\(\)
Problem 9-4
In Sec. 2-1 we discussed the mechanisms for obtaining the mechanical equations of motion from the form of the action \(S\) by obtaining the extremum \(S_{cl}\) under the condition \(\delta S=0\) for variations of coordinates, \(\delta\mathbf{q}\). Show how Maxwell’s equations can be derived from the action \(S\) defined in Eq. (9-23) by requiring \(\delta S=0\) for first-order variations of \(\mathbf{A}\) and \(\phi\).
( 解答 ) 以下は J.J.Sakurai：Advanced Quantum Mechanics の § 1-3 からの抜粋である．
連続的な場合に於ける変分原理を公式化するために,「ラグランジアン密度」を \(\mathscr{L}\) として次を考える：
\begin{equation}
\def\mb#1{\mathbf{#1}}
\def\pdiff#1{\frac{\partial}{\partial #1}}
\def\reverse#1{\frac{1}{#1}}
\def\Bppdiff#1#2{\frac{\partial^{2} #1}{\partial #2^{2}}}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
L=\int \mathscr{L}\,dx,\quad\rightarrow\quad
\delta \int_{t_1}^{t_2}L\,dt=\delta\int_{t_1}^{t_2}\,dt\int dx\,\mathscr{L}
\left(\eta,\dot{\eta},\ppdiff{\eta}{x}\right)
\tag{1}
\end{equation}
\def\mb#1{\mathbf{#1}}
\def\pdiff#1{\frac{\partial}{\partial #1}}
\def\reverse#1{\frac{1}{#1}}
\def\Bppdiff#1#2{\frac{\partial^{2} #1}{\partial #2^{2}}}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
L=\int \mathscr{L}\,dx,\quad\rightarrow\quad
\delta \int_{t_1}^{t_2}L\,dt=\delta\int_{t_1}^{t_2}\,dt\int dx\,\mathscr{L}
\left(\eta,\dot{\eta},\ppdiff{\eta}{x}\right)
\tag{1}
\end{equation}
\(\eta\) の変分 \(\delta\eta\) は, \(t_1\) と \(t_2\) でゼロであり, また空間積分の末端に於いてもゼロであると仮定する． (場の理論では, この後者の要件は明示的に述べられることはない．なぜなら, 無限遠に於いては十分速やかにゼロとなる場を通常は考えるからである)．他の点は, 変分の性質は完全に任意である．作用積分の変分は次となる：
\begin{align}
\delta \int L\,dt&=\int dt \int dx\,\left\{\ppdiff{\mathscr{L}}{\eta}\delta\eta
+\ppdiff{\mathscr{L}}{(\partial \eta/\partial x)}\delta\left(\ppdiff{\eta}{x}\right)
+\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\delta\left(\ppdiff{\eta}{t}\right)\right\}\notag\\
&=\int dt \int dx\,\left\{\ppdiff{\mathscr{L}}{\eta} -\pdiff{x}\left(\ppdiff{\mathscr{L}}
{(\partial\eta/\partial x)}\right)-\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\right\}
\delta\eta
\tag{2}
\end{align}
\delta \int L\,dt&=\int dt \int dx\,\left\{\ppdiff{\mathscr{L}}{\eta}\delta\eta
+\ppdiff{\mathscr{L}}{(\partial \eta/\partial x)}\delta\left(\ppdiff{\eta}{x}\right)
+\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\delta\left(\ppdiff{\eta}{t}\right)\right\}\notag\\
&=\int dt \int dx\,\left\{\ppdiff{\mathscr{L}}{\eta} -\pdiff{x}\left(\ppdiff{\mathscr{L}}
{(\partial\eta/\partial x)}\right)-\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\right\}
\delta\eta
\tag{2}
\end{align}
ただし, 最後の 2 項は以下のような「部分積分」を行った結果である．そのとき,「空間積分と時間積分の端点では \(\delta\eta\) がゼロである」ことを利用している：
\begin{align*}
\int dx\,\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\delta\left(\ppdiff{\eta}{x}\right)
&=\int dx\,\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\ppdiff{\delta\eta}{x}
=\left[\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\delta\eta\right]_{x_1}^{x_2}
-\int dx\,\pdiff{x}\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\delta\eta\\
&=-\int dx\,\pdiff{x}\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\delta\eta,\\
\int dt\,\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\delta\left(\ppdiff{\eta}{t}\right)
&=\int dt\,\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\ppdiff{\delta\eta}{t}
=\left[\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\delta\eta\right]_{t_1}^{t_2}
-\int dt\,\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\delta\eta\\
&=-\int dt\,\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}\delta\eta
\end{align*}
もし上記の要件を満たす任意の変分に対して式 (2) がゼロとなるならば, \(\{\ \}\) 内がゼロであればよい．従って次式が成立すべきである：
\begin{equation}
\pdiff{x}\left(\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\right)
+\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}-\ppdiff{\mathscr{L}}{\eta}=0
\tag{3}
\end{equation}
\pdiff{x}\left(\ppdiff{\mathscr{L}}{(\partial\eta/\partial x)}\right)
+\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\eta/\partial t)}-\ppdiff{\mathscr{L}}{\eta}=0
\tag{3}
\end{equation}
この式は,「オイラー=ラグランジュの方程式」と呼ばれている．
上の議論は容易に3次元に一般化できる．場 \(\phi(\mb{x},t)\) は各々の時空点 \((\mb{x},\,t)\) で定義される実関数であるとしよう．すると, ラグランジアン密度 \(\mathscr{L}\) は \(\phi,\,\partial\phi/\partial x_k\,(k=1,2,3)\), そして \(\partial\phi/\partial t\) に依存する．従って, オイラー=ラグランジュの方程式は次となる：
\begin{equation}
\sum_{k=1}^{3}\pdiff{x_k}\left(\ppdiff{\mathscr{L}}{(\partial\phi/\partial x_k)}\right)
+\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\phi/\partial t)}-\ppdiff{\mathscr{L}}{\phi}=0
\tag{4}
\end{equation}
\sum_{k=1}^{3}\pdiff{x_k}\left(\ppdiff{\mathscr{L}}{(\partial\phi/\partial x_k)}\right)
+\pdiff{t}\ppdiff{\mathscr{L}}{(\partial\phi/\partial t)}-\ppdiff{\mathscr{L}}{\phi}=0
\tag{4}
\end{equation}
以上のことをこの場合に当てはめて考える．以下は, テル・ハール：「解析力学」の § 8.2 の記述をこの本に合わせて書き直したものである．
作用 \(S_1\) には場の変数 \(\mb{A}\) と \(\phi\) を含んでいないので, この場合のラグランジアン密度は次となる：
\begin{align}
S&=S_2+S_3=\int dt\int d^{3}\mb{r}\,\mathscr{L}(\mb{A},\phi),\notag\\
\mathscr{L}&=\reverse{8\pi}(\mb{E}^{2}-\mb{B}^{2})-\rho\dot{\phi}+\reverse{c}\mb{j}\cdot\mb{A}\notag\\
&=\reverse{8\pi}\left(E_x^{2}+E_y^{2}+E_z^{2}\right)-\reverse{8\pi}\left(B_x^{2}+B_y^{2}+B_z^{2}\right)
-\rho\dot{\phi}+\reverse{c}(j_xA_x+j_yA_y+j_zA_z)\notag\\
&=\reverse{8\pi}\left|-\nabla\phi-\reverse{c}\ppdiff{\mb{A}}{t}\right|^{2}-\reverse{8\pi}
|\nabla\times\mb{A}|^{2}-\rho\dot{\phi}+\reverse{c}\mb{j}\cdot\mb{A},\notag\\
&\quad\mathrm{where}\quad E_x=-\ppdiff{\phi}{x}-\reverse{c}\ppdiff{A_x}{t}=-\ppdiff{\phi}{x}
-\reverse{c}\dot{A}_x,\quad B_x=\pdiff{y}A_z-\pdiff{z}A_y,\quad \mathrm{etc.}
\tag{5}
\end{align}
S&=S_2+S_3=\int dt\int d^{3}\mb{r}\,\mathscr{L}(\mb{A},\phi),\notag\\
\mathscr{L}&=\reverse{8\pi}(\mb{E}^{2}-\mb{B}^{2})-\rho\dot{\phi}+\reverse{c}\mb{j}\cdot\mb{A}\notag\\
&=\reverse{8\pi}\left(E_x^{2}+E_y^{2}+E_z^{2}\right)-\reverse{8\pi}\left(B_x^{2}+B_y^{2}+B_z^{2}\right)
-\rho\dot{\phi}+\reverse{c}(j_xA_x+j_yA_y+j_zA_z)\notag\\
&=\reverse{8\pi}\left|-\nabla\phi-\reverse{c}\ppdiff{\mb{A}}{t}\right|^{2}-\reverse{8\pi}
|\nabla\times\mb{A}|^{2}-\rho\dot{\phi}+\reverse{c}\mb{j}\cdot\mb{A},\notag\\
&\quad\mathrm{where}\quad E_x=-\ppdiff{\phi}{x}-\reverse{c}\ppdiff{A_x}{t}=-\ppdiff{\phi}{x}
-\reverse{c}\dot{A}_x,\quad B_x=\pdiff{y}A_z-\pdiff{z}A_y,\quad \mathrm{etc.}
\tag{5}
\end{align}
これより \(\phi\) について次が求められる：
\begin{align}
&\ppdiff{\mathscr{L}}{\phi}=-\rho,\quad \ppdiff{\mathscr{L}}{\dot{\phi}}=0,\quad
\ppdiff{\mathscr{L}}{(\partial\phi/\partial x)}
=\ppdiff{\mathscr{L}}{E_x}\ppdiff{E_x}{(\partial\phi/\partial x)}=\frac{2}{8\pi}E_x\times(-1)
=-\reverse{4\pi}E_x,\notag\\
&\ppdiff{\mathscr{L}}{(\partial\phi/\partial y)}=-\reverse{4\pi}E_y,\quad
\ppdiff{\mathscr{L}}{(\partial\phi/\partial y)}=-\reverse{4\pi}E_y
\tag{6}
\end{align}
&\ppdiff{\mathscr{L}}{\phi}=-\rho,\quad \ppdiff{\mathscr{L}}{\dot{\phi}}=0,\quad
\ppdiff{\mathscr{L}}{(\partial\phi/\partial x)}
=\ppdiff{\mathscr{L}}{E_x}\ppdiff{E_x}{(\partial\phi/\partial x)}=\frac{2}{8\pi}E_x\times(-1)
=-\reverse{4\pi}E_x,\notag\\
&\ppdiff{\mathscr{L}}{(\partial\phi/\partial y)}=-\reverse{4\pi}E_y,\quad
\ppdiff{\mathscr{L}}{(\partial\phi/\partial y)}=-\reverse{4\pi}E_y
\tag{6}
\end{align}
よって式 (4) の \(\phi\) に関するラグランジュの運動方程式からは, Maxwell 方程式の式 (9-2) が得られる：
\begin{align}
&\pdiff{x}\ppdiff{\mathscr{L}}{(\partial\phi/\partial x)}
+\pdiff{y}\ppdiff{\mathscr{L}}{(\partial\phi/\partial y)}
+\pdiff{z}\ppdiff{\mathscr{L}}{(\partial\phi/\partial z)}+\pdiff{t}\ppdiff{\mathscr{L}}{\dot{\phi}}
-\ppdiff{\mathscr{L}}{\phi}\notag\\
&\quad=-\reverse{4\pi}\left(\pdiff{x}E_x+\pdiff{y}E_y+\pdiff{z}E_z\right)-(-\rho)
=-\reverse{4\pi}\nabla\cdot\mb{E}+\rho=0,\notag\\
\therefore&\quad \nabla\cdot\mb{E}=4\pi\,\rho
\tag{7}
\end{align}
&\pdiff{x}\ppdiff{\mathscr{L}}{(\partial\phi/\partial x)}
+\pdiff{y}\ppdiff{\mathscr{L}}{(\partial\phi/\partial y)}
+\pdiff{z}\ppdiff{\mathscr{L}}{(\partial\phi/\partial z)}+\pdiff{t}\ppdiff{\mathscr{L}}{\dot{\phi}}
-\ppdiff{\mathscr{L}}{\phi}\notag\\
&\quad=-\reverse{4\pi}\left(\pdiff{x}E_x+\pdiff{y}E_y+\pdiff{z}E_z\right)-(-\rho)
=-\reverse{4\pi}\nabla\cdot\mb{E}+\rho=0,\notag\\
\therefore&\quad \nabla\cdot\mb{E}=4\pi\,\rho
\tag{7}
\end{align}
また \(A_x\) に関するラグランジュの運動方程式は, Maxwell 方程式の式 (9-3) の \(x\)-成分の式が得られる．まず,
\begin{align}
&\ppdiff{\mathscr{L}}{A_x}=\reverse{c}j_x,\quad \ppdiff{\mathscr{L}}{\dot{A}_x}=\ppdiff{\mathscr{L}}{E_x}
\ppdiff{E_x}{\dot{A}_x}=-\reverse{4\pi c}E_x,\quad\ppdiff{\mathscr{L}}{\dot{A}_y}=-\frac{E_y}{4\pi c},
\quad\ppdiff{\mathscr{L}}{\dot{A}_y}=-\frac{E_z}{4\pi c},\notag\\
&\ppdiff{\mathscr{L}}{(\partial A_x/\partial x)}=0,\quad
\ppdiff{\mathscr{L}}{(\partial A_x/\partial y)}
=\ppdiff{\mathscr{L}}{B_z}\ppdiff{B_z}{(\partial A_x/\partial y)}=\frac{B_z}{4\pi},\notag\\
&\ppdiff{\mathscr{L}}{(\partial A_x/\partial z)}
=\ppdiff{\mathscr{L}}{B_y}\ppdiff{B_y}{(\partial A_x/\partial z)}=-\frac{B_y}{4\pi},
\tag{8}
\end{align}
&\ppdiff{\mathscr{L}}{A_x}=\reverse{c}j_x,\quad \ppdiff{\mathscr{L}}{\dot{A}_x}=\ppdiff{\mathscr{L}}{E_x}
\ppdiff{E_x}{\dot{A}_x}=-\reverse{4\pi c}E_x,\quad\ppdiff{\mathscr{L}}{\dot{A}_y}=-\frac{E_y}{4\pi c},
\quad\ppdiff{\mathscr{L}}{\dot{A}_y}=-\frac{E_z}{4\pi c},\notag\\
&\ppdiff{\mathscr{L}}{(\partial A_x/\partial x)}=0,\quad
\ppdiff{\mathscr{L}}{(\partial A_x/\partial y)}
=\ppdiff{\mathscr{L}}{B_z}\ppdiff{B_z}{(\partial A_x/\partial y)}=\frac{B_z}{4\pi},\notag\\
&\ppdiff{\mathscr{L}}{(\partial A_x/\partial z)}
=\ppdiff{\mathscr{L}}{B_y}\ppdiff{B_y}{(\partial A_x/\partial z)}=-\frac{B_y}{4\pi},
\tag{8}
\end{align}
従って,
\begin{align}
&\pdiff{x}\ppdiff{\mathscr{L}}{(\partial A_x/\partial x)}
+\pdiff{y}\ppdiff{\mathscr{L}}{(\partial A_x/\partial y)}
+\pdiff{z}\ppdiff{\mathscr{L}}{(\partial A_x/\partial z)}+\pdiff{t}\ppdiff{\mathscr{L}}{\dot{A}_x}
-\ppdiff{\mathscr{L}}{A_x}\notag\\
&=\reverse{4\pi}\left(\pdiff{y}B_z-\pdiff{z}B_y\right)-\reverse{4\pi c}\pdiff{t}E_x-\reverse{c}j_x=0,\notag\\
&\qquad \therefore\quad (\nabla\times\mb{B})_x=\reverse{c}\ppdiff{E_x}{t}+\frac{4\pi}{c}j_x
\tag{9}
\end{align}
&\pdiff{x}\ppdiff{\mathscr{L}}{(\partial A_x/\partial x)}
+\pdiff{y}\ppdiff{\mathscr{L}}{(\partial A_x/\partial y)}
+\pdiff{z}\ppdiff{\mathscr{L}}{(\partial A_x/\partial z)}+\pdiff{t}\ppdiff{\mathscr{L}}{\dot{A}_x}
-\ppdiff{\mathscr{L}}{A_x}\notag\\
&=\reverse{4\pi}\left(\pdiff{y}B_z-\pdiff{z}B_y\right)-\reverse{4\pi c}\pdiff{t}E_x-\reverse{c}j_x=0,\notag\\
&\qquad \therefore\quad (\nabla\times\mb{B})_x=\reverse{c}\ppdiff{E_x}{t}+\frac{4\pi}{c}j_x
\tag{9}
\end{align}
同様にして, \(A_y\) 及び \(A_z\) に関するラグランジュの運動方程式からは, Maxwell 方程式の式 (9-3) の \(y\)-成分, 及び \(z\)-成分の式が得られる．
Maxwell 方程式の残りの式 (9-4) 及び式 (9-5) は, ポテンシャル \(\mb{A}\) と \(\phi\) を式 (9-7) 及び式 (9-9) によって導入したことから, 当然ながら満足されているはずである．
