Page:PoincareDynamiqueJuillet.djvu/27

 That of :

$$l=k^{-\frac{1}{3}},\quad k=\theta,\quad klr=const..$$

We then find:

$$H=\frac{\varphi\left(\frac{\theta}{k}\right)}{k^{2}r}.$$

found, in different notation (Göttinger Nachrichten, 1902, p. 37)

$$H=\frac{a}{r}\frac{1-\epsilon^{2}}{\epsilon}\log\frac{1+\epsilon}{1-\epsilon},$$

a is a constant. However, in the hypothesis of, we have θ = 1; then:

which defines the function φ.

This granted, imagine that the electron is subject to a binding, so there is a relation between r and φ; in the hypothesis of this relation would be φr = const., in that of  φ²r² = const. We assume in a more general way

$$r=b\theta^{m}\,$$

b is a constant; hence:

$$H=\frac{1}{bk^{2}}\theta^{-m}\varphi\left(\frac{\theta}{k}\right).$$

What is the shape of the electron when the velocity become -εt, if we do not suppose the involvement of forces other than the binding forces? Its form will be defined by the equality:

or

$$-m\theta^{-m-1}\varphi+\theta^{-m}k^{-1}\varphi^{\prime}=0$$

or

$$\frac{\varphi^{\prime}}{\varphi}=\frac{mk}{\theta}.$$

If we want equilibrium to occur so that θ = k, it is necessary that $$\tfrac{\theta}{k}=1$$, the logarithmic derivative of φ is equal to m.

If we develop $$\frac{1}{k}$$ and the right-hand side of (5) in powers of ε, equation (5) becomes:

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

neglecting higher powers of ε.