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Electron Capture and Coulomb's Law

  1. Apr 16, 2015 #1
    Sorry for the brevity of the post, but eloquence has never been my strong point. Here's my question:

    In a proton-surfeited atom, a proton captures an electron to form a neutron. Now, according to Coulomb's law, the magnitude of the electrostatic force field is equal to (kq1q2)/d2, where k is Coulomb's constant, q1 is the charge of the proton and q2 is the charge of the electron, and d is the distance between the proton and electron. Now, given that neither k, q1, nor q2 is changing, shouldn't the magnitude of the electrostatic force field approach infinity as the electron spirals toward the proton? Does it?
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  3. Apr 16, 2015 #2


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    The events you describe happen at a quantum level where things are not as simple as looking at different forces acting on point particles.
  4. Apr 16, 2015 #3
    Ok. Thank you, Orodruin. So, the particles are treated as probability functions?

    Wow, what was I thinking? Sorry, folks. Highly caffeinated and hardly slept. Forgive me.
  5. Apr 16, 2015 #4


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    A fast answer: no.
    The problem in your approach is that you are taking a classical view on the process of electron capture.
    One very simplistic example to see that it's impossible to think of r->0 (that got in my mind right-away) is that if the energy of the particle goes way beyond some limit (eg twice the mass of the particle), you will get the creation of particles+antiparticles etc...I think this results in shielding the particle's charge (vacuum polarization).
    Nevertheless, I am not sure if this explains why this is impossible, but it gives an idea.
    The most precise answer for why this is impossible, is because of the quantum mechanics (so you don't have two balls that are at some distance D and go to some distance 0) as well as the interactions don't happen at points (there are intermediate W bosons, and you are in fact looking for the amplitudes for them to be mediated throughout the whole space).
  6. Apr 16, 2015 #5


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    So you implying that the electron has infinite velocity and kinetic energy when it collides with the proton?

    Not sure what is the workaround this mini paradox, maybe that the proton cant be considered as a point particle because it is not an elementary particle like electron or quarks are.
  7. Apr 16, 2015 #6


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    I think it's better to say that the Coulomb's law is a result of the low-energetic QED ... at higher energies it's not applicable and you can't go the other way.

    And in EC the electron "sees" the quarks since it interacts with them. The process is [itex] e u \rightarrow \nu_e d[/itex] (with a W-boson t-channel)
  8. Apr 16, 2015 #7
    Is it that the uncertainty principle precludes this from happening because infinity is a definite momentum (or is it?) and zero is a definite position?
  9. Apr 16, 2015 #8


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    This is again simplistic , but you could use that... in fact, some introductory books in elementary particles, prefer introducing the intermediate bosons with the uncertainty principle. It's way before reaching infinite, that you can get (as they say due to the uncertainty principle) extra particles appearing.

    The thing is that once you take a limit, you cannot go from that limit and deduce how your initial theory works. Take for example the special relativity, the gamma factor: [itex]\gamma = \frac{1}{\sqrt{1- (u/c)^2}} [/itex], and the momentum [itex]\vec{p}= \gamma m \vec{u}[/itex]. In the low velocity limit ([itex]u \ll c[/itex]) the gamma factor goes to 1, and then the momentum is [itex]p=m u[/itex], how can you say that this holds when you let your velocity get large enough? You already left the SR...

    The same is true for QED, you can only apply the coulomb's law at low energies [non-relativistic limit].
  10. Apr 16, 2015 #9


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    The finite size of the proton is irrelevant - you can ask the same question for positronium which is made out of two elementary particles. The answer is quantum mechanics in both cases.
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