这是一份bath巴斯大学PH40086作业代写的成功案

$$
\left(\overleftrightarrow{P}{\perp s} \mathcal{E}{s}(\omega)\right)=\left(\mathcal{N}{\mathrm{s}}^{\text {tord }}(\omega)\right)^{2} \overleftrightarrow{I}{\mathrm{s}}\left(\overleftrightarrow{P}{\perp \mathrm{s}} \mathcal{E}{\mathrm{s}}(\omega)\right)
$$
for
$$
\overleftrightarrow{I}{\mathrm{s}} \stackrel{\text { def }}{=} \overleftrightarrow{P}{\perp s}\left(\overleftrightarrow{\epsilon}{c}(\omega)\right)^{-1} \overleftrightarrow{P}{\perp s}
$$
Therefore,
$$
\overleftrightarrow{I}{\mathrm{s}}=\frac{1}{\left(\operatorname{Nord}{\mathrm{s}}(\omega)\right)^{2}} \stackrel{\leftrightarrow}{P}{\perp \mathrm{s}} $$ Since and $$ \mathbf{a} \cdot\left(\overleftrightarrow{P}{\perp \mathbf{s}} \mathbf{b}\right)=\left(\overleftrightarrow{P}_{\perp \mathbf{s}} \mathbf{a}\right) \cdot \mathbf{b} \quad \forall \mathbf{a}, \mathbf{b} \in \mathbb{C}^{3}
$$

PH40086 COURSE NOTES ：

fulfills the conditions
$$
\begin{aligned}
&\jmath(\mathbf{x}, t)=0 \text { for }(|\mathbf{x}|, t) \notin[0,+R] \times[0,+\infty) \
&\Longrightarrow \quad \mathcal{R}(\mathbf{x}, t ; \jmath)=0 \quad \text { for } t \notin\left[0, \frac{|\mathbf{x}|-R}{c}\right]
\end{aligned}
$$
and
$$
\hat{H} \tilde{\mathcal{R}}(\mathbf{x}, \omega ; \widetilde{\jmath})=-i \omega \mu_{0} \tilde{\mathcal{J}}(\mathbf{x}, \omega)
$$
for all sufficiently well-behaved $\boldsymbol{\jmath}$, where
$$
\begin{aligned}
\widetilde{\mathcal{R}}(\mathbf{x}, \omega ; \tilde{\jmath}) & \stackrel{\text { def }}{=} \frac{1}{\sqrt{2 \pi}} \int \mathcal{R}(\mathbf{x}, t ; \jmath) e^{+i \omega t} \mathrm{~d} t \
&=\int \overleftrightarrow{r}\left(\mathbf{x}-\mathbf{x}^{\prime}, \omega\right) \tilde{\mathcal{J}}\left(\mathbf{x}^{\prime}, \omega\right) \mathrm{d} V_{\mathbf{x}^{\prime}}
\end{aligned}
$$