Stokes's theorem becomes an obvious truism if applied to an incompressible fluid. Let a _source_ of fluid be a point from which an incompressible fluid is emitted in all directions. Close to the source the stream lines will be radial lines. Let a very small sphere be described round the source, and let the strength of the source be defined as the total flow per second through the surface of this small sphere. Then if we have any number of sources enclosed by any surface, the total flow per second through this surface is equal to the total strengths of all the sources. If, however, we defined the strength of the source by the statement that the strength divided by the square of the distance gives the velocity of the liquid at that point, then the total flux through any enclosing surface would be 4[pi] times the strengths of all the sources enclosed. To every proposition in electrostatics there is thus a corresponding one in the hydrokinetic theory of incompressible liquids. Let us apply the above theorem to the case of a small parallel-epipedon or rectangular prism having sides dx, dy, dz respectively, its centre having co-ordinates (x, y, z). Its angular points have then co-ordinates (x ± ½dx, y ± ½dy, z ± ½dz). Let this rectangular prism be supposed to be wholly filled up with electricity of density [rho]; then the total quantity in it is [rho] dx dy dz. Consider the two faces perpendicular to the x-axis. Let V be the potential at the centre of the prism, then the normal forces on the two faces of area dy·dx are respectively /dV 1 d²V \ /dV 1 d²V \ - ( -- + -- --- dx ) and ( -- - -- --- dx ), \dx 2 dx² / \dx 2 dx² / and similar expressions for the normal forces to the other pairs of faces dx·dy, dz·dx. Hence, multiplying these normal forces by the areas of the corresponding faces, we have the total flux parallel to the x-axis given by -(d²V/dx²)dx dy dz, and similar expressions for the other sides. Hence the total flux is /d²V d²V d²V\ - ( --- + --- + --- ) dx dy dz \dx² dy² dz²/ and by the previous theorem this must be equal to 4[pi][rho]dx dy dz. d²V d²V d²V Hence --- + --- + --- + 4[pi][rho] = 0 (18). dx² dy² dz² This celebrated equation was first given by S.D. Poisson, although previously demonstrated by Laplace for the case when [rho] = 0. It defines the condition which must be fulfilled by the potential at any and every point in an electric field, through which [rho] is finite and the electric force continuous. It may be looked upon as an equation to determine [rho] when V is given or vice versa. An exactly similar expression holds good in hydrokinetics, provided that for the electric potential we substitute velocity potential, and for the electric force the velocity of the liquid.