(Working in units where $c=1$. Also in a gravitational field with spacetime metric is $g_{\mu\nu}$.)
Suppose, in a particular frame of reference (the "laboratory" frame, or the "far-away observer" frame), that a free-falling particle has $4$-velocity $v^\mu$, and that another observer $\mathcal O$ has $4$-velocity $U^\mu$ (at the same spacetime point). Suppose the particle has speed $V_{\mathcal O}$ in $\mathcal O$'s reference frame.
I want to show that in $\mathcal O$'s frame of reference, the particle's gamma factor is $$\gamma = \frac{1}{\sqrt{1-V_\mathcal O^2}} = v_\mu U^\mu = g_{\mu\nu}v^\mu U^\nu \,\, .$$
This looks like a kind of an "addition of velocities" formula for general relativity. How does one see this?
