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.11endnote: 1 See Darwin to Poincaré, 31.07.1901 (§ 3-15-17). This I will explain. If

r=a1[1+e1(13-μ2)]r=a_{1}\left[1+e_{1}\left(\frac{1}{3}-\mu^{2}\right)\right]

is the first approx. to the surface under rotation ω\omega, 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)]r_{1}=a_{1}\left[1+e_{1}\left(\frac{1}{3}-\mu^{2}\right)+f\mathfrak{P}_{4}(\mu% )+g\mathfrak{P}_{6}(\mu)\right]

where ff and gg are at least of order e12e_{1}^{2}. This is to be solved by making E+12Aω2E+\frac{1}{2}A\omega^{2} stationary, where EE is lost energy & AA m[oment] of i[nertia].22endnote: 2 The total energy of the system is E+12Aω2E+\frac{1}{2}A\omega^{2}, or in Poincaré’s terms, U+12Jω2U+\frac{1}{2}J\omega^{2}. If we concentrate the layers represented by ff and gg we see that their contributions to EE are of order f2f^{2} and g2g^{2} and are negligible ex hypothesi; their contributions to AA are nil as being harmonics of orders 4 & 6. Hence we may as well start by omitting ff and gg. (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)r=a(1-e\mu^{2})

where the ee is not the same as the ee 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 ee as a parameter whose cube is to be retained.

If we write M=43πρa03M=\frac{4}{3}\pi\rho a^{3}_{0} as the mass of the body, it is easy to show that

a03\displaystyle a^{3}_{0} =a3(1-e+35e2-17e3);\displaystyle=a^{3}(1-e+\frac{3}{5}e^{2}-\frac{1}{7}e^{3});
whence   a5\displaystyle\text{whence }\qquad a^{5} =a05(1+53e+1132e2+163347e3),\displaystyle=a^{5}_{0}\left(1+\frac{5}{3}e+\frac{11}{3^{2}}e^{2}+\frac{163}{3% ^{4}\cdot 7}e^{3}\right), (1)

a result needed later.

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

E1\displaystyle E_{1} =1615π2ρ2a5\displaystyle=\frac{16}{15}\pi^{2}\rho^{2}a^{5} (2)
E2\displaystyle E_{2} =-23πρ(3a2-r2)r2ρ𝑑r𝑑μ𝑑φ\displaystyle=-\frac{2}{3}\pi\rho\iiint(3a^{2}-r^{2})r^{2}\rho drd\mu d\varphi
=-2πρ23a5[r3a3-15r5a5]ra𝑑μ𝑑φ\displaystyle=-\frac{2\pi\rho^{2}}{3}a^{5}\iint\left[\frac{r^{3}}{a^{3}}-\frac% {1}{5}\frac{r^{5}}{a^{5}}\right]^{a}_{r}d\mu d\varphi
=-2πρ23a5(2eμ2-e2μ4-e3μ6)𝑑μ𝑑φ\displaystyle=-\frac{2\pi\rho^{2}}{3}a^{5}\iint(2e\mu^{2}-e^{2}\mu^{4}-e^{3}% \mu^{6})d\mu d\varphi
=1615π2ρ2a5(-53e+12e2+527e3)\displaystyle=\frac{16}{15}\pi^{2}\rho^{2}a^{5}\left(-\frac{5}{3}e+\frac{1}{2}% e^{2}+\frac{5}{2\cdot 7}e^{3}\right) (3)

Concentrate the layer of neg. density on the sphere a¯\underline{a} & we get surface density as far as e2e^{2} equal to ρa(eμ2-e2μ4)\rho a(e\mu^{2}-e^{2}\mu^{4}). Express this in harmonics retaining only harmonics whose coefficients are of order ee but developing those coefficients as far as e2e^{2} (you will perceive that this is sufficient) & we get surface density

ρa[13e-15e2-(e-67e2)(13-μ2)].\rho a\left[\frac{1}{3}e-\frac{1}{5}e^{2}-\left(e-\frac{6}{7}e^{2}\right)\left% (\frac{1}{3}-\mu^{2}\right)\right].

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

E3=1615π2ρ2a5(29235e2-129357e3)E_{3}=\frac{16}{15}\pi^{2}\rho^{2}a^{5}\left(\frac{29}{2\cdot 3\cdot 5}e^{2}-% \frac{129}{3\cdot 5\cdot 7}e^{3}\right) (4)

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

E4=1615π2ρ2a5(43457e3)E_{4}=\frac{16}{15}\pi^{2}\rho^{2}a^{5}\left(\frac{43}{4\cdot 5\cdot 7}e^{3}\right) (5)

The energy lost in expanding the layer is

E5=1615π2ρ2a5(-547e3)E_{5}=\frac{16}{15}\pi^{2}\rho^{2}a^{5}\left(-\frac{5}{4\cdot 7}e^{3}\right) (6)

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

E=1615π2ρ2a5(1-53e+2235e2-2657e3)E=\frac{16}{15}\pi^{2}\rho^{2}a^{5}\left(1-\frac{5}{3}e+\frac{22}{3\cdot 5}e^{% 2}-\frac{26}{5\cdot 7}e^{3}\right)

Introducing a0a_{0} from (1)

E=1615π2ρ2a05(1-4325e2-1363457e3)E=\frac{16}{15}\pi^{2}\rho^{2}a_{0}^{5}\left(1-\frac{4}{3^{2}\cdot 5}e^{2}-% \frac{136}{3^{4}\cdot 5\cdot 7}e^{3}\right) (7)

The moment of inertia AA is only counted as far as e2e^{2}. For the sphere

A1=815πρa5=12πρ(1615π2ρ2a5)A_{1}=\frac{8}{15}\pi\rho a^{5}=\frac{1}{2\pi\rho}\left(\frac{16}{15}\pi^{2}% \rho^{2}a^{5}\right) (8)

For the layer

A2\displaystyle A_{2} =-ρr2(1-μ2)r2𝑑r𝑑μ𝑑φ\displaystyle=-\iiint\rho r^{2}(1-\mu^{2})r^{2}drd\mu d\varphi
=-15ρa5[r3a3]ra(1-μ2)𝑑μ𝑑φ\displaystyle=-\frac{1}{5}\rho a^{5}\iint\left[\frac{r^{3}}{a^{3}}\right]^{a}_% {r}(1-\mu^{2})d\mu d\varphi
=-15ρa5e(μ2-(1+2e)μ4+2eμ6)𝑑μ𝑑φ\displaystyle=-\frac{1}{5}\rho a^{5}e\iint(\mu^{2}-(1+2e)\mu^{4}+2e\mu^{6})d% \mu d\varphi
=815πρa5(-e+67e2)\displaystyle=\frac{8}{15}\pi\rho a^{5}\left(-e+\frac{6}{7}e^{2}\right) (9)

From (8) and (9)

A\displaystyle A =12πρ(1615π2ρ2a5)(1-e+67e2)\displaystyle=\frac{1}{2\pi\rho}\left(\frac{16}{15}\pi^{2}\rho^{2}a^{5}\right)% \left(1-e+\frac{6}{7}e^{2}\right)
A\displaystyle A =12πρ(1615π2ρ2a05)(1+23e+26327e2)\displaystyle=\frac{1}{2\pi\rho}\left(\frac{16}{15}\pi^{2}\rho^{2}a_{0}^{5}% \right)\left(1+\frac{2}{3}e+\frac{26}{3^{2}\cdot 7}e^{2}\right) (10)

From (7) and (10)

E+12Aω21615π2ρ2a05=1-4325e2-1363457e3+ω24πρ(1+23e+26327e2)\frac{E+\frac{1}{2}A\omega^{2}}{\frac{16}{15}\pi^{2}\rho^{2}a_{0}^{5}}=1-\frac% {4}{3^{2}\cdot 5}e^{2}-\frac{136}{3^{4}\cdot 5\cdot 7}e^{3}+\frac{\omega^{2}}{% 4\pi\rho}\left(1+\frac{2}{3}e+\frac{26}{3^{2}\cdot 7}e^{2}\right)

Making this stationary for variations of ee

ω24πρ23(1+2637e)\displaystyle\frac{\omega^{2}}{4\pi\rho}\cdot\frac{2}{3}\left(1+\frac{26}{3% \cdot 7}e\right) =8325e+1363357e2\displaystyle=\frac{8}{3^{2}\cdot 5}e+\frac{136}{3^{3}\cdot 5\cdot 7}e^{2}
whence  ω22πρ\displaystyle\text{whence}\qquad\frac{\omega^{2}}{2\pi\rho} =835(e-37e2).\displaystyle=\frac{8}{3\cdot 5}\left(e-\frac{3}{7}e^{2}\right).

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 ee really means when we know that the resulting figure is an ellipsoid. If I write ee^{\prime} for the ee of my last letter, I showed that the equation to an ellipsoid was44endnote: 4 In Darwin to Poincaré, 31.07.1901 (§ 3-15-17), ee denoted half the square of ellipsoid eccentricity; see equation (A).

r=a[1-13e-11235e2+(e+57e2)(13-μ2)+32e2σ4]r=a^{\prime}\left[1-\frac{1}{3}e^{\prime}-\frac{11}{2\cdot 3\cdot 5}e^{\prime 2% }+\left(e^{\prime}+\frac{5}{7}e^{\prime 2}\right)\left(\frac{1}{3}-\mu^{2}% \right)+\frac{3}{2}e^{\prime 2}\sigma_{4}\right]

If I put

a\displaystyle a =a(1-9257e2)\displaystyle=a^{\prime}\left(1-\frac{9}{2\cdot 5\cdot 7}e^{\prime 2}\right)
e\displaystyle e =e+57e2\displaystyle=e^{\prime}+\frac{5}{7}e^{\prime 2}
this may be written
r\displaystyle r =a(1-eμ2+32e2σ4)\displaystyle=a\left(1-e\mu^{2}+\frac{3}{2}e^{2}\sigma_{4}\right)

and this is the form with which I have worked.

If η\eta is the eccentricity of ellipsoid we had e=12η2e^{\prime}=\frac{1}{2}\eta^{2}. Hence

e=12η2+547η4.e=\frac{1}{2}\eta^{2}+\frac{5}{4\cdot 7}\eta^{4}.

Hence

ω22πρ=435(η2+17η4).\frac{\omega^{2}}{2\pi\rho}=\frac{4}{3\cdot 5}\left(\eta^{2}+\frac{1}{7}\eta^{% 4}\right).

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.\frac{\omega^{2}}{2\pi\rho}=\sqrt{1-\eta^{2}}\sum_{1}^{\infty}\frac{2n-1!\eta^% {2n}}{(2n+1)(2n+3)(n-1)!^{2}2^{2n-4}}.

Taking the first two terms of this series we have55endnote: 5 Cf. Thomson & Tait’s formula (1879, § 771).

ω22πρ\displaystyle\frac{\omega^{2}}{2\pi\rho} =1-η2(435η2+2357η4)\displaystyle=\sqrt{1-\eta^{2}}\left(\frac{4}{3\cdot 5}\eta^{2}+\frac{2\cdot 3% }{5\cdot 7}\eta^{4}\right)
=435η2(1-12η2)(1+927η2)\displaystyle=\frac{4}{3\cdot 5}\eta^{2}\left(1-\frac{1}{2}\eta^{2}\right)% \left(1+\frac{9}{2\cdot 7}\eta^{2}\right)
=435(η2+17η4).\displaystyle=\frac{4}{3\cdot 5}\left(\eta^{2}+\frac{1}{7}\eta^{4}\right).

Thus the result is correct.

It appears however that unless the approximation can be carried as far as e4e^{4} there is no way of determining what meaning is to be attributed to ee.

In looking over the investigation, I see that E1E_{1}, E2E_{2}, E3E_{3} and AA can be found for one more power of ee,66endnote: 6 Variant: “can be found as far as e4e^{4}”. but I do not see how E4E_{4} and E5E_{5} 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: " 4.05.2019 00:12"

Notes

  • 1 See Darwin to Poincaré, 31.07.1901 (§ 3-15-17).
  • 2 The total energy of the system is E+12Aω2E+\frac{1}{2}A\omega^{2}, or in Poincaré’s terms, U+12Jω2U+\frac{1}{2}J\omega^{2}.
  • 3 Darwin to Poincaré (§ 3-15-17).
  • 4 In Darwin to Poincaré, 31.07.1901 (§ 3-15-17), ee denoted half the square of ellipsoid eccentricity; see equation (A).
  • 5 Cf. Thomson & Tait’s formula (1879, § 771).
  • 6 Variant: “can be found as far as e4e^{4}”.

References

  • W. Thomson and P. G. Tait (1879) Treatise on Natural Philosophy. Cambridge University Press, Cambridge. Link Cited by: endnote 5.