scraps

This is the Title PhL 1.11.15 Note that page numbering is turned on in this template and view is 125%, located in phil/roaming/microsoft/templates size about 219K. 3. Divergence This section is very similar to the preceding one, but now we start with a 1-form. See // comments above. α = Σifi dxi // 1-form *α = Σifi *dxi = f *dx = Σifi (-1)i-1 dx1 ^ dx2 ^ ... [dxi]... ^ dxn d(*α) = Σi (-1)i-1Σj (∂jfi) dxj ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn = Σi (-1)i-1Σj (∂jfi) δi,j dxj ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn = Σi (∂ifi) (-1)i-1dxi ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn = Σi(∂ifi) dx1 ^ dx2 ^ ... dxi... ^ dxn = Σi(∂ifi) dx1 ^ dx2 ^ ...^ dxn *(d(*α)) = Σi(∂ifi) *(dx1 ^ dx2 ^ ...^ dxn) = Σi(∂ifi) 1 = ( f) *(d(*α)) = ( f ) Apply Stokes' Theorem : ∫M dβ = ∫∂M β ∫M d(*α) = ∫∂M *α ∫M f dx1 ^ dx2 ^ ...^ dxn = ∫∂M f *dx ∫M f d = ∫∂M f *dx ********************** 2. Laplacian α = f // 0-form α ↔ f dα = Σi (∂if) dxi // (10.3.3) dα ↔ f *(dα) = [Σi (∂if)] (*dxi) // for next two lines, see above (10.3.1) on Hodge * = [Σi (∂if)] (-1)i-1 dx1 ^ dx2 ^ ... [dxi]... ^ dxn // means dxi is missing // (sign)I,k = (-1)a+b+..+q (-1)k(k+1)/2 = (-1)i (-1)1(1+1)/2 = (-1)i+1 = (-1)i-1 d(*dα) = Σi(-1)i-1 d[(∂if)] ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn // next line is (10.3.3) f → ∂if = Σi(-1)i-1 Σj(∂j∂if)] dxj ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn // but j ≠ i 0 by (7.2.5), so = Σij(-1)i-1(∂j∂if) δi,j dxj ^ dx1 ^ dx2 ^ ... [dxi]... ^ dxn // symmetric sum Σij (10.3.8) = Σi(∂2if) (-1)i-1 dxi ^ dx1 ^ dx2 ^ .. [dxi]... ^ dxn // use up δi,j = [Σi(∂2if)] dx1 ^ dx2 ^ ... dxi... ^ dxn // slide dxi into position, i-1 swaps = [Σi(∂2if)] dx1 ^ dx2 ^ ...^ dxn *(d(*dα)) = [Σi(∂2if)] *(dx1 ^ dx2 ^ ...^ dxn) = [Σi(∂2if)] 1 = 2f *(d(*dα)) = 2f ************ Now consider ( x f )i = εijk ∂jfk dAi = (1/2) εiab dxa ^ dxb Note back up to dα = Σi<j (∂ifj - ∂jfi) dxi ^ dxj = ( x f ) *dx Clean this up tomorrow!! Proof: *dA1 = dx1 ^ dx2 Apply Stokes' Theorem with β = α ∫M dβ = ∫∂M β ∫M dα = ∫∂M α ∫M Σi<j (∂ifj - ∂jfi) dxi ^ dxj = ∫∂M f dx ************************ Can you just install these into Stokes' theorem? 3. Divergence ∫M dβ = ∫∂M β I want to see this for the divergence theorem ∫M ( f) = ∫∂M f dA So I need β = f dA = f12 dx1 ^ dx2 + f13 dx1 ^ dx3 + f23 dx2 ^ dx3 dβ = ( f) = *(d(*α)) But β = f12 dx1 ^ dx2 + f13 dx1 ^ dx3 + f23 dx2 ^ dx3 = F3 dx1 ^ dx2 + F2 dx3 ^ dx1 + F1 dx2 ^ dx3 = F1 dx2 ^ dx3 + F2 dx3 ^ dx1 + F3 dx1 ^ dx2 = F dA dβ = dF1 ^ dx2 ^ dx3 + dF2 ^ dx3 ^ dx1 + dF3 ^ dx1 ^ dx2 = Σi(∂iF1) dxi ^ dx2 ^ dx3 + Σi(∂iF2) dxi ^ dx3 ^ dx1 + Σi(∂iF3) dxi ^ dx1 ^ dx2 = (∂1F1) dx1 ^ dx2 ^ dx3 + (∂2F2) dx2 ^ dx3 ^ dx1 + (∂3F3) dx3 ^ dx1 ^ dx2 = [ ∂1F1+ ∂1F1+ ∂1F1] dx1 ^ dx2 ^ dx3 = ( F) dx1 ^ dx2 ^ dx3 ∫M dβ = ∫∂M β ∫M ( F) dx1 ^ dx2 ^ dx3 = ∫∂M F dA So how does this generalize to n dimensions. I have missed something here! I showed already far above that *α = Σifi *dxi = f *dx d(*α) = ( f ) dx1 ^ dx2 ^ ...^ dxn So then try Stokes this way ∫M dβ = ∫∂M β ∫M d(*α) = ∫∂M *α ∫M ( f ) dx1 ^ dx2 ^ ...^ dxn = ∫∂M f *dx // divergence theorem I did all this once, where is it? I now think this is a good idea to do all this stuff localized in a new appendix and reference it from the motivation area. Then main is not cluttered. ************ Now specialize to the case of R3, in which case there are only three terms in this sum : i=1 j = 2: (-1)1+2+1(∂1f2 - ∂2f1) dx3 // dx1 and dx2 are missing i=1 j = 3: (-1)1+3+1(∂1f3 - ∂3f1) dx2 // dx1 and dx3 are missing i=2 j = 3: (-1)2+3+1(∂2f3 - ∂3f2) dx1 . // dx2 and dx3 are missing Therefore *(dα) = (∂1f2 - ∂2f1) dx3 - (∂1f3 - ∂3f1) dx2 + (∂2f3 - ∂3f2) dx1 = (∂1f2 - ∂2f1) dx3 + (∂3f1 - ∂1f3) dx2 + (∂2f3 - ∂3f2) dx1 = ( x f )3 dx3 + ( x f )2 dx2 + ( x f )1 dx1 = ( x f ) dx . So we have the following "Hodge correspondence" *(dα) ↔ ( x f) Now back up to dα = Σi<j (∂ifj - ∂jfi) dxi ^ dxj Write out the three terms 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 = ( x f)1 dA1 + ( x f)2 dA2 + ( x f)3 dA3 = ( x f) dA Apply Stokes' Theorem with β = α ∫M dβ = ∫∂M β ∫M dα = ∫∂M α ∫A ( x f) dA = ∫C f dx When both sides are converted to regular calculus integrals (two definitions of Section 10.11), we get ∫A ( x f) dA = ∫C f dx = C f dx which is the traditional Stokes' Theorem where C is the boundary of the area A. Back up now to the general Rn result above, α = Σjfj dxj = f dx dα = Σi<j (∂ifj - ∂jfi) dxi ^ dxj ∫M dα = ∫∂M α ∫M Σi<j (∂ifj - ∂jfi) dxi ^ dxj = ∫∂M f dx In R2 this becomes ∫M (∂1f2 - ∂2f1) dx1 ^ dx2 = ∫∂M [ f1dx1 + f2dx2] When both sides are converted to regular calculus integrals (two definitions of Section 10.11), we get ∫A (∂1f2 - ∂2f1) dx1dx2 = ∫C [ f1dx1 + f2dx2] Setting x1 = x, x2 = y, f1 = f and f2 = g one gets ∫A (∂xg - ∂yf) dxdy = ∫C [ fdx + gdy] which is known as Green's Theorem in a plane.