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from the æther at the instant the negative cell passes out through the surface, but perhaps rather at the instant that it comes out from the molecule. In other words, energy may be regarded as flowing in through the surface, but is only absorbed and converted into kinetic energy when it passes into a molecule.

9. Maxwell has calculated the enormous pressure along the lines of force and the equal tension at right angles to them required, according to his theory, to account for the attraction of the earth by the sun. A minimum value for the amount of energy in one cubic centimetre of gravitational æther can be calculated by finding the negative energy-density on the surface of the sun considered as a body of uniform density.

The volume of the energy-cells at any point is



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The intensity of the gravitational field on the surface of the sun is H2= where M is the mass and r the radius of this body. If p is the density of the sun the volume of an energy-cell is

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The amount of energy lacking in each of these cells is half a unit. Hence one cubic centimetre of æther lacks p2 units. These are static units of energy. One dyne equals 1.544 x 10' static units of force, therefore one erg equals 1.544 x 10 static units of energy. Take as 6370000 × 100 x 110 centim. and p as 1:39. Then the density of negative energy at the surface of the sun will be

2 π × 1.392 × 6372 × 1102 × 1012


1.544 x 10'

ergs per c.c.

= 4·289 × 1011 ergs per c.c.

= 42.89 × 106 joules per c.c.

= 16 horse-power hours per c.c.

This would seem to mean that, at a distance from all gravitating matter, a cubic centimetre of æther contains at least this amount of


10. That the theory may require an even greater energy-. density than this is seen as follows. Suppose the energy of the gravitational æther to be due to vortex filaments or tubes of directed energy interlacing in every direction. Take at any point three axes at right angles to one another.

The irregular distribution can be replaced in imagination by six equal sets of filaments parallel to the positive and negative directions of these three axes.

In space at a great distance from gravitational matter each of these sets of filaments will contain the same amount of energy. Let 6a be the amount of energy in one cubic centimetre of free æther. Let the positive direction of the Xaxis at a point correspond with the positive direction of the gravitational line of force at that point. Let b be the volume of the gravitational energy-cell. Then five sets of filaments in the cell may contain 5ab units of energy, and the sixth, which has the direction of the X-axis, ab-1 units. Altogether the cell will contain 6ab-1 units, while the same volume at a distance from gravitational matter will contain 6ab units of energy.

This would require the existence of 96 or say 100 horsepower hours in every cubic centimetre.

11. It is an interesting question whether this theory of energy-distribution in gravitational æther is or is not the simplest and most probable. That other distributions are conceivable appears evident. For example, the other along the straight line joining the centres of two bodies might be regarded as a stretched elastic cord, the laws of contraction being of course quite different from those which hold for ordinary elastic bodies.

Another explanation which seems possible in the case of two gravitational particles is as follows. In the energy diagram for two equal electrified particles with opposite sign, suppose each infinitesimal tube of polarization to be divided into two equal tubes with opposite directions. If they are regarded as vortex filaments, each vortex will work in with its neighbours rotating the other way, like one frictionpulley on another.

Then it is clear that there is no way in which we can regard the two particles as opposed in sign, and yet if each of the two sets of vortex filaments acts as electrostatic tubes always act, the particles will be attracted together.

The energy-flow in this case will be as indicated in fig. 1, and the distribution of energy-density as in fig. 3. It has not been found possible to map the energy field in this manner for three or more particles. Indeed, the difficulties are such that it seems improbable that this method can be applied.

12. In conclusion it may be well to notice again the assumption on which this paper is based: that the energy-cells preserve their identity, and carry the same energy with them throughout their path. It is not clear that this is necessarily

true. It may be that energy passes from the potential into the kinetic form in the æther itself, and not merely on the surface of the molecules. Kinetic energy may consist of the motion of the whole system of energy-cells. This would lead us very near the theory which regards the molecule as being nothing but the mathematical centre from which forces proceed, or perhaps, from another point of view, as having infinite



XXXV. On a Simple Form of Harmonic Analyser. GEORGE UDNY YULE, Demonstrator in Applied Mathematics, University College, London".

§ 1.

"The subject of the decomposition of an arbitrary function into the sum of functions of special types has many fascinations. No student of mathematical physics, if he possess any soul at all, can fail to recognize the poetry that pervades this branch of mathematics.”—OLIVER HEAVISIDE.

ABOUT a year ago several instruments for deter

mining the coefficients of a Fourier Series expressing the equation to a given curve were described before this Society by Professor Henrici t. One of them, Professor Henrici's shifting-table analyser, used a planimeter as the integrator; an arrangement that seemed to me very noteworthy from the point of view of simplicity and cheapness. The analyser I am going to describe also uses a planimeter consequently it can also only give the value of one coefficient at a time.

§ 2. Let PQR be the curve to be analysed. Let the base PŘ range from a=-1 to +1, and the equation to the

curve in terms of a Fourier Series be

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and A, is the mean ordinate of the curve. The values of the other coefficients are given by

*Communicated by the Physical Society: read March 8, 1895.

+ Phil. Mag. xxxviii., July 1894; also Catalogue of the Mathematical Exhibition at Munich (1892-93).

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These are the integrals which any harmonic analyser has to evaluate.

Now suppose we have a circular disk, centre K (fig. 1),

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constrained to keep in contact with a straight line XX parallel to PR, and capable of rolling along X X without slip. Further, let XX be capable of motion in the plane in a vertical but not a horizontal direction, so that every point fixed in it describes a perpendicular to PR. Then we can make the point K trace out any arbitrary curve by moving XX and rolling the disk along it.

Bring K over P, and then mark any point D on the horizontal diameter at a distancer from the centre (not necessarily inside the disk). Starting from P carry K right round PQR and back to P again by the motion just described. Supposing the circumference of the disk to be an aliquot part of PR, say 21/n, let us find the area of the curve traced out by D during this operation. As the disk turns through an angle 2n in rolling along a length 21 of X X, it will turn through na/l in a distance ; so if x, y be the coordinates of K at some point on its journey, the corresponding coordinates of D will be

x-r.cos nπ. cos no,

y+r. cos nπ. sin no,

where, to fix the sign, we have assumed D to lie initially to the left of K and X X to lie above the disk, as in the figure. Hence the area traced out by D is

R1=Sydx-r.cos ny d (cos no)

+r.cos na sin no dx − r2 С sin në d (cos nė).

The last two integrals vanish on taking them round a closed curve. Nothing is added to either of the first two by continuing the integration from R back to P, as y is then zero. Therefore, calling the area of the whole curve PQR a, we have

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Similarly, if D had been initially on the vertical instead of on the horizontal diameter, and below K, we should have had

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It will evidently be convenient to taker some multiple of 1/π units of length, say 10. We then have, rewriting the last two equations,


when r=10/π

= a + cos nπ. 10nB,

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R2 = a + cos nπ . 10nА„

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Care must be taken with regard to the sign on the right-hand side if any other initial position of disk and line be assumed than that dealt with above.

§ 3. These two equations contain the whole theory of my instrument; they show how to construct a curve the area of which gives the required coefficients. The geometrical mechanism seems to me to be somewhat interesting, and to be possibly capable of generalization by the use of noncircular disks.

The area of the D-curve (in dealing with a material instrument) might be obtained in two ways. We might put a pencil through the disk at D, draw the curve, and integrate it afterwards or we might attach the pointer of an integrator to D and let the integrating go on simultaneously with the following of the curve.

It is the latter alternative that I have adopted. The former method would have some advantages, but would be slow and would lead to mechanical difficulties.

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