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

[PAllen]

Yes. Why can't inertial mass be negative?

There you go: negative inertial mass would provide shielding. But negative inertial mass is prohibited by ep...

 

[PAllen]

How do you know that the orbit is stable? Is it impossible that the sun will attract the shell the shell will end up in the centre of the spherical sun at infinite time, then by spherical symmetry you will be shielded.

It's fair to wait for absolute equilibrium, since that's what one does in classical electrostatics and a perfect conductor.

I guess we need to be clearer on what shielding is. The model was go inside some magic room on Earth and weigh less or nothing. As opposed to (effectively your solution) : drill to the center of the earth, and hang out there; you certainly won't feel the *earth's* gravity there. You'll still feel the sun's just as strongly.

I guess what I want to exclude is *balancing* gravity (which is what etiher being inside a shell or being at a lagrangian point does) versus shielding gravity: there is a region where gravitational influence from any external source is eliminated or attenuated. It is, ultimately a matter of definition.

 

[JesseM]

Since you're relying on the equivalence principle, don't you need to also add the condition that the "shielding" device isn't massive enough to significantly change the spacetime curvature in the small region of spacetime where the measurements are being performed, since the equivalence principle only works in a region of spacetime that's small enough so that tidal forces in the region are negligible? (perhaps this is only true for the 'strong equivalence principle' but not the weak one, I'm not sure)

 

[PAllen]

Since you're relying on the equivalence principle, don't you need to also add the condition that the "shielding" device isn't massive enough to significantly change the spacetime curvature in the small region of spacetime where the measurements are being performed, since the equivalence principle only works in a region of spacetime that's small enough so that tidal forces in the region are negligible? (perhaps this is only true for the 'strong equivalence principle' but not the weak one, I'm not sure)

Yes, this is an issue. If you care, read the earlier parts of this thread. bcrowell and I discussed this, reaching no firm conclusions.

 

[bcrowell]

This is a follow-up to my correction to my #24, where I was obviously high on crack.

Make the box spherical, and then you can be pretty sure that it doesn't have the unintended side-effect of shielding out gravity.

To see that I was totally wrong here, consider the case of a spherical Faraday cage. It *does* exclude electric fields. The special properties of a sphere (Newton's shell theorem) depend only on the 1/r2 nature of the force, so this aspect of the electrical case is really no different than the gravitational case.

 

[bcrowell]

Maybe something like:

There are no solutions of GR in which spacetime is flat except when there is no matter, or if it is contained in a perfect sphere. Hence there is no shielding in general. However this is a property of the Einstein equations only (really, one wonders about Nordstrom gravity), and this cannot be deduced from the EP. Hence the EP is insufficient to guarantee shielding.

I don't quite follow you here. Are you referring to Birkhoff's theorem? Your "Maybe something like" seems to imply that there are some steps that you're not sure how to fill in...? I don't think Birkhoff's theorem can be applied in this way, for the following reasons.

Say we have flat spacetime from r=0 out to r=a, and then some other, arbitrarily specified spacetime for r>a. The r<a region satisfies the vacuum field equations. Suppose that the r>a region also satisfies the (non-vacuum) field equations, but is not spherically symmetric. At the r=a spherical boundary, the derivatives appearing in the Einstein tensor blow up, which means that we need a certain stress-energy tensor at r=0 that has a delta function factor in it, T_{ab}=\delta(r-a)f_{ab}(\theta,\phi). There is no violation of B's theorem, because B's theorem only says anything about vacuum solutions, and we only claim to have a vacuum solution for r<a.

It might be interesting to see, however, whether we can prove that this stress-energy tensor has to violate certain energy conditions.

 

[bcrowell]

No, it doesn't. The closest you can come is counterbalancing rather than shielding (adding mass at a particular location so you end up at a Lagrangian point).

You might also be able to null out the field better than that. That is, we know we can arrange to have g=0 at the origin by counterbalancing. By choosing a more complicated arrangement of counterbalancing masses, you might also be able to make the all derivatives of the form \partial_{a_1}\ldots\partial_{a_n}g_b vanish up to some n. By analogy with E&M, a Helmholtz coil allows you to null out a uniform external field up to n=2, and a Maxwell coil up to n=6.

In the gravitational case, I think a single pointlike mass overhead gives n=0 nulling, and you can probably achieve n=1 (no tidal forces) by combining one overhead mass and another mass underfoot. The thing that makes this trickier than the E&M version is that you can solve the equations to get nulling up to a certain n, but then you have to check that all your masses came out positive. If they didn't, you may not have a valid solution unless you have a secret supply of exotic matter.

 

[atyy]

I don't quite follow you here. Are you referring to Birkhoff's theorem? Your "Maybe something like" seems to imply that there are some steps that you're not sure how to fill in...? I don't think Birkhoff's theorem can be applied in this way, for the following reasons.

Say we have flat spacetime from r=0 out to r=a, and then some other, arbitrarily specified spacetime for r>a. The r<a region satisfies the vacuum field equations. Suppose that the r>a region also satisfies the (non-vacuum) field equations, but is not spherically symmetric. At the r=a spherical boundary, the derivatives appearing in the Einstein tensor blow up, which means that we need a certain stress-energy tensor at r=0 that has a delta function factor in it, T_{ab}=\delta(r-a)f_{ab}(\theta,\phi). There is no violation of B's theorem, because B's theorem only says anything about vacuum solutions, and we only claim to have a vacuum solution for r<a.

It might be interesting to see, however, whether we can prove that this stress-energy tensor has to violate certain energy conditions.

Yes, I don't know how to fill in the steps. I wasn't thinking of Birkhoff's theorem. I was thinking something like the energy conditions you suggest.

 

[atyy]

There you go: negative inertial mass would provide shielding. But negative inertial mass is prohibited by ep...

Why does the ep prohibit negative inertial mass?

 

[bcrowell]

Since you're relying on the equivalence principle, don't you need to also add the condition that the "shielding" device isn't massive enough to significantly change the spacetime curvature in the small region of spacetime where the measurements are being performed, since the equivalence principle only works in a region of spacetime that's small enough so that tidal forces in the region are negligible? (perhaps this is only true for the 'strong equivalence principle' but not the weak one, I'm not sure)

This is essentially the point I was trying to analyze in my #11. There are at least two different e.p.-based arguments being discussed here, see #1 and #5. I claim that my #11 invalidates the argument in #5, but I don't seem to have convinced PAllen.

Hmmm, maybe shielding in classical electrostatics is quantum mechanical. After all, it requires the existence of conductors - materials which constrain the movement of positive and negative charges to a surface.

So it requires

1) Existence of Gauss's law and curl free fields
2) Existence of a material that can guarantee by its nature isopotentiality on its surface.

I disagree with your definition of what's classical. It seems to me that by your definition, I can't analyze the gear system on a 10-speed bike using classical physics. It's probably true that rigid materials like steel don't exist in purely classical physics, because the iron atoms would collapse by emitting radiation, and only quantum mechanics can prevent this collapse. But that doesn't mean that I need anything beyond classical physics in order to analyze the 10-speed bike.

To get a material that keeps itself at an equipotential, all I need is some classical charged particles, surrounded by an impenetrable box. The box can be described simply as a force of constraint, and you can describe that force using purely classical techniques, e.g., the kind described in a standard textbook like Goldstein's Classical Mechanics.

-Ben

 

[bcrowell]

Why does the ep prohibit negative inertial mass?

I think it prohibits a negative ratio of gravitational to inertial mass. If you had a material with those properties, then you could make an arbitrarily small test particle out of it, and that test particle wouldn't follow the same geodesics as normal test particles, which would violate one form of the e.p. (the weak one, I guess?).

I don't think it does prohibit an object from having both negative inertial mass and negative gravitational mass. That would be exotic matter that violated an energy condition, but not the e.p.

-Ben

 

[atyy]

I think it prohibits a negative ratio of gravitational to inertial mass. If you had a material with those properties, then you could make an arbitrarily small test particle out of it, and that test particle wouldn't follow the same geodesics as normal test particles, which would violate one form of the e.p. (the weak one, I guess?).

I don't think it does prohibit an object from having both negative inertial mass and negative gravitational mass. That would be exotic matter that violated an energy condition, but not the e.p.

-Ben

Sounds good to me. So it seems even in Newtonian gravity, the pure ep is insufficient. One needs an energy condition. Seems qualitatively very similar to what you outlined for GR in one of your posts above.

 

[PAllen]

This is essentially the point I was trying to analyze in my #11. There are at least two different e.p.-based arguments being discussed here, see #1 and #5. I claim that my #11 invalidates the argument in #5, but I don't seem to have convinced PAllen.

-Ben

Well, you've given me doubts, but your argument in #11 doesn't address my argument in #9 and following, which to my mind says: if you assume GR, then you can say that the WEP is the aspect of it which prohibits certain definitiions of shielding (if you also assume a conservation / energy condition).

 

[atyy]

I disagree with your definition of what's classical. It seems to me that by your definition, I can't analyze the gear system on a 10-speed bike using classical physics. It's probably true that rigid materials like steel don't exist in purely classical physics, because the iron atoms would collapse by emitting radiation, and only quantum mechanics can prevent this collapse. But that doesn't mean that I need anything beyond classical physics in order to analyze the 10-speed bike.

To get a material that keeps itself at an equipotential, all I need is some classical charged particles, surrounded by an impenetrable box. The box can be described simply as a force of constraint, and you can describe that force using purely classical techniques, e.g., the kind described in a standard textbook like Goldstein's Classical Mechanics.

Sure. But I think it's not really different since we now seem to agree that an energy condition is required - ie. some additional statement in the classical framework which should presumably fall out of a specific quantum theory of matter.

 

[PAllen]

More and more I see this whole topic is ill defined, starting with what is a shield? I will now argue that shielding is possible both in GR and Newtonian gravity. Note, in passing, that tidal effects are not restricted to GR. (What may be prohibited in both GR and Newton is some specific definition of a universal passive shield, but I am not sure how to precisely define this).

Suppose I have a configuration of large masses each with its own propulsion system and a program controlling them. By active adjustment, as I tool around the solar system with a rocket backpack, they actively adjust to such that the gravitational influence on me of all solar system objects are neutralized to high precision. I can hang out close to Jupiter, and neither orbit it, nor fall into it. Of course, my active cage is consuming enormous power to keep itself in (and change) position, but I don't care, an alien gave it to me.

From my point of view, I have a functioning gravity shield.

 

[bcrowell]

OK, now I think we're homing in on the real issue!

I think you can screen out an arbitrary externally applied gravitational field if and only if you have exotic matter that violates an energy condition. Here "arbitrary" and "screen out" mean that if you tell me some field pattern and some integer n, I can insert masses in such a way as to make the field and all its derivatives up to order n vanish.

To prove the "if" part, simply take the field configuration of a Faraday cage immersed in some external electric field, and compute its divergence, which is the charge density on the cage. Then transform all the electric fields into gravitational fields, and all the charges into masses, and you have gravitational shielding.

I can prove the "only if" in the one-dimensional case. Let the x-axis point down, i.e., we have some externally applied field g(x), which is positive at x=0, the location of our laboratory. Since g is arbitrary, we could have g'(0)>0 (as is the case, for example, in the Earth's field). If it's possible to screen out this field in the region near x=0, then I should be able to do that by adding point masses one at a time. The key is that the field of a point mass is an odd function, but the derivative is even. So if you tell me you want me to null your external field up to n=1, I can't do it. I can start placing positive point masses overhead at x<0, and that will start reducing |g(0)|. (I could also do that by putting negative point masses underfoot at x>0.) However, no matter where I place positive point masses, they will always contribute positively to g'(0), thereby increasing |g'(0)|. The only way to get a decrease in |g'(0)| is by using negative masses.

 

[atyy]

http://en.wikipedia.org/wiki/Negative_mass

So it looks like we're coming close to what Feynman (and common sense) said after all. The purely attractive nature of gravity ie. no negative mass in Newtonian gravity, some energy condition in GR is needed for no shielding (in some sense).

 

[JesseM]

http://en.wikipedia.org/wiki/Negative_mass

So it looks like we're coming close to what Feynman (and common sense) said after all. The purely attractive nature of gravity ie. no negative mass in Newtonian gravity, some energy condition in GR is needed for no shielding (in some sense).

Any negative mass that respected the equivalence principle in the sense I talked about in post #27 would have both negative gravitational and negative inertial mass, which would mean it would still fall in the same direction and at the same rate in an external gravitational field as positive mass (the difference would be that the gravitational field of the negative mass itself would be repulsive). This type of negative mass would be no use for "shielding", agreed?

 

[bcrowell]

Any negative mass that respected the equivalence principle in the sense I talked about in post #27 would have both negative gravitational and negative inertial mass, which would mean it would still fall in the same direction and at the same rate in an external gravitational field as positive mass (the difference would be that the gravitational field of the negative mass itself would be repulsive). This type of negative mass would be no use for "shielding", agreed?

Huh? No, I don't think so. I don't think the inertial mass matters at all when you're building shielding.

 

[JesseM]

Huh? No, I don't think so. I don't think the inertial mass matters at all when you're building shielding.

What part are you saying no to? Do you agree that the sense of the "equivalence principle" I was talking about in post #27 implies that all mass must fall downward in an external gravitational field in the same way? Do you agree this can only be true for an object with negative gravitational mass if it also has (equal) negative inertial mass? (because if an object with negative gravitational mass had positive inertial mass, it would 'fall up' in a gravitational field) If you agree with that, then are you just saying "no" to the idea that this type of negative mass would be useless for "shielding"? How is it supposed to provide shielding if it falls downward exactly like positive mass? If you're assuming a significantly large chunk of negative mass that its own gravitational field becomes significant, I'd say you're no longer talking about the "equivalence principle" by my argument in post #53.

 

[bcrowell]

http://en.wikipedia.org/wiki/Negative_mass

So it looks like we're coming close to what Feynman (and common sense) said after all. The purely attractive nature of gravity ie. no negative mass in Newtonian gravity, some energy condition in GR is needed for no shielding (in some sense).

I still wish I could find what Feynman actually said!

Let me see if I can summarize:

The main question I posed in #1 was whether there is really a satisfactory operational definition of a Lorentz frame. If there is no such operational definition, then it seems to lead to a problem with stating the equivalence principle in terms of the local existence of Lorentz frames. I think we have a satisfactory solution to this, which is to build the shielding against electromagnetic and other nongravitational fields as spherical shells. When you initially put this spherically symmetric shielding in place, its contribution to the internal field is zero (in the limit where the sphere gets small, where the Newtonian approximation is good). There's a potential concern that once you have built the shielding, it will somehow rearrange itself at the microscopic level in such a way as to break its own symmetry and produce some unintended effect on the interior field. This is exactly what happens in the electric case if the shield is a conductor or a dielectric: the external electric fields cause polarization of the material, which we can't prevent. But since we've found that gravitational shielding is essentially impossible without violating an energy condition, I think we can be pretty sure that there are no such gravitational polarization effects, as long as the shielding material doesn't violate an energy condition. This final step is still a little vague, but it at least seems like a decent plausibility argument.

If gravitational torsion exists, then I imagine you might have more worries. The source of gravitational torsion is usually assumed to be the intrinsic spins of particles, and since those are microscopic, they're sort of like the microscopic polarization phenomena in the case of electrical shielding; you can't constrain them with macroscopic forces. Torsion does violate the weak e.p.

 

[PAllen]

OK, now I think we're homing in on the real issue!

I think you can screen out an arbitrary externally applied gravitational field if and only if you have exotic matter that violates an energy condition. Here "arbitrary" and "screen out" mean that if you tell me some field pattern and some integer n, I can insert masses in such a way as to make the field and all its derivatives up to order n vanish.

To prove the "if" part, simply take the field configuration of a Faraday cage immersed in some external electric field, and compute its divergence, which is the charge density on the cage. Then transform all the electric fields into gravitational fields, and all the charges into masses, and you have gravitational shielding.

I can prove the "only if" in the one-dimensional case. Let the x-axis point down, i.e., we have some externally applied field g(x), which is positive at x=0, the location of our laboratory. Since g is arbitrary, we could have g'(0)>0 (as is the case, for example, in the Earth's field). If it's possible to screen out this field in the region near x=0, then I should be able to do that by adding point masses one at a time. The key is that the field of a point mass is an odd function, but the derivative is even. So if you tell me you want me to null your external field up to n=1, I can't do it. I can start placing positive point masses overhead at x<0, and that will start reducing |g(0)|. (I could also do that by putting negative point masses underfoot at x>0.) However, no matter where I place positive point masses, they will always contribute positively to g'(0), thereby increasing |g'(0)|. The only way to get a decrease in |g'(0)| is by using negative masses.

I disagree with this whole argument. The key problem comes from this restriction: " If it's possible to screen out this field in the region near x=0, then I should be able to do that by adding point masses one at a time. ". Who says? I now demonstrate cancellation of g to the nth derivative using 2**(n+1)-1 masses.

1) you can cancel ambient g (zeroth derivative) with one mass, properly placed.

2) Consider a two mass system of a closer smaller mass and a larger, farther mass. Properly adjusted, their respective g cancels, leaving a residual g'. Such a two mass system (scaled and oriented) can function as a g' canceller.

3) Now consider a starting mass, add a mas to cancel g, add two more to cancel g', result a 4 mass system with g=0, g'=0, but residual g''. Scaled and rotated, we have a general g'' canceller.

ad nauseum. Do the math and see that my claim is established, at least in the simple case proposed by bcrowell.

On another note, the standard classification of equivalence principles is that they all include the WEP. Existence and behavior of Lorentz frames are added as additional components. See sections 2.1 and 3.1 of:

http://relativity.livingreviews.org/Articles/lrr-2006-3/

 

[bcrowell]

What part are you saying no to? Do you agree that the sense of the "equivalence principle" I was talking about in post #27 implies that all mass must fall downward in an external gravitational field in the same way?

No, because this version of the e.p. only applies in the limit of small test bodies. See #11.

Do you agree this can only be true for an object with negative gravitational mass if it also has (equal) negative inertial mass? (because if an object with negative gravitational mass had positive inertial mass, it would 'fall up' in a gravitational field)

Yes.

If you agree with that, then are you just saying "no" to the idea that this type of negative mass would be useless for "shielding"?

Anything with negative gravitational mass would be extremely useful for shielding, where "shielding" is defined as in #66.

How is it supposed to provide shielding if it falls downward exactly like positive mass?

Why does it have to fall? Why can't it be nailed down? If you had access to a plate made of a material with negative gravitational mass, you could put it under your feet, and it would at least partially/approximately cancel out the Earth's field.

Or are you thinking of gravitational shielding that passively adjusts itself in the same way that the charges in a Faraday cage automatically adjusts itself so as to exclude electric field lines from the interior? Yes, this would clearly violate the e.p.

What PAllen and I are talking about in the posts around #66 is shielding that requires active control, not passive shielding.

If you're assuming a significantly large chunk of negative mass that its own gravitational field becomes significant, I'd say you're no longer talking about the "equivalence principle" by my argument in post #53.

If I'm understand correctly what you're getting at, then I think #11 addresses this.

 

[bcrowell]

I disagree with this whole argument. The key problem comes from this restriction: " If it's possible to screen out this field in the region near x=0, then I should be able to do that by adding point masses one at a time. ". Who says?

Newtonian gravitational fields combine linearly. Since they combine linearly, and vector addition is commutative, I can add the contributions of all the particles in any order I like.

2) Consider a two mass system of a closer smaller mass and a larger, farther mass. Properly adjusted, their respective g cancels, leaving a residual g'. Such a two mass system (scaled and oriented) can function as a g' canceller.

No, this doesn't work. Suppose that in one dimension I have an externally imposed field g(x)=x. You cannot place two additional point masses on the x-axis so as to produce g'(0)=0 for the total field.

 

[PAllen]

Newtonian gravitational fields combine linearly. Since they combine linearly, and vector addition is commutative, I can add the contributions of all the particles in any order I like.

Of course, but you can't necessarily achieve a particular goal by adding only one mass.

No, this doesn't work. Suppose that in one dimension I have an externally imposed field g(x)=x. You cannot place two additional point masses on the x-axis so as to produce g'(0)=0 for the total field.

Here's the misunderstanding, getting to all the different definitions of shield that have floated around. My scheme is intended to make g=g'=g'' etc. = 0 at *one point*. In 'practice' it might be effective for a tiny region, and generally weak fields. Remember my post starting this was a scheme to allow 'me' to float some distance above Jupiter without falling or orbiting.

 

[JesseM]

No, because this version of the e.p. only applies in the limit of small test bodies. See #11.

#11 doesn't convince me that it's even possible to talk about the "equivalence principle" when the gravitational fields of objects inside the lab have significant effects--isn't the equivalence principle all about an equivalence between measurements in a small lab in curved GR spacetime and measurements made in a similar small lab in SR? In SR, by definition nothing inside the lab can have any gravitational effects!

Why does it have to fall? Why can't it be nailed down? If you had access to a plate made of a material with negative gravitational mass, you could put it under your feet, and it would at least partially/approximately cancel out the Earth's field.

See above. If you're talking about "shielding" where the gravitational field of the shield itself is significant, then I don't think you're talking about a situation where the equivalence principle of GR could possibly apply any more. And in any case, why bother with negative mass? I could just as easily attach an object with very large positive mass to the ceiling (a small black hole or a cosmic string, say), and if the mass and distance were calibrated right the upward pull of this mass would cancel out the downward pull of the Earth (ignoring the issue of tidal differences between the pull on my head and feet, which also must be ignored when talking about a negative mass plate underneath me).

 

[PAllen]

Let me just make sure my limited, active, scheme for cancelling gravity to the nth derivative at one point is understood.

At some point x, we are given g, g', g'' etc. for field strength and itis derivatives. I don't care if these are vectors, and make no assumptions about their direction. Suppose g is down. I put a mass above x, at suitble mass and distance, held actively in place, such that effecitvy g is now zero.

Next, consider that if you have a pair of unequal masses, there is a point between them where g=0, but g' does not equal zero. Such a system can be scaled and rotated to cancel g' of any magnitude and direction at a chosen point. Add such a pair around x, and now I have canceled the given g and g' at x. Continuing as described in my earlier post, you can eliminate to the nth derivative with 2**(n+1)-1 masses.

 

[JesseM]

Which version of the equivalence principle is it that says that, at least to a first-order approximation, the motion of test bodies measured for a brief period of time in a small room accelerating in flat spacetime should be the same as the motion of test bodies measured for a brief period of time in a small room "at rest" in a gravitational field? (of course 'at rest' probably only makes sense in the context of a stationary spacetime like the external Schwarzschild solution, but we could always talk about a room that's accelerating relative to a freely-falling frame in the same region) Assuming there is a version of the equivalence principle that says this (this blog by a pair of physicists states it in about the same way), I think it would imply the impossibility of gravitational shielding, since a gravitationally shielded body in a room in a gravitational field would fail to fall down at the same rate as non-shielded objects, but an object in the middle of an accelerating room should move inertially if no outside forces are acting on it (of course we're excluding a trivial form of 'shielding' that just keeps an object from falling by applying some force to it, like a small rocket strapped to an object), so if the floor is accelerating towards the object from the perspective of an inertial frame, an observer in the room should see it "falling" towards the floor.

Incidentally, this book by physicist Anthony Zee makes pretty much the same argument:

Science fiction writers love the notion of antigravity. But if you believe in the equivalence principle, then antigravity is impossible.

The argument is again ludicrously simple. Drop a bunch of objects. Can any of them fall upward? No. According to the equivalence principle, you might as well be inside a rocket ship accelerating in deep space. From this point on, it is the same old argument. To an observer floating outside, the floor is rushing up to meet the objects you dropped, even though you could swear they are falling toward the floor. Obviously if the equivalence principle is correct, then in a gravitational field also, there can't be an object that falls upward.

And I'm pretty sure I originally heard this argument from some professor in college. Feynman http://books.google.com/books?id=jL9reHGIcMgC&lpg=PA90&dq=feynman%20%22equivalence%20principle%22&pg=PA93#v=onepage&q=feynman%20%22equivalence%20principle%22&f=false states the equivalence principle in terms of the effects of gravity and acceleration being indistinguishable in a small box, so I'd bet that if he did every make an argument involving the equivalence principle and the impossibility of "shielding" gravity, it was something along these lines.

 

[PAllen]

Incidentally, this book by physicist Anthony Zee makes pretty much the same argument:

And I'm pretty sure I originally heard this argument from some professor in college. Feynman http://books.google.com/books?id=jL9reHGIcMgC&lpg=PA90&dq=feynman%20%22equivalence%20principle%22&pg=PA93#v=onepage&q=feynman%20%22equivalence%20principle%22&f=false states the equivalence principle in terms of the effects of gravity and acceleration being indistinguishable in a small box, so I'd bet that if he did every make an argument involving the equivalence principle and the impossibility of "shielding" gravity, it was something along these lines.

The WEP embodies this. Using the formulation from the reference I gave earlier:

"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"

This should be true for all objects, everywhere except, where tidal forces are not significant for the body. Thus it is true in a rocket, on earth, on Jupiter, in a rocket around jupiter, etc.

However, in pushing the limits of this, it really doesn't prevent peculiar forms of gravity shielding (or active gravity balancing). For example, suppose I have a huge box that decreases gravity inside only for objects big enough to be subject to significant tidal forces.

There may be more sophisticated arguments, combining WEP with other postulates, that exclude this, but I don't know them.

 

[PAllen]

Maybe a better example of limited antigravity that is not prevented WEP in any formulation I've seen is suggested by bcrowells points in his post#11:

Imagine a material whose antigravity effectiveness grows exponentially with tidal deviation across a mass of it. When it is small enough, in any location, to function as a test body, it will satisfy WEP, however a large amount of it will have large antigravity effect. If you move it to a more uniform field, it will lose its detectible antigravity, thus not violating WEP (then you will need much more of it for antigravity effect). If you try to come up with a formulation of equivalence principle without this scaling hole, you run up against the fact that two objects subject to tidal effects can have different trajectories influenced by internal structure.

So, for any object, you can find or make some place it will function as a test body, and it must satisfy WEP, but bcrowell has convinced me there is no clear way to preclude that it has some signficant antigravity in the right environment.

 

[bcrowell]

Of course, but you can't necessarily achieve a particular goal by adding only one mass.

Here's the misunderstanding, getting to all the different definitions of shield that have floated around. My scheme is intended to make g=g'=g'' etc. = 0 at *one point*. In 'practice' it might be effective for a tiny region, and generally weak fields. Remember my post starting this was a scheme to allow 'me' to float some distance above Jupiter without falling or orbiting.

I understand that we're only zeroing out n derivatives at one point. What you haven't realized yet is that the idea in your #72 doesn't work. To see why it doesn't work, go ahead and try to do the thing I told you was impossible in #74. You will not be able to find a combination of masses that zeroes out the first derivative of g, *even at one point*.

 

[bcrowell]

#11 doesn't convince me that it's even possible to talk about the "equivalence principle" when the gravitational fields of objects inside the lab have significant effects--isn't the equivalence principle all about an equivalence between measurements in a small lab in curved GR spacetime and measurements made in a similar small lab in SR? In SR, by definition nothing inside the lab can have any gravitational effects!

Well, I'm convinced that you're misunderstanding the e.p. here, but I can't force you to see it my way.

And in any case, why bother with negative mass? I could just as easily attach an object with very large positive mass to the ceiling (a small black hole or a cosmic string, say), and if the mass and distance were calibrated right the upward pull of this mass would cancel out the downward pull of the Earth (ignoring the issue of tidal differences between the pull on my head and feet, which also must be ignored when talking about a negative mass plate underneath me).

See #66 for a general proof that positive mass won't suffice to give shielding in the sense defined in #66 (vanishing of derivatives up to any desired order), and the specific counterexample I gave at the end of #74.

 

[bcrowell]

Incidentally, this book by physicist Anthony Zee makes pretty much the same argument:

Science fiction writers love the notion of antigravity. But if you believe in the equivalence principle, then antigravity is impossible.

The argument is again ludicrously simple. Drop a bunch of objects. Can any of them fall upward? No. According to the equivalence principle, you might as well be inside a rocket ship accelerating in deep space. From this point on, it is the same old argument. To an observer floating outside, the floor is rushing up to meet the objects you dropped, even though you could swear they are falling toward the floor. Obviously if the equivalence principle is correct, then in a gravitational field also, there can't be an object that falls upward.

And I'm pretty sure I originally heard this argument from some professor in college. Feynman http://books.google.com/books?id=jL9reHGIcMgC&lpg=PA90&dq=feynman%20%22equivalence%20principle%22&pg=PA93#v=onepage&q=feynman%20%22equivalence%20principle%22&f=false states the equivalence principle in terms of the effects of gravity and acceleration being indistinguishable in a small box, so I'd bet that if he did every make an argument involving the equivalence principle and the impossibility of "shielding" gravity, it was something along these lines.

I agree 100% with this. But what Zee means by "antigravity" is the existence of objects with a negative ratio of gravitational to inertial mass. This proves nothing at all about gravitational shielding as defined in #66 (being able to null out arbitrarily many derivatives of a given field).

 

[JesseM]

Well, I'm convinced that you're misunderstanding the e.p. here, but I can't force you to see it my way.

What specifically do you think I am misunderstanding? If you disagree that one common form of the equivalence principle talks about all experiments in a sufficiently small region of curved spacetime giving results which are indistinguishable from what would be seen by a suitably accelerating observer in flat SR spacetime, see the discussion on pages 372-375 of this textbook which basically says on p. 373 that as long as there is some finite limit to the precision of your measurements, no matter how small, one can find a region of spacetime sufficiently small that curvature effects are too small for you to detect in that region. ('we can choose U small enough to make all the \Gamma_{jk}^{i} smaller than any given \epsilon &gt; 0 If \epsilon is given by the lower limit of our ability to detect curvature using, say, geodesic deviation, then the result is a region U that we may cautiously consider "flat for practical purposes".') Then after some additional clarification, p. 375 says:

The approximations involved vanish when we take the limit as the size of the box goes to zero, if all the "sufficiently small's" are defined carefully. The principle then says that an observer falling freely under gravity in any space-time finds the same local physical laws - that means the same relationships between the values of sets of measurable quantities and their derivatives, at the points he actually passes through - as an inertial observer studying the behaviour of similar quantities in flat spacetime, in the absence of gravity.

The equivalence between an accelerating box in the absence of gravity and a box at rest in a gravitational field (or between an inertial box in the absence of gravity and a free-falling box in a gravitational field) is how the equivalence principle is usually stated in nontechnical discussions by physicists, such as Zee's which I quoted earlier, so I think it's unlikely to be complete nonsense!

 

[atyy]

Any negative mass that respected the equivalence principle in the sense I talked about in post #27 would have both negative gravitational and negative inertial mass, which would mean it would still fall in the same direction and at the same rate in an external gravitational field as positive mass (the difference would be that the gravitational field of the negative mass itself would be repulsive). This type of negative mass would be no use for "shielding", agreed?

Yes. I had forgotten about the double negative. So it looks like the WEP (inertial mass = gravitational mass) is sufficient to guarantee that there is no shielding in Newtonian gravity comparable to Faraday cage shielding in electrostatics. And it is not required that gravity be purely attractive, since the WEP alone allows negative gravitational mass can exist.

 

[bcrowell]

What specifically do you think I am misunderstanding?

In your earlier posts, IMO you were being too careless about the limiting processes involved. In your #84, you are being more careful.

Consider this quote from an earlier post you made:

In SR, by definition nothing inside the lab can have any gravitational effects!

This is an example of where IMO you were being too sloppy. To see that this is not careful enough, consider that the e.p. is only of interest in GR, and in GR the lab certainly will have gravitational effects. Therefore we need to talk about some kind of limiting process in which the gravitational effects of the lab become negligible.

 

[bcrowell]

So it looks like the WEP (inertial mass = gravitational mass) is sufficient to guarantee that there is no shielding in Newtonian gravity comparable to Faraday cage shielding in electrostatics.

I.e., shielding that passively adjusts itself through a mechanism exactly analogous to a Faraday cage? Yes, that's definitely correct.

However, that doesn't mean that the WEP forbids us from nulling a gravitational field in the sense of #66. Suppose that I have exotic matter that has negative gravitational mass and also negative inertial mass. There is no violation of the WEP, only a violation of an energy condition, but I can use this exotic matter to null out any given externally imposed field in the sense defined in #66 (zeroing out the first n derivatives).

 

[JesseM]

In your earlier posts, IMO you were being too careless about the limiting processes involved. In your #84, you are being more careful.

Consider this quote from an earlier post you made:

This is an example of where IMO you were being too sloppy. To see that this is not careful enough, consider that the e.p. is only of interest in GR, and in GR the lab certainly will have gravitational effects. Therefore we need to talk about some kind of limiting process in which the gravitational effects of the lab become negligible.

Fair enough, but do you agree that in the limit which defines this form of the equivalence principle, the gravitational effects of anything inside the box must approach having no measurable effect, so any "shielding" relying on the gravitational field of the shield itself won't work in this limit? Of course apart from all discussion of the equivalence principle, it's interesting to think about what form of "shielding" is possible in a larger region of spacetime where the gravitational fields of masses inside the region does have a significant effect, but I wanted to distinguish the two topics of discussion.

 

[atyy]

I can prove the "only if" in the one-dimensional case. Let the x-axis point down, i.e., we have some externally applied field g(x), which is positive at x=0, the location of our laboratory. Since g is arbitrary, we could have g'(0)>0 (as is the case, for example, in the Earth's field). If it's possible to screen out this field in the region near x=0, then I should be able to do that by adding point masses one at a time. The key is that the field of a point mass is an odd function, but the derivative is even. So if you tell me you want me to null your external field up to n=1, I can't do it. I can start placing positive point masses overhead at x<0, and that will start reducing |g(0)|. (I could also do that by putting negative point masses underfoot at x>0.) However, no matter where I place positive point masses, they will always contribute positively to g'(0), thereby increasing |g'(0)|. The only way to get a decrease in |g'(0)| is by using negative masses.

What if we restrict g(x) to one generated by non-exotic matter?

 

[bcrowell]

What if we restrict g(x) to one generated by non-exotic matter?

By g(x) do you mean the externally imposed field that we're trying to cancel out? Then the restriction doesn't help, and nulling of the first derivative is still impossible. For example, let the Newtonian gravitational constant be G=1, and let the externally imposed field be created by a non-exotic mass m=+1/2 placed at x=1. Then this externally imposed field has g'(0)=1. Since this g' is positive at x=0, you can't null it at x=0 using further chunks of exotic mass. In one dimension, the gravitational field of any point mass always has g'>0, and therefore you can't use the derivative of one mass's field to null the derivative of another mass's field.

 

[atyy]

By g(x) do you mean the externally imposed field that we're trying to cancel out? Then the restriction doesn't help, and nulling of the first derivative is still impossible. For example, let the Newtonian gravitational constant be G=1, and let the externally imposed field be created by a non-exotic mass m=+1/2 placed at x=1. Then this externally imposed field has g'(0)=1. Since this g' is positive at x=0, you can't null it at x=0 using further chunks of exotic mass. In one dimension, the gravitational field of any point mass always has g'>0, and therefore you can't use the derivative of one mass's field to null the derivative of another mass's field.

I'm having the naive intuition that if you're allowed to actively place mass, then you can always cancel out the field by adding mass so that the total mass distribution in the universe is a spherical shell. Then you just sit inside it.

 

[bcrowell]

Fair enough, but do you agree that in the limit which defines this form of the equivalence principle, the gravitational effects of anything inside the box must approach having no measurable effect, so any "shielding" relying on the gravitational field of the shield itself won't work in this limit? Of course apart from all discussion of the equivalence principle, it's interesting to think about what form of "shielding" is possible in a larger region of spacetime where the gravitational fields of masses inside the region does have a significant effect, but I wanted to distinguish the two topics of discussion.

Sorry, now I'm losing track of what we were originally debating. As the number of posts in a thread approaches the triple digits, it gets really tough to know who's claiming what. Do you think you could try to make a stand-alone post encapsulating what you're claiming, with careful attention to how the relevant limits are taken? For me, phrases like "no measurable effect" are too loose to be useful in inferring what limiting process you have in mind. We have multiple versions of the e.p. running around here, multiple statements of what is meant by "antigravity" or "gravitational shielding," etc.

 

[JesseM]

Sorry, now I'm losing track of what we were originally debating. As the number of posts in a thread approaches the triple digits, it gets really tough to know who's claiming what. Do you think you could try to make a stand-alone post encapsulating what you're claiming, with careful attention to how the relevant limits are taken?

I'm not sufficiently knowledgeable about GR math to do that, I'm just making an argument from authority based on pages 372-375 of the textbook I cited: there's some common form of the equivalence principle that says measurements in a small room in curved spacetime are indistinguishable in some limit from measurements in a small room in flat spacetime, no? (presumably part of this limit is that the volume of the room and the time period in which the measurements are made are both approaching zero, but there are probably some other conditions as well, like maybe one detailing how \epsilon approaches zero in the quote we can choose U small enough to make all the \Gamma_{jk}^{i} smaller than any given \epsilon &gt; 0 If \epsilon is given by the lower limit of our ability to detect curvature using, say, geodesic deviation, then the result is a region U that we may cautiously consider "flat for practical purposes") And by definition nothing inside a room in flat spacetime can have any gravitational effect whatsoever, so whatever the precise nature of this limit, it must be true that there are no measurable gravitational effects from objects in the room in curved spacetime in this limit.

 

[PAllen]

I understand that we're only zeroing out n derivatives at one point. What you haven't realized yet is that the idea in your #72 doesn't work. To see why it doesn't work, go ahead and try to do the thing I told you was impossible in #74. You will not be able to find a combination of masses that zeroes out the first derivative of g, *even at one point*.

Ok, I finally see your point. But then I kept thinking there must be some escape from this conundrum, because a spherical shell of matter cancels its own gravity (not gravity from outside, as argued to atyy) precisely; and clearly, if the field is zero over a large region, its gradients of any degree must also be zero. How to resolve the paradox? With some real effort (it's been a long time) I see the key is Newtonian gravity is a vector field, and that the problem is actually easy to solve if you go outside the artificial restriction to one line.

In particular, to neutralize g and g' (along a given line) from some source at one point on the given line requires only two masses not the three proposed in my erroneous scheme. The key is that the masses are off the line. Say the source is to the left of the point of interest and we want to neutralize g and the gradient along the source to given point line. Then place two masses on either side of this line, equidistant from it, to the right of the point of interest. Each off axis mass to the right has a gradient component along the line opposite that from the source, and the symmetric placement cancels other gradient components. The angle to the off axis masses allows you to adjust the gradient canceling efficiency (near 45 degrees is the most effective), while adjusting the mass gives you a degree or freedom to cancel g.

I lack the expertise to generalize this to efficient solutions for higher derivatives etc., but observe that the following shows there must be a solution (this was a suggestion of atty in a different context):

build a shell including the source as part of it, and you have canceled the field precisely in a large region by adding mass.

Given we only want point canceling of n derivatives to some precision, I expect there are much simpler solutions.

Thus active gravity shielding is possible (active, in the sense that the masses must be held in place by some external force).

 

[atyy]

But then I kept thinking there must be some escape from this conundrum, because a spherical shell of matter cancels its own gravity (not gravity from outside, as argued to atyy) precisely

My argument is to arrange matter so that there is no matter outside the shell, and no matter inside the shell, except for the laboratory, whose gravitational influence is negligible.

 

[atyy]

In Faraday shielding we wait for t=infinity. In our universe, since it is expanding, won't we automatically get shielded at t=infinity?

 

[bcrowell]

Hi, PAllen,

#94 is interesting. My proof was only for 1 dimension, and was only meant to prove that *some* externally imposed fields were impossible to null out. As you point out, one can construct examples in three dimensions that can be nulled out. For example, the field inside a hemispherical shell can be nulled by adding another hemisphere to complete the sphere.

I'm having trouble visualizing the geometry you're describing in #94. Let's say that I have an externally imposed field \textbf{g}=x\hat{\textbf{x}}, and I want to produce \partial g_x/\partial x=0, while leaving the field itself equal to zero there. Can you tell me what masses you would put at what coordinates in order to accomplish this? (Let's work in units where G=1.)

Nulling in the sense of #66 is definitely impossible in general, in three dimensions without some further restriction on the externally imposed fields, if we say that all partial derivatives up to order n have to vanish. The reason is that I could give you an external field \textbf{g} that had a negative divergence at a point P (as a field created by normal matter typically will), and ask you to null all of the field's first derivatives at P. If you could do this by adding on a nulling field \textbf{h}, then we would have \nabla\cdot\textbf{g}&lt;0, \nabla\cdot(\textbf{g}+\textbf{h})=0, so your field \textbf{h} would have to have a positive divergence, which is impossible without exotic matter.

I've added the one-dimensional thing as an example in my book http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.1 (subsection 8.1.3, example 1). There is an acknowledgment to P. Allen at the end of the example. I hope this is OK with you (i.e., you agree that the statements I make in the example are true), and that this is the right form of your name to use.

-Ben

 

[PAllen]

Hi, PAllen,

#94 is interesting. My proof was only for 1 dimension, and was only meant to prove that *some* externally imposed fields were impossible to null out. As you point out, one can construct examples in three dimensions that can be nulled out. For example, the field inside a hemispherical shell can be nulled by adding another hemisphere to complete the sphere.

I'm having trouble visualizing the geometry you're describing in #94. Let's say that I have an externally imposed field \textbf{g}=x\hat{\textbf{x}}, and I want to produce \partial g_x/\partial x=0, while leaving the field itself equal to zero there. Can you tell me what masses you would put at what coordinates in order to accomplish this? (Let's work in units where G=1.)

Nulling in the sense of #66 is definitely impossible in general, in three dimensions without some further restriction on the externally imposed fields, if we say that all partial derivatives up to order n have to vanish. The reason is that I could give you an external field \textbf{g} that had a negative divergence at a point P (as a field created by normal matter typically will), and ask you to null all of the field's first derivatives at P. If you could do this by adding on a nulling field \textbf{h}, then we would have \nabla\cdot\textbf{g}&lt;0, \nabla\cdot(\textbf{g}+\textbf{h})=0, so your field \textbf{h} would have to have a positive divergence, which is impossible without exotic matter.

I've added the one-dimensional thing as an example in my book http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.1 (subsection 8.1.3, example 1). There is an acknowledgment to P. Allen at the end of the example. I hope this is OK with you (i.e., you agree that the statements I make in the example are true), and that this is the right form of your name to use.

-Ben

Hi,

Formalities first: fine if you acknowledge me, P. Allen is my actual (partial name). In some more private communication, I could give you more complete attribution, if you want (I have no relation to Microsoft, though we are in the same industry and the same age; bank balance differs).

I don't really want to work out a precise numeric example, but thinking more since post #94, I can make my argument much tighter, and I hope clearer.

Assume a point mass at x=-10, our point of interest (where we want to null g and g' from the source at x=-10) at x=0, and equal (to each other) masses at candidate positions x=1,y=1, and x=1, y=-1. What I am going to argue is by moving the balancing masses further away or closer (along the same line from the origin), and/or changing their angle to the x-axis at the origin (in all cases keeping their distances from the x-axis the equal to each other), we have more than enough degrees of freedom to get any magnitude ratio of g'/g that we want, and that in all cases, the sign of g' for these side masses (taken together) is the opposite of g' from the source at -10 (as is the sign of g). Given that, we find a choice that matches g'/g magnitude for the source we are canceling, then choose masses to match the magnitude for the source values (the ratio g'/g for the balancing masses depends only on distance and angle not on mass).

Ok, put simply (for me), the problem with colinear masses is that g' is positive no matter the sign or value of g. With the off axis masses placed to the 'right' of the origin, we have net g with the correct sign for cancelation, but g' negative, just what we need. This is easily seen by noting that as you move from the left of these masses to the point directly between them, g goes to zero. So we have g decreasing as a funcion of x rather than increasing as for a colinear balancing mass.

To see that we have freedom to match any g'/g magnitude, note that g'/g goes infinity as the balls are moved closer to perpendicular to the x-axis (g goes to zero effectively as cos(angle to perpendicular), g' as sine (same angle)). Note that for some given angle between the balancing balls and the x-axis at the origin, we can make g'/g approach zero by moving the balls further away at the same angle (because g' goes r**-3, while g as r**-2).

Thus, I believe all elements of my argument are established.

My gut feel is that procedures like this will work for higher derivatives, using more balancing masses. The additivity of everything suggest it should work to neutralize any collection of given point source fields. In Newtonian gravity, motion of the source makes no difference (all is instant possition dependent force) except balance balls must move in some complex way.

In GR, my gut is that all this breaks down for moving sources. One line of thought is that I was led to believe this was possible by the spherical shell case. In Newtonian gravity, if this shell rotates, nothing changes. I suspect (you may know for sure) that in GR you get small frame dragging or similar effects. I see no reason to believe these can be canceled. Of course, there are also gravity waves. For E/M waves, the field goes through +/- sign and you can cancel any wave with a phase shifted wave. For gravity waves, there is no negative gravity influence, so I don't see, offhand, how any gravity wave can be cancelled. Thus moving sources in GR seem to produce several types of uncancellable effects. Of course, for weak, slow sources, the Newtonian scheme should work as well in 'practice', given a large supply of self propelled dense matter balls and software to move them as needed.

 

[bcrowell]

PAllen, your argument in #98 makes sense to me, if all you want to do is null out g_x and \partial g_x/\partial x. However, I proved in #97 that you cannot simultaneously null out all nine first derivatives \partial g_{x_i}/\partial x_j in all cases. Because you haven't specified any restriction that would evade that proof, I don't believe that your method works in general. Basically the way I know that your method can't be as general as you think is that if we don't have exotic matter, then you can't independently vary the three i=j derivatives \partial g_{x_i}/\partial x_i, because there are constraints on the divergence of the field.

 

[PAllen]

PAllen, your argument in #98 makes sense to me, if all you want to do is null out g_x and \partial g_x/\partial x. However, I proved in #97 that you cannot simultaneously null out all nine first derivatives \partial g_{x_i}/\partial x_j in all cases. Because you haven't specified any restriction that would evade that proof, I don't believe that your method works in general. Basically the way I know that your method can't be as general as you think is that if we don't have exotic matter, then you can't independently vary the three i=j derivatives \partial g_{x_i}/\partial x_i, because there are constraints on the divergence of the field.

I am not in a position to seriously investigate what can be achieved with balancing masses, so I don't dispute what you say. All I said was 'gut feel' anyway.

Anyway, this thread has clearly established that the relationship between the e.p. and 'no gravity shield' is much more complex than many sources have claimed, starting with pinning down both e.p. and what you mean by a gravity shield; continuing with additional assumptions (but ones the most would accept: no exotic matter, and conservation rules).