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):

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


$$ \begin{align} \mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\ &=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2\\ &=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp+2s_\perp-2s_\perp) \sinh^2a/2\\ &=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel+s_\perp-2s_\perp) \sinh^2a/2\\ &=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (\mathbf{s}-2s_\perp) \sinh^2a/2\\ &=\mathbf{s} \cosh^2a/2-\mathbf{s}\sinh^2a/2+ \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\ &=\mathbf{s} + \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\ &=\mathbf{s} + \mathbf{r}s_\perp \sinh a +s_\perp (\cosh a -1)\\ &=s_\parallel + ( \cosh a + \mathbf{r} \sinh a )s_\perp \end{align} $$

Is this a Lorentz boost? How do I show that it is?


2 Answers 2


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).

  • $\begingroup$ What kind of object is $x+rt$ in your example? Is $\mathbf{r}$ vector, such that $\mathbf{r}=(r_1,r_2,r_3)$, then $x+rt=(x+r_1t,x+r_2t,x+r_3t)$. I am not familiar with what $x+rt$ represents. $\endgroup$
    – Anon21
    Commented Aug 17, 2020 at 14:54
  • $\begingroup$ It's a split-complex number (en.wikipedia.org/wiki/Split-complex_number). For higher dimensions, the hyperbolic versor is related to split-quaternions (en.wikipedia.org/wiki/Split-quaternion) and biquaternions (en.wikipedia.org/wiki/Biquaternion). $\endgroup$
    – Batiatus
    Commented Aug 17, 2020 at 15:04
  • $\begingroup$ What is the relationship between those and the rotor en.wikipedia.org/wiki/Rotor_(mathematics) where a vector is sandwiched between a rotor and its inverse to make a rotation? It seems similar except the vector is sandwiched and the exponent is that of a bivector $\exp a \mathbf{B}$. A rotation is $X'=\exp (a\mathbf{B}) X \exp (-a\mathbf{B})$. In your case, the transformation is applied to the left (instead of being sandwiched). $\endgroup$
    – Anon21
    Commented Aug 17, 2020 at 16:09
  • $\begingroup$ (con't) In your case, the transformation is applied to the left (instead of being sandwiched), and also the vector is representation as split-complex numbers (instead of as a vector). $\endgroup$
    – Anon21
    Commented Aug 17, 2020 at 18:36

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.

  • $\begingroup$ I'm afraid I still do not see how $\mathbf{s}'=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2$ produces a Lorentz boost :(. Can you help with producing the next line or two? Like what am I looking for to get in the end? $\endgroup$
    – Anon21
    Commented Aug 17, 2020 at 23:45
  • $\begingroup$ Can you take a look at my latest edit? $\endgroup$
    – Anon21
    Commented Aug 18, 2020 at 0:21

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