rewrite of section 2_6 on 4_1_16

2.6 A change in notation : Picture E To clarify the way expansions (2.5.1) work, we shall now cosmetically modify Picture A noted above so that x'-space becomes e-space (E is used because B,C,D are already used up in Tensor , and because E matches e) (2.1.1) (2.6.1) As a first stage we do this heavy-handed conversion as follows X → X(u) where X is any x-space object X' → X(e) where X' is any x'-space object . (2.6.2) Of particular interest are the expansions of vectors in x-space and in x'-space. We have shown the table for x-space expansions in (2.5.1) above, while that for x'-space vectors is given in Tensor (7.13.11)' . These expansions are easy to verify, so we won't do that here. V = Σn Vn un where un V = Vn = [V]n V = Σn Vn un where un V = Vn = [V]n V = Σn V'n en where en V = V'n V = Σn V'n en where en V = V'n V' = Σn V'n e'n where e'n V' = V'n = [V']n V' = Σn V'n e'n where e'n V' = V'n = [V']n V' = Σn Vn u'n where u'n V' = Vn V' = Σn Vn u'n where u'n V' = Vn Notice that each set of coefficients appears twice on the right, once for V and once for V'. This duplication arise because a b = a' b' for any pair of vectors in either space. For example, Vn = un V = u'n V' appearing in lines 1 and 7 The rightmost column just shows that [V]n = Vn and [V']n = V'n so the notation is consistent. Consider then [V]n = un V = the component of vector V in the axis-aligned un basis. [V']n = e'n V' = the component of vector V' in the axis-aligned e'n basis. The equations of all earlier sections of Chapter 2 can be mapped from Picture A to Picture E as follows: In other words, wherever there is a prime, we replace it with superscript (e). It was our original intention also to change x-space to u-space such that X→X(u), but this results in so many (u) superscripts that equations become cluttered. But one should still think of x-space as u-space. For example, one can think of Vn = (V(u))n . In other words, the default of no superscript implies a (u) superscript. Vectors and tensors in x'-space then get (e) superscripts. Examples: Pic A: Rab g g' dx'a = Rabdxb V = Vn un V = V'n en Pic E: Rab g g(e) dx(e)a = Rabdxb V = Vn un V = V(e)n en . (2.6.3) The meaning of V(e)n is [V(e)]n. The superscript (e) goes with the tensor, not with the component index. We are replacing [V']n . Notice that basis vectors en and un would be denoted en(u) and un(u) in a full u-space notation. If we momentarily write Vn = V(u)n (but maintain V as V, etc. ), the four expansions of (2.5.1) become (showing the n sum explicitly) , The expansions (2.5.1) are now written V = Σn Vn un where un V = Vn V = Σn Vn un where un V = Vn V = Σn V'n en where en V = V'n V = Σn V'n en where en V = V'n . (2.5.1) (2.6.4) and for a vector V' in x-space Tensor (7.13.11)' gives us V' = ΣnV'n e'n where e'n V' = V'n V' = ΣnV'n e'n where e'n V' = V'n V' = Σn Vn u'n where u'n V' = Vn V' = Σn Vn u'n where u'n V' = Vn (2.6.6) Notice that, taking the first two equations in each group [V]i = Σn Vn [un]i = Vi [V]i = Σn Vn [un]i = Vi [V']i = ΣnV'n [e'n]i = V'i [V']i = ΣnV'n [e'n]i = V'i Thus, the notation is "self consistent". But what exactly does one of these subscripts MEAN? Rule: If V is a vector in x-space, then [V]i means ui V , which is V dotted into the axis aligned basis. If V' is a vector in x'-space, then [V']i means e'i V , which is V dotted into the axis aligned basis. Restate: A component index always means with respect to the corresponding axis-aligned basis vector in the space of interest . Converting these to Picture E notation is amazingly complicated: V(u) = Σn [V(u)](u)n un(u) where un(u) V(u) = [V(u)](u)n V(u) = Σn [V(u)](u)n un(u) where un(u) V(u) = [V(u)](u)n V(u) = Σn [V(u)](e)n en(u) where en(u) V(u) = [V(u)](e)n V(u) = Σn [V(u)](e)n en(u) where en(u) V(u) = [V(u)](e)n . (2.5.1) (2.6.4) V(e) = Σn [V(e)](e)n e(e)n where e(e)n V(e) = [V(e)](e)n V(e) = Σn [V(e)](e)n e(e)n where e(e)n V(e) = [V(e)](e)n V(e) = Σn [V(e)](u)n u(e)n where u(e)n V(e)= [V(e)](u)n V(e) = Σn [V(e)](u)n u(e)n where u(e)n V(e) = [V(e)](u)n (2.6.6) compare to V = Σn V(u)n un where un V = Vn V = Σn V(u)n un where un V = Vn V = Σn V(e)n en where en V = V'n V = Σn V(e)n en where en V = V'n . (2.5.1) (2.6.4) V' = ΣnV'n e'n where e'n V' = V'n V' = ΣnV'n e'n where e'n V' = V'n V' = Σn Vn u'n where u'n V' = Vn V' = Σn Vn u'n where u'n V' = Vn (2.6.6) Now I think we can simplify things because we know for example that un(u) V(u) = un(e) V(e) so [V(u)](u)n = [V(e)](u)n ≡ [V](u)n = V(u)n Similarly, en(u) V(u) = en(e) V(e) so [V(u)](e)n = [V(e)](e)n ≡ [V](e)n = V(e)n since the first label does not really matter. So what does this do? V(u) = Σn V(u)n un(u) where un(u) V(u) = V(u)n ≡ [V(u)]n V(u) = Σn V(u)n un(u) where un(u) V(u) = V(u)n ≡ [V(u)]n V(u) = Σn V(e)n en(u) where en(u) V(u) = V(e)n ≡ [V(e)]n V(u) = Σn V(e)n en(u) where en(u) V(u) = V(e)n ≡ [V(e)]n V(e) = Σn V(e)n en(e) where en(e) V(e) = V(e)n ≡ [V(e)]n V(e) = Σn V(e)n en(e) where en(e) V(e) = V(e)n ≡ [V(e)]n V(e) = Σn V(u)n un(e) where un(e) V(e) = V(u)n ≡ [V(u)]n V(e) = Σn V(u)n un(e) where un(e) V(e) = V(u)n ≡ [V(u)]n Where did the right column come from? Well, now I did it in the prime no-prime world and everything is consistent and the above Rule still applies. Question: do the superscripts like (e) "go with the vector" or do they "go with the index" ? Answer: from the rightmost column, it seems they "go with the vector" . The index or component is always forced to be that for the corresponding axis-aligned basis vector, there is no choice there, so there is no "marker" saying what kind of component you have. NOW suppose we use the "default rule" that anything with a (u) label has that label suppressed. Then: V = Σn Vn un where un V = Vn ≡ [V]n V = Σn Vn un where un V = Vn ≡ [V]n V = Σn V(e)n en where en V = V(e)n ≡ [V(e)]n V = Σn V(e)n en where en V = V(e)n ≡ [V(e)]n V(e) = Σn V(e)n e(e)n where e(e)n V(e) = V(e)n ≡ [V(e)]n V(e) = Σn V(e)n e(e)n where e(e)n V(e) = V(e)n ≡ [V(e)]n V(e) = Σn Vn u(e)n where u(e)n V(e) = Vn ≡ [V]n V(e) = Σn Vn u(e)n where u(e)n V(e) = Vn ≡ [V]n OK now jump ahead to M = Σab [M(e,u)]ab eaub . (2.10.2) Now what do the component indices mean? (eiuj) M = [M(e,u)]ij Does "the rule" still apply? The basis vectors are both (u) type since they are both in u-space. Thus, I think both indices are with respect to the axis-aligned basis vectors ui . I need to check that!!! But just staring at the above that seems wrong! Does this mean V = Σa [V(e)]a ea(u) [V(e)]a = e(e)a V(e) V(e) = Σn [V(e)]a e(e)a M = Σab [M(e,u)]ab ea(u)ub(u) . (2.10.2) (ei(u)uj(u)) M = [M(e,u)]ij ??? What choices do we have for "component index type" ? Is i an index that is with respect to the axis aligned vectors ui ? Same question for j. Well, lets look at a direct product [M(e,u)]ab = V(e)a U(u)b = [V(e)]a [U(u)]b OK, now do a dot product: M = Σab [M(e,u)]ab ea(u)ub(u) = Σab [V(e)]a [U(u)]b eaub = ( Σa[V(e)]a ea(u)) ( Σb[U(u)]b ub(u)) = ( V(u)) ( U(u)) = V(u) U(u) (ei(u)uj(u)) M = [M(e,u)]ij (ei(u)uj(u)) Σab [V(e)]a [U(u)]b ea(u)ub(u) = [V(e)]i [U(u)]j and [V(e)]i = e(e)i V(e) = projection onto e-space axis-aligned vector = e(u)i V(u) = not relevant We have now a whole OTHER issue. For each of the 8 vector equations you can take "components" of the equation. What exactly does this mean? For example you might write for equation #6 above, [V(e)]i = Σn V(e)n [e(e)n]i Rule: the component indices always mean with respect to the axis-aligned basis vectors in the appropriate space. So then the above would mean [V(e)]i ≡ [e(e)i] [V(e)] = V(e)i since the e(e) are axis aligned in e-space [e(e)n]i = [e(e)i] [e(e)n] = δin Consider for example u(e)n V(e)= [V(e)](u)n . There are three markers on the left, the u and the two e superscripts, but on the right there are only two markers. Three are not needed because the vector and the basis vector have to be in the same space for the dot product to be meaningful, in this case the e space. Here then is a color code showing how these markers work u(e)n V(e)= [V(e)](u)n The three e's are green, the two u's are red. If we agree to make (u) be the default for vector labels and for component labels, and agree to not display these labels, the above become V = Σn [V]n un where un V = [V]n V = Σn [V]n un where un V = [V]n V = Σn [V](e)n en where en V = [V](e)n V = Σn [V](e)n en where en V = [V](e)n V(e) = Σn [V(e)](e)n e(e)n where e(e)n V(e) = [V(e)](e)n V(e) = Σn [V(e)](e)n e(e)n where e(e)n V(e) = [V(e)](e)n V(e) = Σn [V(e)]n u(e)n where u(e)n V(e)= [V(e)]n V(e) = Σn [V(e)]n u(e)n where u(e)n V(e) = [V(e)]n We know that V(e) cannot be dotted into a u-space vector, so u(u)n V(e) = un V' is not meaningful, so the single (e) superscript does the job of two, and we might write u(e)n V(e)= [V(e)](u)n We won't be using these much, which is good because the Picture E notation is a bit messy. Mechanically replace primes by (e), V(e) = Σn V(e)n en(e) where en(e) V(e) = V(e)n en(e) = Ren V(e) = Σn V(e)n en(e) where en(e) V(e) = V(e)n en(e) = Ren V(e) = Σn Vn un(e) where un(e) V(e) = Vn un(e) = Run V(e) = Σn Vn un(e) where un(e) V(e) = Vn un(e) = Run . (2.6.7) STOP! I now have two conflicting notations: V(e)n ≡ en V a component of V in the en basis V(e) = Σn V(e)n en(e) a vector in e-space Are these consistent? Take a component of this last thing to get [V(e)]i = Σn V(e)n en(e)i Nevertheless, in these sums we still have the correlation between the (e) or implied (u) on a vector component and the name of the basis vector u or e. Exercise: Verify that em(e) = Rem shown in the first line of (2.6.7) is consistent with (em(e))n = (en em) in the last line of (2.5.4): (2.3.4) (2.1.9) (2.3.2) em(e) = Rem (em(e))n = Rnj(em)j = RnjRmj = δnm = (en em) QED . We can apply the four equations shown on the right of (2.6.5) sequentially to the basis vectors V = um, um, em, em to obtain a set of 16 equations. The very first equation would be un um = umn. One can then look up the dot product in (2.4.2) to find that un um = δnm and then one gets the result that umn = δnm . The next equation is un um = (um)n and we look up this dot product to find un um = gnm and so (um)n = gnm. Rather than do all these calculations, since the dot products are already listed in (2.5.4), we can just read off the 16 results we want. This then produces the rightmost column of equations in (2.5.4) above, which we transcribe here, (um)n = δmn (um)n = g(u)mn (em)n = Rmn (em)n = Rmn (um)n = gmn (um)n = δmn (em)n = Rmn (em)n = Rmn (um(e))n = Rnm (um(e))n = Rnm (em(e))n = δmn (em(e))n = g(e)mn (um(e))n = Rnm (um(e))n = Rnm (em(e))n = g(e)mn (em(e))n = δmn . (2.6.8) We have replaced g'→g(e) and have made cosmetic changes such as δnm = δmn as well as gmn = gnm since all metric tensors are symmetric. The set (2.6.8) gives the contravariant (up) and covariant (down) components of all eight basis vectors: um, um, em, em and um(e), um(e), em(e), em(e). Notice in (2.6.8) [ col 3 row 3] that (em(e))n = δmn . According to the Comment at the end Section 2.3, this equation applies for an arbitrary set of basis functions em(x). The equation is eminently reasonable. Suppose we expand em = [ Σn (em(e))n en] . We can see that we must have (em(e))n = δmn. Expanding a basis vector on its own basis yields a single term in the sum. Being true for any basis, (em(e))n = δmn must also be true for the u basis, and we see as well that (um(u))n = (um)n = δmn in (2.6.8) [ col 2 row 2]. Let's now review the paradox (2.1) presented at the start of Chapter 2, but in our "improved" notation. We first restate the paradox in covariant notation: Paradox. Let v = Σnvnen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em)n en. This implies that (em)n = δmn. Thus, the only possible basis vectors em allowed in the universe are axis-aligned unit vectors. (2.6.9) The paradox arises because the expansion as stated is ambiguous and in our Picture E notation it is in fact wrong. The expansion should be em = Σn(em(e))n en and it is true that (em(e))n = δmn. This does not put any restriction on (em)n = (em(u))n = Rmn where Rmn can be a set of arbitrary numbers. Conclusion: One must be careful when dealing with tensors and their components to understand the space to which a tensor belongs. In our Picture E notation, em is associated with u-space, while em(e) is associated with e-space. We now restate the transformation rules for rank-1 and rank-2 tensors in our (2.6.1) Picture E context: [V(e)]a = RabVb (2.1.5) [M(e)]ab = Raa' Rbb' Ma'b' . (2.1.7) line 1 (2.6.10) If the above tensors are tensor fields, one then has [V(e)]a(x(e)) = RabVb(x) [M(e)]ab(x(e)) = Raa' Rbb' Ma'b'(x) (2.6.11) where in general Rij = Rij(x) . Comment. The ambiguity of the meaning of a vector component is made very clear in the Dirac notation discussed below in Section 2.11. In that notation one writes for example, (um)n = (um(u))n = <un | um > = the vector | um > projected onto the <un| basis element (um(e))n = <en | um > = the vector | um > projected onto the <en| basis element (2.6.12)