UA OPTI501 電磁波 求解麥克斯韋方程組的Fourier方法3 Coulomb Gauge下討論Maxwell方程

Use the macroscopic equation with D-field, H-field eliminated by bound charge and bound current:
$$\begin{gathered} ??\text{E}=\frac{{\rho }_{total}^{(e)}}{{?}_0} \\ ?\times \text{B}={\mu }_0{\text{J}}_{total}^{(e)}+{\mu }_0{?}_0\frac{?}{?t}\text{E} \\ ?\times \text{E}=?\frac{?}{?t}\text{B} \\ ??\text{B}=0 \end{gathered}$$

Also define scalar potential $\psi$ and $\text{A}$ satisfying
$$\begin{gathered} \text{B}=?\times \text{A} \\ \text{E}=??\psi ?\frac{?\text{A}}{?t} \end{gathered}$$

so that Maxwell 3 and 4 are naturally satisfied. Now assume
$$\begin{gathered} {\rho }_{total}^{(e)}={\rho }_0{e}^{i(\text{k}?\text{r}?wt)},{\text{J}}_{total}^{(e)}={\text{J}}_0{e}^{i(\text{k}?\text{r}?wt)} \\ \text{A}=({\text{A}}_{||}+{\text{A}}_{\perp })e^{i(\text{k}?\text{r}?wt)},\psi ={\psi }_0{e}^{i(\text{k}?\text{r}?wt)} \end{gathered}$$

Under Coulomb gauge, $??\text{A}=0$. Replace $?$ by $i\text{k}$ and $\frac{?}{?t}$ by $?iw$, $i\text{k}?({\text{A}}_{||}+{\text{A}}_{\perp })e^{i(\text{k}?\text{r}?wt)}=0$, ${\text{A}}_{||}=0$. So $\text{A}={\text{A}}_{\perp }{e}^{i(\text{k}?\text{r}?wt)}$. As a result,
$$\begin{gathered} \text{B}=({\text{B}}_{||}+{\text{B}}_{\perp })e^{i(\text{k}?\text{r}?wt)}=i\text{k}\times {\text{A}}_{\perp }{e}^{i(\text{k}?\text{r}?wt)} \\ {\text{B}}_{||}=0,{\text{B}}_{\perp }=i\text{k}\times {\text{A}}_{\perp } \\ \text{E}=({\text{E}}_{||}+{\text{E}}_{\perp })e^{i(\text{k}?\text{r}?wt)}=?i\text{k}{\psi }_0{e}^{i(\text{k}?\text{r}?wt)}+iw{\text{A}}_{\perp }{e}^{i(\text{k}?\text{r}?wt)} \\ {\text{E}}_{||}=?i\text{k}{\psi }_0,{\text{E}}_{\perp }=iw{\text{A}}_{\perp } \end{gathered}$$

Now replace $?$ by $i\text{k}$ and $\frac{?}{?t}$ by $?iw$ in Maxwell 1 and 2.
$$\begin{gathered} i\text{k}?({\text{E}}_{||}+{\text{E}}_{\perp })=k^2{\psi }_0=\frac{{\rho }_0}{{?}_0},{\psi }_0=\frac{{\rho }_0}{{?}_0{k}^2} \\ i\text{k}\times {\text{B}}_{\perp }={\mu }_0{\text{J}}_0?iw{\mu }_0{?}_0({\text{E}}_{||}+{\text{E}}_{\perp }) \end{gathered}$$

LHS is $\begin{array}{rl}i\text{k}\times (i\text{k}\times {\text{A}}_{\perp })& =?\text{k}\times (\text{k}\times {\text{A}}_{\perp })=k^2{\text{A}}_{\perp }\end{array}$ and the second term of RHS is $iw{\mu }_0{?}_0({\text{E}}_{||}+{\text{E}}_{\perp })=iw{\mu }_0{?}_0(?i\text{k}{\psi }_0+iw{\text{A}}_{\perp })={\mu }_{0}w{\rho }_0\text{k}/k^2?(w/c{)}^2{\text{A}}_{\perp }$. Thus,
$$k^2{\text{A}}_{\perp }={\mu }_0({\text{J}}_{0||}+{\text{J}}_{0\perp })?{\mu }_{0}w{\rho }_0\text{k}/k^2+(w/c{)}^2{\text{A}}_{\perp }$$

Consider the continuity equation of charge,
$$\begin{gathered} i\text{k}?{\text{J}}_0?iw{\rho }_0=0 \\ {\text{J}}_{0||}?w\rho \hat{k}/k=0 \end{gathered}$$

Thus,
$$\begin{gathered} k^2{\text{A}}_{\perp }={\mu }_0{\text{J}}_{0\perp }+(w/c{)}^2{\text{A}}_{\perp } \\ {\text{A}}_{\perp }=\frac{{\mu }_0{\text{J}}_{\perp }}{k^2?(w/c{)}^2} \end{gathered}$$

So the E-field and B-field are
$$\begin{gathered} \text{E}=?i\text{k}{\psi }_0+iw{\text{A}}_{\perp }=?i\frac{{\rho }_0\text{k}}{{?}_0{k}^2}+i\frac{{\mu }_{0}w{\text{J}}_{0\perp }}{k^2?(w/c{)}^2} \\ \text{B}=i\text{k}\times {\text{A}}_{\perp }=i\frac{{\mu }_0\text{k}\times {\text{J}}_0}{k^2?(w/c{)}^2} \end{gathered}$$

B-field is the same as the result under Lorenz gauge. However, the E-field looks a little different. The E-field under Lorenz gauge is
$$\text{E}=\frac{?i({\rho }_0/{?}_0)\text{k}+i{\mu }_{0}w{\text{J}}_0}{k^2?(w/c{)}^2}$$

Since
$${\text{J}}_0={\text{J}}_{0||}+{\text{J}}_{0\perp }=w\rho \text{k}/k^2+{\text{J}}_{0\perp }$$

The E-field under Lorenz field can be reformulated as
$$\text{E}=\frac{?i({\rho }_0/{?}_0)\text{k}+i{\mu }_0{w}^2\rho \text{k}/k^2+i{\mu }_{0}w{\text{J}}_{0\perp }}{k^2?(w/c{)}^2}=?i\frac{{\rho }_0\text{k}}{{?}_0{k}^2}+i\frac{{\mu }_{0}w{\text{J}}_{0\perp }}{k^2?(w/c{)}^2}$$

which is the same as the E-field under Couloub gauge.

總結

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