scraps3

This is the Title PhL 1.11.15 Note that page numbering is turned on in this template and view is 125%, located in phil/roaming/microsoft/templates size about 219K. If we make these definitions, (gdn)ab = g(e)ab (gup)ab = g(e)ab // both gdn and gup are symmetric then we know that gup = (gdn)-1 = cof (gdnT)/det(gdn) = cof (gdn)/det(gdn) So this gives an explicit inversion of g(e)ab to get g(e)ab. ************************* Suppose we apply our vector expansions (2.6.2) to basis vectors. Then for example, letting V = ei, ei = Σn (ei)(u)n un where un ei = (ei)(u)n = Rin // (2.6.11) ei = Σn (ei)(u)n un where un ei = (ei)(u)n = Rin // (2.6.11) ei = Σn (ei)(e)n en where en ei = (ei)(e)n = δin // (2.6.7) ei = Σn (ei)(e)n en where en ei = (ei)(e)n = g(e)ni // (2.6.7) (2.6.12) We have suddenly learned a lot more detail about the components of the vector ei in both bases. We will next do the same four expansions for ei to find the components of ei. The result will be that shown above with the i index raised everywhere, and with the idea that δin = gin so raising gives gin and the reverse of this on the last line. Not completely trusting up/down magic (trust but verify), we compute ei = Σn (ei)(u)n un where un ei = (ei)(u)n = Rin // (2.6.11) ei = Σn (ei)(u)n un where un ei = (ei)(u)n = Rin // (2.6.11) ei = Σn (ei)(e)n en where en ei = (ei)(e)n = g(e)im // (2.6.7) ei = Σn (ei)(e)n en where en ei = (ei)(e)n = δin // (2.6.7) (2.6.13) We now repeat the above two calculations for ui and then ui ui = Σn (ui)(u)n un where un ui = (ui)(u)n = δin // (2.6.9) ui = Σn (ui)(u)n un where un ui = (ui)(u)n = g(u)in // (2.6.9) ui = Σn (ui)(e)n en where en ui = (ui)(e)n = Rni // (2.6.11) ui = Σn (ui)(e)n en where en ui = (ui)(e)n = Rni // (2.6.11) (2.6.14) ui = Σn (ui)(u)n un where un ui = (ui)(u)n = g(u)in // (2.6.9) ui = Σn (ui)(u)n un where un ui = (ui)(u)n = δin // (2.6.9) ui = Σn (ui)(e)n en where en ui = (ui)(e)n = Rni // (2.6.11) ui = Σn (ui)(e)n en where en ui = (ui)(e)n = Rni // (2.6.11) (2.6.15) We now have a complete table for all the components of all the basis vectors: (ei)(u)n = Rin (ei)(u)n = Rin e basis vector components (ei)(u)n = Rin (ei)(u)n = Rin (ei)(e)n = δin (ei)(e)n = g(e)im (ei)(e)n = g(e)ni (ei)(e)n = δin (2.6.16) (ui)(u)n = δin (ui)(u)n = g(u)in u basis vector components (ui)(u)n = g(u)in (ui)(u)n = δin (ui)(e)n = Rni (ui)(e)n = Rni (ui)(e)n = Rni (ui)(e)n = Rni (2.6.17) ******************* We now translate a few of the above equations to this new Picture A' notation: First, the metric tensors still raise and lower indices, V(u)a = g(u)abV(u)b V(e)a = g(e)abV(e)b V(u)a = g(u)abV(u)b V(e)a = g(e)abV(e)b from (2.2.1) (2.6.4) The vector transformation rule and its inverse are now V(e)a = RabV(u)b V(e)a = RabV(u)b from (2.1.5) (2.6.5) V(u)b = RabV(e)a V(u)b = RabV(e)a from (2.1.9) (2.6.6) Next, we gather up equations relating to the various basis vectors: en em = g(e)nm en em = δnm en em = g(e)nm from (2.3.2) (2.6.7) en = g(e)ni ei en = g(e)ni ei from (2.3.3) (2.6.8) un um = g(u)nm un um = δnm un um = g(u)nm from (2.4.2) (2.6.9) un = g(u)ni ui un = g(u)ni un from (2.4.3) (2.6.10) en um = Rnm en um = Rnm en um = Rnm en um = Rnm from (2.4.5) (no change) (2.6.11) *************************************** 2.8 Tensor Expansions Redo or at least review this section please. Later show how the general dot rule makes sense with the outer product definition of ab. Also have to redo Section 2.9. Trying to find the right thread here. Maybe outer products has to come before this section in order to talk about tensor expansions! Here are the corresponding tensor expansions onto the ea and eaeb bases. V = Σa [V(e)]a ea (2.8.1) M = Σab [M(e)]ab eaeb (2.8.2) For the first time in Chapter 2 that the symbol appears. Just as the vector V may be expanded in the space V which is spanned by the {ea}, so the rank-2 tensor M can be expanded on the space VV which is spanned by the basis {eaeb} . The tensor is M, and the tensor components are [M(e)]ab in the eaeb basis. We want now to generalize this expansion in the following manner M = Σab [M(e,e')]ab eae'b (2.8.3) where the ea form a basis for vector space V, while the e'b form a basis for vector space W which may have a different dimension from V. How can we "project out" the components [M(e,e')]ab from the tensor M? Consider: M = Σab [M(e,e')]ab eae'b (eie'j) M = Σab [M(e,e')]ab (eie'j) (eae'b) (2.8.4) This dot operator is now operating inside the tensor product space VW, so it is a different than we used earlier in this Chapter, see (2.2.5). We define the action of in VW as follows (aa')(bb') ≡ (a b) (a' b') // = multiplication of two scalars (2.8.5) We can then continue the above analysis, (eie'j) M = Σab [M(e,e')]ab (eie'j) (eae'b) = Σab [M(e,e')]ab (ei ea)(e'j e'b) = Σab [M(e,e')]ab δia δjb // using **** = [M(e,e')]ij . (2.8.6) Thus, we have successfully projected out from M the coefficient [M(e,e')]ab. For the case W = V we get M = Σab [M(e)]ab eaeb (eiej) M = [M(e)]ij . (2.8.7) This methodology works for any tensor expansion. Here for example is a "mixed basis" expansion, M = Σab [M(e,u)]ab eaub // expansion (eiuj) M = [M(e,u)]ij // projection (2.8.8) The up/down indices always match on both sides of the projection. In the expansion, indices are "pseudo" contracted (index with label) so that the rank-2 tensor M has no indices. It is a "total tensor" which can be expanded on any set of components one wants. We can extend the idea to a tensor of any rank. Here is a sample expansion for a rank-3 tensor defined on the space VWX, where X has basis e"n : M = Σab [M(e,u',e")]abc ea u'b e"c (2.8.9) Here we have picked a perverse mixed basis for the expansion. The projection works this way, (ei u'j e"k) M = (ei u'j e"k) Σab [M(e,u',e")]abc ea u'b e"c = Σab [M(e,u',e")]abc (ei u'j e"k) (ea u'b e"c) = Σab [M(e,u',e")]abc (ei ea)(u'j u'b)(e"k e"c) = Σab [M(e,u',e")]abc δia δjb δkc = Σab [M(e,u',e")]ijk . (2.8.10) Here one sees the symbol defined for its action in the space VWX. Comments: 1. Whether the object M is really a "tensor" when the spaces are different, such as for VW, is a semantic question. For example, one would have to define the transformation of such an object in terms of two different R objects associated with two different transformations F and F' operating in V and W. [M(e,e')]ab = Raa' R' bb' [M(u,u')]a'b' // note red prime (2.8.11) 2. Notice how in the notation like [M(e,u)]ab, the label (e,u) "announces" the interpretation of the corresponding component indices which are to follow. The label (e,u) is really associated with the indices, not with M. One could perhaps make this clearer by writing Mab(e,u) where ab(e,u) is the index structure with the index types indicated. This seems a bit clunky so we shall keep the labels with M. 2.9 The Outer Product of Tensors Going back to our economical Picture A notation, consider two vectors which transform in the usual rank-1 tensor manner relative to some underlying transformation F (for which R is the linearization), a'i = Rijaj b'k = Rkmbm from (2.1.5) (2.9.1) Multiplying these equations together gives (a'ib'k) = RijRkm (ajbm) . (2.9.2) But looking at (2.1.6), we see that this object is transforming as a rank-2 tensor, therefore it is a rank-2 tensor, and we can write it as M'ik = RijRkmMjm where Mij ≡ aibj . (2.9.3) The rank-2 tensor Mij = aibj is said to be the outer product of two rank-1 tensors (vectors). This idea can be generalized ad infinitum. For example, if K is a rank-2 tensor and v is a vector, then Mijk = Kijvk (2.9.4) is a rank-3 tensor because it transforms as one, using the same argument shown above. Consider, Mabcde = KabKcd ve . (2.9.5) If K is a rank-2 tensor and v is a rank-1 tensor, then M is a rank-5 tensor. Of course since this is a "true tensor equation, indices may be shuffled any way one wants, such as Mabcde = KabKcd ve . (2.9.6) Just imagine applying g** several times to both sides of (2.9.5) to get (2.9.6). There are so many possibilities for creating outer product tensors that one sometimes forgets that not all tensors can be "factored" into products of lower rank tensors. Now reconsider this earlier expansion M = Σab [M(e)]ab eaeb . (2.9.2) Let's try to make this be an identity by taking a tensor component of both sides in the e-basis [M(e)]ij = { Σab [M(e)]ab eaeb}(e)ij = Σab [M(e)]ab (eaeb)(e)ij (2.9.7) Let us conjecture that the follow is true (a b)ij = aibj (2.9.8) and that this is true in any basis we use to specify for both sides of the equation. For example (a b)(e)ij = a(e)i b(e)j e-basis (a b)(u)ij = a(u)i b(u)j u-basis (2.9.9) After all, the equation (a b)ij = aibj is claiming that (a b)ij are the components of a rank-2 tensor called ab which is the outer product of the rank-1 tensors a and b. We now apply the first equation of (2.9.9) to the vectors a = ea and b = eb. Then if our conjecture (2.9.8) is correct, (eaeb)(e)ij = (ea)(e)i (eb)(e)j . (2.9.10) Looking back at our table in (2.6.4) [ col 3 row 3] we see that (ea)(e)i = δai . (2.9.11) Therefore, (eaeb)(e)ij = δaiδbj. (2.9.12) We can now continue the processing sequence started in (2.9.7) above, [M(e)]ij = { Σab [M(e)]ab eaeb}(e)ij = Σab [M(e)]ab (eaeb)(e)ij = Σab [M(e)]ab δaiδbj = [M(e)]ij . (2.9.13) and we say "it works!". Recall from the Comment at the end of Section 2.3 that we can regard the ea as a completely arbitrary basis. So the self-consistency of (2.9.13) is strong evidence that the conjecture (2.9.8) is quite reasonable. Fact: ab is a rank-2 tensor, because its components (ab)ij = aibj (as conjectured above) transform as a rank-2 tensor. In fact, ab is precisely the outer product of a and b, and we can write (a b)ij = aibj . (2.9.14) Since this is a true tensor equation, indices i and/or j can be lowered on both sides as desired. The component indices i and j can refer to whatever basis one likes, but one must treat both sides the same way! For example, in a perverse mixed basis (a b)(u,e)ij = a(u)ib(e)j . (2.9.15) This notion of the components of the tensor product of two tensors can be generalized with development similar to that done above. For example, (Kv)ijk = Kijvk . (2.9.16) Here we identify Kv with the outer product of the rank-2 tensor K and the rank-1 tensor v. Similarly (KM)abcd = KabMcd . (2.9.17) In another extension of the idea, we can write (abc)ijk = aibjck = the outer product of three vectors = a rank-3 tensor (2.9.18) (aKc)ijkl = aiKjkcl = the outer product of three tensors = a rank-4 tensor (2.9.19) Fact: The operator between tensors creates an outer product when one views components. It provides a convenient name for the outer product tensor. Rather than saying "the tensor M which has components Mijkl = aiKjkcl " we can just say "the tensor aKc ". (2.9.20) One can use this approach to the symbol (that is, writing down components of outer products) to develop the tensor product concept in the first place, as an alternative to the methods shown in Sections 1.1 and 1.2. One declares that X Y ≡ the outer product of the tensors X and Y. The equations like (a b)ij = aibj do have component indices, which is perhaps inelegant, but as noted above, the equations are basis-independent. The equation (a b)ij = aibj is valid in any basis one chooses, as long as that basis is used on both sides of the equation. Reminder: In general all vectors are vector fields, so for example (in Picture A notation) : Mabcde(x) = Kab(x)Kcd(x) ve(x) . (2.9.6) [(a b)(x)]ij = a(x)ib(x)j (2.9.14) 2.10 The Inner Product (Contraction) of Tensors It is easy to show that, due to the orthogonality rules ***, internal index contractions within a tensor structure behave as a scalar, which is to say, behave as if they weren't there at all. Such contractions in a tensor structure reduce the rank of the tensor by two, resulting in an "inner product". The contracting sum must occur on a "tilted pair" of indices. The standard first example to consider is this (as usual, sums on repeated indices), Mij = aibj = a rank-2 tensor, which we now contract to form: s = Mii = aibi = a rank-0 tensor (a scalar) (2.10.1) Using our notation (2.2.5) this is written s = a b (2.10.2) which is an "inner product" of two vectors. This is of course the inner product / scalar product / dot product which makes our vector space be a Hilbert space. In this example, creating an "inner product" of the two vectors ai and bj which has rank-0 is going in the opposite direction of the "outer product" that creates Mij = Mij = aibj of rank-2. The term "contraction" is more often applied to reducing the rank of tensors than is "inner product", and perhaps it is best to reserve the term "inner product" for the above dot product of two vectors. A few other examples of rank reduction by contraction: Mabcd ≡ KabQcd = rank-4 tensor (2.10.3) Tac ≡ Mabcb = KabQcb = rank-2 tensor (2.10.4) In this last example, contraction on the b index happens to occur between the two rank-2 tensors from which M was constructed as an outer product. One more step, S ≡ Taa = KabQab = rank-0 tensor (scalar) (2.10.5)