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Lorentz transformation and paravectors

  1. Jul 12, 2014 #1

    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 [itex]p^{\prime}[/itex] of a paravector [itex]p[/itex] is given by


    Above, the Lorentz transformation [itex]L[/itex] is defined as:

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

    so that if [itex]\boldsymbol{\theta}=\boldsymbol{0}[/itex] we have, using the hyperbolic Euler formula,

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

    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 [itex]\gamma[/itex]'s. Below is my derivation.

    Let [itex]\boldsymbol{\theta}=\boldsymbol{0}[/itex]. Then

    =[\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 [itex]\boldsymbol{w}[/itex]. Scalars commute with everything, so the scalar [itex]p_{0}[/itex] will commute with everything in the square brackets (i.e., all of [itex]L[/itex]). Parallel vectors commute, so [itex]\boldsymbol{p}_{\parallel}[/itex] will also commute with everything in the square brackets. Perpendicular vectors anticommute, so [itex]\boldsymbol{p}_{\perp}[/itex] will commute with the scalar term in square brackets, and anticommute with the vector term. Performing all of these rules and simplifying gives

    =[\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 [itex]\boldsymbol{p}=\boldsymbol{p}_{\parallel}[/itex], i.e., let [itex]\boldsymbol{p}_{\perp}=\boldsymbol{0}[/itex], for convenience. Then, expanding,

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

    =[\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 [itex]p^{\prime}[/itex], and the terms in the second parentheses are the vector (i.e., space) portion of [itex]p^{\prime}[/itex]. 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 [itex]+\gamma \frac{\boldsymbol{v}}{c}[/itex] in the second-to-last line was replaced with a [itex]-\gamma \frac{\boldsymbol{v}}{c}[/itex], which would occur if the [itex]+\hat{\boldsymbol{w}}\sinh (w)[/itex] a little earlier were replaced with a [itex]-\hat{\boldsymbol{w}}\sinh (w)[/itex].

    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
  2. jcsd
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