Show that (a) Γααβ = 1 2g ∂g ∂xβ , where g

(1)

GR problem set 02

We dene covariant derivative of a vector as:

∇_αu^µ≡ u^µ_;α = u^µ_,α+ Γ^µ_αβu^αu^β, where u^µ,α ≡ ∂_αu^µ≡ ∂u^µ

∂x^α. Covariant derivatives of other tensors is dened using the properties:

∇_α(u^µv_µ) = ∂_α(u^µv_µ) (for scalar φ,α ≡ φ_;α)

∇_α(u^µv_ν) = v_ν∇_αu^µ+ u^µ∇_αv_ν (Leibniz' rule)

16. We want ∇αu^µ to be a tensor (with a proper tensor transformation rule). Then, how should the transformation law for Γ^µ_αβ look like?

17. Show (with explicit calculation) that there is one and only one connection (called metric connection) that is symmetric (Γ^γ_αβ = Γ^γ_βα) and fullls ∇αg_µν = 0 condition.

∇_αg_µν = ∂_αg_µν− Γ^ρ_αµg_ρν− Γ^ρ_ναg_ρµ.

18. Find the metric connection components for the Euclidean metric ds² = dx² + dy² in polar coordinates.

19. Show that (a)

Γ^α_αβ = 1 2g

∂g

∂x^β , where g = | det(gαβ)|.

(b) then using (a) show

∇_αv^α = 1

√g∂_α(√

gv^α), ∇^α∇_αΦ = 1

√g∂_α√gg^αβ∂_βΦ for any vector and scalar eld, v^α and Φ respectively.

20. Show that for any curve with a tangent vector v^α such that v^α∇_αv^β = cv^β, where c is an arbi- trary function on the curve, one can nd a parametrization such that in this parametrization v^α∇_αv^β = 0 holds (ane parametrization). Show that two dierent ane parametrizations are connected with a linear transformation.

(2)

(∇_µ∇_ν − ∇_ν∇_µ)u^α is linear in u and does not contain any derivatives of u (please check). Thus we can write:

(∇µ∇ν− ∇ν∇µ)u^α = R^α_βµνu^β, This denes the Riemann tensor R^α_βµν.

We also dene: Rβν = R^α_βαν (Ricci tensor) and R = g^βνR_βν (Ricci scalar).

21. Show the following symmetry properties for the Riemann tensor:

(a) Rαβµν = R_µναβ = −R_αβνµ = −R_βαµν , (b) R^αβµν+ R^α_νβµ+ R^α_µνβ = 0 ,

(c) R^αβµν;λ+ R^α_βλµ;ν+ R^α_βνλ;µ = 0 (Bianchi identities), (d) (R^αβ − 12g^αβR)_;β = 0 (contracted Bianchi identities).

Hint: (a)-(c) can be easily checked in a locally Lorentzian frame (cf problem 15), then (d) follows from (c).

22. Show that, in n-dimensional space the Riemann tensor has (in general) n²(n²− 1)/12 inde- pendent components.

A. Rostworowski http://th.if.uj.edu.pl/ arostwor/