Page:The Meaning of Relativity - Albert Einstein (1922).djvu/36

$\mathbf {i}$ is a vector, because the current density is defined as the density of electricity multiplied by the vector velocity of the electricity. According to the first three equations it is evident that $\mathbf {e}$ is also to be regarded as a vector. Then $\mathbf {h}$ cannot be regarded as a vector.^[1] The equations may, however, easily be interpreted if $\mathbf {h}$ is regarded as a symmetrical tensor of the second rank. In this sense, we write $h_{23},h_{31},h_{12}$ in place of $h_{1},h_{2},h_{3}$ respectively. Paying attention to the skew-symmetry of $h_{\mu \nu }$, the first three equations of (19) and (20) may be written in the form

${\frac {\delta h_{\mu \nu }}{\delta x_{\nu }}}={\frac {1}{c}}{\frac {\delta e_{\mu }}{\delta t}}+{\frac {1}{c}}i_{\mu }$

(19a)

${\frac {\delta e_{\mu }}{\delta x_{\nu }}}-{\frac {\delta e_{\nu }}{\delta x_{\mu }}}=+{\frac {1}{c}}{\frac {\delta h_{\mu \nu }}{\delta t}}$

(20a)

In contrast to $\mathbf {e}$, $\mathbf {h}$ appears as a quantity which has the same type of symmetry as an angular velocity. The divergence equations then take the form

${\frac {\delta e_{\nu }}{\delta x_{\nu }}}=\rho$

(19b)

${\frac {\delta h_{\mu \nu }}{\delta x_{\rho }}}+{\frac {\delta h_{\nu \rho }}{\delta x_{\mu }}}+{\frac {\delta h_{\rho \mu }}{\delta x_{\nu }}}=0$

(20b)

The last equation is a skew-symmetrical tensor equation of the third rank (the skew-symmetry of the left-hand side with respect to every pair of indices may easily be

1. These considerations will make the reader familiar with tensor operations without the special difficulties of the four-dimensional treatment; corresponding cosiderations in the theory of special relativity (Minkowski's interpretation of the field) will then offer fewer difficulties.