Point electron

[Enthalpy]

Hello everybody!

 

You know that up to now, I considered the idea of particle useful to acount for properties like the charge, the spin... which don't split when a particle gets arbitrarily confined. As we have no means to confine a particle to a point, I felt a point particle unnecessary. This has evolved.

 

I still believe that the interaction of a photon and an electron wouldn't need points and occurs over a significant volume: the volume of the orbitals if an atom or molecule absorbs a photon, thousand atoms volume or more if a semiconductor absorbs a photon... There, the photon's size adapts to the smaller (better localized) electron, the interaction is diffuse, and I feel that diffuse electrons and photons would be fine.

 

An electron attracted by an atom's nucleus could be diffuse. We could even pretend that its charge spreads like q*|Ψ|² provided its mass spreads like m*|Ψ|² as well; this would keep Schrödinger's equation and its solutions. Better: the electron's electrostatic self-energy would be finite, saving the need for dressed particles, and with this energy varying like the kinetic energy does, I believe it could be tinkered into the electron's mass - so my ramblings... And anyway, when radiating light (wavelength >> orbital size) or not, an electron around a nucleus behaves like a charge spread as q*|Ψ|².

 

Two electrons (or electron pairs) interacting could still be diffuse. At an atomic force microscope, they interact as complete orbitals.

 

What does need a point electron is the effect a spread charge would have on the orbital's shape. If the electron could repel itself, it would kind of screen itself from the nucleus, so that the outer part of the orbital would extend far more than it does. This would most probably give a spectrum that differs from the observed and accurately predicted ones, and for sure, van der Waals' forces completely different from the observations.

 

So to avoid the auto-screening, it needs all the electron's charge at one point (or at least, much more concentrated than the size of the orbital) to write Schrödinger's equation. As a consequence, the mass must be concentrated as well.

 

----------

 

Or better, we need the charge and mass to be concentrated when considering the action of the electron on itself. For the interaction with other particles, we need an electron capable of concentrating as much as the other particle requires (... until I change my mind), but it acts over its full extension when the other particle spreads more.

 

This holds for the electron because of its charge. Would you know more reasons that would apply to photons, which can usually superimpose without interaction? To neutral fermions? To all bosons?

 

Your thoughts please?

[Sensei]

See the way electrons are created in pair production:

 

y + y -> e+ + e-

Each photon with E >= 510999 eV, which is wavelength equal to Compton wavelength = 2.426*10^-12 m = 2.426 pico meters.

 

or with single photon and nucleus:

 

y -> e+ + e-

E = 1.022 MeV, wavelength half of Compton wavelength.

After pair production we will have electron with half energy of such photon, and positron with another half of that initial energy.

 

Electron particle "bigger" than this wavelength, wouldn't make sense to me..

 

Another argument for small size of electron is positronium:

http://en.wikipedia.org/wiki/Positronium

Electron and positron are orbiting together making exotic atom.

 

You're thinking too much (and overestimating) orbitals.
While you should concentrate on free electrons, not bound to any atom IMHO.

See electron's trace which it's making in Cloud Chamber:

Electron had to be in the center of that trace. Electron accelerated to significant speed of light, with tremendous amount of kinetic energy (see decay energy calculations to find out it's possible maximum kinetic energy for given unstable isotope (that nucleus had mass-energy concentrated in couple femtometers *)).
Passing through medium, it gave part of its kinetic energy to medium, ionized it, and accelerated atom (or its electrons) of medium hit other medium atom - cascade effect happened (and repeat it along path for billions of billions of medium particles). After giving all its kinetic energy, it's stopping moving.
What happened at quantum, picometer or femtometer scale, is therefor visible in our scale by naked eye.
It would be worth to record it in slow motion 1000+ FPS camera, to see how trace is growing with time.

*) According to http://en.wikipedia.org/wiki/Femtometre "radius of a proton is approximately 0.84 - 0.87 femtometres"
Now imagine proton (in nucleus) decaying via beta decay+, to positron
p+ -> n0 + e+ + Ve
or neutron decaying via beta decay-, to electron
n0 -> p+ + e- + Ve
IMHO electron radius should be even smaller than nucleus that decayed (and while before decay was "containing" our electron/positron). Edited September 7, 2014 by Sensei

[Enthalpy]

Waves have no definite size (nor shape). In the case of electron pairs and positronium (which has the size of an atom), the size of the wave would fit perfectly. Just like a photon of 1µm wavelength and square meters or square light-years spread interacts with a 100pm atom.

 

A trace in a cloud chamber is wide, hugely wider than an atom. It doesn't need an electron smaller than an orbital.

 

Nor does electron emission or capture by a nucleus need an electron smaller than the nucleus. It needs the ability for an electron to concentrate to this size; this is the probability that we compute with |Ψ|², and call it "collapse" when the wave shrinks to fit the interacting particle.

 

This ability to change the size wouldn't alone need the particle to be a point. Every interaction is computed over the common volume of both waves, that is, the ability of the bigger wave to shrink to the size of the smaller wave suffices.

 

More: when a photon is absorbed by a semiconductor, it does not act as a point. It has to act on an electron that spans over >1000 atoms before and after the absorption. It's even worse for X-ray diffraction, where the photon must interact with a crystal and not with one atom.

 

A different formulation would be whether we need something smaller than the wave to represent the particle, and my answer is yes for the particle's electric self-energy, and apparently no for the interactions.

[Sensei]

I thought you wanted free discussion ("Your thoughts please?"), but unfortunately it seems I was wrong..

 

[Sensei]

In the case of electron pairs and positronium (which has the size of an atom), the size of the wave would fit perfectly.

But it wouldn't fit to idea of annihilation of electron and positron that happens later.

 

To do so, there is needed that they're more or less point particles, that meet together and finally annihilate.

Until they don't annihilate, they're existing as e+ and e-.

If you can keep them separate (like in magnetic trap), they won't annihilate.

 

When electron accelerated to relativistic velocity collides with positron, there are created other particles.

 

"There are also many examples of conversion of relativistic kinetic energy into rest energy. In 1974, SLAC National Accelerator Laboratory accelerated electrons and positrons up to relativistic velocities, so that their relativistic energy \gamma mc^{2} (i.e. the sum of their rest energy and kinetic energy) is significantly increased to about 1500 MeV each. When those particles collide, other particles such as the J/? meson of rest energy of about 3000 MeV were produced.[30] Much higher energies were employed at the Large ElectronPositron Collider in 1989, where electrons and positrons were accelerated up to 45 GeV each, in order to produce W and Z bosons of rest energies between 80 and 91 GeV. Later, the energies were considerably increased to 200 GeV to generate pairs of W bosons"

 

Quote from http://en.wikipedia.org/wiki/Tests_of_relativistic_energy_and_momentum

 

A trace in a cloud chamber is wide, hugely wider than an atom.

It's result of cascade ionization I mentioned before.

 

Similar ionization happens when charged particle is passing through liquid Hydrogen in f.e. Bubble Chamber.

 

But that's not the point. If electron would be as "wide" as orbital of let's say Hydrogen, it's path in chamber should looks like zig-zag constantly changing directions after hitting nucleus of medium.

 

Alpha particle has a few femtometers and is occasionally hitting medium nucleus and changing directions, like here:

http://www.nuffieldfoundation.org/practical-physics/alpha-particle-tracks-including-collision-helium-nucleus

 

It doesn't need an electron smaller than an orbital.

Free electron is not limited to any orbital...

 

One might ask, which atom orbital, and with what quantity of electrons.

If you have f.e. Gold nucleus with 79 protons, and 1 electron,

ionization energy is approximately 13.6*Z^2 = 13.6*79^2 = 84.8776 keV

 

In other words you must send photon with ~85 keV to remove the last electron from ion ₇₉Au⁷⁸⁺

 

Such molecule is "Hydrogen-like", all charge concentrated in nucleus (+79e), and single electron outside (-1e).

 

Nor does electron emission or capture by a nucleus need an electron smaller than the nucleus. It needs the ability for an electron to concentrate to this size; this is the probability that we compute with |Ψ|², and call it "collapse" when the wave shrinks to fit the interacting particle.

Simply: finding where was our point particle..

 

 

This ability to change the size wouldn't alone need the particle to be a point.

You completely forgot that whole thread is your own hypothesis/speculation that electron is point particle, and now you're defending opposite...

Edited September 8, 2014 by Sensei

[Enthalpy]

An electron-positron pair can annihilate precisely because they overlap. This is the fundamental requirement of any interaction. Annihilation doesn't require small dimensions nor points. Think of two photons which have a locally destructive interference: they do that over some extension, in some experiments wide enough to be seen with the naked eye. No need for points there.

 

Yes, particles are created. This may start in a small volume, for instance a photon created by an atom is initially as small as the atom, about 10,000 times smaller than the wavelength, and the wave extension is even much bigger. But why should this need a point photon? It needs the capability for the photon to be small, as any wave can.

 

More generally, an interaction extends as much as the overlapping of both particles, that is often the smaller particle. This provides no means to tell if a particle is a point. It does tell - and this is a historical reason for QM - that an interaction can occur in a volume much smaller than a particle has been, with some properties of this particle fully available in the small volume, albeit with a saller probability for a small volume.

 

You write: "finding where the point particle was", and, no, an interaction does not reduce the position uncertainty to a point. "Collapse" is often misinterpreted, I'd prefer "reduction" or "adaptation".

 

----------

 

A beta electron is hugely wider than an atom. After hitting an atom and ejecting an electron from there, the beta continues its path from there. If the collision is at small parameter impact, the beta deviates, but this is scarce because the cross section is much smaller than an atom.

 

Simplified: at one hydogen radius, the electrostatic energy is two Ryberg or 26eV. When a collision implies 260keV the distance between both electrons is 1/10,000 of an atom radius, so such a collision happens every 100,000,000 atoms. Most encounters hence happen at a bigger impact parameter, so they imply a much smaller energy, and the deflection of the beta is much smaller - the trajectory is about straight. As the ionizing particle loses energy, its path makes more zigzag since both (a) efficient collisions have a bigger section (b) the particle have less momentum - and this is observed.

 

This is not a reason for a point particle. It does need that some interactions be small, here 1/10,000 of an atom radius for electrons and 260keV, but this is still an extension and not a point. Far better: if an interaction could happen in a point instead of a volume, the particles would leave in all directions as a result. The very relationship between the energy and the angle after a collision tells that every collision has a volume. This is necessary to limit the diffraction.

 

By the way, the scarce efficient collisions, which imply a big momentum change, don't localize both particles more than the smallest was before. Only their relative positions are known if we observe the deflection, but the position of the collision is as uncertain as before - for instance "within the atom" for a target electron and "much larger that that for a beta, resulting in "somewhere in the atom" for the collision. Entanglement, yes. Only some subsequent event, say one collision with a betetr localised nucleus, would refine the position of both particles.

 

----------

 

You put "an alpha has a few fm" and this is only the extension of its constituents. The delocalization of the alpha is huge: it's emitted in any direction before some interaction reduces the uncertainty, for instance when the alpha meets a first atom, or the daughter atom hits an other one. These are different dimensions. Yet an other dimension would be the collision cross-section (sqrt as you wish) with, for instance, a helium nucleus. Still an other, the distance at which the electric field can create virtual electron-positron pairs.

 

The only difference with an electron is that the alpha, the proton, a nucleus, an atom... are composite particles, where the constituents define this extra radius, area or volume where they're located. This extra volume can be smaller or bigger than the delocalisation - extremely delocalised atoms in a Bose-Einstein condensate for instance. Is this any reason for a point electron?

 

I've seen no reason in interactions for a point electron. But in the lack of "self-interaction", yes: the electron doesn't screen itself against other particles.

Edited September 8, 2014 by Enthalpy

[swansont]

Whenever people do experiments that explicitly rely on the electron having a size (or not), they get a very small number, consistent with it being a point particle. If you put an electron in a Penning trap, its size will affect how it interacts. This shows up in g-factor measurements; you can get an upper limit if you assume all of the error is due to the electron size

 

http://gabrielse.physics.harvard.edu/gabrielse/overviews/ElectronSubstructure/ElectronSubstructure.html

http://iopscience.iop.org/1402-4896/1988/T22/016/

[Enthalpy]

Interactions as well demand point electrons, be it between atomic orbitals, in a collision... Changed my mind.

 

In a model where the electrons would be diffuse, with a charge and mass density both multiplied by |Ψ|², the electrostatic interaction energy around a pair of possible positions would be multiplied by |Ψ₁|²*|Ψ₂|² but the mass only by |Ψ₁|² or |Ψ₂|², which would lead to a different hence wrong solution.

 

So diffuse electrons are no alternative to the standard formulation: electrons spread according to Ψ, but around each possible position we shall compute with the complete charge and mass there.

 

This is not patent for electrons that emit or absorb visible light in atoms, I guess because such photons have a wavelength much larger than the orbitals, so that only the electron behaviour averaged over the orbital influences the emission or absorption.

 

----------

 

The argument above does not apply as is to photons.

 

----------

 

Thanks to Swansont for the papers about Landé factor versus the size of a hypothetic composite electron. It puts strong limits on the distance between the hypothetic constituents of an electron.