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

Van der force

  1. Aug 17, 2010 #1
    My lecturer writes the following in his lecture notes:
    “Van der Waal’s : Fluctuating dipole of molecules induces dipoles in molecules or other-
    wise neutral atoms (eg noble gases)”

    Does a fluctuating dipole refer to a dipole that is changing its orientation in space?

    How causes a dipole to be formed within a molecule, provided that there are no nearby fluctuating dipoles?

    Finally, how can the fluctuating dipole of one molecule induces dipoles in molecules or other-
    wise neutral atoms?

    Tahns in advance for any help!
     
  2. jcsd
  3. Aug 17, 2010 #2

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    The dipole can be thought of as being something akin to
    [tex] \mathbf{p} = \alpha (t) \hat{z} [/tex]
    where
    [tex] -d \leq \alpha (t) \leq d[/tex]

    In other words, the fluctuating dipole is like a harmonic oscillator where the positive nucleus lies at the equilibrium position and the electron is moving back and forth in space. In the above I just restricted it along a single dimension for simplicity though. So yes, the dipole is changing its orientation and its strength.

    The original paper by London did not make any assumptions on how these dipoles are fluctuating. In quantum mechanics, the harmonic oscillator has a non-zero energy ground state and is always oscillating. London assumed that the electrons in the polarizable atoms or molecules behave like the quantum oscillator. Thus, the basic assumption assumes that the dipoles will always fluctuate but he did not give a more basic reason.

    One reason for this can be found by the fact that the quantum vacuum contains fluctuating electromagnetic fields even in the ground state. If we use quantum field theory to describe the electromagnetic fields, we find that in the ground state, called the vacuum where there are no photons, the expectation (mean) value of the electric and magnetic fields is zero. However, the fields do fluctuate about the zero point because they are also described as harmonic oscillators. The fluctuating electric and magnetic fields couple with the electrons in an atom or molecule. This coupling causes the fluctuations that give rise to the dipole moment that London assumed in his paper.

    So in short, one way to think about it is that the electrons couple with the vacuum fields which cause them to fluctuate from their normal orbitals and this gives rise to a fluctuating dipole moment.

    Now this new fluctuating dipole moment gives rise to its own electric field. This electric field will act on a nearby atom or molecule and induce a charge separation between the nucleus and orbiting electron and thus cause another dipole moment (this is the same mechanism as how the vacuum fields create the original fluctuating dipole moment).

    The difference with the fluctuating dipole moment that is induced by the neighboring atom versus the vacuum induced moment is that the vacuum induced moment is independent (more or less) of the position of the atom in space. However, the position of the two atoms relative to each other affects the dipole moment that one atom's dipole induces in the other atom. The fact that the induced dipole moment is dependent upon position means that the energy of the system is position dependent and thus there is a force (since force is the gradient of the potential of the system).
     
  4. Aug 17, 2010 #3
    Thanks! Your discussion is very clear.

    I am wondering if the diople moment should equal q times alpha.
     
  5. Aug 17, 2010 #4

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    Eh, depends. The dipole moment is always time-varying. However, a lot of times we work in the frequency domain which assumes a exp(-i \omega t) time dependence. So often when you are given a dipole moment described as \alpha q, they really mean \alpha q exp(-i \omega t). So we can certainly think of the dipole moment as some q\alpha but there still needs to be an inherent time dependence somewhere since the charges are being acted upon by a time-varying force. As for \alpha versus \alpha q, it doesn't matter. I'm just using \alpha as a constant here in which we can absorb the charge. We generally talk about the molecules or atoms have a polarizability of \alpha as opposed to being a direct dipole of q d where d is the distance of separation between two charges of strength q.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Van der force
Loading...