General Relativity Volker Schlue - sorbonne-universiteschlue/grt.pdfof Special Relativity according...

General Relativity

Volker Schlue

Department of Mathematics, University of Toronto, 40 St GeorgeStreet, Toronto, Ontario M5S 2E4, Canada

E-mail address: [email protected]

Preface

I have prepared this manuscript for a course on general relativity that I taughtat the University of Toronto in the winter semester 2013. It is based on notes that Itook as a student at ETH Zurich in a lecture course that Demetrios Christodoulougave in the fall of 2005. (Chapters that are in significant parts transcribed fromthese notes will be marked by a ∗.) His teaching and understanding of generalrelativity made a great impression on me as a student, and I hope that theselecture notes can contribute to a wider circulation of his approach.

I am very grateful to Andre Lisibach, who provided me with most of the exer-cises, as well as complementary notes from a similar course that Prof. Christodoulougave last fall, which were useful for the preparation of the present manuscript.

Toronto, January 2014V. Schlue

iii

Contents

Preface iii

Chapter 1. The equivalence principleand its consequences ∗ 1

1. Classical theory of gravitation 22. Special Relativity 53. Geodesic correspondence 20

Chapter 2. Einstein’s field equations ∗ 471. Einstein’s field equations in the presence of matter 502. Action Principle 663. The material manifold 744. Cosmological constant 79

Chapter 3. Spherical Symmetry 811. Einstein’s field equations in spherical symmetry 812. Schwarzschild solution 863. General properties of the area radius and mass functions 904. Spherically symmetric spacetimes with a trapped surface 92

Chapter 4. Dynamical Formulation of the Einstein Equations ∗ 971. Decomposition relative to the level sets of a time function 972. Slow Motion Approximation 1153. Gravitational Radiation 125

Bibliography 149

v

CHAPTER 1

The equivalence principleand its consequences ∗

There are two motivations for the General Theory of Relativity.

(1) Extension of the principle of Special Relativity (invariance of the physicallaws under change from one inertial system to another — one such systemrelative to another is in a state of uniform motion) to all systems ofreference in any state of motion whatsoever.

(2) Establish a theory of gravitation which is in accordance with SpecialRelativity.

Einstein’s Equivalence Principle relates the two motivations.

Remark. (1) is sometimes referred to as the requirement that the laws ofphysics should be valid in any system of spacetime coordinates. (General Covari-ance.)

In fact, however, (1) refers to the requirement of invariance of the physical lawsunder change of physical description, from one relative to one set of observers toanother relative to another set of such observers.

The history of an observer is represented in relativity by a timelike curve. Afamily of observers is given by a family of timelike curves that do not intersecteach other. Each set of curves is a foliation of a given spacetime region. See Fig. 1.

Figure 1. Two families of observers. The history of each observeris a timelike curve.

1

2 1. THE EQUIVALENCE PRINCIPLE

Notes on Special Relativity

An inertial system is a system of reference relative to which any mass moves on a straightline if no external forces act upon it. In view of Newton’s Laws set in Euclidean space,any such system is equivalent to (is not distinguished from) any other system of referencein uniform relative motion, which leads to the principle of Galilean relativity. The corre-sponding transformation laws leave time unchanged (up to translations) and thus entailsurfaces of absolute simultaneity (level sets of time) which separate the future from thepast. See Fig. 2. Special Relativity is based on the premise that light, which prior to Ein-stein had been assumed to propagate in a medium at rest relative to a distinguished frameof reference, in fact propagates in the absence of any such medium on cocentric sphereswith respect to any inertial system when emitted from a point source; in other words thespeed of light is finite and the same in any inertial system. Einstein realized this obser-vation can only be reconciled if we give up the notion of absolute simultaneity in favourof a notion of simultaneity relative to the observer such that the propagation of light isindependent of the system of reference, c.f. Fig. 2. This is precisely achieved in the theoryof Special Relativity according to which space and time are unified in Minkowski space(R3+1,m), a 3 + 1-dimensional linear space endowed with a quadratic form m of index 1.A frame of reference corresponds to the choice of a unit timelike vector e, m(e, e) = −1,which determines a unique spacelike hyperplane Σe = X : m(e,X) = 0 with positive-definite induced metric m|Σe ≥ 0, a 3-dimensional Euclidean space consisting of all eventsconsidered simultaneous by observers on Σe at rest relative to the frame defined by e,namely observers whose “world-lines” are straight lines with tangent vector e. The setof null vectors L at a point p, m(L,L) = 0, form a “light cone” consisting of all eventsreached by light emitted from and received at p. The light cone takes the role of level setsof absolute time in Galilean physics in the sense that it separates the past from the future:The set of time-like vectors T at a point p, m(T, T ) < 0, the interior of the light cone,has two disconnected components referred to as future- and past-directed time-like vec-tors pointing to events that can possibly be influenced by an event at p, or have possiblyinfluenced p, respectively. The exterior of the cone at p, the set of spacelike vectors X atp, m(X,X) > 0, is “causally disconnected” from p, given that no speed of propagation ofany physical action whatsoever has ever been observed to surpass the speed of light. The“Principle of special relativity” asserts that all physical phenomena occur in one frame ofreference as they do with respect to another, and in principle cannot be used to distinguisha certain frame of reference, i.e. a specific unit time-like vector e. The elements of theorthogonal group with respect to the Minkowski metric m, m(OX,OY ) = m(X,Y ), arecalled Lorentz transformations, and a “proper” (time-orientation preserving) subgroupmediates in particular the change of one frame of reference to another: e 7→ e′ = Oe,m(Oe,Oe) = m(e, e) = −1. Thus the validity of the principle of special relativity requiresthe invariance of all physical laws under Lorentz transformations.

1. Classical theory of gravitation

A postulate of the classical theory (mechanics: motion of bodies in a givenforce field & gravitation: the gravitational force field of a given distribution ofbodies) is the equality (up to a universal constant depending on the choice ofphysical units) of the passive gravitational mass and the inertial mass.

1. CLASSICAL THEORY OF GRAVITATION 3

Figure 2. Propagation of light in Galilean and Special Relativity.

ψ: Newtonian gravitational potential.−mG∇ψ(t, x): Gravitational force acting on a test body of gravitational

mass mG at time t, the instantaneous position of the body at time t ∈ Rbeing x ∈ R3. mG is thus a concept of the classical theory of gravitation.

x = x(t): Motion of a test body.

If a body is subjected to a given force F when at position x and time t, thenits acceleration is:

(1.1)d2x

dt2(t) =

1

mIF (x, t)

This law (Newton’s Second Law of Motion) defines the inertial mass mI . Thepostulate is

(1.2) mG = mI .

In the classical framework there is no explanation for this equality.Set

(1.3) F = −mG∇ψ(t, x(t)).

Then mG = mI implies that these cancel in the equations of motion:

(1.4)d2x

dt2(t) = −∇ψ(t, x(t))

That is, the motion of the test body in a given gravitational field is independentof the test body.

1.1. Tidal forces. Consider the gravitational field in a small space-time re-gion, the neighborhood of an event (t0, x0), then a0 = −∇ψ(t0, x0) is the commonacceleration of all test bodies in this region. Let x = x0(t) be the history of the testmass which we take as a reference mass. In the new reference system x′ = x− x0

the origin is translated at each time to the instantaneous position of the referencemass. (See Fig. 3, 4.) Note that this is an accelerated reference system.


t

x1

x2

x′1

x′2

x = x0(t)

Figure 3. History of a reference mass.

t

x1

x2

x0(t) = 0′x′

x

Figure 4. Reference coordinate system with origin at each timeat the instantaneous postion of the reference mass.

Now consider the motion of another, arbitrary test body in this new referencesystem: x′ = x′(t).

d2x′

dt2(t) = −∇ψ

(t, x0(t) + x′(t)

)− a0(t)

= −∇ψ(t, x0(t) + x′(t)

)+∇ψ

(t, x0(t)

)= −∇2ψ

(t, x0(t)

)· x′(t) +O(|x′(t)|2)

(1.5)

In terms of rectangular components, and to linear order in x′(t) this equationreads

(1.6)d2x′i

dt2(t) = −

3∑j=1

∂2ψ

∂xi∂xj(t, x0(t)

)· x′j(t)

and is called the tidal equation.

2. SPECIAL RELATIVITY 5

R

P ′: Periphery

Figure 5. Measurement of P ′/R in the rotating system.

Remark. This equation governs the distortion of a dust cloud in time: tidaldistortions. Take here the reference mass to be the test particle occupying thecenter of mass of a small dust cloud.

1.2. Equivalence principle.

(1) In a freely falling reference frame the gravitational force itself is elim-inated. Therefore such a system is equivalent to an inertial referencesystem in the absence of gravity. What remains however is the differen-tial of the gravitational field which causes tidal distortions.

(2) Similarly, an accelerated reference frame in the absence of gravitationalfields is equivalent to a stationary reference frame in a gravitational field.

As a consequence of the equivalence principle, a theory extending the theoryof special relativity must include the theory of gravitation.

2. Special Relativity

We shall see that the equivalence (2) leads to Riemannian geometry.

2.1. Uniformly rotating system. Consider a uniformly rotating referencesystem.

2.1.1. Preliminary considerations. We can ignore the space dimension per-pendicular to the rotating plane. We have a stationary system x, and a referencesystem x′ obtained from x by rotation with constant angular velocity ω. Considera circle of radius |x′| = R with periphery P ′. What is P ′/R as measured in therotating system? Suppose we use rods of unit length in the rotating system formeasurement. As seen in the stationary system, the rods layed along the radius arealso of unit length but those along the periphery experience a Lorentz-contractionso their length is only

√1− v2, where v = ωR. (See Fig. 5.) However, in the

stationary system the periphery of |x| = |x′| = R is P = 2πR. Since the unit rodof the rotating system, when laid along the periphery, has less than unit lengthby the factor

√1− v2, it follows that

(2.1) P ′ =P√

1− v2=⇒ P ′

R=

2π√1− v2

> 2π


(x0, y0)

t = 0

Figure 6. History of a rotating observer with respect to the ro-tating frame.

in the rotating system.1 This means that the laws of Euclidean geometrydo not hold in the rotating system!

2.1.2. Uniformly rotating observers in Minkowski space. Consider the situa-tion in Minkowski space with 2 space and 1 time dimensions. In rectangularcoordinates (t, x, y) of the stationary inertial reference system the metric takesthe form:

(2.2) ds2 = −dt2 + dx2 + dy2

It is convenient to use polar coordinates (t, r, ϕ) such that x = r cosϕ, y = r sinϕand thus

(2.3) ds2 = −dt2 + dr2 + r2dϕ2 .

Let us denote by (x0, y0), and (r0, ϕ0) the rectangular and polar spatial coor-dinates, respectively, in the uniformly rotating system.2 A uniformly rotatingobserver has the history

(2.4) t 7−→ (t, r = r0, ϕ = ϕ0 + ωt)

with respect to the stationary system. In the rotating system the observer remainsat its initial position (see Fig. 6.)

(2.5) x0 = r0 cosϕ0 y0 = r0 sinϕ0 ,

and has the history

(2.6) t 7−→ (t, x0 = x0, y0 = y0) .

The arc element ds2 is a geometric invariant and can be expressed in terms ofthe (r0, ϕ0) coordinates; (in agreement with general covariance, compare remarkon page 1). We find

r = r0 ϕ = ϕ0 + ωt(2.7)

dr = dr0 dϕ = dϕ0 + ωdt(2.8)

1This is a sign of negative curvature.2The domain of these coordinates is the disc ω2r2

0 < 1, because no observer can go fasterthan the speed of light.


and thus

ds2 = −dt2 + dr2 + r2dϕ2

= −dt2 + dr20 + r2

0(dϕ0 + ωdt)2(2.9)

or

(2.10) ds2 = −(1− ω2r2

0

)(dt− ωr2

0 dϕ0

1− ω2r20

)2+ dr2

0 +r2

0 dϕ20

1− ω2r20

.

Let us denote by

α =√

1− ω2r20 : a real number(2.11a)

θ = dt− ωr20 dϕ0

1− ω2r20

: a 1-form(2.11b)

dσ2 = dr20 +

r20 dϕ2

0

1− ω2r20

: a Riemannian metric(2.11c)

then we can write

(2.12) ds2 = −α2θ2 + dσ2 .

We can view dσ as an arc length element in space. The rays ϕ0 = const. aregeodesics of dσ and r0 is arc length along these geodesics:

(2.13) dσ = dr0 : along ϕ0 = const.

The curves r0 = const. are the geodesic circles orthogonal to these rays and

(2.14) dσ =r0 dϕ0√1− ω2r2

0

: along r0 = const.

So we find for the perimeter P ′ of a circle of radius r0 = R that

(2.15) P ′ =

∫ 2π

0

r0dϕ0√1− ω2r2

0

=2πr0√

1− ω2r20

=⇒ P ′

R=

2π√1− ω2R2

,

which coincides with (2.1), (and shows that dσ2 is a Non-Euclidean metric).We are led to think that the coordinates (r0, ϕ0) have no physical meaning in

the rotating system. However, what can be measured is the arc length elementdσ in (2.12). Any such measurement of arclength in space is carried out simul-taneously as judged by the rotating observers, which brings us to the question ofsimultaneity in the rotating system.

An inertial observer corresponds to a straight timelike line in Minkowski space.The set of events which the observer considers simultaneous with an event p of hisown is the spacelike hyperplane Σ through p orthogonal (with respect to (2.2)) tothe line. (See Fig. 7.) A family of observers defining an inertial system is a familyof parallel timelike straight lines. The simultaneous hyperplanes Σ will be paralleland the same at each point.

The family of accelerated observers defining the rotating system is (2.4) asshown in Fig. 8. The hyperplane of locally simultaneous events is different at eachpoint.


p

Σ

Figure 7. Set of simultaneous events Σ to p as judged by aninertial observer.

pΣ

t = 0

t > 0

Figure 8. History of a rotating observer.

Consider the vectorfield ∂∂t . Its integral curves are the histories of the uniformly

rotating observers, parametrized by t. In the original (stationary) coordinates(t, r, ϕ) the motion of an observer is (t, r, ϕ) 7→ (t + b, r, ϕ + ωb), while in thecoordinates of the rotating system this motion is simply (t, r0, ϕ0) 7→ (t+b, r0, ϕ0).Let

(2.16) V =( ∂∂t

)r0,ϕ0

=( ∂∂t

)r,ϕ

+ ω( ∂

∂ϕ

)t,r

be the vectorfield generating the above motion (a 1-parameter group of trans-lations), and let ΣV be the hyperplane in the tangent space at a given pointconsisting of all vectors at that point which are orthogonal to V , namely

(2.17) ΣV =X : g(V,X) = 0

.

Here and in the following we now write g = ds2.

Prop 2.1. The 1-form θ is characterized by the following two properties:

(1) θ · V = 1(2) ΣV = ker θ =

X : θ ·X = 0

.


Σ0

C0

P0

P

Figure 9. Projection map related to uniformly rotating observers.

Proof. By (2.11b) and (2.16) we simply have: θ ·V = dt · ( ∂∂t)r0,ϕ0 = 1. Now

dσ2(V, ·) = 0 because dσ2 does not contain dt. Therefore

g(X,V ) = −α2θ(X)θ(V ) = −α2θ(X) = 0

is equivalent to θ(X) = 0, since α > 0.

Both α and ΣV have physical significance:

α: the rate of a uniformly rotating clock relative to a stationary clock:

α =√−g(V, V ) =

(ds

dt

)r0,ϕ0

.

ΣV : local simultaneous space corresponding to an observer with tangentvector V to his history.

While ΣV contains the events locally simultaneous to an event on the historyof an observer with velocity V , we shall now address the question of simultaneityglobally. In other words, given a set of simultaneous events as judged by the rotat-ing observers, which set of events do they correspond to as seen in the stationarysystem? We can rephrase this questions as follows.

Let C0 be a curve in the hyperplane Σ0 = (t, x, y) : t = 0. In polar coordi-nates we have the parametric equations

(2.18) C0 : r0 = r0(λ) , ϕ0 = ϕ0(λ) .

Let P0 be the point on C0 with λ = 0, P0 = (0, r0(0), ϕ0(0))

Question: Find a curve C in spacetime through the point P0 such that Cprojects to C0 and such that the curve C is horizontal.

Projection: Here the projection is a map π of spacetime onto the hyperplaneΣ0 defined as follows: π(p) = p0 if p is an event belonging to a history of one ofthe rotating observers and p0 is the initial position of that observer. In terms ofthe coordinates (t, r0, ϕ0) this is simply π(t, r0, ϕ0) = (r0, ϕ0).

Horizontal: This means that the tangent vector to C at each point belongs toΣV at that point, i.e. it is orthogonal to the history of the observer through thatpoint. This means that the curve C is locally simultaneous as judged by eachobserver.


Σ0

C

V

C0

C

∆t

Figure 10. A horizontal curve C that projects to C0.

Remark. The question addresses the problem of measurement in the rotatingsystem. We can think of C0 as a curve on the rotating disc whose length we wish tomeasure using a ruler on the rotating disc. The curve C describes the collectionof events in spacetime that are considered when the measurement of length ofC0 occurs simultaneously as seen in the rotating system. The condition that Cprojects to C0 ensures that the curve whose length is measured is C0 in the rotatingsystem; the condition that C is horizontal means that this measurement occurslocally simultaneous as judged by each uniformly rotating observer on C0.

The conditions on C are thus simply:

(1) π(C) = C0

(2) X = C ∈ ΣV

By (1) we can assume that the parametric equations for C in the coordinates ofthe rotating system are:

(2.19) C : t = t(λ) , r0 = r0(λ) , ϕ0 = ϕ0(λ)

Now C is a vector with components

(2.20) C =dt

dλ

( ∂∂t

)r0,ϕ0

+dr0

dλ

∂

∂r0+

dϕ0

dλ

∂

∂ϕ0,

and we have shown above

(2.21) C ∈ ΣV ⇐⇒ θ · C = 0 .

Therefore, in view of (2.11b), condition (2) implies

(2.22)dt

dλ(λ)− ωr2

0

1− ω2r20

dϕ0

dλ(λ) = 0 , t(0) = 0 .

We can integrate this equation to find

(2.23) t(λ) = t(0) +

∫ λ

0

ωr20(λ′)

1− ω2r20(λ′)

dϕ0

dλ(λ′)dλ′ .

In the case where C0 is a circle, r0(λ) = R, we obtain

(2.24) t(λ) = t(0) +ωR2

1− ω2R2

(ϕ(λ)− ϕ(0)

),


so the change of t in going around a full circle is

(2.25) ∆t =2πωR2

1− ω2R2.

The curve C does not close! See Fig. 10.Finally, observe that dσ is in fact arc length along C.

Remark. This result is contrary to our intuition of rigid space and time. Ittells us that it is impossible to measure the length of a curve on the rotating discin a globally instantaneous manner: When measuring the length of a closed curveC0 on the disc initiating from P0 — in such a way that each observer on C0 (andat rest with respect to the uniformly rotating disc) agrees that the measurementoccurs instantaneously according to his local standard of simultaneity — then uponreturn to P0 the proper time α∆t has elapsed in the course of the measurement.This shows in particular that no global standard of simultaneity exists in therotating system.

Note on vectorfields, 1-forms and metrics

At any point p on a differentiable manifold M we have a homeomorphism ϕ from aneighborhood of p onto Rn, n = dimM.

Consider the coordinate lines ai : R→ Rn, t 7→ (0, . . . , t, . . . , 0) in Rn. The coordinatevectorfield ei : i = 1, . . . , n at p is defined to be the tangent vector to the curve Ki =ϕ−1 ai:

ei · f =d

dtf Ki

∣∣t=0

for all f ∈ C∞(M) .

Since f Ki = f ai, where f = f ϕ−1 is a function on Rn, and

ei · f =∂f

∂xi|0 , we simply write: ei =

∂

∂xi

∣∣∣p.

These vectors form a basis for the n-dimensional vectorspace TpM: the tangent space toM at p. Any tangent vector X ∈ TpM can be expanded in this basis:

X =

n∑i=1

Xip

∂

∂xi

∣∣∣p.

The dual space of TpM, denoted by T∗pM is the set of linear functions on TpM (eachelement is also called a covector). While a vectorfield X is an assignment p 7→ X(p) of atangent vector at each point, a 1-form is an assignment p 7→ θ(p) of a covector at eachpoint. The tensor product θ2 = θ⊗ θ, or more generally θ⊗ ξ of two 1-forms, is a bilinearfunction on TpM at each p ∈M:(

θ ⊗ ξ)p(Xp, Yp) = θ(Xp)ξ(Yp) .

A special 1-form is given by the differential of function f :M→ R:

df ·X = X · f .


If we denote by xi = ϕi local coordinates in a neighborhood of p then we find that

dxi · ∂

∂xj=∂ϕi

∂xj= δij =

0 i 6= j

1 i = j.

The 1-forms (dx1, . . . ,dxn) form the dual basis in T∗pM, and any covector ξ ∈ T∗pM canbe expressed as

ξ =

n∑i=1

ξi dxi∣∣p.

We have that for any 1-form θ =∑i θidx

i and any vector X =∑j X

j ∂∂xj it holds that

θ ·X =∑i

θiXi .

A metric g on M is an assignment p 7→ gp of a non-degenerate symmetric bilinearform gp in TpM at each p ∈ M. Here symmetric means g(X,Y ) = g(Y,X), and non-degenerate:

gp(Xp, Yp) = 0 for all Yp ∈ TpM =⇒ Xp = 0 .

A symmetric bilinear form is often called a quadratic form. Since g is symmetric andbilinear we can expand g at any point p in the basis dxi ⊗ dxj :

gp =

n∑i,j=1

gij dxi|p ⊗ dxj |p ,

where it is understood that gij = gji.A Riemannian metric is a positive definite metric. It allows us to assign a magnitude

(length) to a vector X,

|X| =√g(X,X) ≥ 0 ,

and an angle α to two non-zero vectors:

g(X,Y ) = |X||Y | cosα .

A Lorentzian metric is an assignment p 7→ gp of a non-degenerate quadratic form ofindex 1 in TpM at each point p ∈ M. Recall that the index of a quadratic form is thelargest dimension of a subspace where the quadratic form is negative definite. There istherefore a vector Tp ∈ TpM such that gp(Tp, Tp) < 0; Tp is a timelike vector at p. Bymultiplying with a suitable non-zero factor we can assume that Tp is unit: gp(Tp, Tp) = −1.Define Σp to be the the orthogonal complement of Tp in TpM. Then gp|Σp is positivedefinite. Choosing an orthonormal basis (E1|p, . . . , En|p) (here n = 3) for Σp any vectorXp ∈ TpM can be expanded as

Xp = X0Tp +

n∑i=1

XiEi|p ,

and we have

gp(Xp, Xp) = −(X0)2

+

n∑i=1

(Xi)2.

The set of vectors Xp for which gp(Xp, Xp) > 0 are called spacelike. A Lorentzian metricdefines a cone gp(Xp, Xp) = 0 at each point, the set of null vectors at the point p.


p0

ψt(p0)

K0

ψt(K0)

Figure 11. Condition of symmetry on the family of observers.

2.2. General accelerated system. In the general situation of a given ac-celerated reference system spacetime is invariant under a 1-parameter group ψtwhose orbits are timelike curves. An orbit t 7→ ψt(p0), with p0 fixed, is preciselythe history of an observer initiating at the event p0. A family of observers defininga reference system satisfies a symmetry condition: Given a spacelike curve K, thearclength of the curve Kt = ψt(K) equals that of K0 = K for all t. We shall seethat the metric takes the form

(2.26) ds2 = −α2θ2 + γ , θ = dt+∑i

Aidxi

and that (2.25) generalises in this setting to

(2.27) ∆t =

∫C0

Ai dxi .

General Situation. Let us now assume we are in the situation where a 1-parameter group of diffeomorphisms ψt acts on the spacetime manifold such that

(1) the orbits t 7→ ψt(p) are timelike curves,(2) ψt acts by isometries.

The second condition ensures that the arclength of a spacelike curve K remainsunchanged under the flow of ψt; cf. Fig. 11. Moreover, (2) implies

(2.28) ψ∗t g = g

and thus

(2.29) −LV g =d

dtψ∗t g

∣∣∣t=0

= 0 ,

where V is the generating vectorfield of ψt, a vectorfield tangential to the orbits:the velocity of the observers. Indeed, given a curve K : λ 7→ K(λ) (λ ∈ [a, b]) withK(a) = p0, K(b) = q0 the condition that the arclength of Kt : λ 7→ ψt(K0(λ)) isindependent of t reads

(2.30) L[Kt] =

∫ b

a

√|g(Kt(λ), Kt(λ))| dλ = L[K0] ;

Since Kt(λ) = dψt · K0(λ) and g(dψt · K0(λ), dψt · K0(λ)) = (ψ∗t g)(K0(λ), K0(λ))we conclude on (2.28) for every t, and thus by the definition of the Lie derivativeon (2.29). (C.f. note on pull-backs on pg. 17.)


H0

p0

ψt(p0)

Figure 12. Transversal hypersurface H0 to family of observers ψt(p0).

Let H0 be a transversal hypersurface for the set of orbits of ψt (each observerintersects H0 exactly once); see Fig. 12. Here H0 is a spacelike hypersurface in3+1-dimensional Minkowski space. Every point p can be written as p = ψt(p0) forsome value t where p0 ∈ H0, and we can assign to p the coordinates (t, x1, x2, x2)where (x1, x2, x3) are the spatial coordinates of p0 in Minkowski space. (We canthink of H0 as the space of observers. A body with fixed coordinates p0 is at restin the accelerated system.) In these coordinates

(2.31) V =( ∂∂t

)x

(where x = (x1, x2, x3) is held fixed)

and the condition (2.29) simply becomes

(2.32)∂gµν∂t

= 0 ,

where gµν = g(∂/∂xµ, ∂/∂xν) (µ, ν = 0, 1, 2, 3) are the metric components in thesecoordinates (here also x0 = t). Then g takes the form

(2.33) g = −α2θ2 + γ

where α = α(x) is a positive function, θ is a 1-form

(2.34) θ = dt+A , A(x) =3∑i=1

Ai(x)dxi ,

and γ = γ(x) is a Riemannian metric in space,

(2.35) γ(x) =

3∑i,j=1

γij(x)dxidxj .

As above, α signifies the rate of clocks carried by the observers whose world linesare the group orbits (relative to the parameter t), and θ defines the local standard


H0

p0 C0

C

p = ψt0(p0)

Figure 13. A horizontal curve C that projects to the closed curveC0 in H0.

of simultaneity for the given system of observers. It is a 1-form characterized asabove – c.f. Prop. on pg. 8 – by the properties that

(1) θ · V = 1 (normalisation conditon)(2) θ · X = 0 if and only if X belongs to the hyperplane ΣV of all vectors

orthogonal to V .

Let us now turn to the question of simultaneity as in the uniformly rotatingcase (c.f. pg. 9): Let C0 be a curve in H0 which starts at p0, parametrized by λsuch that λ = 0 corresponds to p0. We want to find a curve C through p = ψt0(p0)which is horizontal and projects to C0. (See Fig. 13.) The parametric equationsfor C0 are given by xi = xi(λ), i = 1, 2, 3, and those for C by t = t(λ), xi = xi(λ),i = 1, 2, 3 because C projects to C0. The condition that C is horizontal meansthat

(2.36) C =dt

dλ

∂

∂t+

3∑i=1

dxi

dλ

∂

∂xi∈ ΣV ,

which implies by the second property of θ that

(2.37) 0 = θ · C =dt

dλ+

3∑i=1

Ai(x)dxi

dλ.

Consider now the case that C0 is a closed curve so that for some λ0 > 0: x(λ0) =x(0). The start and end point of C0 are thus both p0. By (2.37) we obtain anincrement of

(2.38) ∆t =

∫ λ0

0

dt

dλdλ = −

∫ λ0

0

3∑i=1

Ai(x(λ)

)dxi

dλdλ ,

or

(2.39) ∆t = −∫C0

A .


H0

H ′0

(0, x)

(f(x), x)

Figure 14. Gauge transformations

Remark. In analogy to electromagnetic theory we may view A as the vectorpotential, and introduce the magnetic field

(2.40) B = dA .

(B is a 2-form, the exterior derivative of a 1-form.) Consider now a surface S0 inH0 whose boundary is C0: ∂S0 = C0. Then by Stokes’ theorem

(2.41)

∫S0

B =

∫C0

A .

We may thus interpret the integral (2.39) as the magnetic flux through a surfaceenclosed by C0.

Remark. The result does not depend on the choice of the hypersurface H0. Infact, another transversal hypersurface H ′0 is described as a graph t = f(x) in thecoordinates constructed above (where H0 = t = 0); see Fig. 14. The coordinatetransformation

(2.42) t′ = t− f(x)

leaves θ invariant provided the vector potential A is transformed according to

(2.43) A′ = A+ df ,

or in coordinates

(2.44) A′i(x) = Ai(x) +∂f

∂xi(x) ;

then indeed

(2.45) θ′ = dt′ +A′ = dt+A = θ .


The maps (2.43) are precisely the gauge transformations, which — as in electro-magnetic theory — leave the magnetic field unaltered:

(2.46) B′ = B .

In particular, in view of (2.41), (2.38) does not depend on the choice of H0.

Finally note that γ measures arclength along C. Since C ∈ ΣV we have thatarclength of C w.r.t. g equals arclength of C0 relative to γ, which is

(2.47) L[C] =

∫ λ0

0

√√√√ 3∑i,j=1

γij(x(λ)

)dxi

dλ

dxj

dλdλ .

Note on pull-backs

Let ψ : M → M be a diffeomorphism, and α a 1-form on M (in general the target).Recall that the differential dψ(p) is a linear mapping of TpM into Tψ(p)M, and we candefine the pull-back of α, ψ∗α: a 1-form on M, by

(ψ∗α) ·X = α(ψ(p)) · (dψ(p) ·X) (X ∈ TpM) .

The metric g is a quadratic form in the tangent space at each point, and

(ψ∗g)(X,Y ) = gψ(p)(dψ(p) ·X,dψ(p) · Y ) (X,Y ∈ TpM).

Interpretation in Vector Bundles

The construction of the curve C – as first discussed on page 9 – has a natural interpretationas a horizontal lift in vector bundles.

In the situation (2.26) the spacetime manifold is B = R × N 3 (t, x), where we canthink of (N , γ) as the manifold of observers. We have a projection π : B → N which tellsus which observer q ∈ N passes through a given event p: π(p) = q. The fiber π−1(q) isa timelike curve in B: the history of an observer q. V is the tangent vector field to thesecurves parametrized by t, and θ · V = 1. Recall that the local simultaneous space at p isgiven by the null space of θ:

Σp =X ∈ TpB : θ(p) ·X = 0

Let now Xq be a tangent vector to N at q. Observe that the projection π is surjective.Moreover dπ(p) is an isomorphism of Σp onto TqN . Therefore given any Xq ∈ TqN thereexists a unique vector X]

p ∈ Σp such that

dπ(p) ·X]p = Xq ;

The vector X]p is called the horizontal lift of Xq to p; see Fig. 15. The horizontal lift is

determined by the 2 conditions:

(1) dπ(p) ·X]p = Xq

(2) θ(p) ·X]p = 0 .


q

C0

p

N

Σp

C]0π−1(q)

Xq

X]p

Figure 15. Horizontal lift of vectors and curves.

In local coordinates (t, x) we have:

Xq =∑i

Xiq

∂

∂xi∣∣q∈ TqN and X]

p =(X]p

)t ∂∂t

∣∣p

+∑i

(X]p

)i ∂∂xi

∣∣p∈ TpB .

By (1) we have (X]p)i = Xi

q, and (X]p)t is determined from (2):

θ = dt+∑i

Ai(x)dxi : θ ·X]p =

(X]p

)t+∑i

Ai(x)(X]p

)i= 0

Therefore,

X]p = −

(A(q) ·Xq

) ∂∂t

∣∣p

+∑i

Xiq

∂

∂xi∣∣p.

With the horizontal lift of tangent vectors known, we can define the horizontal lift ofcurves: Let C0 : I → N a curve in the manifold of observers, I ⊂ R an interval con-

taining 0. Let q = C0(0), and p ∈ π−1(q), then there exists a unique curve C]0 suchthat

˙C]0(λ) = (C0(λ))]C0(λ) (λ ∈ I) .

Then the curve C = C]0 is precisely the solution to the problem under consideration;c.f. (2.37). That is to say the horizontal lift C(λ) = (t(λ), x(λ)) is the solution to theequation

dt(λ)

dλ= −

∑i

Ai(C0(λ))dxi(λ)

dλ, t(0) = 0 .

In (2.39) we have shown that the horizontal lift C of a closed curve C0 in N does notclose in B. This is a manifestation of curvature.

2.3. Conclusion. We have studied accelerated systems in comoving coordi-nates (t, p0) such that the histories of the corresponing observers at rest in theaccelerated sytem are given by t 7→ (t, p0). We have seen that following simul-taneous events in the accelerated system starting at (t0, p0) along a closed curveλ 7→ p(λ) with p(0) = p(λ0) = p0 will – in general – take us back to (t1, p0) at


advanced coordinate time t1 with ∆t = t1 − t0 > 0. We conclude that coordinatesdo not have physical meaning beyond labeling spacetime events. The differentialdt is not the actual time increment measured by the clocks of the acceleratingsystem, and the level sets of t are not surfaces of simultaneity. Instead the rateof clocks in the accelerating system is α, and the relevant time differential α dt.Moreover θ determines a set of locally simultaneous events, and we have shownthat it is impossible in general to find a global hypersurface of simultaneity. Simi-larly we have seen that dp0 does not measure elements of arclength in the space ofan accelerating system, but instead it is γ. (E.g. for the uniformly rotating systemdϕ0 does not give the arc element of circles when multiplied with r0, but insteadit is r0dϕ0/

√1− ω2r2

0.) We are led to conclude finally: what does have physi-cal meaning (and can be measured) are not coordinates but elements of arclength,namely the metrical ground form

(2.48) ds2 =3∑

µ,ν=0

gµνdxµdxν .

Remark. A study of the metrical ground form first appeared in 1828 inC.F. Gauss’ “General Investigations of Curved Surfaces” and stood at the be-ginning of the development of Riemannian Geometry. A. Einstein arrived at theabove conclusion in 1915, and General Relativity is formulated in the context ofLorentzian Geometry.

Exercises

(Receding observers in Minkowski spacetime): Consider a set of observers in uniformmotion receding from a center. In other words consider a family of observers whose histories arestraight lines in Minkowski space intersecting in a single point. These are rays contained in thefuture of the origin:

I+0 =

(x0, x1, x2, x3) ∈ R1+3 : −(x0)2 +

3∑i=1

(xi)2 < 0 and x0 > 0

.

(1) Depict the family of receding observers and the set I+0 in diagram.

(2) At each p ∈ I+0 construct the hyperplane of simultaneous events Σp to p as judged by

the observer passing through p.(3) Do there exist global hypersurfaces of simultaneity in I+

0 ?In other words, is the distribution of planes

∆ := Σp : p ∈ I+0

integrable? Hint: Consider the hyperboloids H+τ given by

H+τ :=

(x0, x1, x2, x3) ∈ I+

0 : (x0)2 −3∑i=1

(xi)2 = τ2

and show that for a given event p ∈ H+

τ the tangent plane at p to H+τ coincides with

Σp.(4) Consider the hyperboloid H+

1 . Assign to each event p ∈ H+1 the polar normal

coordinates (χ, θ, φ), where θ and φ are the polar and azimuthal angle of the


standard coordinates on the unit sphere, and χ is the geodesic distance on H+1 from

(1, 0, 0, 0) to the event p. Verify that the induced metric on H+1 is given by

dσ2 = dχ2 + sinh2 χ(dθ2 + sin2 θ dφ2).

Show that the Gauss curvature of H+1 is equal to −1.

(5) Show that the Minkowski metric can be written in I+0 as

ds2 = −dτ2 + τ2dσ2 .

That is to say I+0 can be foliated by the hyperboloids H+

τ and the induced metric onH+τ is precisely τ2dσ2. Also show that the Gauss curvature of H+

τ is equal to

K = −τ−2 .

(6) Consider two receding observers. Suppose a light signal is sent from one to the other.Let

v = tanhχ

be the velocity of the observer emitting the light as measured in the rest frame of theobserver receiving the signal. Show that a redshift is observed with

ωeωr

= eχ ,

where ωe and ωr are the frequencies of the emitted and the received light respectively.

3. Geodesic correspondence

In view of the equivalence principle our conclusions for accelerated systemsof reference in the absence of a gravitational field apply to stationary referenceframes in the presence of a gravitational field. In particular we know that themetrical ground form (2.48) in the presence of a gravitational field is non-trivial.Since we may change to a freely falling reference frame where the gravitationalforce is eliminated locally we are left with the differential of the gravitational field.In the following we shall find the laws that govern the metric (2.48) in the presenceof a gravitational field by an analogy of tidal motions in Newtonian gravity andgeodesic motion of nearby particles on Lorentzian manifolds (with curvature).

3.1. Tidal equation. Recall our discussion in Section 1.1 of tidal forces inthe framework of the Newtonian theory. We took a freely falling particle as areference particle and translated the origin of a new coordinate system at eachtime to the instantaneous position of this particle.

x0(t): motion of reference particlex(t): motion of arbitrary nearby particle

Then we have seen that the displacement

(3.1) y(t) = x(t)− x0(t)

to a nearby freely falling particle satisfies to first order in y (small displacements)the tidal equation (1.6):

(3.2)d2y

dt2= −∇2ψ

(x0(t)

)· y(t)

3. GEODESIC CORRESPONDENCE 21

In matrix notation this simply reads

(3.3)d2y

dt2= −M y , i.e.

d2yi

dt2(t) = −

3∑j=1

Mij(t) yj(t) ,

where M is the Hessian matrix of the gravitational potential:

(3.4) Mij(t) =∂2ψ

∂xi∂xj(x0(t)

).

Suppose that the nearby particle is at rest relative to the reference particle at timet = 0, i.e.

(3.5)dy

dt

∣∣∣t=0

= 0 .

Since the tidal equation is linear we have

(3.6) y(t) = A(t) y(0)

for some matrix A(t) depending on t. Clearly, A(0) = 1 (identity matrix), andthe initial condition (3.5) for y translates into the analogous conditon on A:

(3.7)dA

dt(0) = 0

And inserting (3.6) into the tidal equation (3.3) yields of course an identical equa-tion of motion for the matrix A:

(3.8)d2A

dt2(t) = −M(t)A(t)

Let us introduce a matrix θ(t) by:

(3.9) θ =(dA

dt

)A−1

Then,

dθ

dt=(d2A

dt2

)A−1 +

(dA

dt

) d

dt

(A−1

)=(d2A

dt2

)A−1 −

(dA

dt

)A−1 dA

dtA−1

=(d2A

dt2

)A−1 − θ2 ,

(3.10)

and by the matrix tidal equation (3.8) we obtain the following equation of motionfor θ:

dθ

dt= −θ2 −M(3.11)

θ(0) = 0(3.12)

We have not yet exploited the fact that the Hessian is symmetric:

(3.13) M = M


(For an arbitrary matrix A we denote by A the transpose of A: Aij = Aji.) Takingthe transpose of (3.11) we obtain as well

(3.14)dθ

dt= −θ2 −M

We now decompose

(3.15) θ = σ + ω

into its symmetric part σ and its antisymmetric part ω:

(3.16) σ =1

2(θ + θ) ω =

1

2(θ − θ)

Substracting (3.14) from (3.11) M cancels and we obtain the following equationfor ω:

dω

dt= −1

2(θ2 − θ2) = −1

2

((σ + ω)2 − (σ − ω)2

)= −(σω + ωσ)

(3.17)

This is a linear homogeneous equation for ω considering σ as given. It follows thatif ω vanishes initially, then it vanishes for all time. But here θ(0) = 0, thereforeσ(0) = 0 as well as ω(0) = 0. It follows that with our initial conditions ω vanishesidentically for all time, therefore

(3.18) θ = σ is symmetric.

It follows that

(3.19) tr(θ2) = tr(θθ) =∑i,j

(θij)2 = |θ|2 ≥ 0 ,

and taking the trace of (3.11) we obtain

(3.20)d

dttr θ = −|θ|2 − trM .

Note that

(3.21) trM(t) =∑i

∂2ψ

∂xi2(x0(t)

)= ∆ψ

(x0(t)

),

and according to the Newtonian theory of gravitation:

(3.22) ∆ψ = 0 in vacuum.

The meaning of (3.20) becomes evident by looking at the evolution of volume.Consider the volume of a parallelepiped spanned by linearly independent displace-ments (y1, y2, y2) of nearby particles: Ω(y1, y2, y3). (See Fig. 16.) Let us assumethat three nearby particles are initially at rest and at positions (E1, E2, E3): anorthonormal (positive) basis. Then Ω(E1, E2, E3) = 1. Recall that the displace-ments at time t are then obtained using the linear map A(t): deformation matrix.


E1

E2

E3

y1(t)

y2(t)

y3(t)

Figure 16. Volume of parallelepiped.

In fact, yi(t) = A(t)Ei, i = 1, 2, 3. Therefore, we obtain for the volume of thecorresponding parallelepiped at time t:

Ω(t) = Ω(y1(t), y2(t), y3(t)) = Ω(A(t)E1, A(t)E2, A(t)E3)

= detA(t) Ω(E1, E2, E3) = detA(t) Ω(0) = detA(t)(3.23)

This formula holds for any volume form Ω. The volume of a parallelepiped is infact the unique volume form on a vector space with inner product; c.f. note onvolume form on pg. 25. Now consider the function

(3.24) µ(t) = log Ω(t) = log detA(t) .

Since, by differentiation of a determinant,

(3.25)d

dtdetA(t) = detA(t) tr

(dA

dtA−1

)we have

(3.26)d

dtµ(t) = tr θ(t) , µ(0) = 0 ,

or

(3.27) µ(t) =

∫ t

0tr θ(t′)dt′ .

Moreover, by (3.12) θ(0) = 0 we also have

(3.28)dµ

dt(0) = 0 ,

and

(3.29)d2µ

dt2(0) =

d

dttr θ∣∣t=0

= − trM∣∣t=0

.

Remark. We have shown that tidal forces deform shapes in such a way thatvolumes remain the same up to third order. Note that while equality to first orderis simply a consequence of the fact that all particles constituting the shape areinitially at rest, the equality for volumes in second order is precisely the equationfor the gravitational field in vacuum, namely the Poisson equation (3.22) for theNewtonian potential.


Earth

Elevator

7→

Figure 17. Elevator experiment.

Elevator experiment

Tidal forces are easily visualised in the following experiment: Consider a distribution ofmasses in a freely falling elevator in the gravitational field of the Earth. See Fig. 17.From the point of view of a freely falling observer at the center of the elevator the massesaccelerate towards her horizontally and away from her vertically. An initially sphericaldistribution of nearby masses surrounding the observer is deformed to a prolate spheroid ofthe same volume. The equality of the volumes is precisely the equation of the gravitationalfield.

Note on volume forms in vector spaces

In this note we discuss the notion of a volume form Ω in a n-dimensional vector space,and show that (3.17) holds in general. Moreover, we prove that if Vn is endowed with aninner product, then Ω is in fact unique.

Let X1, . . . , Xn be vectors in an n-dimensional space Vn. A volume form Ω in Vnis a n-linear form in Vn which is totally antisymmetric and nondegenerate. That is tosay Ω(X1, . . . , Xn) depends linearly on Xi if we fix all vectors X1, . . . , Xn except Xi, andchanges sign if we interchange two vectors:

Ω(. . . , Xi, . . . , Xj , . . .) = −Ω(. . . , Xj , . . . , Xi, . . .) .

Moreover it being nondegenerate means there is a n-tuplet of vectors (X1, . . . , Xn) in Vnsuch that

Ω(X1, . . . , Xn) 6= 0 .

If Vn is endowed with an inner product 〈·, ·〉 then this automatically defines a volumeform on Vn. For, we can then choose a basis (E1, . . . , En) for Vn which is orthonormal:

〈Ei, Ej〉 = δij .

We then set

Ω(E1, . . . , En) = 1 .


However, an inner product only determines an orthonormal basis up to rotation. Let(E′1, . . . , E

′n) be another such orthonormal basis. We must verify that also Ω(E′1, . . . , E

′n) =

1. (Otherwise the definition does not make sense.) To be precise we shall show that avolume form is uniquely determined by the inner product and orientation.

Let us first return to Ω(X1, . . . , Xn). We can expand any vector Xi in the basis(E1, . . . , En):

Xi =

n∑j=1

(Xi)jEj : i = 1, . . . , n .

The coefficients (Xi)j : j = 1, . . . , n form the ith column of a matrixX. Then Ω(X1, . . . , Xn)

is determined. Therefore a volume form Ω is uniquely determined by the basis (E1, . . . , En).Let us verify that in the case n = 2:

X1 = (X1)1E1 + (X1)2E2 X2 = (X2)1E1 + (X2)2E2 .

By multilinearity

Ω(X1, X2) =(X1)1(X2)1Ω(E1, E1) + (X1)1(X2)2Ω(E1, E2)

+ (X1)2(X2)1Ω(E2, E1) + (X1)2(X2)2Ω(E2, E2)

=(X1)1(X2)2 − (X1)2(X2)1 = detX

This is in fact the definition of the determinant.Given a linear transformation A of Vn there is a matrix A which represents the linear

map in the basis (E1, . . . , En). Since A · Ei is a vector, it can be expanded in the basis(E1, . . . , En):

AEi =

n∑j=1

Aji Ej .

The coefficients Aji constitute the corresponding matrix, and thus

Ω(AE1, . . . , AEn) = (detA)Ω(X1, . . . , Xn) .

Given a basis (E1, . . . , En) we obtain another basis (E′1, . . . , E′n) by transformation with

an element of the general linear group GL(n,R):

E′i =

n∑j=1

Gji Ej : i = 1, . . . , n where detG 6= 0 .

The set of all bases in Vn has two components, one obtained from (E1, . . . , En) withdetG > 0, the other with detG < 0. Selecting one of the two classes, say the one contain-ing (E1, . . . , En) which one then refers to as positive bases, corresponds to introducing anorientation on Vn. Hence

Ω(E′1, . . . , E′n) > 0

on all positive bases.Consider transformations O which leave the inner product invariant:

〈OX,OY 〉 = 〈X,Y 〉We can represent O in a positive orthonormal basis (E1, . . . , En), then the columns of thematrix

E′i =

n∑j=1

OjiEj : i = 1, . . . , n


are another orthonormal basis:

〈E′i, E′j〉 = 〈O · Ei, O · Ej〉 = 〈Ei, Ej〉 = δij

This condition reads

OO = 1 ,

the defining condition of an orthogonal matrix. (Here O is the transpose of O.) Therefore(detO)2 = 1, or detO = ±1, with detO = +1 if the primed basis is positive.

We defined Ω(E1, . . . , En) = 1. The definition makes sense because if (E′1, . . . , E′n) is

another positive orthonormal basis then

Ω(E′1, . . . , E′n) = (detO)Ω(E1, . . . , En) = 1 , because detO = 1 .

It follows that if Vn is a oriented vector space with an inner product, then Vn has a uniquevolume form Ω.

Exercises

1. (Ratio of volumes): Consider a freely falling reference particle in a gravitationalfield in vacuum. Suppose the particle is surrounded by a sphere of volume Ω which isinitially at rest relative to the reference particle. Show that in fact up to fourth orderthe volume of the sphere does not change in time:

Ω(t)

Ω(0)= 1 +O(t4) .

What is the obstruction to the equality of volumes to higher order in time?2. (Convergence of particles): Consider a freely falling reference particle in vacuum

as in Exercise 1. Assume now in addition that the Hessian matrix of thegravitational potential does not vanish initially in the vicinity of the referenceparticle. Recall that the displacement of a nearby particle y(t) is given in terms of itsinitial displacement y(0) (we assume that the particle is initially at rest relative tothe reference particle) by

y(t) = A(t)y(0) ,

where A is the deformation matrix. Show that there is an initial displacementy(0) 6= 0 and a time t∗ > 0 such that

y(t∗) = 0.

Give a physical interpretation of this result.Hints. Recall the definition of the strain matrix

θ =dA

dtA−1.

To solve Exercise 2 one may proceed as follows:(1) Show that under the assumption that the Hessian of the gravitational field does

not vanish identically initially we have

d3

dt3tr θ∣∣t=0

< 0 .

(2) Deduce from the tidal equation the following ordinary differential inequality:

d

dttr θ ≤ −1

3

(tr θ)2.

(Hint: Decompose θ into its diagonal and its traceless part.)


(3) Show that there exists a time t∗ > 0 such that

tr θ → −∞ as t→ t∗.

(Hint: Derive an equation for u = −1/ tr θ.)(4) Show that there exists a tangent vector y(0) 6= 0 such that

y(t∗) = 0.

(Hint: What is the implication of (3) for A(t) as t→ t∗?)

3.2. Theory of Geodesics. According to the equivalence principle the grav-itational force is eliminated in a freely falling reference frame. A reference systemis given by a family of timelike curves in the spacetime. We have seen — in ourdiscussion of accelerated reference frames in Section 2.2 — that in the presence ofa gravitational field the spacetime is curved ; i.e. it is not Minkowski space as inSpecial Relativity, but a Lorentzian manifold with curvature (as we shall discussin some detail in the next Chapter). Which timelike curves in the spacetime aredescribed by freely falling bodies?

Hypothesis: (Einstein) A freely falling test particle in a (non-trivial) grav-itational field corresponds to a timelike geodesic in a spacetime with cur-vature.

Remark. It is an important feature of the theory that this hypothesis can berecovered from the equations of motion. We shall return to this question in ourdiscussion of the Einstein equations in the presence of matter.

It is thus a consequence of the equivalence principle that the gravitational forceis an aspect of the spacetime itself. The aim of this section is to relate Newton’slaw for the gravitational potential to an equation for the spacetime curvature.Since in a freely falling reference frame only tidal forces can be measured it canonly be the differential of the gravitational field that contributes to spacetimecurvature. To derive the law — the Einstein vacuum equations — we shall look ata family of nearby timelike geodesics, corresponding to the trajectories of nearbyfreely falling particles in the Newtonian theory.

3.2.1. Jacobi equation. Consider a timelike reference geodesic Γ0 (a freelyfalling reference test particle) in a spacetime (M, g). Let Γ0 be parametrizedby arc length t (proper time), and let T be the unit future directed (parallel)tangent vector field along Γ0:

(3.30) ∇TT = 0 .

Let O be an arbitrarily chosen origin on Γ0 where we set t = 0. Consider aspacelike vector X at O orthogonal to T , and let K0 be the spacelike geodesicthrough O with initial tangent vector X at O. Then X extends along K0 as thetangent vector field of K0, and the magnitude of X, |X| =

√g(X,X), is constant

along K0. Here K0 is parametrized by λ = |X|s, where s is the arclength alongK0. Let us now extend T along K0 by parallel transport ; we obtain a unit timelikefuture directed vectorfield along K0, which is orthogonal to X:

(3.31) g(T,X) = 0 along K0 .


O

Γ0

Γλ

T T

X K0

Γ0(t)

K0(λ)

Figure 18. Construction of a timelike geodesic congruence.

At each point K0(λ) along K0, we draw the timelike geodesic towards the futurewith initial tangent vector T . We call this geodesic Γλ, and the family of curvesa timelike geodesic congruence; see Fig. 18. Extend T along Γλ to be the tangentvector field to Γλ; since the tangent vector is parallel transported along a geodsic,we have

(3.32) g(T, T ) = −1 .

Remark. The timelike curves Γλ correspond to the histories of nearby freelyfalling bodies in the Newtonian theory.

Remark. In principle, we may observe three different kinds of behaviour: thedistance to the nearby geodesic remains constant; the distance decreases; or thedistance increases. As we shall describe in detail these cases are related to zero,positive, or negative (sectional) curvature, respectively; see Fig. 19.

We define a timelike 2-dimensional (ruled) surface

(3.33) S =⋃λ

Γλ ,

and for each t ≥ 0 a curve Kt ⊂ S by:

(3.34) Kt(λ) = Γλ(t) .

Then extend the vectorfield X to the surface S to be the tangent field of the curvesKt. The parameters t, λ can be thought of as coordinates on S; with respect tothese coordinates:

(3.35) T =∂

∂t, X =

∂

∂λ,

and thus

(3.36) [T,X] = 0 .


Plane Sphere Hyperboloid

Figure 19. Three characteristic cases for geodesic congruences:Constancy of distance to nearby geodesics (zero curvature); De-crease of distance (positive curvature); Increase of the distance toa nearby geodesic (negative curvature).

φ

(t, λ)

p

Xp

Tp

Γ0

K0

Kt

∂∂t

∂∂λ

Figure 20. Push-forward of coordinate vectorfields to S.

Remark. More precisely, we have a mapping

φ : (t, λ) 7→ φ(t, λ) = Γλ(t) = Kt(λ) ;

see Fig. 20. If p = φ(t, λ) then

Tp = dφ(t, λ) · ∂∂t

∣∣∣(t,λ)

, Xp = dφ(t, λ) · ∂∂λ

∣∣∣(t,λ)

.

In other words the vectorfields T and X on S are the push-forward of coordinatevectorfields in the (t, λ) plane,

T = φ∗∂

∂t, X = φ∗

∂

∂λ.

Therefore,

[T,X] =[φ∗

∂

∂t, φ∗

∂

∂λ

]= φ∗

[ ∂∂t,∂

∂λ

]= 0 .


Prop 3.1. We have that T is orthogonal to X everywhere on S:

g(T,X) = 0 on S .

Proof. We know this is true along K0. Differentiating the function g(T,X)along Γλ we obtain

T(g(T,X)

)= g(∇TT,X) + g(T,∇TX) .

Since the integral curves of T are geodesics the first term vanishes by (3.30). Now,by (3.36):

∇TX −∇XT = [T,X] = 0 ;

therefore

g(T,∇TX) = g(T,∇XT ) =1

2X(g(T, T )

)= 0 ,

since g(T, T ) = −1 is a constant function on S. Hence

T(g(T,X)

)= 0 ,

which says that g(T,X) is constant along Γλ. Since the integral curves of T initiateon K0 where g(T,X) = 0, it follows that

g(T,X) = 0 along Γλ .

Remark. Here we have used that the connection in general relativity is sym-metric3 and compatible with the metric. While having a symmetric connectionmeans

∇XY −∇YX = [X,Y ] ,

for any vectorfields X, Y , metric compatibility is equivalent to

Z(g(X,Y )

)= g(∇ZX,Y ) + g(X,∇ZY ) ,

for any vectorfields X, Y , Z.

The curves Kt for t > 0 are not in general geodesics. The arclength of Kt

between the parameter values 0 and λ is:

L[Kt[0, λ]

]=

∫ λ

0|X|Kt(λ′)dλ′ .

The magnitude |X|Γ0(t) thus measures the deviation of nearby geodesics. Thereforewe have to study

(3.37) X(t) = X|Γ0(t) : the displacement at time t .

Remark. The vectorfield (3.37) corresponds precisely to the displacement(3.1) in the tidal equation.

∇TX: First rate of change of the displacement.∇2TX: Second rate of change of the displacement.

3this is the difference to gauge theories, not the metric.


Since ∇TX = ∇XT and ∇XT = 0 along K0 (T was parallel transported alongK0), we have

(3.38) ∇TX|t=0 = 0 ,

which corresponds in the Newtonian theory to the condition that the nearby testparticles are initially at rest relative to the reference particle. As regards thesecond rate of change we have

∇2TX = ∇T∇TX = ∇T∇XT .

Now recall that the definition of curvature is equivalent to:4

(3.39) [∇X ,∇Y ]Z −∇[X,Y ]Z = R(X,Y )Z

Since

∇2TX = ∇T∇XT = ∇X∇TT + [∇T ,∇X ]T = [∇T ,∇X ]T ,

we obtain, by (3.39), in view also of the antisymmetry of the curvature,

[∇T ,∇X ]T = ∇[T,X]T +R(T,X) · T = −R(X,T ) · T ,the so-called Jacobi equation:

(3.40) ∇2TX = −R(X,T ) · T

This is a second order equation for the vectorfield X along Γ0. The displacementfield X to nearby geodesics is called a Jacobi field5. We have derived the Jacobiequation, or Geodesic deviation equation, by variation through geodesics.

We express (3.40) in components relative to a comoving orthonormal framealong Γ0. First choose an orthonormal basis (E1, E2, E3) for Σ0, the hyperplanein TOM consisting of all vectors orthogonal to T . In particular X(0) ∈ Σ0. Ingeneral, for every t ≥ 0 we denote by

(3.41) Σt =Y ∈ TΓ0(t)M : g(Y, T ) = 0

the hyperplane in the tangent space toM at Γ0(t) defined by all vectors orthogonalto T . Now recall from Prop. 3.1 that X is orthogonal to T along Γ0, and thus

(3.42) X(t) ∈ Σt (t ≥ 0) .

We can propagate the vectors Ei : i = 1, 2, 3 along Γ0 by parallel transport, i.e. wedefine vectorfields Ei : i = 1, 2, 3 along Γ0 such that

(3.43) ∇TEi = 0 : i = 1, 2, 3.

Since the inner products remain unchanged we have an orthonormal basis forΣt at each t ≥ 0 which we denote by (E1(t), E2(t), E3(t)). Complemented withE0 = T we obtain an orthonormal basis (Eµ : µ = 0 . . . , 3) for TΓ0(t)M at each

4C.f. notes on connections and the curvature tensor on pg. 39.5In fact, we have restricted ourselves to the case where K0 is orthogonal to Γ0 at O; X is

then called a normal Jacobi field.


t ≥ 0, i.e. along Γ0. But since X(t) ∈ Σt we can expand X(t) in the basis(Ei(t) : i = 1, 2, 3):

(3.44) X(t) =3∑i=1

Xi(t)Ei .

Remark. The quadruple (t,X1, X2, X3) defines cylindrical normal coordi-nates in a neighborhood of Γ0. We say that p = expΓ0(t)(X(t)) if p lies on the

spacelike geodesic from Γ0(t) with initial tangent vector X(t) ∈ Σt at arclength|X(t)|. We then assign to p the coordinates (t,X1, X2, X3), and p lies in thegeodesic hypersurface

Ht =

expX : X ∈ Σt

= exp

(Σt

).

We have that τ 7→ (t, λ1τ, λ2τ, λ3τ) are spacelike geodesics for each (λ1, λ2, λ3) ∈R3, while τ 7→ (τ, 0, 0, 0) is a timelike geodesic; c.f. note on pg. 37 on Riemanniannormal coordinates at a point. Finally note that the curves Kt constructed abovedo not lie in Ht in general.

Take the covariant derivative (now everything is evaluated on Γ0):

(3.45) ∇TX =

3∑i=1

(dXi

dt

)Ei ,

because ∇TEi = 0 (parallel transport), and

(3.46) ∇2TX =

3∑i=1

(d2Xi

dt2

)Ei .

On the other side,

(3.47) R(X,T ) · T =3∑i=1

Xi Ri0 · E0 .

Consider now the vector

(3.48) Yi = Ri0 · E0 = R(Ei, T ) · T ,which can be expanded in the basis (Eµ : µ = 1, . . . , 3):

(3.49) Yi = Y 0i E0 +

3∑j=1

Y ji Ej .

We find the coefficients by taking inner products with the orthonormal basis vec-tors, in fact

(3.50) Y 0i = −g(Yi, E0) , and Y j

i = g(Yi, Ej) .

Recall the definition of the curvature tensor :

(3.51) R(W,Z,X, Y ) = g(W,R(X,Y ) · Z) .


Thus,

(3.52) Y 0i = −R(T, T,Ei, T ) = 0 .

Remark. Here we have used the symmetry property

(3.53) R(X,Y ) · Z = −R(Y,X) · Z ,which is clear from the definition (3.39). More properties of the curvature tensorare discussed in the note on page 39.

Furthermore,

(3.54) Y ji = R(Ej , T, Ei, T ) .

Remark. Note that the expression (3.54) is symmetric in (i, j) by the pairsymmetry of the curvature tensor; c.f. Notes on page 39.

The term on the right hand side of the Jacobi equation is therefore

R(X,T ) · T =3∑j=1

Xj( 3∑i=1

Ri0j0Ei

).

Thus equating the two sides of (3.40) yields

(3.55)d2Xi(t)

dt2= −

3∑j=1

Ri0j0Xj(t) .

3.2.2. Einstein vacuum equation. In conlusion we have shown that in a paral-lelly propagated orthonormal frame (E1, E2, E3) along the reference geodesic Γ0,i.e. ∇TEi = 0, the components of the displacement vector X(t) =

∑iX

i(t)Ei(t)to a nearby timelike geodesic satisfy the equation

(3.56)d2Xi(t)

dt2= −

3∑j=1

Mij(t)Xj(t) ,

where M is the symmetric matrix

(3.57) Mij = Ri0j0 ,

or indeed the matrix valued function M(t) along Γ0:

(3.58) Mij(t) = R(Ei, T, Ej , T )∣∣Γ0(t)

.

Thus the Jacobi equation (3.56) is identical in form to the Newtoniantidal equation (3.3). In the Newtonian theory we have

(3.59) trM = ∆ψ = 0 : in vacuum .

The vacuum equations of the relativistic theory (of general relativity) should there-fore be

(3.60) trM = 0 .


3.2.3. First variation equation. We conclude our comparison of the tidal equa-tion to the Jacobi equation with a discussion of velocities.

Recall that we have extended T from O along K0 by parallel transport. Thiscorresponds to vanishing initial velocity,

(3.61) 0 = ∇XT |O = ∇TX|O =

3∑i=1

dXi

dt(0) Ei =⇒ dXi

dt

∣∣∣t=0

= 0 : i = 1, 2, 3 .

The curve K0 here lies in the geodesic spacelike hypersurface H0. This is a specialcase of a orthogonal hypersurface to Γ0. Let now more generally Γ0 be a timelikegeodesic normal to a spacelike hypersurface M (not necessarily geodesic). Thenatural way to extend T from O to M in this case is to define T at every pointp ∈M to be the unit future-directed (timelike) normal to M at p. Then let Γp bethe timelike geodesic normal to M at p.

We define the second fundamental form k of a spacelike hypersurface M withunit (future-directed timelike) normal T to be the bilinear form in TpM at eachp ∈M defined by:

(3.62) k(X,Y ) = g(∇XT, Y ) : ∀X,Y ∈ TpM

Remark. In fact, k is symmetric. So k is a quadratic form in TpM at eachp ∈M . To show this, extend X, Y to vectorfields on M tangential to M , then:

k(X,Y ) = −g(T,∇XY ) , since g(T, Y ) = 0 .

Similarly with X and Y interchanged, so taking the difference,

k(X,Y )− k(Y,X) = −g(T,∇XY −∇YX) = −g(T, [X,Y ]) = 0 ,

since [X,Y ] is also tangential to M .

We define a linear transformation k] in TpM at each p ∈M by:

(3.63) g(k] ·X,Y ) = k(X,Y ) : ∀X,Y ∈ TpM ,

in other words

(3.64) k] ·X = ∇XT .In particular at O ∈M :

(3.65) ∇TX|O = ∇XT |O = k] ·X|Ois the initial velocity at t = 0. With respect to the orthonormal basis (Ei : i =1, 2, 3) we define

(3.66) kij = k(Ei, Ej) .

Then

(3.67) k] · Ei =3∑j=1

kijEj ,


since g(k]Ei, Ej) = kij , and we obtain

(3.68) k] ·X|O =

3∑i,j=1

kijXiEj |O .

Therefore

(3.69)dXi

dt

∣∣∣t=0

=3∑j=1

kij |O Xj |t=0 .

We now define – similarly to Kt above – for each t (at least in a neighborhood ofO) a spacelike hypersurface Mt by

(3.70) Mt =

Γp(t) : p ∈M

The vectorfield T extends to spacetime (at least a neighborhood of Γ0) to bethe tangent vector field to the timelike geodesic congruence Γp. Then T remainsorthogonal to each spacelike hypersurface Mt.

Remark. We have already made this observation in the proposition on pg. 30,for its proof only relies on the fact that the geodesic tangent field T is orthogonalto M initially. Indeed, given any vectorfield X on M tangential to M , we denoteby φt the flow generated by T (i.e. φt(p) = Γp(t) for all p ∈M), and define

(3.71) Xq = dφt ·Xp , where q ∈Mt is given by q = φt(p), p ∈M.

The vectorfield X extended this way to spacetime is tangential to each Mt andsatisfies [T,X] = 0. Then

(3.72) T(g(T,X)

)= g(∇TT,X) + g(T,∇TX) =

1

2X(g(T, T )

)= 0 ,

which yields that if g(T,X) = 0 initially then it vanishes on each Mt, thus inspacetime.

Since T is the unit future-directed timelike normal to each Mt we have thatat each t ≥ 0:

(3.73)dXi

dt(t) =

3∑j=1

kij(t)Xj(t) ,

where kij(t) are the components of the second fundamental form of Mt at thepoint Γ0(t) with respect to the parallel transported frame (E1, E2, E3). Recallnow from our discussion of the tidal equation, the dependence on the initial databeing linear, we can write

(3.74) Xi(t) =3∑j=1

Aij(t)Xj(0) ,


r q

Np

p

TpM

M

Figure 21. Polar normal coordinates.

and the deformation matrix A satisfies the same equation as in the Newtoniantheory; c.f. (3.8):

(3.75)d2A

dt2= −MA.

Recall the definition of the rate of strain matrix

(3.76) θ =dA

dtA−1 .

We have, from (3.74) and (3.75):

(3.77)dXi

dt(t) =

3∑j=1

dAijdt

(t)Xj(0) =3∑j=1

θij(t)Xj(t) .

The comparison to (3.73) yields

(3.78) θij = kij

which is also called the first variation equation.

Introductory remarks on normal coordinates and curvature

Consider a (n-dimensional) Riemannian manifold (M, γ). Let us fix a point p ∈ M andconsider all tangent vectors Np of unit length |Np|2 = γ(Np, Np) = 1 at p:

Sn−1p =

Np ∈ TpM : |Np|2 = 1

Any vector Xp ∈ TpM can written as Xp = r Np, for some r ≥ 0 and Np ∈ Sn−1

p . EachXp defines a geodesic, Xp being the initial velocity at p. Recall the velocity of a geodesicis constant. We assign to a point q in a neighborhood of p the polar normal coordinates(r,Np) with origin at p if q is the point at parameter value 1 along the geodesic fromp with initial velocity Xp = rNp. The point q is therefore defined only by Xp. (Seealso figure 21.) To obtain rectangular normal coordinates we choose an orthonormal basis(E1, . . . , En) for TpM. Then any Xp ∈ TpM has a unique expansion

Xp =

n∑i=1

xiEi .


Then we assign to q the rectangular coordinates (x1, . . . , xn). Note that

r = |Xp| =√γ(Xp, Xp) =

√∑i,j

δijxixj =

√∑i

(xi)2 .

Hence the distance from the origin is the same expression as in Euclidean geometry.We now consider the case where M is 2-dimensional and introduce the concept of

Gauss curvature. In two dimensions Np is an element of the unit circle and characterizedby a single angle ϕ; the polar normal coordinates are thus simply (r, ϕ). The Lemma ofGauss says that the geodesic rays corresponding to the lines of constant ϕ are orthogonalto the geodesic circles Sr: the set of all points which are at distance r from a the point p.The metric takes the form:

ds2 = dr2 +R2(r)dϕ2 .

We see that R(r)dϕ is the element of arc length along the geodesic circles Sr. Thecondition that the metric is locally Euclidean at p is simply

R(r)

r→ 1 as r → 0 or

∂R

∂r(0, ϕ) = 1 .

The Gauss curvature K(p) is then defined in polar normal coordinates by

∂2R

∂r2+KR = 0 .

Alternatively consider the circumference C(r) of the geodesic circle Sr:

C(r) =

∫ 2π

0

R(r, ϕ)dϕ .

Then another definition of the Gauss curvature is given by

2π − d

dr=

∫Dr

K =

∫ r

0

∫ 2π

0

KR drdϕ ,

where Dr is the geodesic disc of radius r: the set of all points at distance less or equal rfrom p. We can expand the right hand side to leading order,∫

Dr

K = Kp πr2 + higher order terms,

and on the left hand side similary,

C(r) = 2π(r +

a3

6r3)

+ higher order terms.

Therefore the Gauss curvature appears as the coefficient to the third order correction tothe circumference,

Kp = −a3 .

Finally let us mention Gauss’ theorem for a geodesic triangle: If α, β, γ are the innerangles of a geodesic triangle enclosing the area T ⊂M (see figure 22) then

α+ β + γ − π =

∫T

K .

The generalisation thereof is the Gauss-Bonnet theorem for closed surfaces S:∫S

K = 2πχ ,

where χ = 2(1− g) is the Euler characteristic, and g the genus of S (g = 0 for a sphere,g = 1 for a torus).


α

β

γ

T

Figure 22. A geodesic triangle.

Notes on connections and the curvature tensor in general relativity

Recall that the definition of curvature is equivalent to

R(X,Y ) · Z = [∇X ,∇Y ]Z −∇[X,Y ]Z .

We shall see in the next Chapter that R(X,Y ) has the cyclic property

R(X,Y ) · Z +R(Y, Z) ·X +R(Z,X) · Y = 0 ,

and satisfies the Bianchi identity

(∇XR)(Y,Z) + (∇YR)(Z,X) + (∇ZR)(X,Y ) = 0 .

These identities come from the Jacobi identity for linear operators. The cyclic propertyfollows by considering vectorfields as linear operators on functions,

X : f 7−→ X · f ,while the Bianchi identity follows by considering vectorfields as linear operators on vec-torfields:

X : Y 7−→ ∇XY .Here ∇ denotes the covariant derivative, a linear operator on vectorfields with the prop-erties:

(1) ∇X(fY ) = X(f)Y + f∇XY .(2) ∇fXY = f∇XY .

In general relativity it comes from a connection which is symmetric and compatible withthe metric. The two conditions are:

(1) ∇XY −∇YX = [X,Y ] (symmetry)(2) Zg(X,Y ) = g(∇ZX,Y ) + g(X,∇ZY ) (metric compatibility)

The cyclic property and the Bianchi identity only depend on the connection being sym-metric, and do not rely on the metric property. However, metric compatibility implies theantisymmetry of the curvature tensor in the first two slots:

R(W,Z,X, Y ) = −R(Z,W,X, Y )

(Note that the antisymmetry in the last two slots is an immediate consequence of thedefinition above.) Recall here the definition of the Riemann curvature tensor, a fullycovariant tensor of fourth rank:

R(W,Z,X, Y ) = g(W,R(X,Y ) · Z) .

We can interpret this antisymmetry of the curvature tensor as follows:


The curvature Rp at a point p in the spacetime M is an antisymmetric bilinear formin the tangent space TpM toM at p with values in the space of linear transformations ofTpM. Let use denote in general by L(U, V ) the space of linear maps from a vector spaceU to a vector space V . So if Xp, Yp ∈ TpM then

Rp(Xp, Yp) ∈ L(TpM,TpM)

depends linearly on Xp, Yp, and Rp(Xp, Yp) = −Rp(Yp, Xp). With Xp, Yp both fixedRp(Xp, Yp) · Zp is an element of TpM depending linearly on Zp. The curvature R of thespacetime (M, g) is an assignment p 7→ Rp at each p ∈M. If X, Y , Z are vectorfields onM, then R(X,Y )Z is the vectorfield whose value at p ∈M is Rp(Xp, Yp) · Zp.

Consider, by comparison, a 2-form ω on M, that is an assignment p 7→ ωp of anantisymmetric bilinear real valued form in TpM. Then if Π is a 2-dimensional plane inTpM, and (E1, E2) an orthonormal basis for Π, and (X1, X2) any pair of vectors in Π,we have

Xi = X1i E1 +X2

i E2 : i = 1, 2 ,

and hence

ω(X1, X2) = det[X] ω(E1, E2) , where [X] =

(X1

1 X12

X21 X2

2

).

So if (E′1, E′2) is another orthonormal basis with the same orientation, then

E′i = O1iE1 +O2

iE2 i = 1, 2 ,

where O ∈ SO(2), and it follows that

ω(E′1, E′2) = ω(E1, E2) .

Thus ω can be thought of as assigning a real number to an oriented plane, which we maydenote by ωΠ.

Similarly Rp assigns a linear transformation of TpM to each oriented plane Πp (a 2-dimensional subspace of TpM). This linear transformation can be thought of as follows:Consider an infinitesimal closed curve C on the plane Πp, (see Fig. 23); given any vectorV ∈ TpM we parallel transport V along the curve C, and let us denote by V ′ the vectorobtained upon completing the contour. If the curvature at p is not zero, then V ′ does notcoincide with V , and there is a difference ∆V = V ′ − V ; in fact

∆V

∆A= RΠp · V ,

where ∆A is the area of Πp enclosed by the curve C. The connection being metric meansthat magnitudes of vectors are preserved by parallel transport. So upon completing thecontour C we obtain a vector V ′ which is different from V but of the same magnitude|V ′| = |V |.

Here the metric g is Lorentzian, so |V ′| = g(V ′, V ′) = g(V, V ) =|V | means that V ′ isrelated to V by a Lorentz transformation; i.e. V ′ = ΛV for some Λ ∈ SO(3, 1). Moreover,since V ′ = V +(∆A)RΠ ·V we see that RΠp is an element of the Lie algebra of the Lorentzgroup.

We express this condition in an orthonormal basis (E0, E1, E2, E3) at a point p, suchthat gµν = g(Eµ, Eν) is the Minkowski metric in rectangular coordinates

[gµν ] = [ηµν ] = diag(−1, 1, 1, 1) .


Πp

p

C

V∆V

Figure 23. Illustration of the linear transformation RΠp .

Consider first the condition g(V, V ) = g(V ′, V ′). Expanding V , V ′ in the above basis, anddenoting by [Λνµ] the matrix representing Λ with respect to the above basis,

V =∑µ

V µEµ , V ′ =∑µ

V ′µEµ , ΛEα =

∑β

ΛβαEβ ,

it is straight-forward to show that the condition reads:∑µ,ν

gµνΛµαΛνβ = gαβ .

For an infinitesimal curve C the transformation Λ differs from the identity by an infini-tesimal ϑ:

Λµα = δµα + ϑµα .

Inserting into the condition above implies∑µ,ν

gµν

(ϑµαδ

νβ + ϑνβδ

µα

)= 0 , or equivalently

∑µ

gµβϑµα +

∑ν

gανϑνβ = 0 ,

which is the statement that ϑαβ =∑µ gαµϑ

µβ is antisymmetric in α, β:

ϑαβ + ϑβα = 0 .

The matrix [(RΠp)µν ] representing RΠp with respect to the basis (Eµ : µ = 0, . . . , 3)therefore has the property that∑

µ

gλµ(RΠp)µν = (RΠp)λν

is antisymmetric in λ, ν. [In fact, as we have sketched above, given any matric [Λµν ]such that [Λκν ] is skew symmetric, where Λκν =

∑µ gκµΛµν , then exp Λ is an element

of the Lorentz group SO(3, 1), (and conversely).] This is equivalent to R(W,Z,X, Y ) =−R(Z,W,X, Y ). For, in terms of the basis basis (Eµ : µ = 0, . . . , 3), the linear transfor-mation Rµν = R(Eµ, Eν) is represented as follows:

(Rµν) · Eβ =∑α

(Rµν)αβEα ,


and

Rαβµν = R(Eα, Eβ , Eµ, Eν) = g(Eα, Rµν · Eβ) = g(Eα,∑λ

(Rµν)λβEλ)

=∑λ

(Rµν)λβg(Eα, Eλ) =∑λ

gαλ(Rµν)λβ .

The condition that Rµν is an element of the Lie algebra of the Lorentz group thus reads:

Rβαµν = −Rαβµν .

3.3. Ricci curvature. We conclude this chapter with a discussion of thegeometric meaning of the equations

trM = 0 , where Mij = Ri0j0 .

3.3.1. Sectional curvature. The curvature tensor

(3.79) R(W,Z,X, Y ) = g(W,R(X,Y ) · Z)

is antisymmetric in (W,Z) as well as in (X,Y ). This property is a manifestationof the fact that the connection is compatible with the metric, (c.f. Note on pg. 39).Moreover we have the pair symmetry

(3.80) R(X,Y,W,Z) = R(W,Z,X, Y ) .

This is a consequence of the cyclic identity6:

(3.81) R(X,Y ) · Z +R(Y, Z) ·X +R(Z,X) · Y = 0 .

Let Π be an arbitrary 2-dimensional plane in TpM at a point p ∈M, and (E1, E2)an orthonormal basis for this plane. Then

(3.82) KΠ = R(E1, E2, E1, E2)

is independent of the choice of an orthonormal basis for Π7 and also independentof the orientation of Π. Indeed if (E′1, E

′2) is another orthonormal basis for Π,

E′1 = O11E1 +O2

1E2

E′2 = O12E1 +O2

2E2 ,(3.83)

then

(3.84) R(E′1, E′2, E

′1, E

′2) = (detO)2 R(E1, E2, E1, E2) .

KΠ is called the sectional curvature of the plane Π.

6We will return to the cyclic identity when we discuss the Bianchi identities in the nextchapter.

7This also means that KΠ is invariant under Lorentz transformations of Π.


3.3.2. Jacobi equation. Recall the orthonormal frame (T,E1, E2, E3) that weconstructed in our discussion of the Jacobi equation. If we set

X∣∣t=0

= E1

∣∣t=0

,

then we obtaind2X1

dt2

∣∣∣t=0

= −M11

∣∣t=0

= KΠ01

∣∣t=0

where Π01 is the plane spanned by E0 = T , and E1. In general,

(3.85) X =

3∑i=1

XiEi , |X| =√g(X,X) =

√√√√ 3∑i=1

(Xi)2 ,

also

(3.86)d2|X|dt2

∣∣∣t=0

= −KΠ |X| ,

where Π is the plane spanned by T , X.

Remark. Review here Fig. 19 on page 29. We see that the distance to anearby geodesic remains constant, decreases, or increases according as to whetherthe sectional curvature of the plane spanned by the tangent vector of the referencegeodesic and the displacement vector is zero, positive or negative respectively.

Remark. Given a plane Π ⊂ TpM, then consider the surface HΠ = exp Π.Then KΠ is in fact the Gauss curvature of the surface HΠ at the origin p.

3.3.3. Ricci curvature. Consider then

(3.87) trM =3∑i=1

Mii =3∑i=1

Ri0i0 =3∑i=1

Ki ,

where Ki = KΠi0 is the sectional curvature of the plane spanned by E0 and Ei.We define the (E0, E0) component of the Ricci curvature by

(3.88) Ric(E0, E0) =3∑i=1

Ki .

Given any vector V , if V is not unit length we define V = V|V | . Then Ric(V , V ) is

simply defined as above taking E0 = V , and

(3.89) Ric(V, V ) = |V |2 Ric(V , V ) .

This defines a symmetric bilinear form in TpM at each p ∈M by polarisation:

(3.90) Ric(U, V ) =1

4

Ric(U + V,U + V )− Ric(U − V,U − V )

We shall from now on use the notation:

(3.91) S = Ric .


Alternatively we can express the Ricci curvature in terms of the curvature trans-formation (3.39). In an arbitrary basis we have

(3.92) Sµν = Ric(Eµ, Eν) =3∑

α=0

(Rαν)αµ .

We usually denote

(3.93) (Rµν)αβ by Rαβµν .

In terms of the inverse of g we have

(3.94) (Rαν)αµ =∑β

(g−1)αβRαµβν .

Since

(3.95) Sµν =∑α

Rαµαν =∑α,β

(g−1)αβRαµβν ,

the pair symmetry Rµναβ = Rαµβν implies the symmetry

(3.96) Sνµ = Sµν .

The vacuum Einstein equations are therefore

(3.97) Sµν = 0 .

Exercises

In the following exercises we study cylindrical normal coordinates which correspond to a freelyfalling laboratory. Recall our discussion of the Jacobi equation in cylindrical normalcoordinates: Let Γ0 be a timelike reference geodesic with unit tangent vector field T ,parametrized by arclength x0. We have ∇TT = 0. In the local simultaneous space Σx0 at eachpoint along Γ0,

Σx0 =X ∈ TΓ0(x0)M : g(T,X) = 0

,

we have an orthonormal basis (E1, E2, E3) which is parallel transported along Γ0: ∇TEi = 0.To each point p in a neighborhood of Γ0 we assign cylindrical normal coordinates (x0, x1, x2, x3)if

p = expΓ0(x0)

(X), X = x1E1 + x2E2 + x2E3 ,

namely if p lies on the spacelike geodesic with initial tangent vector X at Γ0(x0) at arclength

distance r from Γ0: r = |X| =√

(x1)2 + (x2)2 + (x3)2. Then on Γ0:

T =∂

∂x0, Ei =

∂

∂xi: i = 1, 2, 3 .

1. (Expansion of the metric to first order): Show that to first order in r, themetric is equal to the Minkowski metric, i.e.

gµν(x0, x1, x2, x3) = ηµν +O(r2).

Hints: Show that as a consequence of metric compatibility

∂µgαβ = gνβΓνµα + gανΓνµβ ,

where Γ are the connection coefficients. The problem thus reduces to showingΓαµν = 0 along Γ0, which can be done as outlined in the following steps:


(1) Given any curve γ(τ) = (x0(τ), x1(τ), x2(τ), x3(τ)) in local coordinates, showthat the geodesic equation ∇γ γ = 0 is equivalent to

d2xµ

dτ2+

3∑α,β=0

Γµαβdxα

dτ

dxβ

dτ= 0 : µ = 0, 1, 2, 3 .

(2) Use the fact that the curve τ 7→ (τ, 0, 0, 0) is a geodesic (namely Γ0) to deducethat Γµ00(x0, 0, 0, 0) = 0 for all x0.

(3) Use the fact that for any vector k = (k1, k2, k3) and any x0 the curveτ 7→ (x0, k1τ, k2τ, k3τ) is a geodesic, and deduce that for all x0 and τ :

kikjΓµij(x0, k1τ, k2τ, k3τ) = 0 .

(4) Use the parallel transport equation for Ei to deduce that for all x0:Γµ0i(x

0, 0, 0, 0) = 0.

2. (Expansion of the metric to second order): Find the second order terms in theexpansion of Exercise 1 in terms of the curvature components Rµνλσ. In particular,prove:

gij = δij −1

3

3∑k,l=1

Rikjl xkxl +O(r3) : i, j = 1, 2, 3 .

Hints: Similarly to Exercise 1, show that:

∂α∂µgνλ = (∂αΓσµν)gσλ + (∂αΓσµλ)gνσ : along Γ0.

The problem therefore reduces to expressing ∂αΓσµν in terms of Rµνσλ, which can bedone as follows:

(1) Note that we have

Rµνκλ = ∂κΓµλν − ∂λΓµκν along Γ0.

Now define

Sµκλν = ∂κΓµλν + ∂λΓµνκ + ∂νΓµκλ

and derive

3 ∂κΓµνλ = Rµνκλ +Rµλκν + Sµκλν : along Γ0.

(2) Use hint (3) from Exercise 1 to show that Sµijk = 0: i, j, k = 1, 2, 3 along Γ0.

(3) Use the hints from Exercise 1 to show that ∂0Γµ00 = ∂0Γµ0i = ∂0Γµij = 0 along Γ0.

(4) Use (1) and (3) to find ∂iΓµ0j and ∂iΓ

µ00 along Γ0.

3. (Riemannian sectional curvature and Gauss curvature): Consider aRiemannian manifold M of dimension n > 2. Let Π ⊂ TpM be a 2-dimensionalplane in the tangent space at a point p ∈M. Let HΠ be the geodesic surface

HΠ = expp(Π) = expp(X) : X ∈ Π ,consisting of all geodesics starting at p with an initial tangent vector X in Π. Showthat the Gauss curvature of the surface HΠ at p is equal to the Riemannian sectionalcurvature RΠ = R(E1, E2, E1, E2), where (E1, E2) is an orthonormal basis for Π.Hints: Introduce normal coordinates (x1, . . . , xn) with origin at p, such that

HΠ = (x1, x2, x3, . . . , xn) : x3 = . . . = xn = 0 .Use the result of Exercise 1 to calculate the circumference Cr of a geodesic circleSr = (x1, x2) : (x1)2 + (x2)2 = r2 in HΠ. Finally recall from the note on pg. 37that the Gauss curvature K appears in the formula

dCrdr

= 2π − πKr2 +O(r3) .

CHAPTER 2

Einstein’s field equations ∗

In the previous chapter we have arrived at a unified theory of space, timeand gravitation according to which spacetime is a 3 + 1 dimensional manifold Mendowed with a metric g of index 1 whose curvature represents tidal gravitationalforces. The Einstein vacuum equations

(0.1) Sµν = 0 ,

were found in analogy to Poisson’s equation for the gravitational potential invacuum:

(0.2) ∆ψ = 0 .

In this chapter we discuss Einstein’s field equations in the presence of matter,which can be thought of as the analogue of

(0.3) ∆ψ = 4πµ ,

where µ is the mass density.1

Remark. A comparison to the theory of electromagnetism provided someguidance for Einstein from early on. The situation can be compared to electro-statics, where the electric potential φ satisfies

(0.4) ∆φ = −4πρ ,

and ρ is the electric charge density: the source of the electric field E = −∇φ.It is known however that moving charges create a magnetic field B, according toAmpere’s law of magnetostatics:

(0.5) ∇×B = 4πJ

Only Maxwell’s theory of electrodynamics unites the sources (ρ, J), the fields(E,B), and the potentials (φ,A) as time and space components of differentialforms on Minkowski space and constitutes a theory in agreement with specialrelativity. In the Newtonian theory the mass density µ is the source of the grav-itational field. If general relativity relates to Newtonian gravitation as Maxwell’stheory does to electrostatics, then the question is what is the source in generalrelativity? We shall see that the energy density ε will play the role of the source ingeneral relativity. In special relativity ε is a component of the energy-momentumstress tensor. While this analogy is described further in the following note, wenext discuss the energy-momentum tensor in general relativity.

1in units where Newton’s consant G = 1.

45

46 2. EINSTEIN’S FIELD EQUATIONS ∗

Notes on Maxwell’s theory

In electrostatics the electric field E is determined from the electric charge density ρ bythe system of equations

E = −∇φ(0.6)

∇ · E = 4πρ , ∆φ = −4πρ .(0.7)

These are complemented by the equations of magnetostatics for the magnetic field B,given an electric current density J (in this note we keep explicitly the speed of light c inthe equations):

∇×B = 4πJ

c(0.8)

∇ ·B = 0 ,(0.9)

The last equation of course states the absence of a magnetic charge density. In a dynamicalsituation Gauss’ law ∇× E = 0 is replaced by Faraday’s induction law:

(0.10) ∇× E +1

c

∂B

∂t= 0 ,

and Maxwell corrected Amperes law to be:

(0.11) ∇×B − 1

c

∂E

∂t= 4π

J

c.

Taking the divergence of this equation we obtain the continuity equation

(0.12)∂ρ

∂t+∇ · J = 0 ,

which expresses the conservation of charge. This is the integrability condition of theMaxwell equations.

Remark. This logic reappears in general relativity: If we introduce a source it hasto satisfy an integrability condition for the field equations.

We can introduce a vector potential A so that

(0.13) B = ∇×A ,as a general solution to (0.9). Then in view of (0.10) we have

(0.14) E = −∇φ− 1

c

∂A

∂t.

Now it turns out that Maxwell’s equations can be unified in the spacetime point ofview of special relativity: Introduce the potential 1-form

(0.15) A =

3∑µ=0

Aµdxµ ,

and consider the electromagnetic field 2-form

(0.16) F = dA ,

where d is the exterior derivative. Then

(0.17) dF = 0 .

2. EINSTEIN’S FIELD EQUATIONS ∗ 47

These are precisely the homogeneous Maxwell equations. Here, in rectangular coordinateswe set x0 = ct so time intervals become length, and the metric is

(0.18) g = −dx0 ⊗ dx0 +

3∑i=1

dxi ⊗ dxi ;

moreover

Aµ = (g−1)µνAν(0.19)

A0 = φ , Ai = Ai : the components of the vector potential.(0.20)

Remark. Given a 1-form θ then dθ is a 2-form defined by

(dθ)(X,Y ) = X(θ(Y ))− Y (θ(X))− θ([X,Y ]) .

Similarly, if ω is a 2-form then dω is a 3-form given by

dω(X,Y, Z) = X(ω(Y,Z)) + Y (ω(Z,X)) + Z(ω(X,Y ))

− ω([X,Y ], Z)− ω([Y,Z], X)− ω([Z,X], Y )

Note if ω = dθ then dω = 0.

We have

F =

3∑µ,ν=0

Fµνdxµ ⊗ dxν(0.21)

Fµν = ∂µAν − ∂νAµ ,(0.22)

(to see this simply take X = ∂∂xµ and Y = ∂

∂xν in F (X,Y ) = dA(X,Y ) and note that[X,Y ] = 0 in rectangular coordinates); and moreover

Fi0 = ∂iA0 − ∂0Ai = − ∂φ∂xi− 1

c

∂Ai

∂t= Ei(0.23)

Fij = ∂iAj − ∂jAi =∂Aj

∂xi− ∂Ai

∂xj= εijk(∇×A)k = εijkBk .(0.24)

Now, the inhomogeneous Maxwell equations are

(0.25) ∇νFµν = 4πJµ

c; where J0 = cρ , J i = J i : electric charge density.

F being antisymmetric the divergence of the left hand side vanishes identically:

(0.26) ∇µ∇νFµν = 0 .

The integrability condition for Maxwell’s equations is therefore:

∇µJµ = 0(0.27)

∂J0

∂x0+

3∑i=1

∂J i

∂xi=∂ρ

∂t+∇ · J = 0 : in rectangular coordinates.(0.28)

In this way space and time components are united in this theory: F unites (E,B), Aunites (φ,A), and J unites (ρ, J). The spacetime current Jµ acts as a source for Maxwell’sequations. We compare the sources:

Electrostatics: ρ Newtonian gravitation: µMaxwell’s theory: Jµ General relativity: ?


We shall see that the energy density ε will play the role of the source in general relativity;(in a quasi-Newtonian situation the leading term is µc2). In special relativity ε is thecomponent T 00 of a contravariant tensorfield Tµν , then energy momentum stress tensor:

T 00 = ε : energy densityT 0i = pi : momentum densityT ij : stress.

In fact, in Maxwell’s theory:

ε =1

4π

1

2

(|E|2 + |B|2

)pi =

1

4π(E ×B)i

T ij =1

4π

(−EiEj −BiBj

)+

1

4πδij

1

2

(|E|2 + |B|2

).

(0.29)

1. Einstein’s field equations in the presence of matter

1.1. Energy-momentum-stress tensor. The energy-momentum-stress ten-sor T is a 2-contravariant symmetric tensorfield onM (a quadratic form in T∗xM,the cotangent space at each x ∈ M) that vanishes in vacuum. In componentsrelative to an arbitrary frame

(1.1) T =

3∑µ,ν=0

TµνEµ ⊗ Eν , Tµν = T νµ .

There is a corresponding 1-covariant-1-contravariant tensorfield with components

Tµλ =∑ν

Tµνgνλ ,(1.2)

so (−T )Eλ = −∑µ

TµλEµ .(1.3)

Consider the hyperboloid H+x of future-directed unit timelike vectors u at x. Such

a vector u is the 4-velocity of an observer at x. Then

(1.4) −T · u ∈ TxMis the energy-momentum density of matter relative to the observer with 4-velocityu at x. We have the following physical requirement : This energy-momentumdensity is a non-spacelike future-directed vector.

Dominant energy condition: The linear transformation −T maps thefuture hyperboloid H+

x into the closure of the open future cone at x, foreach x ∈M. (See Fig. 1.)

We decompose −T · u into a component colinear to u, εu, and a componentorthogonal to u, p, lying in Σu, the local simultaneous space of the observer with4-velocity u:

(1.5) −T · u = εu+ p , p ∈ Σu = X : g(u,X) = 0 .

1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 49

u

−T · u

x

H+x

Figure 1. Dominant energy condition.

ε is called the energy density of matter at x relative to the observer with 4-velocityu at x, and p is called the momentum density of matter at x relative to the sameobserver. We have

(1.6) ε = g(T · u, u) =∑µ,κ,λ

gµκTµλu

λuκ =∑κ,λ

Tκλuκuλ

where

(1.7) Tκλ =∑µ

gµκTµλ =

∑µ,ν

gµκgνλTµν

are the components of a quadratic form in TxM at each x ∈ M, a 2-covariantsymmetric tensorfield on M. We then write:

(1.8) ε = T (u, u) .

The dominant energy condition reads:

(1.9) |p| ≤ ε for every u ∈ H+x .

In particular, ε ≥ 0.1.1.1. Proper energy density. We minimize ε with respect to u, defining the

proper energy density

(1.10) ρ(x) = infu∈H+

x

T (u, u) .

Note that H+x is non-compact, and consequently the infimum is not necessarily

achieved.

Remark. This is in contrast to its analogue in Euclidean space, the unitsphere which is compact. Indeed, consider a quadratic form S on the vector spaceEO of all displacements from an origin O in Euclidean space. We then consider

e = infu∈SO

S(u, u) ,

where SO is the unit sphere in EO: SO = u ∈ EO : |u| = 1. This is the lowesteigenvalue of S. Here, since the unit sphere is compact, the infimum is achievedat some u1 ∈ SO, and also at −u1, i.e. at some pair of antipodal points in SO;however, uniqueness may not hold. The infimum may be achieved on a great


circle, when the first two eigenvalues coincide, or even on the whole sphere, if allthree eigenvalues coincide. In the last case

S(u, u) = e |u|2 : for all u ∈ EO ,i.e. S is then proportional to the Euclidean inner product.

In other words, a minimizing sequence, i.e. a sequence (un) ⊂ H+x such that

T (un, un) → ρ as n → ∞, may run off to infinity in H+x . Then the vectors un

would approach a generator of the future-directed null cone at x. Moreover, evenif the infimum is achieved at some u ∈ H+

x , it may not be unique.

Example: Let K be a future-directed null vector and

T = K ⊗K ,

i.e. Tµν = KµKν . Then ρ = 0 but the limit is not achieved. To see thiswe construct a sequence of timelike vectors un ∈ H+

x as follows: Let E0

be a timelike vector at x, and take un to lie in the plane spanned by E0

and K. There is another null vector K such that g(K,K) = −2; then

un = anK + anK ,

−1 = g(un, un) = 2 anan g(K,K) = −4 anan .

Consider a sequence such that an →∞. So

T (un, un) = (g(K,un))2 = (−2an)2 = 4a2n =

1

4a2n

→ 0 ,

and thus ρ(x) = 0, but it is not achieved.

A material continuum2 is matter whose energy-momentum-stress tensor T hasat each point x where it does not vanish the property that

(i) the infimum is achieved,(ii) it is achieved at a unique ux ∈ H+

x .

Then the vector ux is called the material velocity. (And u is a vectorfield definedon the support of the matter.) It is in particular a critical point of T (u, u) underthe constraint that u ∈ H+

x . So ux is an eigenvector of T :

(1.11) Tµνuνx = −ρ gµνuνx ,

where ρ is the corresponding eigenvalue.

Remark. We may view this as a constraint variational problem: MinimizeTµνu

µuν under the constraint that gµνuµuν = −1 (and u future-directed). By the

method of Lagrange multipliers, the condition for u to be a critical point is

Tµνuν = λgµνu

ν ,

where λ is the Lagrange multiplier, or eigenvalue. Multiplying by uµ we obtainλ = −T (u, u), and so λ = −ρ.

2also “ponderable matter”.


Example: The electromagnetic field is not a material continuum. Sincethe Faraday tensor is antisymmetric,

Fµν = −Fνµ ,there are two conjugate null eigenvectors L, L such that

FµνLν = α gµνL

ν

FµνLν = −α gµνLν

g(L,L) = −2 .

Then

F = αε+ βε⊥ ,

where ε is the area 2-form of the timelike plane Π spanned by L, andL, and ε⊥ is the area 2-form of the spacelike plane which is orthogonalto the timelike plane Π. In rectangular coordinates (t, x, y, z) at a givenpoint,

F = α dt ∧ dx+ β dy ∧ dz ,

that is the electric and magnetic fields E, B only have x-components:

E = −α ∂

∂x, B = β

∂

∂x.

As we shall see (c.f. note on page 2, in particular (0.29)), the energy-momentum-stress tensor in this situation takes the form:

(1.12) T =1

8π

α2 + β2 0 0 0

0 −α2 − β2 0 00 0 α2 + β2 00 0 0 α2 + β2

txyz

The infimum is achieved on the curve H+x ∩Π, and thus in particular not

unique. (C.f. exercises below for the details of this proof.)

1.1.2. Proper stress tensor. Given a material continuum let us now define theproper stress tensor S (not to be confused with Ricci curvature) by

(1.13) S = T − ρ u⊗ u .Then

Sµν = Tµν − ρ uµuν(1.14a)

Sµλ = Sµνgνλ = Tµν − ρ uµuλ , uλ = uνgνλ ,(1.14b)

Sκλ = gκµSµλ = Tκλ − ρ uκuλ ,(1.14c)

and we have

(1.15) Sκλuλ = Tκλu

λ + ρ uκ = −ρuκ + ρuκ = 0 .

S is a 2-covariant symmetric tensor vanishing on u, so S is in fact a tensor onΣx = u⊥x : the local simultaneous space at x. The metric gx restricted to Σx ispositive definite. So we have the standard eigenvalue problem for S on Σx. There


are 3 eigenvalues p1, p2, p3 called the principal pressures, and 3 correspondingeigenvalues E1, E2, E3, which form an orthonormal basis for Σx.

Prop 1.1. For a material continuum, the dominant energy condition is equiv-alent to:

|pi| ≤ ρ for i = 1, 2, 3 .

Proof. We show that this inequality implies the positivity condition, and weleave the converse as an exercise. Choose a basis (Eµ : µ = 0, . . . , 3) such thatE0 = u is the material velocity and Ei : i = 1, 2, 3 are the eigenvectors of theproper stress S. Let v be a timelike future-directed vector,

v = v0E0 +∑i

viEi , v0 >

√∑i

(vi)2 ;

then we need to show that −T ·v is a future directed causal vector. We decomposealso −T · v with respect to the above basis:

−T · v = εu+ p , p ∈ u⊥ .

Then

ε = g(T · v, u) = −ρ g(u, v) = ρv0 ,

and

p = −S · v = −3∑i=1

viS · Ei = −3∑i=1

vipiEi ,

and therefore

|p|2 =

3∑i=1

(vipi

)2 ≤ ρ23∑i=1

(vi)2< ρ2

(v0)2

= ε2 ,

because |pi| ≤ ρ.

Remark. The proper stress tensor S has the following physical meaning: Weknow S lives on Σu, the local simultaneous space of the material (it vanishes onu). Let Π be a plane in Σu through the origin. There is a covector N of unitlength whose null space is Π, i.e. Π = X ∈ Σ : N ·X = 0. Π is oriented, so itdivides Σu into a postive side Σ+

u , and a negative side Σ−u :

Σ+u =

X ∈ Σu : N ·X > 0

, Σ−u =

X ∈ Σu : N ·X < 0

.

The vector

S ·N ∈ Σu

represents the force per unit area exerted by the material across the surface withtangent plane Π.


1.1.3. Perfect fluids. A special case of particular interest is if all principalpressures are equal:

(1.16) p1 = p2 = p3 = p .

Then the material is called a perfect fluid, and the common value of pi : i = 1, 2, 3is called the pressure p. Moreover, in that case

(1.17) S∣∣Σx

= p g∣∣Σx.

If X, Y are arbitrary vectors at x ∈M, then

(1.18) S(X,Y ) = S(ΠX,ΠY ) = p g(ΠX,ΠY ) ,

where Π is the orthogonal projection to Σx. In terms of components in an arbitraryframe (Eµ : µ = 0, 1, 2, 3) we have

X =∑µ

XµEµ , ΠX =∑µ

(ΠX)µEµ ,(1.19a)

ΠX = X + g(u,X)u , g(u, u) = −1 ,(1.19b)

(ΠX)µ = ΠµνX

ν , Πµν = δµν + uµuν ,(1.19c)

and it follows that

(1.20) Sµν = p(gµν + uµuν

).

Thus

(1.21) Tµν = ρ uµuν + Sµν = p gµν + (ρ+ p) uµuν

is the energy-momentum-stress tensor of a perfect fluid.

Exercises

We study the energy-momentum-stress tensor of the electromagnetic field:

Tµν =1

4π

(FµαF

αν − 1

4gµνFαβF

αβ) .Recall that in a given orthonormal frame its components in terms of the electric field E andmagnetic field B are given by (0.29).

1. (Proper energy density of an electromagnetic plane wave) Consider anelectromagnetic plane wave in Minkowski space. We assume that in stationary coordinates(t, x, y, z) the wave propagates in the positve x-direction, and that the components of theelectric field E and the magnetic field B are only a function of u = t− x (here c = 1):

E = Ey(t− x)∂

∂y+ Ez(t− x)

∂

∂z

B = By(t− x)∂

∂y+Bz(t− x)

∂

∂z.

Moreover, we assume E and B are compactly supported in u.

(1) Use Maxwell’s equations to show that

By = −Ez, Bz = Ey.

Sketch E, B in the y-z-plane, and their support on the t-x-plane. Calculate themagnitude of the momentum vector E ×B.


(2) Recall the definition of the proper energy density (1.10). Show that ρ vanishes butthat the infimum is not attained. Give a physical interpretation.Hint: Consider a sequence (un) in Π ∩H+

p , where Π the t-x-plane (and p a point inthe support of E and B).

2. (Proper energy density of the electromagnetic field) Here we shall show that theelectromagnetic field is not a material continuum in general. To find a convenient frame, firstconsider the eigenvalue problem for the Faraday tensor F :

Fµνvν = λ gµνv

ν

Here λ ∈ C is a root of the characteristic polynomial: detA(λ) = 0, where A = (Aµν) is thematrix with components,

Aµν(λ) := Fµν − λ gµν ,formed from the electromagnetic field tensor Fµν and the metric gµν at a point. Recall thatFµν = −Fνµ. The vector v is the corresponding eigenvector, an element v ∈ C4.

(1) Show that if λ is an eigenvalue of F so is −λ and λ.There are 4 complex roots of the characteristic polynomial. Show that the followinggeneric cases are possible:

(a) Four real eigenvalues: 0 < λ1 < λ2, λ3 = −λ1, λ4 = −λ2.(b) Four imaginary eigenvalues: iµ1, iµ2, iµ3 = −iµ1, iµ4 = −iµ2, where µ1, µ2 ∈ R,

0 < µ1 < µ2.(c) Four complex eigenvalues: λ+ iµ, λ− iµ, −λ− iµ, −λ+ iµ, where λ, µ ∈ R.(d) Two real and two imaginary eigenvalues: λ, −λ, iµ, −iµ, where λ, µ ∈ R.

(2) Show that only the case (d) is possible in the physical case of 3 + 1 dimensions.Hints:

(a) Given a real eigenvalue λ the corresponding eigenvector L is also real. Showthat in the case (a) of two real eigenvalues 0 < λ1 < λ2 the correspondingeigenvectors L1, L2 satisfy

g(Li, Li) = 0, g(Li, Lj) = 0 ,

simply by using the antisymmetry of F :

F (Li, Li) = 0, F (Li, Lj) = −F (Lj , Li) .

Then deduce a contradiction.(b) To a complex eigenvalue λ corresponds a complex eigenvectors M . In (b), let

M1 = R1 + iS1 and M2 = R2 + iS2 be the complex eigenvectors correspondingto the imaginary eigenvalues iµ1, iµ2 (Ri, Si real vectors). Similarly to (i) showthat the anti-symmetry of F implies

g(Ri, Si) = 0, g(Ri, Ri) = g(Si, Si)

as well as

g(Ri, Rj) = g(Si, Sj) = 0, g(Ri, Sj) = 0.

Show that this is impossible in 3 + 1 dimensions.(c) For (c) use the fact that if M is a complex eigenvector corresponding to the

complex eigenvalue λ it follows that M is a complex eigenvector correspondingto the complex eigenvalue λ. Then proceed similary to (a) and (b) to rule outcase (c).

(3) Let F be given by

F = α dt ∧ dx+ β dy ∧ dz,

Show that this form of F corresponds to the physical case (d) above and find theeigenvalues in terms of α, β.


Verify that

L =∂

∂t+

∂

∂x, L =

∂

∂t− ∂

∂x,

M =∂

∂y+ i

∂

∂z, M =

∂

∂y− i

∂

∂z,

are the corresponding eigenvectors.Conversly, show that in the physical case (d) generically we can choose the suitableLorentz frame such that F is of the above form.

(4) Finally, verify that in the above frame (ii) the energy-momentum stress tensor takesthe form (1.12). Show that the infimum of T (u, u) is assumed on the curveu ∈ Π ∩H+ where Π is the t-x-plane, and is thus in particular not uniquely achieved.

1.2. Local energy-momentum conservation laws. We now view T as asymmetric 2-covariant tensorfield. The energy-momentum-stress tensor satisfiesthe conservation laws:

(1.22) ∇ · T = 0 .

In terms of components relative to an arbitrary frame (Eµ : µ = 0, 1, 2, 3) we have

(1.23) ∇Eµ · T = (∇µTαβ)Eα ⊗ Eβ ;

and ∇ · T is a vectorfield

(1.24) ∇ · T = (∇νTµν)Eµ .

The energy-momentum conservation laws are therefore

(1.25) ∇νTµν = 0 ,

expressed in components with respect to an arbitrary frame.

Remark. For a material continuum these are the equations of motion of thecontinuum. In particular, this encompasses Newton’s law of motion.

Remark. Note that in the case of the perfect fluid (1.21) the equations ofmotion (1.25) are underdetermined : All in all there are 4 equations for 5 unknowns(ρ, p, uµ: 3 independent components). One possibility is to study a barotropicfluid where

ρ = ρ(p) .

In general, let v be the specific volume per particle, s the specific entropy, and ethe energy per particle (this contains mc2 where m is the rest mass per particle).The mechanical properties of a fluid are specified once we stipulate the caloricequation of state:

(1.26) e = e(v, s)

According to the first law of thermodynamics we have

(1.27) de = −pdv + θ ds ,


where p is the pressure and θ the temperature. Let now be ρ = e/v the energyper unit volume, and n = 1/v the number of particles per unit volume, and definethe particle current :

(1.28) Iµ = nuµ .

Then the additional equation is the differential conservation law of particle num-ber:

(1.29) ∇µIµ = 0 .

1.2.1. Quasi-Newtonian hierarchy. We analyze the components of the energy-momentum-stress tensor in a quasi-Newtonian situation. Let (E0, E1, E2, E3) bean orthonormal frame field, E0 being the 4-velocity of a background system ofobservers in Minkowski space. We express the components of the material velocityin conventional units, (where c denotes the speed of light):

u = u0E0 +3∑i=1

uiEi , g(u, u) = −(u0)2 +3∑i=1

(ui)2 = −1 ,(1.30)

u0 =1√

1− |v|2c2

, ui =vi

c√1− |v|2

c2

(1.31)

Recall (1.14)

(1.32) Tµν = ρuµuν + Sµν ,

where the proper stress tensor S has the property (1.15) that

(1.33) Sµνuν = 0 .

In the quasi-Newtonian limit this reads:

0-component: S00u0 + S0iui = 0 −→ S00 = 0 as c→∞ ,

i-component: Si0u0 + Sijuj = 0 −→ Si0 = 0 as c→∞ ,(1.34)

because

(1.35) u0 −→ −1 , and ui −→ 0 , as c→∞ .

Thus in the non-relativistic limit Sµν reduces to Sij , and the condition (1.15)reduces to

(1.36) Sµ0 = 0 .

Here T has the units of energy density, and ρ is dominated by the contribution ofthe rest mass density µ, i.e.

(1.37) ρ ∼ µc2 as c→∞ .

We thus obtain the quasi-Newtonian hierarchy :

(1.38)

T 00 = ρ(u0)2 + S00 ∼ µc2 : T 00 = O(c2)

T 0i = ρu0ui + S0i ∼ µcvi = cpi : T 0i = O(c)

T ij = ρuiuj + Sij ∼ µvivj + Sij : T ij = O(1)


Here pi = µvi is the spatial momentum density, and µvivj is also referred to askinematic stress.

Notes on the Newtonian limit of the conservation laws

It might be instructive here to also write out the conservation laws (1.25) in the quasi-Newtonian situation (1.38). On Minkowski space (0.18), the components of (1.25) are

∂µTµν = 0 :

∂T 00

∂x0+

3∑i=1

∂T 0i

∂xi= 0 : µ = 0 ,

∂T i0

∂x0+

3∑j=1

∂T ij

∂xj= 0 : µ = i .

However, in view of (1.38) – and noting that x0 = ct – in the limit c→∞ these become:

∂µ

∂t+

3∑i=1

∂pi

∂xi= 0 ,

∂pi

∂t+

3∑j=1

∂T ij

∂xj= 0 , where T ij = µvivj + Sij .

Consider the flow lines of the non-relativistic material velocity,

dxi

dt= vi(t, x(t)) ;

here the velocity vectorfield v is assumed to be known. This defines a map, Ft(y) = x(t, y),where x(t, y) is the flow line with initial condition xi(0) = yi; here x = (x1, x2, x3),y = (y1, y2, y3). Given a “portion” Ω of the material, let us denote by Ωt the samecollection of particles in the material under time evolution, namely

Ωt = Ft(Ω) .

Let us then consider the mass M and momentum P of the material portion:

M(t) =

∫Ωt

µdV

P i(t) =

∫Ωt

pi dV

In general, for any function f on spacetime, we have:

d

dt

∫Ωt

f dV =

∫Ωt

∂f

∂tdV +

∫∂Ωt

f v · dS ,

where dS is the oriented area element of the boundary ∂Ωt, i.e. dS = NdA, where N isthe unit outward normal, and dA is the scalar area element. Thus by Gauss’ divergencetheorem

d

dt

∫Ωt

f dV =

∫Ωt

(∂f∂t

+∇ · (fv))

dV


In rectangular coordinates dV = dx1dx2dx3 and ∇ · (fv) = ∂i(fvi). Therefore:

dM

dt=

∫Ωt

∂µ∂t

+∂(µvi)

∂xi

dV = 0 ,

i.e. conservation of mass is equivalent to the first conservation law. Furthermore,

dP i

dt=

∫Ωt

∂pi∂t

+∂(pivj)

∂xj

dV ,

where pivi = µvivj is the kinematic stress. Now, by Newton’s second law, the rate ofchange of the momentum equals the force F which is exerted on the material by itssurroundings:

dP i

dt= F i .

Newton’s third law asserts in turn that this is equal and opposite the force exerted by theportion of the material on its surroudnings. So

−F i =

∫∂Ωt

SijNj dA =

∫∂Ωt

SijdSj =

∫Ωt

∂Sij

∂xjdV ,

where SijNj is i-th component of the force per unit area across ∂Ωt, and dSj = NjdA. Weconclude that the conservation of momentum corresponds to the differential conservationlaw,

∂pi

∂t+

∂

∂xj(µvivj + Sij

)= 0 .

Exercises

In these exercises we study the equations of motion of a perfect fluid. These are theconservation laws (1.25) and (1.29) for the energy-momentum-stress tensor (1.21) and theparticle current (1.28).

1. (Non-relativistic limit of the conservation laws.) We consider the equations of motionin Minkowski space and take the material velocity u to be of the form (1.31) in rectangularcoordinates. Moreover, we assume that

ρ = µc2 + h

where µ is the mass density and h is the internal energy density; c.f. (1.37).

(1) Derive in the non-relativistic limit c→∞ the conservation of mass law,

∂µ

∂t+∇ · (µv) = 0 .

Hints: Express the components of the energy-momentum-stress tensor explicitly inorders of c, and derive the conservation law from the 0-component of (1.25).

(2) Derive in the nonrelativistic limit c→∞ the energy conservation law

∂ε

∂t+∇ · f = 0 ,

where ε is the total energy density given by ε = 12µ|v|2 + h and f is the total energy

flux given by f i = (ε+ p) vi.Hints: Consider the current

Jµ = T 0µ − µc2uµ ,

and proceed similarly to (1).


2. (Adiabatic condition.) In addition to the equations of motion (1.25) and (1.28) we havethe equation of state (1.26) which expresses the energy per particle e as a function of the volumeper particle v and the entropy per particle s. We have ρ = e/v, n = 1/v, and according to thefirst law of thermodynamics the pressure p and the temperature θ are then given by (1.27).Prove the adiabatic condition

∇us = 0 ,

namely that the entropy is constant along the flow lines of the fluid.Hints: Consider the u component of (1.25), i.e. the equation uν∇µTµν = 0, and use (1.29) inconjuction with the above relations to show that uµ∇µs = 0.

3. (Pressureless fluid.) Show that if p = 0 then the integral curves of u are timelike geodesics.

1.2.2. Considerations for the field equations. Let us now return to Einstein’svacuum equations

(1.39) Sµν = 0 ,

where from now on S denotes again the Ricci curvature:

(1.40) Sµν = Rαµαν .

The component S00 = S(E0, E0) is taken with respect to a unit future-directedtangent field to a reference geodesic Γ0. Recall that Ri0j0 corresponds to (∇2ψ)ijin the Newtonian theory, whence we get the correspondence3:

S00 ∼ 4π T00(1.41)

S00 = trM = 4πµ : in the Newtonian theory.

However, in view of the quasi-Newtonian hierarchy (1.38) there is a certain am-biguity as to what the right hand side of (1.39) in the presence of matter shouldbe. In fact, if we set

(1.42) Sµν = a Tµν + b gµν trT ,

in the presence of matter, then we obtain the Newtonian theory in the limit c→∞for all a, b such that

(1.43) a+ b = 4π .

For, with respect to an orthonormal frame

(1.44) trT = (g−1)µνTµν = −T00 +3∑i=1

Tii = −T00 +O(1) ,

and so in the non-relativisitic limit:

(1.45) S00 = a T00 − b trT = (a+ b)T00 .

This ambiguity will be broken by the requirement that the conservation laws (1.22)are a consequence of the field equations. We shall illustrate that logic first withthe example of Maxwell’s theory; (c.f. note on pg. 48).

3in units where c = 1, and G = 1.


1.2.3. Conservation laws in electromagnetic theory. In Maxwell’s theory theelectromagnetic field F is a 2-form on Minkowski space M. The homogeneousequations are

dF = 0(1.46)

F = dA ,(1.47)

where A is determined from F only up to a differential of a function f , i.e. A′ isequivalent to A if

(1.48) A′ = A+ df (gauge transformations).

In an arbitrary system of local coordinates (1.47) reads

(1.49) Fµν = ∂µAν − ∂νAµ ,

and (1.46) is an equation for the cyclic sum:

(1.50) ∂µFνλ + ∂νFλµ + ∂λFµν = 0 .

The inhomogeneous equations are

∇νFµν = 4πJµ(1.51)

Fµν = (g−1)µκ(g−1)νλFκλ ,(1.52)

where Jµ denotes the electric current density.

Remark. In macroscopic media the inhomogeneous equations are replaced by

(1.53) ∇νGµν = 4πJµ ,

where Gµν is the electromagnetic displacement. Similary to F , which decomposesinto the electric, and magnetic field E, B,

(1.54) F0i = Ei , Fij = εijkBk ,

G decomposes into the electric and magnetic displacement fields D,H:

(1.55) G0i = Di , Gij = εijkHk .

Since

(1.56) ∇µ∇νFµν = 0 : identically,

the inhomogeneous equations (1.51) imply the conservation of the electric current:

(1.57) ∇µJµ = 0 ;

namely local conservation of electric charge.


1.2.4. Conservation laws and the field equations. In general relativity the equa-tions of motion are:

(1.58) ∇νTµν = 0 .

We require that these should be the consequence of the field equations of gravita-tion.

In (1.42) we have seen that

(1.59) Sµν = a(Tµν + κ (g−1)µν trT

)has the correct Newtonian limit provided

(1.60) κ =b

a, a(1 + κ) = 4π .

Taking the trace, we obtain

(1.61) trS = a(1 + 4κ) trT ,

and so

(1.62) aκ trT =κ

1 + 4κtrS .

Inserting this back into (1.59) we have equally:

(1.63) Sµν − λ (g−1)µν trS = a Tµν ,

where λ = κ1+4κ . The requirement that (1.58) are a consequence of the field

equations is therefore satisfied if and only if

(1.64) ∇ν(Sµν − λ(g−1)µν trS

)= 0 : identically.

In the next section we prove that this is satisfied if and only if

(1.65) λ =1

2, i.e. κ = −1

2.

So the Einstein equations in the presence of matter are:

(1.66) Sµν − 1

2(g−1)µν trS = a Tµν

In units where c = 1 and G = 1: a = 8π. The equations are often simply writtenas

(1.67) Eµν = a Tµν ,

where E is the Einstein tensor :

(1.68) Eµν = Sµν − 1

2(g−1)µν trS .

In rationalized gravitational units,

(1.69) 4πG = 1 , a = 2 .

1.3. Bianchi identities. A this Section we prove certain identities for thecurvature tensor, known as the Bianchi identities, which are central to Einstein’sequations.


1.3.1. Jacobi identity. Let A, B, C be linear operators acting on some (linear)space X. Then we have the Jacobi identity :

(1.70) [A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0

If X is vectorfield on M it defines two kinds of linear operators:

(1) X acts on functions f by f 7→ X · f .(2) X acts on vectorfields Y by X 7→ ∇XY .

The first kind (together with the symmetry of the connection) yields the cyclicidentity of the curvature:

(1.71) R(X,Y ) · Z +R(Y, Z) ·X +R(Z,X) · Y = 0 .

The second kind yields the Bianchi identity:

(1.72) (∇XR)(Y,Z) + (∇YR)(Z,X) + (∇ZR)(X,Y ) = 0 .

Let us first return to (1). We have

(1.73) [X, [Y, Z]]f + [Y, [Z,X]]f + [Z, [X,Y ]]f = 0 .

Using the symmetry of the connection we have

(1.74) [X, [Y,Z]] = [X,∇Y Z −∇ZY ] = ∇X∇Y Z −∇X∇ZY +∇[Z,Y ]X .

And thus,

0 = ∇X∇Y Z −∇X∇ZY +∇[Z,Y ]X

+∇Y∇ZX −∇Y∇XZ +∇[X,Z]Y

+∇Z∇XY −∇Z∇YX +∇[Y,X]Z

(1.75)

Now recalling the definition of curvature (3.39) from Chapter 1 we see that group-ing the terms as indicated this implies (1.71). Setting X = Eµ, Y = Eν , andZ = Eβ, where (Eµ : µ = 0, 1, 2, 3) is an arbitrary frame, we can also write

(1.76) Rαβµν +Rαµνβ +Rανβµ = 0 , where Rαβµν = (Rµν)αβ ;

this together with the symmetries

(1.77) Rαβνµ = −Rαβµν , Rβαµν = −Rαβµν ,implies the pair symmetry:

(1.78) Rµναβ = Rαβµν .

Now consider (2). We have

(1.79) [∇X , [∇Y ,∇Z ]]W = ∇X [∇Y ,∇Z ]W − [∇Y ,∇Z ]∇XW= ∇X

(R(Y,Z) ·W +∇[Y,Z]W

)−R(Y,Z) · ∇XW −∇[Y,Z]∇XW

= (∇XR)(Y,Z) ·W +R(∇XY,Z) ·W +R(Y,∇XZ) ·W+R(X, [Y,Z]) ·W +∇[X,[Y,Z]]W .


Therefore,

0 = (∇XR)(W,Z) ·W + (∇YR)(Z,X) ·W + (∇ZR)(X,Y ) ·W+R(∇XY,Z) ·W +R(Y,∇XZ) ·W

+R(∇Y Z,X) ·W +R(Z,∇YX) ·W+R(∇ZX,Y ) ·W +R(X,∇ZY ) ·W+R(X, [Y, Z]) ·W +R(Y, [Z,X]) ·W +R(Z, [X,Y ]) ·W+∇[X,[Y,Z]]W +∇[Y,[Z,X]]W +∇[Z,[X,Y ]]W ;

(1.80)

the last line vanishes identically by the Jacobi identity (1.70), while the remainingterms cancel as indicated. This proves the Bianchi identity (1.72). With respectto arbitrary frame Eµ : µ = 0, . . . , 3, setting X = Eµ, Y = Eν , Z = Eλ, W = Eβ,

(1.81) (∇XR)(Y,Z) ·W = (∇EµR)νλ · Eβ = ((∇µR)νλ)αβEα = (∇µRαβνλ)Eα ,

the Bianchi identity reads:

(1.82) ∇µRαβνλ +∇νRαβλµ +∇λRαβµν = 0 .

1.3.2. Contracted Bianchi identities. Setting α = ν and noting that

(1.83)∑ν

Rνβνλ = Sβλ ,

we obtain after summation

(1.84) ∇µSβλ +∇αRαβλµ −∇λSβµ = 0 ,

or

(1.85) ∇αRαβλµ = ∇λSµβ −∇µSλβthe first contracted Bianchi identity.

Remark. Note that the right hand side is antisymmetric in (λ, µ) and theequation can expressed schematically as

divR = curlS .

Now multiply the contracted Bianchi identity

(1.86) ∇αRαµνλ = ∇νSλµ −∇λSνµ by (g−1)µλ .

Note on the left hand side

(1.87) (g−1)µλRαµνλ = (g−1)αβ(g−1)µλRβµνλ = (g−1)αβSβν = Sαν ,

while on the right hand side:

(g−1)µλSλµ = trS(1.88)

(g−1)λµ∇λSνµ = ∇αSαν .(1.89)


We obtain the second contracted Bianchi identity :

(1.90) ∇µSµν −1

2∇ν trS = 0

1.3.3. Relation to Einstein equations. We have derived that

(1.91) ∇ν(Sµν − 1

2(g−1)µν trS

)= 0 : identically.

As already discussed, we must therefore set λ = 12 in (1.64) to obtain the con-

servation laws (1.25) as a consequence of Einstein equations. The latter thereforeread

(1.92) Sµν − 1

2(g−1)µν trS = a Tµν ,

where a = 2 in rationalized gravitational units.

Remark. A final remark on units: The Einstein tensor has the dimension ofcurvature, namely L−2, and the energy momentum tensor that of energy density, ormass density times c2, i.e. ML−3[c2], where c denotes the speed of light. Newton’sconstant G has dimensions T−2L3M−1 (from Newton’s law for two point masses).Hence

[G

c2] =

L

M, [

Tµν

c2] =

M

L3,

and the Einstein equations in standard units are

(1.93) Eµν =8πG

c4Tµν .

2. Action Principle

In this section we discuss the insight that the equations of the general theorycan be derived from an action principle.

2.1. General framework. We consider a functional A of the metric g andthe matter fields m, on a bounded region U in space time:

(2.1) A[g,m;U ] =

∫UL[g,m] dµg

The form L[g,m]dµg is called a Lagrangian, where L[g,m] depends only locally ong and m.

2.1.1. Matter equations. We consider a variation of the action A with respectto the matter fields mt, and denote

(2.2) m =d

dtmt

∣∣t=0

.

In other words we consider a variation of A while keeping the metric g fixed:

(2.3) A(t) = A[g,mt;U ]

Assume mt is independent of t inM\U , so m = 0 inM\U . Here U is a boundedregion in spacetime. Assume in fact that there is an open set U ′ such that U ′ ⊂ U

2. ACTION PRINCIPLE 65

so that the above holds with U replaced by U ′. So mt is independent of t inM\U ′,m = 0 in M\ U ′. Thus m = 0 in a neighborhood of the boundary of U .

The Euler-Lagrange equations of matter are:

(2.4) A =d

dtA(t)

∣∣∣t=0

= 0

For all such variations m, we can write

(2.5) A∣∣g

=

∫UM · m dµg

so the equations of matter are

(2.6) M = 0 .

2.1.2. Gravitational equations. Let us now consider a variation of the actionwith respect to the metric g while keeping the matter fields m fixed. In otherwords we consider a 1-parameter family of metrics gt agreeing onM\U ′, and set

(2.7) A(t) = A[gt,m;U ] .

Here

(2.8) g =d

dtgt

∣∣∣t=0

vanishes on M\ U ′. We can write

(2.9) A∣∣m

=

∫UG · g dµg

and the Euler-Lagrange equations of gravitation are

A∣∣m

= 0(2.10)

G = 0 .(2.11)

These should be the Einstein equations.

2.2. Gravitational action. We shall show that the Einstein equations arethe Euler-Lagrange equations of the following action. Take

(2.12) A[g,m;U ] = AG[g;U ] +AM [g,m;U ] ,

where AG is the gravitational action given by

(2.13) AG[g;U ] =

∫ULG[g] dµg , LG = −1

4trS .

Here S is the Ricci curvature of g. This observation is due to Hilbert.Consider the variation of (2.13) through a family of metrics gt, and denote as

above by

g =d

dtgt∣∣t=0

,(2.14)

AG =d

dtAG(t)

∣∣t=0

,(2.15)


where here

(2.16) AG(t) =

∫ULG[gt] dµgt .

Let us first calculate

(2.17)d

dtLG[gt]

∣∣∣t=0

.

We have

(2.18) trSt = (g−1t )µν(St)µν

where (St)µν is the Ricci curvature of gt. In local coordinates:

(2.19) Rαβµν = (Rµν)αβ

(2.20) Rµν = ∂µΓν − ∂νΓµ + ΓµΓν − ΓνΓµ

(2.21) (Γµ)αβ = Γαµβ

(2.22) Rαβµν = ∂µΓανβ − ∂νΓαµβ + ΓαµγΓγνβ − ΓανγΓγµβ

(2.23) Sµν = Rαµαν = ∂αΓανµ − ∂νΓααµ + ΓααγΓγνµ − ΓανγΓγαµ

We can now find S. Γαµβ are the components of a tensorfield Γ.

Sµν = ∂αΓανµ − ∂νΓααµ + ΓααβΓβµν + ΓααβΓβµν − ΓανβΓβαµ − ΓανβΓβµα

= ∇αΓαµν −∇νΓααµ(2.24)

Therefore,

(2.25) (trS) = (g−1) µνSµν + (g−1)µν Sµν = (g−1) µνSµν + tr S ,

where

(2.26) tr S = ∇α((g−1)µνΓαµν

)−∇ν

((g−1)µνΓααµ

).

Define the vectorfield Iα by

(2.27) Iα = (g−1)µνΓαµν − (g−1)µαΓννµ ,

then

(2.28) tr S = ∇αIα .Since also

(g−1) = −g−1gg−1(2.29)

(g−1) µν = −(g−1)µα(g−1)νβ gαβ ,(2.30)

we obtain

(2.31) (trS)˙ = ∇αIα − Sαβ gαβ .Next, we calculate

(2.32)d

dtdµgt

∣∣∣t=0

.


Here dµg is the volume form corresponding to a metric g; c.f. note on pg. 69. Since

(det g)˙ = (det g) tr(g−1g)(2.33)

(det g)˙ = (det g)(g−1)µν gµν ,(2.34)

and

(2.35) (√−det g)˙ =

√−det g

1

2(g−1)µν gµν ,

we obtain

(2.36) (dµg)˙ =1

2(g−1)µν gµν dµg .

Thus in view of (2.31) and (2.36) these calculations yield:

(2.37) AG = −1

4

∫U

∇αIα − Eαβ gαβ

dµg

Here

(2.38) Eαβ = Sαβ − 1

2(g−1)αβ trS ,

and

(2.39) Iα = (g−1)µνΓαµν − (g−1)µαΓνµν .

Finally, define the dual 3-form I∗ to the vectorfield I by

(2.40) I∗βγδ = Iα(dµg)αβγδ .

Then by Stokes theorem,

(2.41)

∫U∇αIα dµg =

∫∂UI∗ = 0 ,

since I∗ vanishes in a neighborhood of ∂U . Hence we are left with

(2.42) AG =

∫UG · g dµg ,

where G is the 2-contravariant symmetric tensorfield

(2.43) Gαβ =1

4Eαβ .

Note on the volume form corresponding to a metric

At each x ∈ M (dµg)x denotes the volume form in TxM determined by the orientationof M. If (E0, E1, E2, E3) is a positive orthonormal basis of TxM then

dµg(E0, E1, E2, E3) = 1 .

Suppose now that the coordinate basis ( ∂∂x0 ,

∂∂x1 ,

∂∂x2 ,

∂∂x3 ) is positive. We can expand

∂

∂xµ=∑α

AαµEα .


Then

gµν = g(∂

∂xµ,∂

∂xν) =

∑α,β

AαµηαβAβν ,

since g(Eα, Eβ) = ηαβ . In terms of matrices, we have

g = AηA ,

so taking determinants

det g = det Adet η detA = −(detA)2 ,

and detA > 0 since we assumed that we have a positive basis. Thus

detA =√−det g ,

and

dµg( ∂

∂x0,∂

∂x1,∂

∂x2,∂

∂x3

)= detA dµg(E0, E1, E2, E3) = detA =

√−det g .

By total antisymmetry,

dµg( ∂

∂xµ,∂

∂xν,∂

∂xκ,∂

∂xλ)

=√−det g [µνκλ] ,

where [µνκλ] is the fully antisymmetric 4-dimensional symbol.

2.3. Energy-stress tensor. In the previous section we have discussed thegravitational action:

(2.44) AG[g;U ] =

∫U−1

4trS dµg ,

We have shown that for variations of the metric g,

(2.45) AG =

∫U

1

4

(Sµν − 1

2(g−1)µν trS

)gµν dµg .

Let us now return to (2.12). The action AM contains the matter fields m:

(2.46) AM [g,m;U ] =

∫UL[g,m] dµg

In general, for variations of the matter fields m while keeping the metric g fixed,we have

(2.47) AM∣∣g

=

∫UM · m dµg .

The equations of matter are simply

(2.48) AM∣∣g

= 0 : M = 0 .

The equations of gravitation should be the variation of the action (2.12) withrespect to the metric, while keeping m fixed:

(2.49) A∣∣m

= AG + AM∣∣m

= 0

These are the Einstein equations

(2.50) Sµν − 1

2(g−1)µν trS = 2Tµν .


2.3.1. Definition. We are thus led to the definition of the energy-stress tensorTµν in the context of the action principle:

(2.51) AM∣∣m

= −1

2

∫UTµν gµν dµg

Remark. This corresponds to the principle of virtual work in classical me-chanics.

If LM [g,m](x) depends on g only through g(x) (so does not depend on thederivatives of g), then

(2.52) LM∣∣m

=∂LM∂gµν

gµν .

Recalling that

(2.53) (dµg)˙ =1

2(g−1)µν gµν dµg

we see that

(2.54) A∣∣m

=

∫U

(∂LM∂gµν

+1

2(g−1)µνLM

)gµν dµg .

Comparing to (2.51) we conclude that the definition of Tµν in this case reduces to

(2.55) Tµν = −2∂LM∂gµν

− (g−1)µνLM .

2.3.2. Conservation Laws. The question that we address next is: What is theconnection between the local energy-momentum conservation laws (1.25) and theequations of matter (2.48).

We observe that AM should be a geometric invariant. Let us make this moreprecise: Let U ⊂ M be a spacetime domain and let φ be a diffeomorphism ofU onto itself. In other words, φ is a diffeomorphism of the spacetime manifoldM onto itself such that φ(U) = U . Now (M, φ∗g) is geometrically equivalent to(M, g). But there should also be a notion of “pullback” for the matter fields mso that (M, φ∗g, φ∗m) is physically equivalent to (M, g,m).

Remark. If the matter is an electromagnetic field then m = A is the electro-magnetic potential 1-form and φ∗m = φ∗A is the standard pullback.

Physical requirement:

(2.56) AM [φ∗g, φ∗m;U ] = AM [g,m;U ] .

Consider

(2.57) AM (t) = AM [φ∗t g, φ∗tm;U ] ,

where φt is a 1-parameter group of diffeomorphisms such that φt(U) = U . Thenφt is generated by a vectorfield X, X(p) being the tangent vector to the orbitφt(p) through p. Suppose in fact that φt is the identity outside U ′, where U ′ is a


subdomain with U ′ ⊂ U ; so if p ∈ U \ U ′ then φt(p) = p, so X(p) = 0. Thus Xvanishes in a neighborhood of the boundary ∂U of U . We have by the chain rule:

(2.58)dAM

dt

∣∣∣t=0

= AM∣∣m

+ AM∣∣g,

where in the first term m = 0, and

(2.59) g =d

dtφ∗t g∣∣t=0

= −LXg ,

and in the second term g = 0, and

(2.60) m =d

dtφ∗tm

∣∣t=0

= −LXm.

Then the physical requirement

(2.61)dAM

dt

∣∣∣t=0

= 0

reads

(2.62)

∫U

−1

2Tµν (−LXg)µν +M · −LXm

dµg = 0 ,

for any vectorfield X with support in U ′. Now by Leibniz rule,

(2.63) (−LXg)(Y,Z) = X(g(Y,Z)

)− g(−LXY, Z)− g(Y,−LXZ) ,

and the connection being metric is equivalent to:

(2.64) X(g(Y,Z)

)= g(∇XY,Z) + g(Y,∇XZ) .

Moreover, for any two vectorfields X, Y , we have −LXY = [X,Y ], and since theconnection is symmetric

(2.65) ∇XY −∇YX = [X,Y ] .

Substituting these above we obtain:

(−LXg)(Y,Z) = g(∇XY,Z) + g(Y,∇XZ)− g([X,Y ], Z)− g(Y, [X,Z])

= g(∇YX,Z) + g(∇ZX,Y ) .(2.66)

Given an arbitrary frame Eα : α = 0, . . . , 3 set Y = Eµ, Z = Eν , then

(2.67) (−LXg)µν = g(∇EµX,Eν) + g(∇EνX,Eµ) .

Expanding ∇EµX = (∇µXα)Eα we obtain

(−LXg)µν = gαν∇µXα + gαµ∇νXα

= ∇µXν +∇νXµ ,(2.68)

where Xµ = gµαXα. So

(2.69) −1

2Tµν(−LXg)µν = −1

2Tµν

(∇µXν +∇νXµ) = −Tµν∇νXµ ,

because Tµν is symmetric.


Suppose we are considering a solution of the matter equations M = 0. Thenthe term M · −LXm in (2.62) vanishes and we obtain

(2.70)

∫U−Tµν∇νXµ dµg = 0 ,

for all vectorfields X with compact support in U . Integrating by parts this isequivalent to:

(2.71)

∫U

(∇νTµν

)Xµ dµg = 0 ,

for all such vectorfields X. This is finally equivalent to the local energy-momentumconservation laws

(2.72) ∇νTµν = 0 .

Remark. This is a version of Noether’s theorem. An advanced treatment ofthe action principle in partial differential equations can be found in [Chr00].

Example: The action of the electromagnetic field is given by

(2.73) AE [g,A;U ] =1

16π

∫UFµνFµν dµg

where

(2.74) F = dA , Fµν = ∂µAν − ∂νAµ .For variations with respect to the electromagnetic potential A be obtain

Fµν = ∂µAν − ∂νAµ(2.75)

AE∣∣g

=1

8π

∫UFµνFµν dµg

=1

4π

∫UFµν∇µAν dµg = − 1

4π

∫U

(∇µFµν

)Aνdµg .

(2.76)

So we obtain Maxwell’s equations (1.51) as the matter equations of thematter action

(2.77) AEM = AE +AM ,

if we define the electric current J by

(2.78) AM∣∣g

= −∫UJµAµdµg .

Finally, with the energy-stress tensor defined by (2.51) we obtain from(2.73) explicitly in terms of F :

(2.79) Tµν =1

4π

(FµαF

να − 1

4(g−1)µνFαβFαβ .

The energy-momentum conservation laws (1.25) then automatically holdas discussed above.


f

y

x

f−1(y)

NM

ux

Figure 2. Mapping f from spacetime M to material manifold N .

3. The material manifold

The material manifold is a 3-dimensional manifold N each point of whichrepresents a material particle. N is endowed with those structures which areindependent of the embedding of the material in space.

The dynamics is then described by a mapping f :M→N :4

y = f(x): the particle occupying the event x.f−1(y): the world line in M of a material point y in N .

We require the inverse image f−1(y) of a material point y to be a timelikecurve. There is then a unique future-directed timelike unit vector ux tangent tof−1(y) at x, y = f(x). This is the material velocity. (See Fig. 2.)

Consider the differential (df)x of the mapping f at x. The null space (kernel)of (df)x is the tangent line at x to the world line f−1(y). This is the linear span ofux. The local simultaneous space Σx of the material particle at x is the orthogonalcomplement (relative to gx) of ux in TxM. Then (df)x|Σx (df restricted to Σx)is a (linear) isomorphism of Σx onto TyN , y = f(x).

We also require that this isomorphism is orientation preserving. (M is orientedand time oriented; N is oriented.) The material manifold N is endowed with avolume form ω. In fact, in fluid mechanics there is no other structure on N :∫

R ω: represents the number of particles contained in the material domainR ⊂ N .

Consider the pull-back f∗ω. This is a 3-form on M. In fact, consider theexterior derivative of the 3-form f∗ω: df∗ω. This is a 4-form onM, equivalent toa function φ, df∗ω = φ dµg. However, exterior derivatives commute with pullbacksso

df∗ω = f∗dω .

But dω = 0 because there is no non-trivial 4-form on a 3-dimensional manifold.So we obtain:

(3.1) df∗ ω = 0 .

4This may not be defined on the whole spacetimeM because there could be vacuum regions.

3. THE MATERIAL MANIFOLD 73

Consider f∗ω|Σx . Since (df)x|Σx is an orientation preserving isomorphism thismust be a volume form on Σx. But Σx is already endowed with a volume form,namely dµΣx from the spacetime metric gx; this is the volume form of the positivedefinite metric gx|Σx . Thus there is a function n(x) > 0 such that

(3.2) f∗ω∣∣∣Σx

= n(x) dµΣx .

n(x): is the number of particles per unit volume in the local rest frame ofthe material.

v = 1/n: is the specific volume, or volume per particle.

3.1. Continuum Mechanics. In fluid mechanics the volume per particle vand the entropy per particle s are an element of the thermodynamic state space:

(3.3) (v, s) ∈ R+ × R+ .

Thermodynamic state space: the space of local thermodynamic equi-librium states. For a fluid this is simply R+ × R+.

State function: a function κ(v, s) on the thermodynamic state space.

We shall first discuss the isentropic case, where the entropy per particle is constantthroughout the fluid. This is followed by a more general discussion, which includeschanges in temperature.

3.1.1. Isentropic case. The Lagrangian in fluid mechanics is LM [g, f ] dµg where

(3.4) LM [g, f ](x) = κ(v(x)) ;

here v(x) = 1n(x) and s is a constant.

We shall analyze the dependence of n(x) on gx. Let us first extend the relation(3.2) defining n(x) to the whole of TxM. Note that

(3.5) (f∗ω)(ux, Xx, Yx) = 0 ,

since ux belongs to the null space of (df)x. Let us extend dµΣx to a 3-form dµΣxdefined on all of TxM by

(3.6) dµΣx(Xx, Yx, Zx) = dµΣx(ΠXx,ΠYx,ΠZx) ,

where Π denotes the orthogonal projection to Σx:

(3.7) ΠX = X + g(u,X)u .

Then we have in fact:

(3.8) dµΣx(Xx, Yx, Zx) = dµg(ux, Xx, Yx, Zx) .

The extended relation is therefore:

(3.9) (f∗ω)x = n(x) dµΣx .

The left hand side is independent of the metric. The problem thus reduces toanalyzing the dependence of dµΣx on gx. Let us work in a coordinate frame

( ∂∂xµ : µ = 0, 1, 2, 3); recall that

(3.10) dµg(∂

∂xα,∂

∂xβ,∂

∂xγ,∂

∂xδ) =

√−det g [αβγδ] .


Then we have

(3.11) dµΣx(∂

∂xβ,∂

∂xγ,∂

∂xδ) = uα [αβγδ]

√−det g .

We know the dependence of the volume element on gx, namely

(3.12) (√−det g) =

1

2

√− det g (g−1)µν gµν ,

but how does ux depend on gx? The tangent line to f−1(y) at x is independentof gx. So ux depends on gx only through the normalization condition:

(3.13) gx(ux, ux) = −1

It follows that ux is colinear to ux:

(3.14) uα = λuα

To determine λ we vary the normalization condition (3.13):

(3.15) 2 gµν uµuν + gµνu

µuν = 0 .

Then substitute (3.14) to obtain

(3.16) 2λ gµνuµuν + gµνu

µuν = 0 .

or

(3.17) λ =1

2gµνu

µuν .

Hence:

(dµΣ)βγδ = λ(dµΣ)βγδ +1

2(g−1)µν gµν(dµΣ)βγδ

=1

2

((g−1)µν + uµ uν

)gµν(dµΣ)βγδ .

(3.18)

Now varying the relation (3.9), namely

(3.19) (f∗ω)βγδ = n (dµΣ)βγδ ,

with respect to g, we have:

(f∗ω)βγδ = 0(3.20)

0 = n (dµΣ)βγδ + n (dµΣ)βγδ .(3.21)

Substituting the expression (3.18) for (dµΣ)βγδ we conclude that:

(3.22) n+1

2n(

(g−1)µν + uµuν)gµν .

Now recall our earlier expression (2.55) for the energy stress tensor:

(3.23) Tµν = −2∂LM∂gµν

− (g−1)µνLM .

Here LM = κ(v), and

(3.24)∂LM∂gµν

=∂κ

∂n

∂n

∂gµν= −1

2n∂κ

∂n

((g−1)µν + uµuν

).

3. THE MATERIAL MANIFOLD 75

Therefore

Tµν =(n∂κ

∂n− κ)

(g−1)µν + n∂κ

∂nuµuν

= κuµuν +(n∂κ

∂n− κ)(

(g−1)µν + uµuν).

(3.25)

We recognize the energy-momentum-stress tensor (1.21) of the perfect fluid

(3.26) Tµν = ρ uµuν + p(

(g−1)µν + uµuν),

with

(3.27) ρ = κ , p = n∂κ

∂n− κ .

Note,

ρ: is energy per unit volume,e = ρ/n: energy per particle.

In terms of e and v the relation

(3.28) p = n∂ρ

∂n− ρ

becomes simply

(3.29) p = −∂e∂v

.

This is the first law of thermodynamics (1.27),

(3.30) de = −pdv , when ds = 0 .

3.1.2. General case. In the more general case the state function κ is a functionof n and the entropy per particle s:

(3.31) κ = κ(n, s) .

Here s is considered to be independent of the metric g. The temperature θ isdefined by

(3.32)∂κ

∂s= n θ .

Let us recall (3.9). In an arbitrary frame,

(3.33) (f∗ω)βγδ = n (dµΣ)βγδ = nuα (dµg)αβγδ .

Therefore the 3-form f∗ω is dual to the vectorfield I,

(3.34) Iα = nuα : the particle current,

and the equation (3.1) reads in terms of I:

(3.35) df∗ω = 0 ⇐⇒ ∇µIµ = 0 .

The equations of motion of a fluid are the two conservation laws:

∇µIµ = 0 : local particle conservation law,(3.36)

∇νTµν = 0 : local energy-momentum conservation laws.(3.37)


The action functional for the fluid is:

(3.38) AM [g, f, s;U ] =

∫Uκ(n, s)dµg .

The matter equations are obtained as above from variations with repect to f ,

(3.39) AM∣∣∣g,s

=

∫UM · f dµg ,

while the temperature θ on M is defined according to (3.32) by the variation ofthe action with respect to s:

(3.40) AM∣∣∣g,f

=

∫Unθ sdµg .

We can proceed as in Section 2.3.2 to derive from the invariance of the action thecondition that

0 = AM = AM∣∣∣f,s

+ AM∣∣∣g,s

+ AM∣∣∣g,f

=

∫U

−1

2Tµν gµν +M · f + nθ s

dµg

=

∫U

∇νT ν

µ +Ma ∂µfa + nθ ∂µs

Xµ dµg

(3.41)

where we have used as before that

f = −LXf = X · f = Xµ∂µf , s = −LXs = X · s = Xµ∂µs ,(3.42)

gµν = (−LXg)µν = ∇νXν +∇νXµ .(3.43)

Therefore in view of the conservation laws (3.37),

(3.44) Ma ∂µfa + nθ ∂µs = 0 .

Since df · u = 0, multiplying by uµ this reduces to

(3.45) uµ∂µs = 0 , or ∇us = 0 .

This is the adiabatic condition, which states that the entropy is constant along theflow lines of the fluid.

3.1.3. Dust model. Let us finally consider the trivial case p = 0.Recall that according to the first law of thermodynamics the energy per particle

e, the volume per particle v, and the pressure are related by

(3.46) e = e(v, s) , p = −∂e∂v

.

Thus p = 0 is equivalent to e being just a function of s, e = m(s), m the rest massof the particle. Since the entropy per particle s is constant along the flow lines by(3.45) (as long as the flow is smooth), the particle conservation law (3.36),

(3.47) ∇µIµ = 0 , Iµ = nuµ ,

implies

(3.48) ∇µ(ρ uµ) = ∇µ(ne uµ) = nuµ∇µe+ e∇µ(nuµ) = 0 ,

4. COSMOLOGICAL CONSTANT 77

since we have

ρ = ev : energy density in the local rest frame of the fluid,

n = 1v : number of particles per unit volume, where v is volume per particle.

Moreover, the energy-momentum-stress tensor reduces to:

(3.49) Tµν = ρ uµ uν ,

so the energy-momentum conservation laws immediately imply,

(3.50) ∇νTµν = (∇νuµ)ρ uν + uµ∇ν(ρ uν) = ρ uν∇νuµ = 0 .

This is equivalent to the geodesic equation:

(3.51) ∇uu = 0 .

The flow lines of a pressureless fluid are timelike geodesics.

Remark. This proves in a trivial case the geodesic hypothesis of Chapter 1,Section 3.2.

4. Cosmological constant

An important modification of the Einstein equations was proposed in 1917which amounts to adding a cosmological term:

(4.1) Sµν −1

2gµν trS + Λ gµν = 2Tµν ,

where Λ is the cosmological constant.

Remark. Observe that the conservation laws (1.25) still hold by virtue of theBianchi identities and metric compatibility.

Remark. Note that in the absence of matter (4.1) reduces to

(4.2) Sµν = Λ gµν .

Exercises

On a domain of Minkowski space (R3+1, η) consider the conformally flat metric

gµν = Ω2ηµν , where Ω(x) =1

1 + c4〈x, x〉 , 〈x, x〉 = −(x0)2 + (x1)2 + (x2)3 + (x3)2 ,

and c is a constant.

(1) Show that the Riemann curvature tensor takes the form

Rαµβν = c(gαβgµν − gανgβµ

).

(2) Show that every spacelike plane Π has sectional curvature KΠ = c.(3) Show that every timelike plane Π has sectional curvature KΠ = −c.(4) Consider the case c = 1. Find and sketch the domain in R3+1 where the metric g is

defined. Show that the x0-axis is a geodesic and prove that the point (2, 0, 0, 0) is notreached in finite proper time.


(5) Consider the case c = −1. Again, find and sketch the domain in R3+1 where themetric is defined. Verify that the x0-axis is a geodesic. Show that the past of thepoint with coordinates (2, 0, 0, 0) is the whole hyperplane x0 = 0.Now consider the points p with coordinates (2 + ε, 0, 0, 0) and q with coordinates(−2− ε, 0, 0, 0), for any ε > 0. Show that q and p can be connected by a curve witharbitrarily large arc length.

Remark. The spacetime discussed in (4) is called de Sitter space and the one in (5) anti deSitter (AdS).

CHAPTER 3

Spherical Symmetry

In Section 1 we impose spherical symmetry and thus reduce the Einstein equa-tions,

(0.1) Sµν −1

2gµν trS = 2Tµν ,

to a system of partial differential equations on a 1 + 1-dimensional quotient mani-fold; our exposition here follows [Chr95]. In Section 2 we discuss the Schwarzschildspacetime as the only solution to the vacuum equations. In Sections 3, 4 we study(in the presence of matter) the concepts and questions that emerge in Section 2— in particular the notion of a black hole and the issue of singularities — froman evolutionary point of view. This being only a brief introduction to the studyof spherically symmetric self-gravitating systems in general relativity, I encouragethe reader to consult the original references [Chr95, Chr99b, Daf05].

1. Einstein’s field equations in spherical symmetry

1.1. Spacetime manifold M. A spacetime is spherically symmetric if therotation group SO(3) acts on the spacetime manifold (M, g) by spacelike isome-tries.

The group orbits are then metric 2-spheres (each group orbit is diffeomorphicto S2). Moreover there is an orthogonal family of timelike hypersurfaces whichintersect each sphere exactly once and which are totally geodesic (that is a geodesicof the hypersurface is a geodesic of the spacetime).

Consider on one hand the map determined by the group orbits from one ofthese hypersurfaces H onto another H ′: a point p ∈ H is mapped to a pointp′ ∈ H ′ if and only if p and p′ lie on the same group orbit S; c.f. Fig. 1. Thismapping is an isometry.

On the other hand consider the map of one of the group orbits S onto anotherS′ determined by the orthogonal hypersurfaces: q ∈ S is mapped to q′ ∈ S′ if andonly if q, q′ lie on the same timelike hypersurface H. This mapping is a conformalisometry, in fact the scale factor is constant.

The quotient space is the space of group orbits:

(1.1) Q =M/SO(3) .

In general Q is a 1+1-dimensional manifold with boundary. The boundary of Q,which we denote by Γ, corresponds to the set of fixed points of the group action.It is a timelike geodesic in M.

79

80 3. SPHERICAL SYMMETRY

H ′

H

p

p′

S

H

S

S′

q

q′

Figure 1. Mapping determined by group orbits, and mapping de-termined by orthogonal hypersurfaces, respectively.

The Lorentzian manifold Q is then endowed with a metricQg, and (Q, Qg) is

represented by any one of the isometric timelike hypersurfaces H. The spacetime

metricMg then takes the form

(1.2)Mg=

Qg +

Sg ,

and

(1.3)Sg= r2(x)

γ , (x ∈ Q) ,

where r is a function on Q, andγ is the metric of the standard unit sphere S2 in

R3. In local coordinates (xa : a = 0, 1) on Q, and local coordinates (yA : A = 2, 3)on S2 we have

(1.4)Mg=

Qgab (x)dxadxb + r2(x)

γAB (y)dyAdyB .

Then

(1.5) Area (S) = r2 Area (S2) = 4π r2 .

Note that r vanishes on Γ, the center of symmetry.1

The non-vanishing connection coefficients are Γabc, ΓABC ,

ΓaBC = −r (g−1)ab∂brγBC ,(1.6a)

ΓBaC =1

r∂ar δBC .(1.6b)

Since the Ricci curvature of (S2,γ) is simply

γ,

(1.7)S=

γ ,

1We assume r : Q → R is a smooth function with r > 0 in the interior ofQ, and g(∇r,∇r) > 0near Γ.

1. EINSTEIN’S FIELD EQUATIONS IN SPHERICAL SYMMETRY 81

and the Ricci curvature of (Q, g) is Kg, where K is the Gauss curvature of (Q, g),

(1.8)QS= Kg ,

we readily verify that the components of the Ricci curvature of M are

MSab = K gab −

2

r∇a∇br(1.9a)

MSaA = 0(1.9b)

MSAB =

(1−∇a(r∇ar)

) γAB .(1.9c)

In view of the Einstein equations (0.1) the energy-momentum-stress tensor isalso spherically symmetric, and necessarily takes has the form:

(1.10) Tab(x)dxadxb + r2(x)S(x)γAB (y)dyAdyB ,

where S is a function on Q. Thus the Einstein equations read:

Kgab −2

r∇a∇br = 2

(Tab −

1

2gab(trT + 2S)

)(1.11a)

1−∇a(r∇ar) = −r2 trT .(1.11b)

This is a non-linear wave equation for r on Q. The trace of (1.11a) is

(1.12) K − 1

rr = −2S . ( = ∇a∇a)

Substituting for r from (1.11b) we obtain

(1.13) K =1

r2

(1− ∂ar ∂ar

)+ trT − 2S .

We substitute this in the first equation to obtain the Hessian equations:

(1.14) ∇a∇br =1

2r

(1− (∂ar)(∂ar)

)gab − r

(Tab − gab trT

).

The system (1.13), (1.14) is equivalent to (1.11a), (1.11b).In spherical symmetry the energy-momentum conservation laws reduce to:

(1.15) ∇b(r2 T ab

)= 2r ∂ar S .

This follows easily from

(1.16) ∇νT aν = ∇bT ab +∇AT aA = ∇bT ab + ΓaAµTµA + ΓAAµT

aµ = 0 ,

using (1.6) and (1.10).

Remark. In fact, one can show that the Hessian equations (1.14) in conjuctionwith the energy-momentum conservation laws (1.15) imply the equation for theGauss curvature (1.13). Also the trace of the Hessian equations is (1.11b).

We conclude that the Hessian equations (1.14) together with the energy-momentum conservation laws (1.15) contain all the information in the Einsteinequations (0.1) in spherical symmetry.


Figure 2. Mass content of a sphere.

1.2. Mass function. We define the mass function m on Q by:

(1.17) 1− 2m

r= (g−1)ab ∂ar ∂br .

Remark. m(x) represents the mass-energy content of the sphere (a grouporbit) corresponding to the point x ∈ Q.

Differentiate m and use the Hessian equations to obtain the mass equations:

(1.18) ∂am = r2(Tab − gab trT

)∂br .

Note that S does not enter.

Remark. It turns out that the mass equations together with the trace of theHessian equations form a system equivalent to the original Hessian equations.

Also, in view of (1.17) we can rewrite (1.13) as

(1.19) K =2m

r3+ trT − 2S .

1.3. Null coordinates on Q. We choose a function u whose level curves areoutgoing null curves, and which is increasing towards the future. The function uis then determined up to a transformation of the form

(1.20) u 7→ f(u) ,

where f is an increasing function. We also choose a function v whose level curvesare incoming null curves, and increasing towards the future; then similarly v isdefined up to a transformation

(1.21) v 7→ g(v) ,

where g is an increasing function. We then define:

(1.22) L = 2∂

∂v, L = 2

∂

∂u.

Thus we are using v as a parameter along the outgoing null curves, u as a parameteralong the incoming null curves. L, L are future-directed null vectorfields, outgoingand incoming, respectively. Then set

(1.23) g(L,L) = −2 Ω2 ;

this defines a function Ω > 0 on Q. The metric on Q in u, v coordinates takes theform:2

(1.24)Qg= −Ω2 dudv .

2We assume here and in the following that the coordinates u, v exist in fact globally on Q.This rules out cases where Q is e.g. a cylinder.

1. EINSTEIN’S FIELD EQUATIONS IN SPHERICAL SYMMETRY 83

The transformations (1.20), (1.21) thus correspond to conformal transformations.

Remark. We can thus assume that the image of (u, v) is a bounded subsetof R1+1 (by applying another conformal transformation).

Note

(1.25) guv = −1

2Ω2 .

The only non-vanishing connection coefficients ofQg are:

(1.26) Γuuu =2

Ω

∂Ω

∂u, Γvvv =

2

Ω

∂Ω

∂v.

In null coordinates the Gauss curvature of Q is given by:

(1.27) K =4

Ω2

∂2 log Ω

∂u ∂v

The Hessian equations (1.14) in null coordinates read:

∂2r

∂u2− 2

Ω

∂Ω

∂u

∂r

∂u= −r Tuu(1.28a)

∂2r

∂u ∂v+

1

r

∂r

∂u

∂r

∂v= −Ω2

4r+ r Tuv(1.28b)

∂2r

∂v2− 2

Ω

∂Ω

∂v

∂r

∂v= −r Tvv .(1.28c)

The defining equation of the mass function (1.17) becomes

(1.29) 1− 2m

r= − 4

Ω2

∂r

∂u

∂r

∂v,

and the mass equations are

∂m

∂u=

2r2

Ω2

(Tuv

∂r

∂u− Tuu

∂r

∂v

)(1.30a)

∂m

∂v=

2r2

Ω2

(Tuv

∂r

∂v− Tvv

∂r

∂u

).(1.30b)

Also note the positivity condition on the energy-momentum-stress tensor implies

(1.31) Tuu ≥ 0 , Tuv ≥ 0 , Tvv ≥ 0 .

Indeed, since the positivity condition is equivalent to T (X,Y ) ≥ 0 for any pair offuture directed non-spacelike vectors X, Y at a point, we take (X,Y ) to be (L,L),(L,L), (L,L) and the conditions (1.31) immediately follow.

Remark. Note that (1.28a) (which is the (u, u) component of the Einsteinequations (1.11a)) can also be written as:

(1.32) ∂u

( 1

Ω2

∂r

∂u

)= − r

Ω2Tuu .

Since by (1.31) the right hand side is negative this implies the convexity of r.The equation (1.32) and its analogue for the (v, v) component are also known asRaychaudhuri equations.


2. Schwarzschild solution

The Schwarzschild solution is a spherically symmetric solution of the vacuumequations.3

Let us now assume that Tab = 0. Note that the conservation laws (1.15) implyS = 0. Since the right hand side of the mass equations (1.18) vanishes, m is aconstant. We take m > 0. 4

The (u, v) component (or trace) of the Hessian equations becomes

(2.1)∂2r

∂u ∂v+

1

r

∂r

∂u

∂r

∂v= −Ω2

4r.

Solving (1.29) for Ω2 we substitute in (2.1) to obtain:

(2.2) r∂2r

∂u ∂v=

2m

r − 2m

∂r

∂u

∂r

∂v.

We now define a new unknown r∗ by:

(2.3) r∗ = r + 2m log∣∣ r2m− 1∣∣ ,

so that

(2.4)dr∗

dr=

1

1− 2mr

.

Then the equation satisfied by r becomes in terms of the new unknown r∗:

(2.5)∂2r∗

∂u ∂v= 0 .

The general solution is given by:

(2.6) r∗ = f(u) + g(v) .

Given r∗, if r∗ > 0 there is a unique value of r corresponding to r∗. However, ifr∗ < 0 there are two values of r corresponding to r∗, one greater and one less than2m; see Fig. 3.

We require that the representation in terms of null coordinates be such thatr∗ = −∞ — which corresponds to r = 2m — is contained in the uv plane. Tosatisfy this requirement we take

f(u) = 2m log|u|(2.7a)

g(v) = 2m log|v|(2.7b)

so that

(2.8) f(u) + g(v) = r∗ = 2m log(|u||v|) .Then

(2.9) |uv| = er∗2m = e

r2m

∣∣ r2m− 1∣∣ .

3In fact it is the spherically symmetric solution of the vacuum equations; the precise state-ment of this uniqueness result is the content of Birkhoff’s theorem.

4If m = 0, then Q× SO(3) is Minkowski space.

2. SCHWARZSCHILD SOLUTION 85

r∗

r

2m

Figure 3. “Tortoise coordinate” r∗ and area radius r.

The condition that u, v are increasing towards the future selects the sign, and weobtain, finally:

(2.10) uv = − e r2m

( r

2m− 1).

Thus we have

(2.11) uv

< 0 : r > 2m

= 0 : r = 2m

> 0 : r < 2m.

We distinguish four regions, as depicted in Fig. 4:

I: u < 0 , v > 0 II: u > 0 , v < 0III: u > 0 , v > 0 IV: u < 0 , v < 0 .

The level curves of r are hyperbolas in the uv-plane, timelike for r > 2m, spacelikefor r < 2m. r = 0 is the spacelike hyperbola uv = 1; there are 2 branches, onein region III and one in region IV. Similarly a timelike hyperbola r > 2m has 2branches, one in region I and one in region II.

Recall (1.19), which reduces to

(2.12) K =2m

r3.

Since m > 0 is constant, the curvature blows up at r = 0.

Next let us findQg explicitly. We have

(2.13)∂r∗

∂u=

2m

u,

∂r∗

∂v=

2m

v.

From these we obtain

∂r

∂u=(

1− 2m

r

)2m

u(2.14a)

∂r

∂v=(

1− 2m

r

)2m

v(2.14b)


r = 0

I

III

II

IV

r = 0

u v

H0

r < 2m

r > 2mH+out

H−out

H+

in

I : ∂r∂u < 0 , ∂r∂v > 0 .II : ∂r∂u > 0 , ∂r∂v < 0 .

III : ∂r∂u < 0 , ∂r∂v < 0 .

IV : ∂r∂u > 0 , ∂r∂v > 0 .

H−in

Figure 4. Global causal geometry of the Schwarzschild solution.

and thus

∂r

∂u

> 0 : in II, IV

< 0 : in I, III(2.15a)

∂r

∂v

> 0 : in I, IV

< 0 : in II, III .(2.15b)

In other words,

I: r increases outward, and decreases inward. This is the exterior region.II: r decreases outward.III: r decreases towards the future. This is the trapped region.IV: r increases toward the future. This is the anti-trapped region.

We substitute the solution (2.10) for r in the formula (1.29) for Ω to obtain:

(2.16)Ω2

4= −

(1− 2m

r

)∂r∗∂u

∂r∗

∂v

(2.17) Ω2 =32m3

re−

r2m .

We can use (2.17) to show that the singularity at r = 0 is reached byany observer in the trapped region in finite proper time. Consider thetimelike geodesics γα(λ) which are the rays from the origin in region III: u = αλ,v = α−1λ (α a positive constant). Then

(2.18) uv = λ2 = er

2m

(1− r

2m

).

2. SCHWARZSCHILD SOLUTION 87

As λ→ 1, r → 0, and in fact

(2.19) λ2 = 1−( r

2m

)2+O(r3) ,

so

(2.20)1√

2(1− λ)

r

2m−→ 1 as r → 0 or λ→ 1 .

The arc length of these geodesics from the origin (λ = 0) to r = 0 (λ = 1) issimply

(2.21)

∫ 1

0

√−g(γα(λ), γα(λ))dλ =

∫ 1

0Ω

√du

dλ

dv

dλdλ =

∫ 1

0Ω dλ <∞ ,

because

(2.22) Ω −→ 4m

(2(1− λ))14

as λ→ 1 .

While the timelike hyperbolas of constant r > 2m in regions I and II areactually timelike metric cylinders in M, the hyperbolas of constant r < 2m inregions III and IV are spacelike cylinders in M.

The pair of null lines u = 0, v = 0 are the level curves r = 2m. We have thebranches, (c.f. Fig. 4):

H+out: u = 0 , v > 0 is an outgoing future null cylinder.

H+in: v = 0 , u > 0 is an incoming future null cylinder.

H0: u = 0 , v = 0 is a single sphere.H−out: v = 0 , u < 0 is an outgoing past null cylinder.H−in: u = 0 , v < 0 is an incoming past null cylinder.

H+out is the outer component of the past boundary of III or the future boundary

of I. It is called the outer future event horizon. Regions I and II are asymptoticallyflat. H+

out is the boundary of the region accessible to observations from the infinityof region I. I is also called the domain of outer communications.

The spacelike rays through the origin in the uv plane do not intersect regionsIII, IV, but go from II to region I, as we move outward. They correspond tospacelike hypersurfaces in M which are isometric to each other. In each suchhypersurface H0 is a minimal sphere (the origin in the uv plane is the criticalpoint of the function r on Q). It is also called the bifurcation sphere; see Fig. 5.

In the domain of outer communications we can introduce a coordinate whichis constant on the rays through the origin:

(2.23) t = 4m artanh(u+ v

u− v)


I

II

H0

Figure 5. The bifurcation sphere H0.

Then with r viewed as a function of (u, v) implicitly defined by (2.10) we readilyverify:

−Ω2dudv =Ω2

4uv

[(v du− udv)2 − (v du+ udv)2

]= −

(1− 2m

r

)dt2 +

1

1− 2mr

dr2 ,(2.24)

and the Schwarzschild metric in the exterior region takes the classical form:

(2.25) g = −(

1− 2m

r

)dt2 +

(1− 2m

r

)−1dr2 + r2 γ

Remark. I recommend Chapter 2 of [DR08] for further reading on theSchwarzschild solution and black holes more generally.

3. General properties of the area radius and mass functions

We have discussed in Section 2 the Schwarzschild solution. The question ariseswhether the characteristic properties of this vacuum solution — in particular theexistence of a future event horizon — can form in evolution. Physically speakingthe question is: can black holes form from the collapse of stars?

We are thus interested in solutions to Einstein’s equations which arise frominitial data. In other words we consider spacetimes which admit a Cauchy hyper-surface.

Evolutionary Hypothesis: There is a spacelike curve Σ0 ⊂ Q such thateach past-directed causal curve from any point in Q terminates at a singlepoint on Σ0.

3. GENERAL PROPERTIES OF THE AREA RADIUS AND MASS FUNCTIONS 89

Γ0

Σ0

(u, v)

(u0, v)

Figure 6. Cauchy hypersurface Σ0.

Note that Σ0 is the past boundary of Q. (See figure 6.)We have seen that in region I of the Schwarzschild solution — i.e. the domain

of outer communications — the area radius is increasing outwards and decreasinginwards. Let us assume that this holds along Σ0.

Hypothesis on initial data: Along Σ0,

∂r

∂u< 0 ,

∂r

∂v> 0 .

Under these assumptions we can now show that in general — i.e. for any mat-ter model satisfying the dominant energy condition — the area radius is decreasinginwards everywhere on Q.

Prop 3.1. Under the above hypotheses we have

∂r

∂u< 0 : on Q .

Proof. Recall that we can rewrite (1.28a) as

(3.1)∂

∂u

( 1

Ω2

∂r

∂u

)= −rΩ−2 Tuu .

By the positivity condition (1.31) we thus have that Ω−2∂ur is a nonincreasingfunction of u. In view of the evolution hypothesis the past directed (ingoing) nullline from any point (u, v) ∈ Q intersects Σ0 at a point (u0, v) with u ≥ u0; seeFig. 6. Therefore

(3.2)1

Ω2

∂r

∂u(u, v) ≤ 1

Ω2

∂r

∂u(u0, v) < 0 ,

by assumption on the initial data.

We can now see from (1.29) that

(3.3)∂r

∂v

> 0 if r > 2m

= 0 if r = 2m

< 0 if r < 2m,

which partitions (by Prop. 3.1) the manifold Q into three regions:

Regular region R: Here ∂vr > 0, and r > 2m,Apparent horizon A: Here ∂vr = 0, and r = 2m,Trapped region T : Here ∂vr < 0, and r < 2m.


Next we can show the property that lends its name to the trapped region:

Prop 3.2. The future-directed outgoing null curve from any point in the trappedregion is contained in the trapped region. Furthermore, a future-directed outgoingnull curve from any point on the apparent horizon cannot enter the regular region.

Proof. Exercise.

With these properties of the radius at hand, let us turn to the mass function.Remarkably, we have the following monotonicity properties:

Prop 3.3. In the regular region,

∂m

∂u≤ 0 ,

∂m

∂v≥ 0 .

Proof. In view of Prop. 3.1 these inequalities follow immediately from themass equations (1.30).

Finally, let us prove that under the above hypotheses the mass is a nonnegativefunction on Q.

Prop 3.4. Under the above hypotheses,

m ≥ 0 : on Q .Proof. Let (u1, v1) ∈ Q. We can assume without loss of generality that

(u1, v1) lies in the exterior region, for otherwise 2m ≥ r > 0. By Prop. 3.2 thepast-directed (outgoing) null curve (u1, v), lies in the exterior region as well forall v < v1, until by the evolutionary hypothesis it terminates at a point (u1, v0),v0 < v1, which either lies on Σ0 or Γ. In any case we have by Prop. 3.3 that

(3.4) m(u1, v1) ≥ m(u1, v0) .

Since m vanishes on Γ, 5 and m is nondecreasing outwards along Σ0 by Prop. 3.3,we have m(u1, v0) ≥ 0 in either case.

Remark. Moreover, one can show that equality m = 0 holds at a point x ∈ Qif and only if the energy momentum tensor and the curvature K of Q vanish inthe entire domain of dependence of the sphere x ∈ Q; see [Chr95] for furtherdiscussion.

4. Spherically symmetric spacetimes with a trapped surface

The general properties of Section 3 allow us to introduce unambiguously thenotions of future null infinity and Bondi mass.

Recall that the coordinate transformations u 7→ f(u), v 7→ g(v) correspondto conformal transformations which we can use to map Q into a bounded subsetof R1+1. The depiction of Q in this manner is commonly referred to as the Penrose

5by regularity of Γ, which we assume.

4. SPHERICALLY SYMMETRIC SPACETIMES WITH A TRAPPED SURFACE 91

diagram of M. The boundary of Q in the topology of the plane is called thePenrose boundary which we denote by ∂Q = Q \ Q.6

The unique limit point on ∂Q of Σ0 in the outgoing direction is called spacelikeinfinity, and denoted by i0. Let us then denote by I+ ⊂ ∂Q the set of all pointson the boundary which are the limiting points of outgoing null curves along whichthe area radius tends to infinity; in other words (u∗, v∗) ∈ I+ if

(4.1) limv :(u=u∗,v)∈Q

r(u, v) =∞ ,

and (u∗, v∗) ∈ ∂Q. This set is called future null infinity. Note that I+ may beempty.

Prop 4.1. If I+ is non-empty, then I+ is a connected incoming null curveemanating from i0.

Proof. To see that I+ is a segment of an incoming null curve, we show thatno incoming null curve emanating from Σ0 intersects I+. Let i0 have coordi-nates (u0, v0) and consider v1 < v0, u1 > u0 such that (u1, v1) ∈ Σ0; then byProp. 3.1 we have r(u, v1) < r(u1, v1) < ∞ for all u > u1 with (u, v1) ∈ Q, soI+ does not intersect the incoming null line v = v1. For completeness of I+ let(u∗, v0) ∈ I+ and u0 < u < u∗; again by Prop. 3.1 we have r(u, v) > r(u∗, v) whilelimv→v0 r(u

∗, v) =∞. Thus also limv→v0 r(u, v) =∞, and (u, v0) ∈ I+.

In the following we shall assume that future null infinity is not empty. In fact,we shall include it in our notion of “asymptotic flatness”:

Definition 4.2. We say that Σ0 is asymptotically flat if (i) ∂vr > 0 in aneighborhood of i0, and (ii) the limit of r = ∞ at i0. Note that then m is non-decreasing in a neighborhood of i0, and we require that (iii) it has a finite limitat i0. Finally we require that (iv) I+ is not empty.

Remark. The condition (iv) is strictly speaking not an assumption on theinital data, but it is reasonable to include it in the definition. For if we considerthe case where the energy-momentum tensor is compactly supported on Σ0 weobtain with our argument of Section 2 that a neighborhood of i0 in Q is isometricto the Schwarzschild solution. Similary we expect to infer that I+ is non-emptyfrom sufficiently fast decaying matter fields along Σ0.

The fundamental question is however if future null infinity is complete, whichis conjectured to be true in the context of gravitational collapse (Penrose).

Weak cosmic censorship conjecture: For “generic” asymptotically flatinitial data on Σ0 the maximal future development of solutions to Ein-stein’s field equations in the presence of “appropriate” matter possess acomplete future null infinity.

6Q is a manifold with boundary Γ ∪ Σ0 in the sense of “manifold with boundary”. The“Penrose boundary” refers the boundary of Q as a set in R1+1 which by convention does notinclude Γ ∪ Σ0.


Remark. This conjecture captures the heuristic expectation that disregardingexceptional initial conditions a self-gravitating system cannot develop singulari-ties which are observable from infinity (“naked” singularities), even though theobservation continues for all time.

Consider the causal past of I+, denoted by J−(I+), namely the set of pointsin Q which can be reached from I+ with a past-directed causal curve. As aconsequence of Prop. 3.2 the causal past of I+ in Q is a subset of the regularregion. This domain is also called the domain of outer communcations.

An important observation is that for the completeness of future null infinityit suffices — in cases where the following extension principle holds — that thespacetime contains a single trapped surface, i.e. that the trapped region is notempty [Daf05].

Extension Principle: A matter model is said to satisfy the extensionprinciple if, informally, a “first singularity” emanating from the regu-lar region can only occur on the centre; or more precisely, if p ∈ R \ Γis a point on the boundary of the regular region which is not at thecentre, and q ∈ R ∩ I−(p) a point in its chronological past such thatJ−(p) ∩ J+(q) \ p ⊂ R ∪A then in fact p ∈ R ∪A.

Prop 4.3. If A∪T is non-empty, and the matter model satisfies the extensionprinciple, then I+ is future complete.

Proof. See [Daf05].

Remark. The weak cosmic censorship conjecture has been proven for thecollapse of a self-gravitating scalar field by Christodoulou in a series of papersculminating in [Chr99a]; (see also [Chr99b] for an overview). It is shown in par-ticular that generically singularities are always “preceded” by trapped surfaces7,and the completeness of future null infinity can then be inferred from Prop. 4.3.

Given the completeness of I+ we define the black hole region to be

(4.2) B = Q \ J−(I+) ,

and say, if B non-empty, that the spacetime contains a black hole. The futureboundary of the causal past of I+ in Q is called the future event horizon,

(4.3) H+ = ∂J−(I+) ∩Q .Clearly, H+ is a subset of R ∪ A; however only if the above extension criterionholds then the event horizon terminates at the future end point i+ of I+, andmoreover the following picture of Q emerges: Fig. 7, (see [Daf05] for furtherdiscussion).

Finally, let us mention that the mass function extends to future null infinityas follows: Assuming that the mass function is uniformly bounded along Σ0,

7It is shown that generic perturbations of initial data leading to “naked singularities” giverise to solutions with the stated property.

4. SPHERICALLY SYMMETRIC SPACETIMES WITH A TRAPPED SURFACE 93

I+

Σ0

Γ

H+B

i0

i+

Figure 7. Penrose diagram of a spherically symmetric black hole spacetime.

supΣ0m ≤ M0, we have by Prop. 3.3 that m ≤ M0 on J−(I+) and m extends to

a function M on I+. Again, in view of Prop. 3.3 the Bondi mass M ,

(4.4) M(u) = limv→v∗

m(u, v) ,

is a (not necessarily differentiable) monotone nonincreasing function of retardedtime u, and we call Mf = lim inf M the final Bondi mass. Note that by Prop. 3.4in particular

(4.5) Mf ≥ 0 .

In the proof of Prop. 4.3 a lower bound on Mf is established in terms of thearea radius of the event horizon; a bound of this form is referred to as a Penroseinequality.

Prop 4.4. Under the assumptions of Prop. 4.3 we have

r ≤ 2Mf : along H+ .

In particular, i+ /∈ I+.

Proof. See [Daf05].

A broader discussion of the weak and strong cosmic censorship conjecturesalong with an introduction to the above mentioned scalar field model can befound in [Chr99b].

CHAPTER 4

Dynamical Formulation of the Einstein Equations ∗

In this chapter we discuss the dynamical formulation of general relativity be-ginning with the decomposition of the Einstein equations with respect to the levelsets of a time function. We then discuss a few exciting consequences of the theory,namely the non-linear laws of post-Newtonian physics in Section 2, and gravita-tional waves in Section 3.

1. Decomposition relative to the level sets of a time function

1.1. Choice of a Time Function. A time function on a spacetime manifold(M, g) is a differentiable function t on M such that for every future directedtimelike vector V at any point p ∈M we have V · t > 0. This implies that if q liesin the chronological future of p — that is there is a future-directed timelike curvefrom p to q — then we have

(1.1) t(q) > t(p) ,

so a time function increases toward the future. We consider the vectorfield Mdefined by

(1.2) g(M,V ) = −V · t ,for any vector V ∈ TpM, at any point p ∈ M. In terms of components in anarbitrary coordinate frame:

(1.3) gµνMµV ν = −V µ∂µt ,

so the definition says:

(1.4) Mµ = −(g−1)µν∂νt .

Then M is timelike future-directed,

(1.5) g(M,M) < 0 .

This is equivalent to t being a time function. Let us then define a positive functionφ by:

(1.6) φ =1√

−g(M,M)= − 1√

−(g−1)µν∂µt∂νt.

This is called the lapse function associated to t.Consider the vectorfields colinear to M :

N : the unit future-directed vectorfield colinear to M ,T : the vectorfield colinear to M satisfying T · t = 1.

95

96 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

We have,

(1.7) N =1√

−g(M,M)M = φM ,

and

(1.8) T = φ N ;

for

(1.9) φN · t = φ2M · t =M · t

−g(M,M)=−gµν∂νt∂µt−g(M,M)

= 1 ,

so

(1.10) T = φN = φ2M .

Since the vectorfields M , N , T are colinear, they have the same integral curves.The integral curves of T are parametrized by t. The integral curves of N areparametrized by arclength s (proper time). Along a given integral curve,

(1.11) T =∂

∂t, N =

∂

∂s.

So the relation T = φN reads:

(1.12) ds = φ dt ,

along the given integral curve. Therefore the proper time that elapses along anintegral curve C between the parameter values t1 < t2 is

(1.13) L[C] =

∫ t2

t1

φ|C dt ;

this is the meaning of the “lapse” function.Consider now the hypersurfaces Σt, the level sets of the time function t:

(1.14) Σc =p ∈M : t(p) = c

The tangent space TpΣc is the subspace of TpM consisting of those vectors X ∈TpM such that X · t = 0 ,

(1.15) TpΣc =X ∈ TpM : X · t = 0

.

By the definition of M , N , T , this is equivalent to

(1.16) g(M,X) = 0 , g(N,X) = 0 , g(T,X) = 0 .

So TpΣc is the orthogonal complement of Np in TpM. Thus N is the futuredirected unit normal to Σc. An arbitrary level set of t will be denoted by Σt. Theintegral curves of T are orthogonal to the Σt.

1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 97

Σt

p

T

X

C

Figure 1. Level sets of a time function t and integral curves of T .

1.2. First and second fundamental form of Σt. The first fundamentalform is the induced metric gt on Σt. This is the restriction

(1.17)(gt)p

= gp

∣∣∣TpΣt

of gp to TpΣt. TpΣt is a spacelike subspace, being the orthogonal complementof N . Therefore gt is positive definite. We then say that the hypersurface Σt isspacelike.

The second fundamental form of Σt is a bilinear form kt in TpΣt at each p ∈ Σt

(a 2-covariant tensorfield on Σt). Given a pair of vectors X, Y ∈ TpΣt,

(1.18) kt(X,Y ) = g(∇XN,Y ) .

Note that since N is unit, we have

g(∇XN,N) =1

2X(g(N,N)) = 0 ,

so the vector ∇XN is tangential to Σt, ∇XN ∈ TpΣt.

Prop 1.1. The second fundamental form is symmetric

(1.19) kt(Y,X) = kt(X,Y ) ,

so kt is a quadratic form in TpΣt at each p ∈ Σt.

Proof. Extend X, Y to vectorfields tangential to Σt. Then

kt(X,Y ) = g(∇XN,Y ) = −g(N,∇XY ) ,

since g(N,Y ) = 0. Hence

kt(X,Y )− kt(Y,X) = −g(N,∇XY −∇YX) = −g(N, [X,Y ]) = 0 ,

because [X,Y ] is tangential to Σt, both X, Y being tangential vectorfields toΣt.


Σt

Σt+τ

ψτ

K0

Kτ

T

Figure 2. Illustration of curve Kτ in Στ .

Let ψt be the 1-parameter group of diffeomorphisms generated by T . Then ifp ∈ Σt, ψτ (p) is a point on Σt+τ at parameter value τ along the integral curveof T initiating at p. Given a curve K0 on Σt with end points a ; b, we define thecurve Kτ on Σt+τ by Kτ = ψτ (K0). Let then L(τ) be the arc length of Kτ . Weshall show that

(1.20)d

dτL(τ)

∣∣∣τ=0

=

∫K0

kt(X,X)ds ,

if K0 is parametrized by arclength and X is its unit tangent vectorfield. If K0 isparametrized with an arbitrary parameter λ rather than arclength then

(1.21)d

dτL(τ)

∣∣∣τ=0

=

∫K0

kt(X,X)

|X| dλ ,

where |X| =√gt(X,X) is the magnitude of X = K0, the tangent field of K0.

LetK0 be parametrized by an arbitrary parameter λ, and let K0 be the tangentvectorfield to K0, K0(λ) being the tangent vector at K0(λ). Then Kτ , the tangentvectorfield to Kτ = ψτ K0 is given by:

(1.22) Kτ (λ) = dψτ · K0(λ) .

The arc length L(τ) of Kτ is given by

(1.23) L(τ) =

∫I

√gt+τ

(Kτ (λ), Kτ (λ)

)dλ ;

(here I is the parameter interval). On the surface S = ∪τKτ we may define thevectorfield X by:

(1.24) XKτ (λ) = Kτ (λ)

We see that

(1.25) (ψτ )∗X = X ,


by definition of X; (c.f. notes on push-forward on pg. 101). Hence

(1.26) [T,X] = 0 .

We want to find

(1.27)dL

dτ(τ)∣∣τ=0

=

∫I

ddτ

(gt+τ

(Kτ , Kτ

))∣∣τ=0

2√gt(K0, K0)

dλ .

Now

(1.28)d

dτ

(gt+τ

(Kτ , Kτ

))∣∣τ=0

= T ·(g(X,X)

),

and X is tangential to the surfaces Σt+τ . Therefore

(1.29) T ·(g(X,X)

)= T ·

(g(X,X)

)= 2g(X,∇TX) = 2g(X,∇XT ) ,

because ∇TX − ∇XT = [T,X] = 0, the connection being symmetric. Sinceg(X,∇XN) = k(X,X) by definition of the second fundamental form, while g(X,N) =0 we obtain:

(1.30) T(g(X,X)

)= 2φk(X,X) .

We conclude

(1.31)dL(τ)

dτ

∣∣∣τ=0

=

∫I

φ k(K0, K0)

|K0|dλ

More generally we have the infinitesimal version:

Prop 1.2 (First variational formula). If X, Y are vectorfields tangential to Σt

which are Lie transported by T , that is [T,X] = [T, Y ] = 0, then

T(g(X,Y )

)= 2φ k(X,Y ) .

Push-forward of a vectorfield X by a diffeomorphism ψ

Let ψ be a diffeomorphism ofM onto itself. Let also X be a vectorfield onM. The push-forward of X by ψ, ψ∗X is another vectorfield onM, defined as follows: Let q = ψ−1(p),then (ψ∗X)p = dψ · Xq, a vector at ψ(q) = p. Let Y be another vectorfield, generatingthe 1-parameter group ψs. Then

−LYX = lims→0

X − (ψs)∗Xs

.

A basic result in differential geometry is that −LYX = [Y,X].


1.3. Codazzi equations. A level set Σt together with the first fundamentalform gt is a Riemannian manifold. It has a Riemannian connection and associatedcovariant derivative, denoted by ∇. The covariant derivative of the ambient space(M, g) is denoted by ∇.

Prop 1.3. The covariant derivative ∇ of (Σt, gt) is given by

∇XY = Π∇XY ,(X, Y tangential to Σt), where Π is the orthogonal projection to Σt.

Proof. Define ∇ by the projection of ∇ and show that this defines a connec-tion which is compatible with the metric and symmetric. Since there is a uniqueRiemannian connection corresponding to a given metric, ∇ must coincide withthe connection of the induced metric gt. Let X, Y be vectorfields tangential to Σt

then∇XY −∇YX = Π(∇XY −∇YX) = Π[X,Y ] = [X,Y ]

since the ambient connection is symmetric, and since the commutator of two vec-torfields tangential to Σt is also tangential to Σt. Next, let also Z be a vectorfieldtangential to Σt; then

Z(g(X,Y )) = Z(g(X,Y )) = g(∇ZX,Y ) + g(X,∇ZY ) =

= g(∇ZX,Y ) + g(X,∇ZY ) .

Let θ be a 1-form on Σt. Then ∇Xθ, X being a vectorfield tangential to Σt,is defined by

(1.32) (∇Xθ) · Y = X(θ(Y ))− θ · ∇XYfor any vectorfield Y tangential to Σt. Similarly for covariant tensorfields, inparticular for the second fundamental form kt of Σt. Let X, Y , Z be vectorstangential to Σt at some point, and extend them to vectorfields tangential to Σt.Then

(1.33) (∇Xk)(Y, Z) = X(k(Y,Z))− k(∇XY,Z)− k(Y,∇XZ) .

Now k(Y,Z) = g(∇YN,Z) and hence

(1.34) X(k(Y, Z)) = g(∇X∇YN,Z) + g(∇YN,∇XZ) .

We substitute, interchange X, Y and substract:

(1.35) (∇Xk)(Y, Z)− (∇Y k)(X,Z) =

= g([∇X ,∇Y ]N,Z)− k([X,Y ], Z)− k(Y,∇XZ) + k(X,∇Y Z)

+ g(∇YN,∇XZ)− g(∇XN,∇Y Z)

The first two terms on the right hand side are

(1.36) g([∇X ,∇Y ]N −∇[X,Y ]N,Z) = g(R(X,Y )N,Z) = R(Z,N,X, Y ) .

The remaining four terms cancel because

(1.37) g(∇YN,∇XZ) = g(∇YN,∇XZ) = k(Y,∇XZ) ,


and similarly with X, Y interchanged. For, let W be the vectorfield W = ∇XZ,then ΠW = ∇XZ; but for any vector W ,

(1.38) ΠW = W + g(W,N)N

so g(∇YN,W ) = g(∇YN,ΠW ) because

(1.39) g(∇YN,N) =1

2Y (g(N,N)) = 0 .

In conclusion, we have arrived at the Codazzi equations:

(1.40) (∇Xk)(Y,Z)− (∇Y k)(X,Z) = R(Z,N,X, Y )

for any three vectors X, Y , Z tangent to Σt at a point.

1.4. Convenient frame. Let E0 = N . We supplement E0 to a frame withvectorfields (Ei : i = 1, 2, 3) tangential to the Σt and let Ei be Lie transported byT :

(1.41) [T,Ei] = 0 i = 1, 2, 3 .

In fact we may define the (Ei : i = 1, 2, 3) first just on Σ0 and then extend themto each Σt by the push-forward of ψt. That is to say with q ∈ Σ0, p = ψt(q) ∈ Σt,we set

(1.42) Ei(p) = dψt · Ei(q) ,then Ei thus extended to the spacetime satisfies (ψt)∗Ei = Ei, hence [T,Ei] = 0.

In such a frame

g00 = −1 (= g(N,N))(1.43)

g0i = 0 (= g(N,Ei))(1.44)

gij = gij (= g(Ei, Ej)) ,(1.45)

and the first fundamental formula becomes:

(1.46)∂gij∂t

= 2φ kij

Similarly, the Codazzi equations become:

(1.47) ∇ikjm −∇jkim = Rm0ij

Consider the contraction of these equations with g. While on the left hand side

(1.48) (g−1)imkim = tr k ,

we find on the right hand side:

(1.49) (g−1)imRm0ij = S0j = 2T0j

by the Einstein equations in this frame. The contracted Codazzi equations arethus the “perp-parallel” components of the Einstein equations:

(1.50) ∇ikji −∇j tr k = 2T0j

Remark. We refer to the components of the Einstein equations G = 0 in thedecomposition relative to the level sets of a time function Σt as follows:


⊥⊥: This is simply the equation G00 = 0.⊥‖: The “perpendicular-parallel” components are the equations G0j = 0,E0 = N being perpendicular, and Ej being parallel to Σt.

‖‖: These are the equations Gij = 0.

1.5. Gauss equations. In this section we derive the (⊥⊥)-components ofthe Einstein equations.

We denote by R(W,Z,X, Y ) the curvature tensor of the induced metric g. Inthis Section, W,Z,X, Y are vector fields tangential to Σt, and

(1.51) R(W,Z,X, Y ) = g(W,R(X,Y ) · Z) ,

where R(X,Y ) denotes the intrinsic curvature transformation to Σt,

(1.52) R(X,Y ) · Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z .

What is the relation between the intrinsic curvature tensor R of (Σt, g) and thecurvature tensor R of the ambient spacetime restricted to directions tangential tothe spacelike hypersurface Σt?

Consider any vectorfield U (not necessarily tangential to Σt) on M. Then

(1.53) g(W,ΠU) = g(W,U)

where Π is the orthogonal projection to Σt:

(1.54) ΠU = U + g(N,U)N

Recalling that ∇XY = Π∇XY we thus have

(1.55) g(W,R(X,Y ) · Z) = g(W,∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z) .

Now,

(1.56) ∇X∇Y Z = ∇X(Π∇Y Z) = Π∇X∇Y Z + (∇XΠ) · ∇Y Z ,where

(1.57) (∇XΠ) · U = ∇X(ΠU)−Π∇XU = ∇X[U + g(U,N)N

]−Π · ∇XU

= ∇XU + g(∇XU,N)N + g(U,∇XN)N + g(U,N)∇XN −Π · ∇XU= g(U,∇XN)N + g(U,N)∇XN .

We apply this to the case where U = ∇Y Z.We have

(1.58) g(∇XN,N) =1

2X · g(N,N) = 0 ,

so ∇XN is tangential to Σt. We have

(1.59) g(∇XN,Y ) = k(X,Y )

for any vector Y tangential to Σt. At a given point p ∈ Σt, we consider X as avector at p. Moreover ∇XN is considered a vector at p ∈ Σt tangent to Σt, thatdepends linearly on X:

X ∈ TpΣt 7−→ ∇XN ∈ TpΣt


This is a linear transformation of TpΣt which we shall call k]:

(1.60) ∇XN = k] ·XThe linear transformation k] corresponds to the symmetric bilinear form k:

(1.61) g(k] ·X,Y ) = k(X,Y )

The components of k] with respect to the frame (Ei : i = 1, 2, 3) are

(1.62) (k])mj = (g−1)mikij .

We were considering

(1.63) g(W,∇X∇Y Z) = g(W,Π∇X∇Y Z + (∇XΠ) · U) ,

where U = ∇Y Z. The first term of

(1.64) (∇XΠ) · U = g(U,∇XN)N + g(U,N)∇XNdoes not contribute as g(W,N) = 0. With the above we find

(1.65) g(U,N)g(W,∇XN) = g(U,N)k(X,W ) ,

and,

(1.66) g(U,N) = g(∇Y Z,N) = −g(Z,∇YN) = −k(Y,Z) .

Therefore we obtain

R(W,Z,X, Y ) = g(W,Π∇X∇Y Z −Π∇Y∇XZ −Π∇[X,Y ]Z)

− k(X,W )k(Y, Z) + k(Y,W )k(X,Z)

= R(W,Z,X, Y )− k(X,W )k(W,Z) + k(Y,W )k(X,Z) .

(1.67)

Setting W = Em, Z = Ei, X = En, Y = Ej these equations read:

(1.68) Rminj + kmnkij − kmjkni = Rminj

The first contracted Gauss equations are obtained by contracting these equationswith g, yielding the intrinsic Ricci curvature to Σt on the left hand side:

(1.69) (g−1)mnRminj = Sij

On the right hand side:

(1.70) (g−1)µνRµiνj = Sij = −R0i0j + (g−1)mnRminj .

Also

(g−1)mnkmnkij = tr k kij(1.71)

(g−1)mnkmiknj = (k])niknj .(1.72)

We now denote (k])ni simply by kni. The contracted Gauss equations thereforeread:

(1.73) Sij + tr k kij − kni knj = R0i0j + Sij

The second contaction with g yields the intrinsic scalar curvature of Σt on the lefthand side:

(1.74) trS = (g−1)ijSij


Also,

(1.75) (g−1)ijkniknj = kni kin = (g−1)ij(g−1)mnkimkjn = |k|2 ,

is the sum of the squares of the entries in an orthonormal frame. On the righthand side

(g−1)ijR0i0j = S00(1.76)

−S00 + (g−1)ijSij = (g−1)µνSµν = trS .(1.77)

The second contracted Gauss equations are thus the (⊥⊥)-component of the Ein-stein equations:

(1.78) trS + (tr k)2 − |k|2 = 2S00 + trS = 4 T00

Remark. Recall that T00 is the energy density of matter as seen by the ob-servers whose world lines are the normal lines.

1.6. Acceleration equations. In the dynamical formulation of general rel-ativity we have a formal relation with classical mechanics:

gij: position,kij: velocity.

With the first fundamental form understood as position, the correspondence ofthe second fundamental form with velocity is simply expressed in

(1.79) kij =1

2φ

∂gij∂t

.

The (‖‖)-components of the Einstein equations will precisely be the equations forthe acceleration:

(1.80)1

φ

∂kij∂t

.

Let us however first discuss the acceleration in the real sense, namely the acceler-ation of the observers.

1.6.1. Proper acceleration. Consider a future-directed timelike curve Γ with aunit future-directed tangent field u. The (proper) acceleration is simply ∇uu, avectorfield along the curve, which is orthogonal to the curve since

(1.81) g(∇uu, u) =1

2u(g(u, u)

)= 0 .

“Proper acceleration” is thus geodesic curvature.In Section 1.1 we have discussed a 3-parameter family of timelike curves: the

normal lines. N is here in the role of the velocity u. The acceleration is therefore∇NN ; a vectorfield tangential to Σt. To calculate ∇NN we use the gradientvectorfield M :

Mµ = −(g−1)µν∂νt = −∂µt(1.82)

N = φM T = φN(1.83)


Consider first ∇MM in arbitrary local coordinates:

(1.84) (∇MM)µ = ∂νt∇ν∂µtSo

gµλ(∇MM)µ = ∂νt∇ν∂λt = ∂νt∇λ∂νt

=1

2∂λ(∂νt∂νt

)= φ−3∂λφ ,

(1.85)

where we have used the fact that the Hessian is symmetric (cf. note on p. 107):

(1.86) ∇µ∂νt = ∇ν∂µtWe conclude that

(∇MM)µ = φ−3∂µφ(1.87)

(∇NM)µ = φ−2∂µφ(1.88)

and in view of

(1.89) ∇NN = φ∇NM + (Nφ)M

also

(1.90) (∇NN)µ = φ−1(∂µφ+ (Nφ)Nµ

).

Therefore

(1.91) ∇NN = Π · U ,where

(1.92) Uµ = φ−1∂µφ = ∂µ log φ ,

and Π denotes the orthogonal projection to Σt as above. Since

U = φ−1∇φ = ∇ log φ(1.93)

Π · U = φ−1∇φ = ∇ log φ(1.94)

we finally obtain that

(1.95) ∇NN = ∇ log φ .

Thus in the formal analogy with mechanics, logφ plays the role of the New-tonian potential.

Notes on the Hessian of a function

The Hessian of a function f is defined by

(∇2f)(X,Y ) = (∇df)(X,Y ) = (∇Xdf)(Y ) = X(df ·Y )−df(∇XY ) = X(Y f)−∇XY ·f .It is symmetric because

(∇df)(X,Y )− (∇df)(Y,X) = XY f − Y Xf − (∇XY −∇YX)f = 0

by the symmetry condition of the connection:

∇XY −∇YX = [X,Y ] .


In local coordinates this follows immediately from the symmetry properties of the con-nection coefficients:

∇µ∂νf = ∂µ∂νf − Γαµν∂αf .

1.6.2. Second variational forumula. Let X, Y be vectorfields defined and tan-gential to Σ0. Extend them to M by the condition that they are Lie transportedby T :

(1.96) [T,X] = [T, Y ] = 0 .

Then X, Y are tangential to each Σt. Consider now

(1.97) T ·(k(X,Y )

).

That is:

(1.98) T(g(∇XN,Y )

)= φN

(g(∇XN,Y )

)=

= φg(∇N∇XN,Y ) + g(∇XN,∇NY )

We have

(1.99) ∇N∇XN = ∇X∇NN +R(N,X) ·N +∇[N,X]N

and, Y being tangential to Σt,

g(∇X∇NN,Y ) = g(∇X(φ−1∇φ), Y ) = g(∇X(φ−1∇φ), Y )(1.100)

g(R(N,X) ·N,Y ) = R(Y,N,N,X) = −R(X,N, Y,N) .(1.101)

Now N = φ−1T and [T,X] = 0, so

[N,X] = [φ−1T,X] = −X(φ−1)T = −φX(φ−1)N(1.102)

∇[N,X]N = −φX(φ−1)∇NN = −X(φ−1)∇φ(1.103)

g(∇[N,X]N,Y ) = −X(φ−1)(Y φ) .(1.104)

Therefore,

(1.105) g(∇X(φ−1∇φ), Y ) + g(∇[N,X]N,Y ) = φ−1(∇2φ)(X,Y ) ,

and we conclude

(1.106) g(∇N∇XN,Y ) = φ−1∇2φ(X,Y )−R(X,N, Y,N) .

This is the first term in (1.98). For the second term, we write

(1.107) ∇NY = ∇YN + [N,Y ] = ∇YN + φY (φ−1)N ,

and so

(1.108) g(∇XN,∇NY ) = g(∇XN,∇YN) = g(k] ·X, k] · Y ) .

The result is the second variational formula,

(1.109) (−LTk)(X,Y ) = T(k(X,Y )) =

= (∇2φ)(X,Y ) + φ

−R(X,N, Y,N) + g(k] ·X, k] · Y )

,


N

Σ0

timelike geodesic

X

Figure 3. Geodesic congruence and normal lines to Σ0 .

which expressed in the convenient frame, setting X = Ei, Y = Ej , where

(k] ·X)m = (k])mi = kmi(1.110)

(k] · Y )m = (k])mj = kmj(1.111)

g(k] ·X, k] · Y ) = gmnkmi k

nj = kmi kmj(1.112)

then reads

(1.113)∂kij∂t

= ∇i∇jφ+ φ−Ri0j0 + kmi kmj

.

Remark. In the case φ = 1 the normal lines are timelike geodesics parametrizedby arc length:

(1.114) ∇NN = φ−1∇φ = 0 .

The Σt are the level sets of the distance function from Σ0. k is the second fun-damental form of the hypersurfaces Σt. Here we expressed k] in terms of itscomponents in a Lie transported frame. In Section 3.2.3 we found that its compo-nents in a parallel orthonormal frame form the rate of strain matrix θ associatedto the geodesic congruence of the normal lines, see (3.78).

The initial condition θ = 0 for the Jacobi equation corresponds to paralleltransport of N along a spacelike geodesic. Note that in the case φ = 1: N = T .Here N is not parallel transported along Σ0 but is simply the unit normal to Σ0.Now

(1.115) ∇NX = ∇XN = k] ·X ,

where the left hand side is the velocity V , and X on the right hand side thedisplacement in the Jacobi equation. The initial condition k = 0 corresponds toΣ0 being totally geodesic for the spacelike geodesic congruence orthogonal to Nat a point. Draw all spacelike geodesics orthogonal to a given timelike geodesicthrough a given point on Σ0, as in Fig. 3. The unit tangent field X to this spacelike


geodesic congruence – not defined at the origin – satisfies: ∇XX = 0. Hence

(1.116) k(X,X) = g(∇XN,X) = −g(N,∇XX) = 0 .

At the origin X is any unit vector, and it follows that k = 0 at the origin. Forthis reason the initial condition for the Jacobi equation becomes V = 0, and thetwo conditions indeed coincide.

1.6.3. The (‖‖)-components of the Einstein equations. Substitute in the sec-ond variation formula (1.113) for the curvature from the first contracted Gaussequation (1.73) to obtain:

(1.117)∂kij∂t

= ∇2ijφ− φ

Sij + tr k kij − 2kmi kmj − Sij

The (‖‖)-components of the Einstein equations

(1.118) Sij = 2(Tij −

1

2gij trT

)are thus expressed in the evolution equations

(1.119)∂kij∂t

= ∇2ijφ− φ

Sij + tr k kij − 2kmi kmj − 2Tij + gij trT

.

In the mechanical correspondence, where g plays the role of position and kij thatof velocity related by (1.79), we can think of (1.119) as an equation for the acceler-ation. It captures the dynamics of the gravitational system. The (‖‖) componentsof the Einstein equations are complemented by the (⊥‖) and (⊥⊥) equations(1.50) and (1.78) which are constraints on the inital conditions. They are calledconstraint equations.

1.6.4. Trace of second variation formula. We shall also derive an equation forthe trace of the second fundamental form k. Since

(1.120)∂

∂ttr k =

∂

∂t

((g−1)ijkij

)= (g−1)ij

∂kij∂t− 2φ kijkij ,

we can substitute from (1.119) to obtain

(1.121)∂

∂ttr k = 4φ− φ

trS + (tr k)2 − 3T00 + trT

.

Then substitute for trS from the second contracted Gauss equation (1.78) toobtain

(1.122)∂

∂ttr k = 4φ− φ

|k|2 + T00 + trT

.

1.7. Summary. We have discussed the decomposition of the Einstein equa-tions (in rationalized gravitational units where 4πG = 1; also at this point c = 1)with respect to the level sets Σt of an arbitrary time function t, and associatedlapse function φ:

(1.123) φ−2 = −(g−1)µν∂µt∂νt .

We have discussed the normal lines, a family of timelike curves normal to Σt

(which can be viewed as the histories of a system of observers), parametrized by t.


The tangent vectorfield is denoted by T , T · t = 1, and related to the unit tangentfield N by T = φN .

Moreover we use a frame field (Eµ : µ = 0, 1, 2, 3), where E0 = N and theEi : i = 1, 2, 3 are tangential to Σt and Lie transported by T : [T,Ei] = 0. In thisframe we have:

(1.124) g00 = −1 , gi0 = 0 , gij = gij .

We denote by S the Ricci curvature of (M, g), and by S the Ricci curvatureof (S, g). Let us also introduce

pi = T i0: momentum density relative to the system of observers whose worldlines are the normal lines.

ρ = T 00: energy density relative to the same system of observers.

We have seen that the (⊥‖)- and (⊥⊥)-components of the Einstein equations withrespect to this decomposition are constraint equations, which may referred to asthe momentum constraint and energy constraint respectively:

∇jkij −∇i tr k = 2T0i = −2pi(1.125)

trS − |k|2 + (tr k)2 = 4T00 = 4ρ(1.126)

The first variation formula

(1.127)∂gij∂t

= 2φ kij

can be viewed as the definition of the second fundamental form k, and for thesecond variation formula we have found:

(1.128)∂kij∂t

= ∇2ijφ− φ

Sij − 2kimk

mj + tr k kij

+ 2φ

Tij −

1

2gij trT

Note also that

(1.129) trT = −ρ+ trT ,

where T ij = Tij are the components of the stress tensor relative to the system ofobservers whose histories are the normal lines. The proper acceleration of theseobservers is given by

(1.130) ∇NN = ∇ log φ ,

which suggests that log φ is to be identified with the Newtonian potential.Given the frame (Eα : α = 0, 1, 2, 3) we can define the dual frame (θα : α =

0, 1, 2, 3) by δα(Eβ) = δαβ , and

(1.131) g = gαβθα ⊗ θβ , gαβ = g(Eα, Eβ) .

Here E0 = N = φ−1T = φ−1∂t and so θ0 = φdt, and

g = −θ0 ⊗ θ0 + gij θi ⊗ θj

= −φ2dt2 + gij θi ⊗ θj

(1.132)


Σ0U0

Σt

Figure 4. Variation of a spacelike hypersurface.

If we choose (Ei : i = 1, 2, 3) to be a coordinate frame (which in general we do notneed) Ei = ∂

∂xithen

(1.133) g = −φ2dt2 + gijdxidxj .

1.8. Maximal foliations. We shall now make a choice of a time function:

(1.134) tr k = 0

Consider a limiting case, where the level sets Σt coincide with Σ0 outside adomain U0 ⊂ Σ0 with compact closure in Σ0; see figure 4. Then φ vanishes outsideU0 on Σ0, and outside the corresponding domain Ut = Φt(U0) on Σt. Let V (t) bethe volume of Ut. Then

(1.135)dV

dt

∣∣∣t=0

=

∫U0

d

dtdµgt

∣∣∣t=0

=

∫U0

φ tr k∣∣∣Σ0

dµg0,

by the first variational formula, because

(1.136) V (t) =

∫U0

dµgt =

∫U0

√det gt θ

1 ∧ θ2 ∧ θ3

and

∂

∂t

√det gt =

1

2(g−1)ij

∂gij∂t

√det gt(1.137)

∂gij∂t

= 2φ kij .(1.138)

The condition that the volume of U0 is stationary means that

(1.139)d

dtV (t)

∣∣∣t=0

= 0

for any such variation, that is

(1.140)

∫U0

φ tr k dµg = 0

for any nonnegative function φ vanishing outside U0 in Σ0, and any U0 with com-pact closure in Σ0. This is thus equivalent to the conditon (1.134).

Remark. The above formulas hold in fact for arbitrary variations vanishingoutside U0. A general variation does not correspond to a foliation by the levelsets of a time function, but to a normal differentiable homotopy. In other words,consider a differentiable mapping H : I × Σ0 → M, where I = (−ε, ε) is aparameter interval, such that Hλ×Σ0

is a diffeomorphism onto its image Σλ, aspacelike hypersurface in M, and H0×Σ0

is the identity mapping on Σ0. Such amapping H is called a differentiable homotopy, which is called normal if for any


point p ∈ Σ0, HI×p is a timelike curve in M orthogonal to the Σλ. Along eachsuch timelike curve can define φ = dτ/dλ which generalizes the lapse function ofa foliation and is not in general nonnegative. Then as before

(1.141)

∫U0⊂Σ0

φ tr k dµg = 0⇒ tr k = 0 on Σ0 .

A spacelike hypersurface Σ in M is called maximal if

(1.142) tr k = 0 .

It corresponds to a maximum of the volume under compactly supported varia-tions. The motion is analogous to that of a minimal surface in Euclidean space(Plateau’s problem). Here we consider complete spacelike hypersurfaces, in factCauchy hypersurfaces. In Minkowski spacetime the only complete maximal space-like hypersurfaces are the spacelike hyperplanes. In a general spacetime which isasymptotically flat at spacelike infinity there is a unique maximal hypersurfaceasymptotic to a given hyperplane at spacelike infinity. A time function t shallbe called maximal if each of its level sets Σt is a maximal hypersurface and theassociated lapse function φ satisfies

(1.143) φ −→ 1 at ∞ along each Σt.

Then t becomes proper time at infinity.

1.9. Equations of motion. Let us finally discuss the decomposition of theconservation laws

(1.144) ∇νTµν = 0 .

The connection coefficients Γγαβ are defined by

(1.145) ∇EαEβ = ΓγαβEγ ,

where E0 = N , and the tangential Ei : i = 1, 2, 3 are transported by T = φN :

(1.146) [T,Ei] = 0

We have seen that the acceleration of the observers is

(1.147) ∇E0E0 = ∇NN = φ−1∇φ = φ−1(g−1)ij∇jφ Ei ,and thus

(1.148) Γ000 = 0 Γi00 = φ−1(g−1)ij∇jφ .

Moreover,

(1.149) ∇EiE0 = ∇EiN = kjiEj ,

hence

(1.150) Γ0i0 = 0 Γji0 = kji ;

next,

∇E0Ei = ∇NEi = φ−1∇TEi = φ−1∇EiT= φ−1∇Ei(φN) = φ−1(Eiφ)N + kji Ej

(1.151)


hence

(1.152) Γ00i = φ−1∇iφ Γj0i = kji .

Note that:

(1.153) Γ00i 6= Γ0

i0 .

Finally,

(1.154) ∇EiEj = Γ0ijN + Γmij Em ,

where

(1.155) Γ0ij = −g(∇EiEj , N) = g(Ej ,∇EiN) = kij ,

and

(1.156) Γmij = Γmij

are the connection coefficients of the induced connection on Σt:

∇EiEj = ΓmijEm(1.157)

∇EiEj = Π∇EiEi = Γmij Em ,(1.158)

where Π is the orthogonal projection to Σt.Now the equations of motion are:

(1.159) Eν(Tµν) + Γµνλ Tλν + Γννλ T

µλ = 0

However, for a maximal foliation:

Γνν0 = Γ000 + Γii0 = tr k = 0(1.160)

Γννj = Γ00j + Γiij = φ−1∇jφ+ Γiij .(1.161)

With our previous notation

(1.162) T 00 = ρ , T 0i = pi , T ij = T ij ,

the conservation laws henceforth read:

1

φ

∂ρ

∂t+∇ipi + 2φ−1(∇iφ) pi + kij T

ij = 0(1.163)

1

φ

∂pi

∂t+∇j T ij + φ−1(∇jφ)T ij + φ−1(∇iφ)ρ+ 2kij p

j = 0(1.164)

Remark. The last terms can be compared with the Lorentz force density inMaxwell’s theory:

(1.165) F iνJν = F i0J

0 + F ijJj = Ei σ + εijkJ

jBk ,

where J0 = σ denotes the charge density, and J i the current density.

Exercises

2. SLOW MOTION APPROXIMATION 113

1. (Kasner metric) Let us look at the Kasner metric

g = −dt2 +

3∑i=1

t2pi(dxi)2 .

(1) Calculate the components of the second fundamental form kij .

(2) Calculate the components of the Ricci tensor Sij .(3) Show that the Kasner metric is a solution of the Einstein vaccum field equations if

and only if3∑i=1

pi =

3∑i=1

p2i = 1.

2. (Scalar curvature of maximal hypersurface) Suppose we have a maximal foliation.

Look at the scalar curvature R of a maximal hypersurface. Show that R ≥ 0 with R = 0 at apoint if and only if kij = 0 and ρ = 0 at this point.

2. Slow Motion Approximation

The slow motion approximation consists formally in the limit

c −→∞ .

2.1. Field equations in conventional units. In units where c 6= 1 conven-tional time t is related to the spacetime coordinate x0 by

(2.1) x0 = c t .

We shall keep rationalized units where 4πG = 1, and the Einstein equations are

(2.2) Sµν − 1

2(g−1)µν trS =

2

c4Tµν .

We shall consider the decomposition of the Einstein equations with respect to amaximal time function x0 = ct in these units.

Recall also the quasi-Newtonian hierarchy discussed in Section 1.2.1 accordingto which

T 00 = c2 ρ ρ : mass density

T 0i = c pi pi : momentum density

T ij = T ij : stress

and in particular T 00 T 0i T ij .In these units the constraint equations read

tr k = 0(2.3a)

∇jkij = − 2

c3pi(2.3b)

trS − |k|2 =4

c2ρ ,(2.3c)

and (1.122) becomes the lapse equation:

(2.4) 4φ = φ(|k|2 +

1

c2ρ+

1

c4trT

).


The first and second variation formulas are now

(2.5a)1

c

∂gij∂t

= 2φ kij

(2.5b)1

c

∂kij∂t

= ∇i∇jφ− φ(Sij − 2kimk

mj

)+ 2φ

(1

2

1

c2ρ gij +

1

c4

(Tij −

1

2gij tr T

)).

Finally introducing c into the equations of motion we obtain:

∂ρ

∂t+ φ∇i pi + 2(∇iφ)pi +

1

cφ kij T

ij = 0(2.6a)

∂pi

∂t+ φ∇j T ij + (∇jφ)T ij + c2(∇iφ)ρ+ 2cφ kijp

j = 0(2.6b)

We have to determine φ to order c−2 and kij to order c−1 just to obtain theNewtonian theory. And to obtain the equations of motion correct to order c−2 wehave to determine φ to the order c−4, kij to order c−3, and gij to order c−2.

2.2. Slow motion limits. From the momentum constraint (2.3b) we seethat we may assume that the limit

(2.7) limc→∞

c3kij = lij

exists, so

(2.8) kij =1

c3lij + o

(c−3).

Definition 2.1. We say f = o(c−α), with α ≥ 0, if f and its derivatives up

to certain fixed order, after multiplying by cα, tend to 0 as c→∞.

We also make the basic assumption that

(2.9) limc→∞

φ = 1 ,

which is now seen to be consistent with the lapse equation (2.4). From the firstvariation formula (2.5a) we then have

(2.10)∂gij∂t

=2

c2lij + o

(c−2).

We also assume that on Σ0 the first fundamental form gij tends to the Eu-clidean metric as c → ∞. Choosing rectangular coordinates of the limiting flatmetric we have:

(2.11) limc→∞

gij = δij : on Σ0 .

Hence by (2.10) also for each t > 0:

(2.12) limc→∞

gij = δij : on Σt .

We shall denote the covariant derivative with respect to the limiting Euclideanmetric simply by ∇, and the Laplacian by 4.


Consider the coefficients of φ in the lapse equation:

(2.13) limc→∞

c2(|k|2 +

1

c2ρ+

1

c4tr T

)= ρ

Therefore

(2.14) limc→∞

c2(φ− 1

)= ψ

exists and satisfies

(2.15) 4ψ = ρ .

ψ is the Newtonian potential. The equations of motion (2.6) in the limitc→∞ then become:

∂ρ

∂t+∇ipi = 0(2.16a)

∂pi

∂t+∇j T ij + (∇iψ)ρ = 0(2.16b)

We now consider the second variation equation (2.5b). Multiply (2.5b) by c2

and take the limit c→∞. In view of (2.8) the left hand side vanishes in the limit,and we conclude that

(2.17) limc→∞

c2Sij = Qij

exists and is given by

(2.18) Qij = ∇i∇jψ + ρ δij .

The trace of this equation is

(2.19) trQ = 4ψ + 3ρ = 4ρ

by (2.15). This coincides with the limit of the energy constraint (2.3c). We cansubstitute for ρ in terms of ψ in (2.18) to obtain:

(2.20) Qij = ∇i∇jψ +4ψ δijWe take the trace-free part of that:

Qij = Qij −1

3δij trQ

= ∇i∇jψ −1

3δij4ψ

(2.21)

Now we take the symmetric curl of Q with respect to the limiting Euclidean metric.That is define (c.f. notes on pg. 120):

(2.22) cij = curl Qij =1

2

(ε abi ∇aQbj + ε abj ∇aQbi

)We have

(2.23) ε abi ∇aQbj = ε abi ∇a∇b∇jψ −1

3ε aji ∇a4ψ =

1

3ε aij ∇a4ψ


because the covariant derivatives ∇a with respect to the flat metric δ commute.This expression in antisymmetric in i, j, hence

(2.24) cij = 0 .

Prop 2.2 (Bach’s Theorem). Let (Σ, g) be a 3-dimensional manifold and Sthe Ricci curvature of g. Define the Bach tensor B by

B = curl S .

Then the vanishing of B is necessary and sufficient for the existence of a positivefunction χ on Σ such that

g = χ4e

where e is flat. In other words, then g is conformally flat.

Remarks on Bach’s theorem. Consider first the linearized version: Weconsider a 1-parameter family of metrics gij(λ) such that gij |λ=0 = eij is the flatmetric, and denote

gij =dgijdλ

∣∣λ=0

.

Let Sij(λ) be the Ricci curvature of gij(λ), and

Sij =d

dλSij

∣∣∣λ=0

.

ThenSij = ∇mΓmij −∇jΓmim ,

where ∇ denotes the covariant derivative with respect to e, and

Γmij =d

dλΓmij

∣∣∣λ=0

=1

2(e−1)mn

(∇igjn +∇j gin −∇ngij

).

Suppose now that the metric variation is conformal:

gij = 4χeij , χ =d

dλχ|λ=0 ,

then

Γmij = 2(∇iχ δmj +∇jχ δmi −∇mχ eij

)and

Γmim = 6∇iχ .Substitute in the expression for Sij to obtain:

1

2Sij = 2∇i∇jχ−4χ eij − 3∇i∇jχ = −∇i∇jχ−4χeij

Conversely, Sij corresponds to a conformal variation if there is a solution χ of thisequation:

1

2tr S = −44χ

Substitute then for 4χ in the equation for Sij to obtain:

−∇i∇jχ =1

2

(Sij −

1

4eij tr S

)=

1

2Aij


The integrability condition is:

∇kAij −∇iAkj = 0 .

In the exact version of Bach’s theorem

Aij = Sij −1

4gij trS .

The integrability condition is equivalent to

ε kim ∇kAij = 0 .

Let us decompose Sij into its trace-free part Sij and its trace-part:

Sij = Sij +1

3gij trS

Then

Aij = Sij +1

12gij trS ,

and the anti-symmetric part of the integrability condition above is equivalent to

ε mjl ε kim ∇kAij = 0 ,

or in view of the identity

ε mjl ε kim = (g−1)jkδil − (g−1)jiδkl :

∇jAlj −∇l trA = 0 .

Now substituting for Aij in term of the original Sij this condition reads

∇jSlj −1

2∇l trS = 0 .

This however is an identity, the second contracted Bianchi identity (in 3 dimen-sions). So the integrability condition above reduces to its symmetric part, whichis

curlAij = 0 .

Finally we show that

(curlA)ij = (curl S)ij = Bij

is the Bach tensor. Indeed the additional term in Aij does not contribute to thecurl:

ε abi ∇a(gbj trS) + ε abj ∇a(gbi trS) =(ε ai j + ε aj i

)∇a trS = 0 .

We have seen in the discussion of the linearized version above that once theintegrability condition is satisfied χ is determined by the equation:

(2.25) 4χ = −1

8tr S

In the exact theory this equation is:

(2.26) trS = −8χ−54eχ ,

where Sij is the Ricci curvature of gij = χ4eij .


Now we have shown that the symmetric curl of the trace-free part of thelimiting Ricci curvature (2.17) vanishes (2.24). Therefore by virtue of Bach’stheorem,

(2.27) gij = χ4eij + o(c−2),

or, after linearizing,

(2.28) gij =(1 + 4χ

)eij + o

(c−2),

where in view of (2.25) and (2.19) the function χ satisfies

(2.29) 4χ = −1

2ρ ,

and thus in comparison to (2.15) in fact

(2.30) χ = −1

2ψ .

In conclusion:

(2.31) gij =(

1− 2c−2ψ)eij + o

(c−2)

where e is the flat metric.

Note on symmetric curl operator

In general if θ is a symmetric 2-covariant tensorfield in a 3-dimensional manifold (Σ, g)the symmetric curl of θ is defined by:

(curl θ)ij =1

2

(ε abi ∇aθbj + ε abj ∇aθbi

)where ∇ is the covariant derivative of g and ε is the volume form of g, and indices areraised with respect to g.

Since the Σt are diffeomorphic to R3, e is isometric to the Euclidean met-ric. Consider in general a 1-parameter family of metrics e(t) depending on thetime t which are isometric to the Euclidean metric. Then there are new spatialcoordinates Xk : k = 1, 2, 3 such that

(2.32) eij =∂Xk

∂xi∂X l

∂xjδkl .

Writing ~X = (X1, X2, X3) this is

(2.33) eij =∂ ~X

∂xi· ∂

~X

∂xj.

Let us define now,

(2.34) ~V =∂ ~X

∂t, (x constant).


This is the velocity of the normal lines (the observers) with respect to the ~X-constant coordinate lines (that is the conformally Euclidean coordinate lines).Let us also define

(2.35) βi = ~V · ∂~X

∂xi.

Then differentiating (2.33) with respect to t (at x constant),

(2.36)∂eij∂t

=∂~V

∂xi· ∂

~X

∂xj+∂ ~X

∂xi· ∂

~V

∂xj=∂βj∂xi

+∂βi∂xj− 2~V · ∂2 ~X

∂xi∂xj.

Now ifeΓm

ij are the connection coefficients of eij with respect to the x-coordinatesthen in fact

(2.37)eΓk

ij βk = ~V · ∂2 ~X

∂xi∂xj.

This follows from

(2.38)eΓk

ij=∂xk

∂ ~X· ∂2 ~X

∂xi∂xj,

which is the general formula for the transformation of the connection coefficientsunder change of coordinates in view of the fact that the connection coefficients in

the ~X coordinates vanish. Therefore (2.36) actually is:

(2.39)∂eij∂t

=e∇ βj+

e∇j βi = (−Lβ]e)ij ,

where β] is the vectorfield corresponding to the 1-form β,

(2.40) (β])i = (e−1)ijβj .

These formulas hold for a general 1-parameter family of metrics eij(t).Let us now return to the study of the post-Newtonian approximation. We

differentiate (2.31) with respect to t (at x constant):

(2.41)∂gij∂t

= −2c−2∂ψ

∂tδij +

∂eij∂t

+ o(c−2)

The limit

(2.42) limc→∞

c2βi = αi

exists and

(2.43) limc→∞

c2∂eij∂t

= ∇iαj +∇jαi .

We compare this result to the limit of the first variation formula (2.10) derivedabove and obtain:

(2.44) lij = limc→∞

c2

2

∂gij∂t

= −∂ψ∂tδij +

1

2

(∇iαj +∇jαi

).

The condition (2.3a) implies

(2.45) tr l = 0 ,


which is

(2.46) ∇ · α = 3∂ψ

∂t.

Eliminating ∂tψ in (2.44) then yields:

(2.47) 2lij = ∇iαj +∇jαi −2

3δij∇ · α

The momentum constraint (2.3b) becomes in the limit

(2.48) ∇jlij = −2pi ;

and from the above,

(2.49) 2∇jlij = 4αi +1

3∇i(∇ · α) .

We thus obtain the following elliptic system for α:

(2.50) 4αi +1

3∇i(∇ · α) = −4pi

Taking the divergence gives

(2.51) 4(∇ · α) = −3∇ · p ,which in view of (2.46) really reads:

(2.52) 4(∂ψ∂t

)= −∇ · p

This is the integrability condition of the post-Newtonian system. By virtue of(2.15) the integrability condition (2.52) is equivalent to the continuity equation(2.16a):

(2.53)∂ρ

∂t+∇ · p = 0 .

Prop 2.3. The solution of the elliptic system (2.50) (tending to zero at infin-ity) is:

(2.54) αi(x) =1

8π

∫R3

7δij|x− x′| +

(xi − x′i)(xj − x′j)|x− x′|3

pj(x′) d3x′

Consider now the transformation of coordinates (t, x) to (t,X),

Xa = Xa(t, x)(2.55)

dXa = V adt+∂Xa

∂xidxi .(2.56)

Using (2.32) we obtain

(2.57) eijdxidxj = δab

(dXa − V adt

)(dXb − V bdt

).

Let us then find the spacetime metric to order c−2 (after setting x0 = ct):

(2.58) g = −c2φ2(dt)2 + gij dxidxj ,


where

φ = 1 +1

c2ψ +

1

c4ω + o

(c−4)

(2.59a)

gij = χ4eij + o(c−2)

(2.59b)

χ = 1− 1

2

1

c2ψ + o

(c−2).(2.59c)

Thus:

(2.60) g = −c2dt2 − 2ψdt2 + δabdXadXb

+1

c2

−(ψ2 + 2ω

)dt2 − 2αa dtdXa − 2ψ δabdX

adXb

+ o(c−2)

Here ω is the correction to the Newtonian potential. In fact set:

(2.61) ω = limc→∞

c4(φ− 1− 1

c2ψ)

We revisit the lapse equation (2.4) to find the equation for ω:

4φ = φ[|k|2 +

1

c4

(c2ρ+ tr T

)]Recall that |k| = o

(c−6). We have to determine both sides to order c−4. On

the left hand side, we have to consider (2.59b) because the c−2 contribution tog cancels. However, we may set eij = δij because this is a scalar equation forthe 3-dimensional manifold (Σt, g), (and thus does not depend on the choice of

coordinates on Σt and we may choose the coordinates ~X w.r.t. which eij = δij).In general,

(2.62) 4φ =1√

det g

∂

∂xi

(√det g (g−1)ij

∂φ

∂xj

)Here,

(2.63)√

det g = χ6 , (g−1)ij = χ−4δij ,

so

(2.64) 4φ =1

χ6∇i(χ2∇iφ

)=

1

χ4

(4φ+

2

χ∇χ · ∇φ

),

with respect to the coordinates ~X. Substitute (2.59a) and (2.59c) then using(2.15) we obtain:

(2.65) 4φ+2

χ∇χ · ∇φ =

1

c2ρ+

1

c4

(4ω − |∇ψ|2

)+ o(c−4)

On the right hand side,

(2.66) |k|2 +1

c4

(c2ρ+ tr T

)=

1

c2ρ+

1

c4tr T + o

(c−4)


Taking also into account (2.59c) for the factor χ−4 on the left hand side, and(2.59a) for the factor φ on the right hand side, we finally obtain the equation:

(2.67) 4ω = −ρψ + |∇ψ|2 + tr T

The first two terms on the right hand side are quadratic in Newton’s potential andthus display the non-linearity of the Einstein field equations. This leads to3-body forces.

Exercises

Consider an axially symmetric stationary mass distribution of compact support. That is take ρsuch that

∂ρ

∂t= 0 , supp ρ = Ω

where Ω has compact closure in R3, and take p = ρv with the integral curves of v being circularorbits. Show that the leading order term of α at spatial infinity depends only on the angularmomentum of the mass distribution.

Hints: We may choose rectangular coordinates (x1, x2, x3) such that the total angularmomentum vector

Lk =

∫Ω

εkijxipjdx

lies in the x3-axis. Moreover set r =√∑3

i=1(xi)2, and show that to leading order in 1/r:

1

|x− x′| =1

r+x · x′

r3+ o(r−2) x′ ∈ Ω .

Now:

(1) Use the continuity equation to show that the total momentum vanishes:

P i =

∫Ω

pidx = 0

(2) Define

Dij =

∫Ω

xipjdx ,

and show that D is antisymmetric and moreover

Dij =1

2εijkL

k .

(3) Show that to leading order in 1/r:

αi =1

2π

1

r3εijkL

jxk

Note that this implies that upon introducing polar coordinates (r, θ, φ) the cross term of themetric (2.60) is given by:

− 2

c2αi dtdxi = − 1

πc2|L|r

sin2 θ dtdφ

3. Gravitational Radiation

In this section we discuss gravitational radiation. As a precursor we first derivethe radiation field of scalar fields, and then treat electromagnetic waves using asuitable null decomposition.

3. GRAVITATIONAL RADIATION 123

Figure 5. Example of an axially symmetric and rotating mass configuration.

(t, x)

C−t0(t, x)

Σt0

Figure 6. Truncated light cone C−t0 .

3.1. Scalar wave equation in Minkowski space. Consider the scalar waveequation in Minkowski space,

(3.1) φ = ρ ,

where = (g−1)µν∇µ∇ν = − ∂2

∂t2+∑3

i=1∂2

∂xi2in rectangular coordinates (t, x1, x2, x3).

Here ρ is a given source, and let φ be a solution with trivial initial data at t = t0:

(3.2) φ∣∣t=t0

= 0 ,∂φ

∂t

∣∣t=t0

= 0 .

We denote by C−t0(t, x) the past light cone of the point (t, x) truncated by Σt0 =

(t, x) : t = t0, x ∈ R3, (c.f. figure 6):

C−t0(t, x) =

(t′, x′) : (t− t′)2 − |x− x′|2 = 0, t0 ≤ t′ ≤ t

3.1.1. Representation formula. Then, by Kirchhoff’s formula,

φ(t, x) = − 1

4π

∫x′∈R3

t−|x−x′|≥t0

ρ(t− |x− x′|, x′)|x− x′| d3x′

= − 1

4π

∫C−t0

(t,x)ρ dµC−(t,x)

(3.3)


Letting t0 → −∞ we obtain the retarded solution

(3.4) φ(t, x) = − 1

4π

∫C−(t,x)

ρ dµC−(t,x) = − 1

4π

∫x′∈R3

ρ(t− |x− x′|, x′) d3x′

|x− x′| .

Notes on Kirchhoff’s formula

The representation formula (3.3) is easily obtained with the method of spherical means.We can take, without loss of generality the point (t, x) on the time axis by transla-

tion. Any point in spacetime, away from the axis, can be assigned spherical coordinates(t, r, ϑ), where r =

√(x1)2 + (x2)2 + (x3)2, and ϑ ∈ S2. The Minkowski metric in these

coordinates reads g = −dt2 + dr2 + r2γ and

φ = −∂2φ

∂t2+∂2φ

∂r2+

2

r

∂φ

∂r+

1

r2

4/ φ ,

where4/ denotes the Laplacian on (S2,

γ). Given any function f in spacetime, we denote

by f(t, r) the mean value of f over (St,r, γ). Since γ = r2γ, dµγ = r2dµ

γ, and the area

of St,r is 4πr2 we have

f(t, r) =1

4πr2

∫St,r

f(t, r, ϑ)dµγ =1

4π

∫S2f(t, r, ϑ)dµ

γ,

that is f(t, r) is simply the mean value of f(t, r, ϑ) on S2. Note that4/ φ = 0, because S2

is closed (without boundary). We conclude from (3.1) that φ satisfies

−∂2φ

∂t2+∂2φ

∂r2+

2

r

∂φ

∂r= ρ ,

Now set ψ = rφ, σ = rρ. These functions of (t, r) satisfy the 1 + 1-dimensional waveequation −∂2

t ψ + ∂2rψ = σ, and ψ vanishes on the time axis (r = 0). Setting v = t + r,

and u = t− r this becomes

−4∂2ψ

∂u∂v= σ ,

where the initial data to be trivial on t = t0 by (3.2). Since, by continuity of φ,limr→0 φ(t, r) = φ(t, 0), we obtain φ(t, 0) as an integral of σ over the line v = t:

−2φ(t, 0) = limr→0−4

∂ψ

∂v=

∫ t

2t0−tσ du .

But since σ(t′, r′) is the mean value of ρ(t′, r′) over the sphere St′,r′ , the integral of σ onthe line v = t is really the integral of ρ over the past light cone C−t0(t, 0) of (t, 0) truncatedat the past hypersurface t = t0:∫ t

2t0−tσdu

∣∣v=t

= 2

∫ t−t0

0

rφ(t− r, r)dr =1

2π

∫ t−t0

0

∫St−r,r

φ dµγdr

r.

We have thus obtained, after a suitable space translation, the formula (3.3).


(u+ r, rξ)

C−(u+ r, rξ)

U

Σuξ

Figure 7. Plane approximation of C−(u+ r, rξ) ∩ U .

3.1.2. Radiation field. Set

x = r ξ , ξ ∈ S2 ,(3.5)

r = |x| , t = u+ r .(3.6)

Then with ξ, u fixed, the equations t = u + r, x = rξ are those of a light rayparametrized by r, namely the light ray in the direction ξ which starts at thespatial origin at t = u.

Let us suppose that the source ρ has compact support in a neighborhood ofthe spatial origin. Then for any fixed bounded spacetime region U ,

U ∩ C−(u+ r, rξ)→ U ∩ P (u, ξ) , as r →∞ ,

where P (u, ξ) is a null plane corresponding to the given ray (u, ξ), c.f. figure 7:

(3.7) P (u, ξ) =

(t′, x′) : t′ = u+ ξ · x′ , x′ ∈ R3

Since

|x− x′| =√|x|2 − 2x · x′ + |x|2 =

√r2 − 2rξ · x′ + |x′|2

= r

√1− 2

rξ · x′ + 1

r2|x′| = r − ξ · x′ + o(r−1) ,

(3.8)

we have

(3.9) t− |x− x′| = t− r + ξ · x′ + o(r−1) = u+ ξ · x′ + o(r−1) ,

and it follows that

(3.10) limr→∞

(rφ)(u+ r, rξ) = − 1

4π

∫P (u,ξ)

ρ = − 1

4π

∫x′∈R3

ρ(u+ ξ · x′, x′)d3x′ .


Σt

N

LL ∂

∂t

St,r

Figure 8. Null normals L, L to wave fronts St,r.

Remark. The function thus defined,

(3.11) Ψ(u, ξ) := limr→∞

(rφ)(u+ r, rξ)

is also called the radiation field.

3.1.3. Tangential derivatives. Consider the spheres St,r on Σt of radius r,

(3.12) St,r =

(t, x) : |x| = r

We may think of these spheres as wave fronts. Every St,r has two null normals ateach point, the outgoing null normal L and the incoming null normal L, normalizedso that their ∂t components are equal to 1, c.f. figure 8:

(3.13) L =∂

∂t+N , L =

∂

∂t−N ,

where N is the outward unit normal to St,r in Σt:

(3.14) N =xi

r

∂

∂xi= ξi

∂

∂xi.

We denote by Π the orthogonal projection to the tangent space to St,r at a point.

Let us also introduce the generators Ωi : i = 1, 2, 3 of the rotations about the ith

rectangular coordinate axis:

(3.15) Ωi = εijk xj ∂

∂xk.

We also consider

(3.16) Ωi =1

rΩi = εijk ξ

j ∂

∂xk.

Note that3∑i=1

Ω ai Ω b

i =

3∑i=1

εijaξj εikbξ

k = (δjkδab − δjbδka)ξjξk

= δab − ξaξb = Πab ,

(3.17)

or

(3.18)

3∑i=1

Ωi ⊗ Ωi = Π .


Prop 3.1. Given any function ρ of compact spacelike support, we have

Ω µi (ξ)

∫P (u,ξ)

∇µρ = 0(3.19)

Lµ(ξ)

∫P (u,ξ)

∇µρ = 0 .(3.20)

Proof. Recall (3.7) for the definition of the plane P (u, ξ). We have

∂

∂x′kρ(u+ ξ · x′, x′) = (∇0ρ ξ

k +∇kρ)(u+ ξ · x′, x′)

and since ρ has compact spacelike support∫x′∈R3

∂

∂x′kρ(u+ ξ · x′, x′) d3x′ = 0 .

Hence ∫P (u,ξ)

∇kρ = −ξk∫P (u,ξ)

∇0ρ .

Now recall that

Ω0i = 0 , Ω k

i = εijkξj ,

and thus

Ω µi

∫P (u,ξ)

∇µρ = −εijkξjξk∫P (u,ξ)

∇0ρ = 0 .

Moreover, recall that

L0 = 1 , Lk = ξk ,

and therefore:

Lµ∫P (u,ξ)

∇µρ =

∫P (u,ξ)

(∇0ρ+ ξk∇kρ

)=

∫P (u,ξ)

(∇0ρ− |ξ|2∇0ρ

)= 0 .

The Ωi : i = 1, 2, 3, span the tangent space to St,r at each point, and togetherwith L they represent the tangential derivatives to the surfaces of constant retardedtime u. A simple consequence of the above general proposition is a decay statementfor these tangential derivatives.

Cor 3.2. Let φ be a solution to (3.1, 3.2) with a source ρ of compact spatialsupport. Then

limr→∞

r Lφ = 0 ,(3.21)

limr→∞

r Ωiφ = 0 , i = 1, 2, 3.(3.22)

Proof. We differentiate the equation (3.1) to obtain that ∂µφ : µ = 0, 1, 2, 3,satisfies the wave equation

( ∂φ∂xµ

)=

∂ρ

∂xµ;


in other words the generators of the translations along the coordinate axes com-mute with the wave operator. Therefore

limr→∞

r Lµ∂µφ(u+ r, rξ) = − 1

4π

∫P (u,ξ)

Lµ∂µρ = 0

limr→∞

r Ωµi ∂µφ(u+ r, rξ) = − 1

4π

∫P (u,ξ)

Ωµi ∂µρ = 0

by the previous proposition.

Notes on transversal derivatives and energy flux

Recall the outgoing and incoming (future-directed) null normals L, L to the spheres St,r,(3.13). In terms of retarded and advanced time coordinates, u = t − r, and v = t + r,respectively, we have

L = 2∂

∂v, L = 2

∂

∂u.

We consider now the limit of the transversal derivative Lφ. On one hand, using theidentities in the proofs of Prop. 3.1 and Cor. 3.2,

limr→∞

r Lφ = − 1

4π

∫P (u,ξ)

Lµ∇µρ = − 1

4π

∫P (u,ξ)

∇0ρ− ξk∇kρ = − 1

4π

∫P (u,ξ)

2∇0ρ .

On the other hand, with the already found above, 2∂uΨ = limr→∞ rLφ. Thus

∂Ψ

∂u= − 1

4π

∫P (u,ξ)

∂ρ

∂t.

The transversal derivative is related to the energy radiated to infinity, or energy flux. Inparticular, the power radiated to future null infinity at retarded time u, is given by

G(u) = 2

∫S2

(∂Ψ

∂u

)2

(u, ξ) dµγ(ξ) .

This can be seen using the method of energy currents; c.f. exercises.

Exercises

1. (Tangential and longitudinal decay) Recall the setting and the decay statements ofCorollary 3.2. In this exercise we shall look at the limits to next order in r. Show that for aretarded solution φ to 3.1 with a source ρ of compact spatial support we have:

(1)limr→∞

rΩiφ = ΩiΨ

Hints:(a) Use polar coordinates (θ, ϕ) on S2 such that ξ1 = sin θ cosϕ, ξ2 = sin θ sinϕ,

ξ3 = cos θ and arrange the coordinate axes in such a way that

Ω3 =∂

∂ϕ= x1 ∂

∂x2− x2 ∂

∂x1.

In this setting it suffices to show that

limr→∞

rΩ3φ =∂Ψ

∂ϕ.


(b) To prove (1) make use of the commutation relation [Ωi,] = 0 to infer that

limr→∞

rΩ3φ(u, ξ) = − 1

4π

∫P (u,ξ)

Ω3ρ.

(2)

limr→∞

r2 Lφ = −Ψ

Hints:(a) Consider the scaling vector field S := xµ ∂

∂xµand compute the commutator

[S,] to conclude that

limr→∞

rSφ =

∫P (u,ξ)

(Sρ+ 2ρ).

(b) Next compute that

x′i∂

∂x′iρ(u+ ξ · x′, x′) =

(t∂ρ

∂t+ xi

∂ρ

∂xi

)(u+ ξ · x′, x′)− u∂ρ

∂t(u+ ξ · x′, x′),

and integrate by parts to show that∫P (u,ξ)

(Sρ+ 2ρ) = −Ψ + u∂Ψ

∂u.

(c) Finally note that S = 12(uL+ vL) and use

limr→∞

rLφ = 0, limr→∞

rLφ = 2∂Ψ

∂u

together with u = t− r, v = t+ r to prove (b).

2. (Energy flux) The energy flux through a segment of a null hypersurface of constantadvanced time v with respect to the timelike vectorfield ∂t is given by

F (v)[u1, u2] =

∫ u2

u1

∫S2T (L, ∂t) r

2dµγdu ,

where T denotes the energy momentum tensor of the scalar field, a 2-covariant tensorfield whichis given by Tµν [φ] = ∂µφ∂νφ− 1

2gµν g

αβ∂αφ∂βφ . Show that

limv→∞

F (v)[u1, u2] =

∫ u2

u1

G(u) du ,

where G(u) is defined as in the note on page 130.Hints:

(1) Express ∂t in terms of L, L, and verify T (L,L) = (Lφ)2, and T (L,L) = |∇/φ|2γ , where∇/ denotes the connection on the sphere St,r.

(2) Show that∑3i=1(Ωiφ)2 = r2|∇/φ|2γ and use Corollary 3.2 to compute the limit.

3. (Incoming and outgoing radiation) Consider the incoming radiation field of the retardedsolution of (3.1), namely for fixed advanced time v, and ξ ∈ S2, consider the limit of rφ alongthe past-directed light ray t = v − r, x = rξ, parametrized by r in the direction ξ, as r →∞.

(1) Show that for a spatially compact source, there is no incoming radiation. In otherwords, show that the incoming energy flux vanishes.Hints:

(a) Show that

limr→∞

(rφ)(v, ξ) = − 1

4πlim

u→−∞

∫P (u,ξ)

ρ


(b) The incoming energy flux is given by

F [v1, v2] =

∫ v2

v1

G(v)dv , G(v) = limr→∞

∫Sv−r,r

T (L, ∂t)dµγ ;

c.f. Exercise 2 above. Show that G(v) = 0.(2) Find the conditions on the source so that the total energy radiated to infinity is finite.

Hint: The amount of energy radiated from the past up to retarded time u0 is given by

F−(u0) =

∫ u0

−∞G(u)du .

Make a suitable assumption on the support of ρ, and ∂tρ that ensures the finitenessof F−(∞).

3.2. Maxwell’s equations. Consider now Maxwell’s equations in Minkowskispace,

∇αFβγ +∇βFγα +∇γFαβ = 0(3.23a)

∇νFµν = Jµ(3.23b)

for the Faraday tensor F in the presence of sources J .3.2.1. Wave equations. We take the derivative ∇λ of the inhomogoneous equa-

tion (3.23b), in covariant form, and antisymmetrize in λ, µ:

(3.24) ∇ν(∇λFµν −∇µFλν) = ∇λJµ −∇µJλFrom the homogeneous equations (3.23a) we have

(3.25) ∇λFµν −∇µFλν = ∇λFµν +∇µFνλ = −∇νFλµ ,

and−∇ν∇ν = −Fλµ. We thus obtain a scalar wave equation for each rectangularcomponent of F :

(3.26) Fλµ = −∇λJµ +∇µJλConversely, a solution F to these wave equations whose initial data satisfies

Maxwell’s equations is actually a solution of Maxwell’s equations for all time. For,suppose we have a solution to (3.26) then

(∇αFβγ +∇βFγα +∇γFαβ) =−∇α∇βJγ +∇α∇γJβ−∇β∇γJα +∇β∇αJγ−∇γ∇αJβ +∇γ∇βJα = 0 ,

(3.27)

and, by charge conservation ∇νJν = 0,

(∇νFµν) = −∇ν∇µJν +∇ν∇νJµ = (Jµ) ,

(∇νFµν − Jµ) = 0 .(3.28)


3.2.2. Null decomposition of Faraday tensor. We decompose the Maxwell fieldrelative to the wave fronts into:

α, α: St,r-1-forms, that is 1-forms in Minkowski spacetime which vanish on

the orthogonal space (TpSt,r)⊥ to St,r,

ρ, σ: functions .

The null decomposition of the Faraday tensor is then defined by:

αµ = Π κµ LλFκλ αµ = Π κ

µ LλFκλ(3.29a)

ρ =1

2LkLλFκλ(3.29b)

σ εµν = Π κµ Π λ

ν Fκλ ,(3.29c)

where εij = ξkεijk is a St,r-2-form proportional to the area form of St,r.3.2.3. Radiative properties. We now consider the radiation fields

(3.30) fµν(u, ξ) = limr→∞

(rFµν)(u+ r, rξ) ,

which by (3.10), and (3.26), are given by

(3.31) fµν(u, ξ) =1

4π

∫P (u,ξ)

(∇µJν −∇νJµ) .

Recall the decomposition (3.29) of the Faraday tensor F . We denote thecorresponding decomposition of the limit f by

α[f ] , α[f ] , ρ[f ] , σ[f ] .

Specifically,

α[f ](Ωi) = f(Ωi, L) , α[f ](Ωi) = f(Ωi, L) ,(3.32a)

ρ[f ] =1

2f(L,L) , σ[f ]ε(Ωi, Ωj) = f(Ωi, Ωj) .(3.32b)

Now, by (3.31) and Prop. 3.1 we have that

(3.33) α[f ] = 0 , σ[f ] = 0 ,

and

(3.34) ρ[f ] =1

8πLµLν

∫P (u,ξ)

∇µJν .

Furthermore, since

(3.35) L =∂

∂x0+ ξi

∂

∂xi, L =

∂

∂x0− ξi ∂

∂xi,

we have

(3.36) ρ[f ] =1

8π

∫P (u,ξ)

(∇0J0 − ξi∇iJ0 + ξi∇0Ji − ξiξj∇iJj) ,


and again by Prop. 3.1,

−∫P (u,ξ)

ξi∇iJ0 =

∫P (u,ξ)

∇0J0 ,(3.37) ∫P (u,ξ)

ξiξj∇iJj = −∫P (u,ξ)

ξj∇0Jj .(3.38)

Therefore

(3.39) ρ[f ] =1

4π

∫P (u,ξ)

(∇0J0 − ξiξj∇iJj) .

Now recall the formula (3.17). By Prop. 3.1 we have

(3.40) 0 =

3∑i=1

Ω ai Ω b

i

∫P (u,ξ)

∇aJb =

∫P (u,ξ)

(δab − ξaξb)∇aJb ,

or

(3.41)

∫P (u,ξ)

ξiξj∇iJj =

∫P (u,ξ)

∇iJi .

Hence

(3.42) ρ[f ] =1

4π

∫P (u,ξ)

(∇0J0 −∇iJi) = 0 ,

by the current conservation law:

(3.43) 0 = ∇µJµ = ∇0J0 +∇iJi = −∇0J0 +∇iJi .In conclusion, only the α-component survives in the limit. We denote

(3.44) a = α[f ] ,

which by Prop. 3.1 reduces to

(3.45) aµ(u, ξ) = − 1

4πΠ κµ Lλ

∫P (u,ξ)

∇λJκ .

Since

(3.46) Lλ∫P (u,ξ)

∇λJκ = 0 ,

and

(3.47) L+ L = 2∂

∂x0,

we finally obtain

(3.48) a0 = 0 , ai = − 1

2πΠ ki

∫P (u,ξ)

∂Jk∂t

.


3.2.4. Dipole approximation. If the time scale of variation of the source ismuch larger than its spatial dimension then we can replace the integral over thenull plane P (u, ξ) by an integral over the spacelike plane Σu.

We can then express (3.48) in terms of the dipole moments

(3.49) Di(t) =

∫Σt

xiρ(t, x)d3x ,

where ρ is the charge density. In view of conservation of charge (3.43),

(3.50)∂ρ

∂t+∇kJk = 0 ,

we have, after integration by parts,

(3.51)dDi

dt(t) =

∫Σt

xi∂ρ

∂t(t, x) d3x = −

∫Σt

xi∇kJkd3x =

=

∫Σt

∇kxi Jkd3x =

∫Σt

J id3x .

Therefore, taking another time-derivative,

(3.52)d2Di

dt2(t) =

∫Σt

∂J i

∂td3x .

Thus, we have in this approximation,

(3.53) ai(u, ξ) = − 1

2πΠik(ξ)

d2

dt2Dk(u) .

3.2.5. Polarisation. Note finally that the Maxwell radiation field can be writ-ten as:

(3.54) fµν = −1

2aµLν +

1

2aνLµ

Let us denote by e, and b the limits of the electric and magnetic fields respectively,

(3.55) ei = limr→∞

rEi , bi = limr→∞

rBi .

Then

(3.56) ei = fi0 =1

2ai ,

and

bi =1

2ε jki fjk =

1

2ε jki (−1

2ajξk +

1

2akξj) =

1

2ε jki ξjak

=1

2akε

ki = ekε

ki .

(3.57)

That is to say, the electric and magnetic fields are orthogonal in the limit, c.f. fig-ure 9.


e

b

Figure 9. Electric and magnetic components of the radiation field.

3.3. Gravitational waves. We turn to the gravitational case. We recall theBianchi identities from Section 1.3,

(3.58) ∇αRβγδε +∇βRγαδε +∇γRαβδε = 0 ,

which after the first contraction become

(3.59) ∇αRαβγδ = ∇γSβδ −∇δSβγ .We now substitute for Sµν from the Einstein equations

(3.60) Sµν = 2Tµν , Tµν = Tµν −1

2gµν trT ,

to obtain the inhomogeneous equations

(3.61) ∇αRαβγδ = 2(∇γ Tβδ −∇δTβγ

).

The are analogous to the Maxwell equations (3.23b) and we can proceed as inSection 3.2.1 to derive a wave equation for the curvature tensor. Apply ∇ε to(3.61) and antisymmetrize in (β, ε),

(3.62) ∇ε∇αRαβγδ −∇β∇αRαεγδ =

= 2(∇ε∇γ Tβδ −∇β∇γ Tεδ −∇ε∇δTβγ +∇β∇δTεγ

).

Next we wish to commute the covariant derivatives on the left hand side, whichproduces quadratic terms in the curvature, by its very definition. After commu-tation, however, the l.h.s. becomes by (3.58):

∇α∇εRαβγδ −∇α∇βRαεγδ = ∇α(∇εRαβγδ +∇βRεαγδ

)= −∇α∇αRβεγδ = −Rβεγδ

(3.63)

Therefore we obtain the following wave equation for the curvature tensor:

(3.64) Rαβγδ = 2(∇α∇γ Tβδ −∇β∇γ Tαδ −∇α∇δTβγ +∇β∇δTαγ

)+Qαβγδ

Here Qαβγδ are expressions that are quadratic in the curvature.We shall now impose of a major simplification which is often referred to as the

weak field approximation.

Weak field approximation: We break the connection between the met-ric and the curvature, and consider the wave equations (3.64) and theBianchi identities (3.58), (3.61) as equations for a tensorfield Rαβγδ inthe Minkowski spacetime. Here Rαβγδ is to posses the algebraic sym-metries of the curvature tensor. Moreover we drop the quadratic termsQαβγδ in (3.64).


Set

(3.65) Sβδ = (g−1)αγRαβγδ .

Let us verify that a solution Rαβγδ of (3.64) whose initial data satisfies theEinstein equations is actually a solution of the Einstein equations (3.60). Takethe trace of (3.64), then Sβδ satisfies:

(3.66) (Sβδ − 2Tβδ

)= −2∇β∇αTαδ − 2∇δ∇αTαβ + 2∇β∇δ tr T

Now taking into account the conservation laws,

(3.67) ∇νTµν = 0 ,

we have

(3.68) ∇ν Tµν = −1

2∇µ trT ,

and the right hand side above vanishes:

(3.69) − 2∇β∇αTαδ − 2∇δ∇αTαβ + 2∇β∇δ tr T =

= ∇β∇δ trT +∇δ∇β trT − 2∇β∇δT = 0 .

Therefore, in the weak field approximation, any solution of (3.64) satisfies

(3.70) (Sµν − 2Tµν

)= 0 .

3.3.1. Asymptotics of linearized gravitational waves. The retarded solution ofthe linearized wave equation

(3.71) Rαβγδ = 2(∇α∇γ Tβδ −∇β∇γ Tαδ −∇α∇δTβγ +∇β∇δTαγ

)in rectangular coordinates relative to the Minkowski metric g is given by:

(3.72) limr→∞

rRαβγδ(u+ r, rξ) = qαβγδ(u, ξ) =

= − 1

2π

∫P (u,ξ)

(∇α∇γ Tβδ −∇β∇γ Tαδ −∇α∇δTβγ +∇β∇δTαγ

).

But, by the equations (3.60) we have that Sµν = 2Tµν = 0 outside the support ofthe source, in particular at infinity

(3.73) (g−1)αγqαβγδ = 0 .

3.3.2. Null decomposition of the curvature tensor. Consider the curvature ten-sor of a solution of the Einstein equations in a vacuum region. Let S be a spacelikesurface diffeomorphic to S2 (a wave front). Let L, L be respectively the outgoing,incoming future directed null normals to S, normalized according to g(L,L) = −2.L, L are determined up to the transformation L → aL, L → a−1L where a is apositive function on S. (These correspond to Lorentz transformations in the or-thogonal timelike plane, c.f. figure 10.)

The curvature then decomposes into:


S

LL

M (TpS)⊥

p

LL

Figure 10. Null normals to S.

α, α: symmetric trace-free 2-covariant-tensorfields on S defined by

α(X,Y ) = R(X,L, Y, L)(3.74a)

α(X,Y ) = R(X,L, Y, L) ∀X,Y ∈ TpS .(3.74b)

β, β: 1-forms on S defined by

R(X,Y, Z, L) = ε(X,Y )β(Z)(3.75a)

R(X,Y, Z, L) = ε(X,Y )β(Z) ∀X,Y, Z ∈ TpS ,(3.75b)

where ε denotes the area 2-form of S.ρ, σ: functions defined by

R(X,Y, Z,W ) = ε(X,Y )ε(Z,W )ρ(3.76a)

1

2R(X,Y, L, L) = ε(X,Y ) σ ∀X,Y, Z,W ∈ TpS .(3.76b)

Let us introduce a frame for the spacetime M along S by setting

(3.77) E1, E2 ∈ TS , E3 = L ,E4 = L .

In this frame,

g33 = g44 =0 , g34 = −2 , gAB = δAB , A,B = 1, 2(3.78a)

(g−1)33 = (g−1)44 =0 , (g−1)34 = −1

2, (g−1)AB = δAB .(3.78b)

Moreover, we can write

αAB = α(EA, EB) , βA

= β(EA) ,(3.79a)

αAB = α(EA, EB) , βA = β(EA) ,(3.79b)

and

αAB = RA3B3 , αAB = RA4B4(3.80a)

RABC3 = εABβC , RABC4 = εABβC(3.80b)

RABCD = εABεCDρ ,1

2RAB34 = εABσ ,(3.80c)

and the metric can also be written as

(3.81) g−1 = −1

2E3 ⊗ E4 −

1

2E4 ⊗ E3 +

∑A=1,2

EA ⊗ EA .


To see that α, α are traceless note that, in vacuum,

0 = S33 = (g−1)αγRα3γ3 = −1

2R3343 −

1

2R4333 +

∑A=1,2

RA3A3 = trα ,(3.82a)

0 = S44 = trα .(3.82b)

Also, in vacuum,

(3.83) 0 = SAB = (g−1)αγRαAγB = −1

2

(R3A4B +R4A3B

)+∑C=1,2

RCACB ,

and

(3.84)∑A=1,2

RABAD =∑A=1,2

εABεADρ = δBD ρ ;

that is

(3.85)1

2

(R3A4B +R4A3B

)= ρδAB

Since, by the cyclic property of the curvature,

(3.86) RA3B4 −RB3A4 = RA3B4 +R3BA4 = −RBA34 = RAB34 ,

we have

(3.87)1

2

(RA3B4 −RB3A4

)= εABσ

and we conclude that

(3.88) RA3B4 = ρ δAB + σ εAB .

Furthermore, in vacuum,

(3.89) 0 = SA3 = (g−1)αγRαAγ3 = −1

2R3A43 −

1

2R4A33 +

∑B=1,2

RBAB3 ,

and

(3.90)∑A=1,2

RABA3 =∑A=1,2

βAεAB = β∗

B

so that

(3.91)1

2RA334 = β∗

A.

Similarly,

(3.92)1

2RA434 = −β∗A .

Let us return to the discussion of Section 3.3.1. In view of Proposition 3.1, andthat α, β, ρ do not contain L in their definition, it follows that the corresponding


components of q vanish. Consider now the σ component of q:

(3.93)1

2q(Ωi, Ωj , L, L) = − 1

2πΩαi Ωβ

j LγLδ

∫P (u,ξ)

(∇α∇γ Tβδ −∇β∇γ Tαδ

−∇α∇δTβγ +∇β∇δTαγ)

= 0 .

So σ[q] vanishes as well. Thus only the α component remains:

(3.94) a = α[q]

(3.95) a(Ωi, Ωj) = − 1

2πΩαi L

βΩγjL

δ

∫P (u,ξ)

(∇α∇γ Tβδ −∇β∇γ Tαδ

−∇α∇δTβγ +∇β∇δTαγ)

= − 1

2πΩαi L

βΩγjL

δ

∫P (u,ξ)

∇β∇δTαγ .

Again by Prop. 3.1,

(3.96) Lµ∫P (u,ξ)

∇µρ = 21

2

(Lµ + Lµ

) ∫P (u,ξ)

∇µρ = 2

∫P (u,ξ)

∇0ρ .

Therefore

(3.97) a(Ωi, Ωj) = − 2

πΩαi Ωγ

j

∫P (u,ξ)

∇20Tαγ = − 2

πΩmi Ωn

j

∫P (u,ξ)

∂2Tmn∂t2

.

Since a is trace-free, we conclude that

(3.98) aij = − 2

π

(Π mi Π n

j −1

2ΠmnΠij

)∫P (u,ξ)

∂2Tmn∂t2

.

3.3.3. Quadropole formula. In the case of a slowly varying source we mayreplace the intgral over P (u, ξ) by the integral over Σu.

Now consider the quadropole moment

(3.99) Qij(t) =

∫R3

xixjρ(t, x) d3x .

In the context of the slow motion approximation of Section 2, we have in particularthe limit of the continuity equation (2.16a), which then implies:

(3.100)dQijdt

=

∫R3

xixj∂ρ

∂t(t, x)d3x =

= −∫R3

xixj∇kpkd3x =

∫R3

(xjpi + xipj

)d3x = Dij(t)

Furthermore, we substitute from the equations of motion (disregarding non-linearterms),

(3.101)∂pi

∂t= −∇kT ik ,


to obtain that

(3.102)dDij

dt= −

∫R3

(xi∇kT jk + xj∇kT ik

)d3x =

=

∫R3

(∇kxi T jk +∇kxj T ik

)= 2

∫R3

T ijd3x .

Therefore

(3.103) 2

∫R3

∂2Tmn∂t2

=d4

dt4Qmn(t)

and we conclude that, for a slowly varying source,

(3.104) aij(u, ξ) = − 1

π

(Π mi Π n

j −1

2ΠmnΠij

)(ξ)

d4

dt4Qmn(u)

where Π mi (ξ) = δ mi − ξiξm.

Finally, we note that the radiation field of linearized gravitational waves takesthe form:

(3.105) qαβγδ =1

4

(aαγLβLδ − aαδLβLδ − aβγLαLδ + aβδLαLγ

).

Remark. In [CK93] D. Christodoulou and S. Klainerman published a rig-orous analysis of the exact asymptotic behaviour of gravitational waves. In par-ticular, it turns out that the nonlinearity of the Einstein equations accounts foran effect on gravitational radiation that is on the same level of magnitude as thelinear effects discussed above [Chr91]. Thus the weak field approximation doesnot reflect correctly the nature of gravitational waves.

3.4. Gravitational wave experiments. In this section we explain the ex-perimental implications of the theory of gravitational waves.

3.4.1. Gravitational wave detectors. A gravitational wave detector consists ofthree test masses, a reference mass m0, and two masses m1, m2, arranged in theclassical interferometer configuration depicted in figure 11. Each of the masses m1,m2 is positioned at a distance d0 from the reference mass m0, and any changesin distance are measured by comparing light travel times from the reference massm0 to m1 and m2, respectively, as depicted in the spacetime diagram of figure 12for the light beam that is split at m0, and reflected at m1, m2. The experiment iseither performed on earth, in which case the test bodies are suspended from pen-dulums, or in outer space. On earth, for periods much smaller than the pendulumperiod, the motion can be considered to be free in the horizontal plane, (definedby the three masses). If the experiment is performed in space, then the motion ofthe test masses is free anyway.

The reference mass m0 describes a geodesic Γ0 in spacetime, and m1, m2

are freely falling on nearby geodesics Γ1, Γ2. Let us pick an orthonormal basis(E1, E2, E3) for the spacelike hyperplane Σ0,

Σ0 =X ∈ TΓ0(0)M : g(X, Γ0(0)) = 0

.


m0

m1

m2

laser beam

beam splitter

mirrorhorizontal plane

Figure 11. Gravitational wave detector.

spacetime

m0

m1

m2

hypersurface of constant phase

same phase

Figure 12. Spacetime diagram of a light beam used to measuredistances in a gravitational wave detector.

Γ0

T

Σt

Γ0(t)Ei

Figure 13. Timelike geodesic Γ0 described by m0.

In general, see figure 13, we define,

Σt =X ∈ TΓ0(t)M : g(X,T ) = 0

,

where T = Γ0(t) denotes the unit tangent vector field along Γ0. We now consider


T

X

p

Γ0

Γ0(t)

Figure 14. Exponential map with base points on Γ0.

for each t the spacelike geodesic hyperplane

Ht = expΓ0(t)(Σt) ;

here p = expΓ0(t)(X) is the point at parameter value 1 along the spacelike geodesic

issuing at Γ0(t) with initial tangent vector X, c.f. figure 14. That is to say

(3.106) dist(p,Γ0(t)) := d = |X| , p = expΓ0(t)(X) .

We parallel propagate the basis (E1, E2, E3) along Γ0 to obtain an orthonormalbasis for each Σt. Any event p in a (cylindrical) spacetime neighborhood of Γ0 isthen assigned the coordinates

(t, x1, x2, x3) if p = expΓ0(t)(X) with X = x1E1 + x2E2 + x3E3 .

Since (E1, E2, E3) is an orthogonal basis, we have

(3.107) |X| =√

(x1)2 + (x2)2 + (x3)2 = d .

In this coordinate system, (as shown in the exercises on page 44 on cylindricalnormal coordinates),

(3.108) gµν − ηµν = O(Rd2

),

where R denotes the magnitude of the curvature components along Γ0. As we shallsee, the displacements of the test masses due to the passage of a gravitational wavetrain are O(Rτ2), where τ is the period (time scale) of variation of the curvature.

Assumption: We assume

d0

τ 1 .

This is satisfied for typical sources and detectors: For example for abinary neutron star system τ ∼ 10−3 s = 300 km, and d0 = 3 km forVirgo, so d0/τ = 10−2.

Under this assumption

(1) the speed of light can be taken to be 1 in the above coordinates, sodifferences in distance accurately reflect differences in light travel time.


horizontal plane

spaceE3

wave fronts

Figure 15. Vertical incoming radiation.

(2) we can replace the geodesic equation by the Jacobi equation, (recall herethe discussion in Ch. 1, Section 3):

(3.109)d2Xi

dt2= −RiT jTXj , where RiT jT = R(Ei, T, Ej , T ) .

Let us now suppose that the incoming gravitational waves are coming fromthe vertical direction in space, with respect to the horizontal plane defined by themasses, c.f. figure 15. That means the normal N to the wave fronts in Ht is −E3.Since we have previously chosen (E1, E2, N) to be a postive basis, (E1, E2, E3) isa negative basis, and thus (E2, E1, E3) again a positive basis. Moreover,

(3.110) L = T − E3 , L = T + E3 .

Recall, from Section 3.3,

(3.111) αAB = R(EA, L,EB, L) =aABr

+ o(r−2),

and all other curvature components are o(r−2) at least.Let us denote by xi(A) the ith coordinate of the test mass mA, A = 1, 2. Then,

the Jacobi equation (3.109) reduces to, (see exercises below),

d2x3(A)

dt2= o(r−2)

(3.112a)

d2xB(A)

dt2= − 1

4raBC x

C(A) + o

(r−2),(3.112b)

with initial conditions

limt→−∞

x3(A)(t) = 0 , lim

t→−∞xB(A)(t) = d0δ

BA ,(3.113)

limt→−∞

xi(A)(t) = 0 .(3.114)

To order o(r−1) we then obtain x3(A) = 0: the masses remain on the horizontal

plane. Since the fractional displacements are small, we can substitute on the rightthe initial values for xB(A), to obtain:

(3.115)d2xA(B)

dt2= −d0

4raAB .


m0

m1

m2

Figure 16. Instantaneous acceleration of the test masses.

Remark. Recall aAB is a symmetric trace-free 2×2-matrix. This means thatthe instantaneous acceleration of m2 is obtained from that of m1, by counter-clockwise rotation by 90, c.f. figure 16, (and exercises below).

Integrating once, we obtain the velocities,

(3.116) xA(B) = −d0

4r

∫ t

−∞aAB(u)du .

Integrating again, we obtain the displacement of the masses:

(3.117) ∆xA(B)(t) = xA(B)(t)− xA(B)(−∞) = −d0

4r

∫ t

−∞

∫ t′

−∞aAB(u)du

dt′ .

Set

(3.118) ΞAB(t) = −1

2

∫ t

−∞aAB(u)du

then we have

(3.119) ∆xA(B)(t) =d0

2r

∫ t

−∞ΞAB(u)du .

A quantity of interest, both theoretically and experimentally, is the overall dis-placement of the test masses:

(3.120) limt→∞

∆xA(B)(t) = xA(B)(∞)− xA(B)(−∞)

In the framework of the linear theory this vanishes provided that:

(1) in the initial and final states the astronomical bodies are at rest in theirrespective center of mass frames, and

(2) the final center of mass frame is at rest relative to the initial center ofmass frame.

Indeed, in the context of the weak field approximation discussed in Section 3.3 wehave established the quadrupole formula according to which

(3.121) aAB ∼d4Qmn

dt4.


Therefore

(3.122) ΞAB ∼d3Qmn

dt3,

and the instantaneous displacement results in:

(3.123) ∆xA(B)(t) ∼d2Qmn

dt2(t)− d2Qmn

dt2(−∞)

However, for an N -body configuration the quadropole moment is given by

(3.124) Qij =N∑α=1

MαXiαX

jα

so that

(3.125) Qij = 2N∑α=1

MαXiαX

jα +

N∑α=1

Mα

(XiαX

jα +Xi

αXjα

)takes the same value initially (at t = −∞) and finally (t =∞) provided the bodiesare at rest relative to their center of mass frame, and the final center of mass frameis at rest relative to the initial frame.

Remark. This is not true in the framework of the fully nonlinear theory. Infact, a permanent displacement was proven to occur and is known as the memoryeffect [Chr91]. Let us denote, for each (u, ξ), ξ ∈ S2, by

F (u, ξ) = |Ξ(u, ξ)|2

the radiative power per unit solid angle, in the direction ξ, at retarded time u.Then in the full theory the permanent displacement of the test masses is given by

(3.126) limt→∞

∆xA(B)(t) =d0

2rΣAB ,

where

ΣAB(ξ) =

∫ ∞−∞

ΞAB(u, ξ)du

However, we have an explicit formula,

ΣAB(ξ) =

∫S2

GAB(ξ, ξ′)E(ξ′)dµγ(ξ′)

(where GAB is a certain kernel), which relates the total displacement to the totalenergy radiated per unit solid angle in the direction ξ:∫ ∞

−∞F (u, ξ)du = E(ξ)

Note that the energy radiated in all directions ξ′ ∈ S2 contributes to the displace-ments of masses located in a certain direction ξ from the source. See [Chr91] formore details.

Exercises


1. (Equations of motion) Derive the equations of motion (3.112) from the Jacobi equation(3.109) using the results of Section 3.3.Hint: Use

T =1

2

(L+ L

), E3 =

1

2

(L− L

),

to write all components of the curvature in terms of the null decomposition of Section 3.3.2. Forexample,

R(E3, T, E3, T ) =1

4R(L,L, L, L) = −1

2

3∑A=1

R(L,EA, L, EA) = −ρ = o(r−2)

where we have used the (L,L) component of the vaccuum Einstein equation, and the form(3.81) of the metric.2. (Polarisation) Show that the specific form of the (approximate) equations of motion(3.115) implies that the instantaneous acceleration of m2 is obtained from that of m1, bycounter-clockwise rotation by 90.

Bibliography

[Chr91] Demetrios Christodoulou, Nonlinear nature of gravitation and gravitational-wave ex-periments, Physical Review Letters 67 (1991), no. 12, 1486–1489.

[Chr95] Demetrios Christodoulou, Self-gravitating relativistic fluids, Arch. Rational Mech. Anal.130 (1995), 343–400.

[Chr99a] , The instability of naked singularities in the gravitational collapse of a scalarfield, Annals of Mathematics 149 (1999), 183–217.

[Chr99b] , On the global initial value problem and the issue of singularities, Class. Quan-tum Grav. 16 (1999), A23–A35.

[Chr00] , The action principle and partial differential equations, Annals of MathematicsStudies, vol. 146, Princeton University Press, 2000.

[CK93] Demetrios Christodoulou and Sergiu Klainerman, The global nonlinear stability of theminkowski space, Princeton Mathematical Series, vol. 41, Princeton University Press,1993.

[Daf05] Mihalis Dafermos, Spherically symmetric spacetimes with a trapped surface, Classicaland Quantum Gravity 22 (2005), 2221–2232.

[DR08] Mihalis Dafermos and Igor Rodnianski, Lectures on black holes and linear waves, Pro-ceedings of the Clay Mathematics Summer School, arXiv:0805.3880, 2008.

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