General Relativity 2012 – Solutions

Home , Connection (mathematics), Geodesic deviation, Killing vector field

Yichen Shi Easter 2014

(a) Explain the following terms: covariant derivative, connection components.

A covariant derivative ∇ on a manifold M is a map sending every pair of smooth vector ﬁelds X, Y to a smooth vector ﬁeld ∇X Y , with properties

∇fX+gY Z = f∇X Z + g∇Y Z

∇X (Y + Z) = ∇X Y + ∇X Z

∇X (fY ) = f∇X Y + (∇X f)Y µ In a basis eµ the connection components Γνρ are deﬁned by

µ ∇ρeσ = ∇eρ eσ = Γρσeµ

The Christoﬀel symbols are the coordinate basis components of the Levi-Civita connection, which is deﬁned on any manifold with a metric.

(b) What does it mean for a connection to be torsion-free? Deﬁne the Levi-Civita connection and derive the formula 1 Γµ = gµσ(∂ g + ∂ g − ∂ g ). νρ 2 ρ σν ν σρ σ νρ

A connection ∇ is torsion-free if [∇a, ∇b]f = 0 for any function f. For a torsion-free connection, if X and Y are vector ﬁelds then ∇X Y − ∇Y X = [X,Y ]. On a manifold with a metric, the metric singles out a preferred connection. Let M be a manifold with a metric g. There exists a unique torsion-free connection ∇, the Levi-Civita connection, which is metric compatible, i.e., ∇g = 0. Hence we have

l l ∇igjk = 0 = ∂igjk − Γijgkl − Γikgjl

l l ⇒ ∂igjk = Γijgkl + Γikgjl l l¨¨ l Hl l¨¨ Hl l ⇒ ∂igjk + ∂jgik − ∂kgij = Γijgkl +¨Γikgjl + Γjigkl + ΓjkHgHil −¨Γkigjl − ΓkjHgHil = 2Γijgkl, k since torsion free implies Γ[ij] = 0. 1 ⇒ Γl g = (∂ g + ∂ g − ∂ g ) ij kl 2 i jk j ik k ij 1 ⇒ Γl = gkl(∂ g + ∂ g − ∂ g ). ij 2 i jk j ik k ij

1 (c) A 2d Riemannian manifold has metric

ds2 = dr2 + f(r)2dφ2 where φ is periodically identiﬁed with period 2π (so curves of constant r are circles).

(i) Determine the necessary and sufﬁcient conditions on f(r) for the circle r = r0 to have the property that all vectors are invariant under parallel transport around the circle.

We want that for any vector ﬁeld X,

a b µ ν µ ν ρ ∇V X = 0, or, V ∇aX = 0 ⇒ V ∂µX + V ΓµρX = 0,

∂ for some V indicating “around the circle”. Since ∂φ is tangent to the circle, we take ∂ V = . ∂φ Hence our equation reduces to ν ν ρ ∂φX + ΓφρX = 0. To ﬁnd the Γ’s, we start with the Lagrangian

L =r ˙2 + f(r)2φ˙2.

The equations of motion are

2f 0 f 0 f 2φ¨ + 2ff 0r˙φ˙ = 0 ⇒ φ¨ + r˙φ˙ = 0 ⇒ Γφ = Γφ = ; f rφ φr f

0 ˙2 r 0 r¨ − ff φ = 0 ⇒ Γφφ = −ff . φ φ r ⇒ ν = φ : ∂φX + ΓφrX = 0; r r φ ⇒ ν = r : ∂φX + ΓφφX = 0, which give f 0 ∂ Xφ + Xr = 0; ∂ Xr − ff 0Xφ = 0. φ f φ

r f 2 φ The ﬁrst gives ∂φX = − f 0 ∂φX . Substituting it into the second, we have

2 φ 02 φ ∂φX + f X = 0.

φ 1 2 r And the second gives ∂φX = ff 0 ∂φX . Substituting it into the ﬁrst, we have

2 r 02 r ∂φX + f X = 0.

Hence Xφ(φ) = Xφ(0) cos(f 0φ) + X0φ(0) sin(f 0φ); Xr(φ) = Xr(0) cos(f 0φ) + X0r(0) sin(f 0φ). So we require Xφ(0) = Xφ(2π) and Xr(0) = Xr(2π). This is true iﬀ

0 f = n, n ∈ Z.

2 (ii) Deduce that there is a 2-parameter family of functions f(r) for which all circles of constant r have this property.

From (i), we have f(r) = nr + c, c = const.

(iii) For this 2-parameter family, show that the metric is locally isometric to the Euclidean metric. Is it globally isometric?

Locally isometric means we can ﬁnd coordinates such that the metric looks like

ds2 = dr˜2 +r ˜2dφ˜2, r˜ =r ˜(r), φ˜ = φ˜(φ).

Indeed we have

ds2 = dr2 + (nr + c)2dφ2 = dr˜2 +r ˜2dφ˜2

where we have setr ˜ = r + c/n and φ˜ = nφ. Hence we have a local isometry. We have a global isometry only when n = 1, since now φ˜ is periodically identiﬁed with period 2πn.

Let Ta be tangent to a 1-parameter family of timelike geodesics of the Levi-Civita connection, parame- terized by proper time. Let Sa be a deviation vector for this family.

(a) Explain the term deviation vector. Explain why [S, T ] = 0.

A 1-parameter family of geodesics is a map γ : I × I0 → M where I and I0 both are open intervals in R, such that for ﬁxed s, γ(s, t) is a geodesic with aﬃne parameter t, and s is the parameter that labels the geodesic. Further, the map (s, t) → γ(s, t) is smooth and one-to-one with a smooth inverse.

A deviation vector Sa is the displacement of two objects travelling along two such neighbouring geodesics such that xa(s + δs, t) ' xa(s, t) + δsSa(s, t)

a a a ∂x (s,t) a ∂x (s,t) where S (s, t) = ∂s . Note that T (s, t) = ∂t .

The family of geodesics forms a two-dimensional surface Σ ⊂ M. Treating s and t as coordinates, we can extend them to (s, t, u, ...) deﬁned in a neighbourhood of Σ. This gives a coordinate chart on Σ where S = ∂/∂s and T = ∂/∂t are vector ﬁelds satisfying

[S, T ] = 0.

3 (b) State and prove the geodesic deviation equation. You may assume the definition of the Riemann curvature tensor R(X,Y )Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ]Z.

The geodesic deviation equation is ∇T ∇T S = R(T,S)T if ∇ has vanishing torsion.

Proof: Vanishing torsion implies ∇X Y − ∇Y X = [X,Y ]; we have from above [S, T ] = 0; and we have ∇T T = 0 since T is tangent to aﬃnely parametrised geodesics. Hence

R(T,S)T = ∇T ∇ST − ∇S∇T T − ∇[T,S]T = ∇T ∇T S.

(c) Prove that if Sa and T a are orthogonal at one point along a geodesic γ (belonging to the 1-parameter family) then they are orthogonal everywhere along γ.

1 1 ∇ (T · S) = (∇ T ) · S + T · (∇ S) = T ∇ S = T ∇ T = ∇ (T · T ) = ∇ (-1) = 0 T T T T S 2 S 2 S since T is timelike. Hence there is no change in the value of T · S along the geodesic.

(d) Suppose that spacetime is four-dimensional with Riemann tensor

1 R = R(g g − g g ) abcd 12 ac bd ad bc Show that R must be constant.

We ﬁrst contract the given Riemann tensor to ﬁnd 1 1 R = R(4g − g ) = Rg . bd 12 bd bd 4 bd Using this, we have the Einstein tensor 1 1 G = R − Rg = − Rg . ab ab 2 ab 4 ab a But ∇ Gab = 0. Hence ∇aR = 0, i.e., R is constant.

a a a (e) Assume that S is orthogonal to T . Let f = SaS . Show that, in the above spacetime, 1 K = (∇ S) (∇ S)a − Rf T a T 12 is constant along a geodesic γ in the family.

4 1 1 ∇ K = ∇ (∇ S · ∇ S) − (∇ R)S · S − R∇ (S · S) T T T T 12 T 12 T 1 = 2∇ ∇ S · ∇ S − RS · ∇ S (since R here is constant) T T T 6 T 1 = 2R(T,S)T · ∇ S − RS · ∇ S T 6 T 1 = 2R T bT cSdT e∇ Sa − RSaT e∇ S abcd e 6 e a 1 1 = R(g g − g g )T bT cSdT e∇ Sa − RSaT e∇ S 6 ac bd ad bc e 6 e a 1 1 1 = RT T SdT e∇ Sa − RT T cS T e∇ Sa − RS T e∇ Sa 6 d a e 6 c a e 6 a e 1 1 1 = R(T Sd)T T e∇ Sa − R(T T c)S T e∇ Sa − RS T e∇ Sa 6 d a e 6 c a e 6 a e 1 1 = RS T e∇ Sa − RS T e∇ Sa (since T Sd = 0 by orthogonality and T is timelike) 6 a e 6 a e d = 0.

(f) Obtain a second order differential equation for the evolution of f along γ. Hence deduce that if R > 0 then geodesics which are close initially will diverge exponentially. What happens if R < 0?

∇T ∇T f = ∇T ∇T (S · S)

= 2∇T (S · ∇T S)

= 2(∇T S) · (∇T S) + 2S · ∇T ∇T S 1 = 2(K + Rf) + 2S · R(T,S)T 12 1 1 = 2K + Rf + 2Sa R(g g − g g )T bT cSd 6 12 ac bd ad bc 1 1 1 = 2K + Rf + RSTT cSd − RS T T cSd 6 6 c d 6 d c 1 = 2K + Rf (since SdT = 0). 3 d So the diﬀerential equation for the evolution of f is

d2f 1 − Rf − 2K = 0, dt2 3 from which we can see that if R > 0, the late-time solution is dominated by an exponential growth ∼ exp(pR/3 t), but if R < 0 there are oscillations.

In the study of linearised perturbations of Minkowski spacetime, it is assumed that there exist global µ coordinates x with respect to which the metric has components gµν = ηµν + hµν where the components of hµν have absolute values much smaller than 1.

ρ ¯ 1 µ¯ (a) Let h = h ρ and hµν = hµν − 2 hηµν . By imposing the gauge condition ∂ hµν = 0, derive the linearised Einstein equation in the form ρ ¯ ∂ ∂ρhµν = −16πTµν

5 You may use µ µ µ τ µ τ µ R νρσ = ∂ρΓνσ − ∂σΓνρ + ΓνσΓτρ − ΓνρΓτσ, valid in a coordinate basis.

To first order, the Christoffel symbols are 1 Γµ = ηµσ(∂ h + ∂ h − ∂ h ) νρ 2 ν σρ ρ σν σ νρ And to first order, the Riemann tensors are

µ µ µ R νρσ = ∂ρΓνσ − ∂σΓνρ 1 = ηµα(∂ ∂ h + ∂ ∂ h − ∂ ∂ h − ∂ ∂ h − ∂ ∂ h + ∂ ∂ h ) 2 ν ρ ασ σ ρ αν α ρ νσ ν σ αρ ρ σ αν α σ νρ

1 ⇒ Rρ = (∂ ∂ hρ + ∂ ∂ hρ − ∂ρ∂ h − ∂ ∂ hρ − ∂ ∂ hρ + ∂ρ∂ h ) νρσ 2 ν ρ σ σ ρ ν ρ νσ ν σ ρ ρ σ ν σ νρ 1 1 = ∂ρ∂ h − ∂ρ∂ h − ∂ ∂ h (ν σ)ρ 2 ρ νσ 2 ν σ = Rνσ

Now, 1 1 1 1 R − g R = ∂ρ∂ h − ∂ρ∂ h − ∂ ∂ h − η (∂ρ∂σh − ∂ρ∂ h) µν 2 µν (µ ν)ρ 2 ρ µν 2 µ ν 2 µν σρ ρ 1 1 1 1 1 1 = ∂ρ∂ (h¯ − hη¯ ) + ∂ρ∂ (h¯ − hη¯ ) − ∂ρ∂ (h¯ − hη¯ ) 2 µ νρ 2 νρ 2 ν µρ 2 µρ 2 ρ µν 2 µν 1 1 1 + ∂ ∂ h¯ − η (∂ρ∂σ(h¯ − hη¯ ) + ∂ρ∂ h¯) 2 µ ν 2 µν σρ 2 σρ ρ X1 1 1 ¨ ρ ¯ Xρ X ¯ ρ ¯ ρ ¨¯¨ = ∂ ∂(µhν)ρ − ∂ ∂(µXhηXνX)ρ − ∂ ∂ρhµν + ∂¨∂ρhηµν 2 2 ¨4 H ¨ ¨ 1 H ¯ 1 ρ σ¯ 1 ρ¨ ¯ 1 ρ¨ ¯ + ∂µ∂Hν h − ηµν ∂ ∂ hσρ + ηµν¨∂ ∂ρh − ηµν¨∂ ∂ρh 2 H 2 ¨4 ¨ ¨2 ¨ ¯ 1 ¯ 1 ¯ ¯ where hµν = hµν − 2 hηµν gives hµν = hµν − 2 hηµν and h = −h. Hence from 1 R − g R = 8πT , µν 2 µν µν we get the linearised Einstein equation 1 1 ∂ρ∂ h¯ − ∂ρ∂ h¯ − η ∂ρ∂σh¯ = 8πT . (µ ν)ρ 2 ρ µν 2 µν σρ µν µ¯ Imposing the gauge condition ∂ hµν = 0, the ﬁrst and the third terms disappear and we arrive at

ρ ¯ ∂ ∂ρhµν = −16πTµν .

(b) Consider a vacuum plane gravitational wave solution with

ρ ¯ ikρx hµν = Re(Hµν e ) where Hµν is a constant complex matrix and kρ a constant covector. What restrictions must Hµν and kρ obey in the above gauge?

6 µ¯ In the gauge ∂ hµν = 0, µ k Hµν = 0 i.e., the waves are transverse.

Explain why there is a residual gauge freedom hµν → hµν + 2∂(µξν) provided ξµ satisﬁes a certain condition. Show that this condition is satisﬁed by

ρ ikρx ξµ = Re(Xµe ) where Xµ is constant.

A manifold M with metric g and energy-momentum tensor T is physically equivalent to M with φ∗(g) and energy momentum tensor φ∗(T ) if φ is a diﬀeomorphism. Hence

(φ−t)∗(g) = g + Lξg + ··· = η + h + Lξη + ··· is equivalent to g = η + h, i.e., h and h+Lξη describe physically equivalent metric perturbations. Since (Lξη)µν = ∂µξν +∂ν ξµ = 2∂(µξν) (think about the Killing equation), there is a residual gauge freedom

hµν → hµν + 2∂(µξν).

µ¯ The gauge condition ∂ hµν = 0 needs to be satisﬁed. Note ﬁrst that

ρ h → h + 2∂ ξρ.

1 ⇒ h¯ = h − hη µν µν 2 µν 1 1 → h + 2∂ ξ − hη − η (2∂ρξ ) µν (µ ν) 2 µν 2 µν ρ ¯ ρ = hµν + 2∂(µξν) − ηµν ∂ ξρ. Hence we have µ¯ µ¯ µ µ µ ρ ∂ hµν → ∂ hµν − ∂ ∂µξν −∂ ∂ν ξµ +∂ ∂ ξρηµν µ¯ µ = ∂ hµν − ∂ ∂µξν µ = −∂ ∂µξν (by gauge condition) =! 0.

This condition is satisﬁed by ρ ikρx ξµ = Re(Xµe ) ρ ¯ ρ ¯ since vacuum ⇒ ∂ ∂ρhµν = 0 ⇒ k kρ = 0 with the given solution of hµν .

Now show that Xµ can be chosen to bring Hµν to the form

 0 0 0 0   0 H+ H× 0  Hµν =   .  0 H× −H+ 0  0 0 0 0

Use your results to justify the statements that gravitational waves travel at the speed of light, are transverse, and have 2 independent polarizations.

7 µ ν ¯ First consider the null vector k = ω(1, 0, 0, 1). Then from the transverse gauge condition ∂ hµν = 0, we have here ν 0 3 ik Hµν = ik Hµ0 + ik Hµ3 = 0 ⇒ Hµ0 + Hµ3 = 0. Now, recall that we had ¯ ¯ ρ hµν → hµν + 2∂(µξν) − ηµν ∂ ξρ. ρ ⇒ Hµν → Hµν + i(kµXν + kν Xµ − ηµν k Xρ).

To achieve the longitudinal gauge H0ν = 0, set

ρ k0Xν + kν X0 − η0ν k Xρ := iH0ν , i.e., k0X0 + k1X1 + k2X2 + k3X3 = iH00;

k0X1 + k1X0 = iH01;

k0X2 + k2X0 = iH02;

k0X3 + k3X0 = iH03. Or,       k0 k1 k2 k3 X0 H00  k1 k0 0 0   X1   H10      = i    k2 0 k0 0   X2   H20  k3 0 0 k0 X3 H30

After row reduction, the last equation is 0 on both sides. Hence there exists multiple solutions for Xµ. We µ can impose the traceless condition H µ = 0 to ﬁx the gauge. Combining the results of the transverse and the longitudinal gauge conditions gives H3ν = 0. Hence,

 0 0 0 0   0 H+ H× 0  Hµν =   .  0 H× −H+ 0  0 0 0 0

ρ ¯ ρ Gravitational wave travels at the speed of light since ∂ ∂ρhµν = 0 gives k kρ = 0. It is transverse since, µ¯ µ as mentioned earlier, ∂ hµν = 0 gives k Hµν = 0. And it has two independent polarisations as represented by H× and H+.

(c) (i) Explain briefly why it is not possible to define an energy-momentum tensor for the gravitational field in General Relativity.

The energy in the gravitational ﬁeld should be quadratic in the ﬁrst derivatives of the metric, which in turn can always be set to zero at any point by a change of coordinates.

(ii) Explain how to deﬁne a symmetric tensor tµν that is quadratic in the linearized gravitational ﬁeld µ hµν and conserved, i.e., ∂ tµν = 0.

The Einstein tensor to second order is

(1) (1) (2) (2) Gµν [g] = Gµν [h] + Gµν [h ] + Gµν [h]

Assume that no matter is present, i.e., Gµν [g] = 0 and at ﬁrst order, the linearised Einstein equation is G(1)[h] = 0. At second order, (1) (2) Gµν [h ] = 8πtµν [h]

8 where 1 t [h] := − G(2)[h] µν 8π µν µ Consider the contracted Bianchi identity ∇ Gµν = 0. At ﬁrst order,

µ (1) ∂ Gµν [h] = 0

(2) µ (1) (2) for arbitrary ﬁrst order perturbation h, i.e., if we replace h with h then ∂ Gµν [h ] = 0. At second order, X ¨ µXX(1) (2) µ (2) (1)¨¨ ∂ Gµν X[hXX] + ∂ Gµν [h] + h¨Gµν [h] = 0 where the third term vanished since we assumed that the equation of motion G(1)[h] = 0 is obeyed.

µ ⇒ ∂ tµν = 0

Hence tµν is a symmetric tensor that is (i) quadratic in the linear perturbation h, (ii) conserved if h satisﬁes its equation of motion, and (iii) appears on the RHS of the second order Einstein equation.

(iii) Why is tµν unsatisfactory as a definition of an energy-momentum tensor for hµν ?

tµν is unsatisfactory as it is not invariant under the gauge transformation hµν → hµν + 2∂(µξν). This is how the impossibility of localising gravitational energy arises in linearised theory.

(a) (i) What is a Killing vector ﬁeld?

A vector ﬁeld X is a Killing vector ﬁeld if

LX g = 0,

i.e., ∇aXb + ∇bXa = 0

(ii) Show that if a spacetime admits a Killing vector ﬁeld then along any geodesic there is a conserved quantity.

a a a Let X be a Killing vector field and let V be tangent to an affinely parametrised geodesic. Then XaV is constant along the geodesic. Proof: a b a b a b a ∇V (XaV ) = V ∇b(XaV ) = XaV ∇bV + V V ∇bXa = 0 b a b where V ∇bV = 0 as V is tangent to an affinely parametrised geodesic (it is the geodesic equation!), and b a the second term vanishes as V V is symmetric but ∇bXa is antisymmetric by the Killing equation.

(iii) Write down 3 linearly independent Killing vector ﬁelds of the metric

ds2 = A(z)2(−dt2 + dx2 + dy2) + dz2

where A(z) is a positive function.

∂ Let T := ∂t . We claim that T is a Killing vector. Proof: ρ ρ ρ (LX g)µν = X ∂ρgµν + gρν ∂µX + gρµ∂ν X = 0

9 where the later two terms vanish since Xt = 1 while other Xρ = 0, and the ﬁrst term vanishes since z ∂ρ6=zgµν = 0 as the entries of gµν depend only on z, and for ρ = z, this term vanishes still since X = 0. ∂ ∂ Similarly, X := ∂x and Y := ∂y are Killing vectors.

Note that if a Killing vector ﬁeld X generates isometries, then LX g = 0. And since

L[X,Y ] = LX LY − LY LX = 0, we can find a third Killing vector field by taking the commutator of the first two, given that X and Y are independent Killing vector fields, i.e., [X,Y ] 6= 0.