電磁場のローレンツ変換

真空中のマクスウェル方程式は
$$\begin{eqnarray}
\mathrm{div}\boldsymbol{D}&=&0\\
\mathrm{div}\boldsymbol{B}&=&0\\
\mathrm{rot}\boldsymbol{H}&=&\frac{\partial\boldsymbol{D}}{\partial t}\\
\mathrm{rot}\boldsymbol{E}&=&-\frac{\partial\boldsymbol{B}}{\partial t}
\end{eqnarray}$$
であるが、
$$\begin{eqnarray}
\boldsymbol{D}&=&\varepsilon_0\boldsymbol{E}\\
\boldsymbol{B}&=&\mu_0\boldsymbol{H}\\
c^2&=&\frac{1}{\varepsilon_0\mu_0}
\end{eqnarray}$$
より、\(\boldsymbol{E}\)と\(\boldsymbol{B}\)だけを使って成分で書くと
$$\begin{eqnarray}
\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}&=&0\\
\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z}&=&0\\
\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}&=&\frac{1}{c^2}\frac{\partial E_x}{\partial t}\\
\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}&=&\frac{1}{c^2}\frac{\partial E_y}{\partial t}\\
\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}&=&\frac{1}{c^2}\frac{\partial E_z}{\partial t}\\
\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}&=&-\frac{\partial B_x}{\partial t}\\
\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}&=&-\frac{\partial B_y}{\partial t}\\
\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}&=&-\frac{\partial B_z}{\partial t}
\end{eqnarray}$$
となる。これにローレンツ変換
$$\begin{eqnarray}
t’&=&\frac{t-vx/c^2}{\sqrt{1-(v/c)^2}}\\
x’&=&\frac{x-vt}{\sqrt{1-(v/c)^2}}\\
y’&=&y\\
z’&=&z
\end{eqnarray}$$
を適用することを考える。まず、\(K\)系の電磁場を\(K’\)系の座標で表す。例えば、
$$f(t,x,y,z)\rightarrow f(t’,x’,y’,z’)$$
の時、全微分を使って
$$\begin{eqnarray}
\frac{\partial f}{\partial t}&=&\frac{\partial f}{\partial t’}\frac{\partial t’}{\partial t}+\frac{\partial f}{\partial x’}\frac{\partial x’}{\partial t}+\frac{\partial f}{\partial y’}\frac{\partial y’}{\partial t}+\frac{\partial f}{\partial z’}\frac{\partial z’}{\partial t}=\frac{\partial f}{\partial t’}\frac{1}{\sqrt{1-(v/c)^2}}+\frac{\partial f}{\partial x’}\frac{-v}{\sqrt{1-(v/c)^2}}+0+0=\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)f\\
\frac{\partial f}{\partial x}&=&\frac{\partial f}{\partial t’}\frac{\partial t’}{\partial x}+\frac{\partial f}{\partial x’}\frac{\partial x’}{\partial x}+\frac{\partial f}{\partial y’}\frac{\partial y’}{\partial x}+\frac{\partial f}{\partial z’}\frac{\partial z’}{\partial x}=\frac{\partial f}{\partial t’}\frac{-v/c^2}{\sqrt{1-(v/c)^2}}+\frac{\partial f}{\partial x’}\frac{1}{\sqrt{1-(v/c)^2}}+0+0=\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)f\\
\frac{\partial f}{\partial y}&=&\frac{\partial f}{\partial t’}\frac{\partial t’}{\partial y}+\frac{\partial f}{\partial x’}\frac{\partial x’}{\partial y}+\frac{\partial f}{\partial y’}\frac{\partial y’}{\partial y}+\frac{\partial f}{\partial z’}\frac{\partial z’}{\partial y}=0+0+\frac{\partial f}{\partial y’}\frac{\partial y}{\partial y}+0=\frac{\partial f}{\partial y’}
\end{eqnarray}$$
となるから、マクスウェル方程式は
$$\begin{eqnarray}
\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)E_x+\frac{\partial E_y}{\partial y’}+\frac{\partial E_z}{\partial z’}&=&0\\
\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)B_x+\frac{\partial B_y}{\partial y’}+\frac{\partial B_z}{\partial z’}&=&0\\
\frac{\partial B_z}{\partial y’}-\frac{\partial B_y}{\partial z’}&=&\frac{1}{c^2}\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)E_x\\
\frac{\partial B_x}{\partial z’}-\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)B_z&=&\frac{1}{c^2}\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)E_y\\
\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)B_y-\frac{\partial B_x}{\partial y’}&=&\frac{1}{c^2}\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)E_z\\
\frac{\partial E_z}{\partial y’}-\frac{\partial E_y}{\partial z’}&=&-\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)B_x\\
\frac{\partial E_x}{\partial z’}-\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)E_z&=&-\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)B_y\\
\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial x’}-\frac{v}{c^2}\frac{\partial}{\partial t’}\right)E_y-\frac{\partial E_x}{\partial y’}&=&-\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)B_z
\end{eqnarray}$$
となる。最初の2式は
$$\begin{eqnarray}
v\frac{\partial E_y}{\partial y’}+v\frac{\partial E_z}{\partial z’}&=&-\frac{1}{\sqrt{1-(v/c)^2}}\left(1-\frac{v^2}{c^2}\right)\frac{\partial E_x}{\partial t’}+\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)E_x=-\sqrt{1-(v/c)^2}\frac{\partial E_x}{\partial t’}+\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)E_x\\
v\frac{\partial B_y}{\partial y’}+v\frac{\partial B_z}{\partial z’}&=&-\frac{1}{\sqrt{1-(v/c)^2}}\left(1-\frac{v^2}{c^2}\right)\frac{\partial B_x}{\partial t’}+\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)B_x=-\sqrt{1-(v/c)^2}\frac{\partial B_x}{\partial t’}+\frac{1}{\sqrt{1-(v/c)^2}}\left(\frac{\partial}{\partial t’}-v\frac{\partial}{\partial x’}\right)B_x
\end{eqnarray}$$
となるので、これを3番目の式と6番目の式の右辺に代入して、残り6式をまとめると
$$\begin{eqnarray}
\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial y’}\left(B_z-\frac{v}{c^2}E_y\right)-\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial z’}\left(B_y+\frac{v}{c^2}E_z\right)&=&\frac{1}{c^2}\frac{\partial E_x}{\partial t’}\\
\frac{\partial B_x}{\partial z’}-\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial x’}\left(B_z-\frac{v}{c^2}E_y\right)&=&\frac{1}{c^2}\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial t’}(E_y-vB_z)\\
\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial x’}\left(B_y+\frac{v}{c^2}E_z\right)-\frac{\partial H_x}{\partial y’}&=&\frac{1}{c^2}\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial t’}(E_z+vB_y)\\
\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial y’}(E_z+vB_y)-\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial z’}(E_y-vB_z)&=&-\frac{\partial B_x}{\partial t’}\\
\frac{\partial E_x}{\partial z’}-\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial x’}(E_z+vB_y)&=&-\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial t’}\left(B_y+\frac{v}{c^2}E_z\right)\\
\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial x’}(E_y-vB_z)-\frac{\partial E_x}{\partial y’}&=&-\frac{1}{\sqrt{1-(v/c)^2}}\frac{\partial}{\partial t’}\left(B_z-\frac{v}{c^2}E_y\right)
\end{eqnarray}$$
となる。さて、相対性理論では、マクスウェル方程式はどの系でも同様に成り立つ筈であるから、\(K’\)系の電磁場も
$$\begin{eqnarray}
\mathrm{div}\boldsymbol{D}’&=&0\\
\mathrm{div}\boldsymbol{B}’&=&0\\
\mathrm{rot}\boldsymbol{H}’&=&\frac{\partial\boldsymbol{D}’}{\partial t}\\
\mathrm{rot}\boldsymbol{E}’&=&-\frac{\partial\boldsymbol{B}’}{\partial t}
\end{eqnarray}$$
を満たす筈である。したがって、先程求めた6式と
$$\begin{eqnarray}
\frac{\partial B’_z}{\partial y}-\frac{\partial B’_y}{\partial z}&=&\frac{1}{c^2}\frac{\partial E’_x}{\partial t}\\
\frac{\partial B’_x}{\partial z}-\frac{\partial B’_z}{\partial x}&=&\frac{1}{c^2}\frac{\partial E’_y}{\partial t}\\
\frac{\partial B’_y}{\partial x}-\frac{\partial B’_x}{\partial y}&=&\frac{1}{c^2}\frac{\partial E’_z}{\partial t}\\
\frac{\partial E’_z}{\partial y}-\frac{\partial E’_y}{\partial z}&=&-\frac{\partial B’_x}{\partial t}\\
\frac{\partial E’_x}{\partial z}-\frac{\partial E’_z}{\partial x}&=&-\frac{\partial B’_y}{\partial t}\\
\frac{\partial E’_y}{\partial x}-\frac{\partial E’_x}{\partial y}&=&-\frac{\partial B’_z}{\partial t}
\end{eqnarray}$$
を比較すれば、電磁場のローレンツ変換は、
$$\fbox{\(\begin{eqnarray}
E’_x&=&E_x\\
E’_y&=&\frac{1}{\sqrt{1-(v/c)^2}}(E_y-vB_z)\\
E’_z&=&\frac{1}{\sqrt{1-(v/c)^2}}(E_z+vB_y)\\
B’_x&=&B_x\\
B’_y&=&\frac{1}{\sqrt{1-(v/c)^2}}\left(B_y+\frac{v}{c^2}E_z\right)\\
B’_z&=&\frac{1}{\sqrt{1-(v/c)^2}}\left(B_z-\frac{v}{c^2}E_y\right)
\end{eqnarray}\)}$$
となる。