Page:AbrahamMinkowski1.djvu/23

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§ 12. The ponderomotive force.


Now, since the relative stresses as well as the momentum density in there dependence from the electromagnetic vectors, are determined for every given theory, then the components of the ponderomotive force are given from equations (8). According to (Va) it is

(56) \begin{cases}
X'_{x}=\frac{1}{2}\left\{ \mathfrak{E}'_{x}\mathfrak{D}_{x}-\mathfrak{E}'_{y}\mathfrak{D}_{y}-\mathfrak{E}'_{z}\mathfrak{D}_{z}\right\} +\frac{1}{2}\left\{ \mathfrak{H}'_{x}\mathfrak{B}_{x}-\mathfrak{H}'_{y}\mathfrak{B}_{y}-\mathfrak{H}'_{z}\mathfrak{B}_{z}\right\} ,\\
\\X'_{y}=\mathfrak{E}'_{x}\mathfrak{D}_{y}+\mathfrak{H}'_{x}\mathfrak{B}_{y},\\
\\X'_{z}=\mathfrak{E}'_{x}\mathfrak{D}_{z}+\mathfrak{H}'_{x}\mathfrak{B}_{z}.\end{cases}

From that if follows

(57) \begin{cases}
\frac{\partial X'_{x}}{\partial x}+\frac{\partial X'_{y}}{\partial y}+\frac{\partial X'_{z}}{\partial z}=\mathfrak{E}'_{x}\mathrm{div}\mathfrak{D}+\mathfrak{H}'_{x}\mathrm{div}\mathfrak{B}\\
\\\qquad-\mathfrak{D}_{y}\mathrm{curl}_{z}\mathfrak{E}'+\mathfrak{D}_{z}\mathrm{curl}_{y}\mathfrak{E}'-\mathfrak{B}_{y}\mathrm{curl}_{z}\mathfrak{H}'+\mathfrak{B}_{z}\mathrm{curl}_{z}\mathfrak{H}'\\
\\\qquad-\frac{1}{2}\left\{ \mathfrak{E}'\frac{\partial\mathfrak{D}}{\partial x}-\mathfrak{D}\frac{\partial\mathfrak{E'}}{\partial x}+\mathfrak{H}'\frac{\partial\mathfrak{B}}{\partial x}-\mathfrak{B}\frac{\partial\mathfrak{H'}}{\partial x}\right\} .\end{cases}

If we consider the last row, then the analogy to the left-hand side of (54) leaps out; the expressions only differ by the fact, that it was differentiated with respect to time there, and here with respect to a coordinate.

Since the train of thought that led to relation (54), is not concerned by the meaning of the independent variables, it is given

(57a) \frac{1}{2}\left\{ \mathfrak{E}'\frac{\partial\mathfrak{D}}{\partial x}-\mathfrak{D}\frac{\partial\mathfrak{E'}}{\partial x}+\mathfrak{H}'\frac{\partial\mathfrak{B}}{\partial x}-\mathfrak{B}\frac{\partial\mathfrak{H'}}{\partial x}\right\} =\mathfrak{g}\frac{\partial\mathfrak{w}}{\partial x}+\zeta\frac{\partial\epsilon}{\partial x}+\eta\frac{\partial\mu}{\partial x}

The vectorial generalization of (57) gives as the force contribution stemming from the relative stresses

(58) \begin{cases}
\mathfrak{K}_{1}=\mathfrak{E}'\mathrm{div}\mathfrak{D}+\mathfrak{H}'\mathrm{div}\mathfrak{B}-[\mathfrak{D}\mathrm{curl}\mathfrak{E'}]-[\mathfrak{B}\mathrm{curl}\mathfrak{H'}]\\
\quad-(\mathfrak{g}\nabla)\mathfrak{w}-[\mathfrak{g}\mathrm{curl}\mathfrak{w}]-\zeta\nabla\epsilon-\eta\nabla\mu.\end{cases}

The contribution is added to it, which stems from the electromagnetic momentum

(58a) \mathfrak{K}_{2}=-\frac{\delta\mathfrak{g}}{\delta t}

By the aid of the two first main equations of § 4, we want to transform the vector products arising in (58), into

\begin{array}{l}
-[\mathfrak{D}\mathrm{curl}\mathfrak{E'}]=\frac{1}{c}\left[\mathfrak{D}\frac{\partial'\mathfrak{B}}{\partial t}\right]\\
\\-[\mathfrak{B}\mathrm{curl}\mathfrak{H'}]=\frac{1}{c}[\mathfrak{JB}]+\frac{1}{c}\left[\frac{\partial'\mathfrak{D}}{\partial t}\mathfrak{B}\right]\end{array}