# Page:CunninghamExtension.djvu/17

1909.]
98
The principle of relativity in electrodynamics.

The magnetization

 ${\displaystyle \left.{\begin{array}{rl}-\lambda ^{-2}M_{R}=&m_{r}-\beta v\left(W_{\theta }m_{\theta }+W_{\phi }m_{\phi }\right)/c^{2}\\\lambda ^{-2}M_{N}=&\beta \left(1-vW_{R}/c^{2}\right)m_{n}\end{array}}\right\}.}$ (9')

7. The question now arises as to whether in a ponderable body the transformations above given between (E, H) and (e, h) stand as they are, or whether they should be really between (E, B) and (e, b). Examination shews that relativity is fully maintained if the latter is adopted in accordance with Mirimanoff's paper on the Lorentz-Einstein transformation. We know, changing (H, h) into (B, b) that the transformation obtained leaves invariant the equations

${\displaystyle {\begin{array}{rl}-{\frac {1}{c}}{\frac {\partial b}{\partial t}}=&{\mathsf {curl}}\ e\\\\0=&{\mathsf {div}}\ b.\end{array}}}$

The transformation being now

 ${\displaystyle \left.{\begin{array}{ll}E_{R}=\lambda ^{2}e_{r},&B_{R}=-\lambda ^{2}b_{r}\\E_{\theta }=\lambda ^{2}\beta \left(-e_{\theta }+vb_{\phi }/c\right),&B_{\theta }=\lambda ^{2}\beta \left(b_{\theta }+ve_{\phi }/c\right)\\E_{\phi }=\lambda ^{2}\beta \left(-e_{\phi }-vb_{\theta }/c\right),&B_{\phi }=\lambda ^{2}\beta \left(b_{\phi }-ve_{\theta }/c\right)\end{array}}\right\},}$ (10')

with the use of (8'), (9'), the following equations are deduced:

 ${\displaystyle \left.{\begin{array}{ll}D_{R}=\lambda ^{2}d_{r},&Q_{R}=-\lambda ^{2}q_{r}\\D_{\theta }=\lambda ^{2}\beta \left(-d_{\theta }+vq_{\phi }/c\right),&Q_{\theta }=\lambda ^{2}\beta \left(q_{\theta }+vd_{\phi }/c\right)\\D_{\phi }=\lambda ^{2}\beta \left(d_{\phi }-vq_{\theta }/c\right),&Q_{\phi }=\lambda ^{2}\beta \left(q_{\phi }-vd_{\theta }/c\right)\end{array}}\right\},}$ (11')

where the new quantities are defined by

${\displaystyle {\begin{array}{ll}D=E+P,&d=e+p,\\B=M+H,&b=m+h,\\\\Q=H-{\frac {1}{c}}[PU],&q=h-{\frac {1}{c}}[pu].\end{array}}}$

These together with (6'), (7') will, by comparison with the transformation in its original form, leave unchanged the equations

${\displaystyle {\begin{array}{rl}{\frac {1}{c}}\left({\frac {\partial d}{\partial t}}+\rho u\right)=&{\mathsf {curl\ }}q,\\\\\rho =&{\mathsf {div}}\ d.\end{array}}}$