Sjamaar Ex 2_18 on Maxwells (CH2)

Sjamaar Exercise 2.18 PhL 7.7.15 This has mistakes in it, and the whole method I think is not very good, just archive it. See v2 doc which at least shows that dα = 0. This is a good test to see if I really know what I am doing with manipulating forms. I am flunking the test so far below. So now we are in a separate doc. α ≡ (Eidxi)dx4 + Bi(*dxi) in R4 2-form, sums only forR3 β ≡ - (Hidxi)dx4 + Di(*dxi) in R4 2-form, sums only forR3 γ ≡ Ji(*dxi) dx4/c - ρdV in R4 3-form, sum only forR3 dV= dx1dx2dx3 Now compute dα hopefully in a smart way. Restate the above α ≡ (E dx) dx4 + B (*dx) in R4 2-form, sums only forR3 β ≡ - (H dx)dx4 + D (*dx) in R4 2-form, sums only forR3 γ ≡ J (*dx) dx4/c - ρdV in R4 3-form, sum only forR3 dV= dx1dx2dx3 Now let's define some auxiliary forms δ ≡ E dx ε = B dx φ ≡ J dx κ ≡ D dx θ ≡ H dx *(dδ) = [curl E] dx *(dε) = [curl B] dx *(dθ) = [curl H] dx // PL4 Then our definitions of α.β and γ may be stated α = δ dx4 + *(ε) β = - θ dx4 + *(κ) γ = (1/c) *(φ) dx4 - ρdV Meanwhile, try to translate the Maxwell equations into the forms language: div B = 0 => *[d(*ε)] = 0 where ε ≡ B dx // from PL3 div D = 4πρ => *[d(*κ)] = 4πρ where κ ≡ D dx // from PL3 curl E dx = -(1/c2)(∂4B) dx => *(dδ) = -(1/c2) ∂4ε curl H dx = (4π/c) J dx + (1/c2) ∂4D dx => *(dθ) = (4π/c)φ + (1/c2) ∂4κ OK, so here is what the four Maxwell equations say in form-space *[d(*ε)] = 0 *(dδ) = -(1/c2) ∂4ε *[d(*κ)] = 4πρ *(dθ) = (4π/c)φ + (1/c2) ∂4κ Now pursue the first of these for a while *[d(*ε)] = 0 Now use result (2.12) page 28 which I proved elsewhere which says *(*α) = (-1)kn+k α for α = any k-form in Rn Question: what is the meaning of *0 ? Suppose α = ΣI fI dxI where all fI = 0. Then α = 0. Then *α = ΣI fI (*dxI) = ΣI 0 (*dxI) = 0 so I think this is true *0 = 0. So go back to *[d(*ε)] = 0 Apply * to both sides to get **[d(*ε)] = *0 = 0 Now use little theorem to get **[d(*ε)] = (-1)kn+k d(*ε) where k is the order of the form d(*ε) Thus we get (-1)kn+k d(*ε) = 0 which is a very long winded way of arriving at our result d(*ε) = 0 In other words *[d(*ε)] = 0 d(*ε) = 0 Here then is a restatement of our four Maxwell equations: [d(*ε)] = 0 *(dδ) = -(1/c2) ∂4ε *[d(*κ)] = 4πρ *(dθ) = (4π/c)φ + (1/c2) ∂4κ and here is a restatement of our defined forms α = δ dx4 + *(ε) β = - θ dx4 + *(κ) γ = (1/c) *(φ) dx4 - ρdV One claim Sjamaar makes is that dα = 0. Let's now try to show that to be true. Consider dα = d [δ dx4 + *(ε)] = d(δ dx4) + d(*ε) = d(δ dx4) // since Max [d(*ε)] = 0 = (dδ) dx4 dx4 is constant relative to d But I don't see why this should be 0. We don't know anything about dδ . I would have to show dδ = 0 or d(E dx) = 0 or (dE) dx = 0 or Σi=13 dEidxi = 0 But dEi = Σk=14(∂Ei/∂xk)dxk so get Σi=13 Σk=14(∂Ei/∂xk)dxkdxi = 0 ? I have (sadly) done something wrong so more hours need to be sunk into this example.