3-15-23. George Howard Darwin to H. Poincaré

Aug. 12. 01

Newnham Grange–Cambridge

Dear Monsieur Poincaré,

I enclose your former letter.11endnote: 1 Poincaré had asked Darwin to return his letter (§ 3-15-19), of which he had no copy; see Poincaré to Darwin (§ 3-15-22). I do not understand your “double couche”.

With regard to the integrals they are obviously all reducible to elliptic integrals but the coeffts become complicated. In applying them it is necessary to separate the variables θ\theta and φ\varphi & this leads to a terrible complication. I have worked for a long time at the mass of the layer to cubes of small quantities. I found that I could not get it simpler than in a form involving 44 integrals occurring in pairs each of which will involve EE, FF, Π\Pi. I dare say I am very clumsy at this work but I have given up that attempt & am now arranging the work quite otherwise but less accurate. It will remain very heavy but within the limits of patience.

I am still confident of my correctness & my confidence is largely due to the fact that the problem I have solved is done correctly & the analogy is perfect.

This is my procedure.

Let ω2\omega^{2} be sq. of ang. vel. of the critical Jacobian and ω2+ϖ2\omega^{2}+\varpi^{2} that of the pear. Define the pear as an ellipsoid, touching the stalk of the pear, similar to the critical Jacobian, with a layer of negative matter -ρ-\rho lying on it.

Let ω2\omega^{2} be sq. of ang. vel. of the critical Jacobian and ω2+ϖ2\omega^{2}+\varpi^{2} that of the pear. Define the pear as an ellipsoid, touching the stalk of the pear, similar to the critical Jacobian, with a layer of negative matter -ρ-\rho lying on it.

Let dd be distance of center of inertia of pear from origin.

Let e𝑑v\int_{e}dv, 𝑑v\int_{\ell}dv, denote integration thro’ ellipsoid & thro’ layer and e-𝑑v\int_{e-\ell}dv integration thro’ the pear.

The “lost” energy is

E=12e-Ve-ρ𝑑v+12(ω2+ϖ2)e-[y2+(z-d)2]𝑑vE=\frac{1}{2}\int_{e-\ell}V_{e-\ell}\rho dv+\frac{1}{2}(\omega^{2}+\varpi^{2})% \int_{e-\ell}[y^{2}+(z-d)^{2}]dv

Now22endnote: 2 We insert ρ\rho on the left-hand side and in the final term on the right-hand side of the first equation.

12e-Ve-ρ𝑑v\displaystyle\frac{1}{2}\int_{e-\ell}V_{e-\ell}\rho dv =12eVeρ𝑑v-Vρ𝑑v+12Vρ𝑑v\displaystyle=\frac{1}{2}\int_{e}V_{e}\rho dv-\int_{\ell}V_{\ell}\rho dv+\frac% {1}{2}\int_{\ell}V_{\ell}\rho dv
e-ρ[y2+(z-d)2]𝑑v\displaystyle\int_{e-\ell}\rho[y^{2}+(z-d)^{2}]dv =eρ(y2+z2)𝑑v-ρ(y2+z2)𝑑v-e-(2zd-d2)ρ𝑑v\displaystyle=\int_{e}\rho(y^{2}+z^{2})dv-\int_{\ell}\rho(y^{2}+z^{2})dv-\int_% {e-\ell}(2zd-d^{2})\rho dv

But

e-zρ𝑑v=Md,e-ρ𝑑v=M.\int_{e-\ell}z\rho dv=Md,\qquad\int_{e-\ell}\rho dv=M.

Therefore

e-(2zd-d2)ρ𝑑v=Md2\int_{e-\ell}(2zd-d^{2})\rho dv=Md^{2}

Hence

E=12\displaystyle E=\frac{1}{2} eVeρ𝑑v+12(ω2)e(y2+z2)ρ𝑑v\displaystyle\int_{e}V_{e}\rho dv+\frac{1}{2}(\omega^{2})\int_{e}(y^{2}+z^{2})% \rho dv
-\displaystyle- l[Ve+12ω2(y2+z2)]ρ𝑑v+12Vρ𝑑v\displaystyle\int_{l}\left[V_{e}+\frac{1}{2}\omega^{2}(y^{2}+z^{2})\right]\rho dv% +\frac{1}{2}\int_{\ell}V_{\ell}\rho dv
+\displaystyle+ 12ϖ2e-(y2+z2)ρ𝑑v-12Md2(ω2+ϖ2)\displaystyle\frac{1}{2}\varpi^{2}\int_{e-\ell}(y^{2}+z^{2})\rho dv-\frac{1}{2% }Md^{2}(\omega^{2}+\varpi^{2})

dd is clearly at least of second order & d2d^{2} is of fourth order & negligible.33endnote: 3 Variante: “…fourth order & negligible. The first two terms in EE are stationary because the ellipsoid is a Jacobian”.

Suppose that the volume of the pear is equal to 43πρk03abc\frac{4}{3}\pi\rho k_{0}^{3}abc, where

a=cosyκsiny,b=1-κ2sin2yκsiny,c=1κsinya=\frac{\cos y}{\kappa\sin y},\qquad b=\frac{\sqrt{1-\kappa^{2}\sin^{2}y}}{% \kappa\sin y},\qquad c=\frac{1}{\kappa\sin y}

Then yy and κ\kappa have definite numerical values depending on the eccentricities of the principal sections of the critical Jacobian Ellipsoid, so that aa, bb, cc have known numerical values.

Then it is easy to prove that

12eρVe𝑑v+12ω2eρ(y2+z2)𝑑v=815π2ρ2a3b3c3k5[0duu+a2u+b2u+c2+b2+c2abcω24πρ]\displaystyle\frac{1}{2}\int_{e}\rho V_{e}dv+\frac{1}{2}\omega^{2}\int_{e}\rho% (y^{2}+z^{2})dv\\ \displaystyle=\frac{8}{15}\pi^{2}\rho^{2}a^{3}b^{3}c^{3}\cdot k^{5}\left[\int_% {0}^{\infty}\frac{du}{\sqrt{u+a^{2}\cdot u+b^{2}\cdot u+c^{2}}}+\frac{b^{2}+c^% {2}}{abc}\cdot\frac{\omega^{2}}{4\pi\rho}\right]

In this expression kk is such that kckc is equal to the distance from the origin at centre of the ellipsoid to the end of the stalk of the pear, and it is therefore a function of the unknown parameter or parameters which define the pear. The only variable in this expression is kk, all the other quantities including ω\omega having definite numerical values.

I do not assert that the critical Jacobian is the ellipsoid which most nearly resembles the pear, but I assert that any additional harmonic terms of the second degree must have coefficients of the second order in reference to the parameter which defines the pear.

The pear will also require harmonics of the 1st and 4th degrees to be fully represented, but these harmonics have coefficients of at least the second order, and in the energy only of the fourth order.

We have (as in the case of the Maclaurin ellipsoid the general principle that in developing the energy we need only regard terms of the first order, but the coefficients of those terms must be developed as far as cubes of small quantities in the expression for the energy.

The whole energy as far as material is then:

E=815π2ρ2a3b3c3.k5[0duu+a2.u+b2.u+c2+b2+c2abcω24πρ]\displaystyle{E=\frac{8}{15}\pi^{2}\rho^{2}a^{3}b^{3}c^{3}.k^{5}\left[\int_{0}% ^{\infty}\frac{du}{\sqrt{u+a^{2}.u+b^{2}.u+c^{2}}}+\frac{b^{2}+c^{2}}{abc}% \cdot\frac{\omega^{2}}{4\pi\rho}\right]}
(Energy of ellipsoid with rotation ω\omega within itself) -l(Ve+12ω2(y2+z2))ρ𝑑v\displaystyle{-\int_{l}\left(V_{e}+\frac{1}{2}\omega^{2}(y^{2}+z^{2})\right)% \rho dv} (energy of ellipsoid with rotation ω\omega and layer) +12lVlρ𝑑v\displaystyle{+\frac{1}{2}\int_{l}V_{l}\rho dv} (energy of layer within itself) +12ω¯2e-lρ(y2+z2)𝑑v\displaystyle{+\frac{1}{2}\overline{\omega}^{2}\int_{e-l}\rho(y^{2}+z^{2})dv} (additional kinetic energy of pear due to change of rotation)

All these quantities are expressible in terms of ee and of kk, and by expressing kk in terms of k0k_{0} we get the whole of EE expressed [in] terms of ee.

Only terms of first order are retained but those terms are developed as far as e3e^{3}.

By differentiating with respect to ee we shall get ϖ2\varpi^{2} in terms of ee.

In the case of Maclaurin’s ellipsoid of which I have given the solution, ω\omega is zero so that the terms become:

  • 1

    Energy of sphere with itself

  • 2

    Energy of sphere with layer

  • 3

    Energy of layer with itself

  • 4

    Kinetic energy of ellipsoid.

Pray excuse the untidiness of this letter.

It will interest me much to hear what your opinion is.

Yours very Sincerely.

G. H. Darwin

ALS 6p. Collection particulière, Paris 75017.

Time-stamp: " 4.05.2019 00:12"

Notes

  • 1 Poincaré had asked Darwin to return his letter (§ 3-15-19), of which he had no copy; see Poincaré to Darwin (§ 3-15-22).
  • 2 We insert ρ\rho on the left-hand side and in the final term on the right-hand side of the first equation.
  • 3 Variante: “…fourth order & negligible. The first two terms in EE are stationary because the ellipsoid is a Jacobian”.