Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/202

where $dv_{1},dv_{2}$ are two infinitesimal elements of volume, $({\dot {\xi }}+{\dot {\xi '}})_{1},({\dot {\xi }}+{\dot {\xi '}})_{2}$ the corresponding components of flux, $r$ the distance between the elements, and $\textstyle \sum$ denotes a summation with respect to the coordinate axes. Separating the integrations, we may write for the same quantity

${\tfrac {1}{2}}\textstyle \sum \displaystyle \int ({\dot {\xi }}+{\dot {\xi '}})_{1}\left[\int {\frac {({\dot {\xi }}+{\dot {\xi '}})_{2}}{r}}dv_{2}\right]dv_{1}.$

It is evident that the integral within the brackets is derived from ${\dot {\xi }}+{\dot {\xi '}}$ by the same process by which the potential of any mass is derived from its density. If we use the symbol $Pot$ to express this relation, we may write for the kinetic energy

${\tfrac {1}{2}}\textstyle \sum \displaystyle \int ({\dot {\xi }}+{\dot {\xi '}}){\text{ Pot }}({\dot {\xi }}+{\dot {\xi '}})dv.$

The operation denoted by this symbol is evidently distributive, so that ${\text{Pot }}({\dot {\xi }}+{\dot {\xi '}})={\text{Pot }}{\dot {\xi }}+{\text{Pot }}{\dot {\xi '}}.$ The expression for the kinetic energy may therefore be expanded into

${\tfrac {1}{2}}\textstyle \sum \displaystyle \int {\dot {\xi }}{\text{ Pot }}{\dot {\xi }}dv+{\tfrac {1}{2}}\textstyle \sum \displaystyle \int {\dot {\xi }}{\text{ Pot }}{\dot {\xi '}}dv+{\tfrac {1}{2}}\textstyle \sum \displaystyle \int {\dot {\xi '}}{\text{ Pot }}{\dot {\xi }}dv+{\tfrac {1}{2}}\textstyle \sum \displaystyle \int {\dot {\xi '}}{\text{ Pot }}{\dot {\xi '}}dv.$

But ${\dot {\xi ',}}$ and therefore ${\text{Pot }}{\dot {\xi '}},$ has in every wave-plane the average value zero. Also ${\dot {\xi }},$ and therefore ${\text{Pot }}{\dot {\xi }},$ has in every wave-plane a constant value. Therefore the second and third integrals in the above expression will vanish, leaving for the kinetic energy

${\tfrac {1}{2}}\textstyle \sum \displaystyle \int {\dot {\xi }}{\text{ Pot }}{\dot {\xi }}dv+{\tfrac {1}{2}}\textstyle \sum \displaystyle \int {\dot {\xi '}}{\text{ Pot }}{\dot {\xi '}}dv,$

(3)

which is to be calculated for a time of no displacement, when

${\dot {\xi }}=\pm {\frac {2\pi \alpha }{p}}\cos 2\pi {\frac {u}{l}},$⁠${\dot {\eta }}=\pm {\frac {2\pi \beta }{p}}\cos 2\pi {\frac {u}{l}},$⁠${\dot {\zeta }}=\pm {\frac {2\pi \gamma }{p}}\cos 2\pi {\frac {u}{l}}\cdot$

(4)

The form of the expression (3) indicates that the kinetic energy consists of two parts, one of which is determined by the regular part of the flux, and the other by the irregular part of the flux.

8. The value of ${\text{Pot }}{\dot {\xi }}$ may be easily found by integration, but perhaps more readily by Poisson's well-known theorem, that if $q$ is any function of position in space (as the density of a certain mass),

${\frac {d^{2}{\text{ Pot }}q}{dx^{2}}}+{\frac {d^{2}{\text{ Pot }}q}{dy^{2}}}+{\frac {d^{2}{\text{ Pot }}q}{dz^{2}}}=-4\pi q,$

(5)

where the direction of the coordinate axes is immaterial, provided that they are rectangular. In applying this to ${\text{Pot }}{\dot {\xi }},$ we may place two of the axes in a wave-plane. This will give

${\frac {d^{2}{\text{Pot }}{\dot {\xi }}}{du^{2}}}=-4\pi {\dot {\xi }}.$

(6)

In a nodal plane, ${\text{Pot }}{\dot {\xi }}=0,$ since ${\dot {\xi }}$ has equal positive and negative values in elements of volume symmetrically distributed with respect