Geometric product

  • #1

Main Question or Discussion Point

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:

Answers and Replies

  • #2
11,334
4,806
So how did you get from the next to last step to your solution?
 
  • #3
jhae2.718
Gold Member
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[itex]\newcommand{\a}{\boldsymbol a}\newcommand{\b}{\boldsymbol b}[/itex]
So how did you get from the next to last step to your solution?
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].
 
  • #4
11,334
4,806
Thanks for the explanation, I was thinking the math was more involved with some condition we weren't told about.
 
  • #5
jhae2.718
Gold Member
1,161
20
No problem, it gave me a reason to dig out my copy of New Foundations for Classical Mechanics and play with that. :D
 
  • #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
 
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