M&M chapter 4_5

Margenau and Murphy Chapters 4 and 5 PhL 8.29.08 These are "raw notes" from reading sections of Chapters 4 and 5. See nearby documents for special topics related to these chapters. Contents: Selected Sections from Chapter 4: 1 Section 4.20 Tensors (161) 2 Note Added. 3 Question: is Tab itself a mixed tensor? 5 Question: is δab a mixed tensor? 5 Extra Facts 1,2 and 3 Added: 5 Section 4.21 Addition, Multiplication and Contraction (164) 7 pseudoscalars, etc 7 Outer products, contraction, and inner products. 8 Suppose an index-lowering tensor exists gab: associated components of a vector 9 Section 4.22 Differentiation of Tensors (167) 10 Christoffel symbols and comma notation 11 Summary of Proof of Covariance of Vb,a (raw details in separate document) 11 Selected Sections from Chapter 5: 13 Section 5.1. Curvilinear Coordinates. 13 The metric tensor in a space transformed to from Cartesian space, relation between g'ij and Tab 14 Orthogonal Systems: 16 Section 5.16. Non-orthogonal Curvilinear Coordinates: tensor relations 17 The two sets of basis vectors ei and ei and their dot product relations 17 Linear, Area and Volume Elements 19 Here is my proof that det(g) = v2 21 Question: Does the above presentation apply to a 4D space? 22 Section 5.17. The Differential Operators in Tensor Notation 25 (1) Gradient: 25 (2) Divergence. 26 (3) The Laplacian 27 (4) The curl 27 Now the grand finale: we consider "orthogonal" coordinates 28 Section 5.2. Vector Relations in Curvilinear Coordinates (174) 32 _________________________________________________________________________ Selected Sections from Chapter 4: Section 4.20 Tensors (161) I read these tensor sections through today 8.20.08. I can see I read this earlier, but don't have any notes and don't know when I was here last. M&M first try to motivate why tensors show up in nature. One idea is that if a tensor is needed, and since it describes a physical system, and since you can transform frames, probably it will "transform as a tensor" and not just be a "dead matrix" in math space. Goldstein (which I have now finished) talks about tensors being defined as such by their transformation properties, and we have some mention of them in other parts of the book, but not much. He uses the imaginary time idea which I am sure I can fit into a metric tensor viewpoint when I am done here, but he does not get into up and down indices and all that stuff. So OK, in this chapter he will use overbars for the "overbarred" reference frame, compared to the "starting" reference frame. I prefer primes since much easier to type. Let's start with the contravariant vector idea. It is something that transforms like this: A'a = Ab A'a = Tab Ab where Tab ≡ note "a" in upper partial, "b" in lower partial Notice that ALL indices are up here everywhere except in our shorthand T second index. Something that does the above is a "contra variant vector". We are allowed to define a shorthand in any way we want, so the up/down of its indices are arbitrary right now. ________________________________________________________________________________ (Section corrected 11.7.09. ) M&M make a misleading comment on page 162 under equation (4-51a) which had me fooled. They say that the coordinate vector itself xa transforms as a contravariant vector and we write x'a = Tab xb = xb But this is only true if Tab is not a function of position !!! In general, this is not true. So in the general case, all we can say is that a differential coordinate vector dxa transforms as a tensor, dx'a = Tab dxb = dxb This is always true. Imagine that xa is a vector joining two points in space (these two points could be the origin and the point x, or two points r1 and r2). In general Tab(x) will vary as you move along this vector and will be different for example at the two ends! As an example, suppose (r,θ,φ) are xi'. We know we cannot write θ = Ax + By + cz so x'a = Tab xb cannot possibly be true. Now it is true that much tensor analysis is done with linear transformations which would mean for example that x1' = Ax1 + Bx2 + Cx3 so that in fact there is a matrix x' = M x. In this situation, the matrix M is exactly our matrix T and indeed we can say x'a = Tab xb. In effect we are integrating dx'a = Tab dxb since T is constant in space. Better, we start with our linear transformation x' = M x and differentiate it, and that lets us identify M = T. Examples of linear transformations are rotations or Lorentz transformations where T is a function of some angles or a boost parameter, so T is constant in space. In the case of a Lorentz Transformation we might write Tμν = aμν as in BD vol 1 early chapters. In general relativity, however, I think T is going to vary with space so we must be careful. I think in this case of space-dependent T, when we talk about a things that transforms as a vector, it will be a vector field, and we will have A'a = Tab Ab be something that is true at each point in space separately. Perhaps A'a (x') = Tab(x) Ab(x). Since I am trying to apply all this tensor stuff even to curvilinear coordinates, such as sphericals, we have to realize that T is a function of space. I hope this realization does not invalidate anything major below. M&M are quite aware that T might depend on x, and so also then will gij and g etc. It is exactly this space dependence that causes the extra term in the covariant derivative studied below, and that is why the Christoffel symbols only appear in general relativity, not special relativity. In the rest of this document, I will continue to use xa as if it were a true tensor vector. The reader needs to think of this as dxa to keep general, but I will not edit everything. And of course things are true as stated for linear transformations T. At some point, I need to purge this entire directory to get this misunderstanding removed from all files. ___________________________________________________________________________________ So, how does a covariant vector transform? According to 4-51, it is this: A'a = Ab = Tab Ab where Tab ≡ note "a" in lower partial, "b" in upper partial The partial derivative is exactly upside down. So the summation tilt is \ instead of /, and the prime is now on the bottom. You exactly swap the partials top and bottom. But the indices within the partial derivative are still both "up". In our shorthand notation, we could have used a different letter, but we instead stick with the letter T, but we just regard Tab and Tab as different transformations. The notational idea is to have the first index match the thing on the left of the equation, the second index goes with the thing on the right, and you have a diagonal tilt on the sum. Now here are four properties of the various T elements. In the first row, we are summing on the second index, either sum tilt up or sum tilt down. In the second row, we sum on first index. In all four cases, the two T objects are not the same object -- they always have different tilts and that is how our shorthand notation shows that they are not the same object. Tab Tcb = = δac = Tcb Tab sum on second index of T Tab Tac = = δbc = Tac Tab sum on first index of T where δij = δij = good old δi,j, but we choose to write them with up/down sense that sort of matches the product. Later we will see this has more significance. Here is a way to remember the above rules. (1) choose to sum on either first or second index; (2) have the tilts of the two T's be opposite. _____________________ Note Added. Go back up to x'a = Tab xb = xb We know also that xa = (T-1)ab x'b // = ( x'b = Tab x'b ) as we shall confirm below Notice that this gives xa = (T-1)ab Tbc xc which says (T-1)ab Tbc = δac Now apply Tdc to the right on both sides to get (T-1)ab Tbc Tdc = δac Tdc => (T-1)ad = Tda So we find these facts: Tab = (T-1)ba Tab = (T-1)ba Notice that the tilt changes AND the indices are in reverse order! And of course then we can write: xa = (T-1)ab x'b = Tba x'b So here is the comparison of the forward and backward transformations: x'a = Tab xb x'a = Tab xb xa = Tba x'b = x'b Tba xa = Tba x'b = x'b Tba Similarly we have Q'ab = Tac Tbd Qcd Q'ab = Tac Tbd Qcd Qab = Tca Tdb Q'cd Qab = Tca Tdb Q'cd and so on for any sized tensor. You apply the inversion rule to each index. ______________ So far, Aa and Ab are completely different animals and we don't have a way to relate them to each other. Now here are some contravariant tensor transformation rules: S' = S contravariant rank 0 V'a = Taa' Va' contravariant rank 1 X'ab= Taa' Tbb' Xa'b' contravariant rank 2 U'abc= Taa' Tbb' Tcc'Ua'b'c' contravariant rank 3 Here we have introduced some slightly dangerous notation. The prime was originally used to denote something in the "primed frame" like S', but now we start priming indices so we can see how the indices are matching up. So index a' could have been called α or s, it is just a summation index. So these two uses of the "prime" character ' are completely unrelated! Here are the corresponding covariant tensor transformation rules. All we do is take the above rules and change the up-down-ness of every single index: S' = S covariant rank 0 V'a = Taa' Va' covariant rank 1 X'ab= Taa' Tbb' Xa'b' covariant rank 2 U'abc= Taa' Tbb' Tcc'Ua'b'c' covariant rank 3 The next idea is that you can define a "mixed tensor" with some indices up (contra) and some indices down (co), and you adjust the T's appropriately. We are just defining tensor types. For example: X'ab= Taa' Tbb' Xa'b' mixed high-low rank 2 * U'abc= Taa' Tbb'Tcc' Ua'b'c' mixed high-low-high rank 3 It is quite easy to keep track of things using the tilts and the primed indices. Question: is Tab itself a mixed tensor? See Question #8 in "ponderings" doc. The answer is that Tab is NOT a rank-2 tensor, mixed or otherwise. It is "mixed" in a completely different sense because it has the first index in the x' world and the second index in the x world. Consider the transformation of two true rank 2 tensors: X'ab= Tai Tbj Xij // an arbitrary tensor g'ab= Tai Tbj gij // the metric tensor The rules are these: Use g'ab to raise and g'ab to lower either index on X'ab in any of its index position forms. Use gab to raise and gab to lower either index on Xab in any of its index position forms. [ if x-space happens to be Cartesian, then raising and lowering an index on Xab is free. ] But for T the rules are different: Use g'ab to raise and g'ab to lower the first index of Tab or Tab. Use gab to raise and gab to lower the second index of Tab or Tab. This follows from the definitions Tab ≡ Tab ≡ since in both cases the second index is in the x-space and the first index is in the x' space. Question: is δab a mixed tensor? Yes, it is, provided you think of tensor δ being the same as the metric tensor g. Explanation follows: Copy * from above, then replace X with δ : δ 'ab= Taa' Tbb' δ a'b' = Taa' Tba' = δ ab I think the answer here is that yes, δ is a tensor, but in this sense δij = gij in the x-space δ'ij = g'ij in the x'-space We know that, gij = δi,j and g'ij = δi,j where δi,j is not the same as δ'ij ! Really, this δ tensor is identical to the metric tensor. You have to be careful because, for example δ'ij ≠ δi,j . In general, people often use δij as synonym for gij but I would recommend never using δ'ij . For example, when we are working in q-space, we don't show this prime, and we would then have δij = gij ≠ δi,j !!!!! Extra Facts 1,2 and 3 Added: These "facts" Fact 1: Recall that Tbc ≡ Therefore we have (∂a Tbc) = ∂a = = Sbac // symmetric on a↔c ! This symmetry is not very apparent when you write (∂a Tbc). Fact 2: Let's differentiate our two orthogonality conditions, might be useful later: δac = Tab Tcb sum on second index of T δbc = Tab Tac sum on first index of T Therefore we apply ∂s to get 0 = ∂s(Tab Tcb) = (∂sTab) Tcb + Tab (∂s Tcb) 0 = ∂s(Tab Tac) = (∂sTab) Tac + Tab (∂s Tac) which we can write as Sasb Tcb + Tab (∂s Tcb) = 0 (∂sTab) = Sasb Sasb Tac + Tab (∂s Tac) = 0 where we have put in our visibly-symmetric object Sasb = Sabs . The two results above are different because in the first, we sum on one of the symmetric indices of S, b, while in the second we sum on the other index. Now apply Tae to the first equation, and apply Teb to the second equation to get Tae Sasb Tcb + Tae Tab (∂s Tcb) = 0 => Tae Sasb Tcb + δeb (∂s Tcb) = 0 Teb Sasb Tac + Teb Tab (∂s Tac) = 0 => Teb Sasb Tac + δea (∂s Tac) = 0 So we get these two results: Tae Sasb Tcb + (∂s Tce) = 0 => (∂s Tce) = – Tae Sasb Tcb Teb Sasb Tac + (∂s Tec) = 0 => (∂s Tec) = – Teb Sasb Tac Now redo indices so both are in the same form (∂s Tce) = – Tae Sasb Tcb = – Tae Tcb Sasb (∂s Tce) = – Tcb Sasb Tae = – Tae Tcb Sasb where (∂sTab) = Sasb so these results are identical, and we end up with this possibly useful fact: (∂s Tce) = – Tae Tcb Sasb where recall Sasb = (∂sTab) = symmetric s↔b So, if we have a derivative of T** , we replace with S at once. But derivative of T** gives the messier result on the left. Fact 3: Let's try a similar thing on gab gac = bc which we derive in the next section. Apply ∂s : ∂s (gab gac) = 0 => (∂s gab) gac + gab(∂s gac) = 0 Now apply gbd to both terms to get gbd (∂s gab) gac + gbd gab(∂s gac) = 0 => gbd (∂s gab) gac + (∂s gdc) = 0 so we find that (∂s gdc) = – gdb gca(∂s gab) which says we can "raise" both indices of the (∂s gab) in this manner. Note that you cannot willy-nilly raise and lower indices through a derivative ∂s . For example, if you could the above would say (∂s gdc) = – gdb gca(∂s gab) So our result is very special and says you can do what it says, and notice the minus sign to boot! Section 4.21 Addition, Multiplication and Contraction (164) He notes that a rank 2 tensor can be split into symmetric and antisymmetric terms, but the phrase "skew symmetric" is used instead of the now more commonplace antisymmetric. Now they implicitly start talking about parity, but it is poorly done and illogical to me. You cannot represent a parity transformation by the matrix Tab = - δ ab. Such a transformation would change the sign of any vector. Question: Is there a transformation Tab that takes a vector Ab from a right handed to a left handed coordinate system? Note that Tab = - δ ab would negate all contravariant vectors, so that is not the one. The answer is that there is no matrix T that does this action! pseudoscalars, etc I think in order to deal with parity, you have to have a separate type of transformation that I will just indicate by an arrow. Thus, IF a tensor has the following transformation properties under the parity transformation, THEN it has the name shown. A tensor might be neither regular nor pseudo, but you could probably express it as the sum of two such tensors. S → +S scalar S → -S pseudoscalar V → -V vector (normal polar) V → +V pseudovector (axial) This parity transformation applies to the tensor as a whole, and has nothing to do with contra or covariant indices on the tensor. Now we can consider the cross product of two polar vectors C = A x B → (-A) x (-B) = A x B = +C so the result is an axial vector C. Outer products, contraction, and inner products. The next topic is ways to build a larger tensor from a smaller one. Consider Mab = AaBb M'ab = A'aB'b = (Taa' Aa' )( Tbb' Bb') = Taa' Tbb' Aa' Bb' = Taa' Tbb' Ma'b' This shows that the thing so assembled transforms as a second rank tensor. These assemblies in general are called outer products. You can imagine assembling two huge arbitrary mixed tensors together in this way, and the result is a mixed tensor. Next, suppose we have a large mixed tensor and we sum on one (or more) of the index pairs up/down. Each such summation is called a contraction. The resulting object has lower rank by 2 for each contraction, and transforms as a tensor of that lower rank! Here is an example: recall from above we had U'abc= Taa' Tbb'Tcc' Ua'b'c' Now contract the a and b indices to get U'aac= (Taa' Tab')Tcc' Ua'b'c' = δa'b' Tcc' Ua'b'c' = Tcc' Ua'a'c' = Tcc' (Uaac') so the result is a rank one tensor we could call Qc = U'aac. This contraction can happen between indices of two separate tensors which are assembled together, for example AabcBade = Qbcde = a mixed tensor of rank 4. In this case, we have an "inner product" of the two tensors with respect to the contracted index or indices. The prototype example of this is the following. Start with Mmn = AmBn which is a rank two mixed tensor. Contract the only index pair to get S = Mmm = AmBm = a tensor of rank 2-2 = 0, a scalar. Notice that we can NOT say this is the same as AmBm because at this point, we have no connection between up and down indexed objects. Still, it looks like (A,B) inner product, and (A,A) looks like a norm, and we might start talking Hilbert Space at some point. Authors even go so far as to suggest a cosθ dot product thing. Suppose an index-lowering tensor exists gab: associated components of a vector Now finally we come to the idea of an index-lowering tensor gab. We are told to imagine that some symmetric rank 2 tensor gab exists such that Aa = gabAb . If such a tensor exists, then the two kinds of A vectors are said to be "associated" with each other by this tensor gab. We are not calling this the metric tensor, just some tensor. Now comes the big question. If we have such a gab that lowers an index, can we find a matrix gab that raises an index? As usual, we imply no connection between g with the lower and g with the upper indices. We have gab and we try to construct gab. This suggestion is made: gab = cofactor( gab)/det(gab) Now, luckily, in my matrix binder in section 2 page 3 I have this theorem (from MM Chapter 12) p Apq cof(Apq') = q,q' det(A) which I derive right in those notes. In our current context, we would say p gab cof(Aab') = bb' det(g) so let's define gab ≡ cof(gab)/det(g) If gab is symmetric, then gab is also symmetric from definition of a cofactor, since of course detA = detAT. We know from the above quoted theorem then that gabgab' = bb' and g indices in any order since Certainly gab looks like a symmetric contravariant rank 2 tensor, but a brute force proof would be involved, since det(g') is a function of the transformation T, etc. Now consider gabAb = gab (gbcAc) = ( gabgbc)Ac = Aa so we now learn that gab raises indices. If we really have gabAb = Aa, for any A, then this is a contraction rule, and M&M conclude that therefore gab must be a tensor. So IF an index-lowering tensor gab exists, then of course gab also exists (assume detg ≠ 0 of course) and we can now "reverse tilt" anywhere we like: for example A...a........B.......a... = gab A...b........B.......a... = gab gab' A...b........B.......b'... = δbb' A...b........B.......b'... = A...b........B.......b... = A...a........B.......a... so we have shown that your can pick any contracted index pair and reverse the tilt by raising one and lowering the other, and the result is the same. Clearly you could do this on many at once. All ASSUMING that a lowering gab tensor exists. So, case of interest in this context is AmAm = AmAm to be true, only requires that a symmetric lowering tensor gμν exist so this looks even more like a scalar product and norm. Notice that one application would be Tab = gaa'gbb'Ta'b' which means if you know Tab you can compute Tab. Recall that in general these are unrelated objects. Section 4.22 Differentiation of Tensors (167) You might think that the following would be a tensor ∂aVb Let's just test it: (∂'aV'b) = Tad∂d (TbcVc) = Tad Tbc (∂dVc) + Tad(∂d Tbc)Vc so it goes like a tensor, but there is an extra term! Now for the first time, we are taking note of this fact: Tab = Tab(xμ) that is, the transformation can be a function of the coordinates. This is not the case for special relativity say where T might be a rotation or a boost, but it is the case in general relativity. Up to now, we have not cared about this xμ dependence, but when we talk differentiation, we have to worry about it. Christoffel symbols and comma notation Now let's digress to define the Christoffel symbols [ab,c] ≡ ½ ( ∂agbc + ∂bgac – ∂cgab ) // symmetric in ab {ab,c} ≡ gcd [ab,d] // symmetric in ab Thus, the {...} thing is a "raise" on the last argument of [...], but neither of these things is a tensor, so be careful. In any event, {ab,c} = ½ gcd( ∂agbc + ∂bgac – ∂cgab ) ≠ ½ ( ∂agbd + ∂bgad – ∂dgab ) M&M then go through a huge mess of math to arrive at this result: the following object transforms as a tensor: ∂aVb – {ab,c}Vc ≡ Vb,a where we now see "the comma notation" where the index after the comma says what you are differentiating with respect to in the first term. This {....} stuff is what you have to add to ∂aVbto make it become a tensor, so neither ∂aVb nor {ab,c}Vc is a tensor, only the difference. Summary of Proof of Covariance of Vb,a (raw details in separate document) I did a brute force proof on my own to show that Vb,a transforms like a tensor (see M&M Chapter 5 "support" doc). The proof was very messy, but I just wanted to try it without looking at the M&M solution. I learned many useful things along the way. Here is an outline of my proof; (1) By requiring the tensor transformation, I was quickly led to the need for the following to be true: {ab,c}'V'c = Tad(∂d Tbc)Vc – Tad Tbc{dc,e}Ve (2) I was disturbed by the fact that Tad(∂d Tbc)Vc did not appear symmetric in a,b, so this led me off on a digression where I found that I could rewrite this term in a manifestly symmetric form: Tad(∂d Tbc)Vc = –Tas Tbg Tfc Sfsg Vc where Sfsg = (∂sTfg) = = sym(s,g) Showing this required me to derive and then use a little "fact #2" based on ∂s(Tab Tcb) = 0, namely, (∂s Tce) = – Tae Tcb (∂sTab) = – Tae Tcb Sasb "Fact 2" This "fact" showed how to relate the derivative of T** to that of T**. So the problem then became to show this to be true: {ab,c}'V'c = –Tas Tbg Tfc Sfsg Vc + Tad Tbc{dc,e}Ve LHS = RHS2 + RHS1 (3) I then expanded the LHS and ended up with two big terms LHS1 + LHS2 . LHS1 contained the three terms having factors of the form (∂*g**). I then compared this to RHS1, LHS1 = ½geh * Ve { Taa'Tbb' (∂a'gb'h) + Tba'Tab' (∂a'gb'h) –Tab'Tbc'(∂hgb'c') } RHS1 = ½ Tad Tbc ges ( ∂dgcs + ∂cgds – ∂sgdc )Ve and, by renaming indices, it was easy to show these terms were the same. So what remained to show was that LHS2 = RHS2. At this point then I had: RHS2 = –Tas Tbg Tfc Sfsg Vc LHS2 = + ½Tdh geh * Ve { Taa'Tbb'(∂a'Tdc')gb'c'+ Tba'Tab'(∂a'Tdc')gb'c' } + ½geh * Ve * { Taa'(∂a'Tbb') gb'h + Tba'(∂a'Tab') gb'h – (∂hTab')Tbc'gb'c' – Tab'(∂hTbc')gb'c'} where it took quite a bit of routine but detailed work to get this form for LHS2. There are 6 terms here, and each pair of terms shows the a,b symmetry. BUT, it was hard for me to see how this was going to equal RHS2 because I could see no cancellations of terms. (4) I then processed LHS2 into the following form, exposing the (∂sTdc) type factors to which I knew I could apply Fact #2. This gave: LHS2 = + ½Tdh Vh { TasTbc + TbsTac }(∂sTdc) (∂sTdc) = – Tfc Tdg Sfsg + ½ [Vc Tas – Vs Tac ] (∂sTbc) (∂sTbc) = – Tfc Tbg Sfsg + ½ [ Vc Tbs – Vs Tbc ] (∂sTac) (∂sTac) = – Tfc Tag Sfsg (5) I then installed Fact #2 three times on the right, and was able the express LHS2 as the sum of 6 terms: 1 = – ½Vg TasTbc Tfc Sfsg Tfc Sfsg = = not simple 2 = – ½Vg TbsTac Tfc Sfsg 3 = – ½ Vc Tas Tbg Tfc Sfsg 4 = + ½ Vs Tac Tbg Tfc Sfsg 5 = – ½ Vc Tbs Tag Tfc Sfsg 6 = + ½ Vs Tbc Tag Tfc Sfsg (6) At this point, I made use of the symmetry of Sfsg to rewrite terms 4,5 and 6, then getting 1 = – ½Vg TasTbc Tfc Sfsg 2 = – ½Vg TbsTac Tfc Sfsg 3 = – ½ Vc Tas Tbg Tfc Sfsg 4 = + ½ Vg Tac Tbs Tfc Sfsg 5 = – ½ Vc Tbg Tas Tfc Sfsg 6 = + ½ Vg Tbc Tas Tfc Sfsg Now it became clear that terms 1 and 6 cancel, and also terms 2 and 4 cancel, and terms 3 and 5 are equal, so our total result is this: LHS2 = 3+5 = – Vc Tbg Tas Tfc Sfsg = RHS2 ! and this concluded the proof. (7) Among the "lessons learned" by doing the above, I realized that you cannot willy-nilly raise and lower indices on tensors that are acted upon by a derivative. For example, consider this simple example Psc = Ad (∂sTdc) ≠ Ad (∂sTdc) // where I am doing a "tilt change" In fact, calculation shows that Psc = Ad (∂sTdc) = Ad (∂sgdeTec) = Ad gde (∂sTec) + Ad(∂sgde) Tec = Ae (∂sTec) + Ad(∂sgde) Tec = Ad (∂sTdc) + extra term. Also, not that, just as ∂sVd is not a tensor ( our main proof!) , so also (∂sTdc) is not a tensor. Thus, even though T and A are tensors, Psc is not a tensor. Another more obvious example is this. Ac = (∂sgsc) ≠> Ac = (∂sgsc) If it did, we would conclude from the right equation that Ac = 0 since ∂s(δsc) = 0 and we would then think that Ac = gcdAd = 0, which is not true. Also, the fragment Ac is not a tensor! Selected Sections from Chapter 5: Section 5.1. Curvilinear Coordinates. Repairs shown in red done 9/5/08 This section is the main act of this chapter, it is all on one page 173, and sadly it is presented poorly so I need all these notes to clean it up. I am changing notation to everything is contravariant upper index. Imagine we have our starting Cartesian coordinates xi and we want to use some qi which are related as shown, that is, qi = qi(rj). { later we think qi = x'i) Then of course we know that dxi = Σk(∂xi/∂qk) dqk = Σk(∂xi/∂x'k) dx'k // page 5.1 equation A which we can write in matrix form as ( I don't know the official name for the Tik object, even in Weinberg there is no name, so lets just call it the Partial Derivative Tensor or the Coordinate Transformation Tensor , T for Transformation). Recall our forward and inverse transforms on coordinates, as noted above, x'a = Tab xb transform // see correction note earlier! xa = Tba x'b inverse transform Thus we have dxi = Σk Tki dqk Tki = (∂xi/∂qk) = (∂xi/∂x'k) = the covariant transformation tensor The distance between two close points is given by ds2 = Σi (dxi)2 = Σi (Σk Tki dqk)(Σm Tmi dqm) = Σk Σm (Σi Tki Tmi )dqk dqm = Σk Σm Gkm dqk dqm The metric tensor in a space transformed to from Cartesian space, relation between g'ij and Tab where we have defined our "metric tensor" G (same as g ) according to Gkm = Σi Tki Tmi = Σi (∂xi/∂qk) (∂xi/∂qm) // page 192 5-4a Notice the G is symmetric, so if we separate the diagonal terms we can write, ds2 = Σk Gkk dqk dqk + Σk≠m Gkm dqk dqm = Σk Gkk dqk dqk + 2 Σk<m Gkm dqk dqm We could also write ds2 in dyadic or vector notation in this way ds2 = dq G dq = dqT G dq Now we pause to comment on an immense confusion M&M have caused. They use these symbols: (Qii)2 [ written also as Q2ii] to represent the quantity Gii but Qij to represent the quantity Gij i ≠ j This is completely illogical to anyone who uses matrices, because if Qij = Gij, we expect Qii = Gii. I cannot explain why they did this stupid thing. But given it, we can then say ds2 = Σk Qkk2 (dqk)2 + Σk≠m Qkm dqk dqm = Σk Qkk2 (dqk)2 + 2 Σk<m Qkm dqk dqm and this is what you see in equation 5.3. Now to add more to the confusion, they define Qi = Qii by which they mean Qi = = So now at least we get rid of their horrible Q2ii thing, and we now have ds2 = Σk Qk2 (dqk)2 + Σk≠m Gkm dqk dqm = Σk Qk2 (dqk)2 + 2 Σk<m Gkm dqk dqm It seems pretty clear that if you only allow one dqk to vary, you will have ds2 = Qk2 (dqk)2 = Gkk(dqk)2 so we can talk about a distance without regard to sign as dsk = Qk dqk = dqk // equation 5.5 and 5-5a page 192 For example, in spherical coordinates, if we start at some point (r,θ,φ) and allow only θ to vary, the dsθ is a small distance down along the great circle of radius r in the plane φ = constant. This line is an example of a "coordinate line" and is a curve in the Cartesian space along which only one qi coordinate varies. This curve is of course the intersection between the sphere r = constant, and the plane φ = constant. These surfaces are called coordinate surfaces. Now we come to the next huge confusion. Let's first go back to where we started, dxi = Σk Tki dqk Tki = (∂xi/∂qk) and write as a vector equation dx = T dq In our variation above, let's rewrite and assume that δqk > 0 for a single coordinate only, so dq(k) = (0,0,....δqk....0) or dq(k)i = δi,k δqk In this case we find that dx(k) = T dq(k) or dx(k)m = Σi Tim dq(k)i = Σi Tim (δi,k δqk) = Tkm δqk What is the length of this vector? | dx(k) |2 = Σm (Tkm δqk)2 = (δqk)2 Σm (Tkm)2 = (δqk)2 Gkk = (δqk)2 Qk2 since Gkk = Σm Tkm Tkm . So we learn that | dx(k) | = δqk Qk = δqk . Now, consider doing two separate variations of the form dq(i)m= δmi δqi dq(j)m= δmj δqj where in the first case we vary only qi, and in the second only qj . What is dx(i) dx(j) ? dx(i) dx(j) = Σm dx(i)m dx(j)m = Σm Tmi δqi Tmj δqj = Σm Tmi Tmj δqi δqj = Gij δqi δqj But we know that | dx(j) | = δqj Qj | dx(i) | = δqi Qi Therefore, we can compute the cosine of the angle between our two vectors as cosθij = dx(i) dx(j) / (| dx(j) | | dx(i) |) = Gij δqi δqj / (δqj Qj δqi Qi ) = Gij/ (Qj Qi) and FINALLY we have derived page 173 equation E, since Gij = Qij. Now that we have this result, let's try to parse their enigmatic text. They say, " the cosine of the angle between ds1 and dz is (dz/dq1)(dq1/ds1) ..." In this phrase, there is the implication that ds1 and dz are vectors, otherwise you cannot have an angle between them. But the authors have not written any vectors anywhere for these things. I have to conclude this is just junk text, garbage! Orthogonal Systems: We want to consider systems in which Gij = 0 for off diagonal elements. In other words, a system with a diagonal metric tensor G. Fine. Yes, that means θij = 0. This means that the angles between the coordinate lines at any given point are 90 degrees. For example, in spherical coordinates, our three little directions , and are perpendicular for any point we pick in the space. Now, in an orthogonal system we have this for distance, ds2 = Σk Gkk dqk dqk What can we say about surface elements? We know that dx(i) and dx(j) are perpendicular, so my inclination would be to define dS(ij) = dx(i) x dx(j) // a vector equation but of course the angle is 90 degrees, so we know that | dS(ij)| = | dx(i) | | dx(j) | = δqi Qi δqj Qj = Qi Qj δqi δqj which we can write in magnitudes as dS(ij) = Qi Qj dqi dqj // equation 5.6 Finally, a little volume element would be three of these multiplied together | dx(i) | | dx(j) | | dx(k) | = Qi Qj Qk δqi δqj δqk // equation 5.7 What a battle just parsing a single page of M&M, page 173 ! Notice that if things are not orthogonal, then the area and volume have to have extra factors in them. At this point, I am going to skip to Section 5.16 because it continues on the above discussion. Section 5.16. Non-orthogonal Curvilinear Coordinates: tensor relations OK, they now do what I did above, and they refer to my Gij as gij. Notice that all coordinates now have upper indices, so we are doing contravariant notation as usual, and that word does in fact appear. The two sets of basis vectors ei and ei and their dot product relations Now our first action is to define some "base vectors" like so: e(i) ≡ ∂r/∂qi Note that these are in general not unit vectors, and they are not orthogonal! The direction of e(i)is along the curve where qi varies. That is, we are changing only dqi and leaving all other dqj fixed, so we move in 3D space where this change takes us. For example, we might have e(i)= and it points in the direction of increasing θ such that r and φ remain constant. That is what base vectors are (although not normalized). And then according to page 192 D we have dr = e(i)dqi // bold are the vectors, implied sum Now define upper label basis vectors like so e(i) ≡ ½ εijk e(j) x e(k)/v // sums implied on j and k where v = e(1) e(2)x e(3) = cyclic = -anticyclic so can write e(i) e(j)x e(k) = εijk v. Then examine: e(n) e(m) = e(n) [ ½ εmjk e(j) x e(k)] /v= ( ½ εmjk/v) { e(n) e(j) x e(k)} =( ½ εmjk/v){ εnjk v} = ½ εmjk εnjk = ½ * 2δm,n = δm,n where we use a reduction from our Levi-Civita page. Thus is proved 5.79 on page 193. So what is the point here? The covariant basis vectors are orthogonal to the contravariant ones. How does that help me? Now, let us conjecture the existence of some covariant vector components dqi such that 5-81 is true dr = e(i)dqi earlier we had dr = e(i)dqi It must then be true that e(i)dqi = e(i)dqi (1) Now dot equation (1) on the left with e(j) and, using 5-79 above, we find dqj = [e(j) e(i)] dqi = Gji dqi where Gij ≡ [e(i) e(j)] Or, dot equation (1) from the right with e(j)to get [e(j) e(i)] dqi = dqj = Gji dqi where Gij ≡ [e(i) e(j)] Think about what these things are saying! Suppose you vary only dqi (for some specific i). Then it turns out you will have varied probably all the dqj's. The amount each dqj (covariant!) varies is given by Gji times the initial dqi. So this Gji is a cross effect, and of course we see it is e(j) e(i). So several things have happened all at once here. We have defined two matrices Gij and Gij as shown, and as far as we know they are different. Second, we conjectured the existence of the components dqi, but now we know exactly how to compute them , ie, dqj= Gji dqi , so they must exist. Nothing in the previous paragraph requires our vector to be dr, it could be any vector A, and we would then say A = e(i)ai = e(i)ai and ai= Gij aj and ai= Gij aj // 5-87, 5-88 and A e(i) = ai and A e(i) = ai and A = ai e(i) = (A e(i)) e(i) = ai e(i) = (A e(i)) e(i) // 5-89 just to write down all the things M&M are writing down on page 194. Notice that we are using component names ai and ai rather than Ai and Ai because our basis vectors are not unit vectors, and we want to save the component names Ai and Ai for components against unit vectors! But now let's return to the differential vectors where we have: Note Added: What about derivatives of the basis vectors? ei ≡ ∂r/∂qi ∂ei/∂qj = ∂2r/∂qj∂qi and I don't have a good formula for this. Besides, I am more interested in ∂i/∂qj , so let's quietly drop this subject. Linear, Area and Volume Elements dr = e(i)dqi = e(i)dqi we can compute: ds2 = drdr = [e(i)dqi] [e(j)dqj] = δij dqi dqj = dqi dqi = dqi (Gij dqj) = Gij dqi dqj If we compare this to our definition of the metric tensor gij in 5-3a, we discover that Gij = gij = the metric tensor (with both lower indices) Similarly, we can write AA = e(i)ai e(j)aj = aiai = gij ai aj = gij ai aj In particular, notice that e(i) e(i) = gii // no sum implied |e(i)| = ui = e(i)/ = lower index unit vector We can normalize either set of "basis vectors" as shown in page 194 B and. In terms of these unit vectors, we can write A = e(i)ai = ui (ai) = uiAi and similarly for unit vectors ui . Comments: In this section above, M&M define a set of contravariant basis vectors, and a set of covariant ones as well. You can "expand" a vector onto either set, but your expansion coefficients are the opposite type to the basis vectors. That is to say A = ai e(i) = ai e(i) Because the contra and co basis vectors are orthogonal, you can isolate one of the components in the obvious manner, all as shown above, such as A e(i) = ai. I am not used to having two sets of basis vectors! To which set does belong in sphericals? Surface Area Element: Above we said dr = e(i)dqi for some arbitrary dr. Now define: dsi = e(i) dqi // no sum implied These are small vectors ALONG the non-orthogonal non-unit basis vectors. Then define the vector differential area to be dS1 = ds2 x ds3 or dSi = ½ εijk dsj x dsk // sums implied We know this is a parallelepiped with area dS1 which is dS1 = | e(2)x e(3)| dq2 dq3 But e(2)x e(3) e(2)x e(3) = (e(2) e(2)) (e(3) e(3)) - (e(2) e(3)) (e(3) e(2)) = g22g33- g232 so we conclude that dS1 = dq2 dq3 // 5-6a and in general dSi = ½ εijk dqj dqk // sums implied Volume Element. 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. The claim is that v2 = det(g) so we get dV = dq1 dq2 dq3 I prove this in the next section, and have just ignored the book's proof because I wanted to do it my own way. Trying to sharpen the tools which have become very dull. ______________________________________________________________________________ Here is my proof that det(g) = v2 First, we know that v2 = [e(1) e(2) x e(3)]2 and we know that det(g) = εijk g1ig2jg3k // sums implied gij ≡ e(i) e(j) So we can write det(g) = εijk g1ig2jg3k = εijk (e(1) e(i)) (e(2) e(j)) (e(3) e(k)) so we want to show that det(g) = v2 or εijk (e(1) e(i)) (e(2) e(j)) (e(3) e(k)) = [e(1) e(2) x e(3)]2 or LHS = RHS OK, here comes the brute force method that has to work! For a shorthand notation, let's write e(1)= 1 and e(j) = j , etc. Then we want to show that LHS = RHS or εijk (1 i) (2 j) (3 k) = [1 2 x 3]2 or, showing all the vector components, εijk (1m im) (2n jn) (3s ks) = [1mεmns2n 3s] [1aεabc2b 3c] Now rewrite the second factor on the right as [1aεabc2b 3c] = 1/6 * εijk [ia εabc jb kc] The reason this is right is that [ia εabc jb kc] = Fijk is a totally antisymmetric rank-3 tensor. For example, if we swap i and j, we compensate by swapping a and b, and then we undo that by doing a swap on the ab indices of the ε tensor, but that makes a minus sign. So, the idea is this F123 =(1/6) εijkFijk because all 6 non-zero terms in the sum are identical. So we now have to prove the following: εijk (1m im) (2n jn) (3s ks) = εijk (1m ia) (2n jb) (3s kc) { 1/6 * εmns εabc} where the LHS has sum on mns and ijk, and RHS side has these sums plus an abc sum. Let's now write the following grouping of terms which appears on the RHS Aabcmns ≡ εijk (1m ia) (2n jb) (3s kc) // implied sum on ijk This is a totally antisymmetric rank-3 tensor in abc indices for exactly the reason given above for tensor F. In terms of this Aabc object, we have RHS = Aabcmns { 1/6 * εmns εabc} which means = Σmns εmns [ 1/6 * Σabc ( Aabcmns εabc) ] The outer mns sum has exactly 6 non-vanishing terms due to the εmns factor. Thus, contributing values of mns must be permutations of 123. For a particular contributing set of values mns, look at the inner sum. We know that each of the 6 non-vanishing terms in the inner abc sum is identical, because the summand is the product of two antisymmetric tensors on the abc indices. Therefore, let's pick one term in the sum and multiply by 6. Thus, we pick the term abc = mns where, as just noted, mns is a contributing combination of indices. Then , using the fact that εmns2 = 1 for a contributing term, we have RHS = Σmns εmns [ 1/6 * 6 * ( Amnsmns { εmns}) ] = Σmns Amnsmns = Σmns εijk (1m im) (2n jn) (3s ks) = LHS ! QED. _________________________________________________________________________________ Question: Does the above presentation apply to a 4D space? Assume dqμ in such a space. We can still write out these things: (all contravariant to start with) ds2 = gμνdqμdqν gμν = Σi Tμκ Tνκ Tμν = (∂xμ/∂qν) dsμ = dqμ for distance along the μ curve Then for p192 D we would have dxμ = (∂xμ/∂qκ) dqκ = Tμκ dqκ For p193 A we might try, exactly as before, e(μ)κ = Tμκ A = aμ e(μ) Now comes the first sign of trouble. Before we said e(i) ≡ ½ εijk e(j) x e(k)/v But we could generalize this by saying e(μν) ≡ ½ εμνκσ e(κ) x e(σ)/v but now we are in trouble because there seem to be 6 of these. You need to always use εabcd in this world, so axb is hard to define. Here is another possibility: e(i)a ≡ ½ εijk εabce(j)b e(k)c/v // 3D one in components e(α)a ≡ (1/6) εαβγδ εabcde(β)b e(γ) c e(δ)d /v where we now combine three base vectors to make the contravariant base vector. The v would be v = e(1) e(2)x e(3) = e(1)nεnmse(2)me(3)s => v = e(1)μεμναβe(2)νe(3)α e(4)β = εμναβ e(1)μe(2)νe(3)α e(4)β which seems the right thing and generalizable to more dimensions as well. Now let's check on the all-important rule 5-79: e(m) e(n) = e(m)a e(n)a = (1/6) εmβγδ εabcd e(β)b e(γ) c e(δ)d /v e(n)a = εm123 εabcd e(1)b e(2) c e(3)d /v e(n)a If m and n are different, we get zero unless m = 4. Assuming m = 4 we get e(4) e(n) = ε4123 εabcd e(1)b e(2) c e(3)d /v e(n)a Suppose n = 3. Then we have this grouping e(3)d e(3)a εabcd = 0. So we recover our desired result if m and n are different. If they are the same we get e(m) e(m) = (1/6) εmβγδ εabcd e(β)b e(γ) c e(δ)d /v e(m)a = ε1234 εabcd e(2)b e(3) c e(4)d /v e(1)a // where we pick a term and x6 = v/v = 1 so this does work, very good. I think everything else will go through now, and we will have gμν = e(μ) e(ν) = e(μ)κ e(ν)κ A surface element would now be a tensor instead of a vector since 12 kinds of areas = 4x3 dSμν,ρσ = εμναβ(dsα)ρ(dsβ )σ dsα = e(α)dqα no sum α as in 5-76 = εμναβ e(α)ρ e(β)σ dqα dqβ What is the magnitude of our "area" with label μν ? It is probably this: |dSμν|2 = dSμν,ρσ dSμν,ρσ = εμναβ e(α)ρ e(β)σ dqα dqβ εμνα'β' e(α')ρ e(β')σ dqα' dqβ' = εμναβ εμνα'β' e(α)ρ e(β)σ e(α')ρ e(β')σ dqα dqβ dqα' dqβ' Now remember that gαβ = e(α)ρ e(β)ρ = e(α) e(β) so we can write the above as (ignoring combinatoric factors now! ) |dSμν|2 = εμναβ εμνα'β' gαα' gββ' dqα dqβ dqα'dqβ' There are only four terms in the sum, here is an example |dS12|2 = ε12αβ ε12α'β' gαα' gββ' dqα dqβ dqα'dqβ' and here is one of the four terms probably all the same |dS12|2 = ε1234 ε1234 g33g44 dq3 dq4 dq3 dq4 = g33g44 dq3 dq4 dq3 dq4 so you would say then that |dS12| = dq3 dq4 dq3 dq4 A 3-volume element might be this: dAαμνκ = εαβδγ(dsβ)μ (dsδ)ν(dsβ )κ = εαβδγ (eβ)μ (eδ)ν(eγ )κ dqβ dqδ dqγ I would guess that the magnitude of this 3-area is going to be |dAα|2 = dAαμνκ dAαμνκ = εαβδγ (eβ)μ (eδ)ν(eγ )κ dqβ dqδ dqγ εαβ'δ'γ' (eβ')μ (eδ')ν(eγ')κ dqβ' dqδ' dqγ' = εαβδγ εαβ'δ'γ' (eβ)μ (eβ')μ(eδ)ν(eδ')ν(eγ )κ(eγ')κ dqβ dqδ dqγ dqβ' dqδ' dqγ' = εαβδγ εαβ'δ'γ'gββ'gδδ'gγγ' dqβ dqδ dqγ dqβ' dqδ' dqγ' Now there are 6 terms on the right and they are not all the same. Here is one term |dA1|2 = ε1234 ε123 g22g33g44dq2 dq3 dq4 dq2 dq3 dq4 = g22g33g44dq2 dq3 dq4 dq2 dq3 dq4 Perhaps the sum of all terms is the determinant of the 1-cofactor of the gμν matrix And my guess for the 4-volume is that you get dV = dq1 dq2 dq3 dq4 OK, enough of that. This was just an exercise, and someday I will find the above all written up somewhere, perhaps in Weinberg or perhaps elsewhere. I think the whole idea of covariant and contravariant can be extended to any number of dimensions N, with the "basis vectors" and the metric tensor matrix being an N x N matrix. I don't really know what this subject is called. It should have the words "tensor" and "coordinate systems" or something like that. Obviously the Lorentz world would be a simple case N = 4. So for now, let's move on to the next section. Probably you should learn about better "tools" for doing all this tensor stuff before blindly tackling problems in the area. Comments The material of the above sections needs some serious "integration". I have attempted to do this in a separate document MM Chap 4-5 integration. Section 5.17. The Differential Operators in Tensor Notation We are in the q-space which I called x' earlier and I could put primes on everything, but let's not. NOTE: I have detailed out this section in a separate document "divergence etc.doc" (1) Gradient: Nothing very exciting here, we have φ = ei ∂iφ This is the change of a function φ with respect to a change in coordinate qi , summed. This is the rule we have always used for gradient. Were we to move the qi coordinate by δqi, function changes by ∂iφ and we weight this by the basis vector in this direction. M&M sometimes are not to helpful. (2) Divergence. We have to contract the Christoffel symbol on two indices and we get to the left equality in equation D on page 196. But then we need a trick that I had to go look up here http://books.google.com/books?id=8vlGhlxqZjsC&pg=PA57&lpg=PA57&dq=Christoffel+symbol+contracted&source=web&ots=8rH9dL-iqO&sig=sh5L0ixGcpTJA_8iwfJlrMo3vdA#PPA32,M1 Here is the trick: ∂j ln(g) = [ ∂ ln(g)/∂gik][∂gik/∂qj ] Remember that we have gij(qj), the metric tensor in general is a function of the coordinates. But the determinant g is a function of all the elements of gij. So, determinant g is a function of the gik which are functions of the qj so we are just using the chain rule above. I always forget to think about such things. Now look at the first bracket [ ∂ ln(g)/∂gik] = (1/g) (∂g/∂gik) But this is the derivative they got in equation C where we learned (∂g/∂gik) = ggik, so [ ∂ ln(g)/∂gik] = (1/g) (∂g/∂gik) = (1/g)( ggik) = gik So we then have ∂j ln(g) = gik [∂gik/∂qj ] => ½ ∂j ln(g) = ½ gik [∂gik/∂qj ] But this rightmost combination is {ij,i} from the left equality in D which I am happy with. So this shows that {ij,i} = ½ gik [∂gik/∂qj ] = ½ ∂j ln(g) = ½ (1/g) ∂jg But we can also write this as = ½ ∂j ln(g) = ∂j( ½ ln(g) ) = ∂j ln = (1/)( ∂j) So thanks to this result, we end up with the divergence as shown in 5-20a. An interesting formula: V = (1/) ∂i[Vi] so all you need to know here is and you have it. Since this is a scalar, we don't have those basis vectors flying around at all. (3) The Laplacian is much messier, 5-21a. You now have three things inside the derivative: 2φ = (1/) ∂i[ gij ∂jφ ] (4) The curl discussion is less clear. First, the usual difference thing DOES transform as a true tensor because the {..} symbols cancel out between these two terms. But then they say that some tensor is "invariant to such a transformation". I think they just mean that it keeps its form because it is a true tensor, lousy wording. And yes, it is antisymmetric so there are only three components of Vij. I will try this one on my own. Let unprimed be Cartesian, let primed be some new frame with qi. Then define: Vij = ∂iVj - ∂jVi In the new primed frame this is V'ij = ∂'iV'j - ∂'jV'i But we know that ∂'i = Tim ∂m and V'j = Tjn Vn, so we find that ∂'iV'j = Tim ∂m (Tjn Vn) = Tim (∂m Tjn) Vn + Tim Tjn(∂m Vn) Luckily, I can now quote my "extra facts added" where (∂s Tce) = – Tae Sasb Tcb where (∂sTab) = Sasb or (∂m Tjn) = – Tan Samb Tjb I can install this above to get ∂'iV'j = Tim (∂m Tjn) Vn + Tim Tjn(∂m Vn) = – Tim Tan Samb Tjb Vn + Tim Tjn(∂m Vn) = – Tia Tjb Ttn Stab Vn + Tia Tjb(∂a Vb) Well, all I am doing here is showing that these first terms will cancel and the result transforms like a tensor, but I already know that. That is, I know this: V'ij = (∂'iV'j - ∂'jV'i) = Tia Tjb(∂aVb - ∂bVa) So, this is how you compute the curl in the primed system. Except I have not yet defined the curl. I know that in the unprimed Cartesian system we have curl V = i εijk (∂jVk ) = 1 ε123 (∂2V3 - ∂3V2) + ... = 1 V32 + ... // analysis curl Well, this is really what equation C says, except for an overall minus sign noted in the footnote. So we are doing the "tensorial curl" here, and we will have to add a minus sign to get the "analysis curl".. If we had a unit transformation, we would have our curl as x V = V12e(3) + V23e(1) + V31e(2) = C3 e(3) + V1e(1) + V2e(2) //tensor curl and here we have covariant vector against contravariant base vectors. So in Cartesian what I have just written is the same as their equation C because e(3)= and so on. But I guess they are telling us to start with the other form x V = V12 (e1 x e2) + V23 (e2 x e3)+ V31 (e3 x e1) = ½ Vij (ei x ej) since V is asymmetric We use covariant V and contravariant base vectors. I give up, I cannot justify using this form versus the other form, perhaps I need to think more about what "curl" means. But let's accept this for the moment. It does agree with Cartesian in the unprimed frame, and in the primed frame = the q frame, it is just what it says. If we now replace with the covariant base vectors, we have things like this (e1 x e2) = e3/v from 5-80 where v' = 1/v. Then in terms of these basis vectors we pick up a v = in the denominator and we get x V = (1/) [ V12 e3 + V23 e1+ V31 e2 ] and recall that these lower case basis vectors are the ones aligned with the q curves like or . In general these are not normalized and we know that e3e3 = g33 for example, so I guess you need extra factors in the above if you want to use normalized vectors. Note added: What about my friend the A operator that I think started all this reading? From above we have these two facts: φ = ei ∂iφ A = Akek Thus Aφ = A = Akek ei ∂iφ = Akδki ∂iφ = Ai∂iφ Seems pretty simple which is why M&M never mentioned it. But see below: Now the grand finale: we consider "orthogonal" coordinates so the ei are perpendicular (claim 1). I agree that in this case, e3 = e1 x e2 / v must be parallel to e3 . (claim 2). Since eiei = 1, we know | ei| = 1/| ei| which leads to equation D whose mag says exactly this. We already know the dot products are 0 because this is what "orthogonal" means. Now, gij = (1/g) Gij but Gij = cof(gij) = δij Gii because g is diagonal so only diagonal elements will have non-zero subdeterminants, ie, cofactors. So consider G11= cof(g11) = g22g33 = g/g11 . Therefore g11 = (1/g) G11 = (1/g) (g/g11 ) = g11 (claim 4). that is gii = gii i = 1,2,3 In this case, what to our various formulas say? On the left I put the full result, then simplify for orthogonal systems if possible. φ = ei ∂iφ = (ei/gii) ∂iφ = ( 1/) i ∂iφ V = (1/) ∂i[Vi] // no simplification g = g11g22g33 = Πi gii 2φ = (1/) ∂i[ gij ∂jφ ] = (1/) ∂i[gii ∂iφ ] x V = (1/) [ V12 e3 + V23 e1+ V31 e2 ] // no simplification And so concludes section 5.17 which is showing us the "differential operators" in "q coordinates" in the general case of non-orthogonal coordinates! Note that = Qi of section 5., claim 5. And, = Q1Q2Q3 Well, let's go a little more right in this section to anticipate what 5.2 will say. We want unit vectors, and we now ui = ei/Qi This for gradient we get φ = ( 1/) i ∂iφ = Σi (1/Qi) (∂iφ) i will be 5.9 Now look at divergence. V = (1/) ∂i[V'i] V = Σi ei V'i = Σi ui (QiV'i) = Σi ui Vi Although it was not stated on page 196, the authors assumed that the components of V are defined as I show above and which I now call V'i. In practice, we really want to define "Vi" to be projections onto unit vectors, so you see how I have done this above (see page 194 C). So we have to replace V'i = Vi/ Qi so I then get V = (1/Q1Q2Q3) Σi ∂i[Q1Q2Q3Vi'] = (1/Q1Q2Q3){ ∂1(Q1Q2Q3V1/Q1) + ... = (1/Q1Q2Q3){ ∂1(Q2Q3V1) + ∂2(Q3Q1V2) + ∂3(Q1Q2V3) } which now agrees with page 175 5-20! On page 175, all contravariant indices are written lower as we normally do when not worrying about up and down details. Our orthogonally simplified Laplacian was this: 2φ = (1/) ∂i[gii ∂iφ ] = (1/Q1Q2Q3) ∂i[ Qi-2 Q1Q2Q3 ∂iφ] = (1/Q1Q2Q3){ ∂1[Q2Q3/Q1 * ∂1φ] + ∂2[Q3Q1/Q2 * ∂2φ] + ∂3[Q1Q2/Q3 * ∂3φ] } and this agrees with page 175 5-21. Finally we come to the curl which we show above as x V = (1/) [ V'12 e3 + cyclic ] But again the V's inside the Vij are really the primed ones as discussed in the divergence above, so I put primes above. So consider: (the V component goes with the first index! ) V'12 = ∂2V1 - ∂1V2 We need first to get these V' components to be contravariant, so get V'i = giiV'i = Qi2V'i Then we have V'12 = ∂2( Q12V'1 ) - ∂1 (Q22V'2) But now we have to get into the unit vector components, as above, so we use V'i = Vi/ Qi and this gives V'12 = ∂2( Q1V1 ) - ∂1 (Q2V2) But u3 = e3/Q3, so our curl equation becomes x V = (1/) [ V'12 e3 + cyclic ] = (1/Q1Q2Q3)[ { ∂2( Q1V1 ) - ∂1 (Q2V2)} Q3u3 + cyclic ] Now lower the V indices to go back to conventional non-tensor notation to get x V = (1/Q1Q2Q3)[ { ∂2( Q1V1 ) - ∂1 (Q2V2)} Q3u3 + cyclic ] (they are really still contravariant in meaning). Now rewrite as x V = (1/Q1Q2Q3)[ { ∂2( Q1V1 ) - ∂1 (Q2V2)} Q3u3 + cyclic ] If you define a matrix in this way M1i = Qiui first row M2i = ∂i second row M3i = QiVi third row and if you notice that there are 6 terms in [..] above which can be written as - [...] = εijk M1iM2jM3k = ε321 M13 M22 M31 + ... = ε321 Q3u3 ∂2 Q1V1 + ... = – Q3u3 ∂2 Q1V1 + ... so we get det(M) = - [...] = – Q3u3 ∂2 Q1V1 + ... and we end up with x Vtensor convention = (1/Q1Q2Q3)[ – det(M) ] which gives our desired result x Vanalysis convention = (1/Q1Q2Q3) det(M) where M is the matrix shown on page 176 which is the same as what I wrote above. And noted added on Aφ : Above we had Aφ = A = Akek ei ∂iφ = Akδki ∂iφ = A'i∂iφ But once again, I think we have to remember that the A' are not the A's, just as in the various cases above. So we would write A'i = Ai/ Qi and then we get Aφ = Σi (1/Qi) (∂iφ) / I am pretty sure this is correct! Can I find this stated somewhere? It is just omitted from M&M's list, so look on the web. No one on the web mentions this operator combination! It does show up in the A x ( x B) identity and in (AB) . How about (AA) where my identity says this: (AA) = 2 (A)A + 2 A x ( x A) But I don't even know if this identity is true in curvilinear coordinates. General rule: I suppose in any identity, you need to think about where new Q factors might go! Review of the above: In this section we developed formulas for gradient, divergence, Laplacian and curl based on general non-orthogonal coordinates. Then we specialized these general formulas for orthogonal coordinates. Then we used the Qi notation, made all vector components lower although we mean them as contravariant, and we thus derived the four major results of Section 5.2 which section we have not even read yet! I think it is good to get these main results from the general tensor formalism. Tomorrow I will read 5.2 and some of the following sections! Section 5.2. Vector Relations in Curvilinear Coordinates (174) gaga