show ellipsoidal normals are orthogonal

This is the Title PhL 3.26.05 In retrospect, the first section A is all one should read here. I don't have anything better than that. In Section A I use my ni = (gi) rule on the conicoid equations to compute the normals on the three surfaces, and then I just dot them together and then use the x,y,z equations and find n1n2 = 0 etc. A. Geometric reason why the ellipsoidal level-surface normals are orthogonal. 1 B. An earlier less focused version of A. 3 1. The ellipsoidal level surfaces and the x,y,z equations 3 2. Finding the normal vectors at some point x 3 3. Show that n1 n2 = 0 5 C. Another version trying to make better use of indices. 6 1. The ellipsoidal level surfaces and the x,y,z equations 6 2. Restate the above in a more regular notation 7 3. Finding the normal vectors at some point x 8 3. Show that the three normal vectors are mutually perpendicular. 8 D. Rehash of the "metric tensor method". 11 __________________________________________________________________________________ Overview In Section A I use my ni = (gi) rule on the conicoid equations to compute the normals on the three surfaces at some point x on the surface, and then I just dot them together and then use the x,y,z equations and find n1n2 = 0 etc. Here is a summary. The conicoid equations are gi(x,y,z) ≡ x2/( ξi2-a2) + y2/( ξi2- b2) + z2/(ξi2- c2) - 1 = 0 i=1,2,3 c < b < a The Cramer's rule x,y,z equations are the following, and a,b,c cyclic for y2 and z2 x2 = (ξ12-a2) (ξ22-a2) (ξ32-a2)/ [(a2-c2)(a2-b2)] The normals are ni = (gi) = 2[ x/( ξi2-a2)+ y/( ξi2-b2)+ z/( ξi2-c2)] i = 1,2,3 so the dot product of two normals becomes, when x2 etc are substituted in, ni nj/4 = [x2] / [(ξi2-a2) (ξj2-a2)] + cyclic (x,y,z; a,b,c) = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)(ξi2-a2) (ξj2-a2)] + cyclic (a,b,c) When i and j are different, let k be "the third index". We then get at once ni nj/4 = [(ξk2-a2) / [(a2-c2)(a2-b2)] + cyclic = [ (ξk2-a2) (b2-c2) + cyclic ] /[ (a2-c2) (a2-b2) (b2-c2)] But direct calculation shows that (b2-c2)+ cyclic = 0 AND a2(b2-c2) + cyclic = 0 So the numerator is 0 and we then have ni nj = 0 for i ≠ j. We know that close to some point the three level-surfaces in general form a parallelogram with tangent base vectors ei. But if the three parallelogram face normals ni are orthogonal, then so are the ei and our coordinate system is then orthogonal. Section B is where I started and I give there a first-cut proof that n1n2 = 0 with various extra steps and non-relevant side calculations. It was not obvious to me at the point that n2n3 = 0 for example. In Section C, then, I replaced a,b,c with a1, a2, a3 and x,y,x with x1, x2, x3 . I hoped this more organized approach would shed light on things, but in fact it made things more obscure. In Section D I rehash the metric tensor method, talking about the meaning of the base vectors and drawing the parallelogram picture. But this really added nothing to an understanding of why the normals are orthogonal. I cannot help but wonder if the first person to discover this fact was surprised. Who might that person have been? Web scan reveals nada, same for internal book scan. Here is a little I think I will give my prize to Lamé in 1837. _________________________________________________________________________________ A. Geometric reason why the ellipsoidal level-surface normals are orthogonal. The equations for the three ellipsoidal level surfaces are all the same, having the form for i = 1,2,3 gi(x,y,z) ≡ x2/( ξi2-a2) + y2/( ξi2- b2) + z2/(ξi2- c2) - 1 = 0 c < b < a The nature of the surface (ellipsoid, bloid..) is determined by the value of parameter ξ relative to a,b,c: c < ξ3 < b < ξ2 < q < ξ1 bloid' bloid ellipsoid The normal to any of these surfaces is given by (see my geometry doc) ni = (gi) so consider: ni = ( gi) = 2[ x/( ξi2-a2)+ y/( ξi2-b2)+ z/( ξi2-c2)] n1 = ( g1) = 2[ x/( ξ12-a2)+ y/( ξ12-b2)+ z/( ξ12-c2)] n2 = ( g2) = 2[ x/( ξ22-a2)+ y/( ξ22-b2)+ z/( ξ22-c2)] n3 = ( g3) = 2[ x/( ξ32-a2)+ y/( ξ32-b2)+ z/( ξ32-c2)] Taking the dot product gives, for example, n1 n2/4 = [x2] / [(ξ12-a2) (ξ22-a2)] + [y2] / [(ξ12-b2) (ξ22-b2)] + [z2] / [(ξ12-c2) (ξ22-c2)] We want x,y,z to be not just any point, but a point at the intersection of the three level surfaces with labels ξ1, ξ2 and ξ3. That is to say, we want x,y,z to be a point whose ellipsoidal coordinates are ξ1, ξ2, ξ3. If we write out our three gi equations above, we can solve for x,y,z using a 3x3 Cramer's rule with this result x2 = (ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)] y2 = (ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)] z2 = (ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)] // the "xyz equations" Given any equation, you get the next by doing a cyclic rotation of a,b,c. If we now insert these values into our dot product, we get n1 n2/4 = [x2] / [(ξ12-a2) (ξ22-a2)] // from above + [y2]/ [(ξ12-b2) (ξ22-b2)] + [z2]/ [(ξ12-c2) (ξ22-c2)] = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)]] / [(ξ12-a2) (ξ22-a2)] // Cramer's in + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)]]/ [(ξ12-b2) (ξ22-b2)] + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)]]/ [(ξ12-c2) (ξ22-c2)] = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)(ξ12-a2) (ξ22-a2)] // a rewrite + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)(ξ12-b2) (ξ22-b2)] + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)(ξ12-c2) (ξ22-c2)] = [(ξ32-a2)/ [ (a2-c2)(a2-b2)] // after cancellations + [(ξ32-b2)/ [ (b2-a2)(b2-c2)] + [(ξ32-c2)/ [ (c2-b2)(c2-a2)] => (n1 n2/4) (a2-c2) (a2-b2) (b2-c2) = (ξ32-a2) (b2-c2) + (ξ32-b2) (c2-a2) + (ξ32-c2) (a2-b2) = Aξ32 + B = 0ξ32 + 0 = 0 // just do it! More generally we have ni nj/4 = [x2] / [(ξi2-a2) (ξj2-a2)] + [y2] / [(ξi2-b2) (ξj2-b2)] + [z2] / [(ξi2-c2) (ξj2-c2)] = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)(ξi2-a2) (ξj2-a2)] // Cramer's and rewrite + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)(ξi2-b2) (ξj2-b2)] + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)(ξi2-c2) (ξj2-c2)] If i ≠ j, we always get two cancellations on each line. If t is the letter that i and j are not, then we get ni nj/4 = = [(ξt2-a2)/ [ (a2-c2)(a2-b2)] // after cancellations + [(ξt2-b2)/ [ (b2-a2)(b2-c2)] + [(ξt2-c2)/ [ (c2-b2)(c2-a2)] and this is the same form we had above with t = 3, but the value of t does not matter and we get 0. QED. Suppose i and j are the same. In this case, let r and s be "the other two indices" and we get ni ni/4 = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)(ξi2-a2)2] // the rewrite + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2) (ξi2-b2)2] + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2) (ξi2-c2)2] = [(ξr2-a2) (ξs2-a2) / [ (a2-c2)(a2-b2)(ξi2-a2)] // after cancellation + [(ξr2-b2) (ξs2-b2) / [ (b2-a2)(b2-c2)(ξi2-b2)] + [(ξr2-c2) (ξs2-c2) / [ (c2-b2)(c2-a2)(ξi2-c2)] and this tells us the length2/4 of our normal vector ni as defined by ni = ( gi). It is not very simple. B. An earlier less focused version of A. 1. The ellipsoidal level surfaces and the x,y,z equations Here we have the ellipsoid and the two bloids for the ellipsoidal level surfaces x2/( ξ12-a2) + y2/( ξ12- b2) + z2/(ξ12- c2) = 1 c < b < a < ξ1 ellipsoid - x2/(a2- ξ22) + y2/( ξ22- b2) + z2/(ξ22- c2) = 1 c < b < ξ2 < a bloid - x2/(a2- ξ32) - y2/(b2- ξ32) + z2/(ξ32- c2) = 1 c < ξ3 < b < a bloid' We know that we can obtain the x,y,z equations using Cramer's rule and we find (in "ellipsoidal Cramer's rule...doc") that, for general c, x2 = (ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)] y2 = (ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)] z2 = (ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)] where things are cyclic in a,b,c. 2. Finding the normal vectors at some point x Now, the first conicoid equation above (the ellipsoid) can be written as g1(x,y,z; ξ1)= 0 We know from "surface geometry questions.doc" that the normal to this surface is given by 1 = (g1) / |(g1)| n1 = (g1) // unnormalized where the direction is that of maximum positive change, and where is the usual 3D gradient. How do we compute this gradient? g1(x,y,z) = x2/( ξ12-a2) + y2/( ξ12- b2)+ z2/(ξ12- c2)- 1 ∂x g1(x,y,z) = 2x/( ξ12-a2) ∂y g1(x,y,z) = 2y/( ξ12-b2) ∂z g1(x,y,z) = 2z/( ξ12-c2) n1 = (g1) = 2[ x/( ξ12-a2)+ y/( ξ12-b2)+ z/( ξ12-c2)] What is the magnitude of this normal? n12 = n1 n1 = 4 [x/( ξ12-a2)+ y/( ξ12-b2)+ z/( ξ12-c2)] [x/( ξ12-a2)+ y/( ξ12-b2)+ z/( ξ12-c2)] = 4 [x2/( ξ12-a2)2 + y2/( ξ12-b2)2 + z2/( ξ12-c2)2 ] Now notice this fact ∂ξ1 ( ξ12-a2)-1 = -2ξ1( ξ12-a2)-2 ∂ξ1 g1 = -2ξ1 [x2/( ξ12-a2)2 + y2/( ξ12-b2)2 + z2/( ξ12-c2)2 ] => 2[x2/( ξ12-a2)2 + y2/( ξ12-b2)2 + z2/( ξ12-c2)2 ] = - ∂ξ1 g1/ ξ1 We thus obtain this simple form for n12 (but we shall not make use of it, at least not here) n12 = -2 (∂ξ1 g1)/ ξ1 We can write this another way by using our x,y,z equations so that ( we now set c = 0) n12/4 = x2/( ξ12-a2)2 + y2/( ξ12-b2)2 + z2/( ξ12-c2)2 = x2/( ξ12-a2)2 + y2/( ξ12-b2)2 + z2/( ξ12-c2)2 = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)]]/( ξ12-a2)2 + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)]]/( ξ12-b2)2 + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)]]/( ξ12-c2)2 = [ (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)(ξ12-a2)] + [ (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)(ξ12-b2)] + [(ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)(ξ12-c2) ] which is not very enlightening (we can see that it is still cyclic). The gradients of the other two surfaces are the same with a change in ξi, so here are our three normals at the point (x,y,z): n1 = (g1) = 2x/( ξ12-a2)+ 2y/( ξ12-b2)+ 2z/( ξ12-c2) n2 = (g2) = 2x/( ξ22-a2)+ 2y/( ξ22-b2)+ 2z/( ξ22-c2) n3 = (g3) = 2x/( ξ32-a2)- 2y/( ξ32-b2)+ 2z/( ξ32-c2) or in general ni = (gi) = 2x/( ξi2-a2)- 2y/( ξi2-b2)+ 2z/( ξi2-c2) 3. Show that n1 n2 = 0 Consider n1 n2 = 4 [x/( ξ12-a2)+ y/( ξ12-b2)+ z/( ξ12-c2)] [ x/( ξ22-a2)+ y/( ξ22-b2)+ z/( ξ22-c2)] = 4{ x2/ [(ξ12-a2) (ξ22-a2)] + y2/ [(ξ12-b2) (ξ22-b2)] + z2/ [(ξ12-c2) (ξ22-c2)] } Now once again let's use the x,y,z equations here: n1 n2/4 = [x2] / [(ξ12-a2) (ξ22-a2)] + [y2]/ [(ξ12-b2) (ξ22-b2)] + [z2]/ [(ξ12-c2) (ξ22-c2)] = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)]] / [(ξ12-a2) (ξ22-a2)] + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)]]/ [(ξ12-b2) (ξ22-b2)] + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)]]/ [(ξ12-c2) (ξ22-c2)] = [(ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)(ξ12-a2) (ξ22-a2)] + [(ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)(ξ12-b2) (ξ22-b2)] + [(ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)(ξ12-c2) (ξ22-c2)] = [(ξ32-a2)/ [ (a2-c2)(a2-b2)] + [(ξ32-b2)/ [ (b2-a2)(b2-c2)] + [(ξ32-c2)/ [ (c2-b2)(c2-a2)] (n1 n2/4) (a2-c2) (a2-b2) (b2-c2) = (ξ32-a2) (b2-c2) - (ξ32-b2) (a2-c2) + (ξ32-c2) (a2-b2) This is a sum of 12 terms, and Maple or by hand tells us the result is 0. So we have shown that n1 n2 = 0 Summary: We have computed n12 and found that n1 n2 = 0, which means the ellipsoid normal and the first bload normal are in fact orthogonal at the same point x. We would like to look at these two results and immediately come up with an expression for n22 and n1 n3 , say, but it is not obvious how to do this without reworking all the algebra pretty much from scratch. My feeling is that there should be a better notation which will give us results for ni2 and ni nj. That "better notation" is the subject of the next Part of this doc. C. Another version trying to make better use of indices. 1. The ellipsoidal level surfaces and the x,y,z equations Here we have the ellipsoid and the two bloids for the ellipsoidal level surfaces x2/( ξ12-a2) + y2/( ξ12- b2) + z2/(ξ12- c2) = 1 c < b < a < ξ1 - x2/(a2- ξ22) + y2/( ξ22- b2) + z2/(ξ22- c2) = 1 c < b < ξ2 < a - x2/(a2- ξ32) - y2/(b2- ξ32) + z2/(ξ32- c2) = 1 c < ξ3 < b < a which is written here in a form where the denominators are always positive. We can of course write this in the following form x2/( ξ12-a2) + y2/( ξ12- b2)+ z2/(ξ12- c2) = 1 c < b < a < ξ1 x2/( ξ22-a2) + y2/( ξ22- b2)+ z2/(ξ22- c2) = 1 c < b < ξ2 < a x2/( ξ32-a2) + y2/( ξ32- b2)+ z2/(ξ32- c2) = 1 c < ξ3 < b < a so that all three equations look exactly the same apart from the different ξi parameters. We know that we can obtain the x,y,z equations using Cramer's rule and we find (in "ellipsoidal Cramers rule...doc") that, for general c, x2 = (ξ12-a2) (ξ22-a2) (ξ32-a2)/ [ (a2-c2)(a2-b2)] y2 = (ξ12-b2) (ξ22-b2) (ξ32-b2)/ [ (b2-a2)(b2-c2)] z2 = (ξ12-c2) (ξ22-c2) (ξ32-c2)/ [ (c2-b2)(c2-a2)] where things are cyclic in a,b,c. 2. Restate the above in a more regular notation Suppose we define a1 = a, a2= b, a3 = c x1 = x, x2= y, x3= z Then any of our level surface equations in the second group become x12/( ξi2-a12) + x22/( ξi2- a22)+ x32/(ξi2- a32) = 1 i = 1,2,3 or Σk=13 xk2/ ( ξi2-ak2) = 1 i = 1,2,3 The distinction between the three surface shapes is this ellipsoid a3 < a2 < a1 < ξ1 bloid a3 < a2 < ξ2 < a1 bloid' a3 < ξ3< a2 < a1 The Cramer's Rule results above become x12 = (ξ12-a12) (ξ22-a12) (ξ32-a12)/ [ (a12-a32)(a12-a22)] x22 = (ξ12-a22) (ξ22-a22) (ξ32-a22)/ [ (a22-a12)(a22-a32)] x32 = (ξ12-a32) (ξ22-a32) (ξ32-a32)/ [ (a32-a22)(a32-a12)] which can be combined into one line xi2 = (ξ12-ai2) (ξ22-ai2) (ξ32-ai2)/ [ (ai2-ai-12)(ai2-ai+12)] or xi2 = (ξ12-ai2) (ξ22-ai2) (ξ32-ai2)/ [ Πj≠i(ai2-aj2)] or xi2 = Πk=13(ξk2-ai2) / [ Πj≠i(ai2-aj2)] So to summarize, we have Σk=13 xk2/ ( ξi2-ak2) = 1 i = 1,2,3 surfaces xi2 = Πk=13(ξk2-ai2) / [ Πj≠i(ai2-aj2)] x,y,z equations 3. Finding the normal vectors at some point x If we define gi(x,y,z) ≡ Σk=13 xk2/ ( ξi2-ak2) - 1 Then our level surface equations may be written gi(x,y,z) = 0 i = 1,2,3 We know from "surface geometry questions.doc" that the normal to this surface is given by i = (gi) / |(gi)| ni = (gi) // unnormalized where the direction is that of maximum positive change, and where is the usual 3D gradient. How do we compute this gradient? (gi) = ( Σk=13 xk2/ ( ξi2-ak2) - 1) = Σn=13 n∂n [Σk=13 xk2/ ( ξi2-ak2) - 1] = Σn=13n∂n [Σk=13 xk2/ ( ξi2-ak2) = Σn=13Σk=13 {1/ ( ξi2-ak2)} n∂n (xk2) = Σn=13Σk=13 {1/ ( ξi2-ak2)} n2 xk(∂nxk) = Σn=13Σk=13 {1/ ( ξi2-ak2)} n2 xk δnk = Σn=13{1/ ( ξi2-an2)} n2 xn = 2 Σn=13 xn / ( ξi2-an2) So we are then claiming that, ni = 2 Σk=13 xk / ( ξi2-ak2) where, if we define our usual Cartesian unit vectors to be for k = 1,2,3, we have xk ≡ (x) x = x1 + x2 + x3 So ni is an unnormalized normal vector which is normal to the surface whose label is ξi at point x. 3. Show that the three normal vectors are mutually perpendicular. We compute the dot product: ni nj = [ 2 Σn=13 xn / ( ξi2-an2)] [ 2 Σm=13 xm / ( ξj2-am2) ] = 4 Σn=13 Σm=13 1/[( ξi2-an2) ( ξj2-am2)] xn xm = 4 Σn=13 Σm=13 1/[( ξi2-an2) ( ξj2-am2)] δnmxn2 = 4 Σn=13 xn2/[( ξi2-an2) ( ξj2-an2)] Now into this expression we want to insert our Cramer's rule result which was xi2 = Πk=13(ξk2-ai2) / [ Πj≠i(ai2-aj2)] x,y,z equations or xn2 = Πk=13(ξk2-an2) / [ Πj≠n(an2-aj2)] x,y,z equations or xn2 = Πk=13(ξk2-an2) / [ Πs≠n(an2-as2)] x,y,z equations In other words, we are replacing the x,y,z coordinates of x with their values given in ellipsoidal coordinates. These coordinates are, of course, the same ξi values appearing in ni nj above . We then have ni nj = 4 Σn=13 {xn2}/[( ξi2-an2) ( ξj2-an2)] = 4 Σn=13 { Πk=13(ξk2-an2) / [ Πs≠n (an2-as2)]}/[( ξi2-an2) ( ξj2-an2)] = 4 Σn=13 { Πk=13(ξk2-an2) / [( ξi2-an2) ( ξj2-an2) Πs≠n(an2-as2)] } = 4 Σn=13 { Πk=13(ξk2-an2) / [( ξi2-an2) ( ξj2-an2) Dn } Dn = Πs≠n(an2-as2) Now multiply both sides by (a12-a22) (a22-a32) (a32-a12) (a12-a22) (a22-a32) (a32-a12) ni nj = 4 Σn=13 { Πk=13(ξk2-an2) / [( ξi2-an2) ( ξj2-an2)] * (a12-a22) (a22-a32) (a32-a12) / [Πs≠n(an2-as2) ] } Now consider the second ratio. There are three factors up top, and two down. Now we shall look at each case separately n=1 (a12-a22) (a22-a32) (a32-a12) / [(a12-a22) (a12-a32)] = - (a22-a32) n=2 (a12-a22) (a22-a32) (a32-a12) / [(a22-a12) (a22-a32)] = - (a32-a12) n=3 (a12-a22) (a22-a32) (a32-a12) / [(a32-a12) (a32-a22)] = - (a12-a22) The second two cases are cyclics of the first one. We can then write out the n=1 term only and say (a12-a22) (a22-a32) (a32-a12) ni nj = 4 (a32-a22) { [Πk=13(ξk2-a12)] / [( ξi2-a12) ( ξj2-a12)]} + cyclic Notice that both sides have cyclic symmetry under cycling of the ai indices, and that both sides are symmetric under i↔j. The LHS turns into itself under any cyclic rotation of the ai indices, and so does the RHS when all three terms are included. Now suppose i ≠ j and suppose t is the index that is neither i nor j from the set 1,2,3. Then in the {} ratio, the 2 denominator factors cancel 2 of the 3 numerator factors, leaving just ( ξt2-a12). We then have (1/4)(a12-a22) (a22-a32) (a32-a12) ni nj = (a32-a22) ( ξt2-a12) + cyclic i ≠ j The ξt2 terms are these ξt2 [ (a32-a22) + cyclic ] = ξt2 [ (a32-a22) + (a12-a32) + (a22-a12) ] = ξt2[0] = 0 The other terms are these (a22-a32)a12 + cyclic = (a22-a32)a12 + (a32-a12)a22 +(a12-a22)a32 = 21 - 31 +32 - 12 + 13 - 23 = 0 Therefore, we have shown that ni nj = 0 if i ≠ j So we have achieved our goal of generalizing our Part I proof that n1 n2 = 0 . Now go back and try i = j instead. We then get (a12-a22) (a22-a32) (a32-a12) ni ni = 4 (a32-a22) { [Πk=13(ξk2-a12)] / [( ξi2-a12) ( ξi2-a12)]} + cyclic This time, only one factor top and bottom cancels and we get a fairly ugly result (a12-a22) (a22-a32) (a32-a12) ni ni = 4 (a32-a22) { [Πk≠i(ξk2-a12)] / [( ξi2-a12) ]} + cyclic Therefore ni2 = ni ni = 4 (a32-a22) { [Πk≠i(ξk2-a12)] / [( ξi2-a12) ]} / [(a12-a22) (a22-a32) (a32-a12)] = - 4 { [Πk≠i(ξk2-a12)] / [( ξi2-a12) ]} / [(a12-a22) (a32-a12)] + cyclic = 4 { [Πk≠i(ξk2-a12)] / [( ξi2-a12) ]} / [(a12-a22) (a12-a32)] + cyclic = 4 { [Πk≠i(ξk2-a12)] / [( ξi2-a12) (a12-a22) (a12-a32) ]} + cyclic = 4 { [Πk≠i(ξk2-a12)] / [( ξi2-a12) (a12-a22) (a12-a32) ]} + cyclic Specifically, we have n12 = 4 { (ξ22-a12) (ξ32-a12)] / [( ξ12-a12) (a12-a22) (a12-a32) ]} + cyclic n22 = 4 { (ξ32-a12) (ξ12-a12)] / [( ξ22-a12) (a12-a22) (a12-a32) ]} + cyclic n32 = 4 { (ξ12-a12) (ξ22-a12)] / [( ξ32-a12) (a12-a22) (a12-a32) ]} + cyclic so we have generalized our earlier expression for n12 D. Rehash of the "metric tensor method". In our method above, we used our brute force geometry normal expression to obtain the desired result. Here consider the x,y,z equations again xi2 = Πk=13(ξk2-ai2) / [ Πj≠i(ai2-aj2)] Think of the ξk as the qk of our general curvilinear discussion. The transformation matrix T is this Tki = (eξk)i = ∂xi/∂ξk which is perhaps more intuitively written as δxi = (eξk)i δξk => δxk = (eξk) δξk The meaning is what is important. If you vary only curvilinear coordinate ξk and keep the other two fixed, then δxk is how you move in x,y,z space. You move along the ξk "coordinate line", and eξk is known as the tangent base vector for ξk. Let's take k = 1 to be more explicit. We want to vary ξ1 but keep ξ2 and ξ3 fixed, and see where we go in Cartesian space from a starting point which lies on the intersection of the three surfaces having these labels. So in this case, δx1 is a little displacement off the ξ1 ellipsoid, but we remain ON the ξ2 bloid and we remain on the ξ3 bloid. Suppose we could show that δxk δxj = 0 for any pair k≠j. What would that mean in terms of the three level surfaces which intersect at our starting point? vary ξ1 δx1 ~ e1 lies on surface 2 and 3 vary ξ2 δx2 ~ e2 lies on surface 3 and 1 vary ξ3 δx3 ~ e3 lies on surface 1 and 2 Consider this picture: Since our three level surfaces are presumably smooth, if we look only in a tiny neighborhood of some point of interest, those three surfaces look like the faces of a parallelogram in the most general case. In this picture, the left surface is surface 1 because we stay on it if we vary along e2 or e3. The surface number is the index that does not appear on the two vectors spanning the surface. So the front face is surface 2 and the bottom is surface 3. Basically this says that locally, we have a little parallelogram-shaped coordinate system and the three vectors e1, e2, e3 point along the three axes. If we find that these three axis vectors are orthogonal, then we know that locally we have a Cartesian coordinate system with these things as axes. It is just that simple, and this is then called an orthogonal coordinate system. Now back to our x,y,z equations: xi2 = Πk=13(ξk2-ai2) / [ Πs≠i(ai2-as2)] Let's compute the three vectors (eξj)i = ∂xi/∂ξj = ∂xi2/∂ξj * ∂xi/∂xi2 = (1/2xi) ∂xi2/∂ξj = (1/2xi) ∂ξj { Πk=13(ξk2-ai2) / [ Πs≠i(ai2-as2)] } = (1/2xi) * 1/ [ Πs≠i(ai2-as2)] * ∂ξj { Πk=1(ξk2-ai2) } = (1/2xi) * 1/ [ Πs≠i(ai2-as2)] * 2 ξj Πk≠j(ξk2-ai2) = Πk≠j(ξk2-ai2) (ξj/xi) / [ Πs≠i(ai2-as2)] Then we can say eξjeξj' = (eξj)i(eξj')i = [Πk≠j(ξk2-ai2)] (ξj/xi) / [ Πs≠i(ai2-as2)] * [Πk≠j'(ξk2-ai2)] (ξj'/xi) / [ Πs≠i(ai2-as2)] = (ξj/xi) (ξj'/xi) [Πk≠j(ξk2-ai2)] [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)]2 = (ξj ξj'/xi2) [Πk≠j(ξk2-ai2)] [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)]2 We now have to get rid of xi2 in favor of the ellipsoidal coordinates, so = (ξj ξj' [ Πs≠i(ai2-as2)] / [Πk=13(ξk2-ai2)]) [Πk≠j(ξk2-ai2)] [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)]2 = Σi (ξj ξj' / [Πk(ξk2-ai2)]) [Πk≠j(ξk2-ai2)] [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)] = Σi (ξj ξj' / (ξj2-ai2)) * [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)] = ξj ξj'Σi (1 / (ξj2-ai2)) * [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)] How can this possibly be zero? It must be true that, for j ≠ j', we have Σi=13 (1 / (ξj2-ai2)) * [Πk≠j'(ξk2-ai2)] / [ Πs≠i(ai2-as2)] Let's try an example. Let j = 1 and j' = 2: Σi=13 (1 / (ξ12-ai2)) * [Πk≠2(ξk2-ai2)] / [ Πs≠i(ai2-as2)] = 0 ? Σi=13 (1 / (ξ12-ai2)) * [ (ξ12-ai2) (ξ32-ai2)] / [ Πs≠i(ai2-as2)] = 0 ? Σi=13 (ξ32-ai2)] / [ Πs≠i(ai2-as2)] = 0 ? (ξ32-a12) / [(a12-a22) (a12-a32)] + (ξ32-a22) / [(a22-a12) (a22-a32)] + (ξ32-a32) / [(a32-a12) (a32-a22)] Well let's assume this last line is some object F. Then we have F (a12-a22) (a12-a32) (a22-a32) = (ξ32-a12) (a22-a32) + (ξ32-a22) (a32-a12) + (ξ32-a32) (a12-a22) and this is 0 for the same reason we got above, "just do it". Well, all I have really done here is say that the normals are orthogonal because the metric tensor is diagonal, and to show that I have to calculate the elements gij = eξjeξj' of the metric tensor, but this doesn't explain much that we don't already know. Well this is doing nothing for me whatsoever. Just the metric tensor being diagonal method.