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Geometrical algebra in theoretical physics 
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#1
Dec1812, 01:10 PM

P: 49

Is geometric algebra and geometric calculus worth learning for a theoretical physicist? What are the advantages of this approach against the usual vector calculus ?



#2
Dec1812, 05:16 PM

P: 97

1st question: Possibly. It's really interesting stuff, but it's not very main stream so if you work exclusively in it your work may be completely ignored by the larger physics community. And that's if you can even get it published. David Hestenes has had a lot of trouble getting his papers published (might be easier now, now that more people are working on it).
2nd question: Basically any geometric approach, including differential geometry (which is essentially completely contained in GA), allows you to work without reference to a specific coordinate frame. Calculating rotations is super easy in GA compared to vector arithmetic. Also, all of vectors, tensors, linear algebra, forms, quaternions, octonions, complex variables, etc. are all contained in the formalism of GA in some form or another. This makes it so that virtually all of physics can be worked in in terms of just GA. e.g. vector calculus works fine in E&M and classical mechanics, but it won't work for GR (you need more complicated tensor stuff). If you work with GA, you only need GA. Simple example of GA vs. vectors: How many equations do you need for E&M? With vectors you need 4. With GA you only need 1. 


#3
Dec1912, 03:36 AM

P: 49

So why isn't it mainstream? Is it because it's new or what? Shouldn't it replace ordinary vector calculus if it's easier to work with and provide geometrical approach far superior than ordinary calculus?



#4
Dec1912, 07:26 AM

P: 836

Geometrical algebra in theoretical physics



#5
Dec1912, 10:10 AM

P: 834

I consider it worth learning just for the simplicity it introduces into problems that otherwise would've been extremely tedious or difficult.
It's not uncommon that one will have to manipulate a tensor expression to try to get to a simpler result. GA's generality and identities make this much easier to do than laboring through index notation, in my opinion. Example: find ##\epsilon_{ijk} \epsilon^{ljk}##. I can't speak to proving this in index notation, but in GA, you can keep things grounded and simple. The LeviCivita tensor is just components of the pseudoscalar evaluated on some basis. In this case, we can generalize this problem to an equivalent one: Simplify ##(a \wedge B) i (c \wedge B^{1}) i = (a \wedge B)(B^{1} \wedge c)## for vectors ##a,c## and bivector ##B##. This isn't a hard problem to attack, especially with the power of GA. Projection onto grade and associativity make it rather straightforward. Note that ##ac = a B B^{1} c## and project out some components. $$ \langle a B B^{1} c \rangle_0 = (a \wedge B)\cdot (B^{1} \wedge c) = a \cdot c $$ This is actually so much easier than most identity problems, I was surprised I was done at this point. Usually you have to consider two grades at least, but since the result must be scalar, we're done here. Yes, for the identity we meant to consider, there's a missing factor of 2. I can't quite find itprobably would if I were more methodicalbut it does show that there's some work in converting a tensor expression to a GA one. 


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