Page:The Meaning of Relativity - Albert Einstein (1922).djvu/31

Multiplying by the velocity, a tensor of rank 1, we obtain the tensor equation

$\left(m{\frac {d^{2}x_{\nu }}{dt^{2}}}-X_{\nu }\right){\frac {dx_{\mu }}{dt}}=0.$

By contraction and multiplication by the scalar $dt$ we obtain the equation of kinetic energy

$d\left({\frac {mq^{2}}{2}}\right)=X_{\nu }dx_{\nu }.$

If $\xi _{\nu }$ denotes the difference of the co-ordinates of the material particle and a point fixed in space, then the $\xi _{\nu }$ have the character of vectors. We evidently have ${\frac {d^{2}x_{\nu }}{dt^{2}}}={\frac {d^{2}\xi _{\nu }}{dt^{2}}}$, so that the equations of motion of the particle may be written

$m{\frac {d^{2}\xi _{\nu }}{dt^{2}}}-X_{\nu }=0.$

Multiplying this equation by $\xi _{\mu }$ we obtain a tensor equation

$\left(m{\frac {d^{2}\xi _{\nu }}{dt^{2}}}-X_{\nu }\right)\xi _{\mu }=0.$

Contracting the tensor on the left and taking the time average we obtain the virial theorem, which we shall not consider further. By interchanging the indices and subsequent subtraction, we obtain, after a simple transformation, the theorem of moments,

${\frac {d}{dt}}\left[m\left(\xi _{\mu }{\frac {d\xi _{\nu }}{dt}}-\xi _{\nu }{\frac {d\xi _{\mu }}{dt}}\right)\right]=\xi _{\mu }X_{\nu }-\xi _{\nu }X_{\mu }$

(15)

It is evident in this way that the moment of a vector