Geometry of the 2-sphere

Transcript

1 Geometry of the 2-sphere October 28, 2 The metric The easiest way to find the metric of the 2-sphere (or the sphere in any dimension 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 so restricting to gives ds 2 dr 2 + r 2 dθ 2 + r 2 sin 2 θ dϕ 2 r R const. ds 2 R 2 dθ 2 + R 2 sin 2 θ dϕ 2 This is the metric we will study. As a matrix, ( R 2 g ij R 2 sin 2 θ with inverse ( g ij R 2 R 2 sin 2 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 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 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

2 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 R M 2, with tangent vector u α at each point. Let v i be the components of an arbitrary vector at a point P of the curve, and define the family of vectors v α (λ around the curve as the parallel transport of v i along u i : u i D i v j u i ( i v j + v k Γ j ki Since we have the metric, we can compute Γ j ki, so as soon as we specify the curve, we can solve this equation for v i (λ. Then we can find the angle of rotation, α, by taking the inner product of v i (λ final with v i (λ initial, where we have cos α g ijvv i j (λ final g ij v i vj 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 is given by the square root of the determinant of the metric, g det (g ij, so that ˆ ˆ g A dϕ dθ ˆ ˆ R 2 sin θ dϕ dθ 3 Parallel transport on a non-geodesic circle 3. The curve Now consider a circle around the sphere at constant θ (e.g., constant latitude on the surface of Earth. We can parameterize the curve by the angle ϕ, so the curve is given by x i (θ, ϕ A vector tangent to the curve is t i dxi dϕ (, The length of this tangent vector is given by l 2 g ij t i t j R 2 sin 2 θ 2

3 so the unit tangent is u i R sin θ (, 3.2 The connection We also need the connection. We have Γ i jk g im Γ mjk Γ mjk 2 (g mj,k + g mk,j g jk,m Since the metric only has one non-constant component, g ϕϕ, and that one depends only on θ, the only non-vanishing derivative of the metric is g ϕϕ,θ. This means that the only non-vanishing Γ i jk must have two ϕs and one θ index. Using the symmetry of the connection, we have Γ ϕϕθ Γ ϕθϕ 2 (g ϕϕ,θ + g ϕθ,ϕ g ϕθ,ϕ 2 g ϕϕ,θ R 2 sin θ cos θ Γ θϕϕ 2 (g θϕ,ϕ + g θϕ,ϕ g ϕϕ,θ R 2 sin θ cos θ Raising the first index is easy because the metric is diagonal. We have simply 3.3 Parallel transport The parallel transport equation is Γ ϕ ϕθ Γϕ θϕ gϕϕ Γ ϕϕθ R 2 sin 2 θ R2 sin θ cos θ cos θ sin θ Γ θ ϕϕ g θθ Γ θϕϕ u i D i v j R 2 R2 sin θ cos θ sin θ cos θ ( u i i v j + v k Γ j ki ( ϕ v j + v k Γ j kϕ R sin θ 3

4 There are two components to check. For j θ we have ( ϕ v θ + v k Γ θ kϕ R sin θ ( v θ R sin θ ϕ + vϕ Γ θ ϕϕ ( v θ R sin θ ϕ vϕ sin θ cos θ For j ϕ, ( ϕ v ϕ + v k Γ ϕ kϕ R sin θ ( v ϕ R sin θ ϕ + cos θ vθ sin θ Therefore, we need to solve the coupled equations, vθ ϕ vϕ sin θ cos θ vϕ ϕ + vθ cos θ sin θ Taking a second derivative of the first equation and substituting the second, 2 v θ ϕ 2 vϕ ϕ sin θ cos θ 2 v θ ϕ 2 + vθ cos θ sin θ sin θ cos θ 2 v θ ϕ 2 + vθ cos 2 θ Similarly, differentiating the second equation and substituting the first we have 2 v ϕ ϕ 2 + vθ cos θ ϕ sin θ 2 v ϕ ϕ 2 + vϕ sin θ cos θ cos θ sin θ 2 v ϕ ϕ 2 + vϕ cos 2 θ Each of these is just the equation for sinusoidal oscillation, so we may immediately write the solution, v θ (ϕ A cos αϕ + B sin αϕ v ϕ (ϕ C cos αϕ + D sin αϕ 4

5 where β cos θ Starting the curve at ϕ, it will close at ϕ 2π. Then for v α we have the initial condition v α ( ( v, θ v ϕ, and from the original differential equations we must have v θ ϕ v ϕ ϕ v ϕ sin θ cos θ ϕ v θ cos θ ϕ sin θ These conditions determine the constants A, B, C, D to be v θ (ϕ v θ cos βϕ + vϕ sin θ cos θ β v θ cos βϕ + v ϕ sin θ sin βϕ v ϕ (ϕ v ϕ cos βϕ sin βϕ vθ sin θ sin βϕ 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 v α ( is while the norm of v 2 R 2 ( v θ 2 + R 2 sin 2 θ (v ϕ 2 v 2 R 2 ( v θ cos βϕ + v ϕ sin θ sin βϕ 2 + R 2 sin 2 θ (v ϕ cos βϕ sin βϕ vθ ( sin θ R 2 ( v θ 2 cos 2 βϕ + 2vv θ ϕ sin θ cos βϕ sin βϕ + (v ϕ 2 sin 2 θ sin 2 βϕ ( +R 2 sin 2 θ (v ϕ 2 cos 2 βϕ vv θ ϕ βϕ cos βϕsin + ( v θ 2 sin 2 βϕ sin θ sin 2 θ R 2 ( v θ 2 cos 2 βϕ + 2R 2 v θ v ϕ sin θ cos βϕ sin βϕ + R 2 (v ϕ 2 sin 2 θ sin 2 βϕ +R 2 (v ϕ 2 sin 2 θ cos 2 βϕ R 2 vv θ ϕ sin θ cos βϕ sin βϕ + R 2 ( v θ 2 sin 2 βϕ R 2 ( v θ 2 ( cos 2 βϕ + sin 2 βϕ + R 2 (v ϕ 2 sin 2 ( θ sin 2 βϕ + cos 2 βϕ R 2 ( v θ 2 + R 2 (v ϕ 2 sin 2 θ v 2 which, as claimed, is independent of ϕ. Now we turn to the calculation of the curvature. 2 5

6 3.5 Curvature of the 2-sphere We need the angular deficit and the area of the sphere enclosed by the circular path The angular deficit The angular deficit is given by 2π α where the angle of rotatoin, α, is given by The inner product in the numerator is cos α g ijv i v j (λ final g ij v i vj g ijv i v j (2π g ij v i vj g ij vv i j (2π R 2 ( vv θ 2π θ + v ϕ vϕ 2π sin2 θ ( (v R 2 θ v θ cos βϕ + vv θ ϕ sin θ sin βϕ ( + v ϕ v ϕ cos βϕ sin βϕ vθ sin 2 θ ( sin θ R 2 vv θ θ cos βϕ + vv θ ϕ sin θ sin βϕ + (v ϕ 2 sin 2 θ cos βϕ v ϕ vθ sin θ sin βϕ R 2 ( ( v θ 2 + (v ϕ 2 sin 2 θ cos βϕ v 2 cos βϕ The angle of rotation is therefore α β 2π cos θ. Therefore, the angular deficit is 2π α 2π ( cos θ The area enclosde by the loop The area enclosed by the loop is ˆ A ˆ dϕ dθ g R 2 ˆ 2π ˆ θ dϕ sin θdθ 2πR 2 cos θ θ 2πR 2 (cos θ 2πR 2 ( cos θ 6

7 3.5.3 The curvature The curvature is now given by the limit as we shrink the loop to a point, Curvature lim θ A 2π ( cos θ lim θ 2πR 2 ( cos θ R 2 This increases as the sphere shrinks, which indeed makes the curvature greater. 4 Comparison with the Riemann curvature tensor We can also compute the curvature using the Riemann curvature tensor. We already have the connection, Γ ϕ ϕθ Γϕ θϕ cos θ sin θ Γ θ ϕϕ sin θ cos θ so it is straightforward to compute the curvature using R i jkm Γ i jm,k Γ i jk,m + Γ i nkγ n jm Γ i nmγ n jk Since R ijkm R jikm R ijmk, there is only one independent component. All of the rest follow from the symmetries of the curvature tensor. We can compute any one non-vanishing component. Choose R θ ϕθϕ Γ θ ϕϕ,θ Γ θ ϕθ,ϕ + Γ θ nθγ n ϕϕ Γ θ nϕγ n ϕθ ( sin θ cos θ,θ + Γ n ϕϕ Γ θ ϕϕγ ϕ ϕθ ( cos 2 θ + sin 2 θ ( cos θ ( sin θ cos θ sin θ sin 2 θ The full curvature tensor may be written in terms of the metric and Kronecker delta by including all the necessary symmetries, R i jkm R 2 ( δ i k g jm δ i mg jk Then check that R θ ϕθϕ R 2 ( δ θ θ g ϕϕ δ θ ϕg ϕθ R 2 g ϕϕ sin 2 θ 7

8 Since the expression in terms of the metric has all the right symmetries, and the value of the one independent component is correct, it gives the full curvature tensor. We may find the Ricci tensor by contraction: R jm R i jim R 2 ( δ i i g jm δ i mg ji R 2 (2g jm g jm R 2 g jm The Ricci scalar is the contraction of this. Using the inverse metric, R g jm R jm R 2 gjm g jm 2 R 2 which differs from our angular deficit formula only by an overall constant. 8