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I just did a back-of-the-envelope calculation, which surprised me. I calculated the difference in acceleration (due to repelling like-charges) experienced by two sides of an electron the size of the classical electron radius, when placed one angstrom from another electron. I used purely classical formulas:

$$F = m_{e}\Delta a ~=~ \frac{k_e q_e^2}{r^2} - \frac{k_e q_e^2}{(r+r_e)^2},$$

Where $m_e$, $q_e$ are the mass and charge of the electron, $r_e$ is the classical electron radius, and $\Delta a$ is the difference in acceleration (the tidal effect) between the two sides of the electron.

Using $r = One\,\, Angstrom = 10^{-10}m$, I get: $$\Delta a ~=~ 1.5 \times 10^{18} m/s^2.$$

In other words, assuming I didn't make a mistake, the electromagnetic tidal effect is enormous!

This brings up some questions:

  1. Would this effect be measurable if the electron were not point-like (or far smaller than $r_e$)?

  2. Can we prove a particle is nearly point-like by considering electromagnetic tidal effects like the above?

  3. Are these sorts of effects studied or considered at all in QFT?

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Is that really so enormous? It sure sounds big, but I assume it's a rather neglectable accerelation compared to those due to e.g. strong interaction in nuclei. –  leftaroundabout Mar 20 '12 at 12:09
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It has been clearly shown that the electron does not have a spacial extent as large as the classical radius. Indeed, at this time there is no experimental evidence that electrons have any spacial extent at all (down to about $10^{-18}\text{ m}$). –  dmckee Mar 20 '12 at 18:01

2 Answers 2

This is not a good way of determining if the electron is pointlike because you are using classical forces. If the electron is in the ground state, the force can't do anything to it, because it might not be able to mix the electron with th excited state.

It is analogous to a molecule. If you apply a force to a molecule where the work done over a molecular radius is much much less than the excitation energy of the molecule, the molecule will respond to the force, but it won't jiggle. It can't jiggle, because the jiggling requires energy equal to the splitting between the ground state and the excited state.

Any structure of the electron will show up as excited states of the electron, or electron dissociation. A while ago, people suspected that the muon and tau are evidence of electron sub-structure, but today, people just think this is a sign of something like E8 string theory.

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What you are discussing is elastic scattering of electrons on electrons.

Experimental high energy physics has been measuring the form of the particles through elastic scatterings: electrons on protons, muons on protons, pions on protons etc with all the possible combinations. That is how it was found that the proton has an extent and is not a point particle. The extent is parametrized by the electric and or magnetic form factor. That is what gives us the experimental size of a nucleus and also of a proton and generally molecules. (Form factors are similar in concept as getting the crystal structure of a crystal by observing x-ray scattering symmetries and using Fourier transforms.).

What @dmckee said in his comment is correct. The experimental limit for the size of an electron is way beyond any classical estimate of the size so your numbers cannot be correct.

The experimental limits on the decay of the electron also exclude its being composite.

The fact that the limits on the electron dipole moment are so stringent also indicates a "point" particle.

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But even if you plug the experimental limit into my above equations, you get too large a number. My question boils down to whether you use that as evidence that the electron actually must have a smaller radius than the current experimental bounds. In other words, do we have indirect experimental evidence of the electron size due to the fact that it does not tear apart in the presence of other nearby charges (no need for high/any CM energy here)? I guess we don't really know what holds the electron charge together in the first place, so that might be a problem. –  user1247 Apr 15 '12 at 19:12
    
I have not checked your numbers. If they are correct they have the same weight as the observation that an atom exists: i.e. that the electron does not fall into the nucleus by radiating off its energy continuously as classical calculations expect. This was the foundation of the Quantum Mechanical formulation of the microcosm physics; it became necessary when the Bohr model became necessary. It means that the classical mechanics formulae describing behavior of matter interactions do not hold in the microcosm.The existence of the positronium is proof that at such small distances QM reigns. –  anna v Apr 16 '12 at 4:23
    
continued: the positronium en.wikipedia.org/wiki/Positronium:small distances, small energies a bound state appears, not immediate annihilation, which shows that the interactions at that small distances is not classical. –  anna v Apr 16 '12 at 4:34
    
The positron is not attracting the electron in the way the classical theory expects, with the concomitant tidal effects you calculate if it has an extent. The extent does not enter in this view. –  anna v Apr 16 '12 at 4:40

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