Introduction to General Relativity

(G4040 - Fall 2012)

Solutions: Problem set 6Prof. A. Beloborodov

TA: Alex Chen

1. Start with the general expression for the Riemann tensor:

Rαβγδ = gαρRρβγδ

= gαρ(∂γΓρδβ − ∂δΓργβ + ΓργλΓλδβ − ΓρδλΓλγβ). (1)

Let’s consider local inertial coordinates at a point p, chosen such that the Christoffel symbols vanishat p. In those coordinates the above equation becomes:

Rαβγδ = gαρ(∂γΓρδβ − ∂δΓργβ)

= ∂γ(gαρΓρδβ)− ∂δ(gαρδΓργβ) (2)

since ∂δgαρ = ∇δgαρ = 0, and using the fact that

gαρΓρδβ =

1

2(∂δgβα + ∂βgδα − ∂αgδβ ) (3)

we get

Rαβγδ =1

2

{∂γ(∂δgβα + ∂βgδα − ∂αgδβ )− ∂δ(∂γgβα + ∂βgγα − ∂αgγβ )

}=

1

2(∂γ∂βgδα + ∂δ∂αgγβ − ∂γ∂αgδβ − ∂δ∂βgγα). (4)

Now it’s easy to see that Rαβγδ = −Rαβδγ , Rαβγδ = −Rβαγδ and also Rαβγδ = Rγδαβ . Finally wehave

Rαβγδ +Rαδβγ +Rαγδβ

=1

2(∂γ∂βgδα + ∂δ∂αgγβ − ∂γ∂αgδβ − ∂δ∂βgγα)

+1

2(∂δ∂βgγα + ∂α∂γgδβ − ∂α∂βgδγ − ∂δ∂γgβα)

+1

2(∂γ∂δgβα + ∂β∂αgγδ − ∂δ∂αgγβ − ∂β∂γgδα)

= 0 (5)

As all the 4 equations are tensor equation, they are valid in any coordinate system, this finishes ourproof!

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2. (a) First we claim that the maximum number of independent components of the Riemann tensor is112n

2(n2−1). To prove this let’s begin looking at the first 3 symmetries in the first problem. SinceRαβγδ is symmetric under the exchange of αβ and γδ, there are just K(K + 1)/2 independentcomponets, where K is the number of independent components of the αβ subset. On the otherhand Rαβγδ is anti-symmetric on the first and second pair indices, which implies K = n(n− 1)/2,so the number of independent components of the Riemann tensor is reduced to

n(n− 1)

2(n(n− 1)

2+ 1)/2 =

1

8(n4 − 2n3 + 3n2 − 2n). (6)

The last symmetry appearing in the first problem,

Rαβγδ +Rαδβγ +Rαγδβ = 0, (7)

is actually equivalent to Rα[βγδ] = 0 and R[αβγδ] = 0 if the first 3 symmetries are already imposed.

Then we conclude that there are C4n = n!

4!(n−4)! extra constraints, so the resulting number of

independent components is:

1

8(n4 − 2n3 + 3n2 − 2n)− n(n− 1)(n− 2)(n− 3)/4 =

1

12n2(n2 − 1). (8)

Now we can consider the 2 dimensional case, when n = 2. According to the above equationwe have only 1 independent component, which means the Ricci scalar completely determine theRiemann tensor. And the only possible form is:

Rαβγδ = f(R)(gαγgβδ − gαδgβγ) (9)

R = Rαβαβ = 2f(R) =⇒ f(R) = R/2 (10)

so in 2 dimensional case, we have

Rαβγδ =R

2(gαγgβδ − gαδgβγ) (11)

(b) In 3D we have n = 3, from Eq.(8) there are at most 6 independent components. The Ricci tensoris symmetric and thus has 6 independent components, so we can construct all the Riemann tensorfrom the Ricci tensor. And the only possible from is

Rαβγδ = f(Rαγ)gβδ − f(Rαδ)gβγ + f(Rβδ)gαγ − f(Rβγ)gαδ, (12)

it’s then easy to get f(Rαβ) = Rαβ − R4 gαβ .

3. From the last problem set we have the following metric for the 2-sphere,

gαβdxαdxβ = a2dθ2 + a2 sin θ2dφ2, (13)

and the non-zero Christoffel symbols,

Γθφφ = − sin θ cos θ, Γφθφ = cot θ, (14)

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any other being zero or related to the previous ones by symmetries. From problem 2a. we know thatit is sufficient to calculate just one non-zero component of the Riemann tensor to specify it completely.From 2a, a good component to calculate is Rθφθφ,

Rθφθφ = ∂θΓθφφ − ∂φΓθθφ + ΓθθλΓλφφ − ΓθφλΓγθφ

Rθφθφ = sin2 θ

Rθφθφ = gθθRθφθφ = a2 sin2 θ.

The Ricci tensor, Rµν = gαβRαµβν is then,

Rθθ = 1, Rθφ = Rφθ = 0, Rφφ = sin2 θ.

This finally gives R = gαβRαβ = 2/a2, and

Rαβγδ =1

a2(gαγgβδ − gαδgβγ) (15)

4. (Schutz 6.9: 30) The 3D metric in cylindrical coordinates is written as

gµνdxµdxν = dz2 + ρ2dθ2 + dρ2. (16)

A cylinder in 3D cylindrical coordinates is described by ρ = a =constant. So the matric of a cylindertakes the form

gµνdxµdxν = dz2 + a2dθ2. (17)

Doing the change of coordinates {z, aθ} → {z′, θ′} we get,

gµνdx′µdx′ν = dz′2 + dθ′2, (18)

i.e. the space is flat. All derivative of the metric coefficients are zero and so it is the Riemann tensor.

5. (Schutz 6.9: 31) In local inertial frames covariant derivatives reduce to usual partial derivatives. Sincepartial derivatives obey the chain rule we deduce that in a local inertial frame the covariant derivativesfollow the chain rule. In mathematical language:

∇µ(A⊗B)|local inertial frame = ∇µ(A)|local inertial frame ⊗B + A⊗∇µ(B)|local inertial frame. (19)

The above equation is a tensorial equation so it must be satisfied in any frame,

∇µ(A⊗B) = ∇µ(A)⊗B + A⊗∇µ(B). (20)

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