Talk:Classical electron radius

Link

Please note http://atdotde.blogspot.com/2005/09/natural-scales.html --Pjacobi 20:33, 15 October 2005 (UTC)

Good article

This is a smashing good little article. Clear, consice and to the point. Good work all! Kevmitch 04:11, 19 April 2006 (UTC)

Exact energy of charged sphere

Isn't the formula given the energy required to assemble a charged sphere with all the charge on the surface? If it's constant charge density, I believe a prefixed factor of 3/5 is needed.

Actually the correction factor of 3/5's is for a charged sphere with all the charge uniformly distributed over the volume, not over the surface.

So for a uniform volume distribution you should get

U = 3 / 20 * whatever

and for a surface charge distribution you get (if I remember correctly):

U = 1 / 8 * whatever.

-- Eugene Y.

Thats where "approximately" comes in. For the energy E of a sphere of radius R containing a constant charge density I get:

$E={\frac {3}{5}}\,{\frac {e^{2}}{4\pi \epsilon _{0}R}}$

and for the energy of a sphere of radius R containing a constant surface density I get:

$E={\frac {1}{2}}\,{\frac {e^{2}}{4\pi \epsilon _{0}R}}$

I have changed the text to be more exact. PAR 14:58, 2 November 2006 (UTC)

electron radius

if the electron travels fast enough to affect its mass, would the radius of the electron be smaller or larger???

Well, assuming it is a sphere when at rest, its diameter in the direction of travel would be smaller due to the Lorentz contraction while its radius in the directions perpendicular would be unaffected. It would be an oblate spheroid. PAR 02:06, 16 November 2006 (UTC)

question about derivation

Thanks very much for the article, but I was wondering how you derived the constants in front of the expressions (1/2 or 3/5)? I have tried integrating 1/2*[epsilon]*[E field]^2 dr three times, and get the basic term but not the constant. Nor can I find this on any other website (they all seem to ignore the constant). Please could you add the derivation to the article? I'd really appreciate it - thanks for your help! P.S. why does everyone, even this article, ignore the constants when quoting the radius? Thinkingabout physics (talk) 21:44, 16 April 2008 (UTC)

I don't know where your expression 1/2*[epsilon]*[E field]^2 dr came from, but the way I did it was to realize that bringing in an infinitesimal charge dq from infinity to the surface of a ball of radius r having constant charge density ρ requires the same amount of energy as bringing dq from infinity to a distance r from a point charge q where

$q={\frac {4}{3}}\pi r^{3}\rho$

The energy needed to do that is

$dU={\frac {1}{4\pi \epsilon _{0}}}qdq$

So, to build a shell of thickness dr with the same density on that ball requires bringing a charge $dq=4\pi r^{2}\rho \,dr$ from infinity. (This assumes that the charge on the shell is so small that interactions between charges on the shell add negligible energy). That needs energy

$dU={\frac {1}{4\pi \epsilon _{0}}}\,\,\left({\frac {4}{3}}\pi r^{3}\rho \right)\left(4\pi r^{2}\rho \,dr\right)$

To build a ball from scratch, you integrate from 0 to R, the radius of the ball, remembering that ρ is constant, you get

$U={\frac {1}{4\pi \epsilon _{0}}}{\frac {(4\pi \rho )^{2}R^{5}}{15}}$

replacing ρ with the charge divided by the volume ($4\pi R^{3}/3$) gives

$U={\frac {3}{5}}\,{\frac {1}{4\pi \epsilon _{0}}}{\frac {q^{2}}{R}}$

The reason people ignore the constants is because the classical derivation is not valid at the scale of the electron, its just a hand-waving argument, but it does yield a constant with dimensions of length which is relevant in quantum mechanics as well. The constant multiplier loses significance when going to quantum mechanics, but the function of fundamental constants does not. PLEASE NOTE - the above derivation needs to be checked! Yes, I know, its inconceivable that I would make a mistake, but it has been known to happen on rare occasions. PAR (talk) 15:50, 18 April 2008 (UTC)

I just did the dimensional analysis and you must have made a mistake. Two, in fact, which happened to compensate each other. First you forgot to do the integral in the expression for dU (= W_shell) and then you mistook the power of r before integration. 82.139.87.95 (talk) 07:59, 23 September 2009 (UTC)

Density

Given a (classical) sphere with a radius of 2.818×10⁻¹⁵ m, this yields a spherical volume of 9.378×10⁻⁴⁴ m³. Combined with a mass of 9.109×10⁻³¹ kg, this gives a matter density of 9.718×10⁹ g/cm³. This is about 30 times greater than the approximate density of an atomic nucleus of 3×10⁸ g/cm³. — Loadmaster (talk) 22:52, 9 March 2011 (UTC)

Clarity

I would estimate that the average non-math-savvy user will arrive at this page due to a search for the radius/diameter of an electron, and expect to --somewhere on the page-- see the answer in plain english. I would suggest providing it. SentientSystem (talk) 04:27, 22 September 2011 (UTC)