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Parallel Transport and Geodesics Parallel transport and geodesics February 24, 2013 1 Parallel transport Before defining a general notion of curvature for an arbitrary space, we need to know how to compare vectors at different positions on a manifold. Parallel transport provides a way to compare a vector in one tangent plane to a vector in another, by moving the vector along a curve without changing it. Suppose we have a curve with unit tangent t in flat space and Cartesian coordinates. Then to move a vector v along this curve without changing it simply means holding the components constant. At any point λ along the curve, we may find the transported components by solving (t · r) v = 0 or, in components, i j t @iv = 0 Since the connection vanishes in Cartesian coordinates, this is the same as writing i j t Div = 0 but this expression now holds in any coordinates. The same argument holds in a curved space because close enough to any point we may find Cartesian coordinates, transport infinitesimally, then change coordinates to Cartesian again. At each point, the Cartesian expression may be written covariantly, but the covariant expression is the same at every point of the curve regardless of coordinates. We therefore define parallel α α dxα α transport of a vector v along a curve with tangent u (λ) = dλ to be the solution v (λ) to the equation α β u Dαv = 0 A curve is called autoparallel if it is transported along its own direction, α β v Dαv = 0 Notice that parallel transport preserves the length of the vector because α 2 α µ ν u Dα jvj = u Dα (gµν v v ) α µ ν µ ν µ ν = u (Dαgµν v v + gµν Dαv v + gµν v Dαv ) α µ ν µ α ν = gµν (u Dαv ) v + gµν v (u Dαv ) = 0 2 Example: parallel transport on the 2-sphere Consider the parallel transport of a vector around a θ = θ0 curve on the 2-sphere. The curve itself may i i parameterized using ' as x = (θ0;'), with tangent, t = (0; 1). The length of this tangent vector is given 1 by 2 i j l = gijt t 2 2 = R sin θ0 so the unit tangent is 1 ui = (0; 1) R sin θ0 i θ ' i At ' = 0; let v = v0; v0 , and solve the parallel transport equation for v ('), i j 0 = u Div 1 j = D'v R sin θ0 j k j 0 = @'v + v Γ k' 2.1 The metric Since the metric is given by the line element, ds2 = R2dθ2 + sin2 θd'2, we have, in matrix components, R2 0 g = ij 0 R2 sin2 θ with inverse 1 ij R2 0 g = 1 0 R2 sin2 θ There are intrinsic ways to get this metric. One approach is to specify the symmetries we require – three independent rotations. There are techniques for finding the most general metric with given symmetry, so we can derive this form directly. Alternatively, we could ask for 2-dim spaces of constant curvature. Computing the metric for a general 2-geometry, then imposing constant curvature gives a set of differential equations that will lead to this form. 2.2 The connection Since the only non-constant component of the metric tensor is g'', there are only three nonvanishing connection components, 1 Γθ = gθθ (g + g − g ) '' 2 θ';' θ';' ϕϕ,θ 1 1 = − R2 sin2 θ 2 R2 ,θ = − sin θ cos θ 1 Γ' = Γ' = g'' (g + g − g ) θ' ϕθ 2 ϕϕ,θ ϕθ;' θ';' 1 1 = R2 sin2 θ 2 R2 sin2 θ ,θ cos θ = sin θ 2.3 Parallel transport The parallel transport equation becomes j k j 0 = @'v + v Γ k' j θ j ' j = @'v + v Γ θ' + v Γ '' 2 There are two components to check. For j = θ we have θ ' θ 0 = @'v + v Γ'' @vθ = − v' sin θ cos θ @' 0 0 For j = ', ' θ ' 0 = @'v + v Γθ' @v' cos θ = + vθ 0 @' sin θ0 Therefore, we need to solve the coupled equations, @vθ 0 = − v' sin θ cos θ @' 0 0 @v' cos θ 0 = + vθ 0 @' sin θ0 Taking a second derivative of the first equation and substituting the second, @2vθ @v' 0 = − sin θ cos θ @'2 @' 0 0 2 θ @ v θ cos θ0 = 2 + v sin θ0 cos θ0 @' sin θ0 @2vθ = + vθ cos2 θ @'2 0 Similarly, differentiating the second equation and substituting the first we have 2 ' θ @ v @v cos θ0 0 = 2 + @' @' sin θ0 2 ' @ v ' cos θ0 = 2 + v sin θ0 cos θ0 @' sin θ0 @2v' = + v' cos2 θ @'2 0 Each of these is just the equation for sinusoidal oscillation, so we may immediately write the solution, vθ (') = A cos α' + B sin α' v' (') = C cos α' + D sin α' with the frequency α given by α = cos θ0 α α θ ' Starting the curve at ' = 0, it will close at ' = 2π. Then for v we have the initial condition v (0) = v0; v0 , and from the original differential equations we must have θ @v ' = v0 sin θ0 cos θ0 @' '=0 ' @v θ cos θ0 = −v0 @' '=0 sin θ0 3 These conditions determine the constants A; B; C; D to be v' sin θ cos θ vθ (') = vθ cos α' + 0 0 0 sin α' 0 α θ ' = v0 cos α' + v0 sin θ0 sin α' θ ' ' v0 v (') = v0 cos α' − sin α' sin θ0 This gives the form of the transported vector at any point around the circle, θ θ ' v (') = v0 cos (' cos θ0) + v0 sin θ0 sin (' cos θ0) θ ' ' v0 v (') = v0 cos (' cos θ0) − sin (' cos θ0) sin θ0 Look at the inner product of v with the tangent vector at the same point, u = 1 (0; 1), R sin θ0 θ ' ' v0 u · v = g''t v0 cos (' cos θ0) − sin (' cos θ0) sin θ0 θ ' v0 = R sin θ0 v0 cos (' cos θ0) − sin (' cos θ0) sin θ0 ' θ = v0 R sin θ0 cos (' cos θ0) − v0R sin (' cos θ0) π If the circle is at the equator, θ0 = 2 , then ' u · v = v0 R is constant. On the other hand, near the pole, θ0 1, ' θ u · v = v0 Rθ0 cos ' − v0R sin ' and the transported vector rotates almost completely around the tangent. 3 Geodesics Consider a curve, xα (λ) in an arbitrary (possibly curved) spacetime, with the proper interval given by 2 α β dτ = −gαβdx dx Then the 4-velocity along the curve is given by dxα uα = dτ α dxα and in an arbitrary parameterization, the tangent is t = dλ . Then proper time (or length) along the curve is given by integrating τ q τ = −g dxαdxβ ˆ αβ 0 τ r dxα dxβ = −g dλ ˆ αβ dλ dλ 0 4 A curve of extremal proper length is called a geodesic. We may find an equation for geodesics by finding the equation for the extrema of τ, 0 = δτ τ r dxα dxβ = δ −g dλ ˆ αβ dλ dλ 0 τ 1 dxα dxβ = − δ gαβ dλ ˆ q dxα dxβ dλ dλ 0 2 −gαβ dλ dλ τ 1 dxα dxβ dδxα dxβ dxα dδxβ = − δgαβ + gαβ + gαβ dλ ˆ q dxα dxβ dλ dλ dλ dλ dλ dλ 0 2 −gαβ dλ dλ τ α β 1 dx dx µ = − gαβ,µ δx dλ ˆ q dxα dxβ dλ dλ 0 2 −gαβ dλ dλ τ 0 1 β d 1 dx α + @ gαβ A δx dλ ˆ dλ q dxα dxβ dλ 0 2 −gαβ dλ dλ τ 0 1 α d 1 dx β + @ gαβ A δx dλ ˆ dλ q dxα dxβ dλ 0 2 −gαβ dλ dλ Now choose the parameter λ to be proper time (length) so that dxα dxβ dxα dxβ g = g αβ dλ dλ αβ dτ dτ = −c2 = −1 Then we have τ 1 dxα dxβ d dxβ d dxα 0 = −g δxµ + g δxα + g δxβ dλ 2 ˆ αβ,µ dτ dτ dτ αβ dτ dτ αβ dτ 0 τ 1 d d = −g uαuβ δxµ + g uβ δxα + (g uα) δxβ dλ 2 ˆ αβ,µ dτ αβ dτ αβ 0 τ 1 dxν duβ dxν duα = −g uαuβ δxµ + g uβ + g δxα + g uα + g δxβ dλ 2 ˆ αβ,µ αβ,ν dτ αβ dτ αβ,ν dτ αβ dτ 0 τ 1 duβ duα = −g uαuβ + g uν uβ + g + g uν uα + g δxµdλ 2 ˆ αβ,µ µβ,ν µβ dτ αµ,ν αµ dτ 0 τ 1 duβ duα = g uν uβ + g uν uα − g uαuβ + g + g δxµdλ 2 ˆ µβ,ν αµ,ν αβ,µ µβ dτ αµ dτ 0 The equation for the geodesic is therefore 1 duβ duα 0 = g uν uβ + g uν uα − g uαuβ + g + g 2 µβ,ν αµ,ν αβ,µ µβ dτ αµ dτ 5 1 duβ = (g + g − g ) uαuβ + g 2 µβ,α αµ,β αβ,µ µβ dτ duβ 1 0 = gµν g + gµν (g + g − g ) uαuβ µβ dτ 2 µβ,α αµ,β αβ,µ duν 0 = + Γν uαuβ dτ αβ But this is just duν duν + Γν uαuβ = uµ + uαΓν dτ αβ dxµ αµ µ ν = u Dµu and we have the equation for an autoparallel, duν uµD uν = + Γν uαuβ = 0 µ dτ αβ 4 Example: Geodesics on the 2-sphere Once again consider the 2-sphere, but not look for autoparallels.
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