Riemannian Curvature

Transcript

1 Riemannian Curvature February 6, 013 We now generalize our computation of curvature to arbitrary spaces. 1 Parallel transport around a small closed loop We compute the change in a vector, w, which we parallel transport around a closed loop. The loop starts at a point P, which we take to have coordinates 0, 0, progresses along a direction u a distance λ to a point at λ, 0. Next, we transport along the direction v a distance σ to a point at λ, σ, giving a curve C 1 from 0, 0 to λ, σ. Rather than returning in the direction u by λ then v by σ, we perform a second transport, interchanging the order: first v by σ, then u by λ, giving curve C from 0, 0 to λ, σ. We then compare the expressions for w λ, σ along the two curves. This gives the same result as if we had transported around the closed loop C 1 C, but the computation is easier this way. Start with two directions, u 0, 0 and v 0, 0. Parallel transport each in the direction of the other to get u 0, σ and v λ, 0: and v λ, 0 = v 0, 0 + dv 0, 0 λ dλ = v 0, 0 u β 0, 0 v µ 0, 0 µβ 0, 0 λ u 0, σ = u 0, 0 + du 0, 0 σ dλ = u 0, 0 v β 0, 0 u µ 0, 0 µβ 0, 0 σ Now take a third vector, w 0, 0 and transport it to w λ, σ in two different ways, first along u and then v, then in the opposite order. First, w λ, 0 = w 0, 0 u β 0, 0 w µ 0, 0 µβ 0, 0 λ then w λ, σ = w λ, 0 v β λ, 0 w µ λ, 0 µβ λ, 0 σ = w 0, 0 u β 0, 0 w µ 0, 0 µβ 0, 0 λ w v β 0, 0 u λ 0, 0 v τ 0, 0 β τλ 0, 0 λ µ 0, 0 u 0, 0 w σ 0, 0 µ σ 0, 0 λ µβ 0, 0 + u δ δ µβ = w 0, 0 u β 0, 0 w µ 0, 0 µβ 0, 0 λ v β 0, 0 w µ 0, 0 µβ 0, 0 σ +u λ 0, 0 v τ 0, 0 w µ 0, 0 β τλ 0, 0 µβ 0, 0 σλ +v β 0, 0 u 0, 0 w σ 0, 0 µ σ 0, 0 µβ 0, 0 σλ 1

2 v β 0, 0 w µ 0, 0 u δ δ µβ 0, 0 λσ +u λ 0, 0 v τ 0, 0 w µ 0, 0 β τλ 0, 0 uδ δ µβ 0, 0 λ σ +v β 0, 0 u 0, 0 w σ 0, 0 µ σ 0, 0 u δ δ µβ 0, 0 λ σ u λ 0, 0 v τ 0, 0 β τλ 0, 0 u 0, 0 w σ 0, 0 µ σ 0, 0 µβ 0, 0 λ σ u λ 0, 0 v τ 0, 0 β τλ 0, 0 u 0, 0 w σ 0, 0 µ σ 0, 0 u δ δ µβ 0, 0 λ 3 σ Now that all quantities are expressed at P = 0, 0, we may drop these 0, 0 arguments. Keeping only terms to second order: w C 1 λ, σ = w u β w µ µβλ v β w µ µβσ v β w µ u δ δ µβλσ + u λ v τ w µ β τλ µβσλ +v β u w σ µ σ µβσλ Now repeat in the opposite order just interchange u with v and σ with λ, w C λ, σ = w v β w µ µβσ u β w µ µβλ u β w µ v δ δ µβλσ + v λ u τ w µ β τλ µβσλ +u β v w σ µ σ µβσλ The difference between these gives the change in w under parallel transport around the loop so δw = wc 1 C = wc 1 λ, σ wc λ, σ = w u β w µ µβλ v β w µ µβσ + u λ v τ w µ β τλ µβσλ +v β u w σ µ σ µβσλ v β w µ u δ δ µβλσ w + v β w µ µβσ + u β w µ µβλ v λ u τ w µ β τλ µβσλ u β v w σ µ σ µβσλ + u β w µ v δ δ µβλσ δw = w w u β w µ µβλ + u β w µ µβλ v β w µ µβσ + v β w µ µβσ +w µ u β v δ δ µβ v β u δ δ µβ σλ +w µ u λ v τ β τλ µβ + v β u σ µ σβ v λ u τ β τλ µβ u β v σ µ σβ = w µ u β v ν µβ,ν v ν u β µν,β σλ +w µ u λ v τ β τλ β λτ µβ + v ν u β σ µβ σν u β v ν σ µν σβ λσ +w µ µβ,ν µν,β + σ µβ σν σ µν σβ v ν u β λσ The change in w per unit area is our measure of the curvature, δw λσ = w µ µβ,ν µν,β + σ µβ σν σ µν σβ v ν u β = w µ R µβνu β v ν We see that the change in w depends linearly on w, and also on the orientation of the infinitesimal loop spanned by u and v. The rest of the expression, R µβν µβ,ν µν,β + σ µβ σν σ µν σβ λσ

3 is called the Riemann curvature tensor. It may be thought of as a trilinear operator which takes an oriented unit area element, S βν = 1 u β v ν u ν v β to give a linear operator, T µ = R µβν Sβν. This linear operator gives the infinitesimal change in w per unit area, in the limit of zero area, when w is transported around the area element with orientation S βν. Since all elements of this construction are defined geometrically from within the manifold, R µβν is intrinsic to the manifold. To see that it is also a tensor, we could recompute the same construction in different coordinates. Since the entire construction is perturbative, we would find the components of R µβν changing linearly and homogeneously in the transformation matrix. However, there is an easier way to prove that R µβν is a tensor. Consider two covariant derivatives of the vector w, D µ D ν w = D µ ν w + w β βν = µ ν w + w β βν + ν w + w β βν µ w + w β ν νµ Because we use covariant derivatives, this object is necessarily a tensor. Now take the derivatives in the opposite order and subtract, giving the commutator. This is also a tensor, [D µ, D ν ] w = D µ D ν w D ν D µ w = µ ν w + µ w β βν + w β βν,µ + ν w + w β βν µ w + w β ν νµ ν µ w ν w β βµ w β βµ,ν µ w + w β βµ ν + w + w β ν µν = µ ν w ν µ w + µ w β βν + ν w µ ν w β βµ µ w ν w + w β ν νµ + w + w β ν µν +w β βν,µ + w β βν µ w β βµ,ν w β βµ ν = w β R βνµ Since the result is w β properly contracted with R βνµ, and wβ is a tensor, R βνµ must also be a tensor. Symmetries of the curvature tensor Recall that parallel transport of w preserves the length, w w of w. This means that the transformation, δ µ + T µσλ w µ = w + w µ R µβνs βν σλ = w + δw must be an infinitesimal Lorentz transformation, Λ β = δ β + ε β. An infinitesimal Lorentz transformation satisfies η β = η µν Λ µ Λ ν β = η µν δ µ + ε µ δβ ν + ε ν β = η β + η ν ε ν β + η µβ ε µ + O ε 0 = ε β + ε β so when the indices of an infinitesimal Lorentz transformation are both in the same position, they must be antisymmetric. Therefore, T β = T β and we must have R βµν S µν = R βµν S µν 3

4 for any surface element S µν. Therefore, the Riemann curvature tensor is antisymmetric on the first pair of indices, R βµν = R βµν From the explicit expression for R βµν, the curvature tensor must also be antisymmetric on the last pair of indices, R βµν = βµ,ν βν,µ + σν σ βµ σµ σ βν = βν,µ + βµ,ν σµ σ βν + σν σ βµ = βν,µ βµ,ν + σµ σ βν σν σ βµ = R βνµ Another symmetry follows if we totally antisymmetrize the final three indices, R [βµν] = 1 R 3 βµν + R µνβ + R νβµ 3R [βµν] = βµ,ν βν,µ + σν σ βµ σµ σ βν + µ,νβ µβ,ν + σβ σ µν σν σ µβ + ν,βµ νµ,β + σµ σ νβ σβ σ νµ = βµ µβ +,ν νβ βν +,µ µν νµ + σν σ βµ σ µβ + σβ σ µν σ νµ + σµ σ νβ σ βν = 0 the sum vanishes identically because of the symmetry of µν. This condition ultimately arises because the connection may be written in terms of the metric. It is called the first Bianchi identity. Finally, consider R βµν R µνβ = R βµν R µβν R µβν = R βµν R µβν R βµν = R βµν R βνµ R νµβ R βµν R βνµ = R βµν + R βνµ + R νµβ + R βµν + R βνµ = R βµν + R νµβ + R βνµ = R βµν R νβµ + R βνµ = R µνβ + R βνµ = R µβν + R βνµ Now interchange the names of both β and µν. On the left, this gives two minus signs, but on the right only one: R βµν R µνβ = R µβν + R βνµ,β R βνµ R νµβ = R βνµ + R µβν 1 R βµν R µνβ = +R µβν R βνµ = R βµν R µνβ This difference therefore vanishes, and we have symmetry under interchange of the pairs, R βµν = R µνβ 4

5 Summarizing, we have the following symmetries of the Riemann curvature tensor: R βµν = R βµν R βµν = R βνµ R βµν = R µνβ R [βµν] = 0 We can count the independent components by using these symmetries. Because of the antisymmetry on β, there are only 4 3 = 6 independent values for this pair of indices. The same counting holds for the final pair, µν. Since we have symmetry in these pairs, R [β][µν] = R [µν][β] we may think of R [β][µν] as a 6 6 symmetric matrix, which will have 6 7 = 1 independent components. This makes use of the first three symmetries. To use the final symmetry, note that the three final indices must differ from one another, so there are only possible four cases, R 0[βµ] = 0 R 1[βµ] = 0 R [βµ] = 0 R 3[βµ] = 0 Now suppose one of βµ is the same as the first index, for example, R R R 131 = R R 131 = R 131 R 113 = 0 Then the vanishing is automatic using the previous three symmetries and there is no additional constraint. Therefore, to get any new condition, all four indices must differ. But then, notice that R 013 = R 103 = R 310 = R 310 so that once we have the condition with 0 in the first position, the other three possibilities follow automatically. There is therefore only one condition from the fourth symmetry, R R R 031 = 0 reducing the number of degrees of freedom of the Riemann curvature tensor to 0. 3 Ricci tensor and Ricci scalar Because of the symmetries, there is only one independent contraction of R βµν. We define the Ricci tensor, R µν R µν 5

6 Because of the symmetry between pairs, we have R µν R µν = g β R µβν = g β R βνµ = R νµ = R νµ so the Ricci tensor is symmetric. Exercise: Prove that any other contraction of the curvature is either zero, or a multiple of the Ricci tensor. We also define the Ricci scalar, given by taking contracting the Ricci tensor, R = g µν R µν It is possible to decompose the full Riemann curvature into a traceless part, called the Weyl curvature, and combinations of the Ricci tensor and Ricci scalar, but we will not need this now. 4 The second Bianchi identity and the Einstein equation We have already seen the first Bianchi identity, R [βµν] = 0 This is an integrability condition that guarantees that the connection may be written in terms of derivatives of the metric. There is a second integrability condition, called the second Bianchi identity, guaranteeing that the curvature may be written in terms of a connection. The second Bianchi identity is R β[µν;σ] = 0 To prove this, first consider antisymmetrizing a double commutator: [D β, [D µ, D ν ]] w = D β D µ D ν D β D ν D µ D µ D ν D β + D ν D µ D β w 3 [ D [β, [ ]] D µ, D ν] = [D β, [D µ, D ν ]] + [D µ, [D ν, D β ]] + [D ν, [D β, D µ ]] However, we may also write this as = D β D µ D ν D β D ν D µ D µ D ν D β + D ν D µ D β +D µ D ν D β D µ D β D ν D ν D β D µ + D β D ν D µ +D ν D β D µ D ν D µ D β D β D µ D ν + D µ D β D ν = D β D µ D ν D β D µ D ν D β D ν D µ + D β D ν D µ = 0 0 = 3 [ D [β, [ D µ, D ν] ]] w +D µ D ν D β D µ D ν D β D µ D β D ν + D µ D β D ν D ν D β D µ + D ν D β D µ D ν D µ D β + D ν D µ D β = [D β, [D µ, D ν ]] w + [D µ, [D ν, D β ]] w + [D ν, [D β, D µ ]] w = D β D µ D ν D ν D µ w + D β D ν D µ D µ D ν w +D µ D ν D β D β D ν w + D µ D β D ν D ν D β w 6

7 +D ν D µ D β D β D µ w + D ν D β D µ D µ D β w = D β w R µν + Dβ w R νµ + Dµ w R νβ +D µ w R βν + Dν w R µβ + Dν w R βµ = w R [µν;β] + R µν + R νµ Dβ w + R νβ + R βν Dµ w + R µβ + R βµ Dν w = w R [µν;β] and since w is arbitrary, we have the second Bianchi identity. The second Bianchi identity is important for general relativity because of its contractions. First, expand the identity and contract on µ, 0 = R βµν;σ + R βνσ;µ + R βσµ;ν 0 = R βν;σ + R βνσ; + R βσ;ν Using the definition of the Ricci tensor and the antisymmetry of the Riemann tensor, 0 = R βν;σ + R βνσ; R βσ;ν Now contract on βσ using the metric, 0 = g βσ R βν;σ + g βσ R βνσ; g βσ R βσ;ν = g βσ R βν ;σ gβσ Rβ σν; g βσ R βσ ;ν = R σ ν;σ + R ν; R ;ν We may write this as a divergence, 0 = R ν; 1 R ;ν 0 = D R ν 1 δ ν R or, raising an index, We define the Einstein tensor, D R β 1 gβ R = 0 G β R β 1 gβ R and have now shown that it has vanishing divergence. Since both the Ricci tensor and the metric are both symmetric, we have G β = G β D β G β = 0 These are precisely the properties we require of the energy-momentum tensor, T β. It can be shown that G β is the only tensor linear in components of the Riemann curvature tensor to have these properties. Reasoning that it is the presence of energy that leads to curvature, the only candidate equation consistent with the properties of T β and linear in the curvature hence, a second order differential equation for the metric is G β = κt β This is the Einstein equation for general relativity. 7