General Relativity 2013 – Solutions

Yichen Shi Easter 2014

(a) Let T be a tensor field of type (r, s) and X a vector field. Define the Lie derivative LX T .

The Lie derivative of T with respect to X at p is

((φ−t)∗(T ))p − Tp (LX T )p = lim , t→0 t

where φt : M → M is a diﬀeomorphism, t is a distance parameter, and p ∈ M. It is a map from (r, s) tensors to (r, s) tensors.

(b) Prove that the Lie derivative commutes with contraction and satisﬁes the Leibnitz rule

LX (S ⊗ T ) = (LX S) ⊗ T + S ⊗ (LX T )

where S is a tensor ﬁeld of type (p, q). [You may assume the existence of coordinates (t, x1, x2...) such that X = ∂/∂t].

Note that for a (1, 1) tensor ﬁeld,

µ σ µ ∂y ∂x ρ [(φ∗(T )) ν ]φ(p) = ρ ν (T σ)p ∂x p ∂y p

µ i Since the diﬀeomorphism φ−t sends p with coordinates x = (tp, xp) to the point φ−t(p) with coordinates µ i µ ν µ y = (tp − t, xp), we have ∂y /∂x = δν . So generalising the (1, 1) tensor ﬁeld, we have

[((φ ) (T ))µ1...µr ] = [T µ1...µr ] −t ∗ ν1...νs φ−t(p) ν1...νs p

⇒ [((φ ) (T ))µ1...µr ] = [T µ1...µr ] . −t ∗ ν1...νs p ν1...νs φt(p) i i φt of course, sends p with coordinates (tp, xp) to φt(p) with coordinates (tp + t, xp). Hence if p has coordinates (s, xi) in this chart, then

T µ1...µr (s + t, xi) − T µ1...µr (s, xi) µ1...µr ν1...νs ν1...νs (LX T ) = lim ν1...νs t→0 t ∂ µ1...µr i = T ν1...νs (t, x ) . ∂t (s,xi)

So in this chart, the Lie derivative is simply the partial derivative, which commutes with contraction and satisﬁes the Leibnitz rule.

1 In the chart deﬁned above, ∂ L f = f, X ∂t but we also have ∂ X(f) = f. ∂t

Both sides of this expression are scalars hence this equation is valid in any basis. So LX f = X(f).

Again in the chart deﬁned above, ∂Y µ (L Y )µ = , X ∂t but we also have ∂Y µ [X,Y ]µ = . ∂t

If two vectors have the same components in one basis then they are equal in all bases. So LX Y = [X,Y ].

(d) Let ω be a covector ﬁeld. Prove that, in any coordinate basis,

ν ν (LX ω)µ = X ∂ν ωµ + ων ∂µX .

We have LX (ω(Y )) = (LX ω)Y + ωLX Y. where in coordinates,

µ ν µ ν µ ν LX (ω(Y )) = X ∂µ(ω Yν ) = X (∂µω )Yν + X ω ∂µYν ;

µ (LX ω)Y = (LX ω)µY ; µ ν ν ωLX Y = ω (X ∂ν Yµ − Y ∂ν Xµ).

µ µ ν µ ν µ ν µ ν ⇒ (LX ω)µY = X (∂µω )Yν +Xω ∂µYν −ω X ∂ν Yµ + ω Y ∂ν Xµ ν µ ν µ = X (∂ν ω )Yµ + ω (∂µXν )Y

Since Y is arbitrary, we arrive at

ν ν (LX ω)µ = X ∂ν ωµ + ων ∂µX .

Note though, that

ν ν ν ν ν ρ νρ X ∇ν ωµ + ων ∂µX = X ∂ν ωµ + ων ∂µX −XΓνµωρ +ωνΓµρX . Hence ν ν (LX ω)µ = X ∇ν ωµ + ων ∂µX .

(e) Let T be tensor ﬁeld of type (1, 1). Derive an expression for the components of LX T in an arbitrary coordinate basis. Hence or otherwise prove that

LX LY T − LY LX T = L[X,Y ]T.

We have LX (T (ω, Y )) = (LX T )(ω, Y ) + T (LX ω, Y ) + T (ω, LX Y ). µ ν ρ ν ρ ν ρ ν ρ ⇒ X ∂µ(T ρων Y ) = (LX T ) ρων Y + T ρ(LX ω)ν Y + T ρων (LX Y ) .

2 ν ρ (LX T ) ρων Y ( µ µ ρ µ ν ((((ρ ν µ µ ρ ν µ ρ µ ρ = X ∂µ(T ρ)ων Y +(X(T(ρ∂µ(ων Y ) − T ρ(X∂µων + ωµ∂ν X )Y − T ρων (X∂µY − Y ∂µX ) µ µ ρ µ ν ρ ν µ ρ = X ∂µ(T ρ)ων Y − T ρ∂µ(X )ων Y + T µ∂ρ(X )ων Y .

ν µ ν µ ν ν µ ⇒ (LX T ) ρ = X ∂µT ρ − T ρ∂µX + T µ∂ρX . But

ρ µ µ ρ ρ µ X ∇ρT ν − (∇ρX )T ν + (∇ν X )T ρ X µ ν µ ν ν µ ρ µ σ XXρ Xσ µ µ σρ ρXXσ µ = X ∂µT ρ − T ρ∂µX + T µ∂ρX +XΓρσT ν − X ΓρνXT Xσ −ΓρσX T ν + ΓνσX XT Xρ. Since we have a tensorial expression, we can write

a c a a c c a (LX T ) b = X ∇cT b − (∇cX )T b + (∇bX )T c valid in arbitrary basis.

Now, for a vector ﬁeld Z, by the Jacobi identity,

L[X,Y ]Z = [[X,Y ],Z] = [X, [Y,Z]] − [Y, [X,Z]]

= LX [Y,Z] − LY [X,Z]

= LX LY Z − LY LX Z.

And for a function f,

L[X,Y ]f = X(Y (f)) − Y (X(f))

= X(LY f) − Y (LX f)

= LX (LY f) − LY (LX f).

So for a covector ﬁeld ω, using arbitrary vector ﬁelds Z, by the Leibniz rule,

LX (LY (ω(Z))) − LY (LX (ω(Z))) = LX ((LY ω)Z) + LX (ω(LY Z)) − LY ((LX ω)Z) − LY (ω(LX Z)) ((( hh = (LX LY ω)Z + (L(Y ω()LX Z + (LX ωh)LhYhhZ + ωLX LY Z hh ((( − (LY LX ω)Z − (LX ωh)LhYhhZ − (L(Y ω()LX Z − ωLY LX Z

= LX (LY (ω(Z))) − LY (LX (ω(Z)))

= L[X,Y ](ω(Z)).

Now, recall that all tensors can be represented as a sum of exterior products of vector ﬁelds and covector ﬁelds. Hence our formula applies to tensors too, and in particular, to our (1,1) tensor T .

In the linearised approximation to General Relativity it is assumed that the spacetime manifold is R4 µ i and there exist ‘almost inertial’ coordinates x = (t, x ) such that the metric takes the form gµν = ηµν + hµν ¯ 1 ρ ν ¯ where |hµν | 1. Let hµν = hµν − 2 hηµν where h = h ρ. Then, in the gauge ∂ hµν = 0, the linearised Ein- ρ ¯ stein equation is ∂ ∂ρhµν = −16πTµν .

(a) Assume that matter is conﬁned within a ball of radius d centred on the origin. Assume that the matter moves non-relativistically. Show that, for r = |x| d

2 h¯ (t, x) ≈ I¨ (t − r) ij r ij

3 where Z 3 Iij(t) = d x T00(t, x)xixj.

ρ ¯ We solve ∂ ∂ρhµν = −16πTµν using the retarded Green function as Z T (t − |x − x0|, x0) h¯ = 4 d3x0 µν . µν |x − x0| Since r = |x| >> |x0| ∼ d, 4 Z h¯ (t, x) = d3x0 T (t − r, x0), ij r ij where Z Z 3 ij 3 ik j ik j d x T = d x [∂k(T x ) − (∂kT )x ] Z 3 ik j = − d x (∂kT )x (drop surface terms as the matter is compactly supported) Z 3 i0 j µν = d x (∂0T )x (by conservation of energy-momentum ∂ν T = 0) Z 3 i0 j = ∂0 d x T x Z 1 1 = ∂ d3x ∂ (T k0xixj) − (∂ T k0)xixj 0 2 k 2 k 1 Z = − ∂ d3x (∂ T k0)xixj (drop surface terms) 2 0 k 1 Z = ∂ d3x (∂ T 00)xixj (by conservation) 2 0 0 1 Z = ∂ ∂ d3x T 00xixj 2 0 0 1 := I¨ 2 ij where Z 3 i j Iij(t) = d x T00(t, x)x x .

Hence 2 h¯ (t, x) = I¨ (t − r). ij r ij

(b) Consider a binary system consisting of two stars, each of mass M, in a Newtonian circular orbit of radius R. Assuming that the quadrupole formula applies to this system, calculate the average power emitted in gravitational radiation. You may approximate the energy-momentum tensor of a star at rest at the origin 3 as T00 = Mδ (x), T0i = Tij = 0.

Take the positions of the stars to be ±(R cos Ωt, R sin Ωt, 0) where the radius of the circular orbit is R, and :=1 r GM 2 M = MΩ2R ⇒ Ω = . 4R2 4R3 Now,

3 ⇒ T00 = Mδ (x − x0) = Mδ(x − R cos Ωt)δ(y − R sin Ωt)δ(z) + Mδ(x + R cos Ωt)δ(y + R sin Ωt)δ(z).

4 The second moment of energy distribution is Z 3 i j Iij(t − r) = d x T00(t − r, x)x x .

2 2 ⇒ I11(t − r) = 2MR cos Ω(t − r); 2 2 ⇒ I22(t − r) = 2MR sin Ω(t − r); 2 ⇒ I12(t − r) = 2MR cos Ω(t − r) sin Ω(t − r) = I21(t − r);

⇒ I3i(t − r) = 0 = Ii3(t − r).  cos2 Ωt cos Ωt sin Ωt 0 2 2 ⇒ Iij(t) = 2MR cos Ωt sin Ωt sin Ωt 0 . 0 0 0  sin 2Ωt − cos 2Ωt 0 ˙˙˙ 2 3 ⇒ I ij(t) = 8MR Ω − cos 2Ωt − sin 2Ωt 0 . 0 0 0 The quadrupole formula for energy loss via gravitational wave emission is 1 ˙˙˙ ˙˙˙ 1˙˙˙ ˙˙˙ hP it = I ij I ij − I iiI jj . 5 3 t−r

The trace of ˙˙˙I ij vanishes here. Hence,

1 D˙˙˙ ˙˙˙ E hP it = I ij I ij 5 t−r *1 0 0+ 1 = (8MR2Ω3)2 0 1 0 5   0 0 0 2 M 3 = (8MR2)2 5 43R9 2 M 5 = . 5 R

(c) Explain why the binary systems which emit the most gravitational radiation are tightly bound and involve neutron stars or black holes.

From the above calculation, the smaller the radius and the larger the mass of the system, the more energy would be emitted via gravitational wave. A star cannot be smaller than the Schwarzschild radius 2M (the radius of a black hole of mass M), so the separation d between the two stars cannot be smaller than 4M. An ordinary star has size much larger than M, hence d M for a binary system of such stars. But a neutron star or black hole has size comparable to 2M so M/d can be the largest for binaries involving such compact objects. Hence the statement.

(a) Consider an inﬁnitesimal variation of the spacetime metric gab → gab + δgab. (i) Show that the corresponding change in the Levi-Civita connection is given by

1 δΓa = gad(∇ δg + ∇ δg − ∇ δg ) bc 2 c db b dc d bc where ∇ is the Levi-Civita connection associated to gab.

5 We have 1 Γµ = gµσ(∂ g + ∂ g − ∂ g ). νρ 2 ρ σν ν σρ σ νρ If we introduce normal coordinates at p for the unperturbed connection, we have ∂ρgµν (p) = 0 and µ Γνρ(p) = 0 (which allows ∂ ↔ ∇). Hence for the perturbed connection, at p,

µ 1 µσ 1 µσ δΓ = g (∂ρδgσν + ∂ν δgσρ − ∂σδgνρ) + δg (∂ρgσν+ ···) νρ 2 2 1 = gµσ(∇ δg + ∇ δg − ∇ δg ). 2 ρ σν ν σρ σ νρ µ Now, RHS is a tensorial expression obviously. LHS δΓνρ is the diﬀerence between the components of two a connections, hence are the components of a tensor δΓbc. Since both sides are tensors, the expression is basis independent and we can use abstract indices, giving 1 δΓa = gad(∇ δg + ∇ δg − ∇ δg ). bc 2 c db b dc d bc

(ii) Show that the change in the Ricci tensor is

c c δRab = ∇cδΓab − ∇bδΓac given µ µ µ τ µ τ µ R νρσ = ∂ρΓνσ − ∂σΓνρ + ΓνσΓτρ − ΓνρΓτσ.

Again using normal coordinates, δ(ΓΓ)|p = 2ΓδΓ|p = 0.

µ µ µ µ ⇒ δRνσ = ∂µδΓνσ − ∂σδΓνµ = ∇µδΓνσ − ∇σδΓνµ. Again since this is a tensorial expression, it is basis independent and we can use abstract indices, giving

c c δRab = ∇cδΓab − ∇bδΓac.

(iii) Show that the change in the Ricci scalar is

ab a b c ab δR = −R δgab + ∇ ∇ δgab − ∇ ∇c(g δgab)

First note that ab ab ab δ(g gbc) = 0 = g δ(gbc) + δ(g )gbc. ab bc ad ac bd ⇒ δ(g ) = −g g δ(gdc) = −g g δ(gcd). Hence we have

ab ab ab δR = δ(g Rab) = g δRab + δ(g )Rab ab c c ac bd = g (∇cδΓab − ∇bδΓac) − g g δ(gcd)Rab 1 1 = −Rabδg + gab ∇ gcd(∇ δg + ∇ δg − ∇ δg ) − ∇ gcd(∇ δg + ∇ δg − ∇ δg ) ab c 2 b da a db d ab b 2 c da a dc d ac 1 1 = −Rabδg + ∇ (∇δgbc + ∇ δgac − gabgcd∇ δg ) − ∇ (∇δgbc + gabgcd∇ δg − ∇ δgbd) ab 2 c b a d ab 2 b c a cd d 1 = −Rabδg + (∇ ∇ δgac − ∇c∇ (gabδg ) − ∇a∇ (gcdδg ) − ∇ ∇ δgbd) ab 2 a c c ab a cd b d 1 = −Rabδg + (∇a∇cδg − ∇b∇dδg − ∇c∇ (gabδg ) − ∇a∇ (gcdδg )) ab 2 ac bd c ab a cd ab a b c ab = −R δgab + ∇ ∇ δgab − ∇ ∇c(g δgab)

6 (b) A theory of gravity has action Z 4 √ S = d x −gf(R) + Smatter where f is a smooth function.

(i) Show that varying gab gives the equation of motion Eab = (1/2)Tab where Tab is the energy-momentum tensor of matter and

1 E = f 0R − f 00∇ ∇ R − f 000∇ R∇ R + − f + f 00∇c∇ R + f 000∇cR∇ R g ab ab a b a b 2 c c ab

0 0 ab where f denotes f (R), etc. [You may use without proof δg = gg δgab.]

First note that 1 1 δS δS 1√ T ab = √ matter ⇒ matter = −gT ab. 2 −g δgab δgab 2 Now, we have

δSnon-matter δgab δgab Z √ 4 δ −g √ df(R) δR δSmatter = d x f(R) + −g + δgab δgab dR δgab δgab Z √ 4 1 1 δg √ 0 δR −g ab = d x − √ f(R) + −gf (R) + T δgab 2 −g δgab δgab 2 Z c d c de 4 1√ ab √ 0 ab ∇ ∇ δgcd ∇ ∇c(g δgde) = d x − −gg f(R) + −gf (R) −R + − δgab 2 δgab δgab Z √ 1 δg = d4x −g − gabf(R) − Rabf 0(R) + f 00(R)∇c∇dR + f 000(R)∇cR∇dR cd 2 δgab de 00 c 000 c g δgde − f (R)∇ ∇cR + f (R)∇ R∇cR δgab δgab Z √ 1 = d4x −g − g f − R f 0 + f 00∇ ∇ R + f 000∇ R∇ R − f 00∇c∇ Rg − f 000∇cR∇ Rg δgab. 2 ab ab a b a b c ab c ab

We require δS = δSmatter + δSnon-matter = 0. 1 ⇒ E = T . ab 2 ab

a (ii) Explain why ∇ Eab = 0 for any metric gab. [Hint: This does not require a long calculation.]

From part (i), a a ∇ Eab = 0 ⇔ ∇ Tab = 0. The latter is obviously true by energy-momentum conservation.

(iii) Why does Lovelock’s theorem not apply to this theory?

Lovelock’s theorem says: Let Hab be a symmetric tensor such that 1. in any coordinate chart, at any point, Hµν is a function of gµν , gµν,ρ, gµν,ρσ; a 2. ∇ Hab = 0; 3. either space-time is four-dimensional or Hµν depends linearly on gµν,ρσ.

7 Then there exist constants α, β such that

Hab = αGab + βgab.

00 In any coordinate chart, if f 6= 0 then terms such as ∇a∇bR in Eµν contain functions of gµν,ρστ , which violates condition 1. Hence the theorem doesn’t apply here.

(iv) Determine the necessary and sufﬁcient condition on the function f that ensures that any solution of the vacuum Einstein equation with cosmological constant Λ is also a vacuum solution of this theory.

In vacuum, Tab = 0. So the Einstein equation reads 1 R − Rg + Λg = 0. ab 2 ab ab Taking the trace in four dimensions,

R − 2R + 4Λ = 0 ⇒ R = 4Λ.

Hence Rab = cgab, c = const., 1 and its covariant derivatives vanish. Here we have Eab = 2 Tab = 0 in vacuum, so if f satisﬁes 1 E = f 0cg − fg = 0, ab ab 2 ab then any solution of the vacuum Einstein equation is also a vacuum solution of this theory.