Molecular Simulation/Rotational Averaging

Rotational Averaging


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:

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where $$q$$ is the charge of the particle, $$\mu$$ is the dipole moment, $$\theta$$ is the angle between $$r$$ and the dipole vector, $$\epsilon_0$$ 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:

$$

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where $$\langle M \rangle$$ is the expectation value, $$\mathcal{V}$$ 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 $$\exp \left (\frac{-\mathcal{V}}{k_B\text{T}} \right)$$, 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
 * $$\mathcal{V}(r,\theta) = -\frac{q_{ion} \mu \cos\theta }{4\pi\epsilon_0 r^2}$$

We let $$C = -\frac{q_{ion} \mu}{4\pi\epsilon_0 r^2}$$ This makes $$\mathcal{V}(r,\theta) = -\frac{q_{ion} \mu \cos\theta }{4\pi\epsilon_0 r^2} = C\cos\theta$$

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


 * $$ \langle \mathcal{V}\left(r\right) \rangle = \frac {\int^{\pi}_{0} C \cos\theta \exp \left (\frac{-C\cos\theta}{k_B T}\right)  \sin\theta \textrm{d}\theta}{\int^{\pi}_{0} \exp \left(\frac{-C\cos\theta}{k_B T}\right) \sin\theta \textrm{d}\theta}$$

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

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.
 * $$ \int \exp \left(\frac {-C\cos\theta}{k_B T}\right) \textrm{d}\theta$$

The first order Taylor's series approximations is as follows:
 * $$f(x) \approx f(0) + x \cdot f'(0)$$
 * $$\exp(-x) \approx \exp (-0) - x \cdot \exp (-0)$$
 * $$\exp(-x) \approx 1 - x$$

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

The integral now becomes:
 * $$ \langle \mathcal{V}\left(r\right) \rangle = \frac {\int^{\pi}_{0} C \cos\theta \left [1 - \frac{C\cos\theta}{k_B T}\right]  \sin\theta \textrm{d}\theta}{\int^{\pi}_{0} \left[1 - \frac{C\cos\theta}{k_B T}\right] \sin\theta \textrm{d}\theta}$$

Multiplying into the brackets will give:
 * $$ \langle \mathcal{V}\rangle = \frac {\int^{\pi}_{0} C \cos\theta\sin\theta - C \cos\theta\sin\theta \frac {C\cos\theta}{k_B T} \textrm{d}\theta}{\int^{\pi}_{0} \sin\theta - \sin\theta \frac {C\cos\theta}{k_B T} \textrm{d}\theta}$$

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


 * $$ \langle \mathcal{V}\left(r\right) \rangle = \frac{C \int^{\pi}_{0} \cos\theta\sin\theta \textrm{d}\theta - \frac{C^2}{k_B T} \int^{\pi}_{0} \cos^2\theta\sin\theta \textrm{d}\theta}{\int^{\pi}_{0} \sin\theta \textrm{d}\theta - \frac{C}{k_B T} \int^{\pi}_{0} \sin\theta\cos\theta \textrm{d}\theta}$$

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

First:
 * $$ C\int^{\pi}_{0} \cos\theta\sin\theta \textrm{d}\theta \longrightarrow C\int^{\pi}_{0} \cos\theta\sin\theta \textrm{d}\theta = C \cdot\int^{\pi}_{0} \sin\theta d(\sin\theta) = C\cdot \left[\frac{1}{2} \sin^2\theta\right] \Big|^{\pi}_{0} = C \cdot 0 = 0$$

Second:
 * $$ \frac{C^2}{k_B T} \int^{\pi}_{0} \cos^2\theta\sin\theta \textrm{d}\theta \longrightarrow \frac{C^2}{k_B T} \int^{\pi}_{0} \cos^2\theta\sin\theta \textrm{d}\theta = \frac{C^2}{k_B T} \cdot -\int^{\pi}_{0} \cos^2\theta d(\cos\theta) = \frac{C^2}{k_B T} \left[-\frac{1}{3}\cos^3\theta\right] \Big|^{\pi}_{0} = \frac{C^2}{k_B T}\cdot\frac {2}{3}$$

Third:
 * $$ \int^{\pi}_{0} \sin\theta \textrm{d}\theta \longrightarrow \int^{\pi}_{0} \sin\theta \textrm{d}\theta = -\cos\theta\Big|^{\pi}_{0} = 2$$

Fourth:
 * $$ \frac{C}{k_B T} \int^{\pi}_{0} \sin\theta\cos\theta \textrm{d}\theta \longrightarrow  \frac{C}{k_B T} \int^{\pi}_{0} \sin\theta\cos\theta \textrm{d}\theta = \frac{C}{k_B T} \int^{\pi}_{0} \sin\theta d(\sin\theta) = \frac{C}{k_B T} \cdot \left[\frac{1}{2}\sin\theta \right]\Big|^{\pi}_{0} = \frac{C}{k_B T} \cdot 0 = 0 $$

Plugging each trigonometric solved integral back into the equation gives:
 * $$ \langle \mathcal{V}\left(r\right)\rangle = \frac{C \int^{\pi}_{0} \cos\theta\sin\theta \textrm{d}\theta - \frac{C^2}{k_B T} \int^{\pi}_{0} \cos^2\theta\sin\theta \textrm{d}\theta}{\int^{\pi}_{0} \sin\theta \textrm{d}\theta - \frac{C}{k_B T} \int^{\pi}_{0} \sin\theta\cos\theta \textrm{d}\theta} = \frac{0-\frac{2C^2}{3k_B T}}{2-0} = -\frac{C^2}{3k_B T} $$

Finally replace $$C$$ with $$C = -\frac{q_{ion}\mu}{4\pi\epsilon_0 r^2} $$ to give:

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