Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/381

the current normal to its plane is $u,$ so that the quantity which enters through this triangle is ${\frac {1}{2}}r^{2}{\frac {u}{mn}}.$

The quantities which enter through the triangles $OCA$ and $OAB$ respectively are

${\frac {1}{2}}r^{2}{\frac {v}{nl}},\quad \quad {\mbox{and}}\quad \quad {\frac {1}{2}}r^{2}{\frac {w}{lm}}.$

If $\gamma$ is the component of the velocity in the direction $OR,$ then the quantity which leaves the tetrahedron through $ABC$ is

${\frac {1}{2}}r{\frac {\gamma }{lmn}}.$

Since this is equal to the quantity which enters through the three other triangles,

${\frac {1}{2}}{\frac {r^{2}\gamma }{lmn}}={\frac {1}{2}}r^{2}\left\lbrace {\frac {u}{mn}}+{\frac {v}{nl}}+{\frac {w}{lm}}\right\rbrace ;$

multiplying by ${\frac {2lmn}{r^{2}}},$ we get

$\gamma =lu+mv+nw$

(1)

If we put

$u^{2}+v^{2}+w^{2}=\Gamma ^{2}$

and make $l',m',n'$ such that

$u=l'\Gamma ,\quad v=m'\Gamma ,\quad {\mbox{and}}\quad w=n'\Gamma ;$

then

$\gamma =\Gamma (ll'+mm'+nn').$

(2)

Hence, if we define the resultant current as a vector whose magnitude is $\Gamma ,$ and whose direction-cosines are $l',m',n'$ and if $\gamma$ denotes the current resolved in a direction making an angle $\theta$ with that of the resultant current, then

$\gamma =\Gamma \cos \theta ;$

(3)

shewing that the law of resolution of currents is the same as that of velocities, forces, and all other vectors.

287.] To determine the condition that a given surface may be a surface of flow.

Let

$F(x,y,z)=\lambda$

(4)

be the equation of a family of surfaces any one of which is given by making $\lambda$ constant, then, if we make

$\left.{\frac {\overline {d\lambda }}{dx}}\right|^{2}+\left.{\frac {\overline {d\lambda }}{dy}}\right|^{2}+\left.{\frac {\overline {d\lambda }}{dz}}\right|^{2}={\frac {1}{N^{2}}},$

(5)

the direction-cosines of the normal, reckoned in the direction in which $\lambda$ increases, are

$l=N{\frac {d\lambda }{dx}},\quad \quad m=N{\frac {d\lambda }{dy}},\quad \quad n=N{\frac {d\lambda }{dz}}.$

(6)