3-15-18. George Howard Darwin to H. Poincaré

Aug. 3. 1901

Newnham Grange–Cambridge

Dear Monsieur Poincaré,

Since I wrote last I see that the method of my last letter gives less than I hoped for.11See Darwin to Poincaré, 31.07.1901 (§ 3-15-17). This I will explain. If

r=a1[1+e1(13-μ2)]

is the first approx. to the surface under rotation ω, the problem we should like to solve is to find the next approx.

Suppose the next approx. to be

r1=a1[1+e1(13-μ2)+f𝔓4(μ)+g𝔓6(μ)]

where f and g are at least of order e12. This is to be solved by making E+12Aω2 stationary, where E is lost energy & A m[oment] of i[nertia].22The total energy of the system is E+12Aω2, or in Poincaré’s terms, U+12Jω2. If we concentrate the layers represented by f and g we see that their contributions to E are of order f2 and g2 and are negligible ex hypothesi; their contributions to A are nil as being harmonics of orders 4 & 6. Hence we may as well start by omitting f and g. (I had in fact gone thro’ the work and found that they do disappear entirely before I realised the reason of it).

Suppose then the surface to be

r=a(1-eμ2)

where the e is not the same as the e of my last letter. But we are supposed to be ignorant of the nature of the true figure of equilibrium and therefore we must merely regard e as a parameter whose cube is to be retained.

If we write M=43πρa03 as the mass of the body, it is easy to show that

a03 =a3(1-e+35e2-17e3);
whence   a5 =a05(1+53e+1132e2+163347e3), (1)

a result needed later.

Proceeding as in my last letter,33Darwin to Poincaré (§ 3-15-17).

E1 =1615π2ρ2a5 (2)
E2 =-23πρ(3a2-r2)r2ρ𝑑r𝑑μ𝑑φ
=-2πρ23a5[r3a3-15r5a5]ra𝑑μ𝑑φ
=-2πρ23a5(2eμ2-e2μ4-e3μ6)𝑑μ𝑑φ
=1615π2ρ2a5(-53e+12e2+527e3) (3)

Concentrate the layer of neg. density on the sphere a¯ & we get surface density as far as e2 equal to ρa(eμ2-e2μ4). Express this in harmonics retaining only harmonics whose coefficients are of order e but developing those coefficients as far as e2 (you will perceive that this is sufficient) & we get surface density

ρa[13e-15e2-(e-67e2)(13-μ2)].

Find the lost energy of this by the usual method & we get

E3=1615π2ρ2a5(29235e2-129357e3) (4)

Exactly as in my last the energy lost in distorting the layer is

E4=1615π2ρ2a5(43457e3) (5)

The energy lost in expanding the layer is

E5=1615π2ρ2a5(-547e3) (6)

Adding together (2), (3), (4), (5), (6)

E=1615π2ρ2a5(1-53e+2235e2-2657e3)

Introducing a0 from (1)

E=1615π2ρ2a05(1-4325e2-1363457e3) (7)

The moment of inertia A is only counted as far as e2. For the sphere

A1=815πρa5=12πρ(1615π2ρ2a5) (8)

For the layer

A2 =-ρr2(1-μ2)r2𝑑r𝑑μ𝑑φ
=-15ρa5[r3a3]ra(1-μ2)𝑑μ𝑑φ
=-15ρa5e(μ2-(1+2e)μ4+2eμ6)𝑑μ𝑑φ
=815πρa5(-e+67e2) (9)

From (8) and (9)

A =12πρ(1615π2ρ2a5)(1-e+67e2)
A =12πρ(1615π2ρ2a05)(1+23e+26327e2) (10)

From (7) and (10)

E+12Aω21615π2ρ2a05=1-4325e2-1363457e3+ω24πρ(1+23e+26327e2)

Making this stationary for variations of e

ω24πρ23(1+2637e) =8325e+1363357e2
whence  ω22πρ =835(e-37e2).

This is all that is attainable from this method, but I want to show that it is right & for that purpose I must find what e really means when we know that the resulting figure is an ellipsoid. If I write e for the e of my last letter, I showed that the equation to an ellipsoid was44In Darwin to Poincaré, 31.07.1901 (§ 3-15-17), e denoted half the square of ellipsoid eccentricity; see equation (A).

r=a[1-13e-11235e2+(e+57e2)(13-μ2)+32e2σ4]

If I put

a =a(1-9257e2)
e =e+57e2
this may be written
r =a(1-eμ2+32e2σ4)

and this is the form with which I have worked.

If η is the eccentricity of ellipsoid we had e=12η2. Hence

e=12η2+547η4.

Hence

ω22πρ=435(η2+17η4).

Now I have proved (tho’ I cannot refer you to any book for the result) that

ω22πρ=1-η212n-1!η2n(2n+1)(2n+3)(n-1)!222n-4.

Taking the first two terms of this series we have55Cf. Thomson & Tait’s formula (1879, § 771).

ω22πρ =1-η2(435η2+2357η4)
=435η2(1-12η2)(1+927η2)
=435(η2+17η4).

Thus the result is correct.

It appears however that unless the approximation can be carried as far as e4 there is no way of determining what meaning is to be attributed to e.

In looking over the investigation, I see that E1, E2, E3 and A can be found for one more power of e,66Variant: “can be found as far as e4”. but I do not see how E4 and E5 can be found more exactly.

I conclude from this that the application of the similar method to the pear would throw no light on the further approximation to its figure, but would enable us to determine whether or not the pear corresponds to greater or less m[oment] of m[omentum] and so absolutely determine the question of stability. I am not sure whether or not I shall have the patience to carry out the enormous labour of the investigation. I wish I could see my way to more accurate determination of the figure.

The impasse in which I find myself is quite unexpected by me, although you very probably foresaw it.

I feel quite ashamed to have troubled you by these enormous letters, and I doubt whether you will have the patience to read them.

I remain, Yours very truly,

G. H. Darwin

ALS 6p. Collection particulière, Paris 75017.

Time-stamp: "18.09.2016 01:33"

References