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

For a wave on the atomic chain the phase velocity tells how a single crest moves, $V_p = \omega/k$, while the group velocity tells how a wave packet (energy) travels, $V_g = d\omega/dk$. Compute each from $\omega = (2\vartheta_s/a)\sin(ka/2)$ for small positive $k$.

Crest speed:
\[V_p = \frac{1}{k}\cdot\frac{2\vartheta_s}{a}\sin\!\left(\frac{ka}{2}\right).\]
Packet speed, differentiating the sine:
\[V_g = \frac{2\vartheta_s}{a}\cdot\frac{a}{2}\cos\!\left(\frac{ka}{2}\right) = \vartheta_s\cos\!\left(\frac{ka}{2}\right).\]

The long wavelength limit is $k \to 0$. Then the argument $ka/2$ is tiny, so $\sin(ka/2) \approx ka/2$ and $\cos(ka/2) \approx 1$. Substituting:
\[V_p \approx \frac{2\vartheta_s}{a k}\cdot\frac{ka}{2} = \vartheta_s, \qquad V_g \approx \vartheta_s.\]
The two coincide, which is exactly the acoustic (sound-like) regime where long waves travel without dispersion at the speed $\vartheta_s$. \[\boxed{V_p = V_g}\]