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Why are walls impenetrable?

  1. Jan 16, 2010 #1
    I thought a container wall is impenetrable due to Coulomb repulsions, but I realized that an idealized positively charged cylindrical container is not able to contain positive charges at all.

    So what else is the interaction at the container walls that prohibits particles to espace?

    Is it some effect of both polarities of the charges? Or a dynamical dipole effect? Or even quantum mechanical exchange repulsion?

    Which potential should I take to model the rebounce of neutral or charged particles at boundary walls?
  2. jcsd
  3. Apr 11, 2010 #2
    "but I realized that an idealized positively charged cylindrical container is not able to contain positive charges at all."

    Can you elaborate on this?
  4. Apr 11, 2010 #3


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    Molecules and atoms are generally more or less electrically neutral. So coulombic forces are not a significant factor with inter-molecular forces (ions notwithstanding). Though I would expect that in very close proximities that the shielding effect would make an atom appear slightly negatively charged (and thus you may get coulombic repulsion that way). But I think that the primary forces at play are called the Van der Waals-London forces. There are three forces here, the orientation force, induction force, and London force (also better described as the Casimir-Polder force). The orientation force is the force that occurs between molecules with permanent dipole moments. This occurs because when we have two electric dipoles then the energy of the system is dependent upon the orientation of these dipoles with respect to eachother. When you consider the distribution of energies in a system due to Boltzmann statistics then you find certain orientations are favored over others and there is a net force.

    The induction force is between a molecule of permanent dipole moment and another that is polarizable. That is, the permanent dipole induces a weak dipole in the other molecule. Again this system gives rise to a net force.

    The London force is between two polarizable molecules. However, this only works in quantum mechanics where we assume that the molecules have an oscillating polarization due to the fact that an oscillator has a non-zero ground state energy. London did not explain how this arises in his paper but Casimir and Polder theorized that this arised from the fluctuating fields of the vacuum ground state. Anyway, these fluctuating fields give rise to momentary dipoles in the polarizable molecules and you get your third force that way.

    These forces have both an attraction and repulsion profile. So while it can be used to explain why things are sticky (and how the gecko defies gravity), it also explains how the molecules can repulse. Still, I would imagine that it would depend upon the exact situation to see whether coulombic or Van der Waals-London are dominant.
  5. Apr 12, 2010 #4
    @Born2wire: Thank you for the detailed listing of forces. I need to look them up somewhere. Do they have a well-defined potential? Do I need such thing a exchange interaction?

    @Vodka: I thought I could contain atoms by saying they have some charge distribution within the atom and so do the container walls. Unfortunately this doesn't work, as a container with cylindrical symmetry has absolutely no field anywhere inside. No matter how the radial charge distribution is.
  6. Apr 12, 2010 #5


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    A quick google search yielded this page that has a nice picture of a typical Van der Waals potential.

    http://web.njit.edu/all_topics/Prog_Lang_Docs/html/autodock/AD3.a.0UserGuide.html [Broken]

    You can see how it transitions from a repulsion to an attraction and that there is a minimum to the potential that would correspond to the most likely distance you would observe the two particles. It would be best if you could find a molecular dynamics text. A simple way of modeling molecular interactions, whether it be gas, solid, liquid or Scotch-Irish, is to use an appropriate intermolecular potential to guide a simulation. Depending on the temperature you set, the appropriate Boltzmann statistics should make the molecules behave in the appropriate state.

    The original London force was something like 1/r^6 but there are now more complex potentials and usually what is done is these potentials have parameters that are adjusted to fit empirical measurements. So if you are interested in a particular interaction, you will need to find the appropriate potential and set of parameters.

    For example, F. London's early papers from 1937 and 1941 talk a little bit about this. London's original equation had two parameters, the polarizability of the atoms and the "frequency" of the fluctuations.
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