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Jul
31
comment What's the meaning of the unit bivector i?
@draks that is not geometric algebra. That's Dirac algebra which only makes sense in Dirac theory. Geometric algebra promises (I'm still learning) to be applicable/useful everywhere, not just in the case of $\gamma$s. And an index $i$ is nothing evil.
Jul
31
awarded  Critic
Jul
31
comment What's the meaning of the unit bivector i?
@celtschk OK. But Hestenes maintains quaternions are just an aspect/transformation/... of the more general geometric algebra (at least in sofar they are used in Physics). I can't deduce any geometric meaning from quaternions either, so although insightful, it doesn't help me much (probably why you made it a comment anyways ;-)).
Jul
31
comment What's the meaning of the unit bivector i?
OK. This is part of the meaning I get. The bivector as an operator is a rotor. But the bivector as basis element of the algebra should have a meaning on its own. Is my plane interpretation correct?
Jul
31
comment What's the meaning of the unit bivector i?
It seems you avidly hate Hestenes' formulation. I get the parallel with $\gamma$ algebra the way you see it (I first came into contact with geometric algebra when studying the Dirac equation), but in the pdf referenced, he's not talking about $\gamma$ algebra, he's talking Euclidean geometry, where stuff like $\gamma$-matrices (he's not even talking about matrices) are not even relevant. You're giving a non-geometric interpretation IMHO, which is not quite what I'm after. I'm after the interpretation Hestenes describes.
Jul
30
comment What's the meaning of the unit bivector i?
If someone would be so kind to tag this with a new tag geometric-algebra, I'd appreciate it.
Jul
30
asked What's the meaning of the unit bivector i?
May
17
awarded  Popular Question
Nov
21
accepted Create list of {x,f(x)} pairs
Nov
21
accepted A function of two functions that loses dependence on an argument
Nov
9
awarded  Commentator
Nov
9
comment Create list of {x,f(x)} pairs
Funny how the (IMHO) much worse pops up in comments under my answer :)
Nov
9
answered Create list of {x,f(x)} pairs
Nov
9
asked Create list of {x,f(x)} pairs
Nov
7
comment A function of two functions that loses dependence on an argument
Craig: see edit, my case here is really an addition of sorts, so no undefined things pop up.
Nov
7
revised A function of two functions that loses dependence on an argument
multiplication->addition
Nov
7
asked A function of two functions that loses dependence on an argument
May
26
accepted Mathematica VectorPlot with Piecewise arguments
May
26
comment Mathematica VectorPlot with Piecewise arguments
Yeah, that should be removed. I changed the function to use HeavisideTheta instead of Piecewise, and now it works as it should. Don't really know where I went wrong...
May
26
asked Mathematica VectorPlot with Piecewise arguments