Appendix H

Appendix H : Hodge Star, Differential Operators, Integral Theorems and Maxwell 1 H.1 Properties of the Hodge star operator in Rn 1 H.2 Gradient 5 H.3 Laplacian 5 H.4 Divergence 7 H.5 Curl 8 H.6 Exercise: Maxwell's Equations in Differential Forms 9 Installed on 5.23.16 do not edit here! Appendix H : Hodge Star, Differential Operators, Integral Theorems and Maxwell Here we study the relationship between the d and * operators, differential forms, and the classical differential operators of analysis such as the Laplacian. Some classical integral theorems are derived from the generalized Stokes' Theorem, and the Maxwell Equations are reformulated in terms of differential forms as an exercise. The cosmetic notation dxi is used throughout for the functional λi. Since λi is a functional acting on the Cartesian vector space V = Rn, up and down tensor indices are the same. H.1 Properties of the Hodge star operator in Rn Start with, dx^I = some ordered multi-index wedge product of k dxi in Rn (a basis vector k-form) . (H.1.1) This dx^I has k vectors wedged together in "standard order". The only non-zero n-form in Rn is this, dV ≡ dx1 ^ dx2 ^ ... dxn . (H.1.2) The Hodge dual object *dx^I is defined as (sign is treated below), *dx^I ≡ (sign)I,k dx^Ic // Ic = complement of I (H.1.3) where dx^Ic is the full wedge product dV in which the vectors of dx^I are deleted. Fact: Since dx^I is a k-form, *dx^I is an (n-k)-form which is "dual" to dx^I . (H.1.4) For example, for the k-form dx^I = dxa ^ dxb ^ ....^ dxq // a < b < c.... < q one has dx^Ic = dx1 ^ dx2 ^ ...[dxa] .... [dxb].....[dxq] .....^ dxn (H.1.5) where the notation [dxa] means that dxa is missing. Example: In R3 let dx^I = dx2 . Then dx^Ic = dx1 ^ dx3 . Example: In R6 let dx^I = dx2 ^ dx4 . Then dx^Ic = dx1 ^ dx3 ^ dx5 ^ dx6 . The sign (sign)I,k is selected so that the following is true : dx^I ^ (*dx^I) = dV . (H.1.6) Fact: For a k-form dx^I = dxa ^ dxb ^ ....^ dxq the sign in (H.1.3) is given by (sign)I,k = (-1)a+b+..+q (-1)k(k+1)/2 . (H.1.7) Proof: dx^I ^ (*dx^I) = (sign)I,k dx^I ^ dx^Ic = (sign)I,k ( dxa ^ dxb ^ ....^ dxq ) // first factor has k vectors wedged ( dx1 ^ dx2 ^ ...[dxa] .... [dxb].....[dxq] .....^ dxn ) . (H.1.8) The task is then to slide each of the dxi of the first factor into its corresponding "hole" in the second factor and count up the number of adjacent vector position swaps required : slide dxa to the right, number of swaps = (k-1) + (a-1). then slide dxb to the right, number of swaps = (k-2) + (b-1) then slide dxc to the right, number of swaps = (k-3) + (c-1) .... then slide dxq to the right, number of swaps = (k-k) + (q-1) . (H.1.9) Total swaps then is swaps = Σi=1k (k-i) + (a+b+...+q) - k . (H.1.10) But Σi=1k (k-i) = k Σi=1k [1] - Σi=1k [i] = k * k - k(k+1)/2 = k2/2 - k/2 so Σi=1k (k-i) - k = k2/2 - 3k/2 = (k-3)(k/2) and swaps = (k-3)(k/2) + (a+b+...+q) . (H.1.11) Then since each adjacent pairwise vector swap creates a (-1) factor according to (8.2.4), we get phase = (-1)a+b+...+q (-1)(k-3)k/2 But 1 = (-1)2k = (-1)4k/2 so (-1)(k-3)k/2 = (-1)(k-3)k/2 (-1)4k/2 = (-1)(k+1)k/2 and the result is phase = (-1)a+b+...+q (-1)(k+1)k/2 . (H.1.12) After all these "slides" are completed, equation (H.1.8) says dx^I ^ (*dx^I) = (sign)I,k * phase * dx1 ^ dx2 ^ ... dxn = (sign)I,k * phase dV . According to the requirement (H.1.6) that dx^I ^ (*dx^I) = dV we find (sign)I,k = phase = (-1)a+b+...+q (-1)(k+1)k/2 . QED Example: dx^I = dxi in Rn , so k = 1 and (-1)(k+1)k/2 = (-1)(1+1)1/2 = (-1). Then, (sign)I,k = (-1)i (-1) = (-1)i+1 = (-1)i-1 *dxi = (-1)i-1 dx1 ^ dx2 ^ ... [dxi]... ^ dxn . (H.1.13) Verify: dx^I ^ (*dx^I) = dxi ^ { (-1)i-1 dx1 ^ dx2 ^ ... [dxi]... ^ dxn } = (-1)i-1 dxi ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn = dx1 ^ dx2 ^ ... dxn = dV . One can form an n-component vector from the *dxi objects *dx ≡ ( *dx1, *dx2, ..... *dxn) . (H.1.14) One can think of *dxi = dAi or *dx = dA (H.1.15) as an element of "area" in n-1 dimensions. For R3 we have (cyclic order) *dx1 = dx2 ^ dx3 = dA1 dx^I ^ (*dx^I) = dx1 ^ ( dx2 ^ dx3) = dV *dx2 = dx3 ^ dx1 = dA2 dx^I ^ (*dx^I) = dx2 ^ ( dx3 ^ dx1) = dV *dx3 = dx1 ^ dx2 = dA3 dx^I ^ (*dx^I) = dx3 ^ ( dx1 ^ dx2) = dV (H.1.16) or dAk = *dxk = (1/2) εkij dxi ^ dxj . (H.1.17) Another useful example: Fact: *dV = 1 (H.1.18) Proof: Then dx^I ^ (*dx^I) = dV ^ (*dV ) = dV ^ 1 = dV, satisfying (H.1.6) . Fact: dxi ^ *dxj = δi,j dV (H.1.19) Proof: If i ≠ j, then dxi appears in *dxj since *dxj only has dxj missing. But then dxi appears twice, and so the wedge product must vanish, hence the factor δi,j. And then dxi ^ *dxi = dV by (H.1.6) . Fact: *(*dx^I) = (-1)kn+k dx^I (H.1.20) Proof: The requirement is that dx^I ^ (*dx^I) = dV . (H.1.6) which applied to *dx^I says *dx^I ^ *(*dx^I) = dV . (H.1.21) We know that there exists some sign such that *(*dx^I) = (sign) dx^I (H.1.22) since doing the complement twice restores all the original dxi factors. Thus, (H.1.21) says *dx^I ^ [(sign) dx^I ] = dV or (sign) *dx^I ^ dx^I = dV . Now consider *dx^I ^ dx^I = (sign') dx^I ^ *dx^I = (sign') dV (H.1.23) where (sign') arises from sliding dx^I to the left. Once we find (sign'), we then have dV = (sign) *dx^I ^ dx^I = (sign)(sign') dV so the solution to our problem is then sign = sign'. To find sign' we slide each dxi in dx^I to the left in (H.1.23). Doing so, we pick up a sign (-1)n-k since n-k is the number of vectors in *dx^I . Doing this one at a time for each of the vectors in dx^I one gets, (sign') = (-1)(n-k)k = (-1)kn-k = (-1)kn (-1)k = (-1)kn (-1)k = (-1)kn+k . Therefore *(*dx^I) = (sign) dx^I = (sign') dx^I = (-1)kn+k dx^I . QED Corollary: If α is a k-form in Rn, then *(*α) = (-1)kn+k α. (H.1.24) Proof: α = Σ'I fI dx^I *(*α) = Σ'I fI *(*dx^I) = Σ'I fI (-1)kn+k dx^I = (-1)kn+k Σ'I fI dx^I = (-1)kn+k α in agreement with Sjamaar p 28 Exercise 2.15. H.2 Gradient Start with a simple 0-form and compute dα, α = f // 0-form dα = (∂if) dxi = f dx . // (10.3.3) (H.2.1) One then has the following "Hodge correspondence", α = f 0-form in Rn α ↔ f dα = f dx 1-form in Rn dα ↔ f . (H.2.2) Apply Stokes's theorem (boundary here is two oriented endpoints of a curve C) ∫M dα = ∫∂M α ∫M f dx = ∫∂M f ∫C f dx = f(b) - f(a) . (H.2.3) When both sides are converted to regular calculus integrals (two definitions of Section 10.11), one gets ∫C f dx = f(b) - f(a) (H.2.4) which we shall call the "line integral of a gradient theorem" . H.3 Laplacian Start again with a simple 0-form and compute various interesting objects : α = f // 0-form dα = (∂if) dxi // (10.3.3) (H.3.1) *(dα) = (∂if) *dxi d(*dα) = (∂j∂if) dxj ^ *dxi // (10.3.3) = (∂j∂if) δi,j dV // (H.1.19) = (∂2if ) dV = (2f) dV *(d(*dα)) = 2f (*dV) = 2f . // (H.1.18) (H.3.2) One then has the following "Hodge correspondence", α = f 0-form in Rn α ↔ f *(d(*dα)) = 2f 0-form on Rn *(d(*dα)) ↔ 2f . (H.3.3) Consider now, β ≡ f g (*dx) = f g dA = f (∂ig) (*dxi) // (H.1.15) (H.3.4) dβ = d [ f (∂ig) ] (*dxi) = ∂j [ f (∂ig) ] dxj ^ (*dxi) // (10.3.3) = ∂j [ f (∂ig) ] δi,j dV // (H.1.19) = ∂i [ f (∂ig) ] dV = [ f (∂i2g) + (∂if)(∂ig) ] dV = [ f (2g) + f g ] dV . (H.3.5) Apply Stokes's theorem, ∫M dβ = ∫∂M β ∫M [ f (2g) + f g ] dV = ∫∂M f g dA . (H.3.6) When both sides are converted to regular calculus integrals (two definitions of Section 10.11), one gets ∫V [ f 2g + f g ] dV = ∫S f g dA = ∫S f g [ dA] = ∫S f (∂ng)dA (H.3.7) which is known as Green's 1st identity. Swapping f↔g and subtracting gives ∫V [ f 2g – g 2f] dV = ∫S [f (∂ng) - g (∂nf)] dA (H.3.8) which is Green's 2nd identity. H.4 Divergence Start this time with a 1-form and compute various interesting objects : α = Fi dxi = F dx // 1-form (H.4.1) *α = Fi (*dxi) = F *dx = F dA // (H.1.15) (H.4.2) d(*α) = ∂jFi dxj ^ (*dxi) // (10.3.3) = ∂jFi δi,j dV // (H.1.19) = (∂iFi) dV = (div F) dV (H.4.3) *(d(*α)) = (div F) *dV = (div F) . //(H.1.18) (H.4.4) One then has the following "Hodge correspondence", α = F dx 1-form on Rn α ↔ F *(d(*α)) = (div F) 0-form on Rn *(d(*α)) ↔ div F . (H.4.5) Apply Stokes' Theorem with β = *α : ∫M dβ = ∫∂M β ∫M d(*α) = ∫∂M (*α) ∫M div F dV = ∫∂M F dA . (H.4.6) When both sides are converted to regular calculus integrals (two definitions of Section 10.11), one gets ∫V div F dV = ∫S F dA (H.4.7) which is the divergence theorem in n dimensions. For R3 this is Gauss's Theorem. H.5 Curl Start again with a 1-form and compute objects of interest: α = ΣjFj dxj = F dx // 1-form (H.5.1) dα = Σi<j (∂iFj - ∂jFi) dxi ^ dxj // (10.3.24b), dα written in standard form (H.5.2) *(dα) = Σi<j (∂iFj - ∂jFi) *(dxi ^ dxj) // on next line specialize to R3 : = (∂1F2 - ∂2F1) *(dx1 ^ dx2) + (∂1F3 - ∂3F1) *(dx1 ^ dx3) + (∂2F3 - ∂3F2) *(dx3 ^ dx2) = (∂1F2 - ∂2F1) *(dx1 ^ dx2) + (∂3F1 - ∂1F3) *(dx3 ^ dx1) + (∂2F3 - ∂3F2) *(dx3 ^ dx2) = (curl F)3 dx3 + (curl F)2 dx2 + (curl F)1 dx1 = (curl F) dx . (H.5.3) One then has the following "Hodge correspondence", α = F dx 1-form in R3 α ↔ F *(dα) = [curl F] dx 1-form in R3 *(dα) ↔ curl F . (H.5.4) Apply Stokes' Theorem in Rn to get ∫M dα = ∫∂M α ∫M Σi<j (∂iFj - ∂jFi) dxi ^ dxj = ∫∂M F dx (H.5.5) In R2 there is only one term in the sum on the left and one gets, ∫M (∂1F2 - ∂2F1) dx1 ^ dx2 = ∫∂M [ F1dx1 + F2 dx2 ] . (H.5.6) When both sides are converted to regular calculus integrals (two definitions of Section 10.11), we get ∫A (∂1F2 - ∂2F1) dx1dx2 = ∫C [ F1dx1 + F2dx2] . (H.5.7) Setting x1 = x, x2 = y, F1 = f and F2 = g one gets ∫A (∂xg - ∂yf) dxdy = ∫C [ fdx + gdy] (H.5.8) which is known as Green's Theorem in a plane. In R3 , we can write out the three terms on the left side of (H.5.2) dα = (∂1F2 - ∂2F1) dx1 ^ dx2 + (∂1F3 - ∂3F1) dx1 ^ dx3 + (∂2F3 - ∂3F2) dx2 ^ dx3 = (∂1F2 - ∂2F1) dA3 + (∂1F3 - ∂3F1) [-dA2 ] + (∂2F3 - ∂3F2) dA1 = (∂2F3 - ∂3F2) dA1 + (∂3F1 - ∂1F3) dA2 + (∂1F2 - ∂2F1) dA3 = (curl F)1 dA1 + (curl F)2 dA2 + (curl F)3 dA3 = (curl F) dA . (H.5.9) Then Stokes' Theorem says ∫M dα = ∫∂M α ∫M (curl F) dA = ∫∂M F dx . (H.5.10) When both sides are converted to regular calculus integrals (two definitions of Section 10.11), we get ∫A (curl F) dA = ∫C F dx = C F dx (H.5.11) which is the traditional Stokes' Theorem in R3 where C is the boundary of the area A. Note that the boundary C and its enclosed area A can be non-planar. H.6 Exercise: Maxwell's Equations in Differential Forms This section is based on Sjamaar p 30 Exercise 2.23, but we use SI units instead of cgs units. Maxwell's equations in SI units are, curl H = ∂tD + J Maxwell curl H equation curl E = - ∂tB Maxwell curl E equation div D = ρ Maxwell div D equation div B = 0 . Maxwell div B equation (H.6.1) Write these equations in components and think of time cdt = dx4, so we are working here in spacetime R4. The metric tensor is ±diag(1,1,1,-1) but this fact has no effect on the presentation below. (∂iHj- ∂jHi) - εijk(∂4Dk) = εijk Jk // for example, (∂1H2- ∂2H1) - ∂4D3 = J3 (∂iEj- ∂jEi) + εijk(∂4Bk) = 0 ∂iDi = ρ ∂iBi = 0 . (H.6.2) In the above, indices i,j,k range from 1 to 3 and all implied sums have this range. Define two differential 2-forms α and β as follows, α ≡ (E dx) ^ dx4 + B dA // dA = *dx β ≡ - (H dx) ^ dx4 + D dA (H.6.3) where dx and dA and dV refer to R3 objects as used in earlier sections above. Start with α written in components and compute dα. Again, all implied sums are summed 1 to 3. Then, α = Ej dxj ^ dx4 + Bj *dxj dα = Σi=14 (∂iEj) dxi ^ dxj ^ dx4 + Σi=14 (∂iBj) dxi ^ *dxj // (10.3.6) = (∂iEj) dxi ^ dxj ^ dx4 + (∂iBj) dxi ^ *dxj + (∂4Ej) dx4 ^ dxj ^ dx4 + (∂4Bj) dx4 ^ *dxj . The third term vanishes since there are two dx4 vectors present. In the second term use dxi ^ *dxj = δi,j dxi ^ *dxi = δi,j dV // (H.1.19) so (∂iBj) dxi ^ *dxj = (∂iBj)δi,j dV = (∂iBi) dV . Then dα = (∂iEj) dxi ^ dxj ^ dx4 + (∂iBi) dV + 0 + (∂4Bk) dx4 ^ *dxk = (∂iEj) dxi ^ dxj ^ dx4 + (∂4Bk) dx4 ^ *dxk + (∂iBi) dV . (H.6.4) Recall from (H.1.17) that *dxk = dAk = (1/2) εkij dxi ^ dxj = (1/2) εijk dxi ^ dxj . (H.1.17) Using this fact, and writing the first term in dα as two terms, we find dα = (1/2) (∂iEj - ∂jEi) dxi ^ dxj ^ dx4 + (∂4Bk) (1/2) εijk dx4 ^ dxi ^ dxj + (∂iBi) dV = (1/2) [ (∂iEj - ∂jEi) + εijk(∂4Bk) ] dxi ^ dxj ^ dx4 + (∂iBi) dV . (H.6.5) According to Maxwell's equations (H.6.2) each of these terms vanishes so the result is simply dα = 0 . (H.6.6) The form β in (H.6.3) is the same as α with replacements: E → -H and B → D. We can then convert result (H.6.5) to get dβ = (1/2) [- (∂iHj - ∂jHi) + εijk(∂4Dk) ] dxi ^ dxj ^ dx4 + (∂iDi) dV . (H.6.7) According to Maxwell's equations (H.6.2) we then get (writing the result many ways), dβ = (1/2) [ - εijk Jk ] dxi ^ dxj ^ dx4 + ρ dV = - Jk [ (1/2) εijk dxi ^ dxj] ^ dx4 + ρ dV = - Jk dAk ^ dx4 + ρ dV = - (J dA) ^ dx4 + ρ dV = - Jk *dxk ^ dx4 + ρ dV = - (J *dx) ^ dx4 + ρ dV . (H.6.8) Now compute d(dβ) = d [ - Jk *dxk ^ dx4 + ρ dV ] = - Σj=14 (∂jJk) dxj ^ *dxk ^ dx4 + Σj=14 (∂jρ) dxj ^ dV = - (∂jJk) dxj ^ *dxk ^ dx4 + (∂jρ) dxj ^ dV - (∂4Jk) dx4 ^ *dxk ^ dx4 + (∂4ρ) dx4 ^ dV = - (∂jJk) δj,k dV ^ dx4 + 0 // (H.1.19) - 0 - (∂4ρ) dV ^dx4 = - [(∂jJj) + (∂4ρ) ] dV ^dx4 = - [div J + (∂tρ) ] dV ^dx4 . (H.6.9) But d2β = 0 from (10.3.10) so we conclude that div J + (∂tρ) = 0 (H.6.10) which is the well-known equation of continuity stating that charge is conserved, - ∂t[∫V ρ dV] = ∫S J dS . (H.6.11) "charge enclosed in V decreases at a rate equal to the current flowing out through boundary S" Here then is a summary of our results α ≡ (E dx) ^ dx4 + B dA // dA = *dx dα = 0 curl E = - ∂tB and divB = 0 β ≡ - (H dx) ^ dx4 + D dA dβ = - (J dA) ^ dx4 + ρ dV curl H = ∂tD + J and divD = ρ d2β = 0 div J + (∂tρ) = 0 . // ∂μJμ = 0 (H.6.12) In free space where J = ρ = 0, Maxwell's Equations take this impressively simple form, dα = 0 dβ = 0 . (H.6.13)