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CAF123
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Homework Statement
Consider a system made up of joining together ##N## beads and ##N-1## springs. The positions of the beads is indicated by ##N## real numbers ##\left\{x_i\right\}_{i=1,...N}.## The Hamiltonian which characterises it is $$\mathcal H = \frac{k}{2}\sum_{i=1}^{N-1}(x_{i+1}-x_i)^2,$$ where k is a real and positive constant. Assume that the first bead is fixed at the origin (##x_1=0)##.
Find the partition function of the system. Change variables to ##\left\{\sigma\right\}_{i=1...N}## where ##\sigma_i = x_{i+1}-x_i##.
Homework Equations
Assume the positions of the beads follow Boltzmann distribution ##\text{exp}(-\beta \mathcal H)##
The Attempt at a Solution
N beads and N-1 springs, so two springs connected to each bead. The position of the beads (except the one fixed at the origin) may extend to ##\pm \infty##. Then the partition function is the sum over all possible positions, so can write $$Z_N = \int_{-\infty}^{\infty} \text{d}x_1 \text{d}x_2 ... \text{d}x_N \text{exp}\left(-\frac{\beta k}{2} \sum_{i=1}^{N-1} \sigma_i^2\right) \delta(x_1),$$ where the delta function enforces the fixed position of the first bead. I can write this as $$Z_N = \int_{-\infty}^{\infty} \text{d}x_1 \text{d}x_2 ... \text{d}x_N \prod_{i=1}^{N-1} \text{exp}\left(-\frac{\beta k}{2} \sigma_i^2\right) \delta(x_1),$$ but how to continue?
##d \sigma_1 = dx_2## because ##x_1## is fixed but ##d \sigma_2 = d(x_3-x_2)##.
Thanks!