Translation:Determinatio attractionis quam in punctum quodvis positionis datae exerceret planeta si eius massa per totam orbitam ratione temporis quo singulae partes describuntur uniformiter esset dispertita

by Carl Friedrich Gauss

1.

Secular variations, which the elements of a planetary orbit undergo due to the perturbation of another planet, are independent of the position of the latter in its orbit, and would be the same whether the perturbing planet moves in an elliptical orbit according to Kepler's laws or its mass is conceived to be evenly distributed throughout the orbit, so that equal masses are assigned to parts of the orbit described in equal time intervals, provided that the periods of revolution of the perturbed planet and the perturbing one are not commensurable. Although this elegant theorem has not yet been explicitly stated by anyone, it can at least be demonstrated easily from the principles of physical astronomy. Thus the following problem arises, which is worthy of attention both in itself and because of several techniques required for its solution: to determine exactly the attraction of the planetary orbit, or if one prefers, the elliptical ring, whose thickness at an arbitrary given position is infinitely small and varies according to the law just explained.

2.

Let $e$ denote the eccentricity of the orbit, and let $E$ denote the eccentric anomaly of an arbitrary point on the orbit. Then each element $dE$ of the eccentric anomaly corresponds to an element of the mean anomaly $(1-e\cos E)dE,$ and the element of mass of that portion of the orbit to which those elements correspond, will be to the whole mass (which we take to be unity) as $(1-e\cos E)dE$ is to $2\pi ,$ where $\pi$ denotes the semicircumference of a circle of radius $1.$ Therefore, assuming the distance from the attracted point to the point on the orbit to be $=\rho ,$ the attraction produced by the element of the orbit will be

$$={\frac {(1-e\cos E)\operatorname {d} E}{2\pi \rho \rho }}$$

We will denote the major semi-axis by $a,$ the minor semi-axis by $b,$ and we will adopt that line as the abscissa, and the center of the ellipse as the origin. Thus we will have $aa-bb=aaee,$ the abscissa of the point of the orbit will be $=a\cos E,$ and the ordinate will be $=b\sin E.$ Finally, we will denote the distance from the attracted point to the plane of the orbit by $C,$ and the remaining coordinates parallel to the major and minor axes by $A$ and $B.$ With these preparations, the attraction of an element of the orbit will be decomposed into two components parallel to the major and minor axes and a third perpendicular to the plane of the orbit, so that

${\begin{aligned}{\frac {(A-a\cos E)(1-e\cos E)\operatorname {d} E}{2\pi \rho ^{3}}}&=\operatorname {d} \xi \\{\frac {(B-b\sin E)(1-e\cos E)\operatorname {d} E}{2\pi \rho ^{3}}}&=\operatorname {d} \eta \\{\frac {C(1-e\cos E)\operatorname {d} E}{2\pi \rho ^{3}}}&=\operatorname {d} \zeta \end{aligned}}$

where ${\textstyle \rho ={\sqrt {(A-a\cos E)^{2}+(B-b\sin E)^{2}+CC}}.}$

Integrating these differentials from $E=0$ to $E=360^{\circ }$ yields partial attractions $\xi ,$ $\eta ,$ $\zeta ,$ in directions opposite to the coordinate directions, and thus we obtain the total composite attraction, which can be referred to any other direction by a known method.

3.

The main task now is to replace $E$ with another variable, in order to simplify the radical quantity. To this end, we set

$\cos E={\frac {\alpha +\alpha ^{\prime }\cos T+\alpha ^{\prime \prime }\sin T}{\gamma +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T}},\quad \sin E={\frac {\beta +\beta ^{\prime }\cos T+\beta ^{\prime \prime }\sin T}{\gamma +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T}}$

where the nine coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, etc., are not entirely arbitrary, but must satisfy certain conditions, which must be examined before we can proceed. First we observe that the substitution remains the same if all coefficients are multiplied by the same factor, so that without loss of generality, it is possible to assign a determined value to one of them, e.g. $\gamma =1.$ However, for the sake of elegance, all nine will remain indefinite for the time being. Furthermore, we note that values for which $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, or $\beta ,$ $\beta ^{\prime },$ $\beta ^{\prime \prime }$ are proportional to $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$, respectively must be excluded: otherwise, $E$ would no longer be undetermined. Therefore, $\gamma ^{\prime }\alpha ^{\prime \prime }-\gamma ^{\prime \prime }\alpha ^{\prime },$ $\gamma ^{\prime \prime }\alpha -\gamma \alpha ^{\prime \prime },$ $\gamma \alpha ^{\prime }-\gamma ^{\prime }\alpha$ cannot all vanish simultaneously.

It is clear that the coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, etc., must be related in such a way that the following expression becomes indefinite

$\left.{\begin{array}{r}(\alpha +\alpha ^{\prime }\cos T+\alpha ^{\prime \prime }\sin T)^{2}\\+(\beta +\beta ^{\prime }\cos T+\beta ^{\prime \prime }\sin T)^{2}\\-(\gamma +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T)^{2}\end{array}}\right\}=0.$

It follows that the above function must have the form

$k({\cos T}^{2}+{\sin T}^{2}-1).$

Hence we obtain six conditional equations

$\left.{\begin{alignedat}{5}&-\alpha \alpha &&-\beta \beta &&+\gamma \gamma &&=&k\\&-\alpha ^{\prime }\alpha ^{\prime }&&-\beta ^{\prime }\beta ^{\prime }&&+\gamma ^{\prime }\gamma ^{\prime }&&=&-k\\&-\alpha ^{\prime \prime }\alpha ^{\prime \prime }&&-\beta ^{\prime \prime }\beta ^{\prime \prime }&&+\gamma ^{\prime \prime }\gamma ^{\prime \prime }&&=&-k\\&-\alpha ^{\prime }\alpha ^{\prime \prime }&&-\beta ^{\prime }\beta ^{\prime \prime }&&+\gamma ^{\prime }\gamma ^{\prime \prime }&&=&0\\&-\alpha ^{\prime \prime }\alpha &&-\beta ^{\prime \prime }\beta &&+\gamma ^{\prime \prime }\gamma &&=&0\\&-\alpha \alpha ^{\prime }&&-\beta \beta ^{\prime }&&+\gamma \gamma ^{\prime }&&=&0\end{alignedat}}\right\}$

(I)

From these equations result several others which are worth developing. For the sake of brevity, let us set

$\alpha \beta ^{\prime }\gamma ^{\prime \prime }+\alpha ^{\prime }\beta ^{\prime \prime }\gamma +\alpha ^{\prime \prime }\beta \gamma ^{\prime }-\alpha \beta ^{\prime \prime }\gamma ^{\prime }-\alpha ^{\prime }\beta \gamma ^{\prime \prime }-\alpha ^{\prime \prime }\beta ^{\prime }\gamma =\varepsilon$

(II)

Then from various combinations of the equations (I), the following nine can be easily derived:

$\left.{\begin{alignedat}{2}&\varepsilon \alpha &&=-k(\beta ^{\prime }\gamma ^{\prime \prime }-\gamma ^{\prime }\beta ^{\prime \prime })\\&\varepsilon \beta &&=-k(\gamma ^{\prime }\alpha ^{\prime \prime }-\alpha ^{\prime }\gamma ^{\prime \prime })\\&\varepsilon \gamma &&=+k(\alpha ^{\prime }\beta ^{\prime \prime }-\beta ^{\prime }\alpha ^{\prime \prime })\\&\varepsilon \alpha ^{\prime }&&=+k(\beta ^{\prime \prime }\gamma -\gamma ^{\prime \prime }\beta )\\&\varepsilon \beta ^{\prime }&&=+k(\gamma ^{\prime \prime }\alpha -\alpha ^{\prime \prime }\gamma )\\&\varepsilon \gamma ^{\prime }&&=-k(\alpha ^{\prime \prime }\beta -\beta ^{\prime \prime }\alpha )\\&\varepsilon \alpha ^{\prime \prime }&&=+k(\beta \gamma ^{\prime }-\gamma \beta ^{\prime })\\&\varepsilon \beta ^{\prime \prime }&&=+k(\gamma \alpha ^{\prime }-\alpha \gamma ^{\prime })\\&\varepsilon \gamma ^{\prime \prime }&&=-k(\alpha \beta ^{\prime }-\beta \alpha ^{\prime })\end{alignedat}}\right\}$

(III)

From the first three of these equations, we deduce

${\begin{aligned}\varepsilon \alpha (\beta ^{\prime }\gamma ^{\prime \prime }-\gamma ^{\prime }\beta ^{\prime \prime })+\varepsilon \beta (\gamma ^{\prime }\alpha ^{\prime \prime }-\alpha ^{\prime }\gamma ^{\prime \prime })+\varepsilon \gamma (\alpha ^{\prime }\beta ^{\prime \prime }-\beta ^{\prime }\alpha ^{\prime \prime })\\=-k(\beta ^{\prime }\gamma ^{\prime \prime }-\gamma ^{\prime }\beta ^{\prime \prime })^{2}-k(\gamma ^{\prime }\alpha ^{\prime \prime }-\alpha ^{\prime }\gamma ^{\prime \prime })^{2}+k(\alpha ^{\prime }\beta ^{\prime \prime }-\beta ^{\prime }\alpha ^{\prime \prime })^{2}.\end{aligned}}$

This is equivalent to

$\varepsilon \varepsilon =k(-\alpha ^{\prime }\alpha ^{\prime }-\beta ^{\prime }\beta ^{\prime }+\gamma ^{\prime }\gamma ^{\prime })(-\alpha ^{\prime \prime }\alpha ^{\prime \prime }-\beta ^{\prime \prime }\beta ^{\prime \prime }+\gamma ^{\prime \prime }\gamma ^{\prime \prime })-k(-\alpha ^{\prime }\alpha ^{\prime \prime }-\beta ^{\prime }\beta ^{\prime \prime }+\gamma ^{\prime }\gamma ^{\prime \prime })^{2},$

which, with the help of equations 2, 3, 4 in (I), can be transformed into

$\varepsilon \varepsilon =k^{3}$

(IV)

It is equally easy to derive the following from equations (I):

$\left.{\begin{alignedat}{2}(\beta ^{\prime }\gamma ^{\prime \prime }-\gamma ^{\prime }\beta ^{\prime \prime })^{2}&=-k(k-\alpha ^{\prime }\alpha ^{\prime }-\alpha ^{\prime \prime }\alpha ^{\prime \prime })\\(\gamma ^{\prime }\alpha ^{\prime \prime }-\alpha ^{\prime }\gamma ^{\prime \prime })^{2}&=-k(k-\beta ^{\prime }\beta ^{\prime }-\beta ^{\prime \prime }\beta ^{\prime \prime })\\(\alpha ^{\prime }\beta ^{\prime \prime }-\beta ^{\prime }\alpha ^{\prime \prime })^{2}&=+k(k+\gamma ^{\prime }\gamma ^{\prime }+\gamma ^{\prime \prime }\gamma ^{\prime \prime })\\(\beta ^{\prime \prime }\gamma -\gamma ^{\prime \prime }\beta )^{2}&=+k(k+\alpha \alpha -\alpha ^{\prime \prime }\alpha ^{\prime \prime })\\(\gamma ^{\prime \prime }\alpha -\alpha ^{\prime \prime }\gamma )^{2}&=+k(k+\beta \beta -\beta ^{\prime \prime }\beta ^{\prime \prime })\\(\alpha ^{\prime \prime }\beta -\beta ^{\prime \prime }\alpha )^{2}&=-k(k-\gamma \gamma +\gamma ^{\prime \prime }\gamma ^{\prime \prime })\\(\beta \gamma ^{\prime }-\gamma \beta ^{\prime })^{2}&=+k(k+\alpha \alpha -\alpha ^{\prime }\alpha ^{\prime })\\(\gamma \alpha ^{\prime }-\alpha \gamma ^{\prime })^{2}&=+k(k+\beta \beta -\beta ^{\prime }\beta ^{\prime })\\(\alpha \beta ^{\prime }-\beta \alpha ^{\prime })^{2}&=-k(k-\gamma \gamma +\gamma ^{\prime }\gamma ^{\prime })\end{alignedat}}\right\}$

(V)

We now provide a derivation of the first equation, based on whose example derivations of the others can be found. Specifically, equations 4, 2, 3 in (I) yield

$(\gamma ^{\prime }\gamma ^{\prime \prime }-\beta ^{\prime }\beta ^{\prime \prime })^{2}-(\gamma ^{\prime }\gamma ^{\prime }-\beta ^{\prime }\beta ^{\prime })(\gamma ^{\prime \prime }\gamma ^{\prime \prime }-\beta ^{\prime \prime }\beta ^{\prime \prime })=\alpha ^{\prime }\alpha ^{\prime }\alpha ^{\prime \prime }\alpha ^{\prime \prime }-(\alpha ^{\prime }\alpha ^{\prime }-k)(\alpha ^{\prime \prime }\alpha ^{\prime \prime }-k)$

which, when expanded, immediately yields the first equation in (V).

From the equations (V), we conclude that the value $k=0$ is not admissible in our discussion; otherwise, all nine quantities $\beta ^{\prime }\gamma ^{\prime \prime }-\gamma ^{\prime }\beta ^{\prime \prime }$, etc., would necessarily vanish, i.e. the coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$ and $\beta ,$ $\beta ^{\prime },$ $\beta ^{\prime \prime }$ and $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ would all become proportional. In fact, the quantity $\varepsilon$ cannot vanish, due to equation (IV); hence $k$ must be a positive quantity, since all of the coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, etc., must be real. Combining the first three equations in (III) with the first three in (V), the following new equations emerge, which clearly depend on the non-vanishing value of $k$:

$\left.{\begin{aligned}\alpha \alpha -\alpha ^{\prime }\alpha ^{\prime }-\alpha ^{\prime \prime }\alpha ^{\prime \prime }&=-k\\\beta \beta -\beta ^{\prime }\beta ^{\prime }-\beta ^{\prime \prime }\beta ^{\prime \prime }&=-k\\\gamma \gamma -\gamma ^{\prime }\gamma ^{\prime }-\gamma ^{\prime \prime }\gamma ^{\prime \prime }&=+k\end{aligned}}\right\}$

(VI)

Combining the other equations would produce the same result. Finally, we add the following three equations,

$\left.{\begin{aligned}\beta \gamma -\beta ^{\prime }\gamma ^{\prime }-\beta ^{\prime \prime }\gamma ^{\prime \prime }&=0\\\gamma \alpha -\gamma ^{\prime }\alpha ^{\prime }-\gamma ^{\prime \prime }\alpha ^{\prime \prime }&=0\\\alpha \beta -\alpha ^{\prime }\beta ^{\prime }-\alpha ^{\prime \prime }\beta ^{\prime \prime }&=0\end{aligned}}\right\}$

(VII)

which are easily derived from the equations (III). For example, the second, fifth, and eighth yield

$\varepsilon \beta \gamma -\varepsilon \beta ^{\prime }\gamma ^{\prime }-\varepsilon \beta ^{\prime \prime }\gamma ^{\prime \prime }=-k\gamma (\gamma ^{\prime }\alpha ^{\prime \prime }-\alpha ^{\prime }\gamma ^{\prime \prime })-k\gamma ^{\prime }(\gamma ^{\prime \prime }\alpha -\alpha ^{\prime \prime }\gamma )-k\gamma ^{\prime \prime }(\gamma \alpha ^{\prime }-\alpha \gamma ^{\prime })=0$

Clearly, these equations also require the value $k=0$ to be excluded.^[1]

Since, as mentioned above, we can multiply all coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, etc., by the same factor, the value of $k$ multiplied by the square of this factor will always result. Henceforth, we will always assume

$k=1$,

from which it follows that either $\varepsilon =+1$ or $\varepsilon =-1.$ It is evident, therefore, that the nine coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$ etc., among which there are six conditional equations, must be reducible to three independent quantities. This is most conveniently accomplished using three angles, as follows,

${\begin{aligned}\alpha &=\cos L\tan N\\\beta &=\sin L\tan N\\\gamma &=\sec N\\\alpha ^{\prime }&=\cos L\cos M\sec N\pm \sin L\sin M\\\beta ^{\prime }&=\sin L\cos M\sec N\mp \cos L\sin M\\\gamma ^{\prime }&=\cos M\tan N\\\alpha ^{\prime \prime }&=\cos L\sin M\sec N\mp \sin L\cos M\\\beta ^{\prime \prime }&=\sin L\sin M\sec N\pm \cos L\cos M\\\gamma ^{\prime \prime }&=\sin M\tan N\end{aligned}}$

where the upper signs refer to the case $\varepsilon =+1,$ and the lower ones to the case $\varepsilon =-1.$ Nevertheless, the analytical treatment can be completed for the most part more elegantly without using these angles. Of course, it would not be difficult to assign a geometric interpretation to both these angles and the other auxiliary quantities occurring in this discussion; however, we leave this interpretation, which is not necessary for our purpose, to be explained by the knowledgeable reader.

4.

If the above values are substituted for $\cos E$ and $\sin E$ in the expression for the distance $\rho$, then this expression obtains the form

$\rho ={\frac {\sqrt {G+G^{\prime }\cos T^{2}+G^{\prime \prime }\sin T^{2}+2H\cos T\sin T+2H^{\prime }\sin T+2H^{\prime \prime }\cos T}}{\gamma +\gamma ^{\prime }\cos T^{\prime }+\gamma ^{\prime \prime }\sin T}},$

where the coefficients $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, etc. must be determined in such a way that the six conditional equations are satisfied,

$\left.{\begin{alignedat}{7}&-\alpha \alpha &&-\beta \beta &&+\gamma \gamma &&=&1\\&-\alpha ^{\prime }\alpha ^{\prime }&&-\beta ^{\prime }\beta ^{\prime }&&+\gamma ^{\prime }\gamma ^{\prime }&&=&-1\\&-\alpha ^{\prime \prime }\alpha ^{\prime \prime }&&-\beta ^{\prime \prime }\beta ^{\prime \prime }&&+\gamma ^{\prime \prime }\gamma ^{\prime \prime }&&=&\,-1\\&-\alpha ^{\prime }\alpha ^{\prime \prime }&&-\beta ^{\prime }\beta ^{\prime \prime }&&+\gamma ^{\prime }\gamma ^{\prime \prime }&&=&0\\&-\alpha ^{\prime \prime }\alpha &&-\beta ^{\prime \prime }\beta &&+\gamma ^{\prime \prime }\gamma &&=&0\\&-\alpha \alpha ^{\prime }&&-\beta \beta ^{\prime }&&+\gamma \gamma ^{\prime }&&=&0\end{alignedat}}\right\}$

[1]

and so must be the others which follow from them,

$H=0,\quad H^{\prime }=0,\quad H^{\prime \prime }=0$,

and thus the problem will be generally determined. Therefore, if we denote the denominator of $\rho$ by $t,$ and make the substitution

${\begin{alignedat}{2}&t\cos E&&=\alpha +\alpha ^{\prime }\cos T+\alpha ^{\prime \prime }\sin T\\&t\sin E&&=\beta +\beta ^{\prime }\cos T+\beta ^{\prime \prime }\sin T\\&t&&=\gamma +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T,\end{alignedat}}$

in the following function of the three quantities $t,$ $t\cos E,$ and $t\sin E,$

$(AA+BB+CC)tt+aa(t\cos E)^{2}+bb(t\sin E)^{2}-2aAt.t\cos E-2bBt.t\sin E,$

then we obtain

$G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2}$

It would clearly be equivalent to say that substituting

${\begin{aligned}&x=\alpha u+\alpha ^{\prime }u^{\prime }+\alpha ^{\prime \prime }u^{\prime \prime }\\&y=\beta u+\beta ^{\prime }u^{\prime }+\beta ^{\prime \prime }u^{\prime \prime }\\&z=\gamma u+\gamma ^{\prime }u^{\prime }+\gamma ^{\prime \prime }u^{\prime \prime }\end{aligned}}$

in the following function ($W$) of the three variables $x,$ $y,$ and $z$

$aaxx+bbyy+(AA+BB+CC)zz-2aAxz-2bByz$,

results in the following function of the variables $u,$ $u^{\prime },$ and $u^{\prime \prime }{:}$

$Guu+G^{\prime }u^{\prime }u^{\prime }+G^{\prime \prime }u^{\prime \prime }u^{\prime \prime }$

But from these formulas, with the help of the equations [1], it easily follows that

${\begin{alignedat}{8}&u&&=-&&\alpha x-\beta y+\gamma z\\&u^{\prime }&&=&&\alpha ^{\prime }x+\beta ^{\prime }y-\gamma ^{\prime }z\\&u^{\prime \prime }&&=&&\alpha ^{\prime \prime }x+\beta ^{\prime \prime }y-\gamma ^{\prime \prime }z,\end{alignedat}}$

so it is clear that the function $W$ must be identical to

$G(-\alpha x-\beta y+\gamma z)^{2}+G^{\prime }(\alpha ^{\prime }x+\beta ^{\prime }y-\gamma ^{\prime }z)^{2}+G^{\prime \prime }(\alpha ^{\prime \prime }x+\beta ^{\prime \prime }y-\gamma ^{\prime \prime }z)^{2}.$

This leads us to six equations

$\left.{\begin{array}{rl}aa&=G\alpha \alpha +G^{\prime }\alpha ^{\prime }\alpha ^{\prime }+G^{\prime \prime }\alpha ^{\prime \prime }\alpha ^{\prime \prime }\\bb&=G\beta \beta +G^{\prime }\beta ^{\prime }\beta ^{\prime }+G^{\prime \prime }\beta ^{\prime \prime }\beta ^{\prime \prime }\\AA+BB+CC&=G\gamma \gamma +G^{\prime }\gamma ^{\prime }\gamma ^{\prime }+G^{\prime \prime }\gamma ^{\prime \prime }\gamma ^{\prime \prime }\\bB&=G\beta \gamma +G^{\prime }\beta ^{\prime }\gamma ^{\prime }+G^{\prime \prime }\beta ^{\prime \prime }\gamma ^{\prime \prime }\\aA&=G\gamma \alpha +G^{\prime }\gamma ^{\prime }\alpha ^{\prime }+G^{\prime \prime }\gamma ^{\prime \prime }\alpha ^{\prime \prime }\\0&=G\alpha \beta +G^{\prime }\alpha ^{\prime }\beta ^{\prime }+G^{\prime \prime }\alpha ^{\prime \prime }\beta ^{\prime \prime }\end{array}}\right\}$

[2]

From these twelve equations [1] and [2], we must determine the values of our unknowns $G,$ $G^{\prime },$ $G^{\prime \prime },$ $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime }$, etc.

5.

By combining equations [1] and [2], the following equations are easily derived:

${\begin{aligned}&-\alpha aa+\gamma aA=\alpha G\\&-\beta bb+\gamma bB=\beta G\\&\gamma (AA+BB+CC)-\alpha aA-\beta bB=\gamma G\end{aligned}}$

From this we further obtain:

$\alpha ={\frac {\gamma aA}{aa+G}}$

[3]

$\beta ={\frac {\gamma bB}{bb+G}}$

[4]

$AA+BB+CC-{\frac {aaAA}{aa+G}}-{\frac {bbB}{bb+G}}=G$

We can also express the final equation as follows:

${\frac {AA}{aa+G}}+{\frac {BB}{bb+G}}+{\frac {CC}{G}}=1$

[5]

Similarly, by combining equations [1] and [2], we deduce:

${\begin{aligned}&\alpha ^{\prime }aa-\gamma ^{\prime }aA=\alpha ^{\prime }G^{\prime }\\&\beta ^{\prime }bb-\gamma ^{\prime }bB=\beta ^{\prime }G^{\prime }\\&-\gamma ^{\prime }(AA+BB+CC)+\alpha ^{\prime }aA+\beta ^{\prime }bB=\gamma ^{\prime }G^{\prime },\end{aligned}}$

and hence

$\alpha ^{\prime }={\frac {\gamma ^{\prime }aA}{aa-G^{\prime }}},$

[6]

$\beta ^{\prime }={\frac {\gamma ^{\prime }bB}{bb-G^{\prime }}},$

[7]

${\frac {AA}{aa-G^{\prime }}}+{\frac {BB}{bb-G^{\prime }}}-{\frac {CC}{G^{\prime }}}=1,$

[8]

and in a completely similar manner:

$\alpha ^{\prime \prime }={\frac {\gamma ^{\prime \prime }aA}{aa-G^{\prime \prime }}}$,

[9]

$\beta ^{\prime \prime }={\frac {\gamma ^{\prime \prime }bB}{bb-G^{\prime \prime }}}$,

[10]

${\frac {AA}{aa-G^{\prime \prime }}}+{\frac {BB}{bb-G^{\prime \prime }}}-{\frac {CC}{G^{\prime \prime }}}=1.$

[11]

It is therefore clear that $G,$ $-G^{\prime },$ $-G^{\prime \prime }$ are the roots of the equation

${\frac {AA}{aa+x}}+{\frac {BB}{bb+x}}+{\frac {CC}{x}}=1,$

[12]

which, when properly expanded, takes the form

$x^{3}$$-(AA+BB+CC-aa-bb)xx$$+(aabb-aaBB-aaCC-bbAA-bbCC)x$$-aabbCC=0$

[13]

6.

Regarding the nature of this cubic equation, the following observations are noteworthy.

I. From the last term of the equation, $-aabbCC,$ it follows that the equation has at least one real root, which can either be positive, or in the case $C=0,$ equal to zero. Let us denote this real non-negative root by $g.$

II. Writing equation 12 in the form

$x={\frac {AAx}{aa+x}}+{\frac {BBx}{bb+x}}+CC,$

subtracting the equation

$g={\frac {AAg}{aa+g}}+{\frac {BBg}{bb+g}}+CC$

and dividing by $x-g,$ we obtain the following new equation, which is satisfied by the two remaining roots:

$1={\frac {aaAA}{(aa+x)(aa+g)}}+{\frac {bbBB}{(bb+x)(bb+g)}}.$

When arranged and solved in the usual way, this equation yields

$2x={\frac {aaAA}{aa+g}}$$+{\frac {bbBB}{bb+g}}$$-aa-bb$$\pm {\sqrt {\left(aa-bb-{\frac {aaAA}{aa+g}}+{\frac {bbBB}{bb+g}}\right)^{2}+{\frac {4aabbAABB}{(aa+g)(bb+g)}}}}.$

[14]

Since the quantity under the radical sign is positive, or at least non-negative, this indicates that the two remaining roots are always real.

III. Subtracting the following equations from each other,

${\begin{aligned}&gx={\frac {AAgx}{aa+x}}+{\frac {BBgx}{bb+x}}+gCC\\&gx={\frac {AAgx}{aa+g}}+{\frac {BBgx}{bb+g}}+xCC\end{aligned}}$

and dividing by $g-x,$ we obtain an equation satisfied by the two remaining roots, of the form:

$0={\frac {AAgx}{(aa+g)(aa+x)}}+{\frac {BBgx}{(bb+g)(bb+x)}}+CC$

From this we see that if $g$ is positive, the equation cannot be satisfied by any positive value of $x.$ Therefore, we conclude that our cubic equation cannot have more than one positive root.

IV. Thus, whenever $0$ is not among the roots of our equation, there will always be one positive root and two negative ones. However, when $C=0,$ and hence $0$ is one of the roots, the equation satisfied by the remaining roots will be

$xx-(AA+BB-aa-bb)x+aabb-aaBB-bbAA=0,$

from which these roots are found to be

${\tfrac {1}{2}}(AA+BB-aa-bb)\pm {\tfrac {1}{2}}{\sqrt {(AA-BB-aa+bb)^{2}+4AABB}}.$

Here, three cases need to be distinguished again.

Firstly, if the last term $aabb-aaBB-bbAA$ is positive (i.e., if the attracting point lies within the curve of the attracting ellipse), both roots, being real, will have the same sign. Since they cannot both be positive, they must be negative. Furthermore, this can also be concluded independently of what has already been shown, since the middle coefficient, which can be expressed as

$(aabb-aaBB-bbAA)\left({\frac {1}{aa}}+{\frac {1}{bb}}\right)+{\frac {bbAA}{aa}}+{\frac {aabB}{bb}}$

is clearly positive in this case.

Secondly, if the last term is negative, indicating that the attracting point lies outside the curve, then one root will be positive, and the other negative.

Thirdly, if the last term vanishes, indicating that the attractor lies precisely on the circumference of the ellipse, then the second root will also be $=0,$ while the third will be

$=-{\frac {bbAA}{aa}}-{\frac {aaBB}{bb}},$

i.e. it will be negative. But this case, in which the attraction would become infinitely large, is physically impossible, and we exclude it from our discussion, at least for the moment.

7.

Using equations 1, 3, 4, 6, 7, 9, 10 to determine the coefficients $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime },$ we find that

$\left.{\begin{aligned}\gamma &={\frac {1}{\sqrt {1-\left({\frac {aA}{aa+G}}\right)^{2}+\left({\frac {bB}{bb+G}}\right)^{2}}}}\\\gamma ^{\prime }&={\frac {1}{\sqrt {\left({\frac {aA}{aa-G^{\prime }}}\right)^{2}+\left({\frac {bB}{bb-G^{\prime }}}\right)^{2}-1}}}\\\gamma ^{\prime \prime }&={\frac {1}{\sqrt {\left({\frac {aA}{aa-G^{\prime \prime }}}\right)^{2}+\left({\frac {bB}{bb-G^{\prime \prime }}}\right)^{2}-1}}}\end{aligned}}\right\}$ [15]

Combining these equations with 5, 8, 11, we also obtain:

$\left.{\begin{aligned}\gamma &={\sqrt {\frac {G}{\left({\frac {AG}{aa+G}}\right)^{2}+\left({\frac {BG}{bb+G}}\right)^{2}+CC}}}\\\gamma ^{\prime }&={\sqrt {\frac {G^{\prime }}{\left({\frac {AG^{\prime }}{aa-G^{\prime }}}\right)^{2}+\left({\frac {BG^{\prime }}{bb-G^{\prime }}}\right)^{2}+CC}}}\\\gamma ^{\prime \prime }&={\sqrt {\frac {G^{\prime \prime }}{\left({\frac {AG^{\prime \prime }}{aa-G^{\prime \prime }}}\right)^{2}+\left({\frac {BG^{\prime \prime }}{bb-G^{\prime \prime }}}\right)^{2}+CC}}}\end{aligned}}\right\}$ [16]

Since $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ must be real, these expressions show that none of the quantities $G,$ $G^{\prime },$ $G^{\prime \prime }$ can be negative.

Therefore, in the case where $C$ is not $=0,$ $G$ must be set equal to the positive root of equation 13. Consequently, $-G^{\prime }$ must be equal to one of the negative roots, and $-G^{\prime \prime }$ must be equal to the other.^[2] However, which root to adopt for $-G^{\prime },$ and which for $-G^{\prime \prime },$ is entirely arbitrary.

Whenever $C=0,$ and the attractor lies outside the ellipse, the positive root of equation 13 must be set equal to $G,$ and the negative root must be set equal either to $-G^{\prime },$ with $G^{\prime \prime }=0,$ or to $-G^{\prime \prime },$ with $G^{\prime }=0.$ The coefficient $\gamma ^{\prime \prime }$ or $\gamma ^{\prime }$ can then be found using the formula

${\frac {1}{\sqrt {{\frac {AA}{aa}}+{\frac {BB}{bb}}-1}}}$

However, in the case already excluded, where the attractor is assumed to lie on the circumference of the ellipse, the coefficients $\gamma$ and $\gamma ^{\prime }$, or $\gamma$ and $\gamma ^{\prime \prime }$, would become infinite, indicating that our transformation is not applicable in this case at all.

8.

Although formulas 15 and 16 could suffice to determine the coefficients $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime },$ more elegant formulas can be found. To this end, we will multiply equation [5] by $aabb-GG,$ which yields, after a slight reduction:

${\frac {aaAA(bb+G)}{aa+G}}-AAG+{\frac {bbBB(aa+G)}{bb+G}}-BBG+{\frac {aabbCC}{G}}-CCG=aabb-GG$

But from the nature of the cubic equation, we have:

the sum of roots: $\quad G-G^{\prime }-G^{\prime \prime }=AA+BB+CC-aa-bb$

the product of roots: $\quad GG^{\prime }G^{\prime \prime }=aabbCC$

Hence, the preceding equation transforms into the following,

${\frac {aaAA(bb+G)}{aa+G}}+{\frac {bbBB(aa+G)}{bb+G}}+G^{\prime }G^{\prime \prime }-G(G-G^{\prime }-G^{\prime \prime }+aa+bb)=aabb-GG,$

which can also be expressed as

${\frac {aaAA(bb+G)}{aa+G}}+{\frac {bbBB(aa+G)}{bb+G}}-(aa+G)(bb+G)+(G+G^{\prime })(G+G^{\prime \prime })=0.$

Hence the value of the coefficient $\gamma$ from the first formula in [15] transforms into the following:

$\gamma ={\sqrt {\frac {(aa+G)(bb+G)}{(G+G^{\prime })(G+G^{\prime \prime })}}}$

[17]

A completely similar analysis yields:

$\gamma ^{\prime {\phantom {\prime }}}={\sqrt {\frac {(aa-G^{\prime })(bb-G^{\prime })}{(G+G^{\prime })(G^{\prime \prime }-G^{\prime })}}}$

[18]

$\gamma ^{\prime \prime }={\sqrt {\frac {(aa-G^{\prime \prime })(bb-G^{\prime \prime })}{(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}}$

[19]

Once the coefficients $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ have been found, the remaining $\alpha ,$ $\beta ,$ $\alpha ^{\prime },$ $\beta ^{\prime },$ $\alpha ^{\prime \prime },$ $\beta ^{\prime \prime }$ can be derived from formulas 3, 4, 6, 7, 9, 10.

9.

It is easy to understand that the signs of the radical expressions by which $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ are determined can be taken arbitrarily. However, it is worth considering how the sign of the quantity $\varepsilon$ is related to these signs. To this end, let us consider the third equation in III art. 3.

$\varepsilon \gamma =\alpha ^{\prime }\beta ^{\prime \prime }-\beta ^{\prime }\alpha ^{\prime \prime },$

which through formulas 6, 7, 9, 10 transforms into

${\begin{aligned}\varepsilon \gamma &={\frac {abAB\gamma ^{\prime }\gamma ^{\prime \prime }}{(aa-G^{\prime })(bb-G^{\prime \prime })}}-{\frac {abAB\gamma ^{\prime }\gamma ^{\prime \prime }}{(aa-G^{\prime \prime })(bb-G^{\prime })}}\\&={\frac {ab(aa-bb)AB(G^{\prime \prime }-G^{\prime })\gamma ^{\prime }\gamma ^{\prime \prime }}{(aa-G^{\prime })(aa-G^{\prime \prime })(bb-G^{\prime })(bb-G^{\prime \prime })}}\end{aligned}}$

Now, considering equation 13, we easily deduce that

${\begin{aligned}&(aa+G)(aa-G^{\prime })(aa-G^{\prime \prime })=aa(aa-bb)AA\\&(bb+G)(bb-G^{\prime })(bb-G^{\prime \prime })=-bb(aa-bb)BB\end{aligned}}$

Hence, the preceding equation becomes

$\varepsilon \gamma ={\frac {(aa+G)(bb+G)(G^{\prime }-G^{\prime \prime })\gamma ^{\prime }\gamma ^{\prime \prime }}{ab(aa-bb)AB}}.$

Combined with equation 17, this yields

$\gamma \gamma ^{\prime }\gamma ^{\prime \prime }={\frac {\varepsilon ab(aa-bb)AB}{(G+G^{\prime })(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}$

Hence, it is clear that $-G^{\prime }$ is taken to be the absolutely greatest negative root of the cubic equation, and at the same time all coefficients $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ are taken positively, then $\varepsilon$ will have the same sign as $AB,$ and the same will happen if these four conditions, or any two of them, are negated, but the opposite will happen if one or three of the conditions are negated. It is also worth noting the following relations, which are easily derived from those above:

${\begin{aligned}\alpha \alpha ^{\prime }\alpha ^{\prime \prime }&={\phantom {-}}{\frac {\varepsilon aabAAB}{(G+G^{\prime })(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}\\\beta \beta ^{\prime }\beta ^{\prime \prime }&=-{\frac {\varepsilon abbABB}{(G+G^{\prime })(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}\\\alpha \beta &={\phantom {-}}{\frac {abAB}{(G+G^{\prime })(G+G^{\prime \prime })}}\\\alpha ^{\prime }\beta ^{\prime }&=-{\frac {abAB}{(G+G^{\prime })(G^{\prime }-G^{\prime \prime })}}\\\alpha ^{\prime \prime }\beta ^{\prime \prime }&={\phantom {-}}{\frac {abAB}{(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}\end{aligned}}$

10.

Our formulas can become indeterminate in certain cases, which need to be considered separately. First let us discuss the case where the negative roots $-G^{\prime }$ and $-G^{\prime \prime }$ of the cubic equation become equal, and through formulas 18, 19, the coefficients $\gamma ,$ $\gamma ^{\prime \prime }$ seem to acquire infinite values, which are actually indeterminate.

Setting $g=G$ in formula 14, it is evident that to make the two values of $x,$ i.e., $-G^{\prime }$ and $-G^{\prime \prime }$, be equal, it must be that:

$AB=0,\quad aa-bb-{\frac {aaAA}{aa+G}}+{\frac {bbBB}{bb+G}}=0$

Therefore, since $aa-bb$ is inherently either positive or $=0,$ it is easy to see that we must have

$B=0$

and

$aa-bb={\frac {aaAA}{aa+G}},\quad$ or $\quad aa+G={\frac {aaAA}{aa-bb}}$

Substituting these values into equation 14 gives

$G^{\prime }=G^{\prime \prime }=bb.$

Further substituting $x=-bb$ into the cubic equation 13 yields

$(aa-bb)(CC+bb)=bbAA$

Whenever this conditional equation holds simultaneously with the equation $B=0,$ the case we are discussing arises. And since

$G={\frac {aaAA}{aa-bb}}-aa={\frac {aaCC}{bb}}$

formula 17 provides

$\gamma ={\sqrt {\frac {aabbAA}{(aa-bb)(aaCC+b^{4})}}}={\sqrt {\frac {aaCC+aabb}{aaCC+b^{4}}}}$

and then formulas 3, 4 yield

${\begin{aligned}&\alpha ={\frac {\gamma (aa-bb)}{aA}}={\frac {\gamma bbA}{a(CC+bb)}}={\sqrt {\frac {bb(aa-bb)}{aaCC+b^{4}}}}={\sqrt {\frac {b^{4}AA}{(CC+bb)(aaCC+b^{4})}}}\\&\beta =0\end{aligned}}$

The values of the coefficients $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ remain indeterminate in this case, according to formulas 18, 19, and so do the values of the coefficients $\alpha ^{\prime },$ $\beta ^{\prime },$ $\alpha ^{\prime \prime },$ $\beta ^{\prime \prime }.$ Nevertheless, all five of the remaining coefficients can be expressed in terms of one of them, e.g. by formula 6,

$\alpha ^{\prime }={\frac {\gamma ^{\prime }aA}{aa-bb}}$

and so

$\beta ^{\prime }={\sqrt {(1-\alpha ^{\prime }\alpha ^{\prime }+\gamma ^{\prime }\gamma ^{\prime })}},\quad$$\gamma ^{\prime \prime }={\sqrt {(\gamma \gamma -1-\gamma ^{\prime }\gamma ^{\prime })}},\quad$$\alpha ^{\prime \prime }={\frac {\gamma ^{\prime \prime }aA}{aa-bb}},\quad$$\beta ^{\prime \prime }={\sqrt {(1-\alpha ^{\prime \prime }\alpha ^{\prime \prime }+\gamma ^{\prime \prime }\gamma ^{\prime \prime })}}$;

but a more concise approach is the following. From

$\gamma \gamma =1+\alpha \alpha ,\quad \alpha \alpha ^{\prime }=\gamma \gamma ^{\prime },\quad 1=\alpha ^{\prime }\alpha ^{\prime }+\beta ^{\prime }\beta ^{\prime }-\gamma ^{\prime }\gamma ^{\prime }$

it follows that

$\beta ^{\prime }\beta ^{\prime }+{\frac {\gamma ^{\prime }\gamma ^{\prime }}{\alpha \alpha }}=1-\alpha ^{\prime }\alpha ^{\prime }+{\frac {\gamma \gamma \gamma ^{\prime }\gamma ^{\prime }}{\alpha \alpha }}=1$

and therefore, we can write

$\beta ^{\prime }=\cos f,\quad \gamma ^{\prime }=\alpha \sin f,\quad \alpha ^{\prime }=\gamma \sin f.$

Then from the formulas

$\varepsilon \alpha ^{\prime \prime }=\beta \gamma ^{\prime }-\gamma \beta ^{\prime },\quad$$\varepsilon \beta ^{\prime \prime }=\gamma \alpha ^{\prime }-\alpha \gamma ^{\prime },\quad$$\varepsilon \gamma ^{\prime \prime }=\beta \alpha ^{\prime }-\alpha \beta ^{\prime },\quad$$\varepsilon \varepsilon =1$

we find that

$\alpha ^{\prime \prime }=-\varepsilon \gamma \cos f,\quad$$\beta ^{\prime \prime }=\varepsilon \sin f,\quad$$\gamma ^{\prime \prime }=-\varepsilon \alpha \cos f.$

Here the value of the angle $f$ is arbitrary, and $\varepsilon$ can be arbitrarily set to either $+1$ or $-1.$

11.

If $G^{\prime }$ and $G^{\prime \prime }$ are not equal, then according to formulas 17, 18, 19, the values of coefficients $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ cannot be indeterminate. On the other hand, if one of the quantities $aa-G^{\prime },$ $bb-G^{\prime },$ $aa-G^{\prime \prime },$ $bb-G^{\prime \prime }$ vanishes, then the value of the corresponding coefficient $\alpha ^{\prime },\beta ^{\prime },$ $\alpha ^{\prime \prime },$ $\gamma ^{\prime \prime }$ appears to be indeterminate according to formulas 6, 7, 9, 10. However, a closer look reveals otherwise.

Let us suppose that e.g. $aa-G^{\prime }=0.$ Then according to equation 18, $\gamma ^{\prime }=0,$ and also, according to equation 7, $\beta ^{\prime }=0$ (since $aa\neq bb$). Consequently, we must have $\alpha ^{\prime }=\pm 1.$ However, if at the same time we have $aa=bb,$ then the formula preceding the sixth in art. 5 yields $\alpha ^{\prime }A+\beta ^{\prime }B=0,$ which, when combined with $\alpha ^{\prime }\alpha ^{\prime }+\beta ^{\prime }\beta ^{\prime }=1,$ yields:

$\alpha ^{\prime }={\frac {B}{\sqrt {AA+BB}}},\quad \beta ^{\prime }={\frac {-A}{\sqrt {AA+BB}}}$

These expressions cannot be indeterminate unless $A=0,$ $B=0;$ then we would fall into the case already considered in the previous article.

12.

Having determined the twelve quantities $G,$ $G^{\prime },$ $G^{\prime \prime },$ $\alpha ,$ $\alpha ^{\prime },$ $\alpha ^{\prime \prime },$ $\beta ,$ $\beta ^{\prime },$ $\beta ^{\prime \prime },$ $\gamma ,$ $\gamma ^{\prime },$ $\gamma ^{\prime \prime }$ in full, we proceed to the determination of the differential $\operatorname {d} E.$ Let us set

$t=\gamma +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T,$

[20]

so that

$t\cos E=\alpha +\alpha ^{\prime }\cos T+\alpha ^{\prime \prime }\sin T$

[21]

$t\sin E=\beta +\beta ^{\prime }\cos T+\beta ^{\prime \prime }\sin T$

[22]

From this, we deduce

${\begin{aligned}t\operatorname {d} E&=\cos E\operatorname {d} .t\sin E-\sin E\operatorname {d} .t\cos E\\&=\cos E(\beta ^{\prime \prime }\cos T-\beta ^{\prime }\sin T)\operatorname {d} T-\sin E(\alpha ^{\prime \prime }\cos T-\alpha ^{\prime }\sin T)\operatorname {d} T\end{aligned}}$

and therefore

${\begin{aligned}tt\operatorname {d} E&=(\alpha \beta ^{\prime \prime }-\alpha ^{\prime \prime }\beta )\cos T\operatorname {d} T+(\alpha ^{\prime }\beta -\beta ^{\prime }\alpha )\sin T\operatorname {d} T+(\alpha ^{\prime }\beta ^{\prime \prime }-\beta ^{\prime }\alpha ^{\prime \prime })\operatorname {d} T\\&=\varepsilon \gamma ^{\prime }\cos T\operatorname {d} T+\varepsilon \gamma ^{\prime \prime }\sin T\operatorname {d} T+\varepsilon \gamma \operatorname {d} T=\varepsilon t\operatorname {d} T\end{aligned}}$

or

$t\operatorname {d} E=\varepsilon \operatorname {d} T$

[23]

It is worth noting that the quantity $t$ is always positive when the coefficient $\gamma$ is positive, and always negative when $\gamma$ is negative. For when

$(\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T)^{2}+(\gamma ^{\prime \prime }\cos T-\gamma ^{\prime }\sin T)^{2}=\gamma ^{\prime }\gamma ^{\prime }+\gamma ^{\prime \prime }\gamma ^{\prime \prime }=\gamma \gamma -1,$

it will always be the case that $\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T,$ regardless of the sign, is less than $\gamma .$ Thus we conclude that, whenever $\varepsilon \gamma$ is a positive quantity, the variables $E$ and $T$ always increase together; however, when $\varepsilon \gamma$ is negative, either variable must decrease when the other increases.

13.

The relationship between the variables $E$ and $T$ can be further clarified by the following reasoning. Setting ${\sqrt {\gamma \gamma -1}}=\delta ,$ so that $\delta \delta =\alpha \alpha +\beta \beta =\gamma ^{\prime }\gamma ^{\prime }+\gamma ^{\prime \prime }\gamma ^{\prime \prime },$ we derive from equations 20, 21, 22

${\begin{aligned}&t(\delta +\alpha \cos E+\beta \sin E)\\&\quad =\gamma \delta +\alpha \alpha +\beta \beta +(\gamma ^{\prime }\delta +\alpha \alpha ^{\prime }+\beta \beta ^{\prime })\cos T+(\gamma ^{\prime \prime }\delta +\alpha \alpha ^{\prime \prime }+\beta \beta ^{\prime \prime })\sin T\\&\quad =(\gamma +\delta )(\delta +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T)\end{aligned}}$

Likewise, from equations 21, 22 it follows that

$t(\alpha \sin E-\beta \cos E)=\varepsilon (\gamma ^{\prime }\sin T-\gamma ^{\prime \prime }\cos T)$

If we set

${\frac {\alpha }{\delta }}=\cos L,\quad$${\frac {\beta }{\delta }}=\sin L,\quad$${\frac {\gamma ^{\prime }}{\delta }}=\cos M,\quad$${\frac {\gamma ^{\prime \prime }}{\delta }}=\sin M$,

then these equations take the following form:

${\begin{aligned}t(1+\cos(E-L))&=(\gamma +\delta )(1+\cos(T-M))\\t\sin(E-L)&=\varepsilon \sin(T-M)\end{aligned}}$

From this, through division, due to $(\gamma +\delta )(\gamma -\delta )=1,$ we get:

${\begin{aligned}&\tan {\frac {1}{2}}(E-L)=\varepsilon (\gamma -\delta )\tan {\frac {1}{2}}(T-M)\\&\tan {\frac {1}{2}}(T-M)=\varepsilon (\gamma +\delta )\tan {\frac {1}{2}}(E-L)\end{aligned}}$

Thus, not only have we arrived at the same conclusion as in the end of the previous article, but it has also become evident that if the value of $E$ increases by 360 degrees, the value of $T$ must either increase or decrease, depending on whether $\varepsilon \gamma$ is a positive or negative quantity. Furthermore, by setting $\delta =\tan N,$ $\gamma =\sec N,$ it is clear that

$\gamma -\delta =\tan(45^{\circ }-{\frac {1}{2}}N),\quad \gamma +\delta =\tan(45^{\circ }+{\frac {1}{2}}N)$

14.

Combining equations 20, 21, 22 with the equations from art. 5, we obtain:

${\begin{aligned}&at(A-a\cos E)=\alpha G-\alpha ^{\prime }G^{\prime }\cos T-\alpha ^{\prime \prime }G^{\prime \prime }\sin T\\&bt(B-b\sin E)=\beta G-\beta ^{\prime }G^{\prime }\cos T-\beta ^{\prime \prime }G^{\prime \prime }\sin T\end{aligned}}$

For the sake of brevity, let us therefore set

${\begin{aligned}(\alpha G-\alpha ^{\prime }G^{\prime }\cos T-\alpha ^{\prime \prime }G^{\prime \prime }\sin T)(\gamma -e\alpha +(\gamma ^{\prime }-e\alpha ^{\prime })\cos T+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\sin T)&=aX\\(\beta G-\beta ^{\prime }G^{\prime }\cos T-\beta ^{\prime \prime }G^{\prime \prime }\sin T)(\gamma -e\alpha +(\gamma ^{\prime }-e\alpha ^{\prime })\cos T+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\sin T)&=bY\\C(\gamma +\gamma ^{\prime }\cos T+\gamma ^{\prime \prime }\sin T)(\gamma -e\alpha +(\gamma ^{\prime }-e\alpha ^{\prime })\cos T+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\sin T)&=Z,\end{aligned}}$

so that

$\operatorname {d} \xi ={\frac {\varepsilon X\operatorname {d} T}{2\pi t^{3}\rho ^{3}}},\quad \operatorname {d} \eta ={\frac {\varepsilon Y\operatorname {d} T}{2\pi t^{3}\rho ^{3}}},\quad \operatorname {d} \zeta ={\frac {\varepsilon Z\operatorname {d} T}{2\pi t^{3}\rho ^{3}}}$

But we have

$t\rho =\pm {\sqrt {G+G^{\prime }\cos T^{2}+G^{\prime \prime }\sin T^{2}}}$

where the upper or lower sign holds depending on whether $t$ is positive or negative (since $\rho$ is always considered positive), i.e. depending on whether the coefficient $\gamma$ is positive or negative. Hence,

${\frac {\varepsilon \operatorname {d} T}{2\pi t^{3}\rho ^{3}}}=\pm {\frac {\operatorname {d} T}{2\pi (G+G^{\prime }\cos T^{2}+G^{\prime \prime }\sin T^{2})^{\frac {3}{2}}}}$

where the ambiguous sign depends on the sign of the quantity $\gamma \varepsilon$.

To obtain the values of $\xi ,$ $\eta ,$ $\zeta ,$ we need to integrate these differentials from the value of $T$ corresponding to $E=0,$ up to the value corresponding to $E=360^{\circ },$ or even (which obviously produces the same results) from the value of $T$ corresponding to an arbitrary value of $E,$ up to the value corresponding to this value of $E$ increased by $360^{\circ };$ it is permissible to integrate from $T=0$ to $T=360^{\circ }$ when $\varepsilon \gamma$ is a positive quantity, or from $T=360^{\circ }$ to $T=0$ when $\varepsilon \gamma$ is negative. Consequently, we have

${\begin{aligned}&\xi =\int {\frac {X\operatorname {d} T}{2\pi (G+G^{\prime }\cos T^{2}+G^{\prime \prime }\sin T^{2})^{\frac {3}{2}}}}\\&\eta =\int {\frac {Y\operatorname {d} T}{2\pi (G+G^{\prime }\cos T^{2}+G^{\prime \prime }\sin T^{2})^{\frac {3}{2}}}}\\&\zeta =\int {\frac {Z\operatorname {d} T}{2\pi (G+G^{\prime }\cos T^{2}+G^{\prime \prime }\sin T^{2})^{\frac {3}{2}}}},\end{aligned}}$

where the integrations extended from $T=0$ to $T=360^{\circ },$ independently of the sign of $\varepsilon \gamma .$

15.

It is easy to see that the integrals

${\begin{aligned}&\int {\frac {\cos T\operatorname {d} T}{(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\int {\frac {\sin T\operatorname {d} T}{(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\int {\frac {\cos T\sin T\operatorname {d} T}{(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}},\end{aligned}}$

extended from $T=180^{\circ }$ to $T=360^{\circ },$ yield the same values as those which the same integrals would obtain if extended from $T=0$ to $T=180^{\circ }$, but with opposite signs; therefore, these integrals, when extended from $T=0$ to $T=360^{\circ },$ clearly become $=0.$ Hence, we conclude that

${\begin{aligned}&\xi =\int {\frac {((\gamma -e\alpha )\alpha G-(\gamma ^{\prime }-e\alpha ^{\prime })\alpha ^{\prime }G^{\prime }{\cos T}^{2}-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\alpha ^{\prime \prime }G^{\prime \prime }{\sin T}^{2})\operatorname {d} T}{2\pi a(G+G^{\prime }{cosT}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\eta =\int {\frac {((\gamma -e\alpha )\beta G-(\gamma ^{\prime }-e\alpha ^{\prime })\beta ^{\prime }G^{\prime }{\cos T}^{2}-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\beta ^{\prime \prime }G^{\prime \prime }{\sin T}^{2})\operatorname {d} T}{2\pi b(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\zeta =\int {\frac {((\gamma -e\alpha )\gamma +(\gamma ^{\prime }-e\alpha ^{\prime })\gamma ^{\prime }{\cos T}^{2}+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\gamma ^{\prime \prime }{\sin T}^{2})C\operatorname {d} T}{2\pi (G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\end{aligned}}$

where the integrations extend from $T=0$ to $T=360^{\circ }.$ Therefore, if the values of the integrals, with the same extension,

${\begin{aligned}&\int {\frac {{\cos T}^{2}\operatorname {d} T}{2\pi ((G+G^{\prime }){\cos T}^{2}+(G+G^{\prime \prime }){\sin T}^{2})^{\frac {3}{2}}}}\\&\int {\frac {{\sin T}^{2}\operatorname {d} T}{2\pi ((G+G^{\prime }){\cos T}^{2}+(G+G^{\prime \prime }){\sin T}^{2})^{\frac {3}{2}}}}\end{aligned}}$

are denoted by $P$ and $Q$ respectively, we will have

${\begin{alignedat}{8}a\xi &=((\gamma -e\alpha )\alpha G&&-(\gamma ^{\prime }-e\alpha ^{\prime })\alpha ^{\prime }G^{\prime })P&&+((\gamma -e\alpha )\alpha G&&-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\alpha ^{\prime \prime }G^{\prime \prime })Q\\b\eta &=((\gamma -e\alpha )\beta G&&-(\gamma ^{\prime }-e\alpha ^{\prime })\beta ^{\prime }G^{\prime })P&&+((\gamma -e\alpha )\beta G&&-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\beta ^{\prime \prime }G^{\prime \prime })Q\\\zeta &=((\gamma -e\alpha )\gamma &&+(\gamma ^{\prime }-e\alpha ^{\prime })\gamma ^{\prime })CP&&+((\gamma -e\alpha )\gamma &&+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\gamma ^{\prime \prime })CQ\end{alignedat}}$

and thus our problem is completely solved.

16.

Regarding the quantities $P,$ $Q,$ it is clear that each of them is

$={\frac {1}{2(G+G^{\prime })^{\frac {3}{2}}}}$

whenever $G^{\prime }=G^{\prime \prime },$ but in all other cases, they can only be expressed in terms of transcendental functions. It is well known how they may be expressed in terms of series. We hope it will be pleasing to the reader if we take this opportunity to explain how these and other transcendental functions may be determined through a very efficient special algorithm, which we have frequently used many years ago and about which we intend to elaborate more extensively elsewhere.

Let $m,$ $n$ be two positive quantities, and let us set

$m^{\prime }={\tfrac {1}{2}}(m+n),\quad n^{\prime }={\sqrt {mn}}$

so that $m',$ $n'$ respectively represent the arithmetic and geometric mean of $m$ and $n.$ We will always assume the geometric mean to be positive. Similarly, set

${\begin{array}{ll}m^{\prime \prime }={\frac {1}{2}}(m^{\prime }+n^{\prime }),&n^{\prime \prime }={\sqrt {m^{\prime }n^{\prime }}}\\m^{\prime \prime \prime }={\frac {1}{2}}(m^{\prime \prime }+n^{\prime \prime }),&n^{\prime \prime \prime }={\sqrt {m^{\prime \prime }n^{\prime \prime }}}\end{array}}$

and so on, so that the series $m,$ $m',$ $m'',$ $m''',$ etc., and $n,$ $n',$ $n'',$ $n''',$ etc., converge very rapidly towards a common limit, which we will denote by $\mu .$ We will simply call this the 'arithmetic-geometric mean' of $m$ and $n.$ We will now demonstrate that ${\tfrac {1}{\mu }}$ is the value of the integral

$\int {\frac {\operatorname {d} T}{2\pi {\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}}$

extended from $T=0$ to $T=360^{\circ }.$

Proof. Let us assume that the variable $T$ is expressed in terms of another variable $T^{\prime },$ so that

$\sin T={\frac {2m\sin T^{\prime }}{(m+n){\cos T^{\prime }}^{2}+2m{\sin T^{\prime }}^{2}}}.$

It is easily understood that, as $T^{\prime }$ increases from $0$ to $90^{\circ },$ $180^{\circ },$ $270^{\circ },$ $360^{\circ },$ $T$ increases (although in unequal intervals) from $0$ to $90^{\circ },$ $180^{\circ },$ $270^{\circ },$ $360^{\circ }.$ Having correctly performed this substitution, it is found that

${\frac {\operatorname {d} T}{\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}={\frac {\operatorname {d} T^{\prime }}{\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n'n'{\sin T^{\prime }}^{2}}}}$

and therefore the values of the integrals

$\int {\frac {\operatorname {d} T}{2\pi {\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}},\quad \int {\frac {\operatorname {d} {T}^{\prime }}{2\pi {\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n^{\prime }n^{\prime }{\sin T^{\prime }}^{2}}}}}$

are equal to each other if both variables are extended from $0$ to $360^{\circ }.$ We can then continue in the same way, and it is clear the the values of the integrals will eventually be the same as that of

$\int {\frac {d\Theta }{2\pi {\sqrt {\mu \mu {\cos \Theta }^{2}+\mu \mu {\sin \Theta }^{2}}}}}$

from $\Theta =0$ to $\Theta =360^{\circ },$ which is clearly $={\tfrac {1}{\mu }}.$ Q. E. D.

17.

From the equation expressing the relationship between $T$ and $T^{\prime },$

$(m-n)\sin T.{\sin T^{\prime }}^{2}=2m\sin T^{\prime }-(m+n)\sin T,$

it is easily deduced that

${\begin{array}{l}{\sqrt {mm{\cos T^{\vphantom {\prime }}}^{2}+nn{\sin T^{\vphantom {\prime }}}^{2}}}=m-(m-n)\sin T.\sin T^{\prime }\\{\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n^{\prime }n^{\prime }{\sin T^{\prime }}^{2}}}=m\cot T.\tan T^{\prime },\end{array}}$

and hence, with the aid of the same equation,

${\begin{array}{l}{\phantom {=}}\sin T.\sin T^{\prime }.{\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}+m^{\prime }({\cos T}^{2}-{\sin T}^{2})\\=\cos T.\cos T^{\prime }.{\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n^{\prime }n^{\prime }{\sin T^{\prime }}^{2}}}-{\frac {1}{2}}(m-n){\sin T}^{2}\end{array}}$

Multiplying this equation by

${\frac {\operatorname {d} T}{\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}={\frac {\operatorname {d} T^{\prime }}{\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n^{\prime }n^{\prime }{\sin T^{\prime }}^{2}}}}$

yields

${\frac {m^{\prime }({\cos T}^{2}-{\sin T}^{2})\operatorname {d} T}{\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}=-{\frac {{\frac {1}{2}}(m-n){\sin T^{\prime }}^{2}\operatorname {d} T^{\prime }}{\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n^{\prime }n^{\prime }{\sin T}^{2}}}}+\operatorname {d} .\sin T^{\prime }\cos T$

Multiplying this equation by ${\tfrac {m-n}{\pi }},$ substituting $m^{\prime }(m-n)={\tfrac {1}{2}}(mm-nn),$ $(m-n)^{2}=4(m^{\prime }m^{\prime }-n^{\prime }n^{\prime }),$ ${\sin T}^{2}={\tfrac {1}{2}}-{\tfrac {1}{2}}({\cos T^{\prime }}^{2}-{\sin T^{\prime }}^{2}),$ and integrating from $T$ and $T^{\prime }=0$ to $360^{\circ },$ we have:

${\begin{aligned}&(mm-nn)\int {\frac {({\cos T}^{2}-{\sin T}^{2}).\operatorname {d} T}{2\pi {\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}}\\&=-{\frac {2(m^{\prime }m^{\prime }-n^{\prime }n^{\prime })}{\mu }}+2(m^{\prime }m^{\prime }-n^{\prime }n^{\prime })\int {\frac {({\cos T^{\prime }}^{2}-{\sin T^{\prime }}^{2})dT^{\prime }}{2\pi {\sqrt {m^{\prime }m^{\prime }{\cos T^{\prime }}^{2}+n^{\prime }n^{\prime }{\sin T^{\prime }}^{2}}}}}\end{aligned}}$

Moreover, since it is permissible to transform the definite integral on the right-hand side in the same manner, the integral

$\int {\frac {({\cos T}^{2}-{\sin T}^{2})dT}{2\pi {\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}}$

can be expressed by an infinitely convergent series as follows:

$-{\frac {2(m^{\prime }m^{\prime }-n^{\prime }n^{\prime })+4(m^{\prime \prime }m^{\prime \prime }-n^{\prime \prime }n^{\prime \prime })+s(m^{\prime \prime \prime }m^{\prime \prime \prime }-n^{\prime \prime \prime }n^{\prime \prime \prime })+{\text{etc. }}}{(mm-nn)\mu }}=-{\frac {\nu }{\mu }}$

The numerical calculation is most conveniently carried out using logarithms, if we set

${\tfrac {1}{4}}{\sqrt {mm-nn}}=\lambda ,\quad$${\tfrac {1}{4}}{\sqrt {m^{\prime }m^{\prime }-n^{\prime }n^{\prime }}}=\lambda ^{\prime },\quad$${\tfrac {1}{4}}{\sqrt {m^{\prime \prime }m^{\prime \prime }-n^{\prime \prime }n^{\prime \prime }}}=\lambda ^{\prime \prime }$ etc.

from which it follows that

$\lambda ^{\prime }={\frac {\lambda \lambda }{m^{\prime }}},\quad$$\lambda ^{\prime \prime }={\frac {\lambda ^{\prime }\lambda ^{\prime }}{m^{\prime \prime }}},\quad$$\lambda ^{\prime \prime \prime }={\frac {\lambda ^{\prime \prime }\lambda ^{\prime \prime }}{m^{\prime \prime \prime }}}$ etc.

and

$\nu ={\frac {2\lambda ^{\prime }\lambda ^{\prime }+4\lambda ^{\prime \prime }\lambda ^{\prime \prime }+8\lambda ^{\prime \prime \prime }\lambda ^{\prime \prime \prime }+{\text{etc.}}}{\lambda \lambda }}$

18.

Using the method explained here, it is also possible to compute indefinite integrals (starting from a variable value $=0$) with maximal efficiency. Specifically, if $T^{\prime \prime }$ is assumed to be determined from $m^{\prime },$ $n^{\prime },$ $T^{\prime },$ just as $T^{\prime }$ was determined by $m,$ $n,$ $T,$ and similarly $T^{\prime \prime \prime }$ is determined by $m^{\prime \prime },$ $n^{\prime \prime },$ $T^{\prime \prime }$, and so on, then for any given value of $T,$ then the values of the terms of the series $T,$ $T^{\prime },$ $T^{\prime \prime },$ $T^{\prime \prime \prime }$, etc., converge rapidly to the limit $\theta ,$ and we will have

${\begin{aligned}&\int {\frac {\operatorname {d} T}{\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}={\frac {\theta }{\mu }}\\&\int {\frac {({\cos T}^{2}-{\sin T}^{2})\operatorname {d} T}{\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}=-{\frac {\nu \theta }{\mu }}+{\frac {\lambda ^{\prime }\cos T\sin T^{\prime }+2\lambda ^{\prime \prime }\cos T^{\prime }\sin T^{\prime \prime }+4\lambda ^{\prime \prime \prime }\cos T^{\prime \prime }\sin T^{\prime \prime \prime }+{\text{etc. }}}{\lambda \lambda }}\end{aligned}}$

But it suffices to mention these matters in passing here, as they are not necessary for our purpose.

19.

Now, if we already assume $m={\sqrt {G+G^{\prime }}},$ $n={\sqrt {G+G^{\prime \prime }}},$ the values of quantities $P,$ $Q$ can easily be reduced to transcendentals $\mu ,$ $\nu$. For if $P,$ $Q\operatorname {are}$ are the values of integrals

$\int {\frac {{\cos T}^{2}\operatorname {d} T}{2\pi (mm{\cos T}^{2}+nn{\sin T}^{2})^{\frac {3}{2}}}},\quad \int {\frac {{\sin T}^{2}\operatorname {d} T}{2\pi (mm{\cos T}^{2}+nn{\sin T}^{2})^{\frac {3}{2}}}}$

extended from $T=0$ to $T=360^{\circ },$ it is immediately evident that we have

$mmP+nnQ={\frac {1}{\mu }}$

[24]

Moreover, we have

${\begin{alignedat}{1}{\frac {({\cos T}^{2}-{\sin T}^{2})\operatorname {d} T}{2\pi {\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}}&+{\frac {(mm{\cos T}^{2}-nn{\sin T}^{2})\operatorname {d} T}{2\pi (mm{\cos T}^{2}+nn{\sin T}^{2})^{\frac {3}{2}}}}\\&\qquad \qquad ={\frac {(mm{\cos T}^{4}-nn{\sin T}^{4})\operatorname {d} T}{\pi (mm{\cos T}^{2}+nn{\sin T}^{2})^{\frac {3}{2}}}}\\&\qquad \qquad =\operatorname {d} .{\frac {\cos T\sin T}{\pi {\sqrt {mm{\cos T}^{2}+nn{\sin T}^{2}}}}}\end{alignedat}}$

Integrating this equation from $T=0$ to $T=360^{\circ },$ we get

$-{\frac {\nu }{\mu }}+mmP-nnQ=0$

[25]

Finally, combining equations 24 and 25, we conclude that

$P={\frac {1+\nu }{2mm\mu }},\quad Q={\frac {1-\nu }{2nn\mu }}.$

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1. Perhaps it will not be superfluous to note that we have chosen the preceding analysis carefully and preferred it to another derivation of equations (III)-(VII), which, although it may have appeared somewhat more elegant, upon careful examination, was found to be susceptible to certain doubts that could not be resolved without difficulty.
2. Strictly speaking, it only follows from the preceding analysis that both $-G^{\prime }$ and $-G^{\prime \prime }$ must satisfy equation 13, raising the question of whether it is permissible to assign both $-G^{\prime }$ and $-G^{\prime \prime }$ to the same negative root, with the third root being completely ignored. However, it is easy to see that, if the second and third roots of the equation are unequal, then from $-G^{\prime }=-G^{\prime \prime }$ it follows that $\gamma ^{\prime }=\gamma ^{\prime \prime },$ $\alpha ^{\prime }=\alpha ^{\prime \prime },$ $\ell ^{\prime }=\ell ^{\prime \prime },$ and thus $-a^{\prime }a^{\prime \prime }-b^{\prime }b^{\prime \prime }+\gamma ^{\prime }\gamma ^{\prime \prime }=-a^{\prime }a^{\prime }-b^{\prime }b^{\prime }+\gamma ^{\prime }\gamma ^{\prime }=1,$ which contradicts the fourth equation in [1]. See what is said below regarding the case of two equal roots of equation 13.