Page:PoincareDynamiqueJuillet.djvu/14

The right-hand side becomes by partial integration:

$\int dt\ d\tau \left[-\sum \rho \delta U{\frac {dF}{dt}}-\sum \rho \eta \delta U{\frac {dF}{dy}}-\sum \rho \zeta \delta U{\frac {dF}{dz}}+\sum \rho \xi \delta V{\frac {dF}{dy}}+\sum \rho \xi \delta W{\frac {dF}{dz}}\right].$

Now note, that:

$\sum \rho \xi \delta V{\frac {dF}{dy}}=\sum \rho \zeta \delta U{\frac {dH}{dx}},\quad \sum \rho \xi \delta W{\frac {dF}{dz}}=\sum \rho \eta \delta U{\frac {dG}{dx}}$

If, indeed, we develop Σ on the two sides of these relations, they become identities; and remember that

${\frac {dH}{dx}}-{\frac {dF}{dz}}=-\beta ,\quad {\frac {dG}{dx}}-{\frac {dF}{dy}}=\gamma$

the right-hand side in question will become:

$\int dt\ d\tau \left[-\sum \rho \delta U{\frac {dF}{dt}}-\sum \rho \gamma \eta \delta U-\sum \rho \beta \zeta \delta U\right],$

so that finally:

$\delta J=\int dt\ d\tau \sum \rho \delta U\left({\frac {d\psi }{dx}}+{\frac {dF}{dt}}+\beta \zeta -\gamma \eta \right)=\int dt\ d\tau \sum \rho \delta U\left(-f+\beta \zeta -\gamma \eta \right).$

Equating the coefficient of δU on both sides of (10) we get:

$X=f-\beta \zeta +\gamma \eta \,$

This is equation (2) of the preceding §.

§ 3. — The Lorentz transformation and the principle of least action

Let us see if the principle of least action gives us the reason for the success of the Lorentz transformation. We must look at the transformation of the integral:

$J=\int dt\ d\tau \left({\frac {\sum f^{2}}{2}}-{\frac {\sum \alpha ^{2}}{2}}\right)$

(formula 4 of § 2).

We first find

$dt^{\prime }\ d\tau ^{\prime }=l^{4}dt\ d\tau$

because x', y', z', t' are related to x, y, z, t by linear relations whose determinant is equal to l⁴; then we have:

(1)

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

(formula 9 of § 1), hence:

$l^{4}\left(\sum f^{\prime 2}-\sum \alpha ^{\prime 2}\right)=\sum f^{2}-\sum \alpha ^{2}$