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 '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

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.