Introduction, §§ 1, 9

$\begin{array}{cc}\hfill u\hfill & =\frac{\mathrm{df}}{\mathrm{dt}}+\rho \xi =\frac{d\gamma}{\mathrm{dy}}-\frac{d\beta}{\mathrm{dz}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\alpha =\frac{\mathrm{dH}}{\mathrm{dy}}-\frac{\mathrm{dG}}{\mathrm{dz}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}f=-\frac{\mathrm{dF}}{\mathrm{dt}}-\frac{d\psi}{\mathrm{dx}},\hfill \\ \hfill \frac{d\alpha}{\mathrm{dt}}\hfill & =\frac{\mathrm{dg}}{\mathrm{dz}}-\frac{\mathrm{dh}}{\mathrm{dy}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{d\rho}{\mathrm{dt}}+\sum \frac{d\rho \xi}{\mathrm{dx}}=0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum \frac{\mathrm{df}}{\mathrm{dx}}=\rho ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{d\psi}{\mathrm{dt}}+\sum \frac{\mathrm{dF}}{\mathrm{dx}}=0,\hfill \\ \hfill \square \hfill & =\Delta -\frac{{d}^{2}}{{\mathrm{dt}}^{2}}=\sum \frac{{d}^{2}}{{\mathrm{dx}}^{2}}-\frac{{d}^{2}}{{\mathrm{dt}}^{2}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\square \psi =-\rho ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\square F=-\rho \xi .\hfill \end{array}\}$ | $(1)$ |

$X=\rho f+\rho (\eta \gamma -\zeta \beta ).$ | $(2)$ |

$x\text{'}=\mathrm{kl}(x+\u03f5t),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}t\text{'}=\mathrm{kl}(t+\u03f5x),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}y\text{'}=\ell y,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}z\text{'}=\ell z,$ | $(3)$ |

$k=\frac{1}{\sqrt{1-{\u03f5}^{2}}}.$ |

Now if we put:

$\square \text{'}=\sum \frac{{d}^{2}}{\mathrm{dx}\text{'}{}^{2}}-\frac{{d}^{2}}{\mathrm{dt}\text{'}{}^{2}},$ |

we will have:

$\square \text{'}=\square {\ell}^{-2}.$ |

Let a sphere be carried along with the electron in uniform translation, and let the equation of this mobile sphere be:

$(x-\xi t{)}^{2}+(y-\eta t{)}^{2}+(z-\zeta t{)}^{2}={r}^{2},$ |

and the volume of the sphere be $\frac{4}{3}\pi {r}^{3}$. ${}^{1}$ The transformation will change the sphere into an ellipsoid, the equation of which is easy to find. We thus deduce easily from (3):

$x=\frac{k}{\ell}(x\text{'}-\u03f5t\text{'}),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}t=\frac{k}{\ell}(t\text{'}-\u03f5x\text{'}),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}y=\frac{y\text{'}}{\ell},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}z=\frac{z\text{'}}{\ell}.\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(3\text{'})$ |

The equation of the ellipsoid then becomes:

${k}^{2}(x\text{'}-\u03f5t\text{'}-\xi t\text{'}+\u03f5\xi x\text{'}{)}^{2}+(y\text{'}-\eta \mathrm{kt}\text{'}+\eta k\u03f5x\text{'}{)}^{2}+(z\text{'}-\zeta \mathrm{kt}\text{'}+\zeta k\u03f5x\text{'}{)}^{2}={\ell}^{2}{r}^{2}.$ |

This ellipsoid is in uniform motion; for $t\text{'}=0$, it reduces to

${k}^{2}x\text{'}{}^{2}(1+\xi \u03f5{)}^{2}+(y\text{'}+\eta k\u03f5x\text{'}{)}^{2}+(z\text{'}+\zeta k\u03f5x\text{'}{)}^{2}={\ell}^{2}{r}^{2},$ |

and has a volume:

$\frac{4}{3}\pi {r}^{3}\frac{{\ell}^{3}}{k(1+\xi \u03f5)}.$ |

If we want electron charge to be unaltered by the transformation, and if we designate the new charge density $\rho \text{'}$, we will find:

$\rho \text{'}=\frac{k}{{\ell}^{3}}(\rho +\u03f5\rho \xi ).$ | $(4)$ |

$\begin{array}{c}\multicolumn{1}{c}{\xi \text{'}=\frac{\mathrm{dx}\text{'}}{\mathrm{dt}\text{'}}=\frac{d(x+\u03f5t)}{d(t+\u03f5x)}=\frac{\xi +\u03f5}{1+\u03f5\xi},}\\ \multicolumn{1}{c}{\eta \text{'}=\frac{\mathrm{dy}\text{'}}{\mathrm{dt}\text{'}}=\frac{\mathrm{dy}}{\mathrm{kd}(t+\u03f5x)}=\frac{\eta}{k(1+\u03f5\xi )},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\zeta \text{'}=\frac{\zeta}{k(1+\u03f5\xi )},}\end{array}$ |

whence:

$\rho \text{'}\xi \text{'}=\frac{k}{{\ell}^{3}}(\rho \xi +\u03f5\rho ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\rho \text{'}\eta \text{'}=\frac{1}{{\ell}^{3}}\rho \eta ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\rho \text{'}\zeta \text{'}=\frac{1}{{\ell}^{3}}\rho \zeta .\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(4\text{'})$ |

Here is where I must point out for the first time a difference with Lorentz. In my notation, Lorentz put (l.c., page 813, formulas 7 and 8):

$\rho \text{'}=\frac{1}{k{\ell}^{3}}\rho ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\xi \text{'}={k}^{2}(\xi +\u03f5),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\eta \text{'}=k\eta ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\zeta \text{'}=k\zeta .$ |

In this way we recover the formulas:

$\rho \text{'}\xi \text{'}=\frac{k}{{\ell}^{3}}(\rho \xi +\u03f5\rho ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\rho \text{'}\eta \text{'}=\frac{1}{{\ell}^{3}}\rho \eta ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\rho \text{'}\zeta \text{'}=\frac{1}{{\ell}^{3}}\rho \zeta ;$ |

although the value of $\rho \text{'}$ differs. It is important to notice that the formulas (4) and ( $4\text{'}$) satisfy the condition of continuity

$\frac{d\rho \text{'}}{\mathrm{dt}\text{'}}+\sum \frac{d\rho \text{'}\xi \text{'}}{\mathrm{dx}\text{'}}=0.$ |

To see this, let $\lambda $ be an undetermined coefficient and $D$ the Jacobian of

$t+\lambda \rho ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}x+\lambda \rho \xi ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}y+\lambda \rho \eta ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}z+\lambda \rho \zeta$ | $(5)$ |

$D={D}_{0}+{D}_{1}\lambda +{D}_{2}{\lambda}^{2}+{D}_{3}{\lambda}^{3}+{D}_{4}{\lambda}^{4},$ |

with ${D}_{0}=1$, ${D}_{1}=\frac{d\rho}{\mathrm{dt}}+\sum \frac{d\rho \xi}{\mathrm{dx}}=0$. Let $\lambda \text{'}={\ell}^{4}\rho \text{'}$; ${}^{2}$ then the 4 functions

$t\text{'}+\lambda \text{'}\rho \text{'},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}x\text{'}+\lambda \text{'}\rho \text{'}\xi \text{'},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}y\text{'}+\lambda \text{'}\rho \text{'}\eta \text{'},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}z\text{'}+\lambda \text{'}\rho \text{'}\zeta \text{'}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(5\text{'})$ |

are related to the functions (5) by the same linear relationships as the old variables to the new ones. Therefore, if we denote $D\text{'}$ the Jacobian of the functions ( $5\text{'}$) with respect to the new variables, it follows that:

$D\text{'}=D,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}D\text{'}={D}_{0}\text{'}+{D}_{1}\text{'}\lambda \text{'}+\dots +{D}_{4}\text{'}\lambda \text{'}{}^{4},$ |

and thereby: ${}^{3}$

$D}_{0}\text{'}={D}_{0}=1,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{D}_{1}\text{'}={\ell}^{-4}{D}_{1}=0=\frac{d\rho \text{'}}{\mathrm{dt}\text{'}}+\sum \frac{d\rho \text{'}\xi \text{'}}{\mathrm{dx}\text{'}}.\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\text{Q.E.D.$ |

Under Lorentz's hypothesis, this condition would not be met since $\rho \text{'}$ has a different value. We will define the new vector and scalar potentials in such a way as to satisfy the conditions

$\square \text{'}\psi \text{'}=-\rho \text{'},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\square \text{'}F\text{'}=-\rho \text{'}\xi \text{'}.$ | $(6)$ |

$\psi \text{'}=\frac{k}{\ell}(\psi +\u03f5F),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}F\text{'}=\frac{k}{\ell}(F+\u03f5\psi ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}G\text{'}=\frac{1}{\ell}G,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}H\text{'}=\frac{1}{\ell}H.$ | $(7)$ |

$f\text{'}=-\frac{\mathrm{dF}\text{'}}{\mathrm{dt}\text{'}}-\frac{d\psi \text{'}}{\mathrm{dx}\text{'}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\alpha \text{'}=\frac{\mathrm{dH}\text{'}}{\mathrm{dy}\text{'}}-\frac{\mathrm{dG}\text{'}}{\mathrm{dz}\text{'}}.$ | $(8)$ |

$\frac{d}{\mathrm{dt}\text{'}}=\frac{k}{\ell}(\frac{d}{\mathrm{dt}}-\u03f5\frac{d}{\mathrm{dx}}),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{d}{\mathrm{dx}\text{'}}=\frac{k}{\ell}(\frac{d}{\mathrm{dx}}-\u03f5\frac{d}{\mathrm{dt}}),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{d}{\mathrm{dy}\text{'}}=\frac{1}{\ell}\frac{d}{\mathrm{dy}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{d}{\mathrm{dz}\text{'}}=\frac{1}{\ell}\frac{d}{\mathrm{dz}}$ |

and we deduce thereby:

$\begin{array}{cccccc}\hfill 3f\text{'}\hfill & =\frac{1}{{\ell}^{2}}f,\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}g\text{'}\hfill & =\frac{k}{{\ell}^{2}}(g+\u03f5\gamma ),\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}h\text{'}\hfill & =\frac{k}{{\ell}^{2}}(h-\u03f5\beta ),\hfill \\ \hfill \alpha \text{'}\hfill & =\frac{1}{{\ell}^{2}}\alpha ,\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\beta \text{'}\hfill & =\frac{k}{{\ell}^{2}}(\beta -\u03f5h),\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\gamma \text{'}\hfill & =\frac{k}{{\ell}^{2}}(\gamma +\u03f5g).\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}$ | $(9)$ |

$\frac{d\psi \text{'}}{\mathrm{dt}\text{'}}+\sum \frac{\mathrm{dF}\text{'}}{\mathrm{dx}\text{'}}=0.$ | $(10)$ |

$\frac{\mathrm{df}\text{'}}{\mathrm{dt}\text{'}}+\rho \text{'}\xi \text{'}=\frac{d\gamma \text{'}}{\mathrm{dy}\text{'}}-\frac{d\beta \text{'}}{\mathrm{dz}\text{'}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\frac{d\alpha \text{'}}{\mathrm{dt}\text{'}}=\frac{\mathrm{dg}\text{'}}{\mathrm{dz}\text{'}}-\frac{\mathrm{dh}\text{'}}{\mathrm{dy}\text{'}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\sum \frac{\mathrm{df}\text{'}}{\mathrm{dx}\text{'}}=\rho \text{'}$ |

and it is easy to see that these are necessary consequences of (6), (8) and (10). We must now compare forces before and after the transformation. Let $X$, $Y$, $Z$ be the force prior to the transformation, and $X\text{'}$, $Y\text{'}$, $Z\text{'}$ the force after the transformation, both forces being per unit volume. In order for $X\text{'}$ to satisfy the same equations as before the transformation, we must have:

$\begin{array}{cc}\multicolumn{1}{c}{X\text{'}}& =\rho \text{'}f\text{'}+\rho \text{'}(\eta \text{'}\gamma \text{'}-\zeta \text{'}\beta \text{'}),\hfill \\ \multicolumn{1}{c}{Y\text{'}}& =\rho \text{'}g\text{'}+\rho \text{'}(\zeta \text{'}\alpha \text{'}-\xi \text{'}\gamma \text{'}),\hfill \\ \multicolumn{1}{c}{Z\text{'}}& =\rho \text{'}h\text{'}+\rho \text{'}(\xi \text{'}\beta \text{'}-\eta \text{'}\alpha \text{'}),\hfill \end{array}$ |

or, replacing all quantities by their values (4), ( $4\text{'}$) and (9), and in light of (2):

$\begin{array}{cc}\hfill X\text{'}\hfill & =\frac{k}{{\ell}^{5}}(X+\u03f5\sum X\xi ),\hfill \\ \hfill Y\text{'}\hfill & =\frac{1}{{\ell}^{5}}Y,\hfill \\ \hfill Z\text{'}\hfill & =\frac{1}{{\ell}^{5}}Z.\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}$ | $(11)$ |

${X}_{1}=f+\eta \gamma -\zeta \beta ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{X}_{1}\text{'}=f\text{'}+\eta \text{'}\gamma \text{'}-\zeta \text{'}\beta \text{'},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}X=\rho {X}_{1},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}X\text{'}=\rho {X}_{1}\text{'},$ |

and we obtain the equations:

$\begin{array}{cc}\hfill {X}_{1}\text{'}\hfill & =\frac{k}{{\ell}^{5}}\frac{\rho}{\rho \text{'}}({X}_{1}+\u03f5\sum {X}_{1}\xi ),\hfill \\ \hfill {Y}_{1}\text{'}\hfill & =\frac{1}{{\ell}^{5}}\frac{\rho}{\rho \text{'}}{Y}_{1},\hfill \\ \hfill {Z}_{1}\text{'}\hfill & =\frac{1}{{\ell}^{5}}\frac{\rho}{\rho \text{'}}{Z}_{1}.\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(11\text{'})$ |

Lorentz found (page 813, equation (10) with different notation):

$\begin{array}{cc}\hfill {X}_{1}\hfill & ={\ell}^{2}{X}_{1}\text{\'}-{\ell}^{2}\u03f5(\eta \text{\'}g\text{\'}+\zeta \text{\'}h\text{\'}),\hfill \\ \hfill {Y}_{1}\hfill & =\frac{{\ell}^{2}}{k}{Y}_{1}\text{\'}+\frac{{\ell}^{2}\u03f5}{k}\xi \text{\'}g\text{\'},\hfill \\ \hfill {Z}_{1}\hfill & =\frac{{\ell}^{2}}{k}{Z}_{1}\text{\'}+\frac{{\ell}^{2}\u03f5}{k}\xi \text{\'}h\text{\'}.\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(11")$ |

Before going any further, it is important to locate the source of this significant divergence. It obviously springs from the fact that the formulas for $\xi \text{'}$, $\eta \text{'}$ and $\zeta \text{'}$ are not the same, while the formulas for the electric and magnetic fields are the same.

$X=Y=Z=0$ |

$X\text{'}=Y\text{'}=Z\text{'}=0.$ |

$\varphi \hspace{0.5em}(t,\hspace{0.5em}x,\hspace{0.5em}y,\hspace{0.5em}z,\hspace{0.5em}\xi ,\hspace{0.5em}\eta ,\hspace{0.5em}\zeta ,\hspace{0.5em}{\xi}_{1},\hspace{0.5em}{\eta}_{1},\hspace{1em}{\zeta}_{1})=0$ | $(1)$ |

$t,\hspace{0.5em}x,\hspace{0.5em}y,\hspace{0.5em}z,\hspace{0.5em}\xi ,\hspace{0.5em}\eta ,\hspace{0.5em}\zeta ,\hspace{0.5em}{\xi}_{1},\hspace{0.5em}{\eta}_{1},\hspace{1em}{\zeta}_{1}.$ | $(2)$ |

$\xi ={\xi}_{1},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\eta ={\eta}_{1},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\zeta ={\zeta}_{1};$ |

this is then the case we will examine first, by supposing that these velocities are constant, such that the two bodies are engaged in a common uniform rectilinear translation. We may suppose that the $x-$axis is parallel to this translation, such that $\eta =\zeta =0$, and we will let $\u03f5=-\xi $. If we apply the Lorentz transformation under these conditions, after the transformation the two bodies will be at rest, and it follows that:

$\xi \text{'}=\eta \text{'}=\zeta \text{'}=0.$ |

The components ${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$ should then agree with Newton's law and we will have, apart from a constant factor:

${X}_{1}\text{'}=-\frac{x}{r\text{'}{}^{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{Y}_{1}\text{'}=-\frac{y}{r\text{'}{}^{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{Z}_{1}\text{'}=-\frac{z}{r\text{'}{}^{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}r\text{'}{}^{2}=x\text{'}{}^{2}+y\text{'}{}^{2}+z\text{'}{}^{2}.$ | $(3)$ |

$\begin{array}{cc}\hfill x\text{'}\hfill & =k(x+\u03f5t),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}y\text{'}=y,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}z\text{'}=z,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}t\text{'}=k(t+\u03f5x),\hfill \\ \hfill \frac{\rho \text{'}}{\rho}\hfill & =k(1+\xi \u03f5)=k(1-{\u03f5}^{2})=\frac{1}{k},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum {X}_{1}\xi =-{X}_{1}\u03f5,\hfill \\ \hfill {X}_{1}\text{'}\hfill & =k\frac{\rho}{\rho \text{'}}({X}_{1}+\u03f5\sum {X}_{1}\xi )={k}^{2}{X}_{1}(1-{\u03f5}^{2})={X}_{1},\hfill \\ \hfill {Y}_{1}\text{'}\hfill & =k\frac{\rho}{\rho \text{'}}{Y}_{1}={\mathrm{kY}}_{1}\hfill \\ \hfill {Z}_{1}\text{'}\hfill & ={\mathrm{kZ}}_{1}.\hfill \end{array}$ |

We have in addition:

$x+\u03f5t=x-\xi t,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}r\text{'}{}^{2}={k}^{2}(x-\xi t{)}^{2}+{y}^{2}+{z}^{2}$ |

and

${X}_{1}=\frac{-k(x-\xi t)}{r\text{'}{}^{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{Y}_{1}=\frac{-y}{\mathrm{kr}\text{'}{}^{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{Z}_{1}=\frac{-z}{\mathrm{kr}\text{'}{}^{3}};$ | $(4)$ |

${X}_{1}=\frac{\mathrm{dV}}{\mathrm{dx}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{Y}_{1}=\frac{\mathrm{dV}}{\mathrm{dy}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{Z}_{1}=\frac{\mathrm{dV}}{\mathrm{dz}};\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}V=\frac{1}{\mathrm{kr}\text{'}}.\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(4\text{'})$ |

It seems at first that the indeterminacy remains, since we made no hypotheses concerning the value of $t$, i.e., the transmission speed; and that besides, $x$ is a function of $t$. It is easy to see, however, that the terms appearing in our formulas, $x-\xi t$, $y$, $z$, do not depend on $t$. We see that if the two bodies translate together, the force acting on the attracted body is perpendicular to an ellipsoid, at the center of which lies the attracting body. To advance further, we need to look for the

${x}^{2}+{y}^{2}+{z}^{2}-{t}^{2}.$ |

Let us also put:

$\begin{array}{cccccc}\multicolumn{1}{c}{\xi}& =\frac{\delta x}{\delta t},\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\eta \hfill & =\frac{\delta y}{\delta t},\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\zeta \hfill & =\frac{\delta z}{\delta t},\hfill \\ \multicolumn{1}{c}{{\xi}_{1}}& =\frac{{\delta}_{1}x}{{\delta}_{1}t},\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{\eta}_{1}\hfill & =\frac{{\delta}_{1}y}{{\delta}_{1}t},\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{\zeta}_{1}\hfill & =\frac{{\delta}_{1}z}{{\delta}_{1}t};\hfill \end{array}$ |

we see that the Lorentz transformation will make $\delta x$, $\delta y$, $\delta z$, $\delta t$, and ${\delta}_{1}x$, ${\delta}_{1}y$, ${\delta}_{1}z$, ${\delta}_{1}t$ undergo the same linear substitutions as $x$, $y$, $z$, $t$. Let us regard

$\begin{array}{cccc}\hfill x,\hfill & y,\hfill & z,\hfill & t\sqrt{-1},\hfill \\ \hfill \delta x,\hfill & \delta y,\hfill & \delta z,\hfill & \delta t\sqrt{-1},\hfill \\ \hfill {\delta}_{1}x,\hfill & {\delta}_{1}y,\hfill & {\delta}_{1}z,\hfill & {\delta}_{1}t\sqrt{-1},\hfill \end{array}$ |

as the coordinates of 3 points $P$, $P\text{'}$, $P"$ in space of 4 dimensions. We see that the Lorentz transformation is merely a rotation in this space about the origin, assumed fixed. Consequently, we will have no distinct invariants apart from the 6 distances between the 3 points $P$, $P\text{'}$, $P"$, considered separately and with the origin, or, if one prefers, apart from the two expressions

${x}^{2}+{y}^{2}+{z}^{2}-{t}^{2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}x\delta x+y\delta y+z\delta z-t\delta t,$ |

or the 4 expressions of like form deduced from an arbitrary permutation of the 3 points $P$, $P\text{'}$, $P"$. But what we seek are invariants that are functions of the 10 variables (2). Therefore, among the combinations of our 6 invariants we must find those depending only on these 10 variables, i.e., those that are 0th degree homogeneous with respect both to $\delta x$, $\delta y$, $\delta z$, $\delta t$, and to ${\delta}_{1}x$, ${\delta}_{1}y$, ${\delta}_{1}z$, ${\delta}_{1}t$. We will then be left with 4 distinct invariants:

$\sum {x}^{2}-{t}^{2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{t-\sum x\xi}{\sqrt{1-\sum {\xi}^{2}}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{t-\sum x{\xi}_{1}}{\sqrt{1-\sum {\xi}_{1}^{2}}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{1-\sum \xi {\xi}_{1}}{\sqrt{(1-\sum {\xi}^{2})(1-\sum {\xi}_{1}^{2})}}.$ | $(5)$ |

$T=\sum X\xi ;$ |

we will see that (11) can be written ( $\ell =1$):

$\begin{array}{cccc}\hfill 2X\text{'}\hfill & =k(X+\u03f5T),\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}T\text{'}\hfill & =k(T+\u03f5X),\hfill \\ \hfill Y\text{'}\hfill & =Y,\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}Z\text{'}\hfill & =Z;\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}$ | $(6)$ |

$\sum {X}^{2}-{T}^{2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum \mathrm{Xx}-\mathrm{Tt},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum X\delta x-T\delta t,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum X{\delta}_{1}x-T{\delta}_{1}t.$ |

However, it is not $X$, $Y$, $Z$ that we need, but ${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$, with

${T}_{1}=\sum {X}_{1}\xi .$ |

We see that

$\frac{{X}_{1}}{X}=\frac{{Y}_{1}}{Y}=\frac{{Z}_{1}}{Z}=\frac{{T}_{1}}{T}=\frac{1}{\rho}.$ |

Therefore, the Lorentz transformation will act in the same manner on ${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$, ${T}_{1}$, as on $X$, $Y$, $Z$, $T$, except that these expressions will be multiplied moreover by

$\frac{\rho}{\rho \text{'}}=\frac{1}{k(1+\xi \u03f5)}=\frac{\delta t}{\delta t\text{'}}.$ |

Likewise, the Lorentz transformation will act in the same way on $\xi $, $\eta $, $\zeta $, $1$ as on $\delta x$, $\delta y$, $\delta z$, $\delta t$, except that these expressions will be multiplied moreover by the

$\frac{\delta t}{\delta t\text{'}}=\frac{1}{k(1+\xi \u03f5)}.$ |

Next we consider $X$, $Y$, $Z$, $T\sqrt{-1}$ as the coordinates of a fourth point $Q$; the invariants will then be functions of the mutual distances of the five points

$0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}P,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}P\text{\'},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}P",\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}Q$ |

and among these functions we must retain only those that are 0th degree homogeneous with respect, on one hand, to

$X,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}Y,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}Z,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}T,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\delta x,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\delta y,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\delta z,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\delta t$ |

(variables that can be replaced further by ${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$, ${T}_{1}$, $\xi $, $\eta $, $\zeta $, 1), and on the other hand, with respect to ${}^{5}$

${\delta}_{1}x,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{\delta}_{1}y,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{\delta}_{1}z,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{\delta}_{1}t$ |

(variables that can be replaced further by ${\xi}_{1}$, ${\eta}_{1}$, ${\zeta}_{1}$, 1). In this way we find, beyond the four invariants (5), four distinct new invariants:

$\frac{\sum {X}_{1}^{2}-{T}_{1}^{2}}{1-\sum {\xi}^{2}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{\sum {X}_{1}x-{T}_{1}t}{\sqrt{1-\sum {\xi}^{2}}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{\sum {X}_{1}{\xi}_{1}-{T}_{1}}{\sqrt{1-\sum {\xi}^{2}}\sqrt{1-\sum {\xi}_{1}^{2}}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{\sum {X}_{1}\xi -{T}_{1}}{1-\sum {\xi}^{2}}.$ | $(7)$ |

$\sum {x}^{2}-{t}^{2}={r}^{2}-{t}^{2}=0,$ |

from whence $t=\pm r$, and, since $t$ has to be negative, $t=-r$. This means that the velocity of propagation is equal to that of light. It seems at first that this hypothesis ought to be rejected outright. Laplace showed in effect that the propagation is either instantaneous or much faster than that of light. However, Laplace examined the hypothesis of finite propagation velocity

$\frac{t-\sum x{\xi}_{1}}{\sqrt{1-\sum {\xi}_{1}^{2}}}=0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}t=\sum x{\xi}_{1}.$ |

The propagation velocity is therefore much faster than that of light, but in certain cases $t$ could be positive, which, as we mentioned, seems hardly admissible. ${}^{6}$

$\sum {X}_{1}^{2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\sum {X}_{1}x,$ |

or, by Newton's law, to

$\frac{1}{{r}^{4}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}-\frac{1}{r};$ |

in addition, according to hypothesis (A), the ${2}^{\text{d}}$ and ${3}^{\text{rd}}$ invariants in (5) become:

$\frac{-r-\sum x\xi}{\sqrt{1-\sum {\xi}^{2}}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{-r-\sum x{\xi}_{1}}{\sqrt{1-\sum {\xi}_{1}^{2}}},$ |

that is, for absolute rest,

$-r,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}-r.$ |

We may therefore admit,

$\frac{{(1-\sum {\xi}_{1}^{2})}^{2}}{{(r+\sum x{\xi}_{1})}^{4}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}-\frac{\sqrt{1-\sum {\xi}_{1}^{2}}}{r+\sum x{\xi}_{1}},$ |

although other combinations are possible. A choice must be made among these combinations, and furthermore, we need a ${3}^{\text{rd}}$ equation in order to define ${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$. In making such a choice, we should try to come as close as possible to Newton's law. Let us see what happens when we neglect the squares of the velocities $\xi $, $\eta $, etc. (still letting $t=-r$). The 4 invariants (5) then become:

$0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}-r-\sum x\xi ,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}-r-\sum x{\xi}_{1},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}1$ |

and the 4 invariants (7) become:

$\sum {X}_{1}^{2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum {X}_{1}(x+\xi r),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum {X}_{1}({\xi}_{1}-\xi ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}0.$ |

Before we can make a comparison with Newton's law, another transformation is required. In the case under consideration, ${x}_{0}+x$, ${y}_{0}+y$, ${z}_{0}+z$, represent the coordinates of the attracting body at the instant ${t}_{0}+t$, and $r=\sqrt{\sum {x}^{2}}$. With Newton's law we have to consider the coordinates of the attracting body ${x}_{0}+{x}_{1}$, ${y}_{0}+{y}_{1}$, ${z}_{0}+{z}_{1}$ at the instant ${t}_{0}$, and the distance ${r}_{1}=\sqrt{\sum {x}^{2}}$. We may neglect the square of the time $t$ required for propagation, and proceed, consequently, as if the motion were uniform; we then have:

$x={x}_{1}+{\xi}_{1}t,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}y={y}_{1}+{\eta}_{1}t,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}z={z}_{1}+{\zeta}_{1}t,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}r(r-{r}_{1})=\sum x{\xi}_{1}t;$ |

or, since $t=-r$,

$x={x}_{1}-{\xi}_{1}r,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}y={y}_{1}-{\eta}_{1}r,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}z={z}_{1}-{\zeta}_{1}r,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}r={r}_{1}-\sum x{\xi}_{1};$ |

such that our 4 invariants (5) become:

$0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}-{r}_{1}+\sum x({\xi}_{1}-\xi ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}-{r}_{1},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}1$ |

and our 4 invariants (7) become:

$\sum {X}_{1}^{2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum {X}_{1}[{x}_{1}+(\xi -{\xi}_{1}){r}_{1}],\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sum {X}_{1}({\xi}_{1}-\xi ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}0.$ |

In the second of these expressions I wrote ${r}_{1}$ instead of $r$, because $r$ is multiplied by $\xi -{\xi}_{1}$, and because I neglect the square of $\xi $. For these 4 invariants (7), Newton's law would yield

$\frac{1}{{r}_{1}^{4}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}-\frac{1}{{r}_{1}}-\frac{\sum {x}_{1}(\xi -{\xi}_{1})}{{r}_{1}^{2}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\frac{\sum {x}_{1}(\xi -{\xi}_{1})}{{r}_{1}^{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}0.$ |

Therefore, if we designate the ${2}^{\text{nd}}$ and ${3}^{\text{rd}}$ of the invariants (5) as $A$ and $B$, and the first 3 invariants of (7) as $M$, $N$, $P$, we will satisfy Newton's law to first-order terms in the square of velocity by setting:

$M=\frac{1}{{B}^{4}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}N=\frac{+A}{{B}^{2}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}P=\frac{A-B}{{B}^{3}}.$ | $(8)$ |

${k}_{0}=\frac{1}{\sqrt{1-\sum {\xi}^{2}}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{k}_{1}=\frac{1}{\sqrt{1-\sum {\xi}_{1}^{2}}},$ |

which is justified by analogy with the notation

$k=\frac{1}{\sqrt{1-\sum {\xi}^{2}}}$ |

featured in the Lorentz substitution. In this case, and in light of the condition $-r=t$, the invariants (5) become:

$0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}A=-{k}_{0}(r+\sum x\xi ),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}B=-{k}_{1}(r+\sum x{\xi}_{1}),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}C={k}_{0}{k}_{1}(1-\sum \xi {\xi}_{1}).$ |

Moreover, we notice that the following systems of quantities:

$\begin{array}{cccc}\hfill x,\hfill & y,\hfill & z,\hfill & -r=t,\hfill \\ \hfill {k}_{0}{X}_{1},\hfill & {k}_{0}{Y}_{1},\hfill & {k}_{0}{Z}_{1},\hfill & {k}_{0}{T}_{1},\hfill \\ \hfill {k}_{0}\xi ,\hfill & {k}_{0}\eta ,\hfill & {k}_{0}\zeta ,\hfill & {k}_{0},\hfill \\ \hfill {k}_{1}{\xi}_{1},\hfill & {k}_{1}{\eta}_{1},\hfill & {k}_{1}{\zeta}_{1},\hfill & {k}_{1}\hfill \end{array}$ |

undergo the

$\begin{array}{cc}\hfill {X}_{1}\hfill & =x\frac{\alpha}{{k}_{0}}+\xi \beta +{\xi}_{1}\frac{{k}_{1}}{{k}_{0}}\gamma ,\hfill \\ \hfill {Y}_{1}\hfill & =y\frac{\alpha}{{k}_{0}}+\eta \beta +{\eta}_{1}\frac{{k}_{1}}{{k}_{0}}\gamma ,\hfill \\ \hfill {Z}_{1}\hfill & =z\frac{\alpha}{{k}_{0}}+\zeta \beta +{\zeta}_{1}\frac{{k}_{1}}{{k}_{0}}\gamma ,\hfill \\ \hfill {T}_{1}\hfill & =-r\frac{\alpha}{{k}_{0}}+\beta +\frac{{k}_{1}}{{k}_{0}}\gamma .\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}$ | $(9)$ |

${X}_{1}\xi -{T}_{1}=0,$ |

which becomes, replacing ${X}_{1}$, ${T}_{1}$, ${Z}_{1}$, ${T}_{1}$ with their values in (9) and multiplying by ${k}_{0}^{2}$:

$-A\alpha -\beta -C\gamma =0.$ | $(10)$ |

$\beta =0,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\gamma =-\frac{A\alpha}{C}.$ |

To the adopted order of approximation, we obtain

${k}_{0}={k}_{1}=1,\hspace{0.5em}C=1,\hspace{0.5em}A=-{r}_{1}+\sum x({\xi}_{1}-\xi ),\hspace{1em}B=-{r}_{1},\hspace{0.5em}x={x}_{1}+{\xi}_{1}t={x}_{1}-{\xi}_{1}r.$ |

The ${1}^{\text{st}}$ equation in (9) then becomes

${X}_{1}=\alpha (x-A{\xi}_{1}).$ |

But if the square of $\xi $ is neglected, $A{\xi}_{1}$ can be replaced by $-{r}_{1}{\xi}_{1}$, or by $-r{\xi}_{1}$, which yields:

${X}_{1}=\alpha (x+{\xi}_{1}r)=\alpha {x}_{1}.$ |

Newton's law would yield

${X}_{1}=-\frac{{x}_{1}}{{r}_{1}^{3}}.$ |

Consequently, we must select a value for the invariant $\alpha $ which reduces to $-\frac{1}{{r}_{1}^{3}}$ in the adopted order of approximation, that is, $\frac{1}{{B}^{3}}$. Equations (9) will become:

$\begin{array}{cc}\hfill {X}_{1}\hfill & =\frac{x}{{k}_{0}{B}^{3}}-{\xi}_{1}\frac{{k}_{1}}{{k}_{0}}\frac{A}{{B}^{3}C},\hfill \\ \hfill {Y}_{1}\hfill & =\frac{y}{{k}_{0}{B}^{3}}-{\eta}_{1}\frac{{k}_{1}}{{k}_{0}}\frac{A}{{B}^{3}C},\hfill \\ \hfill {Z}_{1}\hfill & =\frac{z}{{k}_{0}{B}^{3}}-{\zeta}_{1}\frac{{k}_{1}}{{k}_{0}}\frac{A}{{B}^{3}C},\hfill \\ \hfill {T}_{1}\hfill & =-\frac{r}{{k}_{0}{B}^{3}}-\frac{{k}_{1}}{{k}_{0}}\frac{A}{{B}^{3}C}.\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}$ | $(11)$ |

$\frac{1}{{B}^{3}}+(C-1){f}_{1}\hspace{0.5em}(A,\hspace{0.5em}B,\hspace{0.5em}C)+(A-B{)}^{2}{f}_{2}\hspace{0.5em}(A,\hspace{0.5em}B,\hspace{0.5em}C),$ |

where ${f}_{1}$ and ${f}_{2}$ are arbitrary functions of $A$, $B$, $C$. Alternatively, we may forgo setting $\beta $ to zero, but add any complementary terms to $\alpha $, $\beta $, $\gamma $ that satisfy condition (10) and are of second order with respect to the $\xi $ for $\alpha $, and of first order for $\beta $ and $\gamma $. 2° The first equation in (11) may be written:

${X}_{1}=\frac{{k}_{1}}{{B}^{3}C}[x(1-\sum \xi {\xi}_{1})+{\xi}_{1}(r+\sum x\xi )]\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(11\text{'})$ |

and the quantity in brackets itself may be written:

$(x+r{\xi}_{1})+\eta ({\xi}_{1}y-x{\eta}_{1})+\zeta ({\xi}_{1}z-x{\zeta}_{1}),$ | $(12)$ |

$\begin{array}{cccccc}\hfill 3{k}_{1}(x+r{\xi}_{1})\hfill & =\lambda ,\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{k}_{1}(y+r{\eta}_{1})\hfill & =\mu ,\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{k}_{1}(z+r{\zeta}_{1})\hfill & =\nu ,\hfill \\ \hfill {k}_{1}({\eta}_{1}z-{\zeta}_{1}y)\hfill & =\lambda \text{'},\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{k}_{1}({\zeta}_{1}x-{\xi}_{1}z)\hfill & =\mu \text{'},\hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{k}_{1}({\xi}_{1}y-x{\eta}_{1})\hfill & =\nu \text{'};\hfill \\ \hfill \hfill \end{array}\}$ | $(13)$ |

$\begin{array}{cc}\hfill {X}_{1}\hfill & =\frac{\lambda}{{B}^{3}}-\frac{\eta \nu \text{'}-\zeta \mu \text{'}}{{B}^{3}},\hfill \\ \hfill {Y}_{1}\hfill & =\frac{\mu}{{B}^{3}}-\frac{\zeta \lambda \text{'}-\xi \nu \text{'}}{{B}^{3}},\hfill \\ \hfill {Z}_{1}\hfill & =\frac{\nu}{{B}^{3}}-\frac{\xi \mu \text{'}-\eta \lambda \text{'}}{{B}^{3}};\hfill \end{array}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\}$ | $(14)$ |

${B}^{2}=\sum {\lambda}^{2}-\sum \lambda \text{'}{}^{2}.$ | $(15)$ |

Paris, July, 1905.

H. Poincaré

Archives Henri Poincaré (CNRS, UMR 7117)