Page:PoincareDynamiqueJuillet.djvu/25

the transverse electric energy; by

$C={\frac {1}{2}}\int \left(\beta ^{2}+\gamma ^{2}\right)d\tau$

the transverse magnetic energy. There is no longitudinal magnetic energy, since α = α' = 0. We denote by A', B', C' the corresponding quantities in the ideal system. We first find:

$C'=0,\ C=\epsilon ^{2}B$

In addition, we can observe that the actual field depends only on x = εt, y, and x, and write:

$d\tau =d(x+\epsilon t)dy\ dz$

$d\tau '=dy'dy'dz'=kl^{3}d\tau \,$

hence

$A^{\prime }=kl^{-1}A,\quad B^{\prime }=k^{-1}l^{-1}B,\quad A={\frac {lA^{\prime }}{k}},\quad B=klB^{\prime }.$

In Lorentz's hypothesis we have B' = 2A', and A ' (being inversely proportional to the radius of the electron) is a constant independent of the velocity of the real electron; we get for the total energy:

$A+B+C=A'lk\left(3+\epsilon ^{2}\right)$

and for the action (per unit time):

$A+B-C={\frac {3A^{\prime }l}{k}}.$

Now calculate the electromagnetic momentum; we find:

$D=\int \left(g\gamma -h\beta \right)d\tau =-\epsilon \int \left(g^{2}+h^{2}\right)d\tau =-2\epsilon B=-4\epsilon klA^{\prime }.$

But there must be some relation between the energy E = A + B + C, the action per unit time H = A + B - C, and the momentum D. The first of these relations is:

$E=H-\epsilon {\frac {dH}{d\epsilon }},$

the second is

${\frac {dD}{d\epsilon }}=-{\frac {1}{\epsilon }}{\frac {dE}{d\epsilon }};$

hence

(2)	$D={\frac {dH}{d\epsilon }},\quad E=H-\epsilon D.$

The second of equations (2) is always satisfied; but the first is so only if

$l=\left(1-\epsilon ^{2}\right)^{\frac {1}{6}}=k^{-{\frac {1}{3}}},$

that is to say if the volume of the ideal electron is equal to that of the real electron; or if the volume of the electron is constant; that's the hypothesis of Langevin.