aberration as the curvatures are altered are now shown for the selected g and u. I propose next to illustrate the use of these methods by solving a few numerical problems, but instead of employing graphs (which would involve expensive plates) the solutions will be entirely algebraical. In a discussion of principles it it desirable to use 5-figure accuracy, although in practice 4 figures would be ample. I take the case of two glasses for which 1/N=0·658805, 1/n=61721; also p1 = 2.508, p2 = -1.508. These two powers were, as a matter of fact, selected for C-F achromatism, but we have no concern with that. The combination is one which was studied in some detail in Professor Conrady's autumn class last year. I take first a few simple questions on the front lens alone; p must then be put equal to zero in the coefficients; A=2.9063, B=5.4143, D= −8.3206, G=—24·5088, H=33.7267, L=67.7599. Problem (1). If the object is at infinity, what curves give least spherical aberration? Solve equation (3) with u1 and g=0; c=-G/2A=4-2165; c2 is less than c, by 4.8426 in each example, and is here -0.6261; the so-called "crossed lens." The aberration, according to (1), is 16·089. Problem (2). If the lens is given, for what point will its aberration be least? For example, suppose c=30, 18426. Solve (5) with c=3 and g=0; u1=-(3D H)/2B= −0·80942, i. e. the object is about 12 focal lengths in front of the lens, and the aberration is 16.843. Problem (3). Is it possible for a single thin lens to be free from aberration? For correction, parabola I must cross the horizontal axis, so that its vertex must not be above that line. The critical positions are where parabola III crosses the line. Its vertex is given by solving (3) and (5), with g=0; these give u1= Pi P1 2(N-1) =24213, as common-sense shows. For these values the aberration is 16-9400. Parabola III is therefore given by: "excess aberration above 16.9400 square of excess curvature above 2:4213." Where it crosses the line, aberration =0, therefore c110-4309 or -5.5883, and, by (3), u,=4.3396 or -68494. So that aberration cannot be corrected except for object-points within about of the focal length on one side (real) and on the other (virtual), and even then the curves have to be very strong. There is no useful case. As a mere arithmetical exercise, we may take u1=-8; then c1 has to be 8:5342, and c, is -13.377. Problem (4). To correct simultaneously for spherical aberration and for sine-error; in other words, to find the two "aplanatic points." The position of these is well-known, so that the exercise affords a useful check on the foregoing formulæ. Equate the expressions (1) and (2) to zero, and solve simultaneously, with P-41603, Q=-6.6683, S=-184354, and g=0; the solutions are u=48426, c=12.1932, c=73506, or u=-73506, c1=-73506, C2-12-1932. These tally with the familiar solutions u2 = c2=p1/(N-1); u1 = c1 = -p/(N-1). In one case the rays enter the first face normally, in the other they leave the second face normally. The cases are both useless, so far as telescopic work is concerned. = I take next a few examples on the doublet. The coefficients are now—A = 1.22153, B 2.22153, C = −1·68475, D=344306, E=-3-36951, F=4-87751, G=-373215, H = 483548, K = 20 77590, L 1-81770, P= 1.72153, Q=-272153, R= −2·43875, S= −2·66771. = The latus rectum of all parabolas of class I is 1/(1-22153); II, — 1/(2·10720); III, −1/(0·103042); IV, −1/(0·099481); V, 1/(1-19840). Problem (5). With parallel light and a cemented lens, what curves give (algebraically) least aberration? Solve (3) with =0, g=0: c=15277, ca=3=-33149, c=C3 +24315=0·8834; the aberration is -1·0330. Problem (6). Given =0 and g=0, find the aberration for 45. Parabola (I) gives: excess aberration above −1·0330 = 1·22153 × (square of excess curvature above -33149), so that the aberration = — - 1.0330+ 1.7155 =0·6824. Problem (7). Given u=0 and g=0, what curves will remove spherical aberration? We must have 1·0330 = 1·22153(-1.5277), or c1=0·60806 or 24473. Then, as usual, deduct 48426 for c2, and add 2-4315 to cg for c. Problem (8). Repeat (5), with g=0.1. The vertex of the parabola is now given by c1 = 1.6656 instead of 1.5277; indeed, every 0·1 in g means 0·1379 extra in ; so, an axial gap (g=-01) would mean c=13898. With u1 =0, g=0·1, and c1=1-6656, the aberration is 0:4897. As this is positive, no choice of curves with air-gap 01 (and parallel rays) can give freedom from aberration. Yet for a suitable value of u, this could be done. Problem (9). Over what range of values of u could it be done? To answer this, we require parabola III. Its latus rectum has been given above; its vertex is found by solving (3) and (5) with g=0·1; namely, c1 =0·24933, aberration=0·6965. 480 Graphical Methods of correcting Telescopic Objectives. It crosses the axis where 0.6965·103042 (c1 −0·24933)2, i. e. where c=2·8492 and −2·3505, and (3) gives for the corresponding values =0.8398 to -2.8497. Provided that does not lie within this range, an air-gap 01 is not incompatible with spherical correction. Problem (10). What is the largest air-gap which could be substituted for 0.1 in problem (8) so as to make spherical correction possible? This time parabola II is needed. For the vertex, solve (3) and (4) with u,=0; namely, c1 = 4.2164. aberration = =14.201. It crosses the axis where -14.201= -2.1072(c, -4.2164), i. e. where c1=1·6204 or 6·8124, and (3) then gives for g the values 006726 and 3-8317. So that for spherical correction for a distant object the air-gap must not exceed 0.06726 (unless it has an absurdly high value). Problem (11). If u lies outside a certain range, all values of the air-gap are compatible with spherical correction; what is this range? Parabola IV supplies the answer. Its vertex is found by solving (3), (4), and (5) simultaneously; namely, c1=3.6073, u1=-0.42546, and g=19426; aberration= 14-2374. Its equation is: (aberration-14.2374)=0·099481 (c, -3 6073)2. It crosses the axis at c1=15·570 or -8.356, which means, by equations (3) and (4), that u1 =7.931 or -8.783. For stronger convergence or divergence than this, correction is possible with any air-gap, provided the curvatures are chosen properly. Problem (12). Returning to problem (7), let us find the comatic error for the two spherically corrected lenses there given. Expression (2) gives (for c1 =0.60806) the value -162092; and for c=2.4473, the error is 1-5454. Equating (2) to zero, with u,=0 and g=0, we have c1 = 1.5496 to comply with the sine-condition. What the manufacturer would probably do, if he were tied down to these two glasses and a cemented doublet, would be to adopt a compromise; and this again shows the difficulty of dealing with lenses by tables, for no tables can be so voluminous as to provide for compromises. Problem (13). Giving up the idea of a cemented doublet, let us find what air-gap will remove both spherical aberration and coma for a distant object. Equate (1) and (2) to zero, and solve with u1=0. The result is: g=06721 (or 3.8339, which does not matter), c=1.6448, c2=-3.1978, c3=-3.1306, c=-0.6991. These curves, then, form a favourable starting point for calculating an achromatic aplanat. They would be improved by making small allowances for the thickness of the lenses, but this matter must be postponed. It also remains to show what actual residue of error remains in all three respects; this involves trigonometric computation. Most important of all is the question how best to utilise the above methods in order to make a much better attempt for the next trigonometric computation, instead of striking out blindly. Two other similarly computed objectives are: (1) For u1=-0.5 (object equal to image): c=0·8351, C240075, c3=-3.9539, c=-15224. (2) For u-10 (object at focus): c=0.0414, c2 = -4.8012, c3-4·7498, c-2.3183: this may be regarded as corrected for infinity, but with flint leading. LVI. Electrical Theories of Matter and their Astronomical Consequences with special reference to the Principle of Relativity. By A S. EDDINGTON, M.A., F.R.S., Plumian Professor of Astronomy in the University of Cambridge*. R. WALKER'S paper in the Philosophical Magazine arising out of the motions of the perihelia of the planets, and perhaps some remarks on the points raised may be useful. I think that he greatly overestimates the differences of opinion between us; our views seem to coincide on what he calls "the main point at issue," and it is difficult to believe that relativists in general hold any other view. Walker describes my method of dealing with the problem as depending on an unsatisfactory assumption, and concludes therefore that my treatment is invalid; but I am afraid he gives a wrong impression by not mentioning that my argument tended to disprove the assumption in question. I did not advocate the assumption leading to the equations of motion used in my paper; on the contrary I showed that the results disagreed with observation. The suggestion had been made that the famous discordance of Mercury could be accounted for by the variation of mass with velocity, according to the well-known hypothesis m=m。(1-w2/c2)−}, if account be taken of its interaction with the sun's motion through the æther. This hypothesis was examined, and the conclusion was unfavourable, the detailed results being irreconcilable with astronomical observation. That is to say, I attempted to disprove the hypothesis mm(1—w2/c2) ̄ § which Walker rejects more summarily. Sir Oliver Lodge also concludes that "If therefore the theory fails to give all the known perturbations correctly, something must be wrong; * Communicated by Sir Oliver Lodge. and by finding out what is wrong, we may perhaps discover something instructive". We may therefore start from this point of general agreement. The foregoing law of inertia corresponds to the Lorentz electron in steady motion, but the method applies equally to any other law of variation of inertia with velocity; the only difference is a numerical factor which is of no consequence to the argument. A fundamental revision of the theory is therefore necessary. Walker has stepped in with the suggestion that a trial should be made with the more general (and, I consider, more plausible) assumption that the inertia involves the acceleration as well as the velocity. I cannot predict whether his mode of developing this view will lead to an accordance with observation; but I certainly do not undertake to prove a general negative. Walker's thesis that the Lagrangian function depends on the acceleration as well as on the velocity cannot be a point of cleavage between relativity dynamics and non-relativity dynamics-if I have rightly grasped the meaning of the statement. In Newtonian particle dynamics, and also in the quasi-stationary treatment of these problems, the Lagrangian function is supposed to consist of two parts, (1) the kinetic energy, involving the velocity only, and (2) the forcefunction involving position in the field of force. The contention is that this separation is inadmissible, and that there is a cross-term involving both the velocity and the force (or acceleration). Walker's standpoint tends to associate this cross-term with the kinetic energy, so that the kinetic energy differs from the value calculated without regard to the acceleration. Sir Oliver Lodget and the writer find it more natural to group the term with the force-function, and say that the force of gravitation involves a term depending on the velocity. The distinction appears to be purely verbal. It is essential to an out-and-out relativity theory that this cross-term should exist, and it is surprising to find relativists represented as opposed to it. The Nor can the quasi-stationary assumption be regarded as a fundamental point of difference. It would, I think, be absurd either to affirm or deny the quasi-stationary principle irrespective of the particular application proposed. question is whether it is a legitimate approximation in a definite problem. I sympathize with Walker in demanding a justification of this approximation in the cases where it has been used-whether by relativists or others. The problem of Phil. Mag. February 1918, p. 143. † Loc. cit. pp. 155-156. |