Molecular Simulation/Induced Polarization

When a molecule is exposed to an electric field ɛ, electron density is polarized away from the nuclei. A dipole is thus created due to the asymmetric distribution of electron density. The induced molecular dipole created is proportional to the strength of the electric field (ɛ) as well as the polarizability of the molecule, $&alpha;$.

Polarizability allows us to better understand the interactions that occur between non polar atoms and molecules and other electrically charged species. In general, polarizability correlates with the interaction between electrons and the nucleus.

Units of Polarizability
Dimensional analysis of induced polarization.

$$\mu_{ind} = \alpha \epsilon$$

$$ \alpha = \frac{\mu_{ind}}{\epsilon}$$

$$ \alpha = \frac{C m}{V m^{-1}}$$

$$ \alpha = (\mathrm{C} \cdot \mathrm{m}^2 \cdot \mathrm{V}^{-1})$$

For convenience we will let:

$$ \alpha^{'} = \frac{1}{ 4 \pi \varepsilon_0 }\alpha $$

$$ \alpha^{'} = \frac{1}{ C V^{-1} m^{-1} }(\mathrm{C} \cdot \mathrm{m}^2 \cdot \mathrm{V}^{-1}) = m^{3} $$

Polarizability has units of volume, which roughly correspond to the molecular volume. A large polarizability corresponds to a "soft" cloud of electron density, which is easily distorted by electric field.

Polarizability in Intermolecular Interactions
When we consider the effect that a charge (Q), has on a neutral atom at distance r, an induced dipole is created:

$$\mu_{ind} = \alpha \epsilon = q\Delta r$$

Where the electrostatic force is defined by:

$$f = (charge) \times E $$

Charge-induced dipole electrostatic force can be approximated by the electrostatic force between $$Q$$ and $$q^{-}$$ minus the force between $$Q$$ and $$q^{+}$$.

$$f \approx q\Delta r \frac{d \epsilon}{dr}$$

Since the electric field at distance r from a point charge Q:

$$ \epsilon = \frac{Q}{4 \pi \epsilon_{0} r^{2} }$$

Take the derivative

$$\frac{d \epsilon}{dr} = -2 \frac{Q}{4 \pi \epsilon_{0} r^{3} }$$

$$\epsilon \frac{d \epsilon}{dr} = \left(\frac{Q}{4 \pi \epsilon_{0} r^{3} } \right) \left(-2 \frac{Q}{4 \pi \epsilon_{0} r^{3} }\right) $$

$$\epsilon \frac{d \epsilon}{dr} = -2 \left(\frac{Q}{4 \pi \epsilon_{0} } \right)^{2} \frac{1}{r^{5}} $$

Therefore:

$$f= \alpha \epsilon \frac{d \epsilon}{dr}$$

$$f = -2 \alpha \left(\frac{Q}{4 \pi \epsilon_{0} } \right)^{2} \frac{1}{r^{5}} $$

We can thus get the intermolecular potential by integrating this force over r:

$$\mathcal{V}(r) = - \int_\infty^r f(r) dr$$

$$\mathcal{V}(r) = \int_\infty^r -2 \alpha \left(\frac{Q}{4 \pi \epsilon_{0} } \right)^{2} \frac{1}{r^{5}} dr = -\frac{\alpha}{2} \left(\frac{Q}{4 \pi \epsilon_{0} } \right)^{2} \frac{1}{r^{4}} $$