## Split-Complex Numbers: Lorentz Transformations

Before getting into the algebra of the split-complex numbers, I will first derive the Lorentz transformations that motivate their use.

Derivation of Lorentz Transformations
The following ideas express what it means to be an inertial reference frame, and how they relate to each other.

1) The ordered pair $(t,x)$, x and t are independent variables.
2) An object observed to be moving with a constant velocity in one inertial reference frame define the origin of another inertial frame.
3) Transformations between inertial frames are linear.
4) The composition of two transformations is a third transformation.
5) The transformations for equal and opposite velocities are inverses.
6) The transformations are isotropic.

Items 3-5 can be expressed more concisely as the transformations form a linear group. The sixth means that any stretching or shrinking in the transformation must be the same for V and -V. This allows us to express a general transformation as follows:

$\left[\begin{array}{c}t'\\x'\end{array}\right]=\left[\begin{array}{cc}\alpha&\beta\\ \delta&\gamma\end{array}\right]\left[\begin{array}{c}t\\x\end{array}\right]$

Let us consider two frames where $O=(0,0)=(0,0)'=O'$ and the velocity of O’ in the original frame, F, is V. (Thus the velocity of O in F’ is -V). Now let us consider the transformations for the lines through the origin along t and t’.

$\left[\begin{array}{c}t'\\-Vt'\end{array}\right]=\left[\begin{array}{cc}\alpha&\beta\\ \delta&\gamma\end{array}\right]\left[\begin{array}{c}t\\ 0 \end{array}\right]=\left[\begin{array}{c}{\alpha t}\\{\delta t}\end{array}\right]$

$\left[\begin{array}{c}t'\\ 0\end{array}\right]=\left[\begin{array}{cc}\alpha&\beta\\ \delta&\gamma\end{array}\right]\left[\begin{array}{c}t\\Vt \end{array}\right]$

The first gives me $\delta t=-Vt'=-V(\alpha t)$, so $\delta=-V\alpha$. The second gives me $0=\delta t+\gamma Vt$, so $\delta=-V\gamma$. This also means that $\alpha=\gamma$. Thus our general transformation becomes:

$\left[\begin{array}{c}t'\\x'\end{array}\right]=\left[\begin{array}{cc}\gamma&\beta\\ -V\gamma&\gamma\end{array}\right]\left[\begin{array}{c}t\\x\end{array}\right]$

Now, the composition of two transformations must be another transformation. In particular, the diagonal elements must be identical.

$\left[\begin{array}{cc}\gamma&\beta\\ -V\gamma&\gamma\end{array}\right]\left[\begin{array}{cc}\gamma^*&\beta^*\\ -V^*\gamma^*&\gamma^*\end{array}\right]=\left[\begin{array}{cc}{\gamma\gamma^*-\beta V^*\gamma^*}&{\gamma\beta^*+\beta\gamma^*}\\{-V\gamma\gamma^*-\gamma V^*\gamma^*}&{-V\gamma\beta^*+\gamma\gamma^*}\end{array}\right]$

So $\gamma\gamma^*-\beta V^*\gamma^*=-V\gamma\beta^*+\gamma\gamma^*$ or $\frac{-\beta}{V\gamma}=\frac{-\beta^*}{V^*\gamma^*}$. But now this can be any two transforms, so this must be a universal constant, $\chi$. This gives us $\beta=-\chi V\gamma$, and an updated general transformation.

$\left[\begin{array}{c}t'\\x'\end{array}\right]=\left[\begin{array}{cc}\gamma&-\chi V\gamma\\ -V\gamma&\gamma\end{array}\right]\left[\begin{array}{c}t\\x\end{array}\right]=\gamma\left[\begin{array}{cc}1&-\chi V\\ -V&1\end{array}\right]\left[\begin{array}{c}t\\x\end{array}\right]$

Since $\gamma$ represents a stretching or shrinking, it should not depend on the direction, so $\gamma(-V)=\gamma(V)$. Now $T(V)^{-1}=T(-V)$ gives us:

$\frac{1}{\gamma}\left[\begin{array}{cc}1&-\chi V\\ -V&1\end{array}\right]^{-1}=\gamma\left[\begin{array}{cc}1&\chi V\\ V&1\end{array}\right]$
$\frac{1}{\gamma}\frac{1}{1-\chi V^2}\left[\begin{array}{cc}1&\chi V\\ V&1\end{array}\right]=\gamma\left[\begin{array}{cc}1&\chi V\\ V&1\end{array}\right]$

Thus $\gamma=\frac{1}{\sqrt{1-\chi V^2}}$. This gives us a general transformation of:
$\left[\begin{array}{c}t'\\x'\end{array}\right]=\frac{1}{\sqrt{1-\chi V^2}}\left[\begin{array}{cc}1&-\chi V\\ -V&1\end{array}\right]\left[\begin{array}{c}t\\x\end{array}\right]$
When $\chi=0$ we have our familiar Galilean transformations. Unfortunately multiple experiments have shown that $\chi$ is finite, and its value has been found to nearly a dozen digits. This value is $\chi=\frac{1}{c^2}$. But now we have the Lorentz transformations.

The fact that a velocity of c is invariant under these transformations is reduced to a theorem. Scaling time by c now gives the standard form of the Lorentz transformations.
$\left[\begin{array}{c}ct'\\x'\end{array}\right]=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}\left[\begin{array}{cc}1&-\frac{V}{c}\\ -\frac{V}{c}&1\end{array}\right]\left[\begin{array}{c}ct\\x\end{array}\right]$

Normally one would generalize this by showing that the orthogonal directions can not be altered by these boosts. Instead, I will focus on how a boost in one direction is a hyperbolic rotation.
Hyperbolic Rotations
The Lorentz transformations actually define a hyperbolic rotation. The equivalent of an angle is a rapidity, $\varphi$.
$\tanh\varphi=\frac{v}{c}$.
$\gamma=\cosh\varphi$

This is useful because hyperbolic rotations can be represented by split-complex numbers, $z=a+bj$, where $j^2=1$.

$(a+bj)+(c+dj)=(a+c)+(b+d)j$
$(a+bj)(c+dj)=(ac+bd)+(ad+bc)j$
$(a+bj)^*=a-bj$
$N(z)=zz^*=a^2-b^2$

The norm is no longer positive definite, so we can not define the absolute value to be the square root of this value. Further we have zero divisors, which will be shown to correspond to the singular nature of light cones.

Every non-zero divisor can be divided by the square root of the absolute value of its norm to get the equivalent of a unit split-complex number. A unit rotation is proper if it has positive norm and a positive real term.

Just as $u=\cos\theta+i\sin\theta$ represents a rotation, $u=\cosh\varphi-j\sinh\varphi$ represents a proper hyperbolic rotation.

$(ct+xj)u=ct'+x'j$

Again letting $v^2=u$, we can represent this as:
$v(ct+xj)v=ct'+x'j$