An ideal gas consisting of $N$ molecules possessing an electric dipole moment $\mathbf{d}$, placed in a constant electric field intensity $\mathbf{E}$. Need to calculate the average of magnitude of the vector of electric polarizatio $\mathbf{P}$.

My decision. I use the differential for the Helmholtz energy:

$$ F = F(V, T, \mathbf{E}), \quad {d}F = -S {d}T - P {d}V - \mathbf{P} {d} \mathbf{E}. $$

From here find:

$$ \vec{P} = -\Big( \frac{\partial F}{\partial \mathbf{E}} \Big)_{V, T} $$

The Helmholtz energy is expressed using statistical integral:

$$ F = -\theta \ln Z_N, $$ Where $\theta = k T$ -- statistical temperature.

To calculate the statistical integral, we need to know the Hamiltonian systems. For an ideal gas in our case the Hamiltonian has the form:

$$ H(\mathbf{q},\mathbf{p}) = \sum_{i=1}^{N} H_i (\mathbf{q_i}, \mathbf{p_i}) = \sum_{i=1}^{N} \Big ( \frac{\mathbf{p^{2}_{i}}}{2m} + U(\mathbf{q_i}) \Big ) = \sum_{i=1}^{N} \Big ( \frac{\mathbf{p^{2}_{i}}}{2m} - E d \cos \phi \Big ), $$ Where $\phi$ - the angle between the vectors $\mathbf{d}$ and $\mathbf{E}$. Let's start to calculate the statistical integral:

$$ Z_N = \frac{1}{N! h^{3N}} \int\limits_{6N} \exp \Big (- \frac{H(\mathbf{q}, \mathbf{p}) }{\theta} \Big ) {d} \mathbf{q} {d} \mathbf{p} = \frac{1}{N! h^{3N}} \prod_{i=1}^{N} \int\limits_{6} \exp \Big ( -\frac{\mathbf{p^{2}_{i}}}{2m\theta} + \frac{E d \cos \phi}{\theta} \Big ) {d} \mathbf{q_i} {d} \mathbf{p_i} $$ Or

$$ Z_N = \frac{1}{N! h^{3N}} A^N, \quad A = \iiint\limits_{-\infty}^{\infty} \exp \Big( - \frac{p^2_x + p^2_y + p^2_z}{2m \theta}\Big ) {d} p_x {d} p_y {d} p_z \int\limits_{3} \exp \Big( \frac{E d \cos \phi}{\theta} \Big ) {d} x {d} y {d} z. $$

The first factor is the integral in the momenta. It is equal $(2 \pi m \theta)^{3/2}$. The problem with the second integral. How to calculate?

Is there another way to solve this problem?


1 Answer 1


It is better to work with the spherical coordinate system. Then the integral takes the form:
$\int drd\theta d\phi r^2\sin\theta e^{\frac{1}{kT}ED\cos\theta}\\ =\int drd\cos\theta d\phi r^2 e^{\frac{1}{kT}ED\cos\theta}$.
The rest can be done by yourself.


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