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General Relativity Solution #2 10/18/2016 1. Diffeomorphisms We Start by Asking How the Components of a Vector Field Transform U

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General Relativity Solution #2 10/18/2016 1. Diffeomorphisms We Start by Asking How the Components of a Vector Field Transform U General Relativity Solution #2 10/18/2016 1. Diffeomorphisms We start by asking how the components of a vector field transform under the arbitrary reparameterizations xµ = f µ(~x) for a differentiable set of functions ff µg. We are given that @xµ the coordinate basis vectors follow e~α(~x) = @x~α eµ(x). α Starting from the definition of a vector field v = v eα, we see that since the abstract vector field is ignorant of the particular coordinatization used on the manifold: @xµ @x~α v = v~ =v ~α(~x)e~ (~x) = v~αe (x) ) v~α(~x) = vµ(x): α @x~α µ @xµ From this point we can easily ascertain the transformation law for a 1-form, ! 2 T 1(M). We α @x~α µ start with the fact that the dual basis dx~ = @xµ dx . The components thus transform as @x~α @xµ ! =! ~ =! ~ (~x)dx~α = !~ (~x)dxµ ) !~ (~x) = ! (x) α @xµ α α @x~α µ Moving one step up the tensorial ladder, we can ask how the components of the metric, gαβ, α β transform. Let us take a lesson from above and write metric g = gαβdx ⊗ dx . Then from the rules governing the transformation of the dual basis: @x~α @x~β @xµ @xν g = g~ =g ~ dxµ ⊗ dxν ) g~ = g αβ @xµ @xν αβ @x~α @x~β µν What is key here is that the arbitrary relabeling of points on M does not change g. What does this mean for the change of the determinant of the metric g = det jjgµνjj? We know that the det AB = det A det B, and so, we can see that @xµ @xν g~ = det jjg~ jj = det g ) g~ = J 2g αβ @x~α @x~β µν where the transformation in the last step was written as a product of Jacobians given by @xµ J = det : @x~α Lastly, we want to know why the spacetime volume element on M is pjgj. First, we note that what we desire is that the volume of M is invariant under reparameterizations Z d V~M = VM = d x · (?) M What should fill in for `?' is something built out of the metric, since we need a notion of the `size' of regions of M, and can compensate for how the dual basis transforms. We know from basic multivariate calculus that ddx = Jddx~ for J given above. Now given the transformation rules above, we see that what should fill −1 in for the volume element must have a transformation like J to keep VM invariant under diffeomorphisms. The previous calculation shows that p pjgj = jJ −2g~j = J −1pjg~j; 1 General Relativity Solution #2 10/18/2016 which is precisely what we want. Thus, demanding that the volume element be built out of the metric and that the total volume is diffeomorphism invariant leads to pjgj being the unique candidate for the volume element. Simply demanding the transformation of J −1 is not sufficient as any the square root of the determinant of any (0; 2)-tensor transforms this way. This above discussion can be rephrased in the language of differential forms. The components of a k-form are totally anti-symmetric covariant tensors of rank k. On a d-dimensional mani- fold M, differentiable functions f are 0-forms and d-forms, or top forms, are all proportional to the Levi-Civita tensor 1 = dxα1 ^ · · · ^ dxαd d! α1...αd and p α1...αd = jgj[α1 : : : αd]: The beauty of the language of forms is that k-forms can be viewed as integrands of k- dimensional integrals. Thus, is the volume form, and integrating a scalar function (0-form), f, over the manifold is the same as integrating against Z Z ddxpjgjf = f M M Put in other language seen with differential forms, the volume form is given in terms of the Hodge-? operation, which maps p-forms to (d − p)-forms on a d-dimensional manifold, by pjgj ?1 = [α : : : α ]dxα1 ^ ::: ^ dxαd d! 1 d 2. Derivatives and Such a. @α log g: For calculations involving variations of the metric determinant, it is convenient to use the identity det A = exp(Tr log A). As such, it becomes clear that −1 @a(log det A) = @aTr log A = Tr A @aA; and so affecting the trace, we see βδ @α(log g) = g gβδ; α: α p b. Γ αβ = @β log jgj: In a coordinate basis, lets recall the definition of the Christoffel symbols gαγ Γα = g + g − g ; βδ 2 βγ; δ δγ; β βδ; γ 2 General Relativity Solution #2 10/18/2016 and trace over the first two indices gαγ 1 Γα = g + g − g = gαγg : αβ 2 γα, β βγ,α γβ; α 2 αγ; β The last equality used the symmetry of the metric to cancel the last two terms. From part [a.], 1 @ log pjgj = gαγg : β 2 αγ; β α c. rαA : Starting with the definition of the covariant derivative of a vector α α α γ rβA = @βA + Γ βγA : Affecting the trace, over α; β α α α γ rαA = @αA + Γ αγA ; and using the result in part [b.] p 1 p r Aα = @ Aα + @ (log g)Aγ = @ Aα + p Aα@ g; α α γ α g α which can be rearranged to give α − 1 p α rαA = jgj 2 ( gA ),α: d. φ: α First, we rewrite ≡ rαr , and use the results from part [c.] α − 1 p α − 1 p αβ rα r φ = g 2 gr φ),α = g 2 gg @βφ),α where in the last equality we used metric compatibility of the Christoffel connection 0 ragbc = 0 and that raf = @af for any f 2 Ω (M). 3. Fun with Sd To begin, we want to understand the particular coordinatization, i.e. the round metric, of d the unit d-dimensional sphere, S , such that the angles fαig cover the sphere with d−1 2 2 2 2 Y 2 2 ds = dα1 + sin α1dα2 + ::: + sin αidαd: (1) i=1 d d+1 d+1 That is we want to study the embedding S ,! R . Let us define coordinates on R , fx0; : : : ; xdg with the flat Euclidean metric 2 a b dsR = δabdx dx : 3 General Relativity Solution #2 10/18/2016 We want to find the unit Sd embedded in this space, which can be found by defining the d+1 co-dimension 1 hypersurface in R such that each point on the surface is unit distance away from the origin f0;:::; 0g. This is found by the restriction 2 2 2 x0 + x1 + ::: + xd = 1: d d+1 Any coordinatization of a unit S ,! R must satisfy this relation. We can use a general- 2 3 ization of the familiar parameterization of the S ,! R : d−1 d Y Y x0 = cos α1; x1 = sin α1 cos α2; : : : ; xd−1 = sin αi cos αd; xd = sin αi: i=1 i=1 If we compute the line element on this surface, pullback the standard Euclidean metric on d+1 R to get the desired form as in eq. (1). To illustrate this, lets use a low dimensional 2 3 example, S ,! R , and compute the line element. Note that, with the above coordinates on the d=2 hypersurface 2 2 2 dx0 = sin α1dα1 2 2 2 2 2 2 2 dx1 = cos α1 cos α2dα1 + sin α1 sin α2dα2 − 2 cos α1 sin α1 cos α2 sin α2dα1dα2 2 2 2 2 2 2 2 dx2 = cos α1 sin α2dα1 + sin α1 cos α2dα2 + 2 cos α1 sin α1 cos α2 sin α2dα1dα2: a b Plugging this into δabdx dx , we see that 2 a b 2 2 2 ds = δabdx dx = dα1 + sin α1dα2 X: What are the ranges for the αi? To cover the sphere, fαi 2 [0; π] 1 ≤ i ≤ d − 1g and 1 αd 2 [0; 2π]. Note that you can visualize this as a nested fibration of an S over d − 1 intervals I = [0; π]. In this coordinate system if we had chosen αd 2 [0; π] and another αk 2 [0; 2π], then we would not have covered the entirety of the sphere (check this again with a low dimensional example). Alternatively if we had chosen multiple αk 2 [0; 2π], then we would have ended up with a multiple covering of the sphere. For brevity, label Γαi ≡ Γi . Using your favorite package for differential geometry, the αj αk jk i calculation of Γ jk over a number of dimensions is done quickly, and from there a pattern quickly emerges: j j Γ ij = Γ ji = cot αi; i < j i−1 i Y 2 Γ jj = −gjj cot αi csc αk; i < j k=1 with the rest vanishing identically. So, the geodesic equation with affine parameter λ becomes, dαi(λ) withα _ i = dλ , i i j i i k 2 α¨ + 2Γ jiα_ α_ + Γ kk(_α ) = 0; j < i; k > i: We can note that the αd equation can always be written as d−1 Y 2 α_ d = cd csc αk(λ) k=1 4 General Relativity Solution #2 10/18/2016 for some constant cd. How are we going to solve this? We can exploit the symmetry of the background and affect an SO(d + 1) rotation such that any solution we wish to find passes through the `north pole' defined by α1(0) = 0. If from this point we only allow α1 to vary such thatα _ j = 0 for j > 1, then the geodesic equations reduce to α¨i = 0 for all i.
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