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?
Best Answer
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.