Harmonic dipole pattern and period-averaged power in the exact dipole law

Leading fields (a). By [L3], $\frac{d}{du}\cos (\omega u)=-\omega \sin (\omega u)$ and $\frac{d^2}{d{u}^2}\cos (\omega u)=-{\omega }^2\cos (\omega u)$; hence $p^{\mathrm{\prime }}(u)=-p_0\omega \sin (\omega u)e_3$ and $p^{\mathrm{\prime }\mathrm{\prime }}(u)=-{\omega }^2{p}_0\cos (\omega u)e_3$ [L4]. Substituting into the leading field of [L1] with $V=p^{\mathrm{\prime }\mathrm{\prime }}$ gives $E_{\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{d}}={\mu }_{0}n\times (n\times p^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}(4\pi r)=-\frac{{\mu }_0{\omega }^2{p}_0\cos (\omega u)}{4\pi r}n\times (n\times e_3)$ and $B_{\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{d}}=n\times E_{\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{d}}\mathrm{/}c$. For a unit vector $n$, $\mathrm{\mid }n\times (n\times e_3)\mathrm{\mid }=\mathrm{\mid }n\times e_3\mathrm{\mid }=(1-(n\cdot e_3{)}^2{)}^{1\mathrm{/}2}=\sin \theta$, which vanishes exactly when $n=\pm e_3$, i.e. $\theta \in \{0,\pi \}$, and is maximal for $\theta =\pi \mathrm{/}2$ [L3, L6]. This is (a).

Angular power density (b). By [L1] the angular density limit of the dipole law is ${\mu }_0\mathrm{\mid }n\times V{\mathrm{\mid }}^2\mathrm{/}(16{\pi }^{2}c)$; substituting $V=p^{\mathrm{\prime }\mathrm{\prime }}=-{\omega }^2{p}_0\cos (\omega u)e_3$ gives $\mathrm{\mid }n\times V{\mathrm{\mid }}^2={\omega }^4{p}_0^2{\cos }^2(\omega u)\mathrm{\mid }n\times e_3{\mathrm{\mid }}^2={\omega }^4{p}_0^2{\cos }^2(\omega u){\sin }^2\theta$ [step 1.1, L3]. Hence $\frac{dP}{d\sigma }=\frac{{\mu }_0{\omega }^4{p}_0^2{\cos }^2(\omega u)}{16{\pi }^{2}c}{\sin }^2\theta$, which vanishes on the axis and is maximal in the equatorial plane. Integrating with [L2], ${\int }_{S^2}{\sin }^2\theta d\sigma =\frac{8\pi }{3}$, so ${\int }_{S^2}\frac{dP}{d\sigma }d\sigma =\frac{{\mu }_0{\omega }^4{p}_0^2{\cos }^2(\omega u)}{16{\pi }^{2}c}\cdot \frac{8\pi }{3}=\frac{{\mu }_0{\omega }^4{p}_0^2{\cos }^2(\omega u)}{6\pi c}$. This is (b).

Period average (c). By [L3], ${\cos }^2(\omega u)=\frac{1+\cos (2\omega u)}{2}$, whose antiderivative $\frac{u}{2}+\frac{\sin (2\omega u)}{4\omega }$ has endpoint increment $\pi \mathrm{/}\omega$ over $[0,2\pi \mathrm{/}\omega ]$; hence ${\int }_0^{2\pi \mathrm{/}\omega }{\cos }^2(\omega u)du=\frac{1}{2}\cdot \frac{2\pi }{\omega }=\frac{\pi }{\omega }$ by [L4]. Multiplying the total power of step 2.1 by $\omega \mathrm{/}(2\pi )$ gives the period mean $\frac{{\mu }_0{\omega }^4{p}_0^2}{6\pi c}\cdot \frac{1}{2}=\frac{{\mu }_0{\omega }^4{p}_0^2}{12\pi c}$, the average being taken in the retarded-time variable as stated. This is (c).

Realization, scope and error control (d),(e). Choose $\eta$ by the compact cutoff lemma of [L5], and choose a nonnegative spatial cutoff equal to one near the origin and supported in $B_a$. Its integral is positive (it is one on a ball) and finite (it is bounded with compact support); dividing by that integral gives the smooth unit-mass $\sigma$ of [L5]. With these $\eta ,\sigma$, the functions $\rho (y,u)=-p_{\eta }(u)\cdot \mathrm{\nabla }\sigma (y)$ and $J(y,u)=p_{\eta }^{\mathrm{\prime }}(u)\sigma (y)$ are $C^{\mathrm{\infty }}$ and compactly supported in a set $B_a\times [t_0,T]$, and ${\mathrm{\partial }}_u\rho +\mathrm{div}J=-p_{\eta }^{\mathrm{\prime }}\cdot \mathrm{\nabla }\sigma +p_{\eta }^{\mathrm{\prime }}\cdot \mathrm{\nabla }\sigma =0$; the total charge is $\int \rho dy=-p_{\eta }\cdot \int \mathrm{\nabla }\sigma dy=0$, and the moment is $p_i(u)=\int y_i\rho dy=-{\sum }_j{p}_{\eta ,j}(u)\int y_i{\mathrm{\partial }}_j\sigma dy=-p_{\eta ,i}(u)(-\int \sigma dy)=p_{\eta ,i}(u)$ by [L5]. On the selected period $p_{\eta }=p_0\cos (\omega u)e_3$, so that source realizes the harmonic moment there, while by [L1] its actual retarded fields are those of the compact smooth branch constructed in [L6] and carry the full $D(u)$, ${\mathrm{\Gamma }}_E\mathrm{/}{r}^2$ and $({\mathrm{\Gamma }}_B+{\mathrm{\Gamma }}_E\mathrm{/}c)\mathrm{/}{r}^2$ errors around the ideal dipole expressions; the asymptotic total-power deviation from the ideal law is bounded by $R(u)$ of [L1] and its period-averaged absolute deviation is at most the average of $\mathrm{\mid }R(u)\mathrm{\mid }$, by the integral triangle inequality [L7]. To prove the non-equivalence in (e), use a separate compact-source witness. Choose a nonnegative even smooth unit-mass bump $\sigma$ supported in $B_a$: symmetrize any such bump as $({\sigma }_0(y)+{\sigma }_0(-y))\mathrm{/}2$, so substitution $y\mapsto -y$ gives $\int y\sigma (y)dy=0$. Choose a compact time cutoff $\chi =1$ on a neighbourhood of $[-a\mathrm{/}c,a\mathrm{/}c]$, vectors $b,d\in {\mathbb{R}}^3$ with $d\mathrm{\ne }0$ (in C m s${}^{-2}$ and C m s${}^{-3}$ respectively), and $\tilde{p}(v)=\chi (v)(b{v}^2\mathrm{/}2+d{v}^3\mathrm{/}6)$. Set $\tilde{\rho }=-\tilde{p}\cdot \mathrm{\nabla }\sigma$ and $\tilde{J}={\tilde{p}}^{\mathrm{\prime }}\sigma$. The same continuity and integration-by-parts computation gives moment $\tilde{p}$. At $u=0$, on the whole retarded window, ${\tilde{J}}_t(y,v)=(b+dv)\sigma (y)$ and ${\tilde{J}}_{tt}(y,v)=d\sigma (y)$. Hence for every $n\in S^2$, $\tilde{V}(n,0)=\int [b+d(n\cdot y)\mathrm{/}c]\sigma (y)dy=b={\tilde{p}}^{\mathrm{\prime }\mathrm{\prime }}(0),\tilde{D}(0)=\frac{a}{c}\mathrm{\mid }d\mathrm{\mid }\int \sigma (y)dy=\frac{a}{c}\mathrm{\mid }d\mathrm{\mid }>0.$ This is a compact smooth conserved counterexample to $V=p^{\mathrm{\prime }\mathrm{\prime }}\Rightarrow D=0$, independent of the harmonic realization. Conversely $D=0\Rightarrow V=p^{\mathrm{\prime }\mathrm{\prime }}$ follows from [L1]. Clause (d) records the scope: no feed, matching, radiation-resistance, measurement or observation-time power statement is made, and no eternal nonzero harmonic moment is asserted for the compact-time source.