Page:Grundgleichungen (Minkowski).djvu/58

in direction of B*C*.. Now by $OA'/B*D*$ is to be understood the ratio of the two vectors in question.

It is clear that this proposition at once shows the covariant character with respect to a Lorentz-group.

Let us now ask how the space-time filament of F behaves when the material point F* has a uniform translatory motion, i.e., the principal line of the filament of F* is a line. Let us take the space time null-point in this, and by means of a Lorentz-transformation, we can take this axis as the t-axis. Let x, y, z, t, denote the point B, let $\tau ^{*}$ denote the proper time of B*, reckoned from O. Our proposition leads to the equations

(25)	${\frac {d^{2}x}{d\tau ^{2}}}=-{\frac {m^{*}x}{(t-\tau {}^{*})^{3}}},\ {\frac {d^{2}y}{d\tau ^{2}}}=-{\frac {m{}^{*}y}{(t-\tau {}^{*})^{3}}},\ {\frac {d^{2}z}{d\tau ^{2}}}=-{\frac {m{}^{*}z}{(t-\tau {}^{*})^{3}}}$

and

(26)	${\frac {d^{2}t}{d\tau ^{2}}}=-{\frac {m{}^{*}}{(t-\tau {}^{*})^{2}}}{\frac {d(t-\tau {}^{*})}{dt}}$,

where

(27)	$x^{2}+y^{2}+z^{2}=(t-\tau ^{*})^{2}$

and

(28)	$\left({\frac {dx}{d\tau }}\right)^{2}+\left({\frac {dy}{d\tau }}\right)^{2}+\left({\frac {dz}{d\tau }}\right)^{2}=\left({\frac {dt}{d\tau }}\right)^{2}-1$

In consideration of (27), the three equations (25) are of the same form as the equations for the motion of a material point subjected to attraction from a fixed centre according to the Newtonian Law, only that instead of the time t the proper time $\tau$ of the material point occurs. The fourth equation (26) gives then the connection between proper time and the time for the material point.

Now for different values of $\tau$ the orbit of the space-point x, y, z is an ellipse with the semi-major axis a and the eccentricity e. Let E denote the excentric anomaly, T the increment of the proper time for a complete description of the orbit, finally $n{\mathsf {T}}=2\pi$, so that from a properly chosen initial point r, we have the Kepler-equation

(29)	$n\tau =E-e\ \sin E$

If we now change the unit of time, and denote the velocity of light by c, then from (28), we obtain

(30)	$\left({\frac {dt}{d\tau }}\right)^{2}-1={\frac {m^{*}}{ac^{2}}}{\frac {1+e\ \cos E}{1-e\ \cos E}}$.