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

Feynman -- Atomic dipole

  1. Apr 26, 2015 #1

    I am reading the volume 2 of the Feynman's Lectures on Physics, and something is bothering me when he calculates the dipole moment of a single atom induced by an extern field :

    https://books.google.co.uk/books?id...=onepage&q=feynman dipole single atom&f=false

    Indeed, he states that : " p = qex "

    But why ? I would use in general : " p = Zqex " where Z is the number of electrons in the atom.

    x is the displacement of the center of charges of the electrons, and thus x is also the displacement of each electron.

    Could you explain his reasoning ? It is not the first time he uses a single electron charge instead of Z in his calculations, and I do not understand.


    PS : First, I thought that was because the square of the natural pulsation ω0 depended on Z, which means that ω²0(Z) = Zω²0 (Z=1), which would simplify the Z replacing ω²0(Z) by Zω²0 (Z=1) ; but Feynman seems to use ω0 = ω0(Z) and not ω0 (Z=1) everywhere, so it does not matter.
  2. jcsd
  3. Apr 27, 2015 #2
    And qe is obviously equal to the charge of an electron and note equal to Z*(charge), because he uses then e² = q²e/4piε0.

    Moreover, when he the does his calculations for the Helium atom, he keeps using qe = charge of one electron and does not use a new value for Z..
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Feynman -- Atomic dipole
  1. Feynman's Derivation (Replies: 4)

  2. Feynman's paradox (Replies: 12)

  3. Feynman Sprinkler (Replies: 8)

  4. Dipole radiation (Replies: 1)