I'm looking to find the thermodynamic (NVT) free energy of a classical coupled harmonic oscillator system such as the one below:
(image taken from http://openmetric.org/StatisticalPhysics/equilibrium/week3.html)
I would like a solution that allows an arbitrary $N$ number of masses, and ideally I would like to have a fully general expression with arbitrary (different) spring constants and arbitrary (different) masses.
I tried to compute this free energy by hand by computing the partition function:
$$ Z = \int_{-\infty}^{\infty}d\vec{p} \int_{-\infty}^{\infty}d\vec{x}\; e^{-\beta H}$$
where
$$ H = \sum_{i=1}^{N} \frac{p_i^2}{2m_i} + \sum_{i=0}^{N} m_i\omega_i^2(x_{i+1}-x_i)^2 $$
and $x_i$ denotes the displacement from equilibrium of the $i$th block, and $x_0=x_{N+1}=0$ represent the walls at the ends.
I was able to derive the expressions for the free energy $F = -\frac{1}{\beta}\ln Z$ for one, two, and three blocks with identical masses and identical springs (with the hope of seeing an extendible pattern) but sadly no obvious patterns emerged. The calculations are also rather tedious.
I don't doubt that this has already been done many times before — does anyone have a reference to a solution?
Best Answer
Make the change of variable to $\delta_i = x_{i+1} - x_i,\; i=0\dots N-1.$ Then the system is uncoupled and $Z = \prod_i z_i$ with $$ z_i = \int e^{-\beta p^2/2m_i}dp \times \int e^{-\beta m_i\omega_i^2\delta^2}d\delta = \frac{\pi\sqrt{2}}{\beta\omega_i}.$$