Newton's Shell Theorem and charge

[DaveC426913]

I've been discussing Newton's Shell Theorem re: gravity with someone, and thought of the analogy to charge.

1. I think the net effect on a negative charge inside a hollow sphere of positive charge will be zero. i.e. No net attraction. Yes?

2. But what would happen to the magnetic field if the negative charge were set moving?(Basic level question - looking for concepts rather than math)

Thinking: it's got to behave as if there is no field at all. Otherwise that magnetic field would oppose the motion, bringing it to a stop. And that would be weird behavior. You could move it (with effort) to anywhere in the sphere, and wherever you left it, it would resist being moved. It would behave as if the negative charge were suspended in a viscous soup. That can't be right, can it?

 

[Dale]

In the rest frame of the sphere there is no electric field and also no magnetic field. So the full EM tensor is 0. A tensor that is 0 in one frame is 0 in every frame. So it doesn’t matter if it is moving or not. It does not produce an EM field inside.

 

[DaveC426913]

Thanks. That's what I figured. So a magnetic charge analogy is useless here.

I'm (still) trying to figure out a conceptual way of demonstrating that, even with a net zero gravitational force within the sphere, the gravitational potential is still non-zero.

The gravity well outside a hollow sphere is shaped like an inverse bell, while the field inside the sphere will have a flat bottom.

An associate, having trouble with this concept, thinks that the potential inside the sphere should be zero. (i.e. the same potential as would be found at an infinite distance from the sphere.) His graph looks like a moat - with inner "walls" that are vertical! (And thus, zero gravitational time dilation inside the hollow.)

 

[Dale]

An associate, having trouble with this concept, thinks that the potential inside the sphere should be zero. (i.e. the same potential as would be found at an infinite distance from the sphere.) His graph looks like a moat - with inner "walls" that are vertical! (And thus, zero gravitational time dilation inside the hollow.)

Potentials are continuous. Even where the fields are discontinuous the potentials are continuous.

 

[etotheipi]

An associate, having trouble with this concept, thinks that the potential inside the sphere should be zero. (i.e. the same potential as would be found at an infinite distance from the sphere.) His graph looks like a moat - with inner "walls" that are vertical! (And thus, zero gravitational time dilation inside the hollow.)

Assuming the shell is negligibly thin, the potential inside will just be whatever it's value is at the surface of the shell:

This is because the gravitational field goes to zero inside the shell, and $\vec{g} = -\nabla \phi$:

 

[DaveC426913]

Yes but there is still gravitational time dilation within the shell.

 

[etotheipi]

Yes but there is still gravitational time dilation within the shell.

If we have gravitational time dilation $d\tau = \sqrt{1 + \frac{2\phi}{c^2}} dt$ (where $\phi$ is zero at infinity), then this is still consistent because $\phi$ is non-zero inside the shell (see the first diagram). Furthermore, this value of gravitational time dilation is the same everywhere inside the shell.

 

[etotheipi]

Thinking: it's got to behave as if there is no field at all. Otherwise that magnetic field would oppose the motion, bringing it to a stop. And that would be weird behavior. You could move it (with effort) to anywhere in the sphere, and wherever you left it, it would resist being moved. It would behave as if the negative charge were suspended in a viscous soup. That can't be right, can it.

Also I don't understand this part; the charges inside the uniformly charged shell are static so produce no magnetic field. As @Dale notes the whole EM tensor is zero.

If the charged particle inside the shell is moving, then it will produce a magnetic field that loops around its line of motion but it will not feel a force from this magnetic field it itself produces!

 

[Ibix]

An associate, having trouble with this concept, thinks that the potential inside the sphere should be zero. (i.e. the same potential as would be found at an infinite distance from the sphere.) His graph looks like a moat - with inner "walls" that are vertical! (And thus, zero gravitational time dilation inside the hollow.)

Simply reply that $\vec g=-\nabla\phi$ and ask what the gravitational field is at the discontinuity. Loosely, gravitational force is the slope of the graph and gravitational potential is the depth. That's why force has a direction and potential doesn't.

Yes but there is still gravitational time dilation within the shell.

With reference to the diagrams in #5, gravitational time dilation is just a measure of the distance of the graph below the x axis. Or, more precisely, the difference in said distance for two clocks. So there's no time dilation between two clocks inside the shell (unless they're moving with respect to each other) but there is time dilation between a clock inside and a clock outside.

 

[A.T.]

An associate, having trouble with this concept, thinks that the potential inside the sphere should be zero. (i.e. the same potential as would be found at an infinite distance from the sphere.)

Then it would take zero work to lift a mass from the inside to infinity. This makes no sense, because as soon it leaves the shell you obviously have to do work against gravity.

Or conversely, if you drop something from infinity, it will arrive inside the shell with some kinetic energy. Where did that energy come from, if the potential energy didn't change?

 

[DaveC426913]

With reference to the diagrams in #5, gravitational time dilation is just a measure of the distance of the graph below the x axis. Or, more precisely, the difference in said distance for two clocks. So there's no time dilation between two clocks inside the shell (unless they're moving with respect to each other) but there is time dilation between a clock inside and a clock outside.

Yes. My associate thinks that there is zero discrepancy between a clock inside the shell and a clock at infinity.

IOW, the gravitational well looks like a ... moat.

 

[DaveC426913]

Then it would take zero work to lift a mass from the inside to infinity. This makes no sense, because as soon it leaves the shell you obviously have to do work against gravity.

Or conversely, if you drop something from infinity, it will arrive inside the shell with some kinetic energy. Where did that energy come from, if the potential energy didn't change?

The same could be said, though, from one gravity well to another. Say, a thousand miles above Earth to 500 miles above Mars, or thereabouts.

It's not a very intuitive argument to highlight the error, but I think it's on the right track.

 

[Ibix]

Yes. My associate thinks that there is zero discrepancy between a clock inside the shell and a clock at infinity.

IOW, the gravitational well looks like a ... moat.

In which case, as I said, point out that $\vec g=-\nabla\phi$ and that there's a rather nasty infinity in $\vec g$ at that discontinuity.

 

[etotheipi]

It's not a very intuitive argument to highlight the error.

It seems pretty intuitive to me; if I make a tiny little hole in the shell and drop something really far away so that the centre of the shell, the hole and the starting point were collinear, then I would sure expect the object to have kinetic energy once it passed through the hole... and that means its potential energy must have decreased...

Did your associate just guess? It is pretty clear from just the equation $\vec{g} = -\nabla \phi$ that the potential is going to be non-zero.

 

[A.T.]

The same could be said, though, from one gravity well to another.

But the point at infinity is not in any gravity well. Claiming that the potential in the shell is the same as at infinity is like claiming that there is no gravity well around the shell.

 

[Dale]

Yes. My associate thinks that there is zero discrepancy between a clock inside the shell and a clock at infinity.

IOW, the gravitational well looks like a ... moat.

View attachment 264432

What this would mean is that on a spherical shell it would take as much energy to go into the shell as it does to rocket away off to infinity. So if you had a spherical shell and cut a hole in the shell and tried to jump into the hole, then you would get stuck. You would stand there hovering on nothing. It would be a wall of gravity.

This wall would super strong. Even though there is nothing there but gravity, it would be so strong that you could fly the SR71 black bird right into it and it wouldn't be able to get through. Neither could bullets, they would just splat in midair. In order to get in you would need to get on a rocket, accelerate to escape velocity, and steer through the hole. You would still get squished by the sudden deceleration, but your mushy remains would be inside the shell.

Any air on the inside of the shell, as soon as it reached the hole, would immediately get sucked out and it would leave the shell as an escape-velocity hyper wind. If you touched the hole from the inside then your hand and probably a good part of your arm would get yanked off and shot out at escape velocity.

If your friend honestly thinks that all of that is plausible then I suggest you decide either to drop the subject or drop the friend. (Personally, I would drop the subject, good friends are worth keeping even if they believe in some nonsense)

Edit: I see that @etotheipi and I are thinking along the same lines!

 

[jbriggs444]

Personally, I would drop the subject, good friends are worth keeping even if they believe in some nonsense

+1. Freshman year in college was an introduction to the premise that idiots are people too.

 

[DaveC426913]

If your friend honestly thinks that all of that is plausible then I suggest you decide either to drop the subject or drop the friend. (Personally, I would drop the subject, good friends are worth keeping even if they believe in some nonsense)

Not a friend.

I was hoping to find an argument so concrete that even a crank couldn't poke a hole in it.

But, upon reflection, I've realized that is a selfish waste of PF members' time. (Unless you're here for the entertainment.)

 

[Ibix]

I was hoping to find an argument so concrete that even a crank couldn't poke a hole in it.

Arguing with cranks is like shooting zombie fish in a barrel. You can't miss, but direct hits don't even slow 'em down.

 

[etotheipi]

I was hoping to find an argument so concrete that even a crank couldn't poke a hole in it.

Hopefully you agree on the potential outside the shell. And also that the gradient of the potential is zero inside the shell. The point of contention seems to be what happens at the boundary.

If a function $f$ is integrable, then its integral is necessarily continuous. Since a potential pertains to an integral of the field, if the field is integrable then the potential must be continuous. And if the field is not integrable, a potential can't be defined at all. We'd expect the field to be finite at the surface (the field can still be discontinuous, so long as it is finite!), in which case the potential will be continuous.

This is what @Dale alluded to in #4.

 

[DaveC426913]

Hopefully you agree on the potential outside the shell. And also that the gradient of the potential is zero inside the shell. The point of contention seems to be what happens at the boundary.

If a function $f$ is integrable, then its integral is necessarily continuous. Since a potential pertains to an integral of the field, if the field is integrable then the potential must be continuous. And if the field is not integrable, a potential can't be defined at all. We'd expect the field to be finite at the surface (the field can still be discontinuous, so long as it is finite!), in which case the potential will be continuous.

This is what @Dale alluded to in #4.

Yes, but I'm looking for a conceptual answer, rather than a mathematical answer.

 

[Ibix]

Yes, but I'm looking for a conceptual answer, rather than a mathematical answer.

I don't think you can do better than the already-mentioned point that his suggestion implies that the gravitational force must be near-infinite (or ill defined if there's actually a discontinuity in the potential) and outward-directed at the shell.

 

[Dale]

Yes, but I'm looking for a conceptual answer, rather than a mathematical answer.

I think the hole in the shell stuff is pretty conceptual. Being able to stand on nothing over a hole in the shell seems pretty conclusive.

It would also make planets unstable. If there is any vibration or quantum fluctuation in the center of a solid sphere then that would suddenly form a huge potential cliff which would push everything away at the center. As things move it would continue to push out further and further until all planets would be hollow spheres of infinite radius.

 

[DaveC426913]

I see it now.

R is a rocket.
A is a point at infinity (i.e escape velocity)
B is a point inside the sphere
Y is the value of potential at any given point in the system.

The energy required to get rocket R to point A has a magnitude of Y (or energy = -Y). Easy peasey.

But according to this (flawed) diagram, the energy required to get from R to B is also Y (-Y).The rocket would have expend an amount of energy equivalent to reaching escape velocity just to get one inch into the hollow. In other words, it would hit an invisible wall.Thanks guys.

The key take away I got from you guys is the realization that "gravitational potential" is simply synonymous with "potential energy", exactly like lifting a rock off the ground. You can gain it or lose it by moving or being moved with a gravitational well (i.e converting PE to KE and KE to PE).

Now I see how it can be applied to the hollow sphere.

 

[A.T.]

The rocket would have expend an amount of energy equivalent to reaching escape velocity just to get one inch into the hollow.

On the plus side, it would be very easy to escape from the hollow, as you would immediately reach escape velocity.

 

[alantheastronomer]

Summary:: What would happen to a moving -ive charge inside a hollow sphere with uniform +ive charge?

I've been discussing Newton's Shell Theorem re: gravity with someone, and thought of the analogy to charge.

1. I think the net effect on a negative charge inside a hollow sphere of positive charge will be zero. i.e. No net attraction. Yes?

2. But what would happen to the magnetic field if the negative charge were set moving?(Basic level question - looking for concepts rather than math)

Thinking: it's got to behave as if there is no field at all. Otherwise that magnetic field would oppose the motion, bringing it to a stop. And that would be weird behavior. You could move it (with effort) to anywhere in the sphere, and wherever you left it, it would resist being moved. It would behave as if the negative charge were suspended in a viscous soup. That can't be right, can it?

1. You're right

2. There is no magnetic field since the charges on the sphere are not moving

but the situation becomes conceptually more interesting if you allow for the sphere to be a perfect conductor, then

1. The charge distribution will become non-uniform, breaking symmetry and so the shell theorem will no longer apply, and

2, If you allow the sphere to have some thickness so that the charges can respond to the magnetic field created by the moving charge, you should find some interesting reactions and back reactions to explore

 

[jaropat]

Summary:: What would happen to a moving -ive charge inside a hollow sphere with uniform +ive charge?

I've been discussing Newton's Shell Theorem re: gravity with someone, and thought of the analogy to charge.

1. I think the net effect on a negative charge inside a hollow sphere of positive charge will be zero. i.e. No net attraction. Yes?

2. But what would happen to the magnetic field if the negative charge were set moving?(Basic level question - looking for concepts rather than math)

Thinking: it's got to behave as if there is no field at all. Otherwise that magnetic field would oppose the motion, bringing it to a stop. And that would be weird behavior. You could move it (with effort) to anywhere in the sphere, and wherever you left it, it would resist being moved. It would behave as if the negative charge were suspended in a viscous soup. That can't be right, can it?

1. Calculation in electrostatics
2. Calculation in electrodynamics. Electric and magnetic fields are bound.