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I think that last point really needs to start with a thorough reform of basic mathematical education (starting from undergrad calculus) to have any chance of re
by keldaris 8y ago
I think that last point really needs to start with a thorough reform of basic mathematical education (starting from undergrad calculus) to have any chance of real success. The present mish-mash of notations, however disjointed, lends itself very well to teaching people who don't know much beyond basic real analysis and differential equations. You can't just throw them in a GA-based E&M class without completely changing the entire curriculum first.
Your first point got me curious, though. I found E&M in differential forms quite satisfying from an intellectual point of view (although of little practical benefit to the problems I was interested in compared to the classical approach, which I was plenty familiar with already). What advantages do you see for GA vs. differential forms in this context? This might make for a few interesting seminars with our math folks...
- wolfgke 8y ago> You can't just throw them in a GA-based E&M class without completely changing the entire curriculum first. Preparing for the additional requirements of the lectures of the coming semester is whar the semester break is for.
- keldaris 8y agoSure, as long as you don't expect anyone to complete the course or are willing to play catch up for most of the semester. In other words, I'd love to plan courses that way, but no university administrator would ever let me (instead I restrict myself to graduate seminars and beyond).
- jacobolus 8y agoNah, you really want to start students learning the GA formalism in their linear algebra and vector calculus classes, if not in high school. Especially important in high school is to dive into not only the “complex numbers” but also the “split complex numbers”.
- eigenspace 8y agoI should be clear here that my longing for a physics community that embraces GA should not be confused with any level of optimism that it'll actually happen. I agree its a monstrously difficult task and our current formalisms are more or less good enough. A man can dream though! One advantage I see in the GA treatment of E&M is in radiation. If one takes the Faraday bivector for a stationary charge (ie. the Coulumb field) and then just 'rotates' it into a moving frame using the GA rotors you can find the standard radiation fields of E&M quite easily. This is cool because it's not only a neat proceedure mathematically, but its computationally quite efficient, say if you need to find the fields produced by N moving charged particles. I'm sure there are more examples of things like this but it's been a while since I last looked at E&M in GA so none are coming to mind right now.
- keldaris 8y agoThat's a neat example, thanks!
- posix_me_less 8y agoHow do you "just 'rotate'" bivector into accelerated frame? Because if you're talking about Lorentz transformations between inertial frames, those alone can't give you radiation fields. To get radiation fields, you need Maxwell's equations and boundary conditions. When you have those, what is the use of GA?
- eigenspace 8y agoYes, it seems I have oversimplified things. My apologies, it's been a little while. Instead of the prescription I described, here is what one must do: 1) Start with the Coulomb potential A(r) = qγ₀/(4πr) 2) make it covariant by replacing r -> r -> X⋅γ₀ where X = r(γ₀ + γᵣ). This is a trivial transformation in the static frame as one can easily show X⋅γ₀ == r, but it's important as it encodes the causal information you astutely pointed out is vital to obtaining radiation solutions. This gives A(r) = qγ₀/(4π X⋅γ₀) 3) Now, one can 'rotate' this to a frame with (not necessarily constant) velocity v via A'(r) = R A(R̃ x R) R̃ ie. evaluate the rotated A at inverse rotate spacetime position. Then A'(r) = q v/(4π X⋅v) This is the Liénard–Wiechert potential (this can also be obtained by solving ∇²A = J using the retarded Green's function G(r, t) = δ(|r|-t)/(4π|r|). 4) Now once can get the radiation fields by taking the derivative of this and doing some algebra: F = ∇A = (q/4π) (X ∧ v + ½ X(v̇ ∧ v)X)/(X⋅v)³ The term proportional to X ∧ v is just the Coulomb field, but the term with the X(v̇ ∧ v)X is the radiation field which one can show propagates to infinity. This is unfortunately much less elegant and straightforward than I let on initially so I apologize. I still think this has advantages over standard treatments but your milage may vary.