Page:A history of the theories of aether and electricity. Whittacker E.T. (1910).pdf/481

an examination of the circumstances of a collision, in the manner of the kinetic theory of gases, shows that (b − a) must have the form -urχ(r)/l, where l denotes a constant which is closely related to the mean free path of the electrons. In the terms on the right-hand side of the equation, on the other hand, Maxwell's term gives a result different from zero; and in comparison with this we may neglect the terms which arisefrom -uχ(r). Thus we have

$-{\frac {ur\chi (r)}{l}}=\left({\frac {eE}{m}}{\frac {\partial }{\partial u}}+u{\frac {\partial }{\partial x}}\right){\frac {N}{\pi ^{\frac {3}{2}}a^{3}}}e^{-{\frac {r^{2}}{a^{2}}}}$,

or

$u\chi (r)={\frac {lu}{\pi ^{\frac {3}{2}}r}}.e^{-{\frac {r^{2}}{a^{2}}}}.\left\{{\frac {2eNE}{ma^{5}}}-{\frac {d}{dx}}\left({\frac {N}{a^{3}}}\right)-{\frac {2Nr^{2}}{a^{6}}}{\frac {da}{dx}}\right\}$;

and thus the law of distribution of velocities is determined.

The electric current i is determined by the equation

$i=e\textstyle \iiint uf(u,v,w)\ du\ dv\ dw$,

where the integration is extended over all possible values of the components of velocity of the electrons. The Maxwellian term in f(u, v, w) furnishes no contribution to this integral, so we have

$i=e\textstyle \iiint u^{2}\chi (r)\ du\ dv\ dw$.

When the integration is performed, this formula becomes

$i={\frac {2le}{3\pi ^{\frac {1}{2}}}}\left({\frac {2eNE}{ma}}-a{\frac {dN}{dx}}-N{\frac {da}{dx}}\right)$,

or

$E={\frac {3\pi ^{\frac {1}{2}}m}{4le^{2}}}{\frac {a}{N}}i+{\frac {m}{2e}}\left({\frac {a^{2}}{N}}{\frac {dN}{dx}}+a{\frac {da}{dx}}\right)$.

The coefficient of i in this equation must evidently represent the ohmic specific resistance of the metal; so if γ denote the specific conductivity, we have

$\gamma ={\frac {4le^{2}}{3\pi ^{\frac {1}{2}}m}}{\frac {N}{a}}$.

Let the equation be next applied to the case of two metals. A and B in contact at the same temperature T, forming an