The Lorentz Transformation

In the previous section we saw that a combination of a scalar and a vector, a paravector, $\v{A} = a_0 + \v{a}$, has an amplitude-squared with the same metric as that of spacetime and yet can be rotated like a normal vector such that the scalar part is unchanged, while the vector part rotates in space, as $\v{\c{R}}\v{A}\v{R}$, where the rotor $\v{R}$ is itself comprised of a scalar and a bivector, $\v{R} = e^{-\v{xy}\frac{\theta}{2}}$.

We also saw that we can define a normalised paravector as $\v{B} = e^{-\v{b}\frac{\phi}{2}}$, so that $\v{B}\c{\v{B}} = 1$. In this section we will consider rotating the paravector $\v{A}$ by a normalized paravector and see that the result, $\v{B}\v{A}\v{B}$, is equivalent to the standard Lorentz transformation.

A hyperbolic rotation

Similar to the spatial rotation, we can see that $\v{A'} = \v{B}\v{A}\v{B}$ is some type of rotation of $\v{A}$, as even though we are not using the Clifford conjugation of the rotor for the left-hand multilication, the amplitude squared is still unchanged:

$\begin{aligned} \v{\c{A'}A'} &= \c{\v{B}\v{A}\v{B}} \v{B}\v{A}\v{B} \\ &= \c{\v{B}}\c{\v{A}}\c{\v{B}} \v{B}\v{A}\v{B} \\ &= \v{\c{A}A} \end{aligned}$

Evaluating $\v{A'}$, we find

$\begin{aligned} \v{A'} &= \v{B}\v{A}\v{B} \\ &= e^{-\v{b}\frac{\phi}{2}}(a_0 + \v{a})e^{-\v{b}\frac{\phi}{2}} \\ &= a_0 e^{-\v{b}\phi} + e^{-\v{b}\frac{\phi}{2}} \v{a}e^{-\v{b}\frac{\phi}{2}} \end{aligned}$

which is, in fact...

The Lorentz Transformation

A vector in the same direction as the rotation

If we first look at the case where the vector being rotated is in the same direction as the rotation itself, ie. $\v{a}\v{b} = \v{b}\v{a} = |\v{a}|$ and since $\v{b}^2 = 1$, then $\v{a} = |\v{a}|\v{b}$, we can further simplify the rotation:

$\begin{aligned} \v{A'} &= a_0 e^{-\v{b}\phi} + e^{-\v{b}\frac{\phi}{2}} \v{a}e^{-\v{b}\frac{\phi}{2}} \\ &= a_0 e^{-\v{b}\phi} + \v{a}e^{-\v{b}\phi} \\ &= a_0 (\cosh{\phi} - \v{b} \sinh{\phi}) + \v{a}(\cosh{\phi} - \v{b} \sinh{\phi}) \\ &= (a_0 \cosh{\phi} - \v{ab}\sinh{\phi}) + (\v{a}\cosh{\phi} - a_0 \v{b} \sinh{\phi}) \\ &= (a_0 \cosh{\phi} - |\v{a}|\sinh{\phi}) + \v{b}(|\v{a}|\cosh{\phi} - a_0 \sinh{\phi}) \\ &= \cosh{\phi}(a_0 - |\v{a}|\tanh{\phi}) + \v{b}\cosh{\phi}(|\v{a}| - a_0 \tanh{\phi}) \\ \end{aligned}$

Note that $\tanh\phi$ has the property of ranging from -1 to 1 (see figure) and that $\cosh\phi$ and $\tanh\phi$ can be related by:

$\cosh\phi = \frac{1}{\sqrt{1-\tan^2\phi}}$

So with the substitution of $v = \tanh\phi$, we're left with the Lorentz transformation of $\v{A} = a_0 + \v{a}$:

$\v{B}\v{A}\v{B} = \frac{1}{\sqrt{1-v^2}}(a_0 - |\v{a}|v) + \frac{1}{\sqrt{1-v^2}}(|\v{a}| - a_0 v)\v{x}$

A vector perpendicular to the direction of the rotation

Next let us look at the case where the vector being rotated is perpendicular to the rotation, ie. $\v{a}\v{b} = -\v{b}\v{a}$. In this case:

$\begin{aligned} \v{A'} &= a_0 e^{-\v{b}\phi} + e^{-\v{b}\frac{\phi}{2}} \v{a}e^{-\v{b}\frac{\phi}{2}} \\ &= a_0 e^{-\v{b}\phi} + \v{a}e^{+\v{b}\frac{\phi}{2}} e^{-\v{b}\frac{\phi}{2}} \\ &= a_0 e^{-\v{b}\phi} + \v{a} \\ \end{aligned}$

So the vector perpendicular to the direction of the rotation is not affected by the transformation, just as with the Lorentz transformation.