Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/136

first and second positions, and $\tau _{1}$ for that between the second and third, and set $t=0$ for the second position,

for $t=-\tau _{3},$

${\mathfrak {R}}={\mathfrak {R}}_{1},$${\frac {d^{2}{\mathfrak {R}}}{dt^{2}}}=-{\frac {{\mathfrak {R}}_{1}}{r_{1}^{3}}};$

for $t=0,$

${\mathfrak {R}}={\mathfrak {R}}_{2},$${\frac {d^{2}{\mathfrak {R}}}{dt^{2}}}=-{\frac {{\mathfrak {R}}_{2}}{r_{2}^{3}}};$

for $t=\tau _{1},$

${\mathfrak {R}}={\mathfrak {R}}_{3},$${\frac {d^{2}{\mathfrak {R}}}{dt^{2}}}=-{\frac {{\mathfrak {R}}_{3}}{r_{3}^{3}}}\cdot$

We may therefore write with a high degree of approximation

${\begin{aligned}{\mathfrak {R}}_{1}&={\mathfrak {A}}-\tau _{3}{\mathfrak {B}}+\tau _{3}^{2}{\mathfrak {C}}-\tau _{3}^{3}{\mathfrak {D}}+\tau _{3}^{4}{\mathfrak {E}}\\{\mathfrak {R}}_{2}&={\mathfrak {A}}\\{\mathfrak {R}}_{3}&={\mathfrak {A}}+\tau _{3}{\mathfrak {B}}+\tau _{1}^{2}{\mathfrak {C}}+\tau _{1}^{3}{\mathfrak {D}}-\tau _{1}^{4}{\mathfrak {E}}\\-{\frac {{\mathfrak {R}}_{1}}{r_{1}}}&=2{\mathfrak {C}}-6\tau _{3}{\mathfrak {D}}+12\tau _{3}^{2}{\mathfrak {E}}\\-{\frac {{\mathfrak {R}}_{2}}{r_{2}^{3}}}&=2{\mathfrak {E}}\\-{\frac {{\mathfrak {R}}_{3}}{r_{3}^{3}}}&=2{\mathfrak {E}}+6\tau _{1}{\mathfrak {D}}+12\tau _{1}^{2}{\mathfrak {E}}\end{aligned}}$

From these six equations the five constants ${\mathfrak {A,B,C,D,E}}$ may be eliminated, leaving a single equation of the form

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

(1)

where

$A_{1}={\frac {\tau _{1}}{\tau _{1}+\tau _{3}}},$$A_{3}={\frac {\tau _{3}}{\tau _{1}+\tau _{2}}},$

$B_{1}={\tfrac {1}{12}}(-\tau _{1}^{2}+\tau _{1}\tau _{3}+\tau _{3}^{2}),$$B_{2}={\tfrac {1}{12}}(\tau _{1}^{2}+3\tau _{1}\tau _{3}+\tau _{3}^{2})$

$B_{3}={\tfrac {1}{12}}(\tau _{1}^{2}+\tau _{1}\tau _{3}-\tau _{3}^{2}).$

This we shall call our fundamental equation. In order to discuss its geometrical signification, let us set

$n_{1}=A_{1}\left(1+{\frac {B_{1}}{r_{1}^{3}}}\right),$$n_{2}=\left(1-{\frac {B_{2}}{r_{2}^{3}}}\right),$$n_{3}=A_{3}\left(1+{\frac {B_{3}}{r_{3}^{3}}}\right)\cdot$

(2)

so that the equation will read

$n_{1}{\mathfrak {R}}_{2}-n_{2}{\mathfrak {R}}_{2}+n_{3}{\mathfrak {R}}_{2}=0.$

(3)

This expresses that the vector $n_{2}{\mathfrak {R}}_{2}$ is the diagonal of a parallelogram of which $n_{1}{\mathfrak {R}}_{1}$ and $n_{3}{\mathfrak {R}}_{3}$ are sides. If we multiply by ${\mathfrak {R}}_{3}$ and by ${\mathfrak {R}}_{1},$ in skew multiplication, we get

$n_{1}{\mathfrak {R}}_{1}\times {\mathfrak {R}}_{3}-n_{2}{\mathfrak {R}}_{2}\times {\mathfrak {R}}_{3}=0,$$-n_{2}{\mathfrak {R}}_{1}\times {\mathfrak {R}}_{2}+n_{3}{\mathfrak {R}}_{1}\times {\mathfrak {R}}_{3}=0,$

(4)

whence

${\frac {{\mathfrak {R}}_{2}\times {\mathfrak {R}}_{3}}{n_{1}}}={\frac {{\mathfrak {R}}_{1}\times {\mathfrak {R}}_{3}}{n_{2}}}={\frac {{\mathfrak {R}}_{1}\times {\mathfrak {R}}_{2}}{n_{3}}}\cdot$

(5)