Polarizability of Fixed Dipole Interactions

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SUMMARY

The discussion centers on the polarizability of induced-fixed dipole interactions and the fixed dipole-dipole interaction. The polarizability is defined by the equation α = 4πεr³, leading to a potential field for two substances given by U(r) = - (μ₁²α₂ + μ₂²α₁) / (16π²ε²r⁶). The confusion arises regarding the fixed dipole-dipole interaction, where the potential field is expressed as U(r) = - (μ₁²μ₂²) / (48π²ε²KTr⁶), which does not account for polarizability. The Keesom interaction is emphasized as being stronger than the Debye interaction, indicating that neglecting induced polarization effectively sets polarizability to zero.

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  • Understanding of van der Waals forces
  • Familiarity with dipole interactions
  • Knowledge of electromagnetic theory
  • Basic grasp of statistical mechanics (kBT concept)
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  • Study the Keesom and Debye interactions in detail
  • Learn about static and dynamic polarizability
  • Explore the implications of fixed dipole interactions in molecular physics
  • Investigate the role of temperature in dipole interactions
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Researchers in molecular physics, chemists studying intermolecular forces, and students seeking to understand dipole interactions and their implications in various substances.

Caveman11
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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:

\begin{equation}\alpha =4\pi\epsilon r^{3} \end{equation}

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

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

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:

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

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

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

Why is that?

Sorry for the lengthy post and thankyou in advance.

Nick
 
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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.
 
Why is that?

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)
 

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