# Polarizability of Fixed Dipole Interactions

1. Aug 26, 2011

### Caveman11

Hi all,

I've been having a problem with understanding some of the van der Waals forces. I understand that the polarizability of a induced-fixed dipole interation is:

$$\alpha =4\pi\epsilon r^{3}$$

Which leads to the potential field (assuming two different substances):

U(r)=-\displaystyle\frac{\mu_{1}^{2}\alpha_{2}+\mu_{2}^{2}\alpha_{1}}{16\pi^{2}\epsilon^{2}r^{6}}

and therefore the force between them.

My problem lies with the fixed dipole-dipole interaction. Now I understand that the polarizability has to take into account the orientation of the molecule/atom to give:

$$\alpha =4\pi\epsilon r^{3} + \displaystyle\frac{\mu^{2}}{3k_{B}T}$$

This is where I get confused. I have read in "Scanning force microscopy-Dror Sarid" that the potential field between two fixed dipoles does not take into acount there polarizability

U(r)=-\displaystyle\frac{\mu_{1}^{2}\mu_{2}^{2}}{48\pi^{2}\epsilon^{2}KTr^{6}}

Why is that?

Sorry for the lengthy post and thankyou in advance.

Nick

2. Sep 6, 2011

### AM_Ru

Polarization is effect of radiation by bodies of electromagnetic waves in directions distinct from a direction of an external electromagnetic field. It is connected to that that dipoles which radiate, can be built in a body not on a direction of a field and somehow differently. The direction of radiation depends on a direction of these dipoles and does not depend on potential between them.

3. Sep 6, 2011

### alxm

Because the Keesom interaction is larger than the Debye interaction. Permanent dipoles are generally stronger than induced ones. So if you neglect the induced polarization, you're effectively setting the polarizability to zero, and so that's what you end up with.

Note that your polarizability here is itself just a simple linear approximation, and that it has higher orders. (And is frequency dependent as well, so this is actually the _static_ polarizability)