Electric field inside a conductor

[Count Iblis]

That was an email I sent you. It's about the "dynamical Casimir effect" (oopsa, I said a bad word ). I've opened a thread that this effect seems to be an example of. It's in the scepticism and debunking section.

I received your mail, I'll visit that thread later today.

 

[Count Iblis]

Kanato is right.

Gedanken experiment: Suppose, for the sake of argument, that gravity induced a polarization in an object, with one charge on the bottom and the opposite on top. Then the charges on the bottom would feel a feel a force downward and a force upward. No problem there. But the charges on top would feel two forces downward. They would fall. And the end result would be no polarization and no net electric field. This applies to any substance, including aluminium. Therefore, gravity does not induce electric fields.

This argument doesn't make any sense whatsoever. The ions are not very mobile. If the electrons sink downward, there is no way the aluminium ions will follow suit.

 

[Phlogistonian]

This argument doesn't make any sense whatsoever. The ions are not very mobile. If the electrons sink downward, there is no way the aluminium ions will follow suit.

The ions don't fall from top to bottom. Each ion moves a little bit.

And what makes you think the electrons fall without the ions?

 

[Count Iblis]

The ions don't fall from top to bottom. Each ion moves a little bit.

And what makes you think the electrons fall without the ions?

In a conductor the electrons are far more mobile than the ions. If you switch on a weak electric field, then the electrons will move and neutralize the electric field. The ions are not relevant here. But the electrons do that because a force acts on them, they don't care that it is an electric force. The reason why (in the absense of gravity) there is no component of the electric field along the surface of the conductor is because the free elecrons respond to any applied electric field in this direction. The sum of the field produced by the electrons and the applied field is zero.

So, when we apply an electric field we do get polarization. There is a net electric field generated by the metal which cancels the applied electric field, so the total electric field is exactly zero in the metal and just outside the metal he component of the total electric field along the metal's surface is zero.

If we now replace the applied external electric field by the gravitational field, the results will be similar. The metal will become polarized and an electric field will be generated that so that the electron's potential energy as a function of height becomes constant.

If someone disputes this then they should also dispute the standard textbook explanation of why the electric field inside conductors is exactly zero.

 

[Phlogistonian]

...

If we now replace the applied external electric field by the gravitational field, the results will be similar. The metal will become polarized and an electric field will be generated that so that the electron's potential energy as a function of height becomes constant.

Gravity affects the ions too. They don't move much relative to each other, but they do move. The whole lattice will move, like an iron cage sinking in water.

 

[Count Iblis]

Gravity affects the ions too. They don't move much relative to each other, but they do move. The whole lattice will move, like an iron cage sinking in water.

http://prola.aps.org/abstract/PR/v151/i4/p1067_1"

 

[Phlogistonian]

Hmm...

Gravitation-Induced Electric Field near a Metal
L. I. Schiff and M. V. Barnhill

Abstract
A quantum-mechanical formalism is developed to calculate the electric field produced in the vicinity of a metallic object through the influence of the Earth's gravitation. The field is proportional to the gradient of the ground-state energy eigenvalue of the object with respect to the position of a test charge located at the field point. This expression can be reduced to the solution of a problem in classical electrostatics, and is valid as well for a superconductor. Simple explicit results are obtained for the field within a closed metallic shell of arbitrary shape, and outside of a metallic sphere. In the former case, the field is uniform and equal to mg / e, directed so as to exert an upward force on an electron; m and e are the electron mass and charge, and g is the acceleration of gravity. This result is of importance in connection with current experiments on the free fall of electrons and positrons, and leads to the expectation that shielded electrons will not fall, while shielded positrons will fall with acceleration 2g. Some comments are made on the gravitation-induced electric field near a nonconductor, and on the field near a rapidly rotating solid.

I'll think about that.

 

[deldotbee]

Hey, Count, did you guys ever resolve this question or find any papers related to its experimental measurement?

-Spence