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. This I will explain. If

$$r=a_1\left[1+e_1\left(\frac{1}{3}-{\mu }^2\right)\right]$$

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

Suppose the next approx. to be

$$r_1=a_1\left[1+e_1\left(\frac{1}{3}-{\mu }^2\right)+f{𝔓}_4(\mu )+g{𝔓}_6(\mu )\right]$$

where $f$ and $g$ are at least of order $e_1^2$. This is to be solved by making $E+\frac{1}{2}A{\omega }^2$ stationary, where $E$ is lost energy & $A$ m[oment] of i[nertia]. If we concentrate the layers represented by $f$ and $g$ we see that their contributions to $E$ are of order $f^2$ and $g^2$ 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

$$r=a(1-e{\mu }^2)$$

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 $M=\frac{4}{3}\pi \rho a_0^3$ as the mass of the body, it is easy to show that

$a_0^3$ $=a^3(1-e+{\displaystyle \frac{3}{5}}{e}^2-{\displaystyle \frac{1}{7}}{e}^3);$

$\text{whence }\mathit{\hspace{1em}\hspace{1em}}{a}^5$ $=a_0^5\left(1+{\displaystyle \frac{5}{3}}e+{\displaystyle \frac{11}{3^2}}{e}^2+{\displaystyle \frac{163}{3^4\cdot 7}}{e}^3\right),$ (1)

a result needed later.

Proceeding as in my last letter,

$E_1$ $={\displaystyle \frac{16}{15}}{\pi }^2{\rho }^2{a}^5$ (2)

$E_2$ $=-{\displaystyle \frac{2}{3}}\pi \rho {\displaystyle \iiint (3a^2-r^2)r^2\rho 𝑑r𝑑\mu 𝑑\phi }$

$=-{\displaystyle \frac{2\pi {\rho }^2}{3}}{a}^5{\displaystyle \iint {\left[\frac{r^3}{a^3}-\frac{1}{5}\frac{r^5}{a^5}\right]}_r^a𝑑\mu 𝑑\phi }$

$=-{\displaystyle \frac{2\pi {\rho }^2}{3}}{a}^5{\displaystyle \iint (2e{\mu }^2-e^2{\mu }^4-e^3{\mu }^6)𝑑\mu 𝑑\phi }$

$={\displaystyle \frac{16}{15}}{\pi }^2{\rho }^2{a}^5\left(-{\displaystyle \frac{5}{3}}e+{\displaystyle \frac{1}{2}}{e}^2+{\displaystyle \frac{5}{2\cdot 7}}{e}^3\right)$ (3)

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

$$\rho a\left[\frac{1}{3}e-\frac{1}{5}{e}^2-\left(e-\frac{6}{7}{e}^2\right)\left(\frac{1}{3}-{\mu }^2\right)\right].$$

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

$$E_3=\frac{16}{15}{\pi }^2{\rho }^2{a}^5\left(\frac{29}{2\cdot 3\cdot 5}{e}^2-\frac{129}{3\cdot 5\cdot 7}{e}^3\right)$$ (4)

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

$$E_4=\frac{16}{15}{\pi }^2{\rho }^2{a}^5\left(\frac{43}{4\cdot 5\cdot 7}{e}^3\right)$$ (5)

The energy lost in expanding the layer is

$$E_5=\frac{16}{15}{\pi }^2{\rho }^2{a}^5\left(-\frac{5}{4\cdot 7}{e}^3\right)$$ (6)

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

$$E=\frac{16}{15}{\pi }^2{\rho }^2{a}^5\left(1-\frac{5}{3}e+\frac{22}{3\cdot 5}{e}^2-\frac{26}{5\cdot 7}{e}^3\right)$$

Introducing $a_0$ from (1)

$$E=\frac{16}{15}{\pi }^2{\rho }^2{a}_0^5\left(1-\frac{4}{3^2\cdot 5}{e}^2-\frac{136}{3^4\cdot 5\cdot 7}{e}^3\right)$$ (7)

The moment of inertia $A$ is only counted as far as $e^2$. For the sphere

$$A_1=\frac{8}{15}\pi \rho a^5=\frac{1}{2\pi \rho }\left(\frac{16}{15}{\pi }^2{\rho }^2{a}^5\right)$$ (8)

For the layer

$A_2$ $=-{\displaystyle \iiint \rho r^2(1-{\mu }^2)r^2𝑑r𝑑\mu 𝑑\phi }$

$=-{\displaystyle \frac{1}{5}}\rho a^5{\displaystyle \iint {\left[\frac{r^3}{a^3}\right]}_r^a(1-{\mu }^2)𝑑\mu 𝑑\phi }$

$=-{\displaystyle \frac{1}{5}}\rho a^{5}e{\displaystyle \iint ({\mu }^2-(1+2e){\mu }^4+2e{\mu }^6)𝑑\mu 𝑑\phi }$

$={\displaystyle \frac{8}{15}}\pi \rho a^5\left(-e+{\displaystyle \frac{6}{7}}{e}^2\right)$ (9)

From (8) and (9)

$A$ $={\displaystyle \frac{1}{2\pi \rho }}\left({\displaystyle \frac{16}{15}}{\pi }^2{\rho }^2{a}^5\right)\left(1-e+{\displaystyle \frac{6}{7}}{e}^2\right)$

$A$ $={\displaystyle \frac{1}{2\pi \rho }}\left({\displaystyle \frac{16}{15}}{\pi }^2{\rho }^2{a}_0^5\right)\left(1+{\displaystyle \frac{2}{3}}e+{\displaystyle \frac{26}{3^2\cdot 7}}{e}^2\right)$ (10)

From (7) and (10)

$$\frac{E+\frac{1}{2}A{\omega }^2}{\frac{16}{15}{\pi }^2{\rho }^2{a}_0^5}=1-\frac{4}{3^2\cdot 5}{e}^2-\frac{136}{3^4\cdot 5\cdot 7}{e}^3+\frac{{\omega }^2}{4\pi \rho }\left(1+\frac{2}{3}e+\frac{26}{3^2\cdot 7}{e}^2\right)$$

Making this stationary for variations of $e$

${\displaystyle \frac{{\omega }^2}{4\pi \rho }}\cdot {\displaystyle \frac{2}{3}}\left(1+{\displaystyle \frac{26}{3\cdot 7}}e\right)$ $={\displaystyle \frac{8}{3^2\cdot 5}}e+{\displaystyle \frac{136}{3^3\cdot 5\cdot 7}}{e}^2$

$\text{whence}\mathit{\hspace{1em}\hspace{1em}}{\displaystyle \frac{{\omega }^2}{2\pi \rho }}$ $={\displaystyle \frac{8}{3\cdot 5}}\left(e-{\displaystyle \frac{3}{7}}{e}^2\right).$

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 $e$ really means when we know that the resulting figure is an ellipsoid. If I write $e^{\prime }$ for the $e$ of my last letter, I showed that the equation to an ellipsoid was

$$r=a^{\prime }\left[1-\frac{1}{3}{e}^{\prime }-\frac{11}{2\cdot 3\cdot 5}{e}^{\prime 2}+\left(e^{\prime }+\frac{5}{7}{e}^{\prime 2}\right)\left(\frac{1}{3}-{\mu }^2\right)+\frac{3}{2}{e}^{\prime 2}{\sigma }_4\right]$$

If I put

$a$ $=a^{\prime }\left(1-{\displaystyle \frac{9}{2\cdot 5\cdot 7}}{e}^{\prime 2}\right)$

$e$ $=e^{\prime }+{\displaystyle \frac{5}{7}}{e}^{\prime 2}$

this may be written

$r$ $=a\left(1-e{\mu }^2+{\displaystyle \frac{3}{2}}{e}^2{\sigma }_4\right)$

and this is the form with which I have worked.

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

$$e=\frac{1}{2}{\eta }^2+\frac{5}{4\cdot 7}{\eta }^4.$$

Hence

$$\frac{{\omega }^2}{2\pi \rho }=\frac{4}{3\cdot 5}\left({\eta }^2+\frac{1}{7}{\eta }^4\right).$$

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

$$\frac{{\omega }^2}{2\pi \rho }=\sqrt{1-{\eta }^2}\sum _1^{\mathrm{\infty }}\frac{2n-1!{\eta }^{2n}}{(2n+1)(2n+3){(n-1)!}^2{2}^{2n-4}}.$$

Taking the first two terms of this series we have

${\displaystyle \frac{{\omega }^2}{2\pi \rho }}$ $=\sqrt{1-{\eta }^2}\left({\displaystyle \frac{4}{3\cdot 5}}{\eta }^2+{\displaystyle \frac{2\cdot 3}{5\cdot 7}}{\eta }^4\right)$

$={\displaystyle \frac{4}{3\cdot 5}}{\eta }^2\left(1-{\displaystyle \frac{1}{2}}{\eta }^2\right)\left(1+{\displaystyle \frac{9}{2\cdot 7}}{\eta }^2\right)$

$={\displaystyle \frac{4}{3\cdot 5}}\left({\eta }^2+{\displaystyle \frac{1}{7}}{\eta }^4\right).$

Thus the result is correct.

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

In looking over the investigation, I see that $E_1$, $E_2$, $E_3$ and $A$ can be found for one more power of $e$, but I do not see how $E_4$ and $E_5$ 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: "18.09.2016 01:33"