Geometry of the 2-sphere October 28, 2010 1 The metric The easiest way to find the metric of the 2-sphere (or the sphere in any dimen- sion) is to picture it as embedded in one higher dimension of Euclidean space, then restrict to constant radius. The 3-dim Euclidean metric in spherical coordinates is ds2 = dr2 + r2dθ2 + r2 sin2 θ d'2 so restricting to r = R = const: gives ds2 = R2dθ2 + R2 sin2 θ d'2 This is the metric we will study. As a matrix, R2 0 g = ij 0 R2 sin2 θ with inverse 1 ij R2 0 g = 1 0 R2 sin2 θ There are more 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 curva- ture. 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 Curvature: a plan One definition of curvature starts by carrying a vector by parallel transport around a closed loop. In general, the vector returns rotated from its original direction. The difference between this angle and the angle expected in a flat geometry is called the angular deficit. Next, calculate the area enclosed in the 1 loop. Then the curvature at point Q is the limit of the angular deficit per unit area, as the loop shrinks to Q and the area to zero. Explicitly, consider a closed curve C : λ 2 R ! M2, with tangent vector α i u at each point. Let v0 be the components of an arbitrary vector at a point P of the curve, and define the family of vectors vα (λ) around the curve as the i i parallel transport of v0 along u : i j 0 = u Div i j k j = u @iv + v Γki j Since we have the metric, we can compute Γki, so as soon as we specify the curve, we can solve this equation for vi (λ). Then we can find the angle of i i rotation, α, by taking the inner product of v (λfinal) with v (λinitial), where we have g vi vj (λ ) cos α = ij 0 final i j gijv0v0 Then the angular deficit is ∆ = 2π − α since transport around a closed loop in flat space will rotate the vector by 2π. For the area inside the loop, we integrate the 2-dim volume element. This p p is given by the square root of the determinant of the metric, g = det (gij), so that p A = g d' dθ ˆ ˆ = R2 sin θ d' dθ ˆ ˆ 3 Parallel transport on a non-geodesic circle 3.1 The curve Now consider a circle around the sphere at constant θ0 (e.g., constant latitude on the surface of Earth). We can parameterize the curve by the angle ', so the curve is given by i x = (θ0;') A vector tangent to the curve is dxi ti = d' = (0; 1) The length of this tangent vector is given by 2 i j l = gijt t 2 2 = R sin θ0 2 so the unit tangent is 1 ui = (0; 1) R sin θ0 3.2 The connection We also need the connection. We have i im Γjk = g Γmjk 1 Γ = (g + g − g ) mjk 2 mj;k mk;j jk;m Since the metric only has one non-constant component, g'', and that one de- pends only on θ, the only non-vanishing derivative of the metric is gϕϕ,θ. This i means that the only non-vanishing Γjk must have two 's and one θ index. Using the symmetry of the connection, we have 1 Γ = Γ = (g + g − g ) ϕϕθ ϕθ' 2 ϕϕ,θ ϕθ;' ϕθ;' 1 = g 2 ϕϕ,θ = R2 sin θ cos θ 1 Γ = (g + g − g ) θ'' 2 θ';' θ';' ϕϕ,θ = −R2 sin θ cos θ Raising the first index is easy because the metric is diagonal. We have simply ' ' '' Γϕθ = Γθ' = g Γϕϕθ 1 = R2 sin θ cos θ R2 sin2 θ cos θ = sin θ θ θθ Γ'' = g Γθ'' 1 = − R2 sin θ cos θ R2 = − sin θ cos θ 3.3 Parallel transport The parallel transport equation is i j 0 = u Div i j k j = u @iv + v Γki 1 j k j = @'v + v Γk' R sin θ0 3 There are two components to check. For j = θ we have 1 θ k θ 0 = @'v + v Γk' R sin θ0 θ 1 @v ' θ = + v Γ'' R sin θ0 @' θ 1 @v ' = − v sin θ0 cos θ0 R sin θ0 @' For j = ', 1 ' k ' 0 = @'v + v Γk' R sin θ0 1 @v' cos θ = + vθ 0 R sin θ0 @' 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 im- mediately write the solution, vθ (') = A cos α' + B sin α' v' (') = C cos α' + D sin α' 4 where β = cos θ0 Starting the curve at ' = 0, it will close at ' = 2π. Then for vα we have the α θ ' 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 These conditions determine the constants A; B; C; D to be v' sin θ cos θ vθ (') = vθ cos β' + 0 0 0 sin β' 0 β θ ' = v0 cos β' + v0 sin θ0 sin β' ' ' θ sin β' v (') = v0 cos β' − v0 sin θ0 This gives the form of the transported vector at any point around the circle. 3.4 Norm of v We have claimed that the norm of a vector is not changed by parallel transport. We can check this in the current example. The initial squared norm of vα (0) is 2 2 θ2 2 2 ' 2 j~v0j = R v0 + R sin θ0 (v0 ) while the norm of 2 2 2 θ ' 2 2 2 ' θ sin β' j~vj = R v0 cos β' + v0 sin θ0 sin β' + R sin θ0 v0 cos β' − v0 sin θ0 2 θ2 2 θ ' ' 2 2 2 = R v0 cos β' + 2v0v0 sin θ0 cos β' sin β' + (v0 ) sin θ0 sin β' 2 2 2 ' 2 2 θ ' sin β' θ2 sin β' +R sin θ0 (v0 ) cos β' − v0v0 cos β' + v0 2 sin θ0 sin θ0 2 θ2 2 2 θ ' 2 ' 2 2 2 = R v0 cos β' + 2R v0v0 sin θ0 cos β' sin β' + R (v0 ) sin θ0 sin β' 2 ' 2 2 2 2 θ ' 2 θ2 2 +R (v0 ) sin θ0 cos β' − R v0v0 sin θ0 cos β' sin β' + R v0 sin β' 2 θ2 2 2 2 ' 2 2 2 2 = R v0 cos β' + sin β' + R (v0 ) sin θ0 sin β' + cos β' 2 θ2 2 ' 2 2 = R v0 + R (v0 ) sin θ0 2 = j~v0j which, as claimed, is independent of '. Now we turn to the calculation of the curvature. 5 3.5 Curvature of the 2-sphere We need the angular deficit and the area of the sphere enclosed by the circular path. 3.5.1 The angular deficit The angular deficit is given by ∆ = 2π − α where the angle of rotatoin, α, is given by g vi vj (λ ) cos α = ij 0 final i j gijv0v0 g vi vj (2π) = ij 0 i j gijv0v0 The inner product in the numerator is i j 2 θ θ ' ' 2 gijv0v (2π) = R v0v2π + v0 v2π sin θ0 2 θ θ θ ' ' ' θ sin β' 2 = R v0v0 cos β' + v0v0 sin θ0 sin β' + v0 v0 cos β' − v0 sin θ0 sin θ0 2 θ θ θ ' ' 2 2 ' θ = R v0v0 cos β' + v0v0 sin θ0 sin β' + (v0 ) sin θ0 cos β' − v0 v0 sin θ0 sin β' 2 θ2 ' 2 2 = R v0 + (v0 ) sin θ0 cos β' 2 = j~v0j cos β' The angle of rotation is therefore α = β = 2π cos θ0. Therefore, the angular deficit is ∆ = 2π − α = 2π (1 − cos θ0) 3.5.2 The area enclosde by the loop The area enclosed by the loop is p A = d' dθ g ˆ ˆ 2π θ0 = R2 d' sin θdθ ˆ0 ˆ0 2 θ0 = −2πR cos θj0 2 = −2πR (cos θ0 − 1) 2 = 2πR (1 − cos θ0) 6 3.5.3 The curvature The curvature is now given by the limit as we shrink the loop to a point, ∆ Curvature = lim θ0!0 A 2π (1 − cos θ0) = lim 2 θ0!0 2πR (1 − cos θ0) 1 = R2 This increases as the sphere shrinks, which indeed makes the curvature greater. 4 Comparison with the Riemann curvature ten- sor We can also compute the curvature using the Riemann curvature tensor.