If we are in a frame $O$, and we observe two different particles moving with four-velocities $\mathbf{U}$ and $\mathbf{V}$. Can we compute the dot product between these two four-vectors?
Suppose that $\mathbf{U} = \gamma_1 ( 1, \vec{u})$, and $\mathbf{V} = \gamma_2 ( 1, \vec{v})$ are the two four-velocity vectors.
How do I show that their dot product is: $$\mathbf{U} \cdot \mathbf{V} = - \gamma$$ where $\gamma$ is the Lorentz factor between these two velocities.
From @JEB's answer (with missing prime added)
$u^{\mu}u'_{\mu} = \gamma\gamma'(c^2-\vec v \cdot \vec v')$
uses the $(+,-,-,-)$ signature convention
and
translates to
\begin{align}
u^{\mu}u'_{\mu}
&= \cosh\theta\cosh\theta'(c^2-c\tanh\theta\ c\tanh\theta')\\
&= c^2(\cosh\theta\cosh\theta'-\sinh\theta \sinh\theta')\\
&= c^2\ \ (\quad\qquad \cosh(\theta-\theta')\quad\qquad)\\
&= c^2\ \left( \frac{1}{\sqrt{1- \tanh^2(\theta-\theta')}}\right)\\
\end{align}
where the difference of rapidities [the relative rapidity] is an invariant.
postscript: no need to work in a particular frame, if one knows geometry and trigonometry.
Given a geometry problem, do you have to rotate/transform the figure given to you so that you work with (say) only angles with respect to a horizontal axis?
If the exercise allows you to use the fact that the dot product between four vectors is a Lorentz invariant (which isn't difficult to prove), then consider the components of four velocities $\mathbf{U}$ and $\mathbf{V}$ in a frame that is itself moving with four velocity $\mathbf{U}$. In that frame $\mathcal{O}^\prime$ the components are: $$ \mathbf{U}_{\mathcal{O}^\prime} = (1,\vec{0}) $$ $$ \mathbf{V}_{\mathcal{O}^\prime} = (\gamma ,\vec{V}) $$ Where by definition $\vec{V}$ is the relative velocity between the frames that also defines their relative $\gamma$ factor, $\gamma=(1-\vec{V}^{\ 2})^{^{-\frac{1}{2}}}$.
From here you immediately see that $ \mathbf{U}_{\mathcal{O}^\prime}\cdot\mathbf{V}_{\mathcal{O}^\prime} = -\gamma$, assuming one uses the east-coast metric convention. Therefore from the Lorentz invariance $\mathbf{U}\cdot\mathbf{V}=-\gamma$ in every inertial frame, and I think this is basically what you wanted to demonstrate.
Just for fun, note that if we also equate the dot product as computed via the lab frame $\mathcal{O}$ components and the one computed via the defined $\mathcal{O}^\prime$ frame components:
$$ \mathbf{U}_{\mathcal{O}}\cdot\mathbf{V}_{\mathcal{O}}=\mathbf{U}_{\mathcal{O}^{\ \prime}}\cdot\mathbf{V}_{\mathcal{O}^{\ \prime}} $$
We get:
$$ -\gamma_1\gamma_2+\gamma_1\gamma_2\vec{u}\cdot\vec{v} = -\gamma $$ Or, substituting in three velocities: $$ \frac{1-\vec{u}\cdot\vec{v}}{\sqrt{(1-\vec{v}^{\ 2})(1-\vec{u}^{\ 2})}} = \frac{1}{\sqrt{1-\vec{V}^2}} $$
Which is also a nice way, after further simplification, to derive the magnitudinal part (relating $|\vec{u}|$ and $|\vec{v}|$ to $|\vec{V}|$) of the general velocity addition formula.
First of all, -1 for using u, v and then 1, 2. Why?
You got:
$$u_{\mu} = \gamma(c, \vec v) $$ $$u'_{\mu} = \gamma'(c, \vec v')$$
and the inner product is:
$$ u^{\mu}u_{\mu} = \gamma\gamma'(c^2-\vec v \cdot \vec v') $$
which is a Lorentz scalar, idk what the significance is.
If we evaluate it in a rest-frame, maybe something will jump out...say $\vec v'=0$, then:
$$ u^{\mu}u'_{\mu} = \gamma c^2 = \frac{c^3}{\sqrt{c^2-v^2}} $$
which doesn't look special.
Maybe in the CoM ($ \vec v'=-\vec v$)?:
$$ u^{\mu}u'_{\mu} = \gamma^2(c^2 + v^2) = c^2\frac{1+\beta^2}{1-\beta^2} $$
which is a factor significance in Minkowski diagrams, since a moving frame has a tic mark scaling of:
$$ U'/U = \sqrt{\frac{1+\beta^2}{1-\beta^2}}$$
relative to the "rest" frame.
