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Conference decwet::physics

Title:Physics
Notice:On the existence of Schr�dinger's Cat
Moderator:AUSS::GARSON
Created:Mon Oct 17 1988
Last Modified:Thu Jun 05 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:423
Total number of notes:5376

421.0. "Van der waals force" by WARFUT::CHEETHAMD () Wed Apr 02 1997 09:40

     Anyone understand Van der Waal's (sp)) forces? I thought I did until I 
    read the following explanation:
     Van der Waals forces are the results of -ve electron charges of atoms
    being in closer proximity to each other than the +ve charges of the 
    nucleus. With the inverse square law this gives a nett force between
    atoms. Trouble is it seems to me that this should give a repulsive
    force whereas I understand that Van der Waals force is an attractive
    force. ???????
     Just as a postscript the explanation was actually for the strong
    nuclear force being the analogue of Van der Waals force caused by the
    colour charges on quarks and gluons rather than the electric charges on
    nuclei and electrons. The above explanation of Van der Waals force was
    used in the explanation as an illustration.
        
    
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421.1Hope this helps...BSS::GINDHARTWed Apr 02 1997 15:1013
    Van der Waals force is attractive.  It results from a weak attraction
    of one atom's electrons by another atom's nucleus.  The electrons, being 
    mobile will find equilibrium at a position which effectively avoids the 
    other atoms electrons, thus resulting in an overall attractive force. 
    A better description lies in considering the quantum mechanical orbits 
    created by two nearby atom, but this is somewhat messy.
    
    As far as the quarks and gluons, I have never been fully up on 
    that theory.
    
    chas.roc
    
    
421.2It does...WARFUT::CHEETHAMDThu Apr 03 1997 01:247
    re .1 Yes thanks, that makes sense. I was assuming an equal probability
    distribution for the electrons, which would give a repulsive force with
    an inverse square law (I think). The distribution of electrons in a crystal
    structure, where each atom has multiple neighbours, would take a little
    figuring though :-) 
                                   
                                 Dennis
421.3AUSS::GARSONDECcharity Program OfficeThu Apr 03 1997 15:564
    re .2
    
    I would have thought that a uniform (spherical) distribution of electrons
    about the nucleus with an inverse square law would give no net force.
421.4DRAGNS::WALLACEFri Apr 04 1997 12:2013
    If you had 2 atoms with spherical electron shells you would
    indeed get repulsion.  The trouble is that once you start
    bringing the atoms close together the electrical field of each
    atom distorts the electrical field of the other, and the charge
    distibution is no longer spherical.  What you end up with is 2
    dipoles, arranged like:
    
    	- +  - +
    
    In which case it's possible to understand why they attract.
    
    Vince
         
421.5REGENT::POWERSTue Apr 08 1997 12:3321
>    distibution is no longer spherical.  What you end up with is 2
>    dipoles, arranged like:
>    
>    	- +  - +
>    
>    In which case it's possible to understand why they attract.


But that arrangement is not symmetrical.
If we're talking similar atoms, as we would be in a homogeneous gas,
then the charge distribution should be symmetrical about the center point 
of the two atoms.

Maybe like:

    -3 +4 -1    -1 +4 -3

Where some of the negative charge gets pushed to outside edge while
other moves to the inside.

- tom]
421.6AUSS::GARSONDECcharity Program OfficeTue Apr 08 1997 17:1813
re .5
    
>But that arrangement is not symmetrical.
>If we're talking similar atoms, as we would be in a homogeneous gas,
>then the charge distribution should be symmetrical about the center point 
>of the two atoms.
    
    Even so, the arrangement shown in .4 is of lower energy than a
    symmetrical one (as in .5). I doubt that you can arrange those charges
    symmetrically and get a net attractive force.
    
    I believe that spontaneous symmetry breaking can occur in transitioning
    to a lower energy state.
421.7My brane hertz(again)WARFUT::CHEETHAMDWed Apr 09 1997 08:387
re .5  Do Van der Waals forces occur in a gas ?

re .6
>>    I believe that spontaneous symmetry breaking can occur in transitioning
>>    to a lower energy state.

deeeerrrrrrrrrrr ????
421.8DRAGNS::WALLACEMon May 05 1997 12:3630
     >> If you had 2 atoms with spherical electron shells you would
     >> indeed get repulsion.
    
    I wrote this in .4, and it is of course incorrect (shouldn't
    write notes when tired).  A spherical distribution of charge
    looks like a point charge of the same magnitude.
    
    What's important is that when we're talking about electrons
    having a spherical distribution around a nucleus, its the
    quantum mechanical wave function that's spherical, not the
    electrical charge distribution.  Averaged over time the 
    charge should also be spherically distributed, but at any
    given instant it will not be.
    
    Think in terms of a hydogen atom, with 1 proton and 1 electron.
    At any particular moment you have 2 point charges creating a 
    dipole.  When two hydrogen atoms start to approach each other
    there is a tendency for the orbits of the two electrons to
    "synchronize" with each other, so that you get the -+-+ charge
    distribution, which creates a net attraction.  Crude picture
    of synchronized electrons :-)
    
               -                -
    -+-+ =>  + +  =>  +-+-  =>  + +
             -                    -
    
    This tendency for the electrons to "synchronize" provides an
    intuitive explanation for Van der Waals forces.  For the
    real explanation of course, do the QM math :-))))
                                                     
421.9AUSS::GARSONDECcharity Program OfficeMon May 05 1997 16:3915
re .8
    
> its the quantum mechanical wave function that's spherical
    
    I think only for s orbitals.
    
>    At any particular moment you have 2 point charges creating a 
>    dipole.
    
    I don't think that it is reasonable from a QM point of view to say
    this. Instead I think QM would say that you have a superposition of
    many states each of which is a bit like a dipole.
    
    How one works that into an explanation of the van der Waals force I
    don't know.