# Total momentum in linear monoatomic chain

Context: Solid state physics. Monoatomic linear chain.

Question: To prove that the total momentum of the chain is zero.

Attempted solution: I consider the sum:

\begin{align*} p = \sum_{n=1}^{N} m \dot{u_n} \end{align*}

where $u_n$ is the displacement from the equilibrium position of the $n$-th atom. The displacement is given by the formula:

\begin{align*} u_n = u_0 \exp\left[-i\left(\omega t \pm k n a\right)\right] \end{align*}

where $k$ is the wavevector, $n$ is the $n$-th atom and $a$ is the distance between atoms. If I substitute the above formula to the first sum, I get a result of the form:

\begin{align*} p \sim \sum_{n=1}^{N} \exp(i k n a) \end{align*}

I wonder, how could I prove that this is always zero ? If I treat it as the sum of a geometric series with $\alpha_1=\exp(ika)$ and $\lambda=\exp(ika)$, I still get a result that isn't necessarily zero.

\begin{align*} S_{1\to N} = \alpha_1 \frac{\lambda^N-1}{\lambda - 1} = e^{ika} \frac{e^{ikNa}-1}{e^{ika}-1} \end{align*}

If I further require that the first and last atoms are fixed, then $\exp(ikNa) = \exp(ika)$ and $S_{1\to N} = \exp(ika)$. Then

\begin{align*} p = -i\omega u_0 m \exp(-i\omega t) \exp(ika) = -i\omega m \underbrace{u_0 \exp\left[-i(\omega t - k a)\right]}_{u_1} = -i \omega m u_{1} = 0 \end{align*}

since we assumed that the 1st atom is fixed. Does this sound correct ?

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'If I treat it as the sum of the geometric series...' can you show this? – Kvothe Apr 10 '14 at 21:03

With $$u(n)=u_0e^{-i(\omega t+k n a)}$$ we have $$p=\sum_{n=1}^Nm\frac{d}{dt}u(n)=i\omega m u_0e^{-i\omega t}\left(\frac{1-e^{-iakN}}{1-e^{iak}}\right).$$ With cyclic boundary conditions we have $$u(N+1)=u(1)\Rightarrow k=\frac{2\pi j}{Na}\text{ for }j\in\mathbb{Z}$$ and inserting $k$ into $p$ gives $$p=i\omega m u_0e^{-i\omega t}\left(\frac{1-e^{-2 i \pi j}}{1-e^{\frac{2 i \pi j}{N}}}\right)=0.$$ Your method uses a slightly different boundary condition, but I think it's still valid.