Retarded fields and radiation — Examples

Smoothness and speed (a). The map $s\mapsto c^2{s}^2+a^2$ is a polynomial, hence $C^{\mathrm{\infty }}$, with values in $[a^2,\mathrm{\infty })\subseteq (0,\mathrm{\infty })$; by [L1] the square root is $C^{\mathrm{\infty }}$ on $(0,\mathrm{\infty })$, so the composite $s\mapsto \sqrt{c^2{s}^2+a^2}$ is $C^{\mathrm{\infty }}$ and $z$ is $C^{\mathrm{\infty }}$ into ${\mathbb{R}}^3$ [L2]. Its derivative is $z^{\mathrm{\prime }}(s)=(c^{2}s\mathrm{/}\sqrt{c^2{s}^2+a^2},0,0)$ by the chain rule [L1, L2], so $\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }=c^2\mathrm{\mid }s\mathrm{\mid }\mathrm{/}\sqrt{c^2{s}^2+a^2}$ [L2]. If $s=0$ this is $0<c$; if $s\mathrm{\ne }0$, dividing numerator and denominator by $c\mathrm{\mid }s\mathrm{\mid }>0$ gives $\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }=c\mathrm{/}\sqrt{1+a^2\mathrm{/}(c^2{s}^2)}<c$ because $a^2\mathrm{/}(c^2{s}^2)>0$ [L3]. Hence $\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }<c$ for every $s$.

No uniform bound (a, continued). For $s>0$, $\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }=c\mathrm{/}\sqrt{1+a^2\mathrm{/}(c^2{s}^2)}\to c$ as $s\to \mathrm{\infty }$, since $a^2\mathrm{/}(c^2{s}^2)\to 0$ and the square root and reciprocal are continuous on their domains [L1]; the same limit holds as $s\to -\mathrm{\infty }$ by symmetry. Consequently, for every $v_{\ast }<c$ there is $s$ with $\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }>v_{\ast }$: choose $s$ so large that $\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }>v_{\ast }$, possible because the values approach $c$ from below and $v_{\ast }<c$; thus no $v_{\ast }<c$ bounds $\mathrm{\mid }{z}^{\mathrm{\prime }}\mathrm{\mid }$ on all of $\mathbb{R}$.

No retarded root at $(0,0)$ (b) and scope. At $(x,t)=(0,0)$ the equation $t-s=\mathrm{\mid }x-z(s)\mathrm{\mid }\mathrm{/}c$ of [L4] reads $-s=\mathrm{\mid }z(s)\mathrm{\mid }\mathrm{/}c=\sqrt{c^2{s}^2+a^2}\mathrm{/}c$, and the right-hand side is $\ge a\mathrm{/}c>0$; hence any solution would satisfy $s<0$, so both sides are nonnegative and squaring is equivalent. Squaring gives $s^2=(c^2{s}^2+a^2)\mathrm{/}{c}^2$, that is $c^2{s}^2=c^2{s}^2+a^2$ and then $a^2=0$, contradicting $a>0$; therefore no solution $s\in \mathbb{R}$ exists at this event. Since $z$ is pointwise subluminal by step 1.1 but carries no uniform bound by step 2.1, this exhibits a smooth curve outside the hypotheses of the root lemma [L4] for which the retarded equation fails: the uniform bound cannot be dropped, and no theorem is contradicted because the curve is not in its hypothesis class. Clause (c) records that the curve is a mathematical object, not the worldline of a physical charge, and that no measurement is reported.

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.

The radius ODE (a, first part). Along $t\mapsto r(t)$, the orbital energy satisfies $\frac{d}{dt}{E}_{\mathrm{o}\mathrm{r}\mathrm{b}}(r(t))=\frac{d}{dt}(-\frac{\kappa }{2r(t)})=\frac{\kappa }{2}\frac{r^{\mathrm{\prime }}(t)}{r(t{)}^2}$ by the power rule and the chain rule [F3]. Equating this to the adopted balance in (H), $\frac{\kappa r^{\mathrm{\prime }}}{2r^2}=-\frac{q^2{a}^2}{6\pi {\epsilon }_0{c}^3}$, and substituting the centripetal acceleration $a=\kappa \mathrm{/}(m{r}^2)$ gives $\frac{\kappa r^{\mathrm{\prime }}}{2r^2}=-\frac{q^2{\kappa }^2}{6\pi {\epsilon }_0{c}^3{m}^2{r}^4}$, hence $r^{\mathrm{\prime }}=-\frac{q^2\kappa }{3\pi {\epsilon }_0{m}^2{c}^3{r}^2}=-\frac{A}{r^2}$, which is the displayed ODE; here $\kappa >0$ and $r>0$ legitimize the divisions [F1].

The cubic (a, second part). Multiplying the ODE of step 1.1 by $3r^2$ gives $\frac{d}{dt}(r(t{)}^3)=3r^2{r}^{\mathrm{\prime }}=-3A$, a constant right-hand side; applying the fundamental theorem on any interval containing $t_0$ on which $r>0$ [F3] yields $r(t{)}^3-r(t_0{)}^3=-3A(t-t_0)$, that is the displayed cubic. Conversely, differentiating the cubic on that interval recovers the ODE, so the two descriptions are equivalent within the reduced model. This is (a).

Monotone decrease and formal maximal continuation (b). Since $A>0$ and $r>0$, $r^{\mathrm{\prime }}=-A\mathrm{/}{r}^2<0$. For $t_1<t_2$ in $J_t$, step 2.1 gives $r(t_2{)}^3-r(t_1{)}^3=-3A(t_2-t_1)<0$, and positivity implies $r(t_2)<r(t_1)$. Put $g(t):=r_0^3-3A(t-t_0)$, which is positive exactly for $t<t_{\ast }:=t_0+r_0^3\mathrm{/}(3A)$. The positive real-power function $\widehat{r}=g^{1\mathrm{/}3}$ is $C^1$ on $(-\mathrm{\infty },t_{\ast })$ by [F3] and has derivative $-A{g}^{-2\mathrm{/}3}=-A\mathrm{/}{\widehat{r}}^2$. The cubic identity forces $J_t\subset (-\mathrm{\infty },t_{\ast })$ and $r=\widehat{r}$ there: positive cubes are injective since $x^3-y^3=(x-y)(x^2+xy+y^2)$ and the second factor is positive. Every positive solution through $(t_0,r_0)$ satisfies the same cubic by step 2.1, so this is the unique maximal positive solution of the scalar balance ODE; it cannot extend positively across $t_{\ast }$ where $g=0$. As $t\uparrow t_{\ast }$, given $\epsilon >0$, $0<g(t)<{\epsilon }^3$ eventually, whence $0<\widehat{r}(t)<\epsilon$. This formal limit belongs to the extended scalar ODE, not to an arbitrary original validity interval, and no negative radius or positive solution at $t_{\ast }$ is asserted.

Scope and qualifications (c). The deduction is conditional on (H) throughout: the balance is the adopted premise, and the Larmor expression is only the prescribed-source rest-event emitted power of [F2], converted into a mechanical loss rate solely by the assumption (H) and not by any derived equation of motion. The quantified low-speed bound $v\mathrm{/}c\le {\delta }_v$ in (H) fails along the formal continuation as $r\to 0$ because $v^2=\kappa \mathrm{/}(mr)\to +\mathrm{\infty }$ with $\kappa >0$, so the model's own regime of validity excludes the collapse point: $t_{\ast }$ is a formal finite-time limit of the reduced model, not the predicted time at which a physical orbit reaches the centre. No coupled self-force dynamics, no lifetime derivation, no measurement, dataset, statistical model or uncertainty is introduced, and nothing here contradicts a theorem: the scalar balance is an independent hypothesis, and the prescribed-field radiation theorems supply no mechanical balance or coupled motion law. This is (c), and with steps 1.1-3.1 it completes the conditional deduction from (H).