3-15-15. George Howard Darwin to H. Poincaré

June 21.01

Newnham Grange–Cambridge

Dear Monsieur Poincaré,

I do not see how it is possible to solve the question without carrying the investigation through completely11Variant: “almost completely”. to the second order of small quantities. I had come to this conclusion before hearing from you,22See Poincaré to Darwin, § 3-15-14. but when you said that my figure afforded the means of deciding the question of increase of momentum I thought you must be right, and besides forgot my previous reflections on the subject.33See Poincaré to Darwin, § 3-15-12.

As it is possible to carry the theory of the figure of the earth to the second order of small quantities by means of spherical harmonics, so I think it is possible to proceed to the second order here with ellipsoidal harmonics. I believe I see how it is to be done numerically, proceeding from a specified Jacobian ellipsoid; but I do not see much prospect of doing it analytically. But the arithmetic would be very heavy, since it would involve the determination of a considerable number of integrals by quadratures—integrals theoretically perhaps expressible in Elliptic integrals, but with coefficients of stupendous complication. The fundamental difficulty is thus. Suppose we have an inequality on the ellipsoid represented by (curvilinear normal)

δn=p.ε𝔓3(μ)𝖢3(ϕ)[3rdzonal]

(p on tangent plane); then we have to find the mass standing on unit area of the ellipsoid as far as squares of ε.

I have obtained the expression in question. The term in ε is obvious by ellipsoidal analysis — that in ε2 depends on p[𝔓3(μ)𝖢3(ϕ)]2 multiplied by a complicated function of the two coordinates μ, ϕ.

In order to proceed further, this function must be expressed in ellipsoidal harmonics and it will involve harmonics of orders 0, 2, 4, 6…and of even orders of tesserality.44Darwin (1903, 253) linked the even-order harmonics to the figure’s symmetry.

I believe that by quadratures & a sufficient degree of labour I can get out those harmonics – but whether or not I have the patience to do so, is another question.

This result shows that we must start by supposing that the figure of equilibrium consists not only of a 3rd harmonic of amplitude ε (whose square is retained), but also of harmonics of orders 0, 2, 4, 6…& of amplitudes unknown save that they are of the order ε2.55Variant: “save that it is of the second order ε2.” These new harmonics will affect the moment of inertia.

The harmonic 0 is what Kelvin calls a focaloid shell & indicates that the elevation of the 3rd zonal must be computed from a surface confocal to the primitive ellipsoid — otherwise the mass of the pear will not be equal to that of the ellipsoid.66The focaloid shell was defined by Thomson (1879, §494 g) as an infinitely-thin shell bounded by two concentric similar ellipsoidal surfaces. He employed it to prove a theorem attributed to MacLaurin, and used here by Darwin, according to which any two confocal homogeneous solid ellipsoids of equal mass give rise to equal attraction. Clairaut’s theory of heterogeneous masses also employed focaloid shells; see the notes to Darwin to Poincaré, 31.07.1901 (§ 3-15-17). This harmonic then enters in the moment of inertia, as also does that of order 2; the remaining harmonics do not do so.77Variant: “do not do so. Hence if I worked out these two only I could decide the question of the moment of momentum with”.88Cf. Poincaré to Darwin (§ 3-15-19), where Poincaré shows that ξ3 (the coefficient of R2,01=𝔓2 in the notation of Poincaré (1902), here η) and ξ4 (the coefficient of R2,21=𝔓22 in Poincaré’s notation, here ζ) enter the computation with the second order while the third zonal enters with the first order.

If I assume these difficulties overcome, then I think (also with sufficient labour) I can overcome the remainder.

In order to find the gravitational terms I should proceed thus:

Draw a concentric and similar ellipsoid touching the stalk of the pear. It is a Jacobian ellipsoid, but its mass depends on the determination of the exact figure of the pear. Nevertheless I can find the expression for its internal gravity although that expression involves ε.99On Darwin’s solution see Darwin to Poincaré, 12.08.1901 (§ 3-15-23).

Consider the shaded pear to consist of negative matter; then by quadratures I can find the energy due to the gravitation of the larger ellipsoid and this shaded matter.

The mutual energy of the layer as far as it depends on the harmonics 0, 2, 4, 6 …is determinable at once by harmonic analysis. That corresponding to the 3rd zonal is to the first order the same as tho’ it were concentrated on the surface. To go to the second order we must first concentrate in surface density on ellipsoid and then deform that surface density — so as to bring it to stand half way between the ellipsoid & the true surface.1010This is the double layer strategy used later by Poincaré; see Poincaré to Darwin (§ 3-15-25). This last may be done thus: If we have a layer of surface density with potential v(+) externally and potential v(-) internally, the force acting on an element dσ is1111The second factor involves Poisson’s equation (Darwin 1901, 507); the last fraction has been corrected from dv(-1)/dn.

-12(dv(+)dn+dv(-)dn)14π(dv(+)dn-dv(-)dn)dσ.

If the element is carried outwards through a distance 12δn, the work done is

116π[(dv(+)dn)2-(dv(-)dn)2]δndσ

Integrating all over the surface one may obtain the whole work done in deformation.

In this sketch of procedure I have left a good deal vague, but I think the processes involved are legitimate.

I do not see that any analytical skill could make the processes necessary other than very laborious – unless indeed it were possible to obtain expressions for harmonic functions applicable to the pear-shaped figure.

I am afraid this letter is very ill expressed, but I hope it is intelligible and that you will not detect faults of principle.

I remain, Yours sincerely

G.H. Darwin

ALS 8p. Collection particulière, Paris 75017.

Time-stamp: "14.09.2016 18:02"

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