M&M ch4_5 ponderings

M&M Chapter 4,5 Ponderings PhL 3.26.05 Here I wrote down questions that occurred to me, and then my answers to these questions. Question #1: What is meant by the following question: "What does xA look like in curvilinear coordinates? " 1 Comments: 4 Conclusion: 5 Question #1.4 Show that A = B x C is form invariant under rotations. 5 Question #1.5. Show that the equation A =xC is form invariant under rotations. 6 Question #1.6 Show that A = B x C is NOT form invariant under an arbitrary transformation! 6 Question #2: Is εijk really a tensor? Transforms like one? What is ε'ijk in the transformed frame? 7 Question #3: How does "the Jacobian" fit into things? 8 Question #4: How does the volume element transform, and what does this even mean? 8 Question #5: What is det(T) for an arbitrary transformation? 8 Question #6: Are the vector identities true in all coordinates? What does this question mean ? 8 Test #1: (φF) = Fφ + φF 9 Test #2: (AxB) = B(xA) - A(xB) 9 Footnote to Test #2: 11 Question #7: What are the components of the various basis vectors? 12 Cartesian components of the basis vectors 13 Components of the basis vectors with respect to the basis vectors! 14 Question #8. Is Tab a Tensor? How do you raise and lower indices on Tab ? 16 Question #1: What is meant by the following question: "What does xA look like in curvilinear coordinates? " Answer #1. This is the wrong answer! But I did the math, so will keep it here. See Question #1.6 below, and see conclusions at the end of this answer. We are wondering here about the form invariance of the equation B = xA under an arbitrary transformation. We know the answer, but here it is anyway. I compute the thing in Cartesian coordinates. I then do an arbitrary transform to some x' coordinates, and I just write the thing out then in those x' coordinates and that is what it "looks like". Let's treat this as if we had this starting equation B = xA and we want to get everything primed to see this in a "transformed frame". We have seen how this comes through with a simple rotation (see question 1.5 below), but here we shall do a general transformation. Let's carry this out now. In Cartesians we know we have this: xA = (x)(∂yAz- ∂zAz) + cyclic = i εijk ∂jAk(x) B = Bi i Here I am using my knowledge that we really mean contravariant indices on vectors, so I show this now explicitly in the above. And of course ∂j means ∂/xj. Now let's apply our usual "inverse transform rules" to show all this in primed coordinates. Those rules say the following, and I try to vary the name of the dummy index anticipating insertion into the above. Notice that the forward transform is x'a = Tab xb so T has all its usual meanings. xa = Tba x'b xa = Tba x'b Tab ≡ Tab ≡ (i)a = Tba (ei) b ∂j = Tcj ∂'c Ak(x) = Tek A'e(x') εijk = (1/) ε'ijk (see question #2 below) Now, in the starting Cartesian frame we have the obvious basis unit vectors i, but we know that in our transformed frame we will have the ei basis vectors (non-orthogonal and non-unit), where the reverse transformation is this (i)a = Tba (ei) b as shown in the above equation set. Thus we start off with Bi ia = εijk ∂j(Ak) ia Bi = εijk ∂j(Ak) // starting point and we make all these replacements: Bi = Tbi B'b εijk = (1/) ε'ijk ∂j = Tej ∂'e Ak = Trk A'r So install all these things to get: Tbi B'b = (1/) ε'ijk Tej ∂'e(Trk A'r) Now we want to clear the T from the left side, so apply Tsi and sum on i to get δbs so we get B's = (1/) Tsi ε'ijk Tej ∂'e(Trk A'r) "question 1 form" I now know the correct answer to the question which is this: B's = (1/) ε'sjk ∂'j( g'kaA'a) "M&M form" where I put primes on everything (x' = q). In this last "correct" result, vectors A' and B' are those expanded directly on the ei (which are not unit vectors). It is hard to imagine that these results agree. We would have to show the following: Tsi ε'ijk Tej ∂'e(Trk A'r) = ε'sek ∂'e( g'krA'r) Let's now remove all primes keeping in mind that we are really talking q-space q = x'; Tsi εijk Tej ∂e(Trk Ar) = εsek ∂e( gkrAr) I do know the following "covariant forwards" transform Q'ab = Tac Tbd Qcd so apply this to the metric tensor to ge g'ab = Tac Tbd δcd = Tac Tbc => g'kr = Tkt Trt which could be installed on the RHS to give Tsi εijk Tej ∂e(Trk Ar) = εsek ∂e(Tkt Trt Ar) This looks unlikely because we have three T on the left and only two T on the right. But let's continue until we know for sure we are dead meat: Let's swap the t and k summation index names on the RHS Tsi εijk Tej ∂e(Trk Ar) = εset ∂e(Ttk Trk Ar) LHS = RHS Now blow out all the derivatives on both sides LHS = Tsi εijk Tej ∂e(Trk) Ar + Tsi εijk Tej Trk ∂e(Ar) RHS = εset ∂e(Ttk) Trk Ar+ εset Ttk ∂e(Trk) Ar + Tsi εijk Tej Trk ∂e(Ar) Inserted comment: suppose T is constant in space. then have Tsi εijk Tej Trk ∂e(Ar) = Tsi εijk Tej Trk ∂e(Ar) = true. Now A is supposed to be an arbitrary vector field, so the Ar and ∂e(Ar) must separately be equal, Tsi εijk Tej ∂e(Trk) Ar = εset ∂e(Ttk) Trk Ar+ εset Ttk ∂e(Trk) Ar (1) Tsi εijk Tej Trk ∂e(Ar) = εset Ttk Trk ∂e(Ar) (2) Inserted comment: suppose T is constant in space. both are true. since we revert to Ttk Trk= gtr= δtr so (2)R = εser ∂e(Ar), and (2)L will make a detT thing. But again, since they are arbitrary, we need this to be true Tsi εijk Tej ∂e(Trk) = εset ∂e(Ttk) Trk + εset Ttk ∂e(Trk) (1) Tsi εijk Tej Trk = εset Ttk Trk (2) Inserted comment: Again, will say ε = ε, so (2) is still true for rotation The simplest of these is (2), and the number of T's is just wrong! We can absorb three T's into nothing sometimes as a det, and we might absorb two into a g, but I don't see how this can possibly balance 3T against 2T. Something is not right. Comments: In question 1.6 below, I show that the very simple equation B = C x A is form-invariant under the transformation T only if T is a rotation, which means gij = δij. Therefore, we can think without much stretch that B = xA might retain its form in the primed system under these same circumstances, or it might not. If yes, we can surely show that our two results (question 1 and M&M) above agree: question 1: B's = (1/) Tsi ε'ijk Tej ∂'e(Trk A'r) MM: B's = (1/) ε'sjk ∂'j( g'kaA'a) We have g' = 1 and g'ka = δka so write as question 1: B's = Tsi εijt Tej ∂e(Tkt Ak) MM: B's = εsek ∂e(Ak) where we shuffled index names and removed primes just to make it easier to fiddle with. The above equations are the same if T is the identity, at least we have that much! We can write the question 1 version as two terms: B's = Tsi εijt Tej ∂e(Tkt) Ak + Tsi εijt Tej Tkt ∂e(Ak) = (εijt Tsi Tej) ∂e(Tkt) Ak + (εijt Tsi Tej Tkt) ∂e(Ak) Now write Tkt = (T-1)tk = (T-1,T)kt = Tkt if T unitary. Then we have for our question 1 form = (εijt Tsi Tej) ∂e(Tkt) Ak + (εijt Tsi Tej Tkt ) ∂e(Ak) = εsek ∂e(Ak)+ (εijt Tsi Tej) ∂e(Tkt) Ak since t unimodular. So the first term is what we want, but now we have picked up a second term due to the possible spatial dependence of our rotation ! Here is the extra term where I restore primes. extra = (εijt Tsi Tej) ∂'e(Tkt) A'k Tab ≡ so our derivative does not have any special symmetry, and we cannot argue extra = 0. Conclusion: We now know that B = xA is form invariant only under a rotation T which is the same at all points x' so there is no derivative. If this last were not the case, we would only have a rotation at one local point anyway, so they really go together. So, here is what this says. Suppose we have a "vector identity" B = xA. If we do a transformation from the Cartesian world to, say, spherical coordinates even (which are orthogonal), we do not end up with the equation B' = 'xA' . But of course we knew that already. You have to first go into the new coordinate system, and then compute the curl in that system from the definition of the curl! That definition is operationally the same in both Spherical and Cartesian coordinates. This is pretty much what M&M do in equation 5-22a on page 197, and of course what I did in "divergence...doc" Question #1.4 Show that A = B x C is form invariant under rotations. This was much harder than I thought it would be: Here we go. Start here Ai = εijkBjCk Then install Ai = RiaA'a in three places RiaA'a = εijk RjbB'b RkcC'c Now apply Rei and sum i on the left to get A'e = Rei εijk RjbB'b RkcC'c = εijk ReiRjbRkc B'bC'c Now we make our familiar observation that εijk ReiRjbRkc = Febc = totally antisym = εebcF123 so we get A'e = εebcF123 B'bC'c = εebc εijk R1iR2bR3c B'bC'c = εebc det(R) B'bC'c = εebc B'bC'c Therefore, in the new frame we have A' = B' x C', as expected. So this is an example of an equation that is form-invariant under a rotation. Question #1.5. Show that the equation A =xC is form invariant under rotations. There is a very slight change here, but we get the result we want. Let's look at how transforms dxi = Riadx'a // from above dx'j = R-1jidxi d/dxi = dx'j/dxi * d/dx'j =R-1ji d/dx'j ∂i = R-1ji ∂'j = RTji ∂'j = Rij ∂'j // R is unitary So we can compare ∂i = Rij ∂'j Bi = RijB'j Because R is unitary (real orthogonal), we find that transforms as a vector just like B, so our result is the same, and in the new frame we find A' ='xC' So this was an example of a very simple "vector identity" that is form invariant under rotations. ___________________________________________________________________________ Question #1.6 Show that A = B x C is NOT form invariant under an arbitrary transformation! Start with the inverse transforms Ai = Tai A'a Bj = Tbj B'b Ck = Tck C'c Ai = εijkBjCk => Tai A'a = εijk Tbj B'b Tck C'c Now apply Tsi to the left and sum on i to get δas so that we have A's = Tsi εijk Tbj B'b Tck C'c which we can rewrite as A's = εijk Tsi Tbj Tck B'b C'c The first T is not quite right for our det thing, so what can we do about that? We know that Tis = (T-1)si => Tsi = (T-1,T)si So we then have A's = εijk (T-1,T)si Tbj Tck B'b C'c Now if it were true that T-1 = TT we could conclude that A's = εijk Tsi Tbj Tck B'b C'c = Fsbc B'b C'c = εsbcF123 B'b C'c = εsbc εijk T1i T2j T3k B'b C'c = det(Tab) εsbc B'b C'c Conclusion: The equation A = B x C is form invariant under Ai = Tai A'a only if the following things are true about T ≡ Tab (1) T is real orthogonal so T-1 = TT (2) T is unimodular so det(T) = 1 In other words, this equation A = B x C is form invariant only of T is a rotation. Under a general transformation, it is not form invariant. Question #2: Is εijk really a tensor? Transforms like one? What is ε'ijk in the transformed frame? To answer this question, let's "give it the test" : ε'ijk = Tia Tjb Tkc εabc We can see from inspection that ε'ijk is totally antisymmetric, so we can write it as (see Question #3) ε'ijk = εijk ε'123 = εijk T1a T2b T3c εabc = εijk |Tab| = εijk So I think the answer is this: Yes, ε123 is a tensor, and in any frame it is totally antisymmetric, but it picks up a factor of in scale as you move from the Cartesian frame to another frame. Question #3: How does "the Jacobian" fit into things? We can write the Jacobian J in this way J = det( ) = det(Tab) and dV = jac dV' In the support document I have proven this fact: J = as expected. So we now have three different ways to think about the volume change J = = v where J = |Tab| g = |gij| v = e1(e2xe3) So this was a good question to ask I think., Question #4: How does the volume element transform, and what does this even mean? See previous question where all three forms are shown. You would use this in transforming a volume integral from one set of coordinates to another. Question #5: What is det(T) for an arbitrary transformation? I don't think we have any idea what this is. For a rotation it is 1 since TTT = 1 which says detT = ± 1. I think we again must conclude that T must be real orthogonal in order to have det(T+ 0 1 Question #6: Are the vector identities true in all coordinates? What does this question mean ? Ouch, this is the tough one. Here is one possible answer: Answer A: All the vector identities are true in an arbitrary curvilinear system as long as you use the correct forms for these objects: grad, div, Laplacian, curl, A and as long as you interpret the vectors as being distributed onto the curvilinear basis vectors meaning A = Aiei = Aiei. You never want to work with a Cartesian component of any vector if you are in curvilinear coordinates! We interpret the identities as being about the concepts of these operators, so they are always true. But I have my doubts. Let's try to prove some vector identities !!! Test #1: (φF) = Fφ + φF Verify in curvilinear coordinates this vector identity: (φF) = Fφ + φF Now let's bring in our general formulas for grad and div: A = (1/) ∂i(Ai) where A = Aiei (φ) = (∂iφ) ei So here we go LHS = (φF) = (1/) ∂i(φFi) RHS = Fφ + φF = F (∂iφ) ei + φ (1/) ∂i(Fi) = Fi (∂iφ) + φ (1/) ∂i(Fi) Now expand the LHS LHS = (1/) ∂i(φFi) = (1/) ∂i(φ)Fi + (1/) φ∂i(Fi) = ∂i(φ) Fi + φ (1/) ∂i(Fi) And sure enough, we have the LHS = RHS and this vector identity is confirmed in a general curvilinear system. Test #2: (AxB) = B(xA) - A(xB) Verify in curvilinear coordinates this vector identity: (AxB) = B(xA) - A(xB) Here are our general formulas: [xF]i = (1/) εijk ∂jFk F = (1/) ∂i(Fi) where F = Fiei So away we go. Oops. First we need to ask: What is AxB? AxB = Aiei x Bkek = AiBk ei x ek but we know that ei x ej = (1/v) εijk ek = (1/) εijk ek so get AxB = AiBj ei x ej = (1/) εijk AiBj ek So we get this important result we would use in any vector identity AxB = (1/) εijk AiBj ek Now we can say that AxB = [AxB]i ei = Dot both sides with ek to get AxB ek = [AxB]k so that [AxB]a = AxB ea = (1/) εijk AiBj ek ea = (1/) εija AiBj or [AxB]k = (1/) εijk AiBj It sure took me a long time to get this simple result. But just compare it to the curl formula which can be written exactly this way: [xB]k = (1/) εijk ∂iBj and now it looks very very good, they have exactly the same form. Now finally we can apply our divergence formula F = (1/) ∂k(Fk) to get LHS = (AxB) = (1/) ∂k([(1/) εijk AiBj] ) = (1/) εijk ∂k(AiBj) Meanwhile, our curl formula is [xF]i = (1/) εijk ∂jFk so we have RHS = Bi[xA]i - Ai[xB]i // why one index up and one down for dot product ??? = Bi (1/) εijk ∂jAk - Ai (1/) εijk ∂jBk = (1/)εijk { Bi ∂jAk - Ai ∂jBk } // swap j and k in second term .... = (1/)εijk { Bi ∂jAk + Ai ∂kBj } // .... and away goes that minus sign Now the LHS is this: LHS = (1/) εijk ∂k(AiBj) so our vector identity will be proven if we can show that εijk ∂k(AiBj) = εijk { Bi ∂jAk + Ai ∂kBj } Now in the first term on the RHS cycle all the dummy index names ahead one εijk { Bi ∂jAk} = εjki { Bj ∂kAi} = εijk { Bj ∂kAi} so we now have to show that εijk ∂k(AiBj) = εijk { Bj ∂kAi + Ai ∂kBj } but this is obviously true and our vector theorem is finally verified! Footnote to Test #2: In trying to show Test #2, I was erroneously doing this kind of thing: "But we will need to know the contravariant component of this, which is [AxB ]a = (1/) εijk AiBj (ek )a But what is this last factor? (ek )i (ej)i = gkj = Σi Tki Tji // how do I know both have contravariant indices? Regular ? which suggests that (ek )i = Tki Do I believe this. We probably have this fact: (e1)i = Tik( 1)k = Tik δ1k = Ti1 => (ek)i = Tik We seem to disagree here on the tilt direction! Let's just pick one to use for now. Then we have [AxB ]a = (1/) εijk AiBj (ek )a = (1/) εijk AiBj Tka which seems rather unusual and painful. Now we have..." This mistake here is that on the first line, I am specifying a Cartesian component [AxB]a, but what we really want of course is the curvilinear system component [AxB]a and they are very different! Question #7: What are the components of the various basis vectors? Before attempting an answer this question, we have to clarify our various "spaces". For this purpose, let's consider a mapping which is a simple rotation. We draw this picture: All the basis vectors ei and ei and the Cartesian unit vectors we are calling exist only in the Cartesian space on the right. A position change of dq1 in the q-space results in a position change dr in the x-space as shown. We see that a vector in q-space like dq11 is getting rotated +150 (RHR) where it maps into dr. We write this rotation as R(15). This is the "active" rotation matrix if you wanted to write it out. The above picture is based on this rule for mapping vectors: xi = R(15)ijqi = R(15)ijx'i where we do our usual identification of x' = q. However, our standard transform notation which defines our transformation matrix T is this x'a = Tabxb Therefore it follows that Tab = R(–15)ab from which we conclude that T = R-1(+15) and T-1 = R(15) Sometimes this inverse relationship can be confusing, but there it is. Cartesian components of the basis vectors In this section, we use square brackets like [ei]n to denote a Cartesian component. Now, the definition of e1 given in equation A on M&M page 193 says, for example, e1 = ∂r/∂q1 = > e1 = [e1]n = ∂xn/∂q1 = ∂xn/∂x'1 = T1n so we have found that our Cartesian components are given by [ei]n = Tin Notice, making use of Tab = (T-1)ba after the 4th equal sign, that: [e1]n = T1n = Tin δi1 = Tin ()i = (T-1)ni ()i = R(15)ni()i so we recover the fact obvious from our picture that e1 = R(15) . Thus we have ek = R(15) = T-1 So the covariant basis vectors ek is just the Cartesian unit vector transformed by T-1 = R(15) due to the way we have defined our transformation matrix T, as explained above. Of course in Cartesian space where gij = 1, we have [ei]n = (T-1)ni = [ei]n = Tin So when we are talking about Cartesian components of our basis vectors ei, it does not matter whether we put the index up or down. What about the "dual" basis vectors ei ? Well, consider δij = eiej = [ei]n[ej]n = [ei]n (T-1)ni We can then apply from the right the factor Tik and sum on i to get Tjk = [ei]k = [ei]k We can summarize our Cartesian component results as follows: [ei]n = [ei]n = Tin = Tin [ei]n = [ei]n = Tin = Tin See discussion in Question #8 below for why second index on T can be up or down with no difference. It exactly makes sense in this context! Remember that T is not a tensor so these Cartesian component objects are not tensors. Notice that ei ej = [ei]n[ej]n = Tin Tjn = gij // can put any n up or down for free so we have verified the idea that we usually only think about in terms of the basis vectors. A corresponding rule is this: ei ej = [ei]n[ej]n = Tin Tjn = gij // can put any n up or down for free In fact, we can regard this as a tensor equation set ei ej = Tin Tjn = gij Each of these three things is a way to write the same tensor gij and is of course a tensor. Components of the basis vectors with respect to the basis vectors! In this section, we use square brackets like (ei)n to denote basis vector component. First, let's just collect facts obtained in divergence etc.doc ei ej = gij` ei ej = gij = Gij/g G = cofactor(g) g = det(g) ei ei = gii = ei2 = Qi2 ei ei = gii = Gii/g v2 = det(g) v ≡ e3 e1x e2 = e1x e2 e3 = e1 e2x e3 = [e1 e2 e3] e1 ≡ e2 x e3 / v e2 ≡ e3 x e1 / v e3 ≡ e1 x e2 / v ei ej = δij e1 = v e2 x e3 e2 = v e3 x e1 e3 = v e1 x e2 1/v = e3 e1x e2 = e1x e2 e3 = e1 e2x e3 = [e1 e2 e3] = v' Now, expand these basis vectors onto the basis vectors as follows: ei = (ei)k ek => ei en = (ei)k eken => gin = (ei)n ei = (ei)k ek => ei en = (ei)k eken => gin = (ei)n ei = (ei)k ek => ei en = (ei)k eken => gin = (ei)n ei = (ei)k ek => ei en = (ei)k eken => gin = (ei)n All this can be summarized in this single statement: (ei)j = gij then fiddle the indices any way you want. Thus, the component (ei)j is just a representation of the metric tensor gij and as such is itself a rank 2 tensor. This is in stark contrast to the Cartesian component [ei]j which is not a tensor as was shown above. So we end up with this interesting situation ei ej = (ei)j = gij This is really much ado about nothing, but I was confused about "components" and here we have done them every which way. Question #8. Is Tab a Tensor? How do you raise and lower indices on Tab ? Consider the transformation of a tensor Xijgoing from Cartesian space to some x' space, so X'ab = Tai Tbj Xij In the Cartesian space our metric tensor is gij = δij. In the x' space, the metric tensor is something we shall call g'ij ≠ δij In fact we know that g'ab= Tai Tbj δij and this transforms the metric tensor itself from the Cartesian space to the x' space. Now here is the question: which metric tensor should we use to raise and lower indices on tensors? I would like to say this: for a tensor like X'ab which is in the x' space, use g'ij to raise and lower indices. And for a tensor like Xij , use δij to raise and lower indices. According to this idea, we would have Xij = Xij = Xij= Xij and I think this is correct. We are used to saying for a vector that Vi = Vi in a Cartesian space. So this is the equivalent thing for a rank-2 tensor. Then here is the zinger: which metric tensor do you use to raise and lower indices on Tab ??? Or is this thing itself not really a tensor? In looking for an answer, consider this: Tab ≡ Tab ≡ I think the answer is going to be this: (1) for Tab you use g' on the first index, but δ on the second index, just looking at the definition as a partial derivative. Then we would say Tab = Tab but Tab = g'ak Tab 2) for Tab the same is true, the second index can be up or down Tab = Tab but Tab = g'ak Tkb Conclusion: For either Tab or Tab , the second index goes with dxb = dxb and so it can be either up or down with no change. This is only true if T is a transformation from Cartesian space to x' space. More generally, for T being a transformation from x-space to x'-space, the first index of Tab or Tab is raised and lowered by the g' of the x' space, and the second index is raised or lowered by the g of the x space. This is a special object that has one "leg" in each space, so it is NOT a rank-2 tensor.