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

[atyy]

BTW, why isn't being at the centre of a black hole considered to be perfect shielding?

 

[atyy]

Hmm...well, there is certainly shielding in the case of electrostatics/magnetostatics, i.e., no radiation fields. If you throw radiation into the mix, then there are forms of shielding that would shield against UV (sunscreen), visible light (duct tape), x-rays (lead), ... Maybe you could say in more detail what you mean by the assertion that there is no shielding in electrodynamics. Is this something to do with classical EM versus quantized EM?

The shielding of duct tape is partial, since it is frequency dependent. Only the electrostatic shielding is perfect in its domain.

If we allow partial shielding, then putting a huge mass around yourself must surely be considered partial shielding too in GR.

 

[bcrowell]

So I guess you are thinking that the EP would forbid such materials existing for gravity even in a quantum theory of gravity?

Bluh...I don't know. You know a lot more about quantum gravity than I do. I'm not even sure this topic requires relativity, much less quantum gravity.

I was considering only shielding by conductors, ie. Faraday cage in classical EM.

So considering more general forms of shielding, it seems they are all quantum mechanical (even the textbook perfect conductor).

Hmm...well, I'm sure that you need QM to explain why zinc oxide or something absorbs UV...but I'm not convinced that we can't gain deeper insight into the problem from a purely classical perspective. I'm pretty sure that I can make a purely classical model of an electromagnetic shield that will block static E fields, static B fields, and EM radiation. To block E fields, all I need is a perfect conductor, modeled by positive and negative charges that are free to move without friction. To block B fields, I could do the same with magnetic monopoles, or -- if you insist on something that actually exists -- I think I can accomplish it with something else classical, like little current loops mounted on gimbals. Absorbing EM radiation is easy; just add a little dissipation to your classical conductor.

 

[atyy]

Wikipedia has an article!

They recommend http://arxiv.org/abs/gr-qc/0602016

 

[PAllen]

BTW, why isn't being at the centre of a black hole considered to be perfect shielding?

Because it isn't. First, if one has the typical case, with a horizon, from outside the black hole no matter ever crosses the horizon. All matter that falls 'into' the black hole region is subject to gravitation.

If you assume a naked singularity, then all mass that falls in is subject to external fields.

In either case, adding mass increases both gravitational mass and inertial mass of the black hole. No shielding at all.

 

[PAllen]

The shielding of duct tape is partial, since it is frequency dependent. Only the electrostatic shielding is perfect in its domain.

If we allow partial shielding, then putting a huge mass around yourself must surely be considered partial shielding too in GR.

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).

 

[atyy]

Because it isn't. First, if one has the typical case, with a horizon, from outside the black hole no matter ever crosses the horizon. All matter that falls 'into' the black hole region is subject to gravitation.

If you assume a naked singularity, then all mass that falls in is subject to external fields.

In either case, adding mass increases both gravitational mass and inertial mass of the black hole. No shielding at all.

I see, so you are saying even if you are at the singularity, you can still detect the change in topology of the singularity.

 

[atyy]

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).

It is, because if you surround yourself with a huge mass, then moving a tiny little piece of mass outside that huge mass will be essentially undetectable - hence partial shielding.

 

[atyy]

Ok, I guess the strategy is:

In Newtonian gravity, there is no region of uniform gravitational potential unless there is no mass at all or the mass distribution is spherically symmetric around you, hence no shielding in general. The form of Newtonian gravity is guaranteed by the EP, plus what assumptions - hence the EP +? guarantees no shielding in Newtonian gravity.

In GR, there is no region of flat space unless there is no mass at all or the mass distribution is spherically symmetric around you, hence no shielding in general. The form of GR is guaranteed by the EP plus no prior geometry - hence the EP + no prior geometry guarantees no shielding in GR.

 

[PAllen]

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.


So this means that shielding is possible in Newtonian gravity?

Surely not!

As in an earlier post, if you strengthen ep to simply explicitly say 'gravitational mass=ineratial mass', then shielding as I understand it is impossible. This is normally assumed for Newtonian gravity.

Is shielding impossible in Newtonian gravity? In fact shielding is guaranteed. Let's make an impenetrable sherical shell. Matter around the universe will attract other pieces of matter, and try to form a black hole. But they can't, because of the impenetrable sherical shell. The best they can do is crowd around the shell, and equilibrium will be spherical, so shielding is guaranteed. Newtonian gravity obeys an EP.

Here we go with the spherical shell again. Go inside a spherical shell, put the shell in orbit around the sun. If you are shielded, as I understand it, the sun's gravity only pulls on the shell, yet the shell's inertial mass includes you. Thus you will orbit differently than anything else. This is false, will not happen. The shell has no effect whatsoever on the ability of the sun's gravity to pull on you as well as the shell.

 

[atyy]

Here we go with the spherical shell again. Go inside a spherical shell, put the shell in orbit around the sun. If you are shielded, as I understand it, the sun's gravity only pulls on the shell, yet the shell's inertial mass includes you. Thus you will orbit differently than anything else. This is false, will not happen. The shell has no effect whatsoever on the ability of the sun's gravity to pull on you as well as the shell.

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.

 

[atyy]

PAllen, the customary style on PF is to http://en.wikipedia.org/wiki/Top-posting#Bottom-posting". I suppose this comes from the fact that some PF old-timers are refugees from usenet, where bottom-posting is customary. Other than that, there is the fact that God hates top-posters and makes them burn in Hell for all eternity.

But we don't know yet if the EP forbids shielding from hell.

 

[atyy]

As in an earlier post, if you strengthen ep to simply explicitly say 'gravitational mass=ineratial mass', then shielding as I understand it is impossible. This is normally assumed for Newtonian gravity.

Yes. Why can't inertial mass be negative?

 

[PAllen]

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.

So this means that shielding is possible in Newtonian gravity?

Surely not!

Is shielding impossible in Newtonian gravity? In fact shielding is guaranteed. Let's make an impenetrable sherical shell. Matter around the universe will attract other pieces of matter, and try to form a black hole. But they can't, because of the impenetrable sherical shell. The best they can do is crowd around the shell, and equilibrium will be spherical, so shielding is guaranteed. Newtonian gravity obeys an EP.

It is, because if you surround yourself with a huge mass, then moving a tiny little piece of mass outside that huge mass will be essentially undetectable - hence partial shielding.

I don't see it. This small mass pulls on you the same amount as if the huge mass weren't there. What a Farady cage achieves is that coulomb field of an exterior charge doesn't reach into the cage. Here it reaches in exactly as strongly as without the purported shield.

 

[PAllen]

Ok, I guess the strategy is:

In Newtonian gravity, there is no region of uniform gravitational potential unless there is no mass at all or the mass distribution is spherically symmetric around you, hence no shielding in general. The form of Newtonian gravity is guaranteed by the EP, plus what assumptions - hence the EP +? guarantees no shielding in Newtonian gravity.

In GR, there is no region of flat space unless there is no mass at all or the mass distribution is spherically symmetric around you, hence no shielding in general. The form of GR is guaranteed by the EP plus no prior geometry - hence the EP + no prior geometry guarantees no shielding in GR.

yes, this I agree with. I don't know how to make it rigorous, though.

 

[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.