If I remember well, they said that it can't, but I do not know why.

Yes, I meant if gravity can be shielded using something like a Faraday cage (or something else?).

Thank you.

  • $\begingroup$ Uhhh this question doesn't make grammatical sense... you mean "shielded by electromagnetism" ? $\endgroup$ Commented Jan 23, 2012 at 21:46
  • 5
    $\begingroup$ Or "like electromagnetism"? In other words, user6090, are you asking about the possibility of something like a Faraday cage for gravity? $\endgroup$
    – David Z
    Commented Jan 23, 2012 at 22:05
  • 1
    $\begingroup$ @DavidZaslavsky i think that's what the OP meant. $\endgroup$
    – Brandon_R
    Commented Jan 23, 2012 at 22:36
  • 3
    $\begingroup$ Related - Is it theoretically possible to shield gravitational waves? $\endgroup$
    – voix
    Commented Jan 24, 2012 at 4:59
  • $\begingroup$ Is there someone that can add a simpler answer or reformulate the answer in a simpler way, even at the risk of loosing precision? $\endgroup$
    – user6090
    Commented Jan 25, 2012 at 9:30

4 Answers 4


There is a simple argument that says the answer is no, although there are constructions in the recent literature which contradict it. These construction are plagued with ghosts, and the argument suggests that this is because the shielding is forbidden.

The argument for no-shielding comes from the focusing property of gravity--- the basis of the singularity theorems and the area theorem.

Static shielding would allow you to float over the shielding mat, because it blocks the gravity of the Earth. This is forbidden by the equivalence principle, and the positive mass theorem (itself closely related to the focusing property, although this is obscured in most proofs. see here for a simple argument for positive mass: Positive Mass Theorem and Geodesic Deviation ).

In a free falling frame, the gravitational field of the Earth is no longer visible, and the shielding mat is just a gravitationally repelling surface. Such a surface cannot exist, because it would increase the area of a black hole, were it to fall in. The reason is that it would push outgoing light plane outward, allowing it to gain area, which violates the area theorem, the focusing property of null geodesics, which is the weak energy condition.

So such a shielding mat does not exist.

Gravitational Higgs Mechanism

Kakushadze and t'Hooft have relatively recently proposed ways around this by using scalars to perform a Higgs mechanism for gravity. The coordinate invariance of GR says that there is no invariant frame, you can relable the coordinates arbitrarily, so t'Hooft introduces 4 scalar fields which label the points of space time. These are massless fields with gradients, so that the field values uniquely specify every event.

Under these conditions, and with the appropriate kinetic terms for the scalars, t'Hooft was able to reproduce the massive spin-2 theory from the massless one. This construction is interesting, but the scalar which labels the time-coordinate has ghost kinetic terms, and is not physically realizable. Kakushadze gives similar constructions. It is not clear that any of these can be fully successful, because of the weak-energy condition argument.

On the other hand, as both t'Hooft and Kakushadze emphasize, massive spin-2 particles in the strong interaction should arise from the string theory dual, and this suggests that there is some sort of gravitational analog of the Higgs mechanism, although what it is is not exactly clear right now.

(Apologies to Zurab Kakushadze and Gerard t'Hooft for any inaccuracies in my recollection)

  • 2
    $\begingroup$ There are some experiments trying to measure whether antiprotons ( antimatter) is repelled by gravity. Are you saying that if the experiments come out positive ( I do not believe they will btw, I think any such repulsion would have been found when designing proton antiproton colliders) theory has to be drastically changed? $\endgroup$
    – anna v
    Commented Jan 24, 2012 at 7:41
  • 2
    $\begingroup$ @anna v: yes, but this was already pointed out by many people, including here (vocally) by Lubos Motl. It would be completely inconsistent with conservation of energy, the equivalence principle, and thermodynamics if antimatter falls up, but I suppose it will be nice to check. The check for photons (which are their own antiparticles) has already been performed. $\endgroup$
    – Ron Maimon
    Commented Jan 24, 2012 at 8:24
  • $\begingroup$ i don't believe as much as this would break thermodynamics of 'vacuum stability' as is often touted. Please see this answer physics.stackexchange.com/a/16752/955. I think the mistake is to extrapolate the Boltzmann factor expression to negative energies. As long as it becomes small in infinite negative energies, the vacuum is safe $\endgroup$
    – lurscher
    Commented Jan 24, 2012 at 16:34

No, gravity cannot be shielded like electromagnetism -- unless negative mass exists.

Such mass would play the role of the negative charge within conductors that shields electromagnetic fields.

Negative mass (mass that violates the so-called Weak Energy Condition) only occurs in certain highly contrived physical systems whose description requires quantum mechanics.

  • $\begingroup$ You should say what these systems are, because the example you linked to is not a good one. It requires a conformally coupled scalar (and I haven't checked it, so I don't know what's causing the WEC violation anyway). There are known violations on the surface of a black hole, to allow the area to shrink while the BH is evaporating. $\endgroup$
    – Ron Maimon
    Commented Feb 5, 2012 at 11:47

Sheilding is caused by redistribution of an area of some(can be zero) net charge into positive and negative charges, in such a manner as to give zero net field inside a conducting body. The absence of negative mass (more accurately, the absence of anything that behaves like negative mass), makes this impossible. Aside from that, we have nothing that behaves as a conductin body for gravitation (ie one that carries 'gravitationl current'. Gravitational current would require negative mass as well)

Unless you consider dark energy and the cosmological constant (which relate to negative curvature $\implies$ negative pressure). Then all this is possible (though the mechanism will be complicated). Of course, you would still need some material that can do the job of a gravitationally conducting sheet..

  • 1
    $\begingroup$ The second paragraph is a bit misleading. Yes, dark energy can be said to "repel," but this is due to it having a negative pressure; it's energy density is still very much positive. In terms of some standard energy conditions, it saturates null, weak, and dominant, and it only violates strong. [Warning: The rest of that wiki article contains some rather blatant factual inaccuracies itself; I suggest not relying on the other sections.] $\endgroup$
    – user10851
    Commented Jul 31, 2013 at 22:39

Yes, gravity can be neutralized (may be the word "shielded" not that suitable here).

How this could be arranged?

Consider we have a toroidally shaped solenoid which is essentially a coil of tubes filled with very dense superfluid liquid. The torus is placed horizontally perpendicular to the gravity force. Now the liquid is put in motion through the tubes in such a way so that in the inner surface of the torus the liquid moves upwards while in the outer it moves down. If the liquid mass and speed are enough high, a body placed in the center of the torus will feel an acting force that will counterbalance the gravity. This force is due to frame-dragging.

  • $\begingroup$ An easier way to 'neutralize' gravity would be to place a suitably massive object in an overhead position. I don't think that is what OP had in mind. $\endgroup$
    – Johannes
    Commented Aug 1, 2013 at 12:32
  • $\begingroup$ @Johannes that would balance the force, but not shield it. If I put two equal masses beside each other, the force is 2x one mass, but if I put one in front of the other, the force is still 2x. No shading occurs. Why? If gravitons interact with matter, they can only do it once. So masses should shade other masses? $\endgroup$
    – user95006
    Commented Oct 8, 2015 at 22:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.