# Physical size of an electron

1. Dec 7, 2011

### rorix_bw

Hi guys. Sorry if this is in the wrong topic, but what is the physical size of an electron? I understand it can sometimes behave as a wave. But when it is a particle, just how [STRIKE]big[/STRIKE] small is that particle?

2. Dec 7, 2011

3. Dec 7, 2011

### Bill_K

The size of an electron is zero. Period. Quantum field theory describes particles that are pointlike. All experimental evidence confirms this up to the highest collider energy, an order of magnitude 1 TeV, corresponding to a distance of 10-17 cm. All elementary particles, to the best of our knowledge, are pointlike. Nucleons, of course have internal structure, and have a size of the order of a fermi.

When we say a particle "behaves like a wave," we are talking about a wavefunction that gives the probability of finding the pointlike particle at a particular location.

4. Dec 7, 2011

### chrisbaird

Electron wavefunctions tend to take the size of their container. If you place an electron in a quantum well, it will tend to spread out to fill the well. If you bind it to an atom, it takes generally ends up the size of the atom (I know there's a lot more to it than this, but I'm trying to keep it simple). An electron is always partially wave-like and partially particle-like at the same time. It is sometimes more particle-like than at other times. So if you ask what is the radius of the electron when it is acting most like a particle, I would say zero. It acts like a point particle. If you want to pretend an electron is a classical ball of charge, you can come up with the radius 3x10-15 m.

5. Dec 7, 2011

### cmb

I think you should un-period that.

Sure, you can work with models in which the electron is defined without dimension, but this is not the whole story.

One line of research looking at testing the standard model is to determine if the electron is spherical. Clearly, it needs to have a size to be spherical!

http://www3.imperial.ac.uk/newsandeventspggrp/imperialcollege/newssummary/news_26-5-2011-8-58-6

6. Dec 7, 2011

### humanino

This is a very misleading article, shame on the university media responsible for this, and you bringing it up in this conversation may want to take the original article. You can measure something which can only be non-zero if there is a non-spherical finite shape, yet which would be zero if there was a spherical shape, which includes as a degenerate limit the zero size (trivial spherical symmetry). This is what they get : they set up an upper bound on the electric dipole moment. Their result is consistent with zero and does not imply a finite size. It is in fact consistent with zero size (point-like).

Last edited: Dec 7, 2011
7. Dec 7, 2011

### rorix_bw

I ask because Neil Degrasse Tyson (http://en.wikipedia.org/wiki/Neil_deGrasse_Tyson)
recently said:

Q) What never fails to blow your mind in physics?
A) The fact that an electron has no known size — it’s smaller than the smallest measurement we have ever made of anything.

I wasn't sure if he means it has zero size (point particle?) or if just fantastically small size.

8. Dec 7, 2011

### kurros

He means it is so small that its size is experimentally indistinguishable from zero. We of course cannot ever say that it is exactly zero from an experimental perspective. There could always be some ridiculously tiny string structure or something.

9. Dec 7, 2011

### cmb

So are you asking that I take your word for it over taking Imperial College's article at face value? Why would I do that? Do you have references available [public access and peer reviewed] to back up a contradiction of the article?

Are you going to say 'shame on Nature' for publishing their work, too? http://www.nature.com/nature/journal/v473/n7348/full/nature10104.html

Last edited by a moderator: May 5, 2017
10. Dec 7, 2011

### Bill_K

Yes, the wavefunction but not the electron itself, which always remains pointlike. Normal Schrodinger quantum mechanics concerns the motion of a point particle. A quantum object that had a finite size would also have additional degrees of freedom: rotation and vibration. Which are needed for example to describe nuclei. An electron has angular momentum but does not exhibit rigid rotator excitations, as it would if it was finite in size.
The experiment in question was designed to detect if possible an electron's electric dipole moment. The popular idea circulated that this meant "nonspherical", but it was just a very misleading journalist's misunderstanding. A dipole moment says nothing about size. The electron has a magnetic dipole moment and yet remains pointlike.
This is known as the "classical electron radius." It was brought up as part of the 1904 Abraham-Lorentz theory that tried to describe an electron as a charged sphere. When quantum mechanics came along the Abraham-Lorentz model had to be discarded. In addition to being inconsistent with quantum mechanics, it was inconsistent with relativity.

11. Dec 7, 2011

### cmb

Well, I read what they put in the abstract in a different way:

If they had meant to say it is the electron dipole moment that is aspheric, then shouldn't they'd have simply said 'The electron dipole moment is predicted to be slightly aspheric.'?

If the EDM is described as a characterisation of the asphericity of an electron, then why would I assume the EDM and the electron shape are the same thing?

A characterisation of a thing is not the thing!

12. Dec 7, 2011

### humanino

This is a public relation advertising. There is not even mention of "electric dipole moment" which what is measured ! I was reading this very day another article where the author complains about public media University services :
String and M-theory: answering the critics
I do not necessarily recommend the entire reading, but if you take a look at the second FAQ
This is M. J. Duff, from the Imperial College in London. Take his word if you do not believe me that this kind of occurrence is common practice.

The Nature article is very good. This is the article I was referring to when I said you should have read it.

So again : they find that the electron respects spherical symmetry, which is consistent with zero size.

Last edited: Dec 7, 2011
13. Dec 8, 2011

### cmb

See my post #11.

14. Dec 8, 2011

### humanino

Oh I understand now. Initially you thought they measured "the electron", instead of its electric dipole moment...

15. Dec 8, 2011

### cmb

I read that to mean they are measuring the EDM as a proxy for the electron shape. Like measuring the shape of the earth's atmosphere to get an idea of the shape of the earth.

Isn't this what it means?

16. Dec 8, 2011

### humanino

The electric charge and magnetic current distributions are encoded in the electron-photon vertex form factors. Up to linear combinations $^{(1)}$, form factors are Fourier transforms of those distributions. To calculate the electromagnetic current carried by a spin 1/2 particle, you sandwich between the initial and final spinors an operator with a vector index $\Gamma^\mu$, which can be in all generality (Lorentz invariance, current conservation and hermiticity) written as :
$$\Gamma^\mu=\gamma^\mu F_1(q^2) + \frac{\sigma^{\mu\nu} q_{\nu}}{2m}\left[i F_2(q^2) - \gamma^5 F_3(q^2)\right]$$

where $q_\nu$ is the 4-vector difference between initial and final momenta, and is the momentum of the exchanged (virtual) photon (in the one photon exchange approximation) ; the $\gamma$s are the usual Dirac matrices ; the $\sigma$s are i/2 times the commutator of the $\gamma$s (usual spin operator).

Those form factors are measurable directly in elastic scattering. One can show using states closed to rest (non-relativistic)

$e F_1(0) =$ electric charge

$\frac{e}{2m} \left[ F_1(0)+F_2(0) \right] =$ magnetic moment (c.f. magnetic coupling $-\mu \, \vec{\sigma}\cdot\vec{B}$)

$\frac{e}{2m} F_3(0) =$ electric dipole moment (c.f. electric dipole coupling $-d\,\vec{\sigma}\cdot\vec{E}$)

For the electron, up to quantum corrections (loop), $F_1=1$, $F_2=0$ and $F_3=0$. That means it acts as pointlike electric charge down to the smallest scales we could probe (the magnetic current distribution vanishes, and the electric charge distribution is the Fourier transform of a constant, which is a "delta function"). That also means we went at great length to recover the usual electron-photon vertex $\Gamma^\mu=\gamma^\mu$

Note
(1) Which linear combination depends whether one uses the rest frame or an infinite momentum frame. There are technical advantages to using infinite momentum frames, in which case the Dirac and Pauli form factors $F_1$ and $F_2$ are directly Fourier transform of the electric charge and magnetic current distributions. In the rest frame, one needs linear combinations which are usually called Sachs form factors. The Sachs form factors have the advantage to separate neatly in the final cross-section.

source
Mostly Itzikson & Zuber

Last edited: Dec 8, 2011
17. Dec 9, 2011

### A. Neumaier

Being point-like doesn't imply being a point.

(emphasis as in the original; see p.2 of the book
O. Steinmann, Perturbative quantum electrodynamics and axiomatic field theory, Springer, Berlin 2000)

(beginning of Section 3.9 of
G. Scharf, Finite Quantum Electrodynamics: The Causal Approach, 2nd ed., Springer, New York 1995.)

In the book
S. Weinberg,The quantum theory of fields, Vol. I,Cambridge University Press, 1995,
Weinberg defines and explicitly computes in (11.3.33) a formula for the charge radius of a physical electron, though he does not carry the computation far enough to get a numerical value.

18. Dec 9, 2011

### cmb

That's waaaay over my head. I'm guessing next you'll be telling me I have to read 20 books and study for 15 years to understand it.

Bottom line is this - I do not accept that an infinite energy density can exist in a stable state. It is beyond my comprehension that all the energy of an electron can exist in a space of zero size.

Now, if the energy of the electron exists probabilistically in a space around a 'location', and that, actually, that is all there is to an electron and there is no physical structure to it, then, OK, that begins to look like a comprehensible answer. So its size could then be defined as, for example, 99.9999% likely to be within a radius of X from a location. I don't know if that description is reasonable or not, because so far all I have read here is that it is 'point-like' which makes no sense to me. Are there other 'point-like' particles, and if so then why do their properties differ if they have no physical shape or structure that 'causes' the energy around that location to be as it is?

Can someone please provide an 'Einstein' answer that a 6 year old would understand?

19. Dec 9, 2011

### chrisbaird

All elementary particles are point particles, otherwise they wouldn't be elementary. From the wikipedia article on elementary particles:

"In particle physics, an elementary particle or fundamental particle is a particle not known to have substructure; that is, it is not known to be made up of smaller particles... For mathematical purposes, elementary particles are normally treated as point particles, although some particle theories such as string theory posit a physical dimension."

20. Dec 9, 2011

### cmb

I do not see how indivisibility implies an object has zero dimension.

A description devised 'For mathematical purposes..' means nothing, in the context of the question.