Addition of Velocities (Velocity Composition) in Special Relativity

The “Addition of Velocities” formula (more correctly, the “Composition of Velocities” formula) in Special Relativity

$$\frac{v_{AC}}{c}=\frac{ \frac{v_{AB}}{c}+\frac{v_{BC}}{c} }{1 + \frac{v_{AB}}{c} \frac{v_{BC}}{c}}$$

is a non-intuitive result that arises from a “hyperbolic-tangent of a sum”-identity in Minkowski spacetime geometry, with its use of hyperbolic trigonometry.

However, I claim it is difficult to obtain this by looking at the Galilean version of this formula and then motivating the special-relativistic version.

Instead, one should start with the Euclidean analogue (in what could be mistakenly called the “addition of slopes” formula… “composition of slopes” is better),
then do the special-relativistic analogue, then do the Galilean analogue (to obtain the familiar but unfortunately-“our common sense” formula).

In response to “Construct a Diagram that Illustrates The Galilean Law of Addition of Velocities”
I posed a sequence of trigonometry questions (which hints at the above motivation).

Here I provide the answers to those questions.

Euclidean geometry

In the figure below [where O is the center],
express
the “slope of OS [with respect to OP]” (PS/OP)
in terms of
the “slope of ON [with respect to OP or OM]” (MN/OM)
and
the “slope of OS [with respect to ON]” NS/ON).
Everything can be done using ratios of segments
(and one can use some trigonometric intuition to guide you). Note: in this Euclidean geometry, NS is Euclidean-perpendicular to ON
since, for radius vector ON, the segment NS is tangent to the Euclidean-circle.

Answer: (note the minus sign in the second line)$$\begin{align*} (\mbox{slope of OS wrt OP}) &=\frac{PS}{OP}\\ &=\frac{PR+RS}{OM➖MP}\\ &=\frac{MN+RS}{OM-NR}\\ &=\frac{\frac{MN}{OM}+\frac{RS}{OM}}{1-\frac{NR}{OM}}\\ &=\frac{\frac{MN}{OM}+\frac{RS}{OM} }{1-\color{green}{\frac{MN}{OM}\frac{NR}{MN}}}\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{-SR}{OM}\right) } }{1-\color{green}{\frac{MN}{OM}\color{blue}{\left( \frac{-RN}{MN} \right) } }}\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{-SN}{ON}\right) } }{1-\frac{MN}{OM}\color{blue}{\left( \frac{-SN}{ON} \right) } }\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{NS}{ON}\right) } }{1-\frac{MN}{OM} \color{blue}{\left( \frac{NS}{ON} \right) }}\\ &=\frac{(\mbox{slope of ON wrt OM})+(\mbox{slope of OS wrt ON})  }{1- (\mbox{slope of ON wrt OM})(\mbox{slope of OS wrt ON}) } \end{align*}$$
where we used the similarity of triangles OMN and SRN.
The “slopes” don’t add…because the spatial displacements are not parallel in this plane. However, the associated radial angles
[interpreted as intercepted Euclidean-arc-lengths on a unit Euclidean-circle
or as intercepted sector-areas in the unit Euclidean-disk]
do add:  $\phi_{POS}=\phi_{MON}+\phi_{NOS}$
with slopes $\frac{PS}{OP}=\tan\phi_{POS}$, $\frac{MN}{OM}=\tan\phi_{MON}$,  $\frac{NS}{ON}=\tan\phi_{NOS}$.
Thus,
$$\begin{align*} \tan\phi_{POS} &=\tan\left( \phi_{MON}+\phi_{NOS} \right) \\&=\frac{ \tan \phi_{MON}+\tan\phi_{NOS}  }{ 1 – \tan\phi_{MON}\tan\phi_{NOS} } \\ \end{align*}$$

Minkowski – Einstein Special Relativity

Then repeat for this figure in special relativity [where O is the center],.
That is,
express
the “velocity of OS [with respect to OP]” (PS/OP)
in terms of
the “velocity of ON [with respect to OP or OM]” (MN/OM)
and
the “velocity of OS [with respect to ON]” NS/ON).
(The method is almost the same…
You might wish to follow your earlier steps and see what has become of them in this case.
However, you’ll have to accept that
in this [Minkowski spacetime] geometry, NS is Minkowski-perpendicular to ON
since, for radius vector ON, the segment NS is tangent to the Minkowski-circle.)

Answer: (note the plus sign in the second line)$$\begin{align*} (\mbox{velocity of OS wrt OP}) &=\frac{PS}{OP}\\ &=\frac{PR+RS}{OM ➕  MP}\\ &=\frac{MN+RS}{OM+NR}\\ &=\frac{\frac{MN}{OM}+\frac{RS}{OM}}{1+\frac{NR}{OM}}\\ &=\frac{\frac{MN}{OM}+\frac{RS}{OM} }{1+\color{green}{\frac{MN}{OM}\frac{NR}{MN}}}\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{-SR}{OM}\right) } }{1+\color{green}{\frac{MN}{OM}\color{blue}{\left( \frac{-RN}{MN} \right) } }}\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{-SN}{ON}\right) } }{1+\frac{MN}{OM}\color{blue}{\left( \frac{-SN}{ON} \right) } }\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{NS}{ON}\right) } }{1+\frac{MN}{OM} \color{blue}{\left( \frac{NS}{ON} \right) }}\\ &=\frac{(\mbox{velocity of ON wrt OM})+(\mbox{velocity of OS wrt ON})  }{1+ (\mbox{velocity of ON wrt OM})(\mbox{velocity of OS wrt ON}) } \end{align*}$$
where we used the similarity of triangles OMN and SRN.
The “velocities” don’t add…because the spatial displacements are not coplanar (collinear) in spacetime.
Physically, the corresponding observers do not mutually share the same notion of which events are simultaneous. However, the associated radial Minkowski-angles
[interpreted as intercepted Minkowski-arc-lengths on a unit Minkowski-circle
or as intercepted sector-areas in the unit Minkowski -disk]
do add:  $\theta_{POS}=\theta_{MON}+\theta_{NOS}$
with velocities $\frac{PS}{OP}=\theta\phi_{POS}$, $\frac{MN}{OM}=\tan\theta_{MON}$,  $\frac{NS}{ON}=\tan\theta_{NOS}$.
Thus,
$$\begin{align*} \tan\theta_{POS} &=\tan\left( \theta_{MON}+\theta_{NOS} \right) \\&=\frac{ \tan \theta_{MON}+\tan\theta_{NOS}  }{ 1 + \tan\theta_{MON}\tan\theta_{NOS} } \\ \end{align*}$$

Galilean relativity

And, now finally,
repeat for this figure in Galilean relativity.
That is,
express
the “velocity of OS [with respect to OP]” (PS/OP)
in terms of
the “velocity of ON [with respect to OP or OM]” (MN/OM)
and
the “velocity of OS [with respect to ON]” NS/ON).
(The method is almost the same…
You might wish to follow your earlier steps and see what has become of them in this case.
However, you’ll have to accept that
in this [Galilean spacetime] geometry, NS is Galilean-perpendicular to ON
since, for radius vector ON, the segment NS is tangent to the Galilean-circle.
)

Answer: (note the second line, as if there was a zero instead of a sign seen above)$$\begin{align*} (\mbox{velocity of OS wrt OP}) &=\frac{PS}{OP}\\ &=\frac{PR+RS}{OM }\\ &=\frac{MN+RS}{OM}\\ &=\frac{\frac{MN}{OM}+\frac{RS}{OM}}{1}\\ &=\frac{\frac{MN}{OM}+\frac{RS}{OM} }{1}\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{-SR}{OM}\right) } }{1 }\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{-SN}{ON}\right) } }{1 }\\ &=\frac{\frac{MN}{OM}+\color{red}{\left(\frac{NS}{ON}\right) } }{1}\\ &=\frac{(\mbox{velocity of ON wrt OM})+(\mbox{velocity of OS wrt ON})  }{1} \end{align*}$$
where we used the similarity of triangles OMN and SRN.

So, in this [degenerate] case, the “slopes” (the “velocities”) do add…because the spatial displacements are coplanar (collinear) in spacetime.
However, this “additivity of velocities” is actually the exception, rather than the rule. (There is an analogous notion of Galilean angles and Galilean trigonometry [due to I.M. Yaglom]… but I won’t discuss this here.]