Instantaneous-rest Larmor power and uniform low-speed comparison for a prescribed point charge

Field on the emitted sphere (a). At the rest event $z^{\mathrm{\prime }}(s_0)=0$, so at the emitted sphere the retarded data are $\beta =0$, $\kappa =1$, $\dot{\beta }=a\mathrm{/}c$ [F2]; [F1] then gives $E_{\mathrm{v}\mathrm{e}\mathrm{l}}=C(1-0)(n-0)\mathrm{/}(1\cdot r^2)=Cn\mathrm{/}{r}^2$ and $E_{\mathrm{a}\mathrm{c}\mathrm{c}}=Cn\times ((n-0)\times a\mathrm{/}c)\mathrm{/}(c\cdot 1\cdot r)=Cn\times (n\times a)\mathrm{/}(c^{2}r)$, together with $B=n\times E\mathrm{/}c$. Writing ${\rho }_E:=Cn\mathrm{/}{r}^2$ gives $E(x_r,t_r)=Cn\times (n\times a)\mathrm{/}(c^{2}r)+{\rho }_E$ and $\mathrm{\mid }{\rho }_E\mathrm{\mid }=\mathrm{\mid }C\mathrm{\mid }\mathrm{/}{r}^2\le \mathrm{\mid }C\mathrm{\mid }(1+v_{\ast }\mathrm{/}c)(1-v_{\ast }\mathrm{/}c{)}^{-3}{r}^{-2}$, because both factors in the last expression are at least $1$. This is (a).

Angular flux limit (b). At the rest event the remainder is exactly ${\rho }_E=Cn\mathrm{/}{r}^2$ by step 1.1, so $n\times {\rho }_E=0$ for either sign of $C$. The identity $n\cdot (E\times (n\times E))=\mathrm{\mid }n\times E{\mathrm{\mid }}^2$ and $B=n\times E\mathrm{/}c$ give $n\cdot S=\mathrm{\mid }n\times E{\mathrm{\mid }}^2\mathrm{/}({\mu }_{0}c)$. Thus $\mathrm{\mid }n\times E\mathrm{\mid }=\mathrm{\mid }C\mathrm{\mid }\mathrm{\mid }n\times a\mathrm{\mid }\mathrm{/}(c^{2}r)$ and, for every $r>0$, $r^{2}n\cdot S(x_r,t_r)=\frac{C^2\mathrm{\mid }n\times a{\mathrm{\mid }}^2}{{\mu }_0{c}^5}=\frac{C^2\mathrm{\mid }n\times (n\times a){\mathrm{\mid }}^2}{{\mu }_0{c}^5}\mathrm{.}$ The expression is independent of $r$ and continuous in $n$, so its limit is uniform, and its sphere integral is finite by [F3, F6]. In particular the integral limit equals the integral of this density without a sign-dependent remainder estimate. The measures agree by [F10]. This proves (b) for all real charges, including $q=0$.

Total power (c). By [F4], $\mathrm{\mid }n\times (n\times a){\mathrm{\mid }}^2=\mathrm{\mid }n\times a{\mathrm{\mid }}^2=\mathrm{\mid }a{\mathrm{\mid }}^2-(n\cdot a{)}^2$; integrating over $S^2$ with [F3] gives ${\int }_{S^2}\mathrm{\mid }n\times (n\times a){\mathrm{\mid }}^{2}d\sigma =4\pi \mathrm{\mid }a{\mathrm{\mid }}^2-(4\pi \mathrm{/}3)\mathrm{\mid }a{\mathrm{\mid }}^2=(8\pi \mathrm{/}3)\mathrm{\mid }a{\mathrm{\mid }}^2$. Substituting into step 2.1, $P_{\mathrm{e}\mathrm{m}}=\frac{C^2}{{\mu }_0{c}^5}\cdot \frac{8\pi }{3}\mathrm{\mid }a{\mathrm{\mid }}^2$; and $\frac{C^2}{{\mu }_0{c}^5}=\frac{q^2}{16{\pi }^2{\epsilon }_0^2{\mu }_0{c}^5}=\frac{q^2}{16{\pi }^2{\epsilon }_0{c}^3}$ by [F8] (using ${\epsilon }_0{\mu }_0{c}^2=1$), so $P_{\mathrm{e}\mathrm{m}}=\frac{q^2\mathrm{\mid }a{\mathrm{\mid }}^2}{6\pi {\epsilon }_0{c}^3}$. This is (c).

Angular distribution (d). For $a\mathrm{\ne }0$ use the uniquely defined angle of (d), whose existence follows from the intermediate value theorem between the stated cosine endpoints and whose uniqueness follows from strict monotonicity [F11]. By [F4], $\mathrm{\mid }n\times (n\times a){\mathrm{\mid }}^2=\mathrm{\mid }a{\mathrm{\mid }}^2-(n\cdot a{)}^2=\mathrm{\mid }a{\mathrm{\mid }}^2(1-{\cos }^2\theta )=\mathrm{\mid }a{\mathrm{\mid }}^2{\sin }^2\theta$; since ${\sin }^2\theta =0$ exactly for $\theta \in \{0,\pi \}$ [F11], when $q\mathrm{\ne }0$ the pattern vanishes precisely along $\pm a$ and is maximal where ${\sin }^2\theta =1$, i.e. in the plane perpendicular to $a$; its coefficient is $\frac{C^2}{{\mu }_0{c}^5}=\frac{q^2}{16{\pi }^2{\epsilon }_0{c}^3}$ [F8]. If $a=0$ or $q=0$, the displayed density is zero for every $n$, so $dP\mathrm{/}d\sigma \equiv 0$ and $P_{\mathrm{e}\mathrm{m}}=0$ by (c). Finally ${\int }_{S^2}\frac{dP}{d\sigma }d\sigma =P_{\mathrm{e}\mathrm{m}}(s_0)$ is (c). This is (d).

General far-zone limit and weighted power (f, first part). Now let $s_0$ be an arbitrary emission event with ${\beta }_0:=z^{\mathrm{\prime }}(s_0)\mathrm{/}c$, $\mathrm{\mid }{\beta }_0\mathrm{\mid }\le v_{\ast }\mathrm{/}c$, and $a:=z^{\mathrm{\prime }\mathrm{\prime }}(s_0)$; the same computation as steps 1.1-2.1 applies with ${\beta }_0$ in place of $0$ and ${\kappa }_0$ in place of $1$ [F2]. By [F1], $E_{\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{d}}:=Cn\times ((n-{\beta }_0)\times a)\mathrm{/}(c^2{\kappa }_0^{3}r)$ is the acceleration part and $E_{\mathrm{v}\mathrm{e}\mathrm{l}}$ is the remainder with $\mathrm{\mid }{E}_{\mathrm{v}\mathrm{e}\mathrm{l}}\mathrm{\mid }\le D_E{r}^{-2}$; writing $N_{\beta }:=n\times ((n-{\beta }_0)\times a)$ and repeating the flux identity of step 2.1, $r^{2}n\cdot S=\frac{C^2\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2}{{\mu }_0{c}^5{\kappa }_0^6}+R_r(n)$ with $\mathrm{\mid }{R}_r(n)\mathrm{\mid }\le K^{\mathrm{\prime }}{r}^{-1}$ for $r\ge 1$, uniformly in $n$ and with $K^{\mathrm{\prime }}$ independent of $n$; here $1-v_{\ast }\mathrm{/}c\le {\kappa }_0\le 1+v_{\ast }\mathrm{/}c$ keeps every power of ${\kappa }_0$ bounded on the compact $S^2$ [F2, F6]. Hence ${\kappa }_0(n)r^{2}n\cdot S\to H_{\beta }(n):=\frac{C^2\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2}{{\mu }_0{c}^5{\kappa }_0^5}$ uniformly, and the weighted emitted power $P_{\mathrm{e}\mathrm{m}}:={\int }_{S^2}{\kappa }_0{\lim }_{r\to \mathrm{\infty }}(r^{2}n\cdot S)d\sigma =\frac{C^2}{{\mu }_0{c}^5}{\int }_{S^2}\frac{\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2}{{\kappa }_0^5}d\sigma$ is the dominated-convergence limit of the weighted fluxes, the weight ${\kappa }_0$ being bounded and continuous [F6, F7].

Evaluation of the weighted angular integral (f, second part). By [F4], $N_{\beta }=(n\cdot a)(n-{\beta }_0)-(1-n\cdot {\beta }_0)a$, so with $b:=\mathrm{\mid }{\beta }_0\mathrm{\mid }\in [0,v_{\ast }\mathrm{/}c]$ and $u:=n\cdot {\beta }_0\mathrm{/}b$ (any unit vector $e_3$ along ${\beta }_0$ when $b>0$; the case $b=0$ is step 3.1) one gets $\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2=-(1-b^2)(n\cdot a{)}^2+2(1-bu)b{a}_{\parallel }(n\cdot a)+(1-bu{)}^2\mathrm{\mid }a{\mathrm{\mid }}^2,$ where $a_{\parallel }$ is the component of $a$ along ${\beta }_0$. Choose orthonormal spatial coordinates with third unit vector ${\beta }_0\mathrm{/}b$ when $b>0$ (when $b=0$, choose any standard unit vector), and put $a_{\parallel }=a\cdot e_3$ and $a_{\perp }=a-a_{\parallel }{e}_3$; this coordinate change preserves the polar measure by [F10], and $a_{\perp }^2$ below abbreviates $\mathrm{\mid }{a}_{\perp }{\mathrm{\mid }}^2$. Averaging the $\phi$-dependence over the sphere with [F3] (the linear term in the tangential component of $n$ integrates to $0$, and the square of that component integrates to $\pi \int (1-u^2)f$), the parallel coefficient of $\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2\mathrm{/}(1-bu{)}^5$ reduces to $1-u^2$ and the perpendicular coefficient to $(1-bu{)}^2-\frac{1-b^2}{2}(1-u^2)$. Hence, with $J_5:={\int }_{-1}^1(1-u^2)(1-bu{)}^{-5}du$ and $J_3:={\int }_{-1}^1(1-bu{)}^{-3}du$, ${\int }_{S^2}\frac{\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2}{{\kappa }_0^5}d\sigma =2\pi a_{\parallel }^2{J}_5+2\pi a_{\perp }^2[J_3-\frac{1-b^2}{2}{J}_5]\mathrm{.}$ The substitution $y=1-bu$ (for $b>0$, with $du=-dy\mathrm{/}b$) and the second fundamental theorem [F9], with the antiderivatives $-y^{-2}\mathrm{/}2$ and $b^{-3}[(1-b^2)y^{-4}\mathrm{/}4-2y^{-3}\mathrm{/}3+y^{-2}\mathrm{/}2]$, give $J_3=2\mathrm{/}(1-b^2{)}^2$ and $J_5=4\mathrm{/}(3(1-b^2{)}^3)$; at $b=0$ the same formulas follow by direct integration ($J_3=2$, $J_5=4\mathrm{/}3$) and all expressions are continuous in $b$, so no division by $b$ is needed in the conclusion. Therefore ${\int }_{S^2}\frac{\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2}{{\kappa }_0^5}d\sigma =\frac{8\pi }{3}[a_{\parallel }^2(1-b^2{)}^{-3}+a_{\perp }^2(1-b^2{)}^{-2}]=\frac{8\pi }{3}{\gamma }_0^6(\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }{\beta }_0\times a{\mathrm{\mid }}^2),$ because $(1-b^2{)}^{-1}={\gamma }_0^2$, $(1-b^2{)}^{-3}={\gamma }_0^6$ and $\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }{\beta }_0\times a{\mathrm{\mid }}^2=a_{\parallel }^2+(1-b^2)a_{\perp }^2$ by [F4]. Substituting into step 5.1 and converting the constant with [F8] gives $P_{\mathrm{e}\mathrm{m}}={\gamma }_0^6(\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }{\beta }_0\times a{\mathrm{\mid }}^2)\frac{q^2}{6\pi {\epsilon }_0{c}^3}$, the displayed general formula.

Uniform low-speed comparisons (f, third part). Write $P_{\mathrm{L}}=q^2\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}(6\pi {\epsilon }_0{c}^3)$ and $P_{\mathrm{e}\mathrm{m}}={\gamma }_0^6(a_{\parallel }^2+{\gamma }_0^{-2}{a}_{\perp }^2)\frac{q^2}{6\pi {\epsilon }_0{c}^3}$ [step 6.1]. Since ${\gamma }_0\ge 1$, one has ${\gamma }_0^6{a}_{\parallel }^2\ge a_{\parallel }^2$ and ${\gamma }_0^4{a}_{\perp }^2\ge a_{\perp }^2$, so $P_{\mathrm{e}\mathrm{m}}\ge P_{\mathrm{L}}$; and ${\gamma }_0^6{a}_{\parallel }^2+{\gamma }_0^4{a}_{\perp }^2\le {\gamma }_0^6\mathrm{\mid }a{\mathrm{\mid }}^2$, so $P_{\mathrm{e}\mathrm{m}}\le {\gamma }_0^6{P}_{\mathrm{L}}$. If $b=\mathrm{\mid }{\beta }_0\mathrm{\mid }\le \delta <1$ then ${\gamma }_0^2=(1-b^2{)}^{-1}\le (1-{\delta }^2{)}^{-1}$, hence $P_{\mathrm{L}}\le P_{\mathrm{e}\mathrm{m}}\le (1-{\delta }^2{)}^{-3}{P}_{\mathrm{L}}$. Moreover $P_{\mathrm{e}\mathrm{m}}-P_{\mathrm{L}}=[({\gamma }_0^6-1)a_{\parallel }^2+({\gamma }_0^4-1)a_{\perp }^2]\frac{q^2}{6\pi {\epsilon }_0{c}^3}\le ({\gamma }_0^6-1)P_{\mathrm{L}}\le [(1-{\delta }^2{)}^{-3}-1]P_{\mathrm{L}}\le 3{\delta }^2(1-{\delta }^2{)}^{-3}{P}_{\mathrm{L}}$, where the last step is the elementary inequality $(1-t{)}^{-3}-1\le 3t(1-t{)}^{-3}$ at $t={\delta }^2$; all these inequalities are valid also when $a=0$ or $q=0$, with both sides zero, and no division by $P_{\mathrm{L}}$ is performed. For $\delta =0$ the angular densities coincide identically. For $0<\delta <1$, fix $n$ and use coordinates with $n=e_3$: writing $p=(a_1,a_2)$, $h=({\beta }_{0,1},{\beta }_{0,2})$, $t=a_3$ and ${\beta }_{0,3}$, one has $N_0=-p$ and $N_{\beta }=-({\kappa }_{0}p+th)$ by [F4], so with $D:=\mathrm{\mid }{N}_{\beta }{\mathrm{\mid }}^2{\kappa }_0^{-5}-\mathrm{\mid }{N}_0{\mathrm{\mid }}^2$, $D=\mathrm{\mid }p{\mathrm{\mid }}^2({\kappa }_0^{-3}-1)+2{\kappa }_0^{-4}t(p\cdot h)+{\kappa }_0^{-5}{t}^2\mathrm{\mid }h{\mathrm{\mid }}^2\mathrm{.}$ Since $1-\delta \le {\kappa }_0\le 1+\delta$, the mean value theorem [F9] gives $\mathrm{\mid }{\kappa }_0^{-3}-1\mathrm{\mid }\le 3\delta (1-\delta {)}^{-4}$ and $\mathrm{\mid }{\kappa }_0^{-4}\mathrm{\mid }\le (1-\delta {)}^{-4}$, ${\kappa }_0^{-5}\le (1-\delta {)}^{-5}$; also $\mathrm{\mid }h\mathrm{\mid }\le \delta$ and $\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }t\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$, and $2\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }p\mathrm{\mid }\le \mathrm{\mid }t{\mathrm{\mid }}^2+\mathrm{\mid }p{\mathrm{\mid }}^2$. Hence $\mathrm{\mid }D\mathrm{\mid }\le \mathrm{\mid }a{\mathrm{\mid }}^2[4\delta (1-\delta {)}^{-4}+{\delta }^2(1-\delta {)}^{-5}]$, and this is at most $\mathrm{\mid }a{\mathrm{\mid }}^2\delta [(2+\delta )(1-\delta {)}^{-5}+5(1-\delta {)}^{-6}]$ because, after dividing by $\delta (1-\delta {)}^{-4}$, the inequality reduces to $4-3\delta \le 2+\delta +5(1-\delta {)}^{-1}$, whose left side is at most $4$ and whose right side exceeds $5$. Multiplying by $\frac{q^2}{16{\pi }^2{\epsilon }_0{c}^3}$ gives the displayed bound on $\mathrm{\mid }{H}_{\beta }-H_0\mathrm{\mid }$.

Scope and conclusion (e). At a rest event $z^{\mathrm{\prime }}(s_0)=0$ one has ${\kappa }_0(n)=1-n\cdot 0=1$ for every $n\in S^2$ [F2], so the emission-time Jacobian is identically $1$ and no retarded-time conversion factor appears: the limits of (b) are simultaneously the fixed-sphere flux and the emitted power per unit emission time, and the angular and total powers of (c)-(d) are the emitted quantities of the rest event. The computation used only the prescribed worldline, the retarded model fields and the adopted causal selection of Retarded point fields off the worldline: Lienard-Wiechert potentials, the velocity/acceleration split and the uniform far-zone remainder and Adopted retarded solution selection: the physical branch carries no incoming free field; it forms no product of distributions, asserts no self-force, radiation reaction, mass derivation or mechanical evolution law, and makes no empirical or measurement claim, while clause (f) records the general low-speed comparison as a separate quantified statement rather than a loss law. Steps 1.1-7.1 together with this clause prove (a)-(f).