Parallel Transport and Curvature

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1 hysics 171 Fall 015 arallel Transport and urvature 1. arallel transport along an infinitesimal curve onsider a curved spacetime with a metric g µν and connection coefficients Γ µν that are symmetric under the interchange of their two lower indices. The mathematical expression for the parallel transport of vector components along a curve x µ σ is 1 dv dσ +Γ µν vµdxν dσ = 0. 1 At some point, corresponding to σ = σ, we specify the initial condition, v v σ. Using this initial condition, onecanuniquely solve eq. 1 forv σ, the valueofthe parallel transported vector at any point σ along the curve x µ σ. The solution is v σ = v σ σ Γ µνσ v µ σ dxν, where we exhibit the σ dependence of Γ µν and vµ along the curve. uppose that the curve from σ to σ is infinitesimal in extent. Then, we can Taylor expand to first order around the point, v v µ σ v µ µ + [ ] x α σ x α x α, 3 Γ Γ µν σ Γ µν µν [ ] + x α σ x α x α. 4 It is convenient to rewrite eq. 1 in the following form by employing the chain rule, dx ν v dσ x ν +Γ µν vµ = 0. 5 We assume that the curve x µ σ is parameterized such that one continuously moves along the curve as σ increases. This means that dx µ /dσ 0 for all points along the curve, in which case one can divide eq. 5 by dx µ /dσ to obtain, v x ν +Γ µν vµ = ee eq of Ta-ei heng, Relativity, Gravitation and osmology: A Basic Introduction nd edition. 1

2 Evaluating eq. 1 at the point and relabeling the indices µ and ν α, yields, v µ = Γ µ x α α v. 7 Inserting eq. 7 into eq. 3, we obtain, v µ σ v µ Γµ α v [ x α σ x α ]. 8 We now insert eqs. 4 and 8 into eq. and discard terms that are quadratic and higher order in x α x α. The end result is, σ v σ v Γ λα v λ dx ν Γ σ [ ] λν σ x α Γ τν Γτ λα v λ x α σ x α dx ν. σ 9. arallel transport around an infinitesimal closed loop and the Riemann curvature tensor We now ask the following question. uppose we parallel transport the vector v σ along a path which starts and ends at the same point. By assumption, the path in question is an infinitesimal closed path so that we can use the approximate formula given in eq. 9 We denote the vector after it traverses the path and returns to by v. Then v v Γ λα v λ dx ν Γ λν x α Γ τν Γτ λα v λ [x ] α σ x α dx ν dσ, where we have used the chain rule to write dx ν = dx ν /. For a closed loop, dx ν = dx ν = 0, since the starting and ending point of the integral coincides. Hence, we are left with Γ v v λν x α Γ τν Γτ λα v λ x α σ dxν, 10 ince ν and α are dummy variables, we are free to relabel the indices such that ν α, which yields v v Γ λα x ν Γ τα Γτ λν v λ x ν σ dxα, 11

3 We now observe that [x 0 = dx ν x α = ν σ dx α +x α σ dx ν] = It follows that x ν σ dxα = Inserting this result into eq. 10 yields v v Γ λν x α Γ τν Γτ λα x ν σ dxα +x α σ dxν. x α σ dxν. If we add eqs. 11 and 1 and then divide by, we end up with v v 1 Γ λα x Γ λν ν x α +Γ τν Γτ λα Γ τα Γτ λν v λ v λ x ν σ dxα, 1 We recognize the appearance of the Riemann curvature tensor in eq. 13, x ν σ dxα. 13 Hence it follows that R λνα = Γ λα x ν Γ λν x α +Γ τν Γτ λα Γ τα Γτ λν. 14 v v 1 R λνα vλ x ν dx α. 15 We see that in the absence of curvature, v = v. Thus, the presence of curvature at a point can be detected by parallel transporting a vector around a closed infinitesimal loop that starts and ends at the point and showing that the vector changes when it returns to its original position. Ta-ei heng performs the computation above on pp of our textbook cf. Box 13. for the special case of an infinitesimal parallelogram. One can easily compute x ν dx α, 16 for this closed path to derive eq on p. 31 of our textbook. However, it is instructive to examine eq. 16 more carefully to understand its geometric interpretation. To accomplish this, I propose to convert eq. 16 into a surface integral using a generalization of the tokes integral theorem, which is evaluated over the two-dimensional surface bounded by. We denote the surface by the equation x α = x α ξ,η, where ξ and η serve as coordinates defined on the surface. The infinitesimal surface area element can then be expressed in terms of the spacetime coordinates x α via, x dσ α α = ξ x η x ξ 3 x α η dξdη, 17

4 where the expression inside the parenthesis is the Jacobian determinant that is needed to relate the area element expressed in terms of ξ,η and the area element expressed in terms of the x α. Note that we do not take the absolute value of the Jacobian determinant, since it is convenient to allow dσ α to be either positive or negative. The infinitesimal surface area element dσ α is an antisymmetric tensor under general coordinate transformations, 3 dσ α = dσ α. 18 We now invoke the tokes integral theorem in curved spacetime, v dx = ρ v λ dσ D ρλ = 1 D ρ v λ D λ v ρ dσ ρλ, 19 where v is an arbitrary covariant vector, D is the covariant derivative and is any finite two-dimensional surface bounded by the closed curve. In obtaining the final form for eq. 19 we have made use of eq. 18. Note that all terms appearing in eq. 19 transform as a scalar under general coordinate transformations, x = x x. However, as shown in problem 1b on roblem et 3, D ρ v λ D λ v ρ = ρ v λ λ v ρ. Hence, eq. 19 can be rewritten as v dx = 1 ρ v λ λ v ρ dσ ρλ, 0 Let us choose v = δ αxν in eq. 0. Then, That is, x ν dx α = 1 [ ρ δ α λ xν λ δ α ρ xν ] dσ ρλ = 1 Inserting eq. 1 into eq. 15 yields dσ να = v v 1 R λνα δ α λ δρ ν δα ρ δν λ dσ ρλ = dσ να. x ν dx α. 1 vλ dσ να. To make contact with eq on p. 31 of our textbook which must be corrected by multiplying theright handside by 1; cf. theerrata linked to the class webpage, we should A negative Jacobian determinant arises if the old coordinate system is right-handed and the new coordinate system is left-handed or vice versa. 3 ome books denote dσ α = dx α dx. However, this requires an introduction to differential forms, which is beyond the scope of this class. 4

5 identify thefour-vectors a ν andb α. If we parameterize thesurface with coordinates ξ,η, where the coordinate curves are tangent to a ν and b α, then we should identify Eq. 17 then yields a ν = xν ξ, bα = xα η. dσ να = a ν b α a α b ν dξdη. Integrating over the parallelogram spanned by a ν and b α then yields 1 1 dσ να = a ν b α a α b ν dξ dη = a ν b α a α b ν. Inserting this result back into eq. yields v v 1 R λνα vλ a ν b α a α b ν. 0 Finally, using the fact that the Riemann curvature tensor is antisymmetric under the interchange of its last two indices, R ρ λνα = R ρ λαν, it follows after some index relabeling that v v R λνα vλ a ν b α, which reproduces eq on p. 31 of our textbook once the omitted minus sign is correctly restored. AENDIX: A brief look at the tokes Integral Theorem Eq. 0 provides the relevant form for the tokes integral theorem in four-dimensional spacetime. Let s check that we get the expected result in three-dimensional Euclidean space. In this case, we do not have to distinguish between upper and lower indices, so that v i dx i = 1 j v k k v j dσ jk, 3 In three-dimensional Euclidean space, 0 dσ jk = ǫ ijk n i d, where n i are the coordinates of the outward normal to the surface and d is an infinitesimal surface element. Hence, j v k k v j dσ jk = ǫ ijk n i j v k k v j d = ǫ ijk n i j v k d = ˆn vd, where in the penultimate step we relabeled indices and used the fact that ǫ ijk = ǫ ikj. Inserting the above result back into eq. 3 then yields the familiar form for the tokes integral theorem for three-dimensional Euclidean space, v d l = ˆn vd. 5