and this is obviously true of their projections on any plane.

The corresponding theorem for the scalar product, that Sa (B+y)= Saß + Say, is obvious if we regard a as an area made to take the steps B, y.

But there is a very important difference between a vector product and a product of two scalar quantities. Namely, the sign of an area depends upon the way it is gone round; an area gone round counter-clockwise is positive, gone round clockwise is negative. Now if V. ab. ac area abdc, we must have by symmetry V.ac.ab = area acdb, and therefore V. ac. ab=-V. ab. ac, or Vyẞ- Vẞy. Hence the sign of a vector product is changed by inverting the order of the terms. It is agreed upon that Vaß shall be a vector facing to that side from which the rotation from a to B appears to be counterclockwise.

It will be found, however, that Saß = Sẞa, so that the scalar product of two vectors resembles in this respect the product of scalar quantities.

MOMENT OF VELOCITY OF A MOVING POINT.

19

The flux of the moment of velocity of a moving point p about a fixed point o is equal to the moment of the acceleration about o. For suppose that during a certain interval of time the velocity has changed from p to p1, so that P1p is the change of the velocity; then the sum of the moments of p and 1-p is equal to the moment of p1, that is the moment of the change in the velocity is equal to the change in the moment of velocity. Dividing each of these by the interval of time, we see that the moment of the mean flux of velocity is equal to the mean flux of the moment of velocity, during any interval. Consequently the moment of acceleration is equal to the flux of the moment of velocity.

The same thing may be shewn in symbols, as follows, supposing the motion to take place in one plane. We

MOMENT OF ACCELERATION.

may write prei, where r is the length of op, and the angle Xop. Then pre+rẻ. ieie, or the velocity consists of two parts, along op and re perpendicular to it. The

j

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moment of the velocity is the sum of the moments of these parts; but the part along op (radial component) has no moment, and the part perpendicular (transverse component) has moment rẻ. Next, taking the flux of p, we find for the acceleration the value

ï = reio + Ör. ieio + rẻ. ¿eio + rė2.. î2eio + rÖ. ieio

= (ï − rẻ3) eio + (2ṛ0+rë) ieio. Or the acceleration consists of a radial component → — rẻ2, and a transverse component 2r+rë. The moment of the acceleration is r times the transverse component, namely 2rrė +r2. But this is precisely the flux of the moment of velocity r2ė.

Observe that the radial acceleration consists of two parts, due to the change in magnitude of the radial velocity, and – rẻ due to the change in direction of the transverse velocity.

We may also make this proposition depend upon the flux of a vector product. The moment of the velocity is Vpp, and the moment of the acceleration is Vpp; we have therefore to prove that Vpp is the rate of change of Vpp. Now upon referring to the investigation of the flux of a product, p. 64, the reader will see that every step of it applies with equal justice to a product of two vectors, whether the product be vector or scalar. In fact, the only property used is that the product is distributive. Hence the rate of change of Vaß is Vaß + Väß. (Observe that the order of the factors must be carefully kept.) Applying this rule to Vpp, we find that its rate of change is Vpp+ Vpp. Now the vector product of two parallel vectors is necessarily zero, because they cannot include any area; thus Vpp=0. Therefore (Vpp)= Vpp. This demonstration does not require the motion to be in one plane.

The moment of velocity about any point is equal to twice the rate of description of areas about that point. When the motion is in a circle, twice the area

aop being equal to r20, and r constant, its flux is r2ẻ, the moment of velocity. In any other path aq, having the same angular velocity, the area described in the same time is oaq, and the mean flux of area in the two cases is oap and oaq respectively divided by the time. The ratio of their difference to either of these is the

ratio of apq to oap or oaq, which is approximately the ratio of pq to op or oq, and can be made as small as we like by taking p near enough to a. Thus the mean fluxes

in the two cases approach one another without limit as they approach the true fluxes; or the true fluxes are equal. Hence twice the rate of description of areas is always r10, the moment of velocity.

When the acceleration is always directed towards a fixed point o, the moment of velocity is constant, and equal areas are swept out by the radius vector in equal times. If the acceleration of p passes through o, its moment about o is zero; consequently the flux of the moment of velocity is zero, or that moment is constant. Because it is constant in direction, the path is a plane curve; for the plane containing op and the velocity has always to be perpendicular to a fixed line. Because it is constant in magnitude, the rate of description of areas is also constant, or, which is the same thing, equal areas are swept out in equal times.

F

f

E

The following is Newton's proof of this proposition. Let the time be divided into equal parts, and in the first interval let the body describe the straight line AB with uniform velocity. In the second interval, if the velocity were unchanged, it would go to c, if Bc = AB;

UNIFORM DESCRIPTION OF AREAS.

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so that the equal areas ASB, BSc would be completed in equal times.

But when the body arrives at B, let a velocity in the direction BS be communicated to it. The new velocity of the body will be found by drawing cC parallel to BS to represent this addition, and joining BC. At the end of the second interval, then, the body will be at C, in the plane SAB. Join SC, then area SCB = ScB (between same parallels SB and Cc) = SBA.

In like manner, if at C, D, E, velocities along CS, DS, ES are communicated, so that the body describes in successive intervals of time the straight lines CD, DE, EF, etc., these will all lie in the same plane; and the triangle SCD will be equal to SBC, and SDE to SCD, and SEF to SDE.

Therefore equal areas are described in the same plane in equal intervals; and, componendo, the sums of any number of areas SADS, SAFS, are to each other as the times of describing them.

Let now the number of these triangles be increased, and their breadth diminished indefinitely; then their perimeter ADF will be ultimately a curved line; and the instantaneous change of velocity will become ultimately a continuous acceleration in virtue of which the body is continually deflected from the tangent to this curve; and the areas SADS, SAFS, being always proportional to the times of describing them, will be so in this case. Q.E.D.

The constant moment of velocity will be called h. It is twice the area described in one second. If p be the length of the perpendicular from the fixed point on the tangent, we shall have h=vpr20. A path described with acceleration constantly directed to a fixed point is called a central orbit, and the fixed point the centre of acceleration. In a central orbit, then, the velocity is inversely as the central perpendicular on the tangent, for v = h : P, and the angular velocity is inversely as the squared distance from the centre, for 0=h: r2.

RELATED CURVES.

Inverse. Two points p and q so situated on the radius of a circle that cp.cq=ca, are called inverse points in regard to the circle.

If p moves about so as to trace out any curve, q will also move about, and trace out another curve; either of these curves is called the inverse of the other in regard to the circle.

The inverse of a circle is in general another circle; but it coincides with its inverse when it cuts the circle of inversion at right angles, and the inverse is a straight line when it passes through the centre of inversion. We know that

cp.cq=ct,

which proves the second case; the first is easily derived from it; and the third follows from the similarity of the triangles cmp, cqb, which gives

cp.cq=cm. cb, which is constant and therefore

In the second case the circle clearly makes equal angles with cpq at p and q. In general, two inverse curves make equal angles with the radius vector at corre

sponding points. For we can always draw a circle to touch the first curve at p and to pass through q; such a circle is then its own inverse, and makes equal angles at p and q with cpq. Moreover it touches the second curve at q, for as two points of intersection coalesce at P, their two inverse points coalesce at q. Hence the two inverse curves make equal angles with cpq.

cd ca3.

=