Talk:Classical electron radius

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Link[edit]

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

Good article[edit]

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[edit]

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:
and for the energy of a sphere of radius R containing a constant surface density I get:
I have changed the text to be more exact. PAR 14:58, 2 November 2006 (UTC)

electron radius[edit]

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

Counter-intuitively, it gets smaller. See Discussion section in Article.

question about derivation[edit]

By request, I replaced the original derivation with one based on energy-conservation. Both approaches yield the same answer, but the derivation based on energy-conservation seems easier to follow, and answers many of the questions raised. The discussion below relates to the original derivation. PBWilson, 10-3-2016 — Preceding unsigned comment added by 184.98.190.165 (talk) 23:44, 3 October 2016 (UTC)

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
The energy needed to do that is
So, to build a shell of thickness dr with the same density on that ball requires bringing a charge 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
To build a ball from scratch, you integrate from 0 to R, the radius of the ball, remembering that ρ is constant, you get
replacing ρ with the charge divided by the volume () gives

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[edit]

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)

The mass of the electron in the theory of classical radius is all in the field on the outside of the sphere. The total mass inside the sphere must be zero. — Preceding unsigned comment added by 2003:C9:8BDE:3BF8:4AE3:6DDC:2CE9:A3A4 (talk) 17:31, 22 December 2017 (UTC)

Clarity[edit]

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)

Plain English. No one actually knows, because no one is sure if the electron has internal structure or not, as is the case with all elementary particles. Frankly I find it quite odd that a proton has internal structure which is the cause of it's permanent electric and magnetic dipole moment and charge, while we consider an electron not to have internal structure, yet still posses a permanent electric and magnetic dipole moment and charge. When both electric and magnetic dipole moments are quite impossible without a system of charges moving internally in relation to one another. "In physics, the electric dipole moment is a measure of the separation of positive and negative electrical charges in a system of electric charges, that is, a measure of the charge system's overall polarity." See Electric_dipole_moment. "The electron is a charged particle of charge (−1e), where e is the unit of elementary charge. Its angular momentum comes from two types of rotation: spin and orbital motion. From classical electrodynamics, a rotating electrically charged body creates a magnetic dipole with magnetic poles of equal magnitude but opposite polarity." See Electron_magnetic_dipole_moment

So, if it has a uniform electric charge, then the rotation of that uniform charge cannot create two oppositely charged poles. Magnetic monopoles do not exist. It takes a minimum of two charges to create a magnetic dipole, two opposite charges or in reality two charges of differing potential. It's the only known laboratory tested way. "The sources of magnetic moments in materials can be represented by poles in analogy to electrostatics. Consider a bar magnet which has magnetic poles of equal magnitude but opposite polarity. Each pole is the source of magnetic force which weakens with distance. Since magnetic poles always come in pairs, their forces partially cancel each other because while one pole pulls, the other repels. This cancellation is greatest when the poles are close to each other i.e. when the bar magnet is short. The magnetic force produced by a bar magnet, at a given point in space, therefore depends on two factors: the strength p of its poles (the magnetic pole strength), and the vector ℓ separating them." See Magnetic_moment

In other words it must have internal structure composed of both negative and positive charges, or more accurately, charges of higher and lower potential.84.226.185.221 (talk) 11:00, 12 October 2015 (UTC)

Planck's Constant Conflict[edit]

The very final point in this article states " is Planck's Constant", isn't the reduced plancks constant (i.e h/2pi)? Anonymous capybara (talk) 13:38, 2 December 2013 (UTC)

Classical radius of proton[edit]

The article Thompson scattering says that a quantity called classical radius of a charged particle is convenient to be defined. Can also proton have a classical radius defined for for it?--5.2.200.163 (talk) 13:49, 12 February 2016 (UTC)

This should be made much more simple...[edit]

Introducing the classical electron radius in terms of the interaction of positrons and electrons is needlessly obscure and historically inaccurate. The standard, straightforward discussion of where it comes from is in terms of the electrostatic self-energy of a charged sphere. No need to mention the virial theorem, radiative energy loss, and other complexities. Keep it as simple as reasonably possible.