Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/141

Now, since the forces which act on the fluid are derived from the potential function $V$, the work which they do is subject to the law of conservation of energy, and the work done on the whole fluid within a certain space may be found if we know the potential at the points where each line of flow enters the space and where it issues from it. The excess of the second of these potentials over the first, multiplied by the quantity of fluid which is transmitted along each line of flow, will give the work done by that portion of the fluid, and the sum of all such products will give the whole work.

Now, if the space be bounded by a surface for which $V=C$, a constant quantity, the potential will be the same at the place where any line of flow enters the space and where it issues from it, so that in this case no work will be done by the forces on the fluid within the space, and $M=0$.

Secondly, if the space be bounded in whole or in part by a surface satisfying equation (4), no fluid will enter or leave the space through this surface, so that no part of the value of $M$ can depend on this part of the surface.

The quantity $M$ is therefore zero for a space bounded externally by the closed surface $V=C$, and it remains zero though any part of this space be cut off from the rest by surfaces fulfilling the condition (4).

The analytical expression of the process by which we deduce the work done in the interior of the space from that which takes place at the bounding surface is contained in the following method of integration by parts.

Taking the first term of the integral $M$,

$\iiint u{\frac {dV}{dx}}\,dx\,dy\,dz=\iint \Sigma (uV)\,dy\,dz-\iiint V{\frac {du}{dx}}\,dx\,dy\,dz$,

where

$\Sigma (uV)=u_{1}V_{1}-u_{2}V_{2}+u_{3}V_{3}-u_{4}V_{4}\,\!$ + &c.;

and where $u_{1}V_{1}$, $u_{2}V_{2}$, &c. are the values of $u$ and $v$ at the points whose coordinates are $(x_{1},y,z)$, $(x_{2},y,z)$, &c., $x_{1}$, $x_{2}$, &c. being the values of $x$ where the ordinate cuts the bounding surface or surfaces, arranged in descending order of magnitude.

Adding the two other terms of the integral $M$, we find

$M=\iint \Sigma (uV)\,dy\,dz+\iint \Sigma (uV)\,dz\,dx+\iint \Sigma (uV)\,dx\,dy$

$-\iiint V({\frac {du}{dx}}+{\frac {dv}{dy}}+{\frac {dw}{dz}})\,dx\,dy\,dz$.