Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Point-like particles, form factors

  1. Apr 7, 2010 #1
    I'm wondering how it has been determined that the electron is really a point-like particle. Searching for this topic, I came across a post by humanino, saying that one way we know this is that observations of the electron's "form factor" are consistent with what you'd see for a pointlike particle.

    I've never heard of form factors before. I googled the term and couldn't find any basic explanation of it. Could anyone help me to understand-
    1. what a form factor is
    2. In general, how we know that the electron is pointlike

    In particular, suppose I wanted to believe that instead of being a fundamentally pointlike thing whose position was often indefinite, the particle was always a continuously distributed entity that sometimes became localized into very narrow regions. what would be wrong with such an idea?
  2. jcsd
  3. Apr 8, 2010 #2
    I think I described it in detailed once. I will follow Povh, Rith, Scholz, Zetsche "Particle and nuclei" 5.2 "Rutherford cross-section"

    Suppose we scatter an electron off a heavy target. In the Born approximation, the ingoing and outgoing electron waves are plane
    [tex]\Psi_i = \frac{1}{\sqrt{V}}e^{i\vec{p}\vec{x}/\hbar}[/tex] and [tex]\Psi_f = \frac{1}{\sqrt{V}}e^{i\vec{p'}\vec{x}/\hbar}[/tex]
    The cross-section will be given in the lowest order (one photon approximation) by a phase-space factor times the amplitude given by the interaction Hamiltonian
    [tex]<\Psi_f|\mathcal{H_\text{int}}|\Psi_i>=\frac{e}{V}\int\text{d}^3x\, e^{-i\vec{p'}\vec{x}/\hbar}V(\vec{x})e^{i\vec{p}\vec{x}/\hbar}[/tex]
    Next you need to massage this a little bit to
    [tex]<\Psi_f|\mathcal{H_\text{int}}|\Psi_i>=\frac{-e\hbar^2}{V|\vec{q}^2|}\int\text{d}^3x\, \Delta V(\vec{x})e^{i\vec{q}\vec{x}/\hbar}[/tex]

    where [tex]\vec{q}=\vec{p}-\vec{p'}[/tex] is the momentum transfer

    Now use Poisson equation for the laplacian of the potential
    [tex]\Delta V{\vec{x})=\frac{-\rho(\vec{x})}{\epsilon_0}[/tex]
    et voila !
    [tex]<\Psi_f|\mathcal{H_\text{int}}|\Psi_i>=\frac{-e\hbar^2}{\epsilon_0V\vec{q}^2} \int\text{d}^3x\, \rho(\vec{x})e^{i\vec{q}\vec{x}/\hbar}[/tex]
    The definition of the electric form factor, the Fourier transform of the charge distribution :
    [tex]F(\vec{q})=\int\text{d}^3x\, \rho(\vec{x})e^{i\vec{q}\vec{x}/\hbar}[/tex]

    In practice it can become more complicated, especially angular distributions are modified due to spin structures. For instance, HERA in Germany confirmed the electron remains point-like down to a thousandth the size of a proton.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook