- 24,488
- 15,057
This makes use of the fact that the time component of the normalized four-velocity in any frame is ##\gamma##, and the time component is given by the Minkowski product of the corresponding time-like basis vector. If you have given ##U^{\mu}## in a frame ##\sigma## and the four-velocity of frame ##\Sigma'## is ##V^{\mu}## then ##\gamma'=U^{\prime 0}=V^{\mu} U_{\mu}##.Grasshopper said:Just found another one by Tipler (Modern Physics, 4th ed.) that does something similar, but then it takes it into a direction that seems ripe for confusing me: since each momentum component has its own gamma, it expresses γ in terms of the primed coordinates, so you get γ.
$$γ' = \frac{1}{ \sqrt{1 - \frac{u'^2}{c^2}}} = γ \frac{1 - \frac{vu_x}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}}$$where in this case
##γ = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}##
which is part of the derivation of the Lorentz transformation for energy and momentum. This is page 73 (Chapter 2.3) in Tipler, 4th edition.
To translate this to the non-covariant three-velocities just use
$$U^{\mu}=\gamma_u (1,\vec{u}/c), \quad V^{\mu}=\gamma_v (1,\vec{v}/c)$$
to get
$$\gamma'=\gamma_u \gamma_v \left (1-\frac{\vec{v} \cdot \vec{u}}{c^2} \right).$$
So Tipler is indeed correct (as expected).