Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/167

of the ecliptic, which wotdd make $({\mathfrak {E}}_{1}.{\mathfrak {B}}),({\mathfrak {E}}_{2}.{\mathfrak {B}}),({\mathfrak {E}}_{3}.{\mathfrak {B}})$ vanish, except for exceedingly minute quantities depending on the latitude of the sun and the geocentric coordinates of the observatories, if these are included in ${\mathfrak {E}}_{1},{\mathfrak {E}}_{2},{\mathfrak {E}}_{3}.$

The equations (a), (b), (c), which are together equivalent to (7), I would solve as follows, almost in the same way as Fabritius, but relying a little more on interpolation, and less on the convergence of which he speaks, which in special cases may more or less fail.

Setting $r_{1}=r_{2}$ and $r_{3}=r_{2}$ in (a), which thus modified I shall call (a'), and solving this (a') by "trial and error," using $\rho _{2}$ as the independent variable, as soon as I have a value of $\rho _{2}$ which I think will give a residual of (a') of the same order of smallness as the effect of changing ${\frac {1}{r_{1}^{3}}}$ and ${\frac {1}{r_{3}^{3}}}$ into ${\frac {1}{r_{2}^{3}}},$ I determine from this value by (b) and (c), $r_{1}$ and $r_{3},$ and then find the residual of (a), using the values of $r_{1},r_{2},r_{3}$ derived all from the same assumed $\rho _{2}.$ Now using the last value of ${\frac {\Delta {\text{(residual)}}}{\Delta \rho _{2}}}$ in my previous calculations on (a') which indeed applies only roughly to the (a), I would get a value $\rho _{2}$ which I would use for the second "hypothesis" in (a). This will give a second residual in (a), which will enable me to make a more satisfactory interpolation. As many more interpolations may be made as shall be found necessary.

Some such method, which should perhaps be called the method of Fabritius, would, I think, in most cases probably be the best for solution of equation (7).

Of course I am quite aware that the merit of my paper, if any, lies principally in the fundamental approximation (1). I will add a few words on this subject.

The equation may be written more symmetrically

${\begin{aligned}{\frac {\tau _{1}}{\tau _{2}}}\left(1+{\frac {\tau _{2}^{2}+\tau _{3}^{2}-3\tau _{1}^{2}}{24r_{1}^{3}}}\right){\mathfrak {R}}_{1}&-\left(1+{\frac {\tau _{3}^{2}+\tau _{1}^{2}-3\tau _{2}^{2}}{24r_{2}^{3}}}\right){\mathfrak {R}}_{2}\\&{\frac {\tau _{3}}{\tau _{2}}}\left(1+{\frac {\tau _{1}^{2}+\tau _{2}^{2}-3\tau _{3}^{2}}{24r_{3}^{3}}}\right){\mathfrak {R}}_{3}=0.\end{aligned}}$

I

It might be made entirely symmetrical by writing $-\tau _{2}$ for $\tau _{2}.$ If an expression ending with $t^{3}$ had been used, we could still have satisfied two of the conditions relating to acceleration, and should have obtained

${\frac {\tau _{1}}{\tau _{2}}}{\mathfrak {R}}_{1}-\left(1+{\frac {\tau _{1}^{2}-\tau _{2}^{2}}{6r_{2}^{2}}}\right){\mathfrak {R}}_{2}+{\frac {\tau _{3}}{\tau _{2}}}\left(1+{\frac {\tau _{1}^{2}-\tau _{3}^{2}}{6r_{3}^{2}}}\right){\mathfrak {R}}_{3}=0,$

IIa

or

${\frac {\tau _{1}}{\tau _{2}}}\left(1+{\frac {\tau _{2}^{2}-\tau _{1}^{2}}{6r_{1}^{2}}}\right){\mathfrak {R}}_{1}-{\mathfrak {R}}_{2}+{\frac {\tau _{3}}{\tau _{2}}}\left(1+{\frac {\tau _{2}^{2}-\tau _{3}^{2}}{6r_{3}^{2}}}\right){\mathfrak {R}}_{3}=0,$

IIb

or

${\frac {\tau _{1}}{\tau _{2}}}\left(1+{\frac {\tau _{3}^{2}-\tau _{1}^{2}}{6r_{1}^{2}}}\right){\mathfrak {R}}_{1}-\left(1+{\frac {\tau _{3}^{2}-\tau _{2}^{2}}{6r_{2}^{2}}}\right){\mathfrak {R}}_{2}+{\frac {\tau _{3}}{\tau _{2}}}{\mathfrak {R}}_{2}=0.$

IIc