Section F1 intro

Trial Balloon PhL 4.29.15 Plan A F.1 Definition and Interpretation of Γ : Γcab = ec (∂aeb) = Rci(∂aRbi) This is the wrong way to do it! We work here in Picture A Each tangent base vector en(x) in x-space varies with x, and one wonders just how en varies with x. For a small variation dx in the x-space coordinates, one has, d(en)i = ∂j(en)i dxj // ∂j ≡ ∂/∂xi or (den)i = (∂jen)i dxj . // ∂x(fi) = = = (∂xfi) Since both en and (∂jen) are vectors in x-space, and since the ek are known to form a complete basis in x-space, it must be possible to expand (∂jen) on the ek with some appropriate coefficients, call them Γkjn : (∂jen) = Γkjn ek (den)i = Γkjn (ek)i dxj ∂j = ∂/∂xj (F.1.5) STOP! This does define an affine connection Γkjn but it is not the right affine connection!!!! It looks right, but it is wrong! But let's continue with it to see where it leads: Dotting the left equation into em , using em ek = δm,k as in (7.18.1), then doing m→ k gives Γkjn = ek (∂jen) = (ek)i(∂jen)i = Rki(∂jRni) (F.1.6) where we recall again from (7.18.1) that (ek)i = Rki and (en)i = Rni . (7.18.1) These coefficients Γkjn(x) comprise a tensor-like field called the affine connection, so we shall regard (F.1.5) as the definition of Γkjn . With more standard index names, the above becomes Γcab = ec (∂aeb) = Rci(∂aRbi) . (F.1.7) From (7.5.16) we know that, Rba = (∂x'b/∂xa) Rci = (∂x'c/∂xi) Rba = (∂xa/∂x'b) Rbi = (∂xi/∂x'b) ∂aRbi = ∂2xi/∂xa∂x'b Then Γcab = Rci(∂aRbi) = (∂x'c/∂xi) ( ∂2xi/∂xa∂x'b) If I convert this from Picture A to Picture C1, doing x→ξ and then x'→x I get Γcab = (∂xc/∂ξi) ( ∂2ξi/∂ξa∂xb) and this does NOT agree with Weinberg p 100 (4.5.1) which says. Γcab = (∂xc/∂ξi) ( ∂2ξi/∂xa∂xb) Plan B F.1 Definition and Interpretation of Γ : Γcab = ec (∂aeb) = Rci(∂aRbi) Context can be very confusing in a discussion of the affine connection Γ. We start with a modified Picture C in which the quasi-Cartesian space on the right is called ξ-space instead of x(0)-space as in Picture C. The notation ξi for the coordinates of ξ-space seems traditional in general relativity work where the Γ object appears frequently. (F.1.1) Recall from the discussion near (1.10) that the metric tensor G is a diagonal matrix whose elements are independently +1 or -1. Since the metric tensor transforms as a rank-2 tensor, we know from the last line of (7.5.8), adapted from Picture A to Picture C1, that gab = RaiRbjGij. Since G is diagonal, we write this as gab = RaiRbiGii and gab = RaiRbiGii (F.1.2) with a single implied sum on i. If G = 1, then the xi coordinates of x-space are "the curvilinear coordinates" and the ξi are "the Cartesian coordinates". The Role of Pictures in Understanding the Affine Connection (F.1.3) Here we attempt to head off a confusion about the meaning of the symbol en, and confusion about the form taken by the definition of the affine connection. In Picture A the tangent base vectors exist in x-space and are called en. From (7.18.1) we know both that (en)i = Rni and the dot product en em = g'nm where g' is the metric tensor in x'-space on the left. In general en = en(x) as shown for example in Fig (3.4.3) for polar coordinates. In Picture C1 the tangent base vectors exist in ξ-space and we will call them qn. From (7.18.1) we know both that that (qn)i = Rni and that the dot product qn qm = gnm where g is the metric tensor in x-space on the left. In general qn = qn(ξ). However, since x = F(ξ), we are free instead to regard qn = qn(x), and that is what we shall do for Picture C1. Notice that (en)i = Rni and (qn)i = Rni so then en = qn. This is so because the same matrix R appears in both Picture A and Picture C1. Thus, in terms of component values we have qn(x) = en(x) (F.1.4) even though these two objects belong to different Pictures. One must keep in mind, however, that qn qm = gnm within Picture C1 (F.1.5) en em = g'nm within Picture A (F.1.6) In both cases, the metric tensor is the one associated with the "left side" of the Picture. If we move a small amount dx in x-space of Picture C1, qn(x) (a vector in ξ-space) will change by some small amount. We have d(qn)i = ∂j(qn)i dxj // ∂j ≡ ∂/∂xj or (dqn)i = (∂jqn)i dxj . // ∂x(fi) = = = (∂xfi) Since qn and (∂jqn) are both vectors in ξ-space, and since the qk are known to form a complete basis in ξ-space, it must be possible to expand (∂jqn) on the qk with some appropriate coefficients, call them Γkjn : (∂aqn) = Γkan qk (dqn)i = (∂aqn)dxa = Γkan (qk)i dxa . (F.1.5) Dotting the left equation into qm , using qm qk = δmk as in (7.18.1), and then doing m→ k gives Γkan = qk (∂jqn) = (qk)i(∂jqn)i = Rki(∂jRni) (F.1.6) where from (7.18.1) and (7.5.16) adjusted from Picture A to Picture C1 we have used (qk)i = Rki = and (qn)i = Rni = (F.1.7) Inserting the partial derivatives from (F.1.7) into (F.1.6) gives Γkjn = Rki(∂jRni) = ( ∂j ) = (F.1.8) and this form shows that Fact: Γkjn is symmetric on the lower two indices, so Γkjn = Γknj (F.1.9) If we take k→λ, i→α, j→μ and n→ν, then (F.1.9) becomes or Γλμν = and this equation appears as Weinberg p 100 (4.5.1). Recall now from ************************** Plan C F.1 Definition and Interpretation of Γ : Γcab = qc (∂aqb) = Rci(∂aRbi) In this Section we shall use a modified Picture C in which the quasi-Cartesian space on the right is called ξ-space instead of x(0)-space as in Picture C. The notation ξi for the coordinates of ξ-space seems traditional in general relativity work where the Γ object appears frequently. (F.1.1) Recall from the discussion near (1.10) that the metric tensor G is a diagonal matrix whose elements are independently +1 or -1. Since the metric tensor transforms as a rank-2 tensor, we know from the last line of (7.5.8) (adapted from Picture A to Picture C1) that gab = RaiRbjGij. Since G is diagonal, we write this as gab = RaiRbiGii and gab = RaiRbiGii (F.1.2) with a single implied sum on i. If G = 1, then the xi coordinates of x-space are "the curvilinear coordinates" and the ξi are "the Cartesian coordinates". In Picture C1 the tangent base vectors exist in ξ-space and we will call them qn. From (7.18.1) (adapted from Picture A to Picture C1) we know both that (qn)i = Rni and that the dot product qn qm = gnm where g is the metric tensor in x-space. In general qn = qn(ξ). However, since x = F(ξ), we are free instead to regard qn = qn(x), and that is what we shall do for Picture C1. If we move a small amount dx in x-space, qn(x) (a vector in ξ-space) will change by some small amount. We have d(qn)i = ∂j(qn)i dxj // ∂j ≡ ∂/∂xj or (dqn)i = (∂jqn)i dxj . // ∂x(fi) = = = (∂xfi) (F.1.3) Since qn and (∂jqn) are both vectors in ξ-space, and since the qk are known to form a complete basis in ξ-space, it must be possible to expand (∂jqn) on the qk with some appropriate coefficients, call them Γkjn : (∂aqn) = Γkan qk (dqn)i = (∂aqn)dxa = Γkan (qk)i dxa . (F.1.4) Dotting the left equation into qm , using qm qk = δmk as in (7.18.1), and then doing m→ k gives Γkan = qk (∂jqn) = (qk)i(∂jqn)i = Rki(∂jRni) , (F.1.5) where from (7.18.1) and (7.5.16) (adjusted from Picture A to Picture C1) we have used (qk)i = Rki = and (qn)i = Rni = . (F.1.6) Inserting the partial derivatives from (F.1.6) into (F.1.5) gives Γkjn = Rki(∂jRni) = ( ∂j ) = (F.1.7) Notice that (∂jRni) = (∂nRji) // since = (F.1.8) Form (F.1.7) shows that: Fact: Γkjn is symmetric on the lower two indices, so Γkjn = Γknj (F.1.9) If we take k→λ, i→α, j→μ and n→ν, then (F.1.9) becomes or Γλμν = and this equation appears as Weinberg p 100 (4.5.1). We now restate some of the results above with more commonly used indices (∂aqb) = Γcab qc (F.1.4) Γcab = qc (∂jqb) (F.1.5) Γcab = Γcba (F.1.9) Γcab = Rci(∂aRbi) (F.1.5) Γcab = – Rbi (∂aRci) (F.1.10) The last line will be derived in the next section. Recall now from ***********88 Now comes the confusing part. We are going to rename so in terms of component values we have qn(ξ) = en(x). Furthermore, to simplify notation, we shall refer to en[ξ] simply as en. This is then a new meaning for the symbol en. These new en are vectors in ξ-space of Picture C1, whereas the old en were vectors in x-space of Picture A. In both cases, the en are the tangent base vectors for "the space on the right", and in both pictures (en)i = Rni. In Picture C1, and with these new en, we have en em = gnm, where g is the metric tensor for the left space. If it happens that x-space of Picture C1 is Cartesian (g = 1) then en(x) = = constant vectors. In this case, we would certainly conclude that ∂ien(x) = ∂en(x)/∂xi = ∂i = 0. Below we shall define Γcin ≡ ec (∂ien) so if x-space is Cartesian, then certainly Γcab = 0, so Γ = 0 for a Cartesian x-space in the context of Picture C1. (F.1.3) Although Γcin is the dot product of two vectors in ξ-space, it is determined by the nature of x-space. In particular, Γ is determined by the metric tensor gnm of x-space, a fact that is explicitly shown in (F.4.1) below: Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab]. If we were to make a new version of Picture C1 with x'-space on the left, and x' = F'(ξ) with R' and S', we would then use the notation Γ'cin ≡ e'c (∂'ie'n) where this Γ' is determined by the g'nm of x'-space and e'n are the corresponding tangent base vectors in ξ-space given by (e'n)i = R'ni. Here is a comparison: Γcin = ec (∂ien) (en)i = Rni Γ'cin = e'c (∂'ie'n) (e'n)i = R'ni (F.1.4) Later on, we will want to talk about the affine connection in the context of Picture B. (F.1.4B) Γ'cin = ec (∂'ien) (en)i = Rni where en are the usual tangent base vectors of Chapter 3. Notice the primes on Γ' and ∂'i which correspond to the fact that the curvilinear coordinates in Picture B are x'. Since x-space is Cartesian in this Picture, we have Γ = 0 as in (F.1.3). The moral of the story here is that one must be very conscious of the Picture one is currently using.