Causal oscillator media and passivity — Examples

At a real frequency $\omega >0$, set $\mathbf{k}=k\widehat{z}$ and substitute the temporal factor $e^{-i\omega t}$ and spatial factor $e^{ikz}$ of the Statement into [A2]; [F5] differentiates these phases, and [F6] fixes the real-field encoding, extended to this complex wavevector. The two curl equations give $\mathbf{k}\times e=\omega b$ and $\mathbf{k}\times h=-\omega \epsilon (\omega )e$, with $b={\mu }_{0}h$. For a transverse amplitude $\mathbf{k}\cdot e=0$, [F4] gives $(\mathbf{k}\cdot \mathbf{k})e={\mu }_0\epsilon (\omega ){\omega }^{2}e$. Since $e\mathrm{\ne }0$, $k^2={\mu }_0\epsilon (\omega ){\omega }^2$; [F1] and the definition of $Z$ imply $k=(\omega \mathrm{/}c)n$ on the selected square-root branch. Conversely, for this root take $e=E_0\widehat{x}$, $b=(k{E}_0\mathrm{/}\omega )\widehat{y}$ and $h=b\mathrm{/}{\mu }_0$. Then $\mathbf{k}\times e=\omega b$ and $\mathbf{k}\times h=-k^{2}e\mathrm{/}({\mu }_0\omega )=-\omega \epsilon (\omega )e$. Also $\mathbf{k}\cdot e=\mathbf{k}\cdot b=0$, so the divergence amplitudes of $D$ and $B$ vanish. Thus the fields in the Statement realize the source-free transverse harmonic mode. As $\mathrm{Im}k>0$, $e^{ikz}=e^{-\mathrm{Im}(k)z}{e}^{i\mathrm{Re}(k)z}$ attenuates along $+z$.

At $\nu =1$, $1-{\nu }^2-i\nu \mathrm{/}2=-i\mathrm{/}2$, hence $Z(1)=1+2i$. Write the selected root as $n(1)=p+iq$ with $p,q>0$. Equating real and imaginary parts gives $p^2-q^2=1$ and $2pq=2$, hence $(p^2+q^2{)}^2=(p^2-q^2{)}^2+(2pq{)}^2=5$ and $p^2+q^2=\sqrt{5}$. For general $\nu >0$, write $Z(\nu )=x(\nu )+iy(\nu )$; rationalization gives $y(\nu )=(\nu \mathrm{/}2)\mathrm{/}((1-{\nu }^2{)}^2+{\nu }^2\mathrm{/}4)>0$. Put $r(\nu )=\sqrt{x(\nu {)}^2+y(\nu {)}^2}$. Then $r(\nu {)}^2-x(\nu {)}^2=y(\nu {)}^2>0$, so $r(\nu )>\mathrm{\mid }x(\nu )\mathrm{\mid }$ and both $(r+x)\mathrm{/}2$ and $(r-x)\mathrm{/}2$ are positive. Define $p(\nu )=\sqrt{(r+x)\mathrm{/}2}$ and $q(\nu )=\sqrt{(r-x)\mathrm{/}2}$. These satisfy $p^2-q^2=x$ and $(2pq{)}^2=r^2-x^2=y^2$; since $p,q,y>0$, $2pq=y$, so $(p+iq{)}^2=Z$. For any other root $p_0+i{q}_0$ with $p_0,q_0>0$, the same equations imply $(p_0^2+q_0^2{)}^2=x^2+y^2=r^2$, hence $p_0^2+q_0^2=r$ and $p_0^2=(r+x)\mathrm{/}2$, $q_0^2=(r-x)\mathrm{/}2$; positivity gives $p_0=p$, $q_0=q$. The rational functions $x,y$ are continuous for $\nu >0$; the positive square root is continuous at each positive $u_0$, since $\mathrm{\mid }\sqrt{u}-\sqrt{u_0}\mathrm{\mid }=\mathrm{\mid }u-u_0\mathrm{\mid }\mathrm{/}(\sqrt{u}+\sqrt{u_0})\le \mathrm{\mid }u-u_0\mathrm{\mid }\mathrm{/}\sqrt{u_0}$ for $u>0$. Applying this to $r$ and to the positive arguments $(r\pm x)\mathrm{/}2$ proves that $p,q$, hence $n$, are continuous near $1$. Therefore $n(\nu )+n(1)\to 2n(1)\mathrm{\ne }0$, and the difference quotient identity $(n(\nu )-n(1))\mathrm{/}(\nu -1)=(Z(\nu )-Z(1))\mathrm{/}((\nu -1)(n(\nu )+n(1)))$ gives $n^{\mathrm{\prime }}(1)=Z^{\mathrm{\prime }}(1)\mathrm{/}(2n(1))$. Set $D(\nu )=1-{\nu }^2-i\nu \mathrm{/}2$. Viewing $\mathbb{C}$ as ${\mathbb{R}}^2$, the real product rule applied to $D{D}^{-1}=1$ gives $Z^{\mathrm{\prime }}(1)=-D^{\mathrm{\prime }}(1)\mathrm{/}D(1{)}^2=(2+i\mathrm{/}2)\mathrm{/}(-i\mathrm{/}2{)}^2=-8-2i$. Thus $n^{\mathrm{\prime }}=(-4-i)\mathrm{/}n$ at $1$, and $1\mathrm{/}n=(p-iq)\mathrm{/}\sqrt{5}$ gives $\mathrm{Re}(n+n^{\mathrm{\prime }})=((\sqrt{5}-4)p-q)\mathrm{/}\sqrt{5}<0$, since $p,q>0$ and $\sqrt{5}<4$.

From $k=(\omega \mathrm{/}c)n(\nu )$ and $\nu =\omega \mathrm{/}{\omega }_0$, the chain and product rules give $\frac{d}{d\omega }\mathrm{Re}k=\frac{1}{c}\mathrm{Re}(n+\nu n^{\mathrm{\prime }})$. At $\nu =1$, this is negative by step 1.2 and $c>0$. Thus the conventional narrowband group-delay derivative is negative for this lossy causal model at that frequency. Its value does not give the support-front time of a pulse, which depends on the full time-domain signal; no front-speed or universal group-speed claim follows.

In the first state $P_t=0$, so [F1] gives zero oscillator energy. In the second state $P_t=v\mathrm{\ne }0$, so [F1] gives $\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}(2a)>0$ by [F2] and $a>0$. Both states have $E=0$ and $P=0$, hence the same $D={\epsilon }_{\mathrm{\infty }}E+P=0$, and both have $B=H=0$; they therefore have identical instantaneous $(E,D)$ data but different total stored energies.

The comparison is between allowed model states at one time; it is not an observed outcome or a claim about microscopic material energy. Hence the full oscillator energy is not a function of the instantaneous pair $(E,D)$ alone.