The total repulsion between the two bodies is

(V1V2+R ̧R ̧)b−(V ̧R2+V ̧R1) c−(V2+R1) Ma−(V2+R) M ̧a,+M ̧M ̧r1

19

The first term of this expression, with its sign reversed, represents the attraction of gravitation, and the second term represents the observed electric action, but the other terms represent forces of a kind which have not hitherto been observed, and we must modify the theory so as to account for their non-existence.

One way of doing so is to suppose bc and a1 = a,= 0. The result of this hypothesis is to reduce the condition of saturation to that of the equality of the two fluids in the body, leaving the amount of each quite undetermined. It also fails to account for the observed action between the bodies themselves, since there is no action between them and the electric fluids.

The other way is to suppose that S1 = S, 0, or that the sum of the quantities of the two fluids in a body always remains the same as when the body is saturated. This hypothesis is suggested by Priestley in his account of the two-fluid theory, but it is not a dynamical hypothesis, because it does not give a physical reason why the sum of these two quantities should be incapable of alteration, however their difference is varied.

The only dynamical hypothesis which appears to meet the case is to suppose that the vitreous and resinous fluids are both incompressible, and that the whole of space not occupied by matter is occupied by one or other of them. In a state of saturation they are mixed in equal proportions.

The two-fluid theory is thus considerably more difficult to reconcile with the facts than the one-fluid theory.

NOTE 2, ARTS. 27 AND 282.

The problem of the distribution, in a sphere or ellipsoid, of a fluid, the particles of which repel each other with a force varying inversely as the nth power of the distance, has been solved by Green*. Green's method is an extremely powerful one, and allows him to take account of the effect of any given system of external forces in altering the distribution.

If, however, we do not require to consider the effect of external forces, the following method enables us to solve the problem in an elementary manner. It consists in dividing the body into pairs of corresponding elements, and finding the condition that the repulsions of corresponding elements on a given particle shall be equal and opposite.

(1) Specification of Corresponding Points on a line.

Let 4,4, be a finite straight line, let P be a given point in the line, and let Q and Q, be corresponding points in the segments AP and PA, respectively, the condition of correspondence being

It is easy to see that when Q, coincides with A1, Q, coincides with A2, and that as Q, moves from A, to P, Q2 moves in the opposite direction from A to P, so that when Q, coincides with P, Q, also coincides with P.

Let Q and Q, be another pair of corresponding points, then

If the points Q, and Q,' are made to approach each other and ultimately

"Mathematical Investigations concerning the laws of the equilibrium of fluids analogous to the electric fluid, with other similar researches," Transactions of the Cambridge Philosophical Society, 1833. Read Nov. 12, 1832. See Mr Ferrers' Edition of Green's Papers, p. 119.

to coincide, Q,Q,' ultimately becomes the fluxion of Q, which we may write Q, and we have

or corresponding elements of the two segments are in the ratio of the squares of their distances from P.

density of the redundant fluid at Q1 is and at Q2 is

Let us now suppose that A,PA, is a double cone of an exceedingly small aperture, having its vertex at P; let us also suppose that the Pai then since the areas of the sections of the cone at Q, and Q, are as the squares of the distances from P, and since the lengths of corresponding elements are also, by (5), as the squares of their distances from P, the quantities of fluid in the two corresponding elements at Q, and Q, are as p,Q,P to PPQ. If the repulsion is inversely as the nth power of the distance, the condition of equilibrium of a particle of the fluid at P under the action of the fluid in the two corresponding elements at Q, and Q, is

We have now to show how this condition may be satisfied by one and the same distribution of the fluid when P is any point within an ellipsoid or a sphere. We must therefore express p so that its value is independent of the position of P.

Multiplying the corresponding members of equations (1) and (7) and omitting the common factor AP. PÃ ̧,

Let us now suppose that 4,4, is a chord of the ellipsoid, whose equation is

then the product of the segments of the chord at Q, is to the product of the segments at Q, as the values of p at these points respectively, or

We may therefore write, instead of equation (9),

where C is constant, every particle of the fluid within the ellipsoid will be in equilibrium.

We have in the next place to determine whether a distribution of this kind is physically possible.

Let E be the quantity of redundant fluid in the ellipsoid,

Let Po be the density of the redundant fluid if it had been uniformly spread through the volume of the ellipsoid, then

and if p is the actual density of the redundant fluid,

(17)

(18)

2

When n is not less than 2, there is no difficulty about the interpretation of this result.

The density of the redundant fluid is everywhere positive.

When n = 4 it is everywhere uniform and equal to p..

When n is greater than 4 the density is greatest at the centre and is zero at the surface, that is to say, in the language of Cavendish, the matter at the surface is saturated.

When n is between 2 and 4 the density of the redundant fluid at the centre is positive and it increases towards the surface. At the surface itself the density becomes infinite, but the quantity collected on the surface is insensible compared with the whole redundant fluid.

When n is

equal to 2, r (222)

becomes infinite, and the value of p is zero for all points within the ellipsoid, so that the whole charge is collected on the surface, and the interior parts are exactly saturated, and this we find to be consistent with equilibrium.

When n is less than 2 the integral in equation (15) becomes infinite. Hence if we assume a value for C in the interior parts of the ellipsoid, we cannot extend the same law of distribution to the surface without introducing an infinite quantity of redundant fluid. We might therefore conclude that if the quantity of redundant fluid is given, we must make C = 0, and suppose the redundant fluid to be all collected at the surface, and the interior to be exactly saturated. But, on trying this distribution, we find that it is not consistent with equilibrium. For when n is less than 2, the effect of a shell of fluid on a particle within it is a force directed from the centre. If, therefore, a sphere of saturated matter is surrounded by a shell of electric fluid, the fluid in the sphere will be drawn towards the shell, and this process will go on till the different parts of the interior of the sphere are rendered undercharged to such a degree that each particle of fluid in the sphere is as much attracted to the centre by the matter of the sphere as it is repelled from it by the fluid in the sphere and the shell together. This is the same conclusion as that stated by Cavendish.

Green solves the problem, on the hypothesis of two fluids, in the following manner.

Suppose that the sphere, when saturated, contains a finite quantity, E, of the positive fluid, and an equal quantity of the negative fluid, and let a quantity, Q, of one of them, say the positive, be introduced into the sphere.

Let the whole of the positive fluid be spread uniformly over the surface of the sphere whose radius is a, so that if P is the surfacedensity,

Green then considers the equilibrium of fluid in an inner and concentric sphere of radius b, acted on by the fluid in the surface whose radius is a, and shows that if the density of the fluid is

there will be equilibrium of the fluid within the inner sphere.

The value of p is evidently negative if n is less than 2.

Green then determines, from this value of the density, the whole quantity of fluid within the sphere whose radius is b, and then by equating this to E, the whole quantity of negative fluid, he determines the radius, b, of the inner sphere, so that it shall just contain the whole of the negative fluid.