3 ms·

I 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

I 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 ago
That's a neat example, thanks!
posix_me_less 8y ago
How 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 ago
  Yes, 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.

keldaris 8y ago

That's a neat example, thanks!

posix_me_less 8y ago

How 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 ago
Yes, 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.