Em Causal Response And Radiation Prerequisites — Examples

The kernel $K(s)=-e^{-s}$ is of class $C^1$ on $[0,\mathrm{\infty })$ with $K^{\mathrm{\prime }}=e^{-s}=-K$ by [F1] and [F6]; $K(s)<0$ for every $s$, and $\mathrm{\mid }K(s)\mathrm{\mid }=e^{-s}$, $\mathrm{\mid }{K}^{\mathrm{\prime }}(s)\mathrm{\mid }=e^{-s}$, $\mathrm{\mid }sK(s)\mathrm{\mid }=s{e}^{-s}$. For $N>0$, the functions $G(s)=-e^{-s}$ and $u(s)=s$ are smooth on $\mathbb{R}$, so their restrictions are continuous on $[0,N]$, differentiable there (relative to the interval at endpoints), and have continuous, hence Riemann-integrable derivatives. Thus [F2] applies to $G$ and, for integration by parts, to $u$ and $v=G$: it gives ${\int }_0^N{e}^{-s}ds=1-e^{-N}$ and ${\int }_0^{N}s{e}^{-s}ds=1-(1+N)e^{-N}$, and converts these to Lebesgue integrals; the nonnegative exponential series gives $e^N\ge N^2\mathrm{/}2$ by [F1], so $e^{-N}\to 0$ and $N{e}^{-N}\le 2\mathrm{/}N\to 0$; since the integrands are nonnegative, [F3] lets integer $N\to \mathrm{\infty }$, giving ${\int }_0^{\mathrm{\infty }}{e}^{-s}ds=1$ and ${\int }_0^{\mathrm{\infty }}s{e}^{-s}ds=1$. Hence $K,K^{\mathrm{\prime }},sK\in L^1(0,\mathrm{\infty })$ and, by [F5], $K$ is locally absolutely continuous on $[0,\mathrm{\infty })$.

A conserved smooth source with nonzero radiation. For the constants of clause (ii), put $u_{\ast }:=(t_0+T)\mathrm{/}2$, $b:=(T-t_0)\mathrm{/}8$ and $ϵ:=\min (a\mathrm{/}2,bc\mathrm{/}2)>0$. By [F13] choose $\psi :{\mathbb{R}}^3\to [0,1]$ smooth, equal to $1$ on ${\bar{B}}_{ϵ\mathrm{/}2}$ and supported in $B_{ϵ}$, and $\theta :\mathbb{R}\to [0,1]$ smooth, equal to $1$ on $[-b,b]$ and supported in $(-2b,2b)$. Define $f(v):=(v-u_{\ast }{)}^2\theta (v-u_{\ast })$, $\rho (y,v):=-{\mathrm{\partial }}_3\psi (y)f(v)$ and $\mathbf{J}(y,v):=(0,0,\psi (y)f^{\mathrm{\prime }}(v))$. These functions are smooth by [F13], their spatial support is contained in $B_{ϵ}\subset B_a$, and their temporal support is contained in $(u_{\ast }-2b,u_{\ast }+2b)\subset (t_0,T)$, so (H1) holds. Direct coordinate differentiation gives ${\rho }_t=-{\mathrm{\partial }}_3\psi f^{\mathrm{\prime }}$ and ${\mathrm{div}}_y\mathbf{J}={\mathrm{\partial }}_3\psi f^{\mathrm{\prime }}$, proving (H2). By [F14] and monotonicity, $C_{\psi }:={\int }_{B_a}\psi (y)dy$ satisfies $0<\lambda (B_{ϵ\mathrm{/}2})\le C_{\psi }\le \lambda (B_{ϵ})<\mathrm{\infty }$.

Therefore the integrable branch of clause (c) of [F9] applies to $K$: $P(t)={\int }_0^{\mathrm{\infty }}K(s)E(t-s)ds$ is a bounded causal linear response on bounded continuous histories by clause (a), $\chi (z)={\int }_0^{\mathrm{\infty }}K(s)e^{izs}ds$ is well defined and holomorphic on $\mathrm{Im}z>0$, continuous on the closed upper half-plane, and its boundary values satisfy the dispersion relations of clause (c).

Positive fixed-retarded-time flux. If $y\in \mathrm{supp}\psi$ and $n\in S^2$, then $\mathrm{\mid }n\cdot y\mathrm{\mid }\mathrm{/}c\le \mathrm{\mid }y\mathrm{\mid }\mathrm{/}c<ϵ\mathrm{/}c\le b\mathrm{/}2$, so $\theta =1$ near $u_{\ast }+n\cdot y\mathrm{/}c-u_{\ast }$ and $f^{\mathrm{\prime }\mathrm{\prime }}(u_{\ast }+n\cdot y\mathrm{/}c)=2$. Thus $\mathbf{V}(n,u_{\ast })=2C_{\psi }p$ for every $n$, where $p:=(0,0,1)$. Orthogonal invariance in [F14] makes the three integrals ${\int }_{S^2}{n}_j^{2}d\mathrm{\Omega }$ equal; their sum is ${\int }_{S^2}\mathrm{\mid }n{\mathrm{\mid }}^{2}d\mathrm{\Omega }=4\pi$, so each is $4\pi \mathrm{/}3$. The cross-product coordinates give $\mathrm{\mid }n\times p{\mathrm{\mid }}^2=n_1^2+n_2^2=1-n_3^2$, hence ${\int }_{S^2}\mathrm{\mid }n\times p{\mathrm{\mid }}^{2}d\mathrm{\Omega }=8\pi \mathrm{/}3$. Clause (c) of [F10] now gives the fixed-retarded-time limit $\frac{{\mu }_0}{16{\pi }^{2}c}4C_{\psi }^2\frac{8\pi }{3}=\frac{2{\mu }_0{C}_{\psi }^2}{3\pi c}>0$.

Computation of the transfer function: fix $z$ with $\mathrm{Im}z\ge 0$ and put $\lambda :=iz-1$, so $\lambda \mathrm{\ne }0$ and $\mathrm{\mid }{e}^{\lambda N}\mathrm{\mid }=e^{-N}{e}^{-N\mathrm{Im}z}\le e^{-N}\to 0$ by [F8]. For every $N>0$ the function $s\mapsto e^{\lambda s}$ is continuous, so its Lebesgue integral over $(0,N)$ agrees with its Riemann and parametric line integral over $[0,N]$ by [F2] and [F7], and the first fundamental theorem of [F7] applied to the primitive $F(s)=e^{\lambda s}\mathrm{/}\lambda$ gives ${\int }_0^N{e}^{\lambda s}ds=(e^{\lambda N}-1)\mathrm{/}\lambda$; the truncated integrands $s\mapsto -e^{\lambda s}{\mathbf{1}}_{(0,N)}(s)$ converge pointwise to $s\mapsto -e^{\lambda s}$ and are dominated by $e^{-s}\in L^1(0,\mathrm{\infty })$, so [F4] gives $\chi (z)={\int }_0^{\mathrm{\infty }}(-e^{\lambda s})ds=-{\lim }_{N\to \mathrm{\infty }}(e^{\lambda N}-1)\mathrm{/}\lambda =1\mathrm{/}\lambda =-1\mathrm{/}(1-iz)$, where for $\mathrm{Im}z>0$ the first equality is the defining integral of clause (c) of [F9] and for real $z$ it is its continuous boundary value.

Boundary values and the dissipative part: for real $\omega$, step 3.1 gives $\chi (\omega )=-1\mathrm{/}(1-i\omega )=-(1+i\omega )\mathrm{/}(1+{\omega }^2)$ by multiplication with the conjugate denominator, so $\mathrm{Im}\chi (\omega )=-\omega \mathrm{/}(1+{\omega }^2)<0$ for every $\omega >0$; for $d=1$ and $e\mathrm{\ne }0$, $e^{\ast }\chi (\omega )e=\mathrm{\mid }e{\mathrm{\mid }}^2\chi (\omega )$ and $e^{\ast }M(\omega )e=\mathrm{\mid }e{\mathrm{\mid }}^2\mathrm{Im}\chi (\omega )<0$, so the Hermitian dissipative part $M(\omega )=\mathrm{Im}\chi (\omega )$ is strictly negative and in particular not positive semidefinite. By clause (d) of [F9] the monochromatic period-averaged work equals $({\epsilon }_0\omega \mathrm{/}2)\mathrm{\mid }e{\mathrm{\mid }}^2\mathrm{Im}\chi (\omega )$, which is strictly negative for every $e\mathrm{\ne }0$ and every ${\epsilon }_0>0$, while steps 2.1 and 3.1 exhibit the response as causal with dispersion relations. Passivity is therefore an additional condition and not a consequence of causality.

Clause (ii): fix $t\in \mathbb{R}$, $n\in S^2$ and $r>\max \{2a,a+c(t-t_0)\}$, so $r>2a$ and $r-a>c(t-t_0)$. For every $y\in B_a$ the triangle inequality of [F11] gives $R=\mathrm{\mid }rn-y\mathrm{\mid }\ge r-\mathrm{\mid }y\mathrm{\mid }\ge r-a>c(t-t_0)$, hence $t-R\mathrm{/}c<t_0$; since (H1) makes $\rho$ and $\mathbf{J}$ vanish on the open set ${\mathbb{R}}^3\times ((-\mathrm{\infty },t_0)\cup (T,\mathrm{\infty }))$, all their partial derivatives vanish there too, so $\rho ,{\rho }_t,\mathbf{J},{\mathbf{J}}_t$ all vanish at $(y,t-R\mathrm{/}c)$ for every $y\in B_a$, and the integrands of the field formulas of clause (a) of [F10] vanish pointwise on ${\mathbb{R}}^3$. Hence $\mathbf{E}(rn,t)=\mathbf{B}(rn,t)=0$, so $n\cdot \mathbf{S}(rn,t)={\mu }_0^{-1}n\cdot (\mathbf{E}\times \mathbf{B})(rn,t)=0$ for every $n\in S^2$, and [F12] gives ${\int }_{S^2}{r}^{2}n\cdot \mathbf{S}(rn,t)d\mathrm{\Omega }(n)=0$ for all such $r$.

For every admissible source, step 5.1 gives an eventually zero flux integral at fixed observation time. For the admissible source of step 1.2, step 2.2 gives a strictly positive limit along $t=u_{\ast }+r\mathrm{/}c$. Consequently a fixed-observation-time limit cannot replace the fixed-retarded-time limit as a theorem about all these sources; zero or nonradiating sources may give zero in both limits. Together with step 4.1, this proves both claimed hypothesis boundaries.