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物質内のMaxwell方程式

投稿日時: 11/04 システム管理者

Maxwell方程式 [c.g.s.単位系使用] 
$$rot\overrightarrow{E}=-\displaystyle \frac{1}{c}\displaystyle \frac{ \partial \overrightarrow{B } }{ \partial t}$$    (1a)    電場$$\overrightarrow{E}$$
$$rot\overrightarrow{H}=\displaystyle \frac{1}{c}\displaystyle \frac{ \partial \overrightarrow{D } }{ \partial t}+\displaystyle \frac{4\pi }{c}\sigma \overrightarrow{E}$$ (変位電流)    (1b)    磁場$$\overrightarrow{H}$$
$$div\overrightarrow{B}=0$$    (1c)    磁束密度$$\overrightarrow{B}$$
$$div\overrightarrow{D}=4\pi \rho$$ (空間電荷)    (1d)    電束密度$$\overrightarrow{D}$$

弱い場では,線形応答が成り立つ: 
$$\overrightarrow{B}=\mu\overrightarrow{H}$$ (透磁率$$\mu=1$$の物質を対象とする)    (2a)    magnetic susceptibility$$\mu$$
$$\overrightarrow{D}=\varepsilon \overrightarrow{E}$$    (2b)    dielectric constant$$\varepsilon $$
$$\overrightarrow{j}=\sigma \overrightarrow{E}$$    (2c)    current density$$\overrightarrow{j}$$,conductivity$$\sigma $$
---------------------------------------------------- 
参考) [MKSA単位系の場合]
$$ \bigtriangledown \times E=-\displaystyle \frac{ \partial B}{ \partial t}$$
$$ \bigtriangledown \times H=\displaystyle \frac{ \partial D}{ \partial t}+j$$
$$ \bigtriangledown \cdot B=0$$
$$ \bigtriangledown \cdot D=\rho $$
ベクトルポテンシャル,スカラーポテンシャルの定義
$$E=- \bigtriangledown \cdot \phi -\displaystyle \frac{ \partial A}{ \partial t}$$
$$B= \bigtriangledown \times A$$
---------------------------------------------------- 
公式 $$[ \bigtriangledown \times \left[ \bigtriangledown \times E \right] ]= \bigtriangledown \left( \bigtriangledown \cdot E \right) - \bigtriangledown ^{2}E$$を使うと,
(1a,b) → $$ \bigtriangleup \overrightarrow{E}=\displaystyle \frac{\varepsilon }{c^{2 } }\displaystyle \frac{ \partial ^{2}\overrightarrow{E } }{ \partial t^{2 } }+\displaystyle \frac{4\pi \sigma }{c^{2 } }\displaystyle \frac{ \partial \overrightarrow{E } }{ \partial t}$$, $$ \bigtriangleup \overrightarrow{H}=\displaystyle \frac{\varepsilon}{c^{2 } }\displaystyle \frac{ \partial ^{2}\overrightarrow{H } }{ \partial t^{2 } }+\displaystyle \frac{4\pi \sigma }{c^{2 } }\displaystyle \frac{ \partial \overrightarrow{H } }{ \partial t}$$ (3)
(3)の解  → $$\overrightarrow{E}=\overrightarrow{E}_{0}exp\left[ i\left( \omega t-2\pi \overrightarrow{k}\overrightarrow{r} \right) \right] $$,$$\overrightarrow{H}=\overrightarrow{H}_{0}exp\left[ i\left( \omega t-2\pi \overrightarrow{k}\overrightarrow{r} \right) \right] $$    (4)
解(4)は,条件(5)の下で成立する.もし,変位電流がなければ,(5)の右辺第2項はない. 
$$\left( \displaystyle \frac{2\pi kc}{\omega } \right) ^{2}=\varepsilon -i4\pi \left( \displaystyle \frac{\sigma }{\varepsilon } \right) $$ (5)
(4)を(1a,b)に代入($$\sigma =0$$の物質を対象とする.): 
$$\overrightarrow{k} \times \overrightarrow{E}_{0}=\displaystyle \frac{\omega }{2\pi c}\overrightarrow{H}_{0}$$ $$ \Rightarrow $$ $$\left[ \begin{array}{@{\,} c @{\, } }
0 \\[0mm]
kE_{x} \\[0mm]
0
\end{array} \right] =\displaystyle \frac{\omega }{2\pi c}\left[ \begin{array}{@{\,} c @{\, } }
0 \\[0mm]
H_{y} \\[0mm]
0
\end{array} \right] $$ (6a)
$$\overrightarrow{k} \times \overrightarrow{H}_{0}=-\displaystyle \frac{\omega }{2\pi c}\overrightarrow{E}_{0}$$ $$ \Rightarrow $$ $$\left[ \begin{array}{@{\,} c @{\, } }
-kH_{y} \\[0mm]
0 \\[0mm]
0
\end{array} \right] =-\displaystyle \frac{\omega }{2\pi c}\left[ \begin{array}{@{\,} c @{\, } }
E_{x} \\[0mm]
0 \\[0mm]
0
\end{array} \right] $$ (6b)
$$\displaystyle \frac{c}{n}=\displaystyle \frac{\omega }{2\pi k}$$ (7)
$$\displaystyle \frac{H_{y } }{E_{x } }=\displaystyle \frac{\omega \varepsilon }{2\pi ck}=\displaystyle \frac{2\pi ck}{\omega }\left( =\displaystyle \frac{\varepsilon }{n}=n \right) $$ (8)
$$n^{2}=\varepsilon $$, $$\displaystyle \frac{H_{y } }{E_{x } }=\sqrt{\varepsilon }$$ (9)

境界条件

物質1と物質2の境界では,電束密度の界面垂直成分,電場の界面平行成分が連続することがMaxwell方程式から導ける.
$$D_{1z}=D_{2z}$$ $$ \Leftarrow $$ $$div\overrightarrow{D}=0$$ (10a)
$$\displaystyle \int_{}^{}(D_{2z}-D_{1z})dxdy=\displaystyle \lim_{ \mit\Delta z \to 0}\displaystyle \int_{}^{}D_{z}df_{z}=\displaystyle \lim_{ \mit\Delta z \to 0}\displaystyle \int_{}^{}div\overrightarrow{D}dV=0$$
$$E_{1x}=E_{2x}$$ $$ \Leftarrow $$ $$div\overrightarrow{B}=0$$ (10b)

$$\displaystyle \int_{}^{}(E_{1x}-E_{2x})dx=\displaystyle \lim_{ \mit\Delta z \to 0}\displaystyle \int_{}^{}(rotE)_{y}df_{y}=\displaystyle \lim_{ \mit\Delta z \to 0}-\displaystyle \frac{1}{c}\displaystyle \int_{}^{}\displaystyle \frac{ \partial B_{y } }{ \partial t}df_{y}=$$

$$=\displaystyle \lim_{ \mit\Delta z \to 0}-\displaystyle \frac{1}{c}\displaystyle \frac{ \partial }{ \partial t}\displaystyle \int_{}^{}div\overrightarrow{B}dV=0$$

垂直入射
$$E_{1x}+E_{1x}^{R}=E_{2x}$$
$$H_{1y}+H_{1y}^{R}=H_{2y}$$ $$ \Rightarrow $$ $$n_{1}\left( E_{1x}-E_{1x}^{R} \right) =n_{2}E_{2x}$$
$$\displaystyle \frac{H_{1y } }{E_{1x } }=\sqrt{\varepsilon _{1 } }=n_{1}$$, $$-\displaystyle \frac{H_{1x}^{R } }{E_{1x}^{R } }=\sqrt{\varepsilon _{1 } }=n_{1}$$, $$\displaystyle \frac{H_{2y } }{E_{2x } }=\sqrt{\varepsilon _{2 } }=n_{2}$$
振幅反射率 $$\displaystyle \frac{E_{1x}^{R } }{E_{1x } }=\displaystyle \frac{n_{1}-n_{2 } }{n_{1}+n_{2 } }$$, $$n_{1}<n_{2}$$で位相反転
振幅透過率 $$\displaystyle \frac{E_{2x } }{E_{1x } }=\displaystyle \frac{2n_{1 } }{n_{1}+n_{2 } }$$
--------------------------------------------------------
$$n_{1}, n_{2}$$が実数なら,$$\displaystyle \frac{\left| E_{1x}^{R} \right| ^{2 } }{\left| E_{1x} \right| ^{2 } }+\displaystyle \frac{\left| E_{2x} \right| ^{2 } }{\left| E_{1x} \right| ^{2 } }=1$$
-------------------------------------------------------- 
$$\displaystyle \frac{ \partial \rho }{ \partial t}+div\overrightarrow{j}=0$$
$$\displaystyle \frac{ \partial }{ \partial t}div\overrightarrow{D}=4\pi \displaystyle \frac{ \partial \rho }{ \partial t}=-4\pi div\overrightarrow{j}$$
$$div\left( \displaystyle \frac{ \partial \overrightarrow{D } }{ \partial t}+4\pi \overrightarrow{j} \right) =0$$
$$rot\overrightarrow{H}=\displaystyle \frac{\varepsilon }{c}\displaystyle \frac{ \partial \overrightarrow{E } }{ \partial t}+\displaystyle \frac{4\pi }{c}\sigma \overrightarrow{E}$$
第1項:第2項の寄与率=$$\varepsilon \omega :4\pi \sigma $$
$$\omega \gg \displaystyle \frac{4\pi \sigma }{\varepsilon }$$ $$\to $$ 誘電体