Sjamaar Ex 2_18 on Maxwells v2 (CH2)

Sjamaar Exercise 2.18 PhL 7.7.15 Part 1: Show dα = 0 by brute force. 1 Part 2: Show dα = 0 in a reasonable manner 3 3. Conclusion 5 Part 1: Show dα = 0 by brute force. 1. Write α = (E1dx1 + E2dx2 + E3dx3)dx4 + (B1dx2dx3+ B2dx3dx1 + B3dx1dx2 ) 2. Apply operator d to the two fields shown, dEi = ∂jEidxj + ∂4Ei dx4 dBi = ∂jBidxj + ∂4Bi dx4 Then dE1 = ∂jE1dxj + ∂4E1 dx4 = ∂1E1dx1 + ∂2E1dx2 + ∂3E1dx3 + ∂4E1 dx4 dE2 = ∂jE2dxj + ∂4E2 dx4 = ∂1E2dx1 + ∂2E2dx2 + ∂3E2dx3 + ∂4E2 dx4 dE3 = ∂jE3dxj + ∂4E3 dx4 = ∂1E3dx1 + ∂2E3dx2 + ∂3E3dx3 + ∂4E3 dx4 dB1 = ∂jB1dxj + ∂4B1 dx4 = ∂1B1dx1 + ∂2B1dx2 + ∂3B1dx3 + ∂4B1 dx4 dB2 = ∂jB2dxj + ∂4B2 dx4 = ∂1B2dx1 + ∂2B2dx2 + ∂3B2dx3 + ∂4B2 dx4 dB3 = ∂jB3dxj + ∂4B3 dx4 = ∂1B3dx1 + ∂2B3dx2 + ∂3B3dx3 + ∂4B3 dx4 3. Insert these into dα dα = (dE1dx1 + dE2dx2 + dE3dx3)dx4 + (dB1dx2dx3+ dB2dx3dx1 + dB3dx1dx2 ) = ( ∂1E1dx1 + ∂2E1dx2 + ∂3E1dx3 + ∂4E1 dx4) dx1dx4 + (∂1B1dx1 + ∂2B1dx2 + ∂3B1dx3 + ∂4B1 dx4) dx2dx3 + ( ∂1E2dx1 + ∂2E2dx2 + ∂3E2dx3 + ∂4E2 dx4) dx2dx4 + (∂1B2dx1 + ∂2B2dx2 + ∂3B2dx3 + ∂4B2 dx4) dx3dx1 + (∂1E3dx1 + ∂2E3dx2 + ∂3E3dx3 + ∂4E3 dx4) dx3dx4 + ( ∂1B3dx1 + ∂2B3dx2 + ∂3B3dx3 + ∂4B3 dx4) dx1dx2 I have marked in red terms that vanish. So rewrite with all these terms removed dα = ( ∂2E1dx2 + ∂3E1dx3) dx1dx4 + (∂1B1dx1+ ∂4B1 dx4) dx2dx3 + ( ∂1E2dx1+ ∂3E2dx3) dx2dx4 + (∂2B2dx2 + ∂4B2 dx4) dx3dx1 + (∂1E3dx1 + ∂2E3dx2 ) dx3dx4 + ( ∂3B3dx3 + ∂4B3 dx4) dx1dx2 We still have 12 terms. Next, bring in the dx trailing factors dα = ( ∂2E1dx2 dx1dx4 + ∂3E1dx3 dx1dx4) + (∂1B1dx1dx2dx3+ ∂4B1 dx4dx2dx3) + ( ∂1E2dx1dx2dx4+ ∂3E2dx3dx2dx4) + (∂2B2dx2dx3dx1 + ∂4B2 dx4dx3dx1) + (∂1E3)dx1 dx3dx4 + ∂2E3dx2 dx3dx4) + ( ∂3B3dx3 dx1dx2+ ∂4B3 dx4dx1dx2) Now put all triple factors into standard order with apropos signs dα = ( - ∂2E1dx1dx2 dx4 - ∂3E1dx1dx3 dx4) + (∂1B1dx1dx2dx3+ ∂4B1 dx2dx3dx4) + ( ∂1E2dx1dx2dx4 - ∂3E2dx2dx3dx4) + (∂2B2dx1dx2dx3 - ∂4B2 dx1dx3dx4) + (∂1E3)dx1dx3dx4 + ∂2E3dx2dx3dx4) + ( ∂3B3dx1dx2dx3 + ∂4B3 dx1dx2dx4) Now group like terms dα = dx1dx2dx3 [ ∂3B3 + ∂2B2 + ∂1B1] + dx1dx2dx4 [ - ∂2E1 + ∂1E2 + ∂4B3] + dx1dx3dx4 [ - ∂3E1- ∂4B2 + ∂1E3] + dx2dx3dx4 [ ∂4B1 - ∂3E2 + ∂2E3 ] Now put terms into a reasonable ordering dα = dx1dx2dx3 [ ∂1B1 + ∂2B2 + ∂3B3] + dx1dx2dx4 [ ∂1E2- ∂2E1 + ∂4B3] - dx1dx3dx4 [ ∂3E1- ∂1E3 + ∂4B2 ] + dx2dx3dx4 [ ∂2E3 - ∂3E2 + ∂4B1] Do some more dα = dx1dx2dx3 [ div B] + dx1dx2dx4 [ [curl E]3 + ∂4B3] - dx1dx3dx4 [ [curl E]2 + ∂4B2 ] + dx2dx3dx4 [ [curl E]1 + ∂4B1] Here are the two homogeneous Maxwell equations, div B = 0 [curl E]k + ∂4Bk = 0 Thus, when we apply these two Maxwell equations ONLY, we obtain the desired result dα = 0 Part 2: Show dα = 0 in a reasonable manner 1. Starting definitions α ≡ (Eidxi)dx4 + Bi(*dxi) in R4 2-form, sums only for R3 β ≡ - (Hidxi)dx4 + Di(*dxi) in R4 2-form, sums only for R3 γ ≡ Ji(*dxi) dx4/c - ρdV in R4 3-form, sum only for R3 dV= dx1dx2dx3 NOTE: (*dxi) is always a 3D object as if we were in spatial R3. 2. Maxwell's Curl Equations curl E = -(1/c)∂tB curl H = (4π/c) J + (1/c)∂tD [curl E]i = -(1/c)∂tBi [curl H]i = (4π/c) Ji + (1/c)∂tDi Now use x4 = ct so dx4 = c dt so ∂4 = (1/c)∂t . Then we have [curl E]i = - ∂4Bi [curl H]i = (4π/c) Ji + ∂4Di Now write out [curl E]1 = - ∂4B1 = [∂2E3- ∂3E2] or [∂2E3- ∂3E2] = - ∂4B1 = - ε23k∂4Bk I think this then means [∂iEj- ∂jEi] = - εijk∂4Bk or [∂iEj- ∂jEi] + εijk∂4Bk = 0 or [∂jEi- ∂iEj] + εjik∂4Bk = 0 or [∂jEi- ∂iEj] + εjik∂4Bk = 0 (*) 3. Maxwell's Div Equations ∂iEi = 4πρ ∂iBi = 0 In all of the above, sums are i = 1 to 3. 4. d applied to fields dEi = ∂jEidxj + ∂4Ei dx4 j = 1,2,3 dBi = ∂jBidxj + ∂4Bi dx4 j = 1,2,3 5. Compute dα: α ≡ (Eidxi)dx4 + Bi(*dxi) dα = (dEidxi)dx4 + dBi(*dxi) = ( ∂jEidxj + ∂4Ei dx4)dxidx4 + (∂jBidxj + ∂4Bi dx4)(*dxi) = ( ∂jEidxj)dxidx4 + (∂jBidxj )(*dxi) + ∂4Bi dx4 (*dxi) The factor dxj(*dxi) vanishes unless i = j. For example, dx1(*dx2) = 0. Therefore the second term can be written (∂iBi)dxi(*dxi) = (∂iBi)dV = 0 dV = 0. So we are then down to dα = ( ∂jEidxj)dxidx4 + ∂4Bi dx4 (*dxi) = [ ∂jEidxjdxi + ∂4Bk(*dxk) ] dx4 Now how about this idea (*dxk) = (1/2) εkjidxjdxi For example: (*dx1) = (1/2) ε1jidxjdxi = ε123dx2dx3 = dx2dx3 (*dx2) = (1/2) ε2jidxjdxi = ε231dx3dx1 = dx3dx1 etc Then we have dα = [ ∂jEidxjdxi + ∂4Bk (1/2) εkjidxjdxi ] dx4 Then write ∂jEidxjdxi = (1/2) ( ∂jEi - ∂iEj)dxjdxi Then we get dα = [ ∂jEidxjdxi + ∂4Bk (1/2) εkjidxjdxi ] dx4 = [(1/2) ( ∂jEi - ∂iEj)dxjdxi + ∂4Bk (1/2) εkjidxjdxi ] dx4 = (1/2) [ ∂jEi - ∂iEj + ∂4Bk εkji ] dxjdxi dx4 = (1/2) [ ∂jEi - ∂iEj + εjik∂4Bk ] dxjdxi dx4 But [...] = 0 by (*) above, so we have shown dα = 0. 3. Conclusion I have had enough with this "exercise" . I showed dα = 0 two ways above. The whole set of results seems just a curiosity to me right now.