Consider a particle that moves with speed u relative to the inertial reference frame I and with speed u' relative to the inertial reference frame I'. Let g(u), g(u') and g(V) be the orresponding gamma factors (V the relative speed of I and I'). m(0) stands for its rest mass, E(0) for its rest energy, p and E for its momentum and enegy measured by observers from I. It is obvious that(adsbygoogle = window.adsbygoogle || []).push({});

p=g(u')g(V)m(0)(V+u')=g(u')g(V')E(0)(V+u')cc

E=g(u')g(V)E(0)(1+Vu'/cc)

I consider that such a presentation presents some (pedagogical) advantages showing clearly what observers from the I frame measure in the case when u'=0 and when u' and V are both equal to zero.

Even if I know that the concept of relativistic mass ispersona non grataon the Forum I would also suggest for the relativistic mass

m=g(u')g(V)E(0)(1+Vu'/cc)=g(u')g(V)m(0)(1+Vu'/cc).

The oppinion of those who teach or learn special relativity theory is highly appreciated of course in the spirit of

sine ira et studio

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# Transformation equations presented in a different way

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