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Has been installed, you can delete this file. 2.6 A change in notation : Picture E Top 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 Ref ** , and because E matches e) (2.1.1) (2.6.1) The equations of all earlier sections of Chapter 2 can be mapped from Picture A to Picture E as follows: X → X where X is any x-space object X' → X(e) where X' is any x'-space object . (2.6.2) 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. 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) , V = Σn Vn(u) un where un V = Vn(u) V = Σn Vn(u) un where un V = Vn(u) V = Σn V(e)n en where en V = V(e)n V = Σn V(e)n en where en V = V(e)n . (2.5.1) (2.6.4) This reveals our motivation for using Picture E: in each expansion, the superscript label on the component matches the letter u or e in the basis vector. We now hide the (u) superscripts and rewrite the above, V = Σn Vn un where un V = Vn V = Σn Vn un where un V = Vn V = Σn V(e)n en where en V = V(e)n V = Σn V(e)n en where en V = V(e)n . (2.5.1) (2.6.5) It is shown in Ref *** that there are four similar expansions for the vector V' in x'-space, V' = ΣnV'n e'n where e'n V' = V'n e'n = Ren [ (e'n)i = Rij(en)j ] V' = ΣnV'n e'n where e'n V' = V'n e'n = Ren V' = Σn Vn u'n where u'n V' = Vn u'n = Run V' = Σn Vn u'n where u'n V' = Vn u'n = Run (2.6.6) We won't be using these much, which is good because the Picture E notation is a bit messy. Mechanically replace primes by (e), except e'n → en(e) = the e-space version of en . 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) 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)