Question 8.22: With reference to the wavepacket expressed by Eq. (8-173), p......
With reference to the wavepacket expressed by Eq. (8-173), propagating in free space, find
(a) \mathscr{H}( z, t) , and
(b) \left\langle \pmb{S} \right\rangle .
\mathscr{E}( z, t) = \pmb{a}_{x} E_{o} \cos (\omega^{-}t – \beta ^{-}z) + \pmb{a}_{x} E_{o} \cos (\omega^{+}t – \beta ^{+}z) \\ \quad \quad \quad = \pmb{a}_{x} 2 E_{o} \cos (\Delta \omega t – \Delta \beta z) \cos (\omega _{o} t – \beta _{o} z) (8-173)
(a) With the help of the relations \left|\mathscr{H}\right| = \left|\mathscr{E}\right| \sqrt{\varepsilon _{o} / \mu _{o}} and \pmb{a}_{k} = \pmb{a}_{E}\times \pmb{a}_{H}, we obtain
\mathscr{H} ( z, t) = \pmb{a}_{y} 2 E_{o} \sqrt{\varepsilon _{o}/ \mu _{o}} \cos (\Delta \omega t – \Delta \beta z) \cos (\omega _{o} t – \beta _{o} z) (8-177)
The temporal period of the phase is given by \tau _{1} = 2\pi / \omega _{o}, while that of the envelope is given by \tau _{2} = 2\pi / \Delta \omega , where \tau _{1} \ll \tau _{2}.
(b) Poynting vector is
\pmb{S} = \mathscr{E} \times \mathscr{H} \\ \quad =\pmb{a}_{z} 4 E^{2}_{o} \sqrt{\varepsilon _{o}/ \mu _{o}} \cos^{2} (\Delta \omega t – \Delta \beta z) \cos^{2} (\omega _{o} t – \beta _{o} z) (8-178)
Time-average power density is
\left\langle\pmb{S}\right\rangle = \frac{1}{T} \int_{-T/2}^{T/2}{\pmb{S}dt} : \tau _{1} \ll T \ll \tau _{2} (8-179)
Inserting Eq. (8-178) into Eq. (8-179), we obtain
\left\langle\pmb{S}\right\rangle = \pmb{a}_{z} 2 E^{2}_{o} \sqrt{\varepsilon _{o}/ \mu _{o}} \cos^{2} (\Delta \omega t – \Delta \beta z) (8-180)
It is evident from Eq. (8-180) that the time-average power density of the wavepacket propagates with the group velocity Δω / Δβ .