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and if no other force intervenes except the binding forces, the shape of that electron, when it is given a uniform velocity, may be such that the ideal electron corresponds to a sphere, except the case where the binding is such that the volume is constant, in conformity with the hypothesis of Langevin.

We are led in this way to pose the following problem: what additional forces, other than the binding forces, are necessary to intervene to account for the law of Lorentz or, more generally, any law other than that of Langevin?

The simplest hypothesis, and the first that we should consider, is that these additional forces are derived from a special potential depending on the three axes of the ellipsoid, and therefore on θ and on r; let F(θ, r) be the potential; in which case the action will be expressed:

$J=\int \left[H+F(\theta ,r)\right]dt$

and the equilibrium conditions are written:

(8)	${\frac {dH}{d\theta }}+{\frac {dF}{d\theta }}=0,\quad {\frac {dH}{dr}}+{\frac {dF}{dr}}=0.$

If we assume r and θ are connected by r = bθ^m, we can look at r as a function of θ, consider F as depending only on θ, and retain only the first equation (8) with:

$H={\frac {\varphi }{bk^{2}\theta ^{m}}},\quad {\frac {dH}{d\theta }}={\frac {-m\theta }{bk^{2}\theta ^{m+1}}}+{\frac {\varphi ^{\prime }}{bk^{3}\theta ^{m}}}$

For k = θ we need equation (8) to be satisfied; which gives, taking into account equations (7):

${\frac {dF}{d\theta }}={\frac {ma}{b\theta ^{m+3}}}+{\frac {2}{3}}{\frac {a}{b\theta ^{m+3}}}$

where:

$F={\frac {-a}{b\theta ^{m+2}}}{\frac {m+{\frac {2}{3}}}{m+2}}$

and in the hypothesis of Lorentz, where m = -1:

$F={\frac {a}{3b\theta }}.$

Now suppose that there is no connection and, considering r and θ as independent variables, retain the two equations (H); it follows:

$H={\frac {\varphi }{k^{2}r}},\quad {\frac {dH}{d\theta }}={\frac {\varphi ^{\prime }}{k^{3}r}},\quad {\frac {dH}{dr}}={\frac {-\varphi }{k^{2}r^{2}}},$

Equations (8) must be satisfied for k = θ, r = bθ^m; which gives:

(9)	${\frac {dF}{dr}}={\frac {a}{b^{2}\theta ^{2m+2}}},\quad {\frac {dF}{d\theta }}={\frac {2}{3}}{\frac {a}{b\theta ^{m+3}}}.$