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A colleague asked me an interesting question: If you have a copper bar, let's say 1 meter long, and stand it upright: Is there an electrical potential difference between top and bottom due to gravity?

My gut feeling is "yes", but you would have trouble measuring it, because when you attach multimeter leads to the top and bottom, they would be subject to the same effect, and your measurement is 0.

You could do the experiment horizontally: Accelerate the bar with $10\,\text{m}/\text{s}^2$ to simulate gravity (centrifuge, maybe?), and measure with a stationary multimeter.

Would the effect depend on the material, so that you could build something like a thermocouple?

I'm having trouble estimating the size of the effect. A back-of-the-envelope calculation gives me $50\,\text{pV}$, but I'm not sure if my approach is correct ($ q U = m g h $).

Qmechanic
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Martin J.H.
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3 Answers3

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As the electrons in the conductor are free, they will experience acceleration due to the force of gravity. This force, will cause the electrons to move downwards, this means that the new charge distribution will be one that causes a net electrical emf.

The electrons will continue to move until the electrostatic force created by the new distribution exactly cancels out the gravitational force and then the electrons will stop.

In this new steady state there are a net zero force on all particles and hence a net emf of zero (assuming a uniform electrical field which is a sensible assumption).

This means that $$F_{E} + F_{g} = 0$$

$$-eE + mg = 0$$

$$eE=mg$$

$$eEd=mgd$$

$$Ed = \frac{mgd}{e}$$

Since a uniform electric field.

$$|\epsilon_{E}| = \frac{mgd}{e}$$

This means that if a voltmeter measures purely electrical fields and not gravitational ones, there will be an emf. This will however not cause a current

jensen paull
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The gravitation-induced electric field in and near a metal has been extensively studied by researchers trying to measure the gravitation force on positrons and antiprotons.

In their original 1966 paper on "Gravitation-Induced Electric Field near a Metal", Schiff and Barnhill concluded that the gravitation sagging of electrons inside a metal lattice produced an electric field

$$E_\mathrm{SB} \sim \frac{m_e g}{e} \sim 5 \times 10^{−11}\,\mathrm{V/m}$$

such that:

free electrons are not expected to fall under gravity if they are within a closed metallic shell of arbitrary shape, since their weight is exactly balanced by the gravitation-induced electric field produced by the metal. In similar fashion, free positrons, should, fall with acceleration $2g$

In their 1968 paper on "Gravitationally Induced Electric Fields in Conductors", however, Dessler, Michel, Rorschach, and Trammell estimated that when lattice compressibility is taken into account, the induced electric field is actually

$$E_\mathrm{DMRT} \sim \frac{M g}{e} \sim 5 \times 10^{−7}\,\mathrm{V/m\;(for\,copper)}$$

where $M$ is the mass of a metal atom. This field is in the opposite direction from Schiff and Barnhill, so positrons would be accelerated up and electrons accelerated down.

This very large induced field (along with other issues) makes direct measurement of the gravitational acceleration of individual electrons and positrons effectively impossible. This is why current efforts to directly measure the effect of gravity on anti-matter focus on anti-protons and anti-hydrogen.

David Bailey
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I think g is too small for the small mass of e. But there is an experiment: you turn a conducting disk very fast, and can measure the voltage between midpoint and circumference. I am sorry I forgot the name of the experiment. So yes there will be a very small voltage, but then the E feld is against the gravitation .

trula
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