5.11. Найти инвариант $F^{ik} F_{ik}$ электромагнитного поля. Используя формулы преобразования полей $\vec E$ и $\vec H$, найденные в предыдущей задаче, доказать инвариантность величины $(\vec E \cdot \vec H)$ при переходе из одной инерциальной системы отсчета в другую.

$$F^{\mu\nu}=\frac{\partial A^{\nu}}{\partial x_{\mu}}-\frac{\partial A^{\mu}}{\partial x_{\nu}}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}=-F^{\nu\mu}$$ $$F^{\mu\nu}=\left(\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & -H_{z} & H_{y}\\ E_{y} & H_{z} & 0 & -H_{x}\\ E_{z} & -H_{y} & H_{x} & 0 \end{array}\right)$$ $$F'^{\mu\nu}=L_{i}^{\mu}L_{j}^{\nu}F^{ij}$$

$$g^{ij}=g_{ij}=g_{ij}^{-1}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)$$ $$A^{i}=\left(A_{0},A_{x},A_{y},A_{z}\right), \ \ A_{j}=A^{i}g_{ij}=\left(A_{0},-A_{x},-A_{y},-A_{z}\right)$$ Инварианты — скаляры, $$A^{i}A_{i}= A_{0}^{2}-A_{x}^{2}-A_{y}^{2}-A_{z}^{2}=inv$$ $$A'^{i}A'_{i}=A'^{i}A'^{j}g_{ij}=A^{\mu}L_{\mu}^{i}A^{\nu}L_{\nu}^{j}g_{ij}=$$ $$A^{\mu}A^{\nu}L_{\mu}^{i}L_{\nu}^{j}g_{ij}=A^{\mu}A^{\nu}g_{\mu\nu}=A^{\mu}A_{\mu}$$ $$F^{\mu\nu}F_{\mu\nu}=F^{\mu\nu}F^{ij}g_{i\mu}g_{j\nu}$$ $$F^{\mu\nu}=\frac{\partial A^{\nu}}{\partial x_{\mu}}-\frac{\partial A^{\mu}}{\partial x_{\nu}}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$ $$\left(\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & -H_{z} & H_{y}\\ E_{y} & H_{z} & 0 & -H_{x}\\ E_{z} & -H_{y} & H_{x} & 0 \end{array}\right)=F^{\mu\nu}\to $$ $$F_{\mu\nu}=\left(\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & -H_{z} & H_{y}\\ -E_{y} & H_{z} & 0 & -H_{x}\\ -E_{z} & -H_{y} & H_{x} & 0 \end{array}\right)$$ $$F^{\mu\nu}F_{\mu\nu}=2\left(-E^{2}+H^{2}\right)=inv$$ $$\left\{ \begin{array}{c} H_{x}=H_{x}'\\ H_{y}=\gamma\left(H_{y}'-\beta E_{z}'\right)\\ H_{z}=\gamma\left(H_{z}'+\beta E_{y}'\right) \end{array}\right.$$ $$\left\{ \begin{array}{c} E_{x}=E_{x}'\\ E_{y}=\gamma\left(E_{y}'+\beta H_{z}'\right)\\ E_{z}=\gamma\left(E_{z}'-\beta H_{y}'\right) \end{array}\right.$$ $$H^{2}-E^{2}= H_{x}'^{2}+\gamma^{2}\left(H_{y}'-\beta E_{z}'\right)^{2}+\gamma^{2}\left(H_{z}'+\beta E_{y}'\right)^{2}-$$ $$E_{x}'^{2}-\gamma^{2}\left(E_{y}'+\beta H_{z}'\right)^{2}-\gamma^{2}\left(E_{z}'-\beta H_{y}'\right)^{2}$$