Question

The dispersion relation for a one-dimensional monoatomic lattice chain is given by the equation \[\omega = \frac{2}{a}\,\vartheta_s\left|\sin\!\left(\frac{ka}{2}\right)\right|,\] where \(a\) is the interatomic spacing, \(k = \dfrac{2\pi}{\lambda}\), and \(\vartheta_s\) has the dimension of velocity. The relation between the phase velocity \(V_p\) and group velocity \(V_g\) in the long wavelength limit is given by:

• A. \(V_p = V_g\)
• B. \(V_p = 2 V_g\)
• C. \(V_p = V_g/2\)
• D. \(V_p = 4 V_g\)

Hint:

Compute \(V_p=\omega/k\) and \(V_g=d\omega/dk\), then let \(k\to 0\) using \(\sin x\approx x\), \(\cos x\approx 1\).

Answer 1

The Correct Option is A

Step 1: The phase velocity is \(V_p = \omega/k\) and the group velocity is \(V_g = d\omega/dk\). Use the given dispersion relation (taking \(k > 0\) so the sine is positive): \[\omega = \frac{2\vartheta_s}{a}\sin\!\left(\frac{ka}{2}\right).\]
Step 2: Phase velocity: \[V_p = \frac{\omega}{k} = \frac{2\vartheta_s}{ak}\sin\!\left(\frac{ka}{2}\right).\]
Step 3: Group velocity by differentiating: \[V_g = \frac{d\omega}{dk} = \frac{2\vartheta_s}{a}\cdot\frac{a}{2}\cos\!\left(\frac{ka}{2}\right) = \vartheta_s\cos\!\left(\frac{ka}{2}\right).\]
Step 4: Long wavelength means \(\lambda \to \infty\), so \(k \to 0\) and \(ka/2 \to 0\). Use \(\sin x \approx x\) and \(\cos x \approx 1\): \[V_p \to \frac{2\vartheta_s}{ak}\cdot\frac{ka}{2} = \vartheta_s, \qquad V_g \to \vartheta_s\cdot 1 = \vartheta_s.\]
Step 5: Both velocities equal \(\vartheta_s\), so the medium is non-dispersive in this limit. \[\boxed{V_p = V_g}\]