Page:PoincareDynamiqueJuillet.djvu/28

 By differentiating, we get:

$$-\epsilon\varphi^{\prime}\left(1-\frac{\epsilon^{2}}{2}\right)=\frac{2}{3}\epsilon a.$$

For ε = 0, that is to say when the argument of φ is equal to 1, these equations become:

We must therefore have $$m=-\tfrac{2}{3}$$ in conformity with the hypothesis of.

This result should come nearer to that which is connected to the first equation (a), and from which actually it does not differ. Indeed, suppose that every element dτ of the electron is subjected to a force Xdτ parallel to the x-axis, X is the same for all elements; we will then have, in conformity with the definition of momentum:

$$\frac{dD}{dt}=\int Xd\tau.$$

In addition, the principle of least action gives us:

$$\delta J=\int X\delta U\ d\tau\ dt,\quad J=\int H\ dt,\quad\delta J=\int D\delta U\ dt,$$

δU is the displacement of the center of gravity of the electron; H depends on θ and on ε if we assume that r is related to θ by the equation of binding; we have thus:

$$\delta J=\int\left(\frac{\partial H}{\partial\epsilon}\delta\epsilon+\frac{\partial H}{\partial\theta}\right)dt.$$

In addition $$\delta\epsilon=-\tfrac{d\delta U}{dt}$$; where, by integrating by parts:

$$\int D\delta\epsilon\ dt=\int D\delta U\ dt$$

or

$$\int\left(\frac{\partial H}{\partial\epsilon}\delta\epsilon+\frac{\partial H}{\partial\theta}\delta\theta\right)dt=\int D\delta\epsilon\ dt,$$

hence

$$D=\frac{\partial H}{\partial\epsilon},\quad\frac{\partial H}{\partial\theta}=0.$$

But the derivative $$\tfrac{dH}{d\epsilon}$$, contained in the right-hand side of equation (2), is the derivative taken by supposing θ as a function of ε, so that

$$\frac{dH}{d\epsilon}=\frac{\partial H}{\partial\epsilon}+\frac{\partial H}{\partial\theta}\frac{d\theta}{d\epsilon}.$$

Equation (2) is therefore equivalent to equation (6).

The conclusion is that if the electron is subject to a binding between its three axes,