comments on the Jacobian

Comments on the Jacobian 6.3.09 Where is my proof that the Jacobian is the correct "volume transformation factor" when you change coordinates? The answer is that my proof is strewn around in various documents and is not really collected in any one place, so I will try to do that here. 1. In my notes "M&M chapter 4_5.doc" I review M&M's treatment of distance, area and volume in their discussion of curvilinear coordinates. In that discussion, the new coordinates are x'i = qi and the metric tensor is gij = Gij. (a) The distance situation is this: ds2 = drdr = [e(i)dqi] [e(j)dqj] = δij dqi dqj = dqi dqi = dqi (Gij dqj) = Gij dqi dqj where the main point is this: ds2 = gij dqi dqj (b) The area situation is this dS1 = ds2 x ds3 or dSi = ½ εijk dsj x dsk // sums implied dS1 = dq2 dq3 // 5-6a and in general dSi = ½ εijk dqj dqk // sums implied (c) The volume situation is this: dV = ds1 dS1 = ds1 ds2 x ds3 = e(1) e(2) x e(3) dq1 dq2 dq3 = v dq1 dq2 dq3 where v is the volume of the parallelepiped v = [e(1) e(2) x e(3)]. We start in Cartesian coordinates where dV = dx1dx2dx3 and we go to our curvilinear coordinates (which means any new coordinate system!) and "v" is the volume of the little volume parallelepiped with unit vector edges. We then scale each edge by its dqi and we get dV = v dq1 dq2 dq3, which I think is extremely clear so there is no confusion about what "v" means. I show a picture of this parallelepiped at the start of "MM Chap 4_5 integration.doc" which I replicate here: 2. Now, the claim is made that "v", which is the volume scaling factor, is given by this simple formula: v = ≡ // both sides are rotational scalars where we now use symbol "g" both to mean matrix gij and to mean the determinant of this matrix! I give a proof of this fact in a dedicated section of "M&M chapter 4_5.doc". M&M give their own proof as well. Here is a rough outline of my proof that g = v2 . We first write g = det(g) = εijk g1ig2jg3k = εijk (e(1) e(i)) (e(2) e(j)) (e(3) e(k)) and then we write v2 = [e(1) e(2) x e(3)]2 where now everything is in terms of those lower-case covariant unit vectors. I then compact down on the notation and write our desired theorem g = v2 in this way { e(1)= 1 and e(j) = j , etc. } εijk (1 i) (2 j) (3 k) = [1 2 x 3]2 which can be written out "in components" as εijk (1m im) (2n jn) (3s ks) = [1mεmns2n 3s] [1aεabc2b 3c] I then show that these two sides are equal using an argument based on totally antisymmetric tensors, see "M&M chapter 4_5.doc" for further details. 3. But how does this all relate to the Jacobian? So far we have said nothing about the Jacobian which we know is the determinant of the transformation matrix! J ≡ det(T) = | | = | Tki | If we "start in Cartesian coordinates" and move to non-Cartesians, we know that (various places) gkp = Tki Tpi = so at least we have some connection between J and v2 = det(g). The problem is to now prove that this is true: J = = // since we know from (2) above that = v I prove this in " M&M chap5 support.doc " and here is a rough outline of my proof that J2 = g. We start by writing these expressions for the items of interest, J = εstuT1sT2tT3u = jac1 J = εijkTi1Tj2Tk3 = jac2 J2 = jac2 * jac1 = = (εijk Ti1 Tj2 Tk3)( εstu T1s T2t T3u) On the other hand, we write, making use of the above fact that gkp = Tki Tpi, g = εijk g1ig2jg3k = εijk T1s Tis T2t Tjt T3u Tku //6 sums and 6 T's You can see that it is at least feasible that J2 = g since both sides involve sums of products of six T factors. I show that the equality is in fact true using an antisymmetric tensor argument which is similar to that used in the previous proof. 4. So now we gather up all the above information and state our conclusions: J = = = v // our Big Double Theorem ! where J ≡ |det(T)| = | | = | Tki | = εijkTi1Tj2Tk3 // Jacobian dV = v dq1 dq2 dq3 = dx1dx2dx3 // volume element gkp = Tki Tpi = // metric tensor so our volume transformation can be written: dx1dx2dx3 = | | dq1 dq2 dq3 // = dV which is certainly easy to remember since it looks like this single variable formula dx = ( ) dq Here are more standard notations for the Jacobian: which in my notation would be dx1dx2dx3 = | | dq1 dq2 dq3 = J(q1, q2, q3) dq1 dq2 dq3 5. I have expressed most results here for the case n=3, but at least the volume-related results are valid in any number of dimensions, though M&M don't treat the general case. Eventually I will find a book that does all this stuff right. I think these would be some of the generalizations: v2 = [1 2 x 3]2 = εijk (1 i) (2 j) (3 k) → εijkl... (1 i) (2 j) (3 k) (4l) ... = εi1i2i3..in (e(1) e(i1)) (e(2) e(i2)) ...... (e(n) e(in)) J ≡ |det(T)| = | | = | Tki | = εstuv...T1sT2tT3uT4v .... gkp = Tki Tpi = // Σi=1n implied g = det(gij) = εijkl... g1ig2jg3kg3l ... dx1dx2dx3... = | | dq1 dq2 dq3... // = dV I would first show that g = v2 as in item (2) above, then I would show that J = as in (3) above. 6. Does the above formula for v work in the case n = 2? It had better. The generalized formula for v given above applied to the case n = 2 says this: v2 = εij(1 i) (2 j) ?? We know that v = sinθ because v2 = (1 x 2 )2 . Ie, the area of a parallelogram with unit vector edges is just sinθ. But if we write out the above expression for v2, we get v2 = εij(1 i) (2 j) = (1 1) (2 2) – (12)2 = 1 - cos2θ = sin2θ QED. In the case n = 4, I don't offhand know a simple "geometric formula" for the volume of the 4D parallelepiped, so I then have to rely on the general formula.