of the two other points, but generally to determine any two of the six co-ordinates from the other four. Next let us assume that the points Pa which the ray cuts planes a and b, are given, and that the point is yet unknown in which it cuts plane c. This point let us call p., to distinguish it from other points in the same plane. Take any point p, in plane c, and consider two auxiliary rays, one of which goes from Pa to Pe and the other from Po to p.. In Fig. 26 these are again shewn dotted, while the primary ray is shewn full. Let Ta and T be the times of passage of these auxiliary Fays. Then the value of the difference T-Te depends on the position of p. in plane c. Among the various ac and Po (Fig. 26), in Рь Pa Fig. 26. values obtained by giving to p, various positions in the neighbourhood of p', the maximum must be that obtained by making p, coincide with p'.. For in that case the ray passing from p. to p, cuts the plane b in the given point p,, and is therefore made up of the ray which passes from p. to p, and of that which passes from p, to p.. Accordingly we may put Pa Tac = Tas + Toe Pa Hence the required difference is given in this special case by If on the other hand p. does not coincide with p', then the ray which passes from Pa to Pe does not coincide with the two which pass from Pa to Ро and from p, to pc; and since the direct ray between P. and Po travels in the shortest time, we must have and therefore we have in general for the required difference the inequality Tac-Toe <Tab This difference is thus generally smaller than in the special case where p. lies in the continuation of the ray which passes from pa to P, and this special value of the difference thus forms a maximum*. Hence we have the two following conditions: Pa If we lastly assume that the points p, and p. in the planes b and c are given beforehand, while the point in which the ray cuts the plane a is still unknown, we obtain by an exactly similar procedure the two following conditions: d (Tac-Tab) =0; d (Tac-Tab) = 0........(3). In this way we arrive at three pairs of equations, each of which serves to express the corresponding relation between the three points in which a ray cuts the three planes a, 3, c; so that if two of the points are given the third can be found, or, more generally, if of the six co-ordinates of the three points four are given the other two may be determined. § 6. Relation between Corresponding Elements of Surface. We will now take the following case. Given on one of the planes, say a, a point P., and on another, say b, an element of surface ds; then if rays pass from pa to the different points of ds, and if we suppose these rays produced till they cut the third plane c, they will all cut that plane in general within another indefinitely small element of surface, which we will call ds. (Fig. 27). Let us now determine the relation between ds, and ds.. *In Kirchhoff's paper (p. 285) it is stated of the quantity there considered, which is essentially the same as the difference here treated of, except that it refers to four planes instead of three, that it must be a minimum. This may possibly be a printer's error, and in any case an interchange of maximum and minimum in this place would have no further influence, because the principle used in the calculations which follow, viz., that the differential coefficient=0, holds equally for a maximum or a minimum. In this case, of the six co-ordinates which relate to each ray (viz. those of the three points in which the ray cuts the three planes) two, viz. x and y., are given beforehand. If we now take any values we please for and ყ.. the co-ordinates x, and y are in general thereby determined. Thus in this case we may consider x and y. as two functions of x, and y. As the form of the element ds, may be any whatever, let it be a rectangle da, dy, and let us find the point in plane c corresponding dsc Pa ds. Fig. 27. to every point in the outline of this rectangle. We shall then have on plane c an indefinitely small parallelogram which forms the corresponding element of surface. The magnitude of this parallelogram is determined as follows. Let be the length of the side which corresponds to the side da, of the rectangle in plane b, and let (xx) and (y) be the angles which this side makes with the axes of co-ordinates. Then Again, let u be the other side of the parallelogram, and let (ux) and (uy) be the angles it makes with the axes. Then we have dy。 = dy Let (u) be the angle between the sides λ and μ. Then we have cos (μ) = cos(xx) cos (ux) + cos (λy) cos (μy.) Now to determine the area ds. of the parallelogram, we may write Here we may substitute for cos (u) the expression just given, and for λ and the following expressions derived from the above equations : Then several terms under the root cancel each other, and the remainder form a square as follows: This quadratic equation can therefore be solved at once. But it must be observed that the difference within the brackets may be either positive or negative, and, as we have only to do with the positive root, we will denote this by putting the letters v.n. (valor numericus) before this difference. We can then write . To ascertain how x, and y, depend upon x, and y, we must apply one of the three pairs of equations in § 5. We will first choose equations (1). If we differentiate those equations according to x, and y,, remembering that each of the quantities denoted by T contains two of the three pairs of co-ordinates xa, Ya, Xo, Yo, xc, Y., as denoted by the indices; and if in differentiating we treat x, and y, as functions of If by help of these equations we determine the four differential coefficients dx dx dy dye and substitute с dx dy dx' dy the values thus found in equation (4), we obtain the required relation between ds, and ds. To be able to write the result more briefly, we will use the following symbols: Then the required relation may be written as follows: is Again, if we suppose in like manner that a point pe given on plane c (Fig. 28), and find on plane a the element ds, which corresponds to the given element ds, on plane b, then the result can be derived from that last given by simply interchanging the indices a and c. If for brevity we write dTab dᎢ . d2Tab dx dx dy dy dx dy dy)...(9), a de Tab dyadx |