Page:Electromagnetic phenomena.djvu/7

${\mathfrak {a}}'$ for the instant, at which the local time in P is t' , we must take $\varrho$ and the vector $\varphi ,\ {\mathfrak {u}}'$, such as they are in the element dS at the instant at which the local time of that element is $t'-{\frac {r'}{c}}$.

§ 6. It will suffice for our purpose to consider two special cases. The first is that of an electrostatic system, i.e. a system having no other motion but the translation with the velocity w. In this case ${\mathfrak {u}}'=0$, and therefore, by (12), ${\mathfrak {a}}'=0$. Also, φ' is independent of t' , so that the equations (11), (13) and (14) reduce to

${\begin{cases}\triangle '\varphi '=-\varrho ',\\{\mathfrak {d}}'=-grad'\ \varphi ',\ {\mathfrak {h}}'=0.\end{cases}}$

(19)

After having determined the vector b' by means of these equations, we know also the electric force acting on electrons that belong to the system. For these the formulae (10) become, since u'=0

${\mathfrak {f}}_{x}=l^{2}{\mathfrak {d}}_{x}^{'},\quad {\mathfrak {f}}_{y}={\frac {l^{2}}{k}}{\mathfrak {d}}_{y}^{'},\quad {\mathfrak {f}}_{z}={\frac {l^{2}}{k}}{\mathfrak {d}}_{x}^{'}$.

(20)

The result may be put in a simple form if we compare the moving system Σ with which we are concerned, to another electrostatic system Σ' which remains at rest and into which Σ is changed, if the dimensions parallel to the axis of x are multiplied by kl, and the dimensions which have the direction of y or that of z, by l, a deformation for which (kl, l, l) is an appropriate symbol. In this new system, which we may suppose to be placed in the above mentioned space S' , we shall give to the density the value $\varrho '$, determined by (7), so that the charges of corresponding elements of volume and of corresponding electrons are the same in Σ and Σ' . Then we shall obtain the forces acting on the electrons of the moving system Σ, if we first determine the corresponding forces in Σ' , and next multiply their components in the direction of the axis of x by l², and their components perpendicular to that axis by ${\frac {l^{2}}{k}}$. This is conveniently expressed by the formula

${\mathfrak {F}}\left(\Sigma \right)=\left(l^{2},\ {\frac {l^{2}}{k}},\ {\frac {l^{2}}{k}}\right){\mathfrak {F}}\left(\Sigma '\right)$

(21)

It is further to be remarked that, after having found ${\mathfrak {d}}'$ by (19), we can easily calculate the electromagnetic momentum in the moving system, or rather its component in the direction of the motion. Indeed, the formula