# Solid as a large molecule

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1. May 23, 2015

### A_s_a_d

In the discussion of calculating specific heat for a solid, it is assumed that the whole solid body is a molecule with N atoms and the Hamiltonian of this solid is similar to that of a molecule with N atoms, i.e.
$\mathcal{H}_1=\mathcal{V}^{*}+\sum_{j=1}^{3n} \big(\frac{\widetilde{p}_s^2}{2m}+\frac{K_s}{2} \widetilde{u_s}^2\big)$. How this assumption of taking solid as a huge molecule is justified?

Last edited: May 23, 2015
2. May 23, 2015

### The_Duck

That's the Hamiltonian of *any* system of N particles, not just an N-atom molecule.

3. May 23, 2015

### A_s_a_d

Thanks, modified the Hamiltonian in the question
.