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

[atyy]

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.

 

[bcrowell]

It's cool to see this thread drawing some intense discussion!

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.

This occurs only in electrostatics, and not in electrodynamics.

I think GR is "closer" to electrodynamics than to electrostatics.

If there is no shielding in electrodynamics in which there is no EP (I think), then it is clear there can be factors other than the EP that forbid shielding.

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?

 

[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?

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).

So what guarantees the existence of such materials in the classical theory? Or what permits the existence. So I guess you are thinking that the EP would forbid such materials existing for gravity even in a quantum theory of gravity?

 

[PAllen]

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).

So what guarantees the existence of such materials in the classical theory? Or what permits the existence. So I guess you are thinking that the EP would forbid such materials existing for gravity even in a quantum theory of gravity?

It's not so much what guarantees their existence as that no principle in Maxwell+(SR/GR) precludes their existence (that I know of). Meanwhile, EP in GR seems to preclude such in GR, but it seems not so easy to make the argument tight. I think I have come closest with the idea that if you assume GR+EP, you can make any given matter configuration function as a test body for the purposes of EP to any desired level of precision; this, then seems to exclude shielding that universally shields gravity without affecting inertial mass.

I don't know how to make this argument work as robustly without assuming GR. I also don't know how to preclude a purported shield that only works in the presence of tidal forces, or even more radically, a shield that specifically shields tidal forces.

Finally, it seems only conservation of mass/energy (which is very tricky in GR) prevents a shield that hides mass for all purposes.

 

[atyy]

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.

 

[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