An antigravity device for electrons

There is a very simple way to nullify the effects of the Earth's gravity on electrons, without using any external power source. Simply put the electrons inside a long, thin, vertical metal tube which you have fixed in position. While inside the tube the electrons will not be accelerated at the usual rate

g = - 9.8 m/s2

(minus sign for downwards!), but will float about at near constant velocity (until, of course, they bump into something or fly out of the tube). It helps if you evacuate the tube, since otherwise the electrons will just collide with a molecule in the air very quickly and not float about at all.

Well I could stop here, having added the references

"Gravitation-Induced Electric Field near a Metal", L. I. Schiff and M. V. Barnhill, Phys. Rev. 151 p. 1067 (1966)
(work supported in part by the USAF)

and

"Experimental Comparison of the Gravitational Force on Freely Falling Electrons and Metallic Electrons", F. C. Witteborn and W. M. Fairbank, Phys. Rev. Lett. 19 p. 1049 (1967)
(work supported by NASA)

but you'd be asking "What's this bullshit about"? So, let me explain.

In fact, gravity is still exerting a force on the electrons fg = g me. However, inside the bulk of the tube and (thanks to Poisson's equation) in the space enclosed, there is a (nearly) uniform electric field of magnitude |E| = |g me/e| directed downwards. Remembering that the charge on the electron e is negative, the electric force is fe = - g me, thus the total force, and acceleration, vanish to a very good approximation. This was experimentally tested in the second reference. Conversely, a positron in the vicinity of the tube will suffer an acceleration of 2g downwards -- falling twice as fast as usual.

Now you may be saying "Fine, but you just pulled that electric field out of a hat! We learnt in freshman electrostatics that the field inside a conductor vanishes -- method of images and all that -- and here you are telling us that a bit of metal sitting at rest automatically generates a non-zero field?" And this would be a good question.

First, I hope you won't be shocked to learn that physics is a bit more complicated than what gets taught to 18-year-olds. But of course, condescension is not an answer. You have to think about the internal structure of a metal. There are positively charged ions which vibrate gently about the vertices of the crystal lattice and there are conduction electrons that whizz about unimpeded and give the metal its exciting electrical properties.

Now consider the situation where a bit of metal is fixed in place within a gravitational field. The ions are, on average, held stationary by their mutual repulsion. As the conduction electrons fly about, they also feel the force of gravity, so they begin to accelerate downwards on average. They also feel lots of other forces, of course, but to begin with these average out to zero. Very soon the conduction electrons occupy a position somewhat lower than the rest of the metal. Charge separation and polarization have taken place due to gravity! There is now a net negative charge at the bottom of the piece of metal and a net positive at the top. Hence, just like inside a capacitor, there's got to be an electric field in between. In equilibrium, the size of this field is of course exactly that needed to stop the conduction electrons from drooping any farther, viz. g me/e. The authors of the first paper of course had a much more rigorous and mathematical derivation, starting from first principles, but this is what it boils down to.

Now this is the field inside the metal; but assuming that E vanishes at infinity, we can solve the wave equation to find the field outside. For an enclosed space surrounded by metal, the solution is just a constant field, the same as the field inside the metal. Inside a sufficiently long, thin tube, it's almost exactly constant except near the ends of the tube.