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Given an account of the propagation of atomic vibrations along a monatomic chain of atoms, mass $m$, in which the spacing between atoms is $a$ and in which atoms, distance $na$ from each other are connected by a force constant $k_n$,

Show that phonons in a 1-D chain with nearest, and next-nearest, harmonic interactions have frequencies given by: $$ \omega^2=\frac4m\left(k_1\sin^2(qa/2)+k_2\sin^2(qa)\right) $$

I have seen some examples, but I don't understand why sometimes only equation of motion for 1 atom is needed while sometimes 2 atoms are needed? For this question, there must be 4 terms on the right hand side of Newton's second law, so do I need to write down the motion of the $(n+1)^\text{th}$ atom? Why there will there be $k_1$ and $k_2$ instead of $k_n$?

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When there is only one type of atom, we only need one equation of motion. When there are two types of atoms, we need two equations, one for each type. For example, a FCC lattice with the corners occupied by one type of atom, and the face centers occupied by another type requires two equations.

The number of neighbors does not increase the number of equation, only the number of terms on the right-hand side, as you have noted.

In your problem, we only have one type of atom, so just one equation. We do have $4$ neighbors (two nearest, two next nearest), so there are four terms on the right-hand side.

Particularly, we write the equation for $x_n$, $$ m \ddot x_{n} = -k_1 \, (x_n - x_{n+1}) -k_1 \, (x_n - x_{n-1}) - k_2 \, (x_n - x_{n+2}) -k_2 \, (x_n - x_{n-2}). $$ Now using the normal mode equation $$ x_n = A \, e^{i q n a - i \omega t}, $$ we have $$ - m \, \omega^2 \, x_{n} = -\left[ k_1 \, (1 - e^{iqa}) + k_1 \, (1 - e^{-iqa}) + k_2 \, (1 - e^{2iqa}) + k_2 \, (1 -e^{-2iqa}) \right] x_n, $$ which yields $$ m \, \omega^2 = 4 \, k_1 \sin\left(\frac{qa}{2}\right) + 4 \, k_2 \sin\left(qa\right). $$

So there is no need to write down the equation of motion of $x_{n+1}$, for it would be virtually the same as that for $x_n$, with only a phase difference $e^{iqa}$.

There is no $k_n$, because we only have the nearest-neighbor interaction (represented by $k_1$) and the next nearest-neighbor interaction (represented by $k_2$).

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  • $\begingroup$ Thank you for the nice demonstration. But there is a square missing on both sinus. Must read: 𝑚𝜔^2= 4𝑘1 sin^2 (𝑞𝑎/2)+ 4𝑘2 sin^2(𝑞𝑎) since 1-cos(x) = 2sin^2(x/2), $\endgroup$ – Stefan Haacke Dec 13 '19 at 17:06

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