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Geometric product

  1. Aug 21, 2014 #1
    Hey JO,
    I'm reading a book on geometric algebra and in the beginning (there was light, jk) a simple calculation is shown:
    Geometric product is defined as:
    [itex]ab = a \cdot b + a \wedge b[/itex]
    or
    [itex]ba = a \cdot b - a\wedge b[/itex]


    Now
    [itex](a\wedge b)(a \wedge b)=(ab-a \cdot b)(a\cdot b - ba)[/itex]
    [itex]=-ab^{2}a-(a \cdot b)^{2}+a \cdot b(ab+ba)[/itex]
    [itex]=(a \cdot b)^{2}-a^{2}b^{2}[/itex]
    [itex]=-a^{2}b^{2}sin^{2}(\phi)[/itex]

    I think this term [itex]a \cdot b(ab+ba)[/itex] has to vanish somehow, but it is [itex]
    (a\cdot b)^{2}[/itex] and that doesn't make sense :( Any suggestions?

    Ok I know the answer, the term is [itex]2(a\cdot b)^{2}[/itex]. But thank you for your attention !
     
    Last edited: Aug 21, 2014
  2. jcsd
  3. Aug 22, 2014 #2

    jedishrfu

    Staff: Mentor

    So how did you get from the next to last step to your solution?
     
  4. Aug 22, 2014 #3

    jhae2.718

    User Avatar
    Gold Member

    [itex]\newcommand{\a}{\boldsymbol a}\newcommand{\b}{\boldsymbol b}[/itex]
    From the original post I'm going to assume the question is answered, so here are the steps to the final solution for anyone interested (noting that I've only done a bit of work with geometric algebra after seeing a talk by David Hestenes...):

    The quantity [itex]\a\b + \b\a[/itex] reduces to [itex]2\a\cdot\b[/itex] as stated, so the last term becomes [itex](\a\cdot\b)(2\a\cdot\b)[/itex] or [itex]2(\a\cdot\b)^2[/itex]. The whole RHS then reduces to [itex](\a\cdot\b)^2 - \a^2\b^2[/itex].

    The contraction property states that [itex]\a\a = \a^2 = |\a|^2[/itex], so this becomes [itex]|\a\cdot\b|^2 - a^2b^2[/itex]. The dot operator is just the inner product, so [itex]|\a\cdot\b| = ab\cos\phi[/itex]. Using the Pythagorean identity (there may be a more geometric algebra way to do this, but it's late and I am tired), the whole thing then reduces to [itex]-a^2b^2\sin^2\phi[/itex].
     
  5. Aug 23, 2014 #4

    jedishrfu

    Staff: Mentor

    Thanks for the explanation, I was thinking the math was more involved with some condition we weren't told about.
     
  6. Aug 23, 2014 #5

    jhae2.718

    User Avatar
    Gold Member

    No problem, it gave me a reason to dig out my copy of New Foundations for Classical Mechanics and play with that. :D
     
  7. Aug 26, 2014 #6
    Hi, again I have a problem with the geometric product:

    In the book the unit trivector is defined like this: [itex](e_{1}e_{2})e_{3}=e_{1}e_{2}e_{3}[/itex]
    But that would mean [itex](e_{1}e_{2})e_{3}= (e_{1} \wedge e_{2})\cdot e_{3}+(e_{1} \wedge e_{2} \wedge e_{3})[/itex] But I thougt it is just [itex]e_{1} \wedge e_{2} \wedge e_{3}[/itex]? I could somehow imagine in my head that the plane spanned by [itex]e_{1} \wedge e_{2}[/itex] is perpendicular to the line [itex]e_{3}[/itex] but I'm not sure that it works like that. Is that right? Would make sense.
    Ok but why is this true [itex](e_{1}\wedge e_{2})e_{1}=(-e_{2}e_{1})e_{1}=-e_{2}e_{1}e_{1}=-e_{2}[/itex] because [itex](e_{1}\wedge e_{2})e_{1}=(e_{1} \wedge e_{2})\cdot e_{1}+ e_{1}\wedge e_{2} \wedge e_{1}[/itex] where the last term is zero. I don't know how the term [itex]e_{1} \wedge e_{2}[/itex] interacts as dot product with [itex]e_{1}[/itex]. Similar is [itex](e_{1}\wedge e_{2})(e_{2}\wedge e_{3})=e_{1}e_{3}[/itex] why isn't this just zero? Because [itex](e_{1}\wedge e_{2})(e_{2} \wedge e_{3})=(e_{1}\wedge e_{2}) \cdot (e_{2} \wedge e_{3})+e_{1}\wedge e_{2} \wedge e_{2} \wedge e_{3}[/itex] any help?
    Greets
     
    Last edited: Aug 26, 2014
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