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Geometric algebra, rotations

  1. Sep 5, 2014 #1
    Hi, I want to calculate a rotation of a vector GA style with this formula [itex]e^{-B \frac{\pi}{2}}(2e_{1}+3e_{2}+e_{3})e^{B\frac{\pi}{2}}[/itex]. Now since no book/pdf on GA exists where a calculation is explicitly done with numbers, I wounder how to calculate this. Should I substitude [itex]e^{-B \frac{\pi}{2}}=cos (\frac{\pi}{2})-Bsin(\frac{\pi}{2})[/itex](same for the other term) and see what I get, or should I convert the vector into a complex number and then convert it into the exponential form and then multiply it? Or how is this done? By the way, how do I convert a vector in [itex]R^{3}[/itex] into a complex number? Multiply on the left with [itex]e_{1}e_{2}[/itex]? In [itex]R^{2}[/itex] it is just multiplication by [itex]e_{1}[/itex] on the left.
  2. jcsd
  3. Sep 5, 2014 #2
    Is [itex]B[/itex] the imaginary unit?

    Your question is unclear. I am particularly confused by your conjugation by "[itex]e^{-B\frac{\pi}{2}}[/itex]," since I'm not sure how it is acting.

    Would you please clarify?
  4. Sep 5, 2014 #3
    Hi Pond Dragon. Sry, the general formula for a rotation is defined as [itex]a^{,}= RaR^{\dagger}[/itex] Where [itex]R=nm[/itex] and [itex]R^{\dagger}=mn[/itex]. Now one uses [itex]B= \frac{m\wedge n}{sin \phi}[/itex], [itex]B^{2}=-1[/itex] (unit bivector) and rewrites [itex]R=cos \frac{\phi}{2} -B sin \frac{\phi}{2}=e^{-B \frac{\phi}{2}}[/itex] and [itex]R^{\dagger}= cos \frac{\phi}{2} +B sin \frac{\phi}{2}=e^{B \frac{\phi}{2}}[/itex]. Hope that helps !
    Thx for reply.
  5. Sep 5, 2014 #4
    So, you don't know how to compute the product? It looks like you could just apply definitions.
  6. Sep 5, 2014 #5
    Hi again, so you think this is the right way

    [itex](cos \frac{\pi}{2}-Bsin\frac{\pi}{2})(2e_{1}+3e_{2}+e_{3})=-e_{1}B2-e_{2}B3-e_{3}B[/itex]

    then from the right
    It should be a rotation of 90 degrees, but the two vectors are not orthogonal hm.
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