any mass should shade any other mass, but I don't think this is what we observe. Instead, we have the picture of dents in a rubber sheet, and so gravitons must be incorrect. – no comprende Oct 8 at 22:26

Nature of gravitation is not understood completely yet.

But I believe you can shield of sorts using black hole ;-)

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Hah, good answer actually! – Noldorin Jan 13 '11 at 13:29
Nope, GWs would get scattered or deflected by the BH, not shielded. – Jeremy Jan 13 '11 at 16:50
-1 same reason as Jeremy. – Marek Jan 13 '11 at 17:12
Actually it depends on the geometry of the situation, I would think. – Noldorin Jan 14 '11 at 20:57
Gravitational radiation can be shielded not only by a BH but by anything. See my answer below. – Anixx Feb 9 '12 at 1:20