3 ms·

How do you "just 'rotate'" bivector into accelerated frame? Because if you're talking about Lorentz transformations between inertial frames, those alone can't g

• 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.