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(b1) Area. (general N) We saw at the end of Section 11 the notion of a cross product in N dimensions. For example, in (6') we were able to express the reciprocal base vectors in terms of the tangent base vectors this way ei = [1/(N-1)!] εiab... ea x eb x ec ... /v (6') for example e1 = e2 x e3 x e4 ... x eN / v This suggests the following form for our differential area element #1 : dS1 = ds2 x ds3 x ds4 .... x dsN where dsk = ek dqk where then dS1 is orthogonal to all the differential distance vectors from which it is formed, as shown at the end of Section 11. We could then write this as dS1 = (dq2 e2) x (dq3e3) x (dq4e4) .... x (dqN eN) = (dq2dq3dq4....dqN) e2 x e3 x e4 ... x eN = (dq2dq3dq4....dqN) [ v e1] The magnitude dS1 can be determined from (dS1)2 = dS1 dS1 = (dq2dq3dq4....dqN)2 v2 e1e1 = (dq2dq3dq4....dqN)2 v2 so that dS1 = (dq2dq3dq4....dqN)v = (dq2dq3dq4....dqN) = ( Πi≠1 dqi) In general, then, we have dSa = [1/(N-1)!] εabc... dsb x dsc ...... dSa = | dSa| = ( Πi≠k dqi) In N dimensions for N≠3, the notation q = a x b has no meaning so we must replace it with something that does have meaning. Consider the N=3 case again for two vectors called a(1) and a(2). The cross product is written this way (repeated indices are summed unless stated otherwise) q = a(1) x a(2) q = A x B qk = εkii a(1)i a(2)i qk = εkab Aa Bb The generalization to N dimensions is this (there must be N-1 vectors crossed together, so there will then be N-2 cross's ) q = a(1) x a(2) x a(3) x .... x a(N-1) q = A x B x C x ... x X qk = εkiii....i a(1)i a(2)i .... a(N-1)i qk = εkabc..x Aa BbCc ... Xx where this ε is the totally antisymmetric tensor in N dimensions (it has N indices), meaning that if any two indices are swapped, it changes sign. It is easy to show that this N-vector q is perpendicular to all the a(i) , just as we are used to having when N = 3: a(2) q = a(2)k qk = a(2)k εkiii....i a(1)i a(2)i .... a(N-1)i = εkiii....i a(1)i [ a(2)k a(2)i ] .... a(N-1)i In terms just of indices k and i2 and their sums, we can thing of ε as being Aki , an Antisymmetric tensor in these indices. Then we have Aki[ a(2)k a(2)i ] = – Aik[ a(2)k a(2)i ] = – Aki[ a(2)k a(2)i ] and something that is minus itself must be 0. In the last step we just swapped the names of the dummy summation indices. In general, AijSij = 0 if A is antisymmetric and S is symmetric. Therefore, we have shown that a(2) q = 0, and in general q is perpendicular to all the a(m) which appear in the cross product. We are then led to this obvious generalization of the formula for differential area in N dimensions dS1 = ds2 x ds3 x ds4 .... x dsN where as desired this area element is perpendicular to all the vector distances dsk from which it is constructed. The general case can be written dSk = Πx;i≠k(x) dsi where the x indicates the mulitplicative operation indended by Π. As for N=3, the vectors dsi skaffold an N dimensional parallelogram in N-space, and each of these vectors can be written in terms of its tangent base vector ei dsi = ei dqi i = 1...N so we can write the area element this way dSk = Πx;i≠k(x) dsi = (Πx;i≠k ei) (Πi≠k dqi) For example, in N=5 dimensions we would have dS3 = e1 x e2 x e4 x e5 dq1dq2dq4dq5 We are interested in the magnitude dSk so we have to compute the magnitude2 of (Πx;i≠k ei ): (Πx;i≠k ei ) (Πx;i≠k ei) = (Πx;i≠k ei )j (Πx;i≠k ei )j Now using our cross product ε formula from above we can write (Πx;i≠k ei )j = εjiii....i (e1)i (e2)i .... (eN)i where ε is missing index ik and the product is missing (ek)i ; the reader will just have to keep this in mind so we don't have to show it in the notation. Then we have (Πx;i≠k ei )j (Πx;i≠k ei )j = (εjiii....i (e1)i (e2)i .... (eN)i )(εjmmm....m (e1)m (e2)m .... (eN)m) = εjiii....i εjmmm....m [(e1)i(e1)m][(e2)i(e2)m] ..... [(eN)i(eN)m] Now we have to deal with this εε product. So we pause to consider: Each of the "other terms" has the same form as the first term, but the set of mi labels has been permuted (and the ii labels stay fixed). Such a term has a sign ± depending on whether this permutation is obtained by an even or odd number of index swaps from the original order. Including the first term, the total number of permutations is (N-1)!, so the portion "other terms" contains (N-1)! - 1 terms. For N=3, there is only one "other term". Notice that no mi in any term (including the first term) ever takes the value k needs repair Therefore, we have shown that (Πx;i≠k ei ) (Πx;i≠k ei) = [e1 e1] [e2 e2]..... [eN eN] + other terms where again [ek ek] is missing. In those other terms, the second set of e's is permuted. We recognize that dot products as metric tensor elements, so we have (Πx;i≠k ei ) (Πx;i≠k ei) = g'11 g'22..... g'NN + other terms Now that we have a relatively simple product, we can account for the signs due to the swaps mentioned above with and ε symbol having N-1 indices. Thus (Πx;i≠k ei ) (Πx;i≠k ei) = g'1m g'2m..... g'Nm εmmm....m // mk missing For example, if N=4 and k=2 then each index mi in the implied sums only take the values mi = 1,3,4 because we can never have In the implied sums here over the mi, we are listing off the permutations of {mi...} = { 1,2....N} where k is missing from the list. Therefore the sums are all of the form Σm≠m . We can add this reminder to the notation by writing (Πx;i≠k ei ) (Πx;i≠k ei) = Σm≠m g'1m g'2m..... g'Nm εmmm....m // mk missing So there are two issues here. (1) ε is missing index mk and g'km is missing from the product of g' objects. (2) each sum excludes mk We have therefore shown that dSk = Πx;i≠k(x) dsi = (Πx;i≠k ei) dSk = [Σm≠m g'1m g'2m..... g'Nm εjmmm....m]1/2(Πi≠k dqi) // mk missing As a check on this result, for N=2 it says dS1 = [Σm≠1g'2m g'3m εmm]1/2 dq2 dq3 = [g'2m g'3m εmm]1/2 dq2 dq3 g'2m g'3m εmm ********************************************************* Contradiction: (1) In the N=3 area section I write dSi = ½ εijk | ej x ek | dqj dqk We go off then and grab our vector identity, (A x B) (A x B) = A2B2 – (AB)2 so that | ej x ek | 2 = (ej x ek) (ej x ek) = (ejej) (ekek) - (ejek)2 = g'jjg'kk - g'jk2 giving the result dSi = ½ εijk dqj dqk (2) An example of the above would be dS1 = | e2 x e3 | dq2 dq3 = dq2 dq3 (3) But I can do this another way and get a different answer: e1 ≡ e2 x e3/v | e1|2 = e1 e1 = g'11 => | e1| = And therefore dS1 = | e2 x e3 | dq2 dq3 = v | e1| dq2 dq3 = dq2 dq3 And this disagrees with result (2) above. This is an N=3 disagreement!!! This is going to be another 20 hour problem, I see it coming. Contradiction B. Boil it down to something simpler. (1) On the one hand, we have | e2 x e3 | 2 = (e2 x e3) (e2 x e3) = (e2e2) (e3e3) - (e2e3)2 = g'22g'33 - g'232 (2) On the other hand, we have | e2 x e3 | 2 = | v e1 |2 = v2 e1 e1 = v2 g'11 = det(g') g'11 where det(g') has lower index g'ij in it, at least that is how I derived it. Is it possible that g'11 =