3-15-17. George Howard Darwin to H. Poincaré

July 31 01

Newnham Grange–Cambridge

Dear Monsieur Poincaré,

I have now devised a method which I think overcomes all difficulties in carrying out my proposed work, although it certainly shows that the work will be very heavy.

In order to see how the process works I have applied it to the ellipsoid of revolution.11Darwin seeks to determine the energy lost in passing from a sphere to an ellipsoid of revolution, up to e3, where 2e is the square of the eccentricity. He follows a similar procedure to get from the critical Jacobian ellipsoid to the pear-shaped figure. I unfortunately made a mistake in my work and have spent nearly fortnight in discovering where I was wrong. It was in an elementary point with which I need not trouble you, but I passed it by over and over again in going over my work.

If 2e is the square of the eccentricity of an ellipsoid of revolution of density ρ, mass M, and semi-major axis a, it is known that the “lost energy” is

35M2asin-12e2e=35M2a(1+13e+310e2+514e3)

I want to get the same result by harmonic analysis.22In what follows, Darwin sums energies of the sphere and the layer to find the same expression for lost energy (up to e3), an approach employed in Darwin (1887, 423).

I consider the ellipsoid to be built up by a sphere of radius a and density ρ, and a (shaded) layer of density -ρ.

If

σ2 =13-μ2,
σ4 =μ4-67μ2+357
σ6 =μ6-3511μ4+511μ2-53711

where μ is cosine of colatitude.

The equation to the ellipsoid is33Darwin’s equation expresses the deformation of the figure with respect to the original sphere. Observing that r=x2+y2+z2 for a surface point (x,y,z), we have (ra)2=x2a2+y2a2+z2a2; thus, ra is precisely zero for surface points. Darwin builds here on an approach taken by Legendre and Laplace, who considered the layers of the figure as spheroids deviating from the sphere by small first-order quantities (Oppenheimer 1922). The equation of such a spheroid is r=a(1+ef) where r is the Euclidean distance from the origin to a surface point, a is the radius of a sphere with center at the origin, e is a small quantity with negligible square, and f is a function of the polar coordinates θ and ψ developed in a series of spherical functions Y0+Y1+Y2+. Supposing a to be the radius of the sphere with volume equal to the volume of the spheroid, one obtains Y0=0, and by supposing the center of inertia to lie at the origin of the coordinate system, it is shown that Y1=0. Laplace gave an expression for the equilibrium condition in terms of these functions, complicated by the consideration of supplementary forces. For the Earth, however, forces other than centrifugal may be neglected, and since the force expression involves only a spherical function of the second order, P2, only Y2 must be retained, and in Y2, only the coefficient corresponding to P2=cos2θ-13 (the function labeled σ2 by Darwin). Comparison with astronomical data indicated a need to introduce higher powers of e, motivating Darwin and others to consider more elaborate forms of the equation; see Wiechert (1897), Callandreau (1897), Darwin (1899).

ra=1-13e-11235e2-1032357e3+(e+57e2+927e3)σ2+32(e2+1911e3)σ4-52e3σ6 (1)

I now form (ra)3-15(ra)5 and drop all terms which are of order higher than e3 or which will vanish when integrated over the sphere. Taking this expression between the limits a and r as given in (A), I find

[r3a3-15r5a5]ra=23e+28325e2+6613257e3-(e2+177e3)(σ2)2+e3(σ2)3.

It is easy to show that

𝑑μ𝑑ϕ=4π;(σ2)2𝑑μ𝑑ϕ=16π45;(σ2)3𝑑μ𝑑ϕ=-64π3357.

Hence I find

[r3a3-15r5a5]ra𝑑μ𝑑ϕ=8π3(e+45e2+710e3).

Now to make use of this.

The lost energy of the sphere, say E1, is

35(43πρa3)2a=1615π2ρ2a5

The lost energy of the sphere and layer, say E2, is the potl of the sphere multiplied by density of layer and integrated through the layer. Hence it is

-23πρ(3a2-r2)ρr2𝑑r𝑑μ𝑑ϕ

with the limits of r: a to r (as above).

But

ra(3r2a3-r4a5)𝑑r=[r3a3-15r5a5]ra

Hence applying the proceeding result

E1+E2=1615π2ρ2a5(1-53e-43e2-76e3)

Now since44For comparison, Thomson & Tait (1879, § 776, eq. 7) found M=43πρa311-2e.

M=43πρa31-2e

We have, on division by 1-2e

E1+E2=35M2a(1+13e-23e2-52e3)

This result may be obtained more shortly, but I have purposely done it as if I was compelled to integrate.

It remains to find the lost energy of the layer. This I consider in three steps.

  • 1.

    Concentrate the layer on the sphere and find its lost energy E3

  • 2.

    Distort the layer of surface density by carrying it to a position halfway between the sphere and ellipsoid, and let E4 be the energy lost

  • 3.

    Expand the surface density until it fills the layer, and let E5 (essentially negative) be the lost energy.

(1) The equation to the ellipsoid (retaining squares only) may be written55Variant after first equation: “-e3(4μ2-6μ4)”.

ra =1-eμ2-e2(2μ2-32μ4).
r3a3 =1-3μ2(e+2e2)+152e2μ4
=1-e-12e2+3σ2(e-17e2)+152e2σ4
13[r3a3]ar =13e+16e2-σ2(e-17e2)-52e2σ4

The mass which stands on the element a2dμdϕ is

-raρr2a2𝑑r=-ρa[13e+16e2-σ2(e-17e2)-52e2σ4]

and this is the surface density of the concentration (1). Proceeding in the usual way to find the lost energy

E3 =1615π2ρ2a5(29235e2+1672357e3)
=35M2a(29235e2+191257e3)
Then
E1+E2+E3 =35M2a(1+13e+310e2+857e3)

This is already right up to e2.

(2) To the first order of small quantities, the surface density of the concentration is

ρae(σ2-13)

The external and internal potentials are

V0 =4πρa2e(15a3r3σ2-13ar)
Vi =4πρa2e(15r2a2σ2)

The mean value of dV0dr and dVidr (r=a) is then

2πρae(-15σ2+13)

This represents the attraction on an element of the surface which bears density ρae(σ2-13), and is to be carried a distance 12ae(σ2-13).

Hence

E4 =2πρae(13-15σ2)12ρa2e2(σ2-13)2a2𝑑μ𝑑φ
=πρ2a5e3μ4(13-135+15μ2)𝑑μ𝑑φ
=4π2ρ2a5e35(17+435)
=35M2a(43e3457)

And

E1+E2+E3+E4=35M2a(1+13e+310e2+1547e3)

(3) Consider a plane surface coated with surface density 2hρ, and find the gain of energy in expanding it to be a slab of thickness 2h — half being carried one way and half the other.

Imagine the expansion of one half carried out until h-x is filled with density ρ & h+x remains unexpanded. The force per unit area is66For the right-hand side, read -4πρx. 2πρ(h-x)-2πρx-2πρh=-4πρhx.

The next element carried up has mass ρdx & is carried thro’ a distance x. Hence the loss of energy in the whole process is

20h2πρ2x2𝑑x

per unit area or 13πρ2(thickness)3.

In the case we are now considering the thickness is ae(σ2-13).

Thus

E5 =13πρ2a3e3(σ2-13)3a2𝑑μ𝑑φ
=-13πρ2a5e3μ6𝑑μ𝑑φ
=-4π2ρ2a5e337
=-35M2a(547e3).

Adding we get

E1+E2+E3+E4+E5=35M2a(1+13e+310e2+514e3).

This is the correct result.

You will see that the whole of this can be done with Jacobi’s ellipsoid & the pear, but if this is somewhat laborious how much more will that be so.77Gharnati (1996, 67, n. 92) recovers Darwin’s result starting from MacLaurin’s exact formula.

Yours sincerely,

G. H. Darwin

ALS 8p. Collection particulière, Paris 75017.

Time-stamp: "22.01.2016 01:24"

References

  • O. Callandreau (1897) Sur la théorie de la figure des planètes; applications à Jupiter et à la Terre. Bulletin astronomique 14, pp. 214–217. External Links: Link Cited by: 3-15-17. George Howard Darwin to H. Poincaré.
  • G. H. Darwin (1887) On figures of equilibrium of rotating masses of fluidv/spPhllosI2n> .m4rans_centestd>

    Bulletin astronomique On figures of eass="ltx_ref ltx_bib_external">Link Cited by: 3-15-17. Georg3 Howard Darwin to H. Poincaré.

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  • G. H. Darwirnati (On figures of equilibrium of rotating masses oftype/spP.Dy="insisb_journal">Bulletin astronomique ultx_he ltUniitécorreNclay 2b_journal">Bulletin astronomique lace/sNclays="ltx_text ltx_bib_title">On figures of eass="ltx_ref ltx_bib_external">Link Cited by: 3-15-17. Georg5 Howard Darwin to H. Ple"ss="ltx_p.
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