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**a**but why do we do the T.S about

**r=a**?

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The answer as to why we do a Taylor expansion at all is because physics is hard. Indeed, for the Lennard Jones potential we use to describe lattice dynamics exact solutions are altogether intractable. So we do what we typically do with potentials in such cases, we Taylor expand around the minima, ei: the equilibrium position of the atoms. Keeping only the first term results in the static approximation of the crystal (ie: no phonons, all atoms rigidly fixed in place). Adding in the second term is equivalent to approximating the vibrations with a Hooke's law term and leads to the harmonic cross-section where single phonons are stationary states that propogate with infinite lifetime through the lattice (unless they hit a crystal defect or the crystal surface). Additional terms are referred to as anharmonic terms and are responsible for multi-phonon creation and annihilation processes. These higher order terms account for increasing number of multi-phonon processes. For this reason, the terms are often classified as the elastic (zero phonon) cross-section for the first term, the one phonon for the quadratic term, two-phonon for the cubic (1 in -> 2 out, or 2 in -> 1 out), three-phonon for the quartic (1 in -> 3 out, 2 in -> 2 out, 3 in -> 1 out). And on up you go, however it is rare to ever go beyond the quartic term when modeling properties of solids since the transition rates for such higher order processes is extremely low. In a nutshell, these 'x in - y out' selection rules result from writing the higher order terms as linear combinations of products consisting of the raising and lowering quantum mechanical operators derived for the harmonic approximation (ie: keeping only the Hooke's law term which results in a more or less direct mapping of the lattice to the good old fashioned quantum mechanical oscillator problem). The selection rules drop out of the details associated with this process.I know that each atom is separated byabut why do we do the T.S aboutr=a

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$$(r-a)\frac{dW(a)}{dr}$$

is zero (because the slope of the potential is zero at the minimum). You can choose to expand around any point you want (call it ##r_0##), but if you don't choose an extremum, you'll have to deal with that extra linear term (##\propto r-r_0##) in your potential. The physics is ultimately the same, but the math is unnecessarily complicated.

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