I can't remember why I initially thought of this situation, but now it's bothering me. You take a loop of wire and place a negative charge Q outside of the loop. ((broken link)) Now surround one half of the loop with a conductive surface so that no net electric field exists inside. In the diagram, the electron e1 on the left will experience a force due to the charge -Q and will want to flow in one direction around the loop. Another electron e2 inside the Faraday cage will not experience a force in either direction from the charge -Q.

Do you see the problem?

If we integrate the electric field along the left and right "sides" of the loop, they should not sum to zero because we can adjust the Faraday cage such that the electric field on the right side is zero along the whole path. This means we should expect a current in the wire. (This is just Ohm's law)

On the other hand, the second law of thermodynamics says that there cannot be a current in the wire unless its resistance is exactly 0; otherwise we'd be dissipating energy.

My problem is that I cannot think of an actual electrodynamic reason for the current not to flow. Of course if we integrate the [static] electric field over any closed loop we must get zero because it has zero curl, but intuitively I do not see why this must be the case here. So I suppose this is my real problem.

Can anyone explain this?

isnt the answer that the faraday cage gets slightly hotter therfore keeping the laws of thermodynamics intact?

A funny situation indeed. Basically if you use a conductive material to separate the other electron from any external electromagnetic fields, it means that:

1) either the conductive material forms a "box" that cuts the wire so that the part inside the conductive box isn't connected to the outer part of the wire, and therefore there is no circuit and the electrons can't create current (electrons would simply move to the other end of the wire, which got a negative charge if Q was positive, and in the end there was an equilibrium), or

2) the conductive box touches the wire, and because both the wire and the conductive box conduct electricity, a circuit is formed and despite the fact that the electric field could not get inside the "box" the electrons could flow trough the circuit (but they would not), or

3) there is a hole in the conductive "box" so that it doesn't touch the wire, but the wire is not cut; in this case the electric field can get inside the "box" and affects the other electron as well (and again we have a circuit).

Then you should consider this, if a circuit is formed: if you put a charge Q outside the wire, it shouldn't probably create a current on the wire. Why? Because it makes some electrons move either towards it or to the other direction, and at the same time it makes other electrons move in the direction opposite of that (because it can't, at the same time, push some electrons away and pull others towards itself; one of the basic "rules" of quantum mechanics is that you can't tell an electron from another, so if we consider all electrons identical, they can't act differently from one another). There is a force between the charge and the electrons in the conductive wire, however small, but neither of them can move, and it shouldn't create any electric current in the wire. Physically the wire has resistance, so if there was a current in it, it meant the charge Q would do work (and therefore consume energy) if it created a current in the wire, and that would mean a forever loop without external power sources, which is considered impossible as of now.

Theoretically it's more or less easy to find situations where the laws of physics seem to fall, but usually it's just a matter of thinking with common sense to find the reason why this won't happen. The theory of relativity is even more often used to create bizarre or paradoxal situations, but in the end most of them are based on one thing: some of the assumptions made is impossible.

No. If we consider this as an isolated system, then, among other things, raising the temperature of the Faraday cage will radiate energy from the system without putting any in, which is not allowed.

This is the closest to what I meant. The wire is insulated and does not make electrical contact with the cage.

Of course we'd expect current not to flow intuitively, but that is not quite a good enough reason. The point was that on the left side, there is impetus for the electrons to move, while on the right there is no such force to counteract this (there is zero electric field inside the cage, this is a result of elementary electrostatics), and since the forces are unbalanced there must be an acceleration of the charges around the circuit. The problem is that I can't think of a decent reason why not.

And of course we can't distinguish individual electrons since they carry no history beyond their last interaction, but this is definitely overcomplicating things; in a macroscopic system we can consider the charge as continuous instead of discrete (the so-called correspondence principle) and use classical electrostatics (Maxwell).

Can you think of an example? I don't think I can.

"Do you see the problem?"

probably.

bringing thermodynamics into this discussion is misplaced. there is no such thing in the real world as zero resistance. the most perfect conductor in the world still has some amount of resistance. if the conductor is of adequate size for the current, and losses to heat can be safely ignored. heat happens, pondering how it is created, flows, or spreads in this context is irrelevant.

there are two surfaces to a faraday cage. the inside and the outside. a negative charge on the outside surface creates an positive on the inside surface, and vice versa. isolating the inside _area_ from external influence.

it's not a magic box that halts all current flow inside.

the inside area of the cage may not see the external charge, but that has no bearing on the charge being applied to the external wire and whether or not current is flowing.

Look, either the conductive cage touches the wire, in which case we have a closed circuit via the external surface of the cage, where there is an electric field present. Or, if there is a gap between the cage and the wire (if for instance the wire is insulated), an electric field is present inside the cage, causing charge carriers to move along the wire. Case closed dude!!

No, it isn't misplaced. The second law says that you cannot create energy (it goes on to say a lot of other things too), and it is always true. If you isolate the system from the surroundings, the ohmic losses will heat the thing up. This would mean it's creating energy.

I am not making any presumptions about what the resistance of the wire is. As an aside, there is such thing as a wire with zero resistance (there is a whole class of such materials, called superconductors).

No, it's only a magic box that shields its entire inner volume from the external electric field. This is why your computer lives in a conductive, metal box.

There is no electric field in the cage regardless of whether there are holes in it.

So far I don't think anyone has understood the problem.

Can you explain to me why? If the cage is closed surface and there are no large charge concentrations in the cage, then yes, there will be no electric field in the cage because that's the only solution to the Poisson equation (nabla^2 V = rho/e0, where V is the potential function and rho is the charge density function).
But if the cage is NOT closed, the solution to Poisson's equation will be different and will allow an electric field to be present inside (where is that btw if we have an open box?) the cage.
The crux of the solution is, is that "inside the box" is not well defined anymore if you have holes in the surface. If I have a closed box, and I keep on cutting away some of it's surface untill all that I have left is a single wall. On which side of the wall will I still be "inside the box"?

I agree that as soon as you put a hole in the box it's no longer closed, that then you really only have one surface and so inside/outside the box becomes less well defined. If you surround something with a solid conductive surface, then in the presence of an external electric field, the free charges on the conductor will rearrange themselves such that their electric fields exactly cancel the external field inside the volume enclosed by the surface. (this is a fact!) When you start poking holes in the surface, this still holds for most practical purposes provided the areas of the holes are much smaller than the area of the surface.

So if we know that the field inside is zero, can you convince me that the path integral of E around the closed [wire] loop should be zero? This is what I'm stuck on!

So you are saying that your nearly closed cage has a small enough hole for the electric field to be nearly zoro ("for most practical purposes"), but is big enough for the wire to get through without touching the cage.
Okey, well, lets calculate the electric field along the wire shall we? We have a one-dimensional situation, so Poisson's equation reduces to (d^2/dx^2)V=0 (rho=0 everywhere), which admits solutions in the form of V(x)=ax+b, where x is the position along the wire.

Now, if we were a little farther away from the wire, we would have the edge conditions V(x=0)=V0 and V(x=L)=V0, where L is the length of the cage and V0 is the potential of the cage with respect to some arbitrary point. (And since the cage is a conductor, the potential is the same everywhere on the cage). This give a solution that V(x)=V0 and thus E=dV/dx=0.

Now we look just out side the wire. Now suddenly the above edge conditions do not apply anymore and thus solutions where E does not vanish become realisitic. So my conclusion is that, even if the hole is small, the electric field just outside the wire will not vanish, allowing the path integral along the wire to become zero.

In higher dimensions this is harder to see, but my guess would be this still holds. Why don't you run simulations where you caluclate the electric field in a 2d plane with a charge e somewhere and a nearly closed cage with two holes somewhere else. My bet is that you will find that the electric field does not vanish "inside" the cage.

Basically what I'm saying is that if your holes are small, the electric field in the cage will be nearly zero nearly everywhere, execpt right at the holes and the imaginary lines connecting the holes inside the cage. Do your calculations and simulations to varify this! And because our wire must go through exactly these holes, the electric field along the wire will not vanish, no matter how small the holes are.