Question 3 of 7: Phonon Dispersion Relation for a Monatomic Chain

Question 3: Phonon Dispersion Relation for a Monatomic Chain (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 1-D monatomic chain of lattice planes, spacing $a$, each plane of atomic mass $M$, coupled to its two nearest-neighbour planes only by a spring constant $C$ (Fig. P3). Trial travelling-wave solution $u_s(t)=u\exp[i(sKa-\omega t)]$.

Find. (a) derive $\omega(K)$; (b) plot it across the first Brillouin zone $-\pi/a\le K\le\pi/a$; (c) characterize the wave motion at the zone boundary.

Approach. Write Newton's second law for plane $s$ using the paper's own eqs. (10)–(11), substitute the travelling-wave trial solution, and solve the resulting algebraic equation for $\omega(K)$; then examine the group velocity $d\omega/dK$ at $K=\pm\pi/a$.

1. Part (a) — Newton's law for plane $s$. With only nearest-plane coupling, the net restoring force on plane $s$ is (eq. 10/11 on the paper): $$M\frac{d^2u_s}{dt^2}=C(u_{s+1}-u_s)-C(u_{s-1}-u_s)=C(u_{s+1}+u_{s-1}-2u_s)$$ Substituting $u_s=u\exp[i(sKa-\omega t)]$, so that $u_{s\pm1}=u_s\exp(\pm iKa)$ and $d^2u_s/dt^2=-\omega^2u_s$: $$-M\omega^2u_s=Cu_s\left[\exp(iKa)+\exp(-iKa)-2\right]$$ Dividing through by $u_s$ (nonzero) and using eq. (1), $\cos\theta=\tfrac12[\exp(i\theta)+\exp(-i\theta)]$: $$-M\omega^2=C\bigl[2\cos(Ka)-2\bigr]=-2C\bigl[1-\cos(Ka)\bigr]$$ $$\omega^2=\frac{2C}{M}\bigl[1-\cos(Ka)\bigr]$$ Applying the paper's own half-angle identity, eq. (1), $\sin^2\theta=\tfrac12(1-\cos2\theta)$ with $\theta=Ka/2$, gives $1-\cos(Ka)=2\sin^2(Ka/2)$, so $$\omega^2=\frac{4C}{M}\sin^2\!\left(\frac{Ka}{2}\right)$$ Taking the positive square root (angular frequency is non-negative): $$\boxed{\omega(K)=\sqrt{\frac{4C}{M}}\,\left|\sin\!\left(\frac{Ka}{2}\right)\right|}$$ — exactly the target relation, confirmed algebraically.
2. Part (b) — plot over the first Brillouin zone. $\omega(K)$ is periodic in $K$ with period $2\pi/a$ and even in $K$ (only $|\sin|$ enters), so the first zone $-\pi/a\le K\le\pi/a$ shows one full "hump": $\omega=0$ at $K=0$, rising smoothly and monotonically in $|K|$ to a maximum $\omega_{max}=\sqrt{4C/M}$ at the zone boundaries $K=\pm\pi/a$, with zero slope ($d\omega/dK=0$) at both $K=0$ and $K=\pm\pi/a$.

3. Part (c) — waves at the zone boundary. Differentiating, the group velocity is $v_g=d\omega/dK=\tfrac{a}{2}\sqrt{4C/M}\,\cos(Ka/2)\,\text{sgn}[\sin(Ka/2)]$, which vanishes exactly at $K=\pm\pi/a$ (where $\cos(Ka/2)=\cos(\pi/2)=0$). Zero group velocity means the wave carries no net energy flux: the $+K$ and $-K$ travelling waves at the zone edge are Bragg-reflected into one another ($\Delta K=G=2\pi/a$, eq. 7) and combine into a standing wave, with alternate lattice planes oscillating exactly out of phase, $u_s\propto(-1)^s$.