old Section 15 g

This is Section 15 (g) retired on this day PhL 3.27.12 (g) vector Laplacian (Section 13) This differential operator is going to take a bit more work. In Section 13 the vector Laplacian is written (instead of 2) and in Cartesian coordinates is defined by B ≡ grad(div B) – curl (curl B) In components this reads (notice that ∂aεbde = εbde∂a ) (B)n = ∂n(∂jBj) - εnab∂a [curl B]b = ∂n(∂jBj) - εnab∂a (εbde∂dBe) In Cartesian space up and down index position does not matter and we are just jockeying the indices in search of a tensorized form with contracted indices where possible. Writing the above in comma notation gives (B)n = (Bj,j),n – εnab(εbdeBe,d),a There are several "technical difficulties" visible here. First, the object (εbdeBe,d) is a vector density of weight -1 and if we replace ,a with ;a, the object enclosed is not a true tensor and things don't work right at all. We repair this problem by adding a factor g-1/2 which is a scalar density of weight +1, to get (B)n = (Bj,j),n – εnab(g-1/2εbdeBe,d),a The next problem is that the first term, upon semicolonization, will become a normal vector, but the second term will be a vector density of weight -1 due to the outside εnab factor. The repair is the same and then one has (B)n = (Bj,j),n – g-1/2εnab(g-1/2εbdeBe,d),a Of course all this time g = 1 in the Cartesian space, so these added factors "do no harm". Having threaded ourselves to this point, we are ready to change commas to semicolons to get (B)n = (Bj;j);n – g-1/2εnab(g-1/2εbdeBe;d);a Since Γ = 0 in x-space, this still agrees with the starting point. But now we have a valid tensor equation and such equations are covariant as in Section 7 (u), so in x'-space the equation will be (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2ε'bdeB'e;d);a (*) Using our usual en expansion for a normal vector, one obtains B = (B)'nen What remains now is to process the expression for (B)'n and show that it agrees with what was found in Section 13. First, consider this object which appears in (*) C'b = (g'-1/2ε'bdeB'e;d) C'b is a normal vector, and from the examples in Appendix F (i) we know that C'b;a = ∂'a C'b – Γ 'nbaC'n = C'b,a – Γ 'nbaC'n . Adding the external ε'nab factor which appears in (*), ε'nab C'b;a = ε'nab C'b,a – ε'nab Γ 'nbaC'n we learn that the second term vanishes because ε'nab (the permutation tensor) is antisymmetric on a↔b while Γ 'nba is symmetric. Therefore ε'nab(g'-1/2ε'bdeB'e;d);a = ε'nab(g'-1/2ε'bdeB'e;d),a and (*) now becomes, (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2ε'bdeB'e;d),a By the exact same argument ( or see Appendix D (h) ) one has ε'bdeB'e;d = ε'bdeB'e,d so then (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2ε'bdeB'e,d),a Finally, the Appendix F (i) example that says D;n = D,n tells us that (B'j;j);n = (B'j;j),n , since the object B'j;j is a scalar like D. Finally then one has (B)'n = (B'j;j),n – g'-1/2ε'nab(g'-1/2ε'bdeB'e,d),a What the left hand putteth in, the right hand taketh away, for three of the four semicolons in equation (*) above. Now we install ε'bde = g' ε'bde ( Appendix D (e)) in the second term to get (B)'n = (B'j;j),n – g'-1/2ε'nab(g'-1/2 g' ε'bde B'e,d),a = (B'j;j),n – g'-1/2ε'nab ε'bde (g'1/2 B'e,d),a since ε'bde is a constant (the permutation tensor). The object (B'j;j) we know from section (c) above [divB] = B'j;j = (1/) ∂'j (B'j) so the final result replacing the commas is this (B)'n = ∂'n [(1/) ∂'j (B'j)] – (1/) ε'nab ε'bde ∂'a( ∂'dB'e) and changing j→i, a→c, b→x , then d,e→a,b, and finally x→ d we get, (B)'n = ∂'n [(1/) ∂'i (B'i)] – (1/) ε'ncd ε'dab ∂'c( ∂'aB'b) and this agrees with the results of Section 13 which we now quote: B = G – V(1) G = ∂'n{ (1/) ∂'i ( B'i)} en B = B'nen V(1) = [ (1/) ε'ncd ε'dab ∂'c{∂'aB'b}] en