Molecular Simulation/Rotational Averaging

Rotational Averaging

An ion interacts with a polar molecule. Vector geometry can be used to describe the energy of this interaction.

Rotational averaging describes the contribution to the potential energy from the rotational orientation of a charge-dipole interaction. Expectation values are utilized to give a single optimal value for the system's potential energy due to rotation.

For example, take a charged particle and a molecule with a permanent dipole. When they interact, the potential energy of this interaction can easily be calculated. For a dipole of length $l$, with a radius of $r$ between the dipole centre and the charged particle, the energy of interaction can be described by:

Potential Energy of a Charge Dipole Interaction $𝒱(r,\theta )=-\frac{q_{ion}\mu \cos \theta }{4\pi {ϵ}_0{r}^2}$

where $q$ is the charge of the particle, $\mu$ is the dipole moment, $\theta$ is the angle between $r$ and the dipole vector, ${ϵ}_0$ is the vacuum permittivity constant, and $r$ is the radius between the particle and dipole.

Geometrically, this interaction is dependant on the radius and the length of dipole, as well as the orientation angle. If the radius $r$ between the ion and dipole is taken to be a fixed value, the angle $\theta$ still has the ability to change. This varied orientation of $\theta$ results in rotation of the dipole about its center, relative to the interacting charged particle. The weights of various orientations are described by a Boltzmann distribution expectation value, described generally by:

Expectation value (discrete states) $⟨M⟩=\frac{\sum _i{M}_i\exp \left(\frac{-{𝒱}_i}{k_B\text{T}}\right)}{\sum _i\exp \left(\frac{-{𝒱}_i}{k_B\text{T}}\right)}$

Expectation value (continuous states) $⟨M⟩=\frac{\int M(r)\exp \left(\frac{-𝒱(r)}{k_{B}T}\right)\text{d}r}{\int \exp \left(\frac{-𝒱(r)}{k_{B}T}\right)\text{d}r}$

where $⟨M⟩$ is the expectation value, $𝒱$ is the energy value for a particular configuration, $k_B$ is Boltzmann's constant, and T is temperature. This Boltzmann-described weighting is the sum over the quantum mechanical energy levels of the system. Therefore, the probability $P$ is directly proportional to $\exp \left(\frac{-𝒱}{k_B\text{T}}\right)$, indicating that at a specific temperature lower energy configurations are more probable. An equation can then be derived from this general expression, in order to relate it to the geometry and energy of a charge-dipole interaction.

Derivation of Rotationally-Averaged Charge-Dipole Interaction Potential

The orientationally averaged potential energy is the expectation value of the charge-dipole potential energy averaged over $\theta$

Starting with the potential energy of a charge-dipole interaction

$𝒱(r,\theta )=-\frac{q_{ion}\mu \cos \theta }{4\pi {ϵ}_0{r}^2}$

We let $C=-\frac{q_{ion}\mu }{4\pi {ϵ}_0{r}^2}$

This makes $𝒱(r,\theta )=-\frac{q_{ion}\mu \cos \theta }{4\pi {ϵ}_0{r}^2}=C\cos \theta$

The average over the dipole orientation using the expectation value in classical statistical mechanics is:

$⟨𝒱\left(r\right)⟩=\frac{\underset{0}{\overset{\pi }{\int }}C\cos \theta \exp \left(\frac{-C\cos \theta }{k_{B}T}\right)\sin \theta \text{d}\theta }{\underset{0}{\overset{\pi }{\int }}\exp \left(\frac{-C\cos \theta }{k_{B}T}\right)\sin \theta \text{d}\theta }$

Note: When integrating over an angle, the variable of integration becomes $\sin \theta \text{d}\theta$

To solve this integral we must first use first order Taylor's series approximation because integrals of exponential of $\cos \theta$ do not have analytical solutions.

$\int \exp \left(\frac{-C\cos \theta }{k_{B}T}\right)\text{d}\theta$

The first order Taylor's series approximations is as follows:

$f(x)\approx f(0)+x\cdot f^{\prime }(0)$

$\exp (-x)\approx \exp (-0)-x\cdot \exp (-0)$

$\exp (-x)\approx 1-x$

Using the Taylor series with $\int \exp \left(\frac{-C\cos \theta }{k_{B}T}\right)\text{d}\theta$ gives: $\exp \left(\frac{C\cos \theta }{k_{B}T}\right)\approx 1-\frac{C\cos \theta }{k_{B}T}$

The integral now becomes:

$⟨𝒱\left(r\right)⟩=\frac{\underset{0}{\overset{\pi }{\int }}C\cos \theta [1-\frac{C\cos \theta }{k_{B}T}]\sin \theta \text{d}\theta }{\underset{0}{\overset{\pi }{\int }}[1-\frac{C\cos \theta }{k_{B}T}]\sin \theta \text{d}\theta }$

Multiplying into the brackets will give:

$⟨𝒱⟩=\frac{\underset{0}{\overset{\pi }{\int }}C\cos \theta \sin \theta -C\cos \theta \sin \theta \frac{C\cos \theta }{k_{B}T}\text{d}\theta }{\underset{0}{\overset{\pi }{\int }}\sin \theta -\sin \theta \frac{C\cos \theta }{k_{B}T}\text{d}\theta }$

All terms that do not depend on $\text{d}\theta$ are constants and can be factored out of the integral. The terms can be expressed as 4 integrals:

$⟨𝒱\left(r\right)⟩=\frac{C\underset{0}{\overset{\pi }{\int }}\cos \theta \sin \theta \text{d}\theta -\frac{C^2}{k_{B}T}\underset{0}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta \text{d}\theta }{\underset{0}{\overset{\pi }{\int }}\sin \theta \text{d}\theta -\frac{C}{k_{B}T}\underset{0}{\overset{\pi }{\int }}\sin \theta \cos \theta \text{d}\theta }$

We must use trigonometric integrals to solve each of these for integrals:

First:

$C\underset{0}{\overset{\pi }{\int }}\cos \theta \sin \theta \text{d}\theta ⟶C\underset{0}{\overset{\pi }{\int }}\cos \theta \sin \theta \text{d}\theta =C\cdot \underset{0}{\overset{\pi }{\int }}\sin \theta d(\sin \theta )=C\cdot \left[\frac{1}{2}{\sin }^2\theta \right]{|}_0^{\pi }=C\cdot 0=0$

Second:

$\frac{C^2}{k_{B}T}\underset{0}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta \text{d}\theta ⟶\frac{C^2}{k_{B}T}\underset{0}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta \text{d}\theta =\frac{C^2}{k_{B}T}\cdot -\underset{0}{\overset{\pi }{\int }}{\cos }^2\theta d(\cos \theta )=\frac{C^2}{k_{B}T}[-\frac{1}{3}{\cos }^3\theta ]{|}_0^{\pi }=\frac{C^2}{k_{B}T}\cdot \frac{2}{3}$

Third:

$\underset{0}{\overset{\pi }{\int }}\sin \theta \text{d}\theta ⟶\underset{0}{\overset{\pi }{\int }}\sin \theta \text{d}\theta =-\cos \theta {|}_0^{\pi }=2$

Fourth:

$\frac{C}{k_{B}T}\underset{0}{\overset{\pi }{\int }}\sin \theta \cos \theta \text{d}\theta ⟶\frac{C}{k_{B}T}\underset{0}{\overset{\pi }{\int }}\sin \theta \cos \theta \text{d}\theta =\frac{C}{k_{B}T}\underset{0}{\overset{\pi }{\int }}\sin \theta d(\sin \theta )=\frac{C}{k_{B}T}\cdot [\frac{1}{2}\sin \theta ]{|}_0^{\pi }=\frac{C}{k_{B}T}\cdot 0=0$

Plugging each trigonometric solved integral back into the equation gives:

$⟨𝒱\left(r\right)⟩=\frac{C\underset{0}{\overset{\pi }{\int }}\cos \theta \sin \theta \text{d}\theta -\frac{C^2}{k_{B}T}\underset{0}{\overset{\pi }{\int }}{\cos }^2\theta \sin \theta \text{d}\theta }{\underset{0}{\overset{\pi }{\int }}\sin \theta \text{d}\theta -\frac{C}{k_{B}T}\underset{0}{\overset{\pi }{\int }}\sin \theta \cos \theta \text{d}\theta }=\frac{0-\frac{2C^2}{3k_{B}T}}{2-0}=-\frac{C^2}{3k_{B}T}$

Finally replace $C$ with $C=-\frac{q_{ion}\mu }{4\pi {ϵ}_0{r}^2}$ to give:

The Orientational Average of the Charge-Dipole Interaction Energy $⟨𝒱\left(r\right)⟩=-\frac{1}{3k_{B}T}{\left(\frac{q_{ion}\mu }{4\pi {ϵ}_0}\right)}^2\frac{1}{r^4}$