 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021/11/v/solution

Apply the dispersion relation separately to the two components and subtract:

m (ω_{0} + δ ω) = (k_{0} + δ k)^{2}, m (ω_{0} - δ ω) = (k_{0} - δ k)^{2} ⟹ \frac{δ ω}{δ k} = \frac{2 k _{0}}{m} .

In the narrow-band limit this is the group velocity, since

ω (k) = k^{2} / m

gives

v_{g} = \frac{d ω}{d k} = \frac{2 k}{m} = 2 v_{p} .

Equivalently on the positive branch

k = mω

d k / d ω = m / (2 k)

and its reciprocal is

2 k / m

. The factor of two applies to signed velocities on either nonzero propagation branch.

There is a small but useful finite-band qualification. Adding the two component relations gives

m ω_{0} = k_{0}^{2} + (δ k)^{2}

. Thus the carrier formed by their exact midpoints has phase velocity

ω_{0} / k_{0} = k_{0} / m + (δ k)^{2} / (m k_{0})

, and its finite-band envelope-to-carrier velocity ratio is

\frac{δ ω / δ k}{ω _{0} / k _{0}} = \frac{2 k _{0}^{2}}{k _{0}^{2} + ( δ k ) ^{2}} .

The exact derivative relation

v_{g} = 2 v_{p}

refers to a single central point on the dispersion relation; the two-tone midpoint ratio approaches two as

δ k / k_{0} \to 0

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