Page:PoincareDynamiqueJuillet.djvu/16

Note that

$\xi ^{\prime }={\frac {\xi +\epsilon }{1+\xi \epsilon }},\quad \eta ^{\prime }={\frac {\eta }{k\left(1+\xi \epsilon \right)}}$

It follows, by replacing δt' by its value

$kl\left(1+\xi \epsilon \right)\delta U=\delta U^{\prime }\left(1+\xi \epsilon \right)+\left(\xi +\epsilon \right)kl\epsilon \delta U,$ $l\left(1+\xi \epsilon \right)\delta V=\delta V^{\prime }\left(1+\xi \epsilon \right)+\eta l\epsilon \delta U.$

If we recall the definition of k, we draw from this:

$\delta U={\frac {k}{l}}\delta U^{\prime }+{\frac {k\epsilon }{l}}\xi \delta U^{\prime },$ $\delta V={\frac {1}{l}}\delta V^{\prime }+{\frac {k\epsilon }{l}}\eta \delta U^{\prime },$

and also

$\delta W={\frac {1}{l}}\delta W^{\prime }+{\frac {k\epsilon }{l}}\zeta \delta U^{\prime };$

hence

(3)	$\sum X\delta U={\frac {1}{l}}\left(kX\delta U^{\prime }+Y\delta V^{\prime }+Z\delta W^{\prime }\right)+{\frac {k\epsilon }{l}}\delta U^{\prime }\sum X\xi .$

Now, in virtue of equations (2) we must have:

$\int \sum X^{\prime }\delta U^{\prime }dt^{\prime }\ d\tau ^{\prime }=\int \sum X\delta U\ dt\ d\tau ={\frac {1}{l^{4}}}\sum X\delta U\ dt^{\prime }\ d\tau ^{\prime }.$

By replacing ΣXδU by its value (3) and by identifying, it follows:

$X^{\prime }={\frac {k}{l^{5}}}X+{\frac {k\epsilon }{l^{5}}}\sum X\xi ,\quad Y^{\prime }={\frac {1}{l^{5}}}Y,\quad Z^{\prime }={\frac {1}{l^{5}}}Z.$

These are the equations (11) of § 1. The principle of least action leads us to the same result as the analysis of § 1.

If we turn to formulas (1), we see that Σf² - Σα² is not affected by the Lorentz transformation, except one constant factor; it is not the case with expression Σf² + Σα² which represents the energy. If we confine ourselves to the case where ε is sufficiently small, so that the square can be neglected so that k = 1, and if we also assume l = 1, we find:

$\sum f^{\prime 2}=\sum f^{2}+2\epsilon \left(g\gamma -h\beta \right),$ $\sum \alpha ^{\prime 2}=\sum \alpha ^{2}+2\epsilon \left(g\gamma -h\beta \right),$

or by addition

$\sum f^{\prime 2}+\sum \alpha ^{\prime 2}=\sum f^{2}+\sum \alpha ^{2}+4\epsilon \left(g\gamma -h\beta \right).$

§ 4. — The Lorentz group

It is important to note that the Lorentz transformations form a group.

Indeed, if we set:

$x^{\prime }=kl(x+\epsilon t),\quad y^{\prime }=ly,\quad z^{\prime }=lz,\quad t^{\prime }=kl(t+\epsilon x),$