Page:DeSitterGravitation.djvu/8

or

$mc{\frac {d(\kappa )}{dt}}=m{\frac {d}{dt}}\left({\frac {dct}{d\tau }}\right)=\mathrm {T} =(\mathrm {T} ){\sqrt {1-\phi ^{2}}},$

(12′)

where

$\mathrm {T} =\mathrm {X} \xi +\mathrm {Y} \eta +\mathrm {Z} \zeta .$

Minkowski gives the name "kinetic energy" to the quantity

${\begin{aligned}\mathrm {E} &=mc^{2}\left[(\kappa )-1\right]={\tfrac {1}{2}}mc^{2}\phi ^{2}+{\tfrac {3}{8}}mc^{2}\phi ^{4}+\dots ,\\{\text{or}}\qquad \mathrm {E} &={\tfrac {1}{2}}m{\mathsf {\mathrm {v} }}^{2}\left(1+{\tfrac {3}{4}}\phi ^{2}+\dots \right).\end{aligned}}$

The equation (12′) thus turns out to be the equation of energy,

${\frac {d\mathrm {E} }{dt}}=\mathrm {X} {\frac {dx}{dt}}+\mathrm {Y} {\frac {dy}{dt}}+\mathrm {Z} {\frac {dz}{dt}}.$

If we use Minkowskian velocities and forces, and put

$(\mathrm {E} )={\tfrac {1}{2}}mc^{2}(\phi )^{2},$

we find similarly from (11′)

${\frac {d(\mathrm {E} )}{d\tau }}=(\mathrm {X} ){\frac {dx}{d\tau }}+(\mathrm {Y} ){\frac {dy}{d\tau }}+(\mathrm {Z} ){\frac {dz}{d\tau }}.$

The law of the force must be such that the form of the equations of motion is not changed by a Lorentz-transformation. Therefore

$(\mathrm {X} ),\ (\mathrm {Y} ),\ (\mathrm {Z} ),\ (\mathrm {T} )$

must be transformed by the same formulæ as

${\begin{array}{cccc}x,&y,&z,&t,\\(\xi ),&(\eta ),&(\zeta ),&(\kappa ),\end{array}}$

or, in other words, $\mathrm {(X),(Y),(Z)}$ must be linear functions of

${\begin{array}{ccccccc}x_{1},&y_{1},&z_{1},&&x_{2},&y_{2},&z_{2},\\(\xi )_{1},&(\eta )_{1},&(\zeta )_{1},&&(\xi )_{2},&(\eta )_{2},&(\zeta )_{2},\end{array}}$

the coefficients being invariants of the transformation.

For zero velocities the equations of Newtonian mechanics must be reproduced, therefore the coordinates can only enter by their differences $x_{1}-x_{2}$, etc.

Introducing further the condition that the resulting equations must not contain the velocities in the first degree, and certain other simplifications, Poincaré is led to take (l.c., page 174)—

$(\mathrm {X} )_{1}={\frac {k^{2}m_{1}m_{2}}{B_{1}^{3}}}\left\{x_{2}-x_{1}-(\xi )_{2}{\frac {\mathrm {A} _{1}}{\mathrm {C} }}\right\}.$

(14)