Solution

For a real propagating positive-frequency branch, require

ω^{2} > a^{2}

and

c \neq = 0

. If

a

is the usual nonnegative cutoff, the source condition

ω > a

gives this branch. With an arbitrary real negative

a

, that inequality alone is insufficient:

a = - 2

and

ω = 1

satisfy it but make the displayed square root imaginary.

On the positive-frequency real branch, the phase velocity

v_{P} = ω / k

gives

v_{P} = \frac{c ω}{ω ^{2} - a ^{2}}, k = \frac{ω ^{2} - a ^{2}}{c} .

Differentiate the dispersion relation to obtain the group velocity:

\frac{d k}{d ω} = \frac{ω}{c ω ^{2} - a ^{2}}, v_{G} = \frac{d ω}{d k} = \frac{c ω ^{2} - a ^{2}}{ω} .

Therefore

v_{G} = \frac{c ^{2}}{v _{P}}, v_{P} v_{G} = c^{2} .

For

c > 0

and a nonzero cutoff,

v_{P} > c

while

v_{G} < c

; when

a = 0

, both equal

c

. The constant

c

is the velocity scale in this assumed medium, and need not independently be identified with the vacuum speed of light. A group velocity describes a narrow-band wave packet on the chosen differentiable branch, rather than automatically fixing a signal front velocity in every dispersive model.

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