Lorentz boost expressed as Hyperbolic versors At this link https://en.wikipedia.org/wiki/Versor#Hyperbolic_versor it is claimed that an hyperbolic versor, defined as:
$$
\exp(a \mathbf{r})=\cosh a+\mathbf{r}\sinh a
$$
where $||\mathbf{r}||=1$ correspond to a Lorentz boost. But I cannot work out a proof. Can anyone help?

I assume one starts by applying the exponential to a 4-vector $\mathbf{s}$ as follows:
$$
\mathbf{s}'=\exp(\frac{a}{2} \mathbf{r})\mathbf{s}\exp(-\frac{a}{2} \mathbf{r})
$$
Then I get
$$
\begin{align}
\mathbf{s}'&= (\cosh a/2+\mathbf{r}\sinh a/2 )\mathbf{s} (\cosh a/2 -\mathbf{r}\sinh a/2)\\
&= (\cosh a/2+\mathbf{r}\sinh a/2 ) (\mathbf{s}\cosh a/2 -\mathbf{s}\mathbf{r}\sinh a/2)\\
&=\mathbf{s} \cosh^2a/2-\mathbf{s}\mathbf{r}\cosh a/2\sinh a/2+ \mathbf{r}\mathbf{s}\sinh a/2\cosh a/2 - \mathbf{r}\mathbf{s}\mathbf{r}\sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+\cosh a/2\sinh a/2(-\mathbf{s}\mathbf{r}+ \mathbf{r}\mathbf{s}) - \mathbf{r}\mathbf{s}\mathbf{r}\sinh^2a/2
\end{align}
$$

edit (based on answer):
$$
\begin{align}
\mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2
\end{align}
$$

edit2:
$$
\begin{align}
\mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp+2s_\perp-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel+s_\perp-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (\mathbf{s}-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2-\mathbf{s}\sinh^2a/2+ \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\
&=\mathbf{s} + \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\
&=\mathbf{s} + \mathbf{r}s_\perp \sinh a +s_\perp (\cosh a -1)\\
&=s_\parallel + ( \cosh a + \mathbf{r} \sinh a )s_\perp
\end{align}
$$
Is this a Lorentz boost? How do I show that it is?
 A: It isn't a Lorentz boost by itself, but it can be used to easily derive one (at least in 1+1 dimensions). It's similar to the expression of spatial rotations in term of Euler's formula:
$$ x'+iy'=e^{i\phi}(x+iy),\ \left[e^{i\phi}=\cos\phi+i\sin\phi,\ i^{2}=-1\right]$$
which gives
$$ x'+iy'=x\cos\phi-y\sin\phi+iy\cos\phi+ix\sin\phi$$
separating real and imaganiry parts gives:
$$\begin{aligned}x'=x\cos\phi-y\sin\phi\\
y'=y\cos\phi+x\sin\phi
\end{aligned}$$
Instead of complex numbers we now use the hyperbolic versor together with split complex numbers, and replace y by t:
$$ x'+rt'=e^{r\eta}(x+rt),\ \left[e^{r\eta}=\cosh\eta+r\sinh\eta,r^{2}=+1\right] $$
which gives
$$ x'+rt'=x\cosh\eta+t\sinh\eta+rt\cosh\eta+rx\sinh\eta $$
separating real and split-complex parts gives:
$$\begin{aligned}x'=x\cosh\eta+t\sinh\eta\\
t'=t\cosh\eta+x\sinh\eta
\end{aligned}$$
In order to use hyperbolic versors in more dimensions, see split-quaternions (2+1 dimensions) and biquaternions (3+1 dimensions).
A: You can write $\mathbf s$ as a sum of parallel and perpendicular parts which commute and anticommute respectively with $\mathbf r$. Then you have $\mathbf r\mathbf s-\mathbf s\mathbf r = 2\mathbf r\mathbf s_\perp$ and $\mathbf r\mathbf s\mathbf r = s_\|-s_\perp$ and I think you'll get the Lorentz transformation.
If you're interested in this then I'd encourage you to work with Clifford algebras instead of the special-case algebras. You can think of a Clifford algebra as having one imaginary unit for each element of an orthonormal basis of the space; the units square to $\pm1$ depending on the signature, and always anticommute with each other. Vectors represent reflections, and therefore products of even numbers of vectors represent rotations. All of the special-case algebras that were introduced historically are even subalgebras of a Clifford algebra. For example, $a+b\mathbf x\mathbf y$ is the complex numbers, $a+b\mathbf x\mathbf t$ is the split-complex numbers, etc. Clifford algebras also have a natural notation for the vectors that the rotations act on, which the special-case algebras don't.
