I have never encountered this before (encountered it now in Sean Carroll's GR text when discussing the benefit of solving problems in locally inertial references frame).

Let $\gamma$ be the Lorentz factor, $g_{\mu\nu}$ be the metric tensor, and let $U^\mu$ and $V^\mu$ be 2 four-velocities.

Can anyone explain the following definition?

$$\gamma = -g_{\mu\nu}U^\mu V^\nu $$


The Lorentz factor $\gamma=\frac{1}{\sqrt{1-v^2}}$ can expressed in terms of the rapidity as $\gamma=\cosh\theta$, where $\theta$ is the rapidity between 4-velocities (which are timelike unit-vectors). The relative-spatial-velocity is $v=\tanh\theta$.

In terms of 4-vector operations (using the $(-+++)$ signature convention), $$\gamma= - \hat U\cdot \hat V=-(1)(1)\cosh\theta,$$ which should remind you of the Euclidean analogue: $\hat A \cdot \hat B= (1)(1)\cos \theta_{\scriptsize \mbox{between $\hat A$ and $\hat B$}}$.

Using the metric tensor explicitly, $$\gamma= - \hat U\cdot \hat V = - g_{\mu\nu} U^\mu V^\nu.$$

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  • $\begingroup$ Kind user: meanwhile +1. I am not used to editing the answers of the best users. Is it preferible $\theta=\widehat{\hat{A}\hat{B}}$ or $\theta$ is the angle measured between $\hat{A}$ and $\hat{B}$? $\endgroup$ – Sebastiano Aug 3 '19 at 12:18
  • $\begingroup$ Ah, beautiful. Thank you! $\endgroup$ – Lopey Tall Aug 3 '19 at 14:41

If you consider one of the velocities to be your rest frame, and the other to be something moving in that frame, then, the Lorentz factor between the frames is given by that expression in all frames (with $c=1).

So if $U^{\mu}$ is the rest frame:

$$ U^{\mu} = (1, \vec 0) $$

then the moving frame is:

$$ V^{\mu} = \gamma(1, \vec v) $$

and their inner product is:

$$ g_{\mu\nu}U^{\mu}V^{\mu} = \gamma(1 - \vec 0 \cdot \vec v) = \gamma $$

and that relationship hold in all other frames.

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