outer product section v3

Outer Products v 2 PhL 9.27.15 This is a covariant version of v2. Starts below at "start". 1.3 Components (a) the metric tensor and covariant notation Up to now we have dealt with vectors like v ϵ V and vw ϵ VW without saying much about the components of these vectors. A vector in R3 might have components (x,y,z) or (r,θ,φ), so the subject of "components" is dependent on one's "coordinate system" and that in turn means one must pay attention to the metric tensor gij of the space V. In order to talk about components, the space V must have a metric tensor, so then V is not only a vector space, it is a metric space. As shown for example in ***, in non-Cartesian spaces a vector v has two kinds of vector components called vi (contravariant) and vi (covariant), where vi = Σjgijvj and vice versa vi= Σjgijvj . In Cartesian space where gij = gij = δi,j, one has vi = vi and the distinction goes away. This is a very large topic and we cannot possibly describe all the details here, but they can be found in Ref **. When space V is given a metric tensor gij, it naturally acquires a norm, a metric, and an inner product, and it becomes a full-blown Hilbert space. One has || v ||2 = Σij gijvivj where the norm || v || is the "length" of the vector v. Then one has [d(v1,v2)]2 = Σij gij (v1-v2)i(v1-v2)j where the metric d(v1,v2) gives the "distance" between two vectors v1 and v2. Finally <v1|v2> ≡ v1 v2 ≡ Σij gij(v1)i(v2)j is the inner (scalar, dot) product of the vectors v1 and v2 which indicates something about the "projection of one vector onto the other". If one assumes that g is a matrix with components gij and vectors have upper indices, one can write the above equations in a matrix/vector notation in this manner: || v ||2 = vTg v [d(v1,v2)]2 = (v1-v2)T g (v1-v2) <v1|v2> ≡ v1 v2 ≡ v1T g v2 (b) the dual basis We often write v = Σiviei and say that the coefficients vi of this expansion of v are "the components of vector v in the ei basis". Given any such basis ei, it is possible to find another basis will we call qi for which the following is true qi ej = δi,j where has the meaning shown in the previous section, the covariant dot product. There is a fairly mechanical process for computing the vectors qi given the vectors ei, see ****. In matrix notation one has qiTg ej = δi,j Σrs grs(qi)r(ej)s = δi,j The basis qi is called the dual basis of ei . It happens that qi = Σijgij ej and for this reason, one usually writes qi as ei in analogy with vi = Σijgijvj . However, one must be conscious of the big distinction between these last two equations: vi is a component of the vector v, so i is a "component index". On the other hand, on ei and ei the i is a "label" and not an index. Components of ei are (ei)r or (ei)r and similarly for ei. So the dual basis is ei and we rewrite the above as ei ej = δi,j // true for any metric tensor gij (ei)Tg ej = δi,j Σrs grs(ei)r(ej)s = δi,j On the other hand, it turns out that ei ej = gij If one has gij = fiδi,j then ei ej = fi δi,j gij = fiδi,j orthogonal and the basis {ei} is said to be orthogonal. If gij = δi,j then ei ej = δi,j gij = δi,j orthonormal STOP. If g = 1 we get ei = ei . Is this correct? If so, then the dual basis and the basis are identical, but I know that is not the case! In Cartesian space, the reciprocal and base tangent vectors are not the same. With a metric tensor gij one can speak of the "length" of a vector v being || v || = Σijgijxi Once V has a metric tensor, it suddenly becomes not just a vector space, but also a metric space where d2(x,y) = Σij gij(x-y)i(x-y)j where d(x,y) is the metric which measures the "distance" between two vectors x and y. This in turn allows one to talk about the "length" of a vector || x || where || x || = d(x,x) = Σijgijxixj and now our vector space is also a normed linear space with norm || x ||. . The metric tensor for Cartesian space is gij = δi,j and in this case there is no distinction between contravariant and covariant vector components vi and vi, but for general gij this topic cannot be avoided. The dual basis. For any basis {ei} of vector space V, one can define a dual basis {qk} such that qiTej = δi,j. or ( (ej)1, (ej)2...) = or Σi=1n (qi)n(ej)n = δi,j When one writes v = Σiviei one implies that vi "the component of v in the ei basis". In covariant notation, this is more correctly written v = Σiviei ******************************* restart again ***************** start here 1.3 Components (a) components of a vector in V 1. Metric tensor. We now assign to vector space V a metric tensor gij . In doing so, the vector space V becomes several other space types: a normed linear space, a metric space, and a Hilbert space. It is a normed linear space because we can define the norm || a || of a vector a ϵ V in this manner || a || 2 = Σij gijaiaj where || a || is the "length" of vector v. It is a metric space because we can then define a metric d(a,b) according to [d(a,b)]2 = || a - b || 2 = Σij gij (a-b)i(a-b)j where d(a,b) is the "distance between" a and b. It is a Hilbert space because we can define an inner (scalar, dot) product of two vectors in V by a b = Σij gij aibj We are using the standard covariant notation described for example in ***. In that notation, upper indices indicate contravariant components and lower indices covariant components. Also, the Kronecker delta function is written in various ways, δij = δij = δi,j. 2. Cartesian space. In all that follows, we shall assume that the metric tensor gij = δi,j. If V were Rn, one would say that V was a Cartesian space with the Cartesian metric tensor δi,j. Although the work of this document can be generalized for an arbitrary gij, we shall be content with the Cartesian form δi,j. For this metric tensor, up and down covariant indices are the same. Therefore the above equations become || a || 2 = Σiai2 = aTa // norm [d(a,b)]2 = || a-b || 2 = Σi(a-b)i2 = (a-b)T(a-b) // metric a b = Σiaibi = aTb = bTa // dot product 3. The un basis. The components ai and bi appearing in these equations, are components relative to an orthonormal set of basis vectors un which are unit vectors which "line up with the Cartesian axes". These vectors may be written un or un, both notations being the same since g = 1. Note that n is a label, not a component index. If V were Rm , one might denote these in bold font as un or even for n = 1,2,3...m. These basis vectors are orthonormal and they are complete ( any basis must be complete) : (un)i = δn,i // unit basis vectors un um = δn,m = Σi(un)i(um)i // orthonormality Σk (uk)n (uk)m = δn,m // completeness Notice that the last two lines can be trivially verified by inserting the first line. 4. Vector components in the un basis. A vector a ϵ V can be expanded on these basis vectors in the following manner a = Σi ai ui // ai = ai since g = 1. Dotting both sides with uk and using * one finds that ai = a ui Since the components ai are components in the {ui} basis, and if we are going to be considering some other bases, it would be clearer (albeit clumsy) to write the last two equations in this manner] a = Σi ai(u) ui ai(u) = a ui To be consistent, we should then rewrite our earlier equations in this manner, || a || 2 = Σi (ai(u)) 2 = (a(u))T a(u) // norm [d(a,b)]2 = || a-b || 2 = Σi [(ai(u)- bi(u))2 // metric a b = Σiai(u)bi(u) = (a(u))T b(u) // dot product (un)i(u) = δn,i // unit basis vectors un um = Σi(un)i(u) (um)i(u) = δn,m // orthonormality of the un Σk (uk)n(u) (uk)m(u) = δn,m // completeness of the un covariant changes start here 5. Introduction of an arbitrary en basis. If en is a basis, it must be a linear combination of the un, so we let the coefficients be called cij and we write ei = Σj cij uj = Σj cij uj = Σj cij uj => cij = cij Dot both sides with uk to show then that uk ei = Σj cij(uk uj) = Σj cijδkj = cik = cik so that ei uj = cij // loops more covariant now Dot both sides of ** instead with ek to get ek ei = ek ( Σj cij uj) = Σj cij (ek uj) = Σj cij ckj ≡ hki = hik Notice in passing that ek ei = δki = hki . // as usual for a metric tensor Going the other direction we can define coefficients Cij according to, ui = Σj Cij ej . Cij = Cij first index up or down Dot both sides with uk to show then that uk ui = δk,i = Σj Cij (uk ej) = Σj Cij cjk Write this as uk ui = δik = Σj Cij (uk ej) = Σj Cij cjk = (Cc)ik //up-tilt mm Not surprisingly we find that Cc = 1 so that C = c-1 as an up-tilt matrix. Finally, dot both sides of ** with ek ek ui = ek ( Σj Cijej) = Σj Cij(ek ej) = Σj Cij hjk = (Ch)ik But the left side is cki so we find that cki = Σj Cij hjk and then cki = Σj Cij hjk = Cik so cki = Cik cki = cki Cij = Cij continue here, I don't know where this is going. cT = Ch = c-1h and therefore ccT = h, which agrees with a result previously found. We add one more matrix H to this little zoo as follows, H ≡ CTC = (c-1)T(c-1) = (ccT)-1 = h-1 Here then are the conclusions on the relationship between the ei and the ui : ei = Σj cij uj ui = Σj Cij ej ei uj = cij ek ei = hki = hik ui uj = δi,j Cc = 1 Hh = 1 h = ccT H = CTC Since the ui are known, once we are handed a basis ej we may compute cij = ei uj and then we know C = c-1 and h = ccT and H = h-1 so all is known about this new basis ej. The matrices, c,C,h,H are all square n x n matrices so there is no issue of left and right inverses being different or not existing. Since we know that both c and C exist, since C = c-1 we know that det(c) ≠ 0 and since c = C-1 we know that det(C) ≠ 0. Then det(h) = det(ccT) = [det(c)]2 ≠ 0 and also det(H) ≠ 0. So all the matrices are non-singular. 6. Vector components in the en basis. Using the notation suggested earlier, we expand a vector a ϵ V as follows a = Σi ai(e) ei ai(e) = a ei which we compare to the corresponding expansion in the un basis a = Σi ai(u) ui ai(u) = a ui Therefore we find that ai(e) = a ei = a [ Σj cij uj] = Σj cij ( a uj) = Σj cij aj(u) Going the other way ai(u) = a ui = ( Σj aj(e) ej) ui = Σj aj(e)(ej ui) = Σj aj(e)cji Thus, in matrix notation, a(e) = c a(u) and a(u) = C a(e) (**) ai(e) = cij aj(u) ai(u) = cji aj(e) Now suppose you do both these at the same time ai(e) = cij aj(u) = cij[ ckj ak(e)] = Σk [Σj(cijckj)]ak(e) so that Σj(cijckj) = δik Does this agree with earlier result? Σj(cijckj) = Σj(Cjickj) = Σj(ckjCji) = (cC)ki = δki yes agrees. What corresponds to the basis change matrix cij in tensor doc? Perhaps it is Rij for the transformation from u-space to x-space as I was pursuing elsewhere. Note in passing that, cij cji = Cji cji Applying the projection rules shown in (**) to the various basis vectors uj and ej we can relate the components of these basis vectors in the two basis to our matrices c and h: (uj)a(e) = uj ea = caj (ej)a(e) = ej ea = hja (uj)a(u) = uj ua = δj,a (ej)a(u) = ej ua = cja How is a dot product like a b expressed in en components? a b = (a(u))T b(u) = [C a(e)]T [Cb(e)] = a(e)T CTC b(e) = a(e)T H b(e) = Σij Hij ai(e)bi(e) = a(e)T H b(e) . In Section 1 above we wrote covariant metric tensors with overbars such as where = g-1. So momentarily identifying H with and h = H-1 = -1 = g, the above dot product looks like this a b = Σij ij ai(e)bi(e) = H g = h Comparing this to **, we see that in the ei basis, it is as if we have a covariant metric tensor = H which is needed to compute dot products. 7. Interpretation of H as a metric tensor in e-space. On can put this into the context of tensor doc by considering this picture: The picture is intended for a general non-linear transformation x' = F(x) from x-space to x'-space where x has coordinates xi in x-space and x' has coordinates x'i in x'-space. In our application of this picture, the transformation is linear [v(e) = c v(u)] , so x'-space is the same as x-space and is the vector space we call V. We then identify un as the basis for x-space on the right and en as the basis for x'-space on the left. The coordinates of a vector would be vi = vi(u) on the right, and v'i = vi(e) on the left. The space on the right has the Cartesian metric tensor gij = δi,j, whereas the space on the left has g'ij = Hij. We might relabel the above picture for our present application in this way (and to our "developmental notation") In the above, we discussed ui and ei as being bases for the vector space V which has dim(V) = n. Imagine the same discussion for vector space W and call those basis vectors u'i and e'i . So V dim(V) = n un = Cartesian basis en = general basis c,C,h,H W dim(W) = n' u'n = Cartesian basis e'n = general basis c',C',h',H' A transformation picture analogous to the above but for W space is then 8. Components of the tensor product. Clearly (un u'm) is a vector in the vector space VW. We assert that the u,u'-space components of this vector are given by, (ui u'j)ab(u,u') = (ui)a(u) (u'j)(u')b = δi,a δj,b . This seems the logical extension to the direct product space of the V space results that (ui)a(u) = δi,a in V (u'b)b(u') = δj,b in W The object on the left of *** can be written as Tab(u,u'). These are the components of a rank-2 tensor T in the u-u' basis. This tensor can be transformed to e-e' space using the two Picture B drawings above, Tab(e,e') = Σcd cac c'bd Tcd(u,u') Recall that the strong meaning of the word "tensor" is always connected with some underlying transformation, and here we see that transformation explicitly stated. As an application of this rank-2 tensor transformation, consider (ui u'j)ab(e,e') = Σcd cac c'bd (ui u'j)cd(u,u') = Σcd cac c'bd (ui)c(u) (u'j)d(u') // from ***** = [ Σc cac (ui)c(u)] [ Σd c'bd (u'j)d(u')] = [ cui(u)]a [ c'u'j(u')]b // vector notation = (ui)a(e) (u'j)b(e') // from ** Therefore we have demonstrated that (ui u'j)ab(u,u') = (ui)a(u) (u'j)b(u') (ui u'j)ab(e,e') = (ui)a(e) (u'j)b(e') This shows that one can write the equation as (ui u'j)ab = (ui)a (u'j)b with the understanding that the components must be taken in the same basis set of the two sides. With that understanding, this unadorned equation is true in all bases (of the type we have been dealing with). Recall that ui and u'i are specific bases, but ei and e'i are arbitrary bases. Consider next the components of the tensor product of two vectors v ϵ V and w ϵ W, (vw)ab(u,u') = ( [Σivi(u)ui] [Σjwj(u')u'j] )ab(u,u') = Σijvi(u)wj(u') (uiu'j)ab(u,u') = Σijvi(u)wj(u') (ui)a(u) (u'j)b(u') = [Σi vi(u) (ui)a(u) ] [ Σj wj(u') (u'j)b(u') ] = va(u)wb(u') . We can repeat the above lines with u→e and u→ e' everywhere to conclude that (vw)ab(e,e') = va(e)wb(e') Not surprisingly, we find that we can write the components of the tensor product of two vectors in the generic manner (vw)ab = vawb without the clutter of basis indicators. We just have to use the same basis set on both sides. Finally, consider the components of a general element of VW in the u-u' basis: Tab(u,u') = [Σij Fij(u,u') (ui u'j)(u,u')]ab = Σij Fij(u,u')(ui)a(u) (u'j)b(u') = Σij Fij(u,u')δi,a δj,b = Fab(u,u') Thus, in the u-u' basis the tensor coefficients Fab(u,u') are the same as the tensor components Tab(u,u'). In the other basis we find instead Tab(e,e') = [Σij Fij(e,e') (ei e'j)(e,e')]ab = Σij Fij(e,e')(ei)a(e) (e'j)b(e') = Σij Fij(e,e') hia hjb = Σij hai hjb Fij(e,e') = ( h F(e,e')h) Maybe I need to use full covariant notation !!!!!! 3. Suppose ei is a general basis for V, not necessarily orthogonal. Then en em = Σi (en)i(em)i = enTem = emTen = w'nm where matrix w' is some symmetric matrix defined as shown. ok to here STOP!!! Below I say (en)i = δn,i so it must be then that w'nm = δn,m . 4. Given the basis vectors {ei}, one can find an alternate set of basis vectors {qi} such that the following is true qn em = Σi (qn)i(em)i = qnTem = δn,m . The vectors qn are given by qn = Σi W'ni ei where the matrix W' is the inverse of the matrix w'. Here is a verification of this claim : qn em = [Σi W'ni ei] em = Σi W'ni (ei em) = Σi W'niw'im = (Ww)n,m = δn,m The basis {qn} is called the dual basis to the basis {ei). In fact, either basis is the dual of the other basis. 5. Notice that if the ei change, then the qi change. To make this dependence of qi on the ei more explicit, it is convenient to give the qi a new name: ei ≡ qi // {ei) is the dual basis to {ei} Then we have from above en em = δn.m en = Σj W'niei 6. If it happens that en em = w'nm = δn,m then w' = 1, W = 1, and one has en = en so the basis and its dual basis are the same. In this case, the en are said to be orthonormal. If it happens that en em = w'nm = fnδn,m, then W'nm = (1/fn)δn,m and so en = (1/fn)en . In this case the basis en is said to be merely orthogonal. We shall generally assume that the basis en is nether orthogonal nor orthonormal. 7. We often expand a vector v ϵ V onto the ei basis by writing v = Σiviei If we dot both sides of this equation into ej we find ej v = ej (Σiviei) = Σivi (ej ei) = Σiviδj,i = vj The conclusion then is vi = ei v The components vi are specific to the basis ei . For some different ei the ei will be different and so the vi will be different. To make the above expansion clearer, we really should say something like this, v = Σivi(e)ei with vi(e) = ei v to indicate that the coefficients vi(e) are associated with the basis ei. However, if we stick with a particular (generally non-orthogonal) basis {ei}, we can dispense with this extra baggage in the notation. But, when we take a component of some equation, we must then understand that the component is with respect to the {ei} basis. Consider ** applied to the vector v = en : (en)i = ei en = δi,n The above equation says: "the ith component of vector en in the en basis is given by δi,n". This should not be confused with the fact that en ei = w'ni which says that "the projection of en onto ei is given by w'ni". Consider again our original vector expansion v = Σiviei We can take the rth component of both sides either by dotting each side into er, or more simply, by putting a subscript r on all vectors appearing in the equation, vr = Σivi(ei)r // = Σiviδi,r = vr PAUSE. But normally one would think of en ei = w'ni as being "the component of en taken along the ei axis". What is going on here? Well en ei = w'ni is the projection of en onto the ei axis. This says that the component of vector en taken along the en axis is 1, and along the other ei axes is 0. PAUSE. What does it now mean to say vr = Σivi(e)(ei)r with vi(e) = ei v Something is wrong here! If the subscript implies the ei basis, then we should have vr(e) = Σivi(e)(ei)r But this seems to imply that (ei)r = δi,r and then ei is an axis aligned unit vector??? ********************* PAUSE. Suppose I eventually decide to write [a b]ij = aibj I guess this must really be saying that [a b]ij(u) = ai(u)bj(u) I do something like this in (E.7.1) and thereabouts in tensor doc. Here I put the (u) label closer to the subscripts which is perhaps a better way to go. I might change tensor doc in that regard. Now what happens when I do this for some other basis ei : a = Σi ai(e) ei ai(e) = a ei Let's "play through" with this full notation and see where it leads. I would have [a b]ij(e) = ai(e)bj(e) [a b]ij(u) = ai(u)bj(u) So the "form" of this equation is the same in either basis. Is there a way to derive this equation? Maybe we start with basis vectors to write [ (un) (um) ]ij(u) = (un)i(u)(um)j(u)) = δn,i δm,j But how would you justify this equation? the first = seems "reasonable". What happens to all our previous equations? Well, we have to find how things are related. For example, we know there is some matrix of coefficients cij such that