# Lorentz transformation and paravectors

1. Jul 12, 2014

### HJ Farnsworth

Greetings,

I have been reading William Baylis - "Electrodynamics - A Modern Geometric approach". For anyone who has the book, the question I am about to ask is related to what's on pages 42-44.

The Lorentz-transformed paravector $p^{\prime}$ of a paravector $p$ is given by

$p^{\prime}=LpL^{\dagger}$.

Above, the Lorentz transformation $L$ is defined as:

$L=\exp [(\boldsymbol{w}-i\boldsymbol{\theta})/2]$,

so that if $\boldsymbol{\theta}=\boldsymbol{0}$ we have, using the hyperbolic Euler formula,

$L=\cosh (w/2)+\hat{\boldsymbol{w}}\sinh (w/2)$,

which represents a Lorentz boost. I have seen this form of the Lorentz boost in other contexts, and been fine with it, but I want to make sure that I understand it well in the paravector formalism, so I attempted a derivation to relate it to the more familiar form of the Lorentz transformation with $\gamma$'s. Below is my derivation.

Let $\boldsymbol{\theta}=\boldsymbol{0}$. Then

$p^{\prime} =LpL^{\dagger}\\ =[\cosh (w/2)+\hat{\boldsymbol{w}}\sinh (w/2)](p_{0}+\boldsymbol{p})[\cosh (w/2)+\hat{\boldsymbol{w}}\sinh (w/2)]\\ =[\cosh (w/2)+\hat{\boldsymbol{w}}\sinh (w/2)](p_{0}+\boldsymbol{p}_{\parallel}+\boldsymbol{p}_{\perp})[\cosh (w/2)+\hat{\boldsymbol{w}}\sinh (w/2)]$.

Above, parallel and perpendicular refer to directions relative to $\boldsymbol{w}$. Scalars commute with everything, so the scalar $p_{0}$ will commute with everything in the square brackets (i.e., all of $L$). Parallel vectors commute, so $\boldsymbol{p}_{\parallel}$ will also commute with everything in the square brackets. Perpendicular vectors anticommute, so $\boldsymbol{p}_{\perp}$ will commute with the scalar term in square brackets, and anticommute with the vector term. Performing all of these rules and simplifying gives

$p^{\prime} =[\cosh (w/2)+\hat{\boldsymbol{w}}\sinh (w/2)]^{2}(p_{0}+\boldsymbol{p}_{\parallel})+\boldsymbol{p}_{\perp}$.

For the rest of the derivation, let $\boldsymbol{p}=\boldsymbol{p}_{\parallel}$, i.e., let $\boldsymbol{p}_{\perp}=\boldsymbol{0}$, for convenience. Then, expanding,

$p^{\prime} =[\cosh ^{2}(w/2)+\sinh ^{2}(w/2)+\hat{\boldsymbol{w}}2\cosh (w/2)\sinh (w/2)](p_{0}+\boldsymbol{p})\\ =[\cosh (w)+\hat{\boldsymbol{w}}\sinh (w)](p_{0}+\boldsymbol{p})$.

Identifying $w$ as the rapidity, we have $\cosh (w)=\gamma$ and $\hat{\boldsymbol{w}}\sinh (w)=\gamma \frac{\boldsymbol{v}}{c}$, so that

$p^{\prime} =[\gamma +\gamma \frac{\boldsymbol{v}}{c}](p_{0}+\boldsymbol{p})\\ =(\gamma p_{0}+\gamma \frac{\boldsymbol{v}}{c}\boldsymbol{p}) + (\gamma \boldsymbol{p} + \gamma \frac{\boldsymbol{v}}{c}p_{0})$,

where the terms in the first parentheses are the scalar (i.e., time) portion of $p^{\prime}$, and the terms in the second parentheses are the vector (i.e., space) portion of $p^{\prime}$. So, I almost get the Lorentz boost, except that the plus in each parentheses should be replaced by a minus, which would occur if the $+\gamma \frac{\boldsymbol{v}}{c}$ in the second-to-last line was replaced with a $-\gamma \frac{\boldsymbol{v}}{c}$, which would occur if the $+\hat{\boldsymbol{w}}\sinh (w)$ a little earlier were replaced with a $-\hat{\boldsymbol{w}}\sinh (w)$.

So, I think I may have made a simple algebraic mistake somewhere, but I have been staring at it for a long time and cannot find my mistake (I'm probably just too close to it). Can anyone else spot it?

Thanks for any help that you can give.

-HJ Farnsworth