Lorentz Transformation Proof Using Hyperbolic Numbers

What are complex numbers? What is i? It's defined by i^(2)=-1. Complex numbers are just numbers of the form a+bi. Or pretty much a vector a+bi~(a,b)

What are hyperbolic numbers? What is j? It's defined by j^(2)=1. Hyperbolic numbers are just numbers of the form a+bj. Or pretty much a vector a+bj~(a,b)

About complex numbers:

Every complex number has a conjugate, (a+bi)* = (a-bi) . And a (squared) norm |z|=root(zz*).

Any z is uniquely written as z=e^(iθ)|z|, where θ is in [0,2π). By substituting e^(iθ)=ω and µ=|z| we see: Any z is uniquely written as z=ωµ where |ω|=1 and µ is real positive. I'll call ω=e^(iθ) the unitary part of z.

So: Two complex numbers, with equal norm and unitary part, are equal.

I'll show deriving the Lorentz Transformation is as easy as deriving the "Rotation Transformation".

What are the transformations that preserve the Euclidean norm? Rotations. What is a 2D rotation? It's exactly equivalent to multiplication by e^(iθ) (a unitary complex number).

What is the norm of a hyperbolic number? It's root(a^(2)-b^(2)). Wow, this is exactly the the square of the Minkowski Norm! (check it)

What is a Lorentz Transformation? A transformation that preserves the Minkowski Norm, ie the "Hyperbolic Norm". What are the transformations that preserve the norm of a hyperbolic number? Multiplication by "e^(jθ)".So a Lorentz Transformation is exactly equivalent to multiplication by a unitary hyperbolic number!!! Now we only need to figure out which unitary hyperbolic number, ω.

Say I'm an observer at rest so my coordinates are (t,x)=(t,0). Say another inertial observer passes by the origin, his coordinates are (t,x)=(1,v)t=(1+jv)t. In his own frame of reference, by definition, he has coordinates (t',0). Let z'=(t',0)=t' and z=(t,x)=(1+jv)t.

We know T(z)=z', by our discovery T(z)=ωz=z'. By rewriting ω(1+jv)t=t'. If they're equal their unitary part is equal which can be obtained dividing by the norm, root(z*z), on both sides obtaining (work it out): ω√(1-v^(2))/(1-jv)=1, or ω=(1-jv)/√(1-v^(2)).

If it's not clear check that multiplication by ω=(1-jv)/√(1-v^(2)) is exactly equivalent to Lorentz Transformations.