Question 3 of 7: Lattice Vibrations — Monatomic-Basis Dispersion Relation and Group Velocity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — Phys-A6: Solid State Physics — May 2015 (3 hours; closed book; useful equations and physical constants annexed to the paper). Any FIVE of the SEVEN questions constitute a complete exam paper (first five as they appear in the answer book are marked); all seven are solved below as a complete study resource.
Reference texts: C. Kittel, Introduction to Solid State Physics, 8th ed. (Ch. 1–3, 4, 6, 8, 14, 18); N. W. Ashcroft & N. D. Mermin, Solid State Physics, 1st ed. (Ch. 2, 4–7, 22, 28, 31–32).
Question 3: Lattice Vibrations — Monatomic-Basis Dispersion Relation and Group Velocity (20 marks)
Given. Equation of motion (paper's own, from Newton's law on plane $s$ with only nearest-plane coupling): $M\omega^2=-C\left[e^{iKa}+e^{-iKa}-2\right]$.
Find. (a) $\omega(K)$ in closed form, plotted over the first Brillouin zone; (b) $v_g$ at the zone edge and for $Ka\ll1$, with physical interpretation.
Fig. 3 — the monatomic-basis dispersion relation $\omega(K)=\sqrt{4C/M}\,|\sin(Ka/2)|$ over the first Brillouin zone $-\pi/a\le K\le\pi/a$: it is the single acoustic branch (no optical branch exists with only one atom per basis), flattening ($v_g\to0$) at both zone edges.
Approach. (a) rewrite the given equation using the paper's own identity (2)–(3) for $\cos\theta$ to reduce the exponentials to a $\sin^2$ form; (b) differentiate the closed-form $\omega(K)$ and evaluate the two limits.
Part (a) — reducing to the closed form. Using the paper's eq. (2), $e^{iKa}+e^{-iKa}=2\cos(Ka)$, so
$$M\omega^2=-C[2\cos(Ka)-2]=2C[1-\cos(Ka)]$$
Using the paper's eq. (1), $1-\cos(Ka)=2\sin^2(Ka/2)$, this becomes
$$M\omega^2=4C\sin^2(Ka/2)\ \Longrightarrow\ \omega^2=\frac{4C}{M}\sin^2(Ka/2)$$
Taking the positive square root (frequency is non-negative, and $\sin(Ka/2)$ can be negative over the zone, hence the absolute value):
$$\boxed{\omega(K)=\sqrt{\frac{4C}{M}}\,\bigl|\sin(Ka/2)\bigr|}$$
Over the first Brillouin zone $-\pi/a\le K\le\pi/a$, $Ka/2$ ranges over $[-\pi/2,\pi/2]$, so $\omega$ rises smoothly from $0$ at $K=0$ (uniform translation, no restoring force) to its maximum $\omega_{\max}=\sqrt{4C/M}$ at the zone boundary $K=\pm\pi/a$ (plotted above) — a single branch since a monatomic basis has only one atom to displace per cell (no optical branch).
Part (b) — group velocity at the two limits. Differentiating the closed form for $K>0$ (where $\sin(Ka/2)\ge0$),
$$v_g=\frac{d\omega}{dK}=\sqrt{\frac{4C}{M}}\cdot\frac a2\cos(Ka/2)=a\sqrt{\frac CM}\,\cos(Ka/2)$$
At the zone edge ($K=\pi/a$): $\cos(Ka/2)=\cos(\pi/2)=0$, so
$$\boxed{v_g(K=\pi/a)=0}$$
The dispersion curve is flat (horizontal tangent) at the zone boundary: the wave there is a standing wave (successive planes vibrate exactly out of phase, $e^{i\pi}=-1$), carrying no net energy flux — a Bragg-reflection condition, the 1-D analogue of a wave being reflected back on itself.
For long wavelengths ($Ka\ll1$, using the paper's eq. (3) small-angle form $\cos\theta\approx1$):
$$\boxed{v_g(Ka\ll1)=a\sqrt{\frac CM}}$$
a CONSTANT, independent of $K$ (and hence of frequency) — this is exactly the ordinary speed of sound in the medium: long-wavelength lattice vibrations are non-dispersive elastic (acoustic) waves, all travelling at the same speed regardless of pitch, just as in a continuous elastic solid.
