Use the result of Prob. 10.34 to determine the power radiated by an ideal electric dipole, p(t), at the origin. Check that your answer is consistent with Eq. 11.22, in the case of sinusoidal time dependence, and with Prob. 11.26, in the case of quadratic time dependence
\langle P\rangle=\int\langle S \rangle \cdot d a =\frac{\mu_{0} p_{0}^{2} \omega^{4}}{32 \pi^{2} c} \int \frac{\sin ^{2} \theta}{r^{2}} r^{2} \sin \theta d \theta d \phi=\frac{\mu_{0} p_{0}^{2} \omega^{4}}{12 \pi c} (11.22)
The fields of a nonstatic ideal dipole (Problem 10.34) contain terms that go like 1 / r, 1 / r^{2}, \text { and } 1 / r^{3} , for fixed t_{0} \equiv t-r / c ; radiation comes from the first of these:
E _{r}=-\frac{\mu_{0}}{4 \pi r}[\ddot{ p }-\hat{ r }(\hat{ r } \cdot \ddot{ p })], \quad B _{r}=-\frac{\mu_{0}}{4 \pi r c}(\hat{ r } \times \ddot{ p })where the dipole moments are evaluated at timet_{0} \text {. The } 1 / r^{2} term in the Poynting vector is
S _{r}=\frac{1}{\mu_{0}}\left( E _{r} \times B _{r}\right)=\frac{\mu_{0}}{16 \pi^{2} c r^{2}}[\ddot{ p }-\hat{ r }(\hat{ r } \cdot \ddot{ p })] \times(\hat{ r } \times \ddot{ p })=\frac{\mu_{0}}{16 \pi^{2} c r^{2}}\{\ddot{ p } \times(\hat{ r } \times \ddot{ p })-(\hat{ r } \cdot \ddot{ p })[\hat{ r } \times(\hat{ r } \times \ddot{ p })]\}
=\frac{\mu_{0}}{16 \pi^{2} c r^{2}}\left\{\hat{ r } \ddot{p}^{2}-\ddot{ p }(\hat{ r } \cdot \ddot{ p })-(\hat{ r } \cdot \ddot{ p })[\hat{ r }(\hat{ r } \cdot \ddot{ p })-\ddot{ p }]\right\}=\frac{\mu_{0}}{16 \pi^{2} c r^{2}}\left[\ddot{p}^{2}-(\hat{ r } \cdot \ddot{ p })^{2}\right] \hat{ r }.
Setting the z axis along \ddot{ p },\left[\ddot{p}^{2}-(\hat{ r } \cdot \ddot{ p })^{2}\right]=\ddot{p}^{2} \sin ^{2} \theta , and the power radiated is
P=\oint S _{r} \cdot d a =\frac{\mu_{0}}{16 \pi^{2} c} \ddot{p}^{2} \oint \frac{1}{r^{2}} \sin ^{2} \theta r^{2} \sin \theta d \theta d \phi=\frac{\mu_{0}}{8 \pi c} \ddot{p}^{2} \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{\mu_{0}}{6 \pi c} \ddot{p}^{2}.
For the sinusoidal case, p = p _{0} \cos \omega t, \text { we have } \ddot{ p }=-\omega^{2} p _{0} \cos \omega t, \ddot{p}^{2}=\omega^{4} p_{0}^{2} \cos ^{2} \omega t , and its time average is (1 / 2) p_{0}^{2} \omega^{4} , so
\langle P\rangle=\frac{\mu_{0}}{12 \pi c} \ddot{p}_{0}^{2} \omega^{4}in agreement with Eq. 11.22. And in the case of quadratic time dependence our result agrees with the answer to Problem 11.26.