# E.p. implies no gravitational shielding?; Feynman?

1. Sep 26, 2010

### bcrowell

Staff Emeritus
I seem to recall that Feynman had a nice discussion somewhere, possibly in the Feynman lectures, of why there is no such thing as gravitational shielding. IIRC it appealed to the microscopic picture of a Faraday cage and the purely attractive nature of the gravitational force. I can't seem to find this in my own copy of the Feynman lectures. Can anyone help me to locate this?

It seems to me that there are more fundamental reasons why gravitational shielding is impossible.

MTW has a nice illustration on p. 18 of a pedagogical device actually build by Waage at Princeton in order to illustrate the distinction between a Lorentz (free-falling, Einstein-inertial) frame and a non-Lorentz frame. Basically you have a box with guns inside, and you verify that the bullets' trajectories inside the box are straight. This constitutes an operational definition of a Lorentz frame.

There are a couple of issues that are not explicitly addressed by MTW. (1) The box has to be initially calibrated under conditions where we know the box's frame is Lorentzian. (2) In order to carry out this initial calibration, we need to shield the box against any non-gravitational forces. It can't be touching anything external. It has to be put inside a Faraday cage to shield against E fields, a layer of mu-metal to block B fields, paraffin wax to exclude neutrons, etc.

If it *were* possible to shield against gravitational attractions, gravitational waves, etc., then all this other shielding would presumably have at least some unintended effect on gravity as well. Therefore it seems to me that the whole operational definition of Lorentz frames depends on the assumption that gravity can't be shielded against. Since one form of the equivalence principle is the statement that local Lorentz frames exist, it seems to me that the e.p. implies the impossibility of gravitational shielding.

Is this correct?

2. Sep 26, 2010

### pervect

Staff Emeritus
It's not intuitively obvious why the box "has to be" calibrated under conditions when the frame was known to be Lorentzian. What if it couldn't be built for practical reasons of convenience, rather than being theoretically impossible with enough effort?

3. Sep 26, 2010

### bcrowell

Staff Emeritus
Is there another good method for calibrating the box? We could, for example, whip out our perfectly rigid rulers, but that brings in a whole bunch of other issues.

I'm not an extreme operationalist. For example, it's not practical to shield the box against neutrinos, but I don't see that as a huge issue in practice. But I would find it disquieting if I couldn't define, even in principle, some limiting process in which the box would approach a perfect Lorentz-frame-detector.

To make this more concrete, isn't it true that the walls of the box absorb some extremely small fraction of the energy of incident gravitational waves? Is that a problem?

4. Sep 27, 2010

### pervect

Staff Emeritus
For what it's worth, I'd probably use something like the Forward Mass Detector (the rotating cruciform gravity gradiometer) to look for gravitational gradients rather than the box you describe. (I gather that there are commercial services with planes and ships to gather such data on the Earth. http://www.bellgeo.com/tech/technology_system.html [Broken] for instance - mainly for use for oil exploration efforts).

I'm not sure this is very helpful to your thought processes though - apologies in advance.

Last edited by a moderator: May 4, 2017
5. Sep 28, 2010

### PAllen

Maybe I'm being to simpleminded, but it seems to me that no gravity shielding follows trivially from even the 'Weak Equivalence Principle' (see, for example: http://arxiv.org/abs/gr-qc/0504086 a discussion of flavors of equivalence principle).

From this source:
"test bodies fall with the same acceleration independently of their
internal structure or composition (Weak Equivalence Principle, or
WEP) " Note that this is *much* weaker than the Einstein equivalence principle or the strong equivalence principle.

This trivially excludes shielding, because a test body consisting of a shield and shielded object would follow a different trajectory than an unshielded object. (The shield+shielded object would have different gravitational mass than inertial mass).

Last edited: Sep 28, 2010
6. Sep 29, 2010

### bcrowell

Staff Emeritus
This is an interesting argument, and I'm not saying that it's obviously wrong, but it doesn't seem obviously right to me, either. The term "test bodies" in that statement of the e.p. implies a limiting process in which the mass of the body approaches zero. This form of the e.p. fails if you relax that requirement. If the body is sufficiently massive, then it radiates gravitational waves, and that gives it a trajectory that is different than that of a less massive body. For example, the neutron stars in the Hulse-Taylor binary are *not* following geodesics.

I think what is directly forbidden by this form of the e.p. is gravitational shielding that is too effective. For instance, suppose you take a steel sphere of radius r, and surround it with a concentric spherical shell of gravitational shielding with outside radius 2r, the result being that the whole thing's ratio of gravitational to inertial mass if reduced by a factor of $1-\epsilon$ relative to what it would have been if the steel sphere had been placed outside the shielding. Then this form of the e.p. simply requires that $\epsilon\rightarrow 0$ as $r\rightarrow 0$. This doesn't seem like a particularly stringent limit on the possible properties of gravitational shielding materials. If that were the *only* such limit, then I'm sure I could still have flying antigravity cars, etc.

7. Sep 29, 2010

### PAllen

I think it is not that weak. Can anyone detect the decay of Jupiter's orbit due to gravitational waves? I think not. Jupiter can be considered a test object for the sun's field for most any plausible measurement. If you can make a car weightless inside a shield with weight of empire state building, and dropped this (on the moon, for vaccuum), the difference from trajectory of any other object would still be very detectable and would by taken by any experimentalist as a violation of WEP (which is verified to better than 10**-11, last I heard).

Thus, it seems to me, WEP is all about ruling out any form gravitational shielding due to magic internal structure of a test object.

8. Sep 29, 2010

### PAllen

Hey, on the other hand, what WEP certainly doesn't prohibit is an 'inertia shield', that decreases apparent mass of an object for all purposes. Then again, I'm not sure any aspect of GR prevents this except to the extent it incorporates conservation of mass/energy (which I believe is a somewhat dicey boundary issue if gravitational potential energy is to be accounted for).

9. Sep 29, 2010

### PAllen

Let me try this with a slightly different definition of WEP that better defines its applicability:

"An alternative statement of WEP is that the trajectory of a freely falling “test” body (one not acted upon by such forces as electromagnetism and too small to be affected by tidal gravitational forces) is independent of its internal structure and composition. "

from: http://relativity.livingreviews.org/Articles/lrr-2006-3/ [Broken]

Combine this with the known fact you can make special arrangements of masses (in principle) that produce an arbitrarily large region that has uniform gravity to an arbitrary precision. Thus, in principle, you can construct a test region for which any mass you want to test is a valid "test particle".

Put these together, and I claim WEP rules out gravitational shielding.

Last edited by a moderator: May 5, 2017
10. Sep 29, 2010

### PAllen

Ok, arguing a little on the other side, my last argument does not rule out a proposed flavor of gravitational shielding that works only where tidal forces are signficant for a given body. You could even say that it only filters out tidal effects.

I have no idea of a tight argument to exclude such gravitational shield.

An additional caveat about my prior argument is that doesn't really accomplish much in the absence of a particular theory of gravity (where the second part is not known to be achievable). I guess a way of phrasing my a argument is:

Given GR, WEP is the feature of it that precludes a 'universal' gravity shield.

11. Sep 29, 2010

### bcrowell

Staff Emeritus
I think it's a little more complicated than this. The only preferred scales in classical GR are G (the Newtonian gravitational constant) and c. Classical GR doesn't have any preferred length or mass scale, so we can't just set some limit on the length or mass of test particles, and say that that's small enough. A supercluster of galaxies might qualify as small enough in some cases. In other cases, an atomic nucleus might be too big. For linear size L, the relevant comparison is probably the object's size compared to the length scale Lo defined by some measure of the local spacetime curvature. We could then say that trajectories of test particles should be universal in the limit of $L/L_o\rightarrow0$, with some kind of error bound that can be expressed in terms of $L/L_o$, because tidal effects scale down in a certain way. Similarly, you can probably define some scale Mo such that gravitational radiation effects scale down appropriately with $M/M_o$. But no matter what the values of Lo and Mo are, the basic requirement is that you need the error to go to zero in the limits where L and M go to zero. If it does, then I don't think there is any violation of the e.p.

Suppose that gravitational shielding of thickness h reduces the gravitational mass of the shielded interior by a factor of f=1-kh (in the limit of small h), where k is a constant. This is the low-h limit you see with many other types of shielding (magnetic, neutron, ...). Then I don't see any obvious way to prove violation of the e.p., since scaling down the L and M of the test object will presumably involve scaling down h as well, with the result that f goes to 1, and the object's trajectory approaches a geodesic. They key is that this difficulty in prosecuting the defendant for e.p. violation is independent of the value of k. Therefore I don't see how to use this line of reasoning to prove the nonexistence of materials with k so large that you could use them to build flying cars.

Ah, interesting! I think this might be a more promising line of attack. If I have an asteroid possessing a gravitational field, and I then wrap the asteroid in gravitational shielding, then it might appear from the outside as if the asteroid's mass had decreased over time. I doubt that you really need integrated mass-energy for this; I think it might simply appear to an observer exactly as if the stress-energy tensor had picked up a nonvanishing divergence, due to an anomalous time derivative that would enter into the Einstein tensor. I'm not sure about this, though. Putting a charged particle inside a Faraday cage doesn't violate Gauss's law as seen by an outside observer based on fields emanating from the system (particle+cage); all it does is ensure that if outside fields act on the system, it's the cage that feels the effect, not the particle.

I wonder if gravitational shielding would violate any energy conditions?

Last edited: Sep 29, 2010
12. Sep 29, 2010

### atyy

But isn't the inside of a perfectly spherical shell shielded?

13. Sep 29, 2010

### PAllen

No way. Do you think that if I coat a canonball with neoprene, gravity only acts on the neoprene, while the inertia include the canonball? Thus this would fall at much less than 9.81 m/sec**2 ??

14. Sep 29, 2010

### atyy

Let's say the universe contains only a perfectly spherical canonball and you. I think if you are inside the canonball, you can't tell the difference whether it's coated with neoprene or not.

15. Sep 29, 2010

### PAllen

You're talking about something completely different. The topic here is whether you can shield an object from gravity. You can't. What you are confusing with this is that the gravitational field inside spherical a shell of matter *due only to that shell of matter* is null (approximately, for GR, exact for Newtonian gravity). However, outside gravity isn't shielded. If this shell is sitting on the earth, you will feel all of the earth's gravity.

The gravity of the shell by itself has *nothing* to do with shielding. It is simply a matter of forces balancing.

16. Sep 29, 2010

### atyy

Well, so you can shield yourself from objects. Just put more matter around them, until the distribution is perfectly spherical around yourself.

17. Sep 29, 2010

### PAllen

No. Goal: allow myself not to experience earth's gravity. Try a few things:

1) Enclose myself in a sperical shell sitting on the earth. If you believe you will be waitless, well I have some things to sell you ...

2) Enclose the whole earth plus you in a shell. Same result . You feel all of the earth's gravity.

What you are talking about is no different from positioning myself between to masses where their gravity balances (except it is more stable). It is nothing like the way, e.g. a superconductor expells an external magneting field.

18. Sep 29, 2010

### atyy

Make a shell enclosing you, such that the earth is part of the shell?

19. Sep 29, 2010

### atyy

BTW, there's no shielding in electrodynamics either.

20. Sep 29, 2010

### PAllen

This is getting silly. Sit yourself at any of the Lagrangian points for earth-moon. You 'feel' no gravity from either. Is this a shield? No. Niether is a spherical shell of matter.

21. Sep 29, 2010

### atyy

No! It was silly from the start

22. Sep 29, 2010

### PAllen

I disagree. A conductive shell really does shield from outside em forces (up to a point). A matter shell provide *no* shielding whatsoever from outside gravitational forces.

Bcrowell is professor of physics who has taught relativity. He didn't start this thread because he believed any gravity shield was possible in practice. He wanted to discuss exactly why and to what degree it is prohibited by GR.

23. Sep 29, 2010

### atyy

Since there's no shielding in electrodynamics, the fair comparison is between Newtonian gravity and Coulomb's law.

In Newtonian gravity, the purely attractive nature of the force is what guarantees no shielding for situations with no spherical symmetry. The EP says gravitational mass = inertial mass. If we postulate that inertial mass is never negative, then that would with the EP guarantee the purely attractive nature of gravity.

24. Sep 29, 2010

### bcrowell

Staff Emeritus
In my OP, I'd been visualizing the box of electrical, magnetic, ... shielding as a rectangular box. Atyy's point about spheres does kind of eliminate the issue I'd raised there. Make the box spherical, and then you can be pretty sure that it doesn't have the unintended side-effect of shielding out gravity.
[EDIT] Gah, have to stop posting on PF when I'm high on crack. This is totally wrong. Correction in #55.

I'm not completely clear on what you're saying here, but it may be similar to what I was just thinking on the way home from the supermarket. You actually *can* have gravitational shielding, in some sense, although it's not necessarily practical. To shield you from the earth's gravity, I can position a hunk of neutron-star matter above your head so as to cancel out the earth's gravitational field.

So maybe it would be relevant to start by trying to define more precisely what we mean when we say that gravitational shielding isn't possible. The obvious differences I see between the neutron-star-of-Damocles and a Faraday cage are: (1) The Faraday cage is passive, whereas the gravitational version requires active control. (2) In the gravitational version, to shield A from B's gravitational force, we have to position the "shielding" C such that C is not between A and B, but rather A is between B and C. (I suppose this could change if we were allowed to violate various energy conditions.)

Another thing to consider is that I'm not even sure this is a relativity question, although it might have relativistic implications. That is, if I ask, "Why can't we have gravitational shielding?" to someone in 1890, they might be able to give a better answer than any we've achieved so far in this thread. On the other hand, there might be some deeper GR-based insight as well.

Last edited: Sep 30, 2010
25. Sep 29, 2010

### PAllen

Why do you keep saying there is no EM shielding? A conductive shell provides real shielding for many cases. The Meissner effect for superconductors provides real shielding for static magnetic fields. There is *no* analog of these for gravity.

The EP as to why no shielding for gravity is exactly the point of this thread. I have been trying to push that, as a practical, the WEP can be taken to prohibit shielding; Bcrowell keeps raising issues that in theory, it is just not so simple.

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