Black Hole Radiation and Energy-Momentum

Tensor

Student: Yang LuSupervisor: Prof. G. ’t Hooft

Department of Physics Theoretical PhysicsUtrecht University

Diploma Thesis for Theoretical Physics inUtrecht University

· 2010 August ·

Abstract

In this thesis the formation of a black hole from collapsing matter isdiscussed in the first part. Then the derivation of Hawking Radiation isreviewed. After this, we focus on finding a renormalized result of the Energy-Momentum of Hawking Radiation.

Contents

Acknowledgments ii

1 Introduction 1

2 Einstein Equation and Energy-momentum Tensor 4

3 Collapsing Matter 7

4 Hawking Radiation 12

5 The Energy-Momentum Tensor for Hawking Radiation 17

Appendix 1 26

Appendix 2 28

Bibliography 29

i

Acknowledgments

During the one year working on this thesis, I received a lot of help fromTimothy Budd, Anne Franzen and other people. It was great pleasure tolearn from and talk to them.

My fellow student Wilke van der Schee and I worked on related subjects,we had a lot of discussion and I was frequently inspired and corrected byhim. I am impressed by his smartness and quick thinking.

I have to express full respect and appreciation to my supervisor Prof.G. ’t Hooft. I was in most time struggling to follow his talk and the limitedunderstanding of his words impressed me with his broad knowledge and deepunderstanding to physics problems. I will always remember his emphasis ongrasping the physics picture underlying the mathematics formulea.

I would also like to thank Mr. Peter Vleming, who helped to correct themistakes in my English writing.

In the end, I would like to thank Sani, Ivano, Josh, Truus and Hajo, whogave constant encouragement that enabled me to finally finish this thesis. Iwish you all the best in the future!

ii

Chapter 1

Introduction

The establishment of a General Relative theory of gravity enables scientiststo study large scale objects, such as in the cosmological processes. Oneprocess of particular interest is the formation of black holes from collapsingmatter.

The study of Black Holes has been through three main phases. Immedi-ately after the discovery of a Black Hole solution to Einstein’s equation, i.e.the appearance of the horizon, people were confused and felt uneasy aboutthe singularity of the metric at the horizon. Some even attacked the validityof General Relativity for this reason [5]. Then it is found that the spacetimeat the horizon actually behaves quite regularly. The apparent singuarity isjust due to a poor choice of coordinates. With a suitable set of coordinates,one can show that the geometry near the horizon is locally flat. But notbefore long scientists became once more perplexed by the so called ”No HairTheorem”, which states that the three parameters, Mass, Charge and Angu-lar Momentum, together describe a black hole completely, which means thatthe black hole as the final stage of the collapsed matter has no memory inits history of formation. Moreover, the mathematical similarity between theformula describing the black holes and that of the first law of Thermal Dy-namics suggests strongly that black holes have a thermal property, namely,they bear a temerature and the area of the horizon should be identified asthe entropy of a black hole.

The second phase of the black hole study begun when Hawking discov-ered that due to the quantum effect, black holes are in fact always emittingparticles there. This discovery surprised the community and initiated greatenthusiam on the study of black holes in a quantum context. One importantissue is the so called ”Information Loss”, which can be illustrated with theexample of the baryon number [8]. To be precise, Hawking radiation canbe taken as a special blackbody radiation, which favors none of particle orantiparticle. As a result the baryon number in the radiaton sums statisti-cally to zero, while the baryon number of matter plunged into a black hole

1

2

is always positive.The total result will then be that the black hole is chargedwith arbitrarily large baryon number over time!

It is String theory that brought the black hole study into its third phase.Black holes have entropy, but what is the origin of this entropy? We knowthat the entropy is responsible to the number of micro states of the thermalsystem. Therefore, to understand the entropy of a black hole one must firstidentify the micro states of it. An important model, without reffering tostring theory, to approach an understanding of the black hole microstatesis the ”Brick Wall” model [8]. Unfortunately, it seems impossible to ob-tain a finite result in the context of quantum field theory.Hoewever, stringtheory provides a unique way to approach this problem. By identifying themicrostates of a black hole as the excitations of elementary string, one cancount the states and get a very similar result, but not identical to a classicalone for extremal black holes [10–12,14]. In both the ”Brick Wall” model andstring approach, the entropy of the black hole is no longer precisely corre-sponding to the horizon area. Instead of that, it corresponds to the area ofsurface near the horizon, the so called Streched horizon.

The above text is just a sketch of the development of the theory on blackholes. Two questions that arised in the first and second phases were notmentioned. First, How can we understand the formation of black holes?Originally, black holes were simply found as solutions to Einstein Equations.However, can we know something about the cosmological process in whichthe black hole is formed? A simple model to study this problem is to assumea dust shell moving inwardly with the speed of light. Then we try to findout how the geometry of the spacetime behaves along the movement of thedust shell. Indeed, it is found that the final stage of the shell is a blackhole with the same mass [8]. In this thesis, we conclude the formation ofa black hole in a different way by showing that the density of the Energyafter a certain time will become infinite. Nevertheless, we have to add theremark here, this is just a macroscopic understanding. We don’t know yetwhat micro processes are involved to form a black hole. Another question,to be asked in the second stage, is about the Hawking Radiation. The blackhole solutions, for exmple, Schwartzchild solution, is found for zero source,i.e. for the case of vacuum. However, now the quantum effect says thatit emits particles constantly, which means that the surrounding of a blackhole is not empty, which indicates a disputable Energy-momentum tensor.Should this tensor be not zero, will it cause change to the geometry of thespacetime? Many physicists believe that it does [8, 13], which correspondsto the study of ”Back Reaction” of Hawking Radiation. Will this tensor bezero or not anyway? Can we caclulate it? Some physists suggest that had wetaken into account the contributions from both fermions and bosons, withthe assumption of supersymmetry that each particle has its super-partner,the two parts will always be cancelled out and sum to zero [5]. However,

3

if we do not assume the existence of supersymmetry, a direct calculation ofthe Energy-momentum tensor generally leads to divergence. Then methodssuch as ”Normal Ordering” the operator of the Energy-momentum tensorand subtracting the background are also suggested which may lead to afinite source. In this thesis study we propose a specific way of doing thesubtraction and find a finite result for the Energy-momentum tensor forHawking Radiation. The result indicates that at the horzion the divergenceremains, although perhaps less severe, a very interesting property to discuss.Obviously, this is not the end of the story. There is still a long way to go forthe study of ”Back Reaction”, particularly, some conceptual questions needto be clarified.

Questions raised in the context of string frame are of course also there.Unfortunately they are beyond the scope of this thesis, we will not refer tothem.

Chapter 2

Einstein Equation andEnergy-momentum Tensor

We start with deriving the expession for the energy-momentum tensor fromEinstein-Hilbert action to lay a background for the later text [6]. The defi-nition for the energy-momentum tensor is a bit complicated according to [7].However we are going to adopt the most straitfoward definition.

The Einstein-Hilbert action for gravity with matter fields reads:

S =

∫d4x√−g(

1

2κR + Lm) (2.1)

where g is the determinant of the metric tensor gµν ; κ = 8πGc4

with G theNewton constant and c the speed of light; R = gµνRµν is the Ricci scalar;and Lm is the Lagrangian of the matter field. For simplicity we choose thenon-interacting scalar field with mass m. Because we are going to adopt forthe metric the sign convention (−,+,+,+), the proper Lagrangian densityfor the scalar field should be(because the mass term in the Halmitonian musthave a postive sign):

Lm = −1

2∂µφ∂

µφ− 1

2m2φ2.

The variation of this action with respect to gµν (the inverse of gµν) leads tothe Einstein equation with the Energy-momentum tensor due to the matterfield. Its variation with respect to the field φ leads to the equation of motionof the field in spacetime with metric gµν . We present a brief derivation ofthese results in the following text.

Variation with respact to gµν :

δS =

∫d4x[δ

√−g(

1

2κR + Lm) +

√−g(

1

2κδR + δLm)] (2.2)

A. Variations of each term are given specifically as follows:

4

5

δ√−g = − 1

2√−g

δg

= − 1

2√−g

δe− lngµν

= −√−g2

gµνδgµν

δR = δ(gµνRµν)

= Rµνδgµν + gµνδRµν

= Rµνδgµν + gµν [∇ρΓ

ρµν −∇νΓ

ρµρ]

= Rµνδgµν +∇ρ(g

µνΓρµν)−∇ν(gµνΓρµρ)

and

δLm = −1

2∂µφ∂νφδg

µν .

Where in the second variation we applied the Platini identity:

δRρσµν = ∇µΓρσν −∇νΓ

ρσµ

to the case Rµν = Rρµρν and used the compatibility of the metric and covariant

derivative ∇ρgµν = 0. Then we get

δS =

∫d4x[−1

2gµν(

1

2κR + Lm) +

1

2κRµν −

1

2∂µφ∂νφ]

√−gδgµν + (2.3)

+1

2κ

√−g[∇ρ(g

µνΓρµν)−∇ν(gµνΓρµρ)] (2.4)

=

∫d4x[ 1

2κ(Rµν −

1

2gµνR) + (−1

2∂µφ∂νφ−

1

2gµνLm)]

√−gδgµν +(2.5)

+1

2κ

√−g[∇ρ(g

µνΓρµν)−∇ν(gµνΓρµρ)] (2.6)

=

∫d4x[(

1

2κGµν +

1

2Tµν)√−gδgµν + boundary terms] (2.7)

(2.8)

which eventually leads to Einstein equation:

Gµν = κTµν (2.9)

with

Gµν = Rµν −1

2gµνR (2.10)

6

andTµν = ∂µφ∂νφ+ gµνLm. (2.11)

B. Variation with respect to φ:

δS =

∫d4x√−gδLm (2.12)

Specifically,

δLm = δ(−1

2∂µφ∂

µφ− 1

2m2φ2)

= −1

2δ(∇µφ∇µφ)−m2φδφ

= −∇µφ∇µ(δφ)−m2φδφ

= (∇µ∇µ −m2)φδφ−∇µ(∇µφδφ)

The second term will contribute as a boundary term in the integration.Therefore, the equation of motion for the field becomes

(∇µ∇µ −m2)φ = 0. (2.13)

Now we have established the foundation based on which we are going to illus-trate the notion of Hawking radiation and calculate the energy-momentumtensor for the Hawking radiation.

Chapter 3

Collapsing Matter

Suppose in the space matter is distributed and suppose all the particles atthe initial moment are stationary. Then according to the Newtonian theoryof gravity, the matter will collapse and form a star. However, it is not clearwhether or not a black hole will be formed in Newtonian theory. In thischapter we are going to discuss the evolution of the matter. We adopt thecomoving coordinates, which is well illustrated in Weinberg’s book [1].Theidea is like this,the enormous number of particles that are distributed in thespace, are dense enough to form a dynamic grid and each particle carries itsown clock. In this way each particle provides a unique set of coordinates fora local event. Thus, one can argue and find that the metric in the comovingcoordinates has such a form:

gµν =[−1, f(τ, ε), g(τ, ε), g(τ, ε) sin2 θ

](3.1)

where τ is the time coordinate and ε is the space coordinate, θ, together withφ is the regular angular coordinates. And we can define a radius via:

g(τ, ε) = r(τ, ε)2

For this metric, the non-vanishing components of the connection are:Γτεε = 1

2f , Γτθθ = 1

2g, Γτφφ = 1

2g sin2 θ,

Γεεε = f ′

2f, Γεθθ = − g′

2f, Γεφφ = −g′ sin2 θ

2f, Γετε = f

2f,

Γθφφ = − sin θ cos θ, Γθθε = g′

2g, Γθθτ = g

2g,

Γφφτ = g2g

, Γφφε = g′

2g, Γφφθ = cos θ

sin θ,

where dot stands for the derivative with respect to τ , while prime to ε

Therefore, the non-vanishing Ricci-tensor components are:

Rττ = − f2f− g

g+(f2f

)2

+ 12

(gg

)2

,

Rτε = − g′

g+ g′g

2g2+ g′f

2fg,

7

8

Rεε = f2− g′′

g+ 1

2

(g′

g

)2

+ f g2g− f2

4f+ f ′g′

2fg,

Rθθ = g2

+ f ′g′−2fg′′

4f2 + f g4f

+ 1,

Rφφ =[g2

+ f ′g′−2fg′′

4f2 + f g4f

+ 1]

sin2 θ.

With the above ingredients, we can write down the Einstein equation. Weadopt the following expression:

Rµν =8πG

c4

(Tµν −

1

2gµνT

),

where the energy-momentum tensor reads:

Tµν = [ρ(τ, ε), 0, 0, 0]

and T = gµνTµν = −ρ is the trace of above tensor.If we define

S(τ, ε) ≡ 8πG

c4ρ(τ, ε), (3.2)

we get the following four equations from Einstein equation:

Rττ = − f

2f− g

g+

(f

2f

)2

+1

2

(g

g

)2

=1

2S (3.3)

Rτε = − g′

g+g′g

2g2+g′f

2fg= 0 (3.4)

Rεε =f

2− g′′

g+

1

2

(g′

g

)2

+f g

2g− f 2

4f+f ′g′

2fg=

1

2f ∗ S (3.5)

Rθθ =g

2+f ′g′ − 2fg′′

4f 2+f g

4f+ 1 =

1

2g ∗ S. (3.6)

The equation corresponding to Rφφ would be identical to that correspondingto Rθθ, so we could omit it. In addition, there is one more equation due toenergy-momentum conservation (∇µT

µν = 0):

S + (f

2f+g

g)S = 0. (3.7)

Solving the Equations

It is easy to see that one can find special solutions by separating the argu-ments, by assuming f(τ, ε) = R(τ)X(ε) [1]. However, we are going to findthe most general solutions. We start with the simplest equation, Eq.(3.4). Itcan be rewritten in the following way:

g′

g′=

1

2(f

f+g

g)

9

which leads to by integrating over τ :

|g′| = |g ∗ f ∗ d(ε)|12 (3.8)

where d is an arbitrary function of ε;We can manipulate the other equations this way, mainly to cancel most

of the derivatives with respect to ε:Eq.(3.6) ∗ 2/g − Eq.(3.3)− Eq.(3.5)/f leads to:

2g

g+

2

g− 1

2f

g′2

g2= 0 (3.9)

which together with Eq.(3.8) leads to:

4g − g2

g= d− 4 ≡ l(ε) (3.10)

This equation is solved by (ref. Appendix 1):

g =k2

4(coshφ− 1)2 (3.11)

where k = k(ε) is also an arbitrary function and θ satisfies an implicit func-tion relationship

sinhφ− φ =

√l

kτ + n(ε)

where n(ε) is another arbitrary function of ε.Also, Eq.(3.3) + Eq.(3.5) gives:

− gg

+1

2

g2

g2+f ′g′ − 2fg′′

2f 2g+

1

2f

g′2

g2+

f g

2fg= S (3.12)

Eq.(3.4) helps to simplify the above equation into:

−g +gg′

g′+f ′g′ − 2fg′′

2f 2+

g′2

2fg= gS. (3.13)

Eq.(3.4) helps to simplify Eq.(3.3) into:

− g′

g′− g

2g+g′g

gg′− g2

4g2=

1

2S. (3.14)

Now the remaining work to do is solve Eq.(3.7),Eq.(3.12) and Eq.(3.13).However, it turns out that Eq.(3.4) and Eq.(3.8) together simplify Eq.(3.12)to an identical one as Eq.(13), which means that Eq.(3.12)is equivalent toEq.(3.13). This should not suppirse us because the four Einstein equations

10

obey the Bianchi identity.Eq.(3.7) is solved by

S = (f12 g)−1c(ε). (3.15)

where c = c(ε) is again an arbitrary function. However, we can see that c(ε)actually is very much closely related to the density of the matter distribution,since c(ε) = 0 means there is no matter.

In the end, when the derivatives of g in Eq.(3.12) are replaced by theexplicit expressions, with the help of Eq.(3.8)and Eq.(3.14) we get finally

(lk)′ = 2d12 c (3.16)

So,Eq(3.8),(3.10) and (3.14) consist of the most general solutions to the orig-inal equations(Eq.(3.3),(3.4),(3.5),(3.6)and (3.7))as long as Eq.(3.15) is sat-isfied.

There are three arbitrary functions of ε,i.e. (l(ε) = d(ε)−4, k(ε)andn(ε)).Eq.(3.15)fixes one given a specific mass distribution. The remaining two areyet to be determined according to the boundary condition.

Suppose we have a uniform distribution of matter, then a solution can befound with proper boundary condition. However, noticing that in Eq.(3.11)when the time τ evolves, at a certain moment, φ acquires a value such thatcoshφ − 1 equals to zero, therefore, g(τ, ε) = 0. From Eq.(3.14), this resultleads to a infinite energy momentum density. Then we conclude that all thematter converge at a single point. Hence a black hole is formed.

We can also understand the above solution in a different way:We first rewrite the metric in a neat form as follows

ds2 = −dτ 2 +r′(τ, ε)2

4κ(ε)+ r(τ, ε)2dΩ2 (3.17)

where r(τ, ε)2 = g(τ, ε), and the second term can be derived according toEq.(3.8).

The above expression of the metric can be recovered by a coordinatetransformation from a Schwartschild metric. Suppose the matter has alreadycollapsed, then a black hole is formed. According to the ”No Hair” theorem,it is simply a Schwartzchild black hole. Then in the normal coordiates system,the line element is:

ds2 = −(1− 2M

r)dt2 + (1− 2M

r)−1dr2 + r2dΩ2 (3.18)

If we describe the same spacetime in term of the geodesics associated withvarious energy, the energy and eigentime can be a pair of coordinates whilethe angular coordinates remain unchanged. One can write down the time

11

and radial commponents of the geodesic equations. After performing the in-tergration over eigentime τ we get a conserved quantity ε which is interpretedas the energy of the moving object. In terms of these two parameters, thesame metric becomes:

ds2 = −dτ 2 +r′2(τ, ε)

ε2dε2 + r2(τ, ε)dΩ2 (3.19)

Apparently, κ(ε) in Eq.(3.16) is identified as ε2,up to a parametrization.This coincidence can be understood in the following way: A. In the intial

stage, when τ = 0, we set drdτ

= 0 which, from the Geodesic Equation,leadsto r(τ = 0, ε) = 2M

1−ε2 . Also we set t(τ = 0) = 0.

B. At this moment the invariant line element is in the original coordi-nates system expressed as:ds2 = (1 − 2M

r(τ=0,ε))−1dr2|τ=0. While in the τ − ε

coordinates system it is expressed as:ds2 = r′(τ=0,ε)2

ε2dε2.

C. Had we written dr2|τ=0 = r′2(τ = 0, ε)dε2, we refer that 1 − 2Mr(τ=0,ε)

coincides ε2. Indeed,1− 2M

(τ=0,ε)= ε2.

Chapter 4

Hawking Radiation

Hawking radiation can be derived in several ways. In Hawking’s originalpaper, he tracked the collapsing matter to explain where the radiation origi-nates from. Soon, Unruh and other physists worked in Rindler spacetime anddeveloped a mathematically more clear formalism to illustrate the Hawkingradiation. In this chapter we present a brief introduction of both methods.

To understand Hawking radiation we must firstly clarify the notion ofparticles.

In Minkovskian spacetime, there is no confusion. Take a scalar field forexample: We solve the equation of motion and find the eigenmodes withpositive energy with respect to the time parameter. Since the theory isLorentzian invariant, the sign of the frequency in the eigenmode will notchange after one performs a Lorentz transformation,i.e. the positive energymodes stay positive. Therefore, we have a well defined notion of particle.However, in the curved spacetime we are confronted with a much differentsituation. In general, the metric is no longer static. If we insist on that thetheory should only be covariant, then we might end up in a case that theenergy of a ”particle” becomes negative after a coordinates transformation[2]. Fortunately, for a spacetime equipped with a general static metric, thereis a well defined way of quantization. Nevertheless, in such a case the notionof a particle is quite different than the normal sense we gained in Minkovskianspacetime.

Suppose we have a manifold equipped with a static metric. The metricwill be unchanged under a time translation, which corresponds to a timelikeKilling vector. Then the eigenmodes are defined as having positive energywith respect to this time-like Killing vector. These eigenmodes are orthonor-mal and complete in the region of the submanifold, with a proper boundarycondition, i.e. their value on the boundary of a maximal smooth submanifold.Then the field is quantized as follows:

φ =

∫dσ(fσaσ + f ∗σa

†σ) (4.1)

12

13

where σ is the index of a particular eigenmode, and fσ has a positive fre-quency. Therefore, the operator aσ and a†σ can be interpreted as such thatannihilate and create a particle in that mode in the sub-region of the space-time.

Following this way, by working in a two dimentional Minkovskian space-time, Stephen Fulling [2] quantized the scalar field in the normal Cartesiancoordinates and in Rindler coordinates respectively. Two sets of annihilationand creation operators were defined. In terms of these operators, Fullingderived the following result

aσ =

∫ ∞−∞

dσ′(Uσ,σ′ bσ′ + Vσ,σ′ b

†σ′

) (4.2)

where a denotes operators defined in Rindler coordinates while b in Cartesiancoordinates, and U and V are some non-vanishing functions of σ and σ

′. This

result suggests that an annihilation operator in Rindler spacetime (whichcovers only a wedge of the full Minkovskian spacetime) will not annihilate thevacuum defined in a full Minkovskian spacetime. Instead, it creates a bunchof particles. In another word, a particle to an observer in Rindler spacetimeis a composite of many particles in the original Minkovskian spacetime.

To fully appreciate this result, we must notice that the Rindler space-time obtained by a coordinate transformation is actually not physicallyequal to the orignal one. It is not geometrically complete, which meansthat some geodesics extending in the full Minkovskian spacetime will haveto be abrupted due to meeting the edge of Rindler spacetime, which func-tions somehow as a black hole event horizon [4]. Similarly, the formation ofthe event horizon of a real black hole tears spacetime up into two sections.Outside the event horizon, the eigenmodes defined in this region are notequivalent to those defined for the whole region. As a result, an stationaryobserver in this region (to whom the operator representing a measurementconsists of annihilation and creation operators defined outside the horizon)will see many particles from the original vacuum.

In Hawking’s paper about pair production [3], he adopted more or less thesame way to derive the notion of black hole radiaton. His arguing involvesmainly the incoming (fj) and outgoing modes(pi) which are decomposed withrespect to advanced and retarded times defined as follows:

v = t+ r + 2M log | r2M− 1| (4.3)

u = t− r − 2M log | r2M− 1| (4.4)

The incoming modes are determined according to their data in the infinitepast that applies to the whole spacetime. The outgoing modes are determinedaccording to their value in the region outside the event horizon. Since they

14

are belonging to different sets of quantization, there is a realtion as follows:

pi =∑j

(αijfj + βijf∗j )

which corresponds to the relation between operators:

ci =∑j

α∗ijdj − β∗ijd†j,

where cj is the operator that annihilates a particle in the outgoing modewhile dj annihilates a particle in the incoming mode. Therefore, from theinitial vacuum for the incoming mode the one will observe particles in theoutgoing modes with the probability

∑j |βij|2. Then, the task is to find βij

Physically, there is a latest time v0, leaving after which an incoming modewill be trapped by the event horizon and can not propagate to the infinitefuture.So, to escape from the horizon, the wave should have left earlier andarrive at the position of the horizon before the horizon is formed. Hence itwould meet the collapsing matter somewhere outside the surface where thehorizon will be located. The critical point is that the wave leaves just at thev0 and meets the collapsing matter at the horizon. For those incoming wavethat barely escapes from the event horizon and comes out as a pure outgoingmode, it should have a phase in the infinite past like

ω

κlog(

v0 − vconst.

) (4.5)

where ω is the frequency of the mode and κ is the surface gravity of theblack hole. Since v is very close to v0, the phase becomes divergent. Thisdivergence corresponds to the waves that acquire an effectively extremelyhigh frequency when they meet and propagate through the collapsing matternear the horizon. These waves contribute most part of the origin of theHawking radiation. After some complicated calculation Hawking was able toprove that the number of particles in a wave packet mode that propagate tothe infinite future is proportional to exp(2πωκ−1)− 1), which suggests thatthe black hole radiates like a black body with temperature κ/2π = 1/8πM.This temperature is found in various ways [8] and in general they agreewith each other. In the case we take the back reaction of the radiation intoaccount, the temperature would be slightly smaller, hence a grey factor isdifined [13].

We now follow a different way to understand Hawking radiation and de-rive the main result based on which the later calcuation for the Energy-momentum tensor is done. What we present here can be found in a reviewarticle by ’t Hooft [8].

It is well known that near the horizon of a Schwartzchild black hole withmass M, the metric around the equator approximate one in flat space time:

15

ds2 = −dT 2 + dX2 + dY 2 + dZ2 (4.6)

with the relation to the Schwartschild coordinate as follows:

T 2 = (r

2M− 1) exp(

r

2M) sinh2(

t

4M)

Z2 = (r

2M− 1) exp(

r

2M) cosh2(

t

4M)

X = 2Mθ −Mπ

Y = 2Mφ

The same spacetime (then one must allow the ρ in the following equationsto be negative) can also be described with Rindler coordinates by performingthe following coordinate transformation

Z = ρ cosh τ

T = ρ sinh τ

which will lead to the metric

ds2 = −ρ2dτ 2 + dρ2 + dX2 + dY 2 (4.7)

The field of course still obeys Eq.(2.13). But now we need to treat thisequation carefully because three components of the connection in this systemis not zero, hence the covariant derivative is no more trival. As a result thefield equation simplifies into:

[(ρ∂ρ)2 − ∂2

τ + ρ2(∂2X + ∂2

Y −m2)]φ = 0 (4.8)

Now we follow the general method to quantize φ. To find the eigenmodes weassume a general form for a particular solution:

φω = f(ρ)ei(kXX+kY Y )−iωτ (4.9)

which leads to the following equation:

(ρ∂ρ)2 + ω2 − ρ2(k2

X + k2Y +m2)f(ρ) = 0 (4.10)

The following function solves this equation:

K(ω,1

2µρ) =

∫ ∞0

ds

ssiωe−

iµρ2

(s− 1s

)

=

∫ ∞−∞

dxeiωxe−iµρ sinhx (4.11)

16

where a change of variable, s = exp(x), is made in the second line andµ =√k2 +m2. Then full solutions to Eq.(4.8) are like:

φω = K(ω,1

2µρ)ei(kXX+kY Y )−iωτ (4.12)

However, so far ω is allowed to be negative, which corresponds to negativeenergy mode. To cure this problem, a Bogolyubov transformation is appliedand eventually we end up with the following quantization of the field in theregion of ρ > 0:

φ =

∫ ∞0

dωe−iωτ∫

d2keik∗x√2(2π)4

K(ω,1

2µρ)√

1− e−2πωa+(k, ω) + h.c. (4.13)

In the region ρ < 0 the field can be quantized analogously, with operatorsfunctioning in that particular region. These two sets of operators can beexpressed in terms of the annihilation and creation operators in Minkovskianspacetime if one traces back. The way they operate on Minkovskian vacuumcan be derived, hence Minkovskian vacuum can be expressed in terms oflinear combinations of many-particle states in Rindler spacetime:

|Ω >= Πk,ω

√1− e−2πω

∞∑n=0

|n >+ |n >− e−πnω (4.14)

Now suppose one is living the Rindler spacetime outside the horizon (withρ > 0). His performance of a measurement is mathematically expressedin terms of an operator consisting of a+(k, ω) and a†+(k, ω) acting on theMinkovskian vacuum. Therefore, one sees that the vacuum is no longerempty to an observer in Rindler spacetime. In fact, it becomes a thermalsource with the same temperature we have found previously.We have omitted a lot of technical details which can be found in the orginalarticle [8]. The results we presented here will be enough for further calcula-tion in the following chapter.

Chapter 5

The Energy-Momentum Tensorfor Hawking Radiation

Now we can start the calculation of the Energy-momentum tensor for theHawking radiation. In the quantum context, Einstein’s equations is suggestedwith a semiclassical form [7]:

Gµν =< Tµν > .

The reason to do this is that after quantization of the matter field, Tµνbecomes an operator while Gµν remains the metric with c-number value. Onemay argue that, if we now take gµν as an object which is to be determinedwith the source operator Tµν , Gµν will turn out to be operator in terms of thematter field, too. However, we don’t follow this way. Instead, we take theexpectation value of both side of Einstein equation over the quantum statethat the physical system sits in and then the above equation is obtained,which of course is an approximation.

Then a crucial question comes about. Over which quantum state shouldwe take the expectation value of the two sides of the equation? As we haveseen in previous chapter, there are at least two sets of quantum states de-fined for the local space time near the horizon. One is defined in Minkovskianspacetime and the other in Rindler spacetime. Which is legistimate? Here wemake an assumption that, the fundamental quantum states in which an eventoccurs should be defined in a coordinate-system that covers the neighbour-hood of that point. Since the coordinates with Minkovskian metric extendacross the horizon smoothly, we conclude that at the horizon the fundamen-tal quantum states are the Minkovskian states. And we assume that theparticluar state that the spacetime near horizon sits in is the Minkovskianvacuum. This dicussion should be sufficient to lay a cornerstone for the fur-ther calculation. According to the result we have obtained previously theenergy-momentum tensor that reads:

T ′µν = ∂µφ∂νφ+ gµνLm = ∂µφ∂νφ−1

2gµν(∂µφ∂

µφ+m2φ2), (5.1)

17

18

where the prime is used to distinguish from the later modified version.We are going to compute < Ω|T ′µν |Ω >. First, the derivatives of the fieldappearing in the stress-tensor are

∂τφ =

∫ ∞0

dω(−iω)e−iωτ∫

d2keik∗x√2(2π)4

K(ω,1

2µρ)√

1− e−2πωa+(k, ω) + h.c.

∂ρφ =

∫ ∞0

dωe−iωτ∫

d2keik∗x√2(2π)4

∂ρK(ω,1

2µρ)√

1− e−2πωa+(k, ω) + h.c.

∂Xφ =

∫ ∞0

dωe−iωτ∫d2k(ikX)eik∗x√

2(2π)4K(ω,

1

2µρ)√

1− e−2πωa+(k, ω) + h.c.

∂Y φ =

∫ ∞0

dωe−iωτ∫d2k(ikY )eik∗x√

2(2π)4K(ω,

1

2µρ)√

1− e−2πωa+(k, ω) + h.c.

Before proceeding with the caculation, we stop for a while to talk aboutthe symmetry of the tensor defined in Eq.(5.1). Unfortunately we find thethe derivatives of the field do not completely commute, which implies thatthe expectation value for the their products in some particular states aredependent on the way that the particular product is formed. For example,some basic algebra shows:

[∂τφ, ∂ρφ] =

∫ ∞0

dω(iω)(1− e−2πω)

∫d2k

4π∂ρ[K(ω,

1

2µρ)K∗(ω,

1

2µρ)] (5.2)

which does not vanish (it is actually infinite). Therefore, such defined tensorcan not be symmetric in all cases. To cure this problem, we redefine thetensor in the following way:

Tµν =1

2(∂µφ∂νφ+ ∂νφ∂µφ) + gµνLm (5.3)

This one is evidently symmetric although artificial. In the following text allthe caculation is done for the new tensor.Now we take for example to compute < Ω|∂τφ∂τφ|Ω > to get a sense of thegeneral structure of the expectation value.

< Ω|∂τφ∂τφ|Ω > =

∫ ∞0

dω(−iω)e−iωτ∫

d2keik∗x√2(2π)4

K(ω,1

2µρ)√

1− e−2πω

∗∫ ∞

0

dω′(iω′)eiω′τ

∫d2k′e−ik

′∗x√2(2π)4

K∗(ω′,1

2µ′ρ)

√1− e−2πω′

[< Ω|a+(k, ω)a†+(k′, ω′)|Ω > + < Ω|a†+(k′, ω′)a+(k, ω)|Ω >]

19

We then have to compute the expectation value of a+(k, ω)a†+(k′, ω′). Notic-ing that |Ω > consists of states of equal number of particles from regionsdenoted by (+) and (-) respectively, we find that to get a non-zero expeca-tion value, k, ω must be identical to k′, ω′. Specifically,

< Ω|a+(k, ω)a†+(k′, ω′)|Ω >= δ(2)(k − k′)δ(ω − ω′) < Ω|a+(k, ω)a†+(k, ω)|Ω >(5.4)

where the operators with tilde are normalized version of the correspondingones according to:

< 0|a+(k, ω)a†+(k′, ω′)|0 >= δ(2)(k − k′)δ(ω − ω′) < 0|a+(k, ω)a†+(k, ω)|0 >

and

< Ω|a+(k, ω)a†+(k, ω)|Ω > =∑n=0

−< n|+< n|a†+(k, ω)a+(k, ω)(∑n′=0

|n′ >+ |n′ >− e−π(n+n′)ω)(1− e−2πω)

=∑

n=0,n′=0

−< n|+< n|n′ + 1|n′ >+ |n′ >− e−π(n+n′)ω)(1− e−2πω)

=∑n=0

(n+ 1)e−2nπω(1− e−2πω)

=1

1− e−2πω(5.5)

Therefore

< Ω|a+(k, ω)a†+(k′, ω′)|Ω >= δ(2)(k − k′)δ(ω − ω′) 1

1− e−2πω(5.6)

and

< Ω|a†+(k, ω)a+(k′, ω′)|Ω > = < Ω|a†+(k, ω)a+(k′, ω′)− δ(2)(k − k′)δ(ω − ω′)|Ω >

= δ(2)(k − k′)δ(ω − ω′) e−2πω

1− e−2πω. (5.7)

We then find

< Ω|∂τφ∂τφ|Ω >=

∫ ∞0

dωω2(1 + e−2πω)

∫d2k√

4πK(ω,

1

2µρ)K∗(ω,

1

2µρ)

(5.8)

However, this integral is divergent. One can see this by first extracting afactor like exp[iω(x − y)] from KK∗, the integration over ω will then leadto a second order derivative of the delta function. Placing this derivative ofdelta function back to complete the integration over x and y in KK∗ onesees immediately the divergence (more details are given in later text)! Thisleads then to the conclusion that < Ω|Tττ |Ω > is divergent(although we have

20

not finished the whole calculation, it is realized that this divergece can notbe cancelled because the remaining part of < Tττ > has an overal factor gττwhich is ρ dependent). We inevitably come to the universal problem thatthe expectation value of the Energy-momentum tensor for Hawking radia-tion is divergent! One may wonder where the divergence comes from. Wenotice that the contribution from < Ω|a†+(k, ω)a+(k′, ω′)|Ω > will not leadto divergence in the integral over ω. Then the divergent part must comefrom < Ω|a+(k, ω)a†+(k′, ω′)|Ω >.A closer analysis of the calulation of thisterm shows that the divergence comes from the expectation value in the vac-uum state. In Eq.(5.5) we can see that, if n > 0 the sub-terms are alwayssuppressed by a factor e−2nπω. As soon as n = 0 we get a contribution of1 − exp(−2πω), which eventually causes the occurence of divergence. Thisn = 0 contribution corresponds to the expectation the operator in Rindlervacuum.

This fact also reminds us that in the quantum field theory in flat space-time, the vacuum expectation of Halmitonian is divergent. In fact, it is of thesame nature of the divergence in the above case. People did not bother muchabout the infinity of the vacuum energy because it was simply impossible tomeasure it. Now we wonder if this vacuum expectation causes gravitationalcurvature in the spacetime. One may argue that the vacuum expectationserves as a the cosmological constant in Einstein’s equation. However, ac-cording to Einstein’s general relativity the space time is just Minkovskianif there is no source particles. Therefore, it is reasonable to assume thatthe contribution in < Ω|Tµν |Ω > from Rindler vacuum is just a ”silent”background and does not affect the metric of the spacetime. Then, what isessential is this expectation subtracted by the expectation in Rindler vac-uum. There is another interpretation to this idea [3, 5, 9]. Had we normalodered the operators in Tµν ,, we would also obtain a finite but different result.However, the divergence is removed in essentially the same way in both ways.In the following text we are going to adopt the more physical interpretation,i.e. subtraction of the background and show that this assumption leads to afinite expectation value of the Energy-momentum tensor for Hawking Radi-ation.

The caculation of < 0|Tµν |0 > can be done by following the same methodfor calculating < Ω|Tµν |Ω >. The corresponding quantities are as follows:

< 0|a†+(k, ω)a+(k′, ω′)|0 >= δ(2)(k − k′)δ(ω − ω′) (5.9)

21

hence,

< 0|∂τφ∂τφ|0 > =

∫ ∞0

dω(−iω)e−iωτ∫

d2keik∗x√2(2π)4

K(ω,1

2µρ)√

1− e−2πω

∗∫ ∞

0

dω′(iω′)eiω′τ

∫d2k′e−ik

′∗x√2(2π)4

K∗(ω′,1

2µ′ρ)

√1− e−2πω′

[< 0|a+(k, ω)a†+(k′, ω′)|0 > + < 0|a†+(k′, ω′)a+(k, ω)|0 >](5.10)

The second expectation in the parentheses is equal to zero. Inserting thecommutator of a†+ and a+ into this equation we get:

< 0|∂τφ∂τφ|0 >=

∫ ∞0

dωω2(1− e−2πω)

∫d2k√

4πK(ω,

1

2µρ)K2(ω,

1

2µρ)

(5.11)Combining Eq.(5.8) and Eq.(5.11) we have:

< ∂τφ∂τφ >s ≡ < Ω|Tττ |Ω > − < 0|Tµν |0 >

= 2

∫ ∞0

dωω2e−2πω

∫d2k

4πK(ω,

1

2µρ)K∗(ω,

1

2µρ)

=

∫ ∞0

dωω2e−2πω

∫d2k

2πK(ω,

1

2µρ)K∗(ω,

1

2µρ) (5.12)

Applying the same method to the other components of the tensor, we obtainthe following results:

< ∂ρφ∂ρφ >s =

∫ ∞0

dωe−2πω

∫d2k

2π∂ρK(ω,

1

2µρ)∂ρK

∗(ω,1

2µρ)(5.13)

< ∂Xφ∂Xφ >s =

∫ ∞0

dωe−2πω

∫d2kk2

X

2πK(ω,

1

2µρ)K∗(ω,

1

2µρ) (5.14)

< ∂Y φ∂Y φ >s =

∫ ∞0

dωe−2πω

∫d2kk2

Y

2πK(ω,

1

2µρ)K∗(ω,

1

2µρ) (5.15)

Now we turn to terms < ∂µφ∂νφ + ∂νφ∂µφ >s with µ 6= ν. It is easyto see that, when µ = X or Y , the integral over k gives zero due to thesymmetry of the integrand and that of the domain of the integration. Thenonly < ∂τφ∂ρφ+ ∂ρφ∂τφ >s remains to be evaluated, and one finds

< ∂τφ∂ρφ+ ∂ρφ∂τφ >s=

∫ ∞0

dω(iω)e−2πω

∫d2k

2π[∂ρKK

∗ −K∂ρK∗] (5.16)

One can prove that this is actually zero for the K functions with the givenform (ref. Appendix 2). To eventually complete the computation, we still

22

have one more term to consider: < φ2 >s, which can be done easily byfollowing the same procedure:

< φ2 >s=

∫ ∞0

dωe−2πω

∫d2k

2πK(ω,

1

2µρ)K∗(ω,

1

2µρ) (5.17)

Then we perform the last step of the computation before giving specifi-cally the components of the tensor. Namely,

< Lm >s = −1

2[gµν < ∂µφ∂νφ >s +m2 < φ2 >s]

= −1

2[gρτ < ∂ρφ∂τφ+ ∂τφ∂ρφ >s +gµµ < ∂µφ∂µφ >s +m2 < φ2 >s]

= −1

2[gµµ < ∂µφ∂µφ >s +m2 < φ2 >s]

=1

2ρ2

∫ ∞0

dωω2e−2πω

∫d2k

2πKK∗ + (5.18)

−1

2

∫ ∞0

dωe−2πω

∫d2k

2π[(k2 +m2)KK∗ + ∂ρK∂ρK

∗] (5.19)

So far, we have caculated all the ingredients necessary for computing theexpectation value of the stress tensor. Since the metric gµν is diagonized, andwe have found that < ∂µφ∂νφ + ∂νφ∂µφ >s= 0 with µ 6= ν, the non-trivalcomponents of the tensor are only those diagonal ones. It is very surprisingthat integrals in the above quantities can be performed completely and weare given a neat final result. Let’s take < ∂τφ∂τφ >s as an example and seehow this is possible.

< ∂τφ∂τφ >s =

∫ ∞0

dωω2e−2πω

∫d2k

2πK(ω,

1

2µρ)K∗(ω,

1

2µρ)

=

∫ ∞0

dωω2e−2πω

∫d2k

2π

∫ ∫dxdyeiω(x−y)e−iµρ(sinhx−sinh y)

=

∫ ∞−∞

dxdy

∫d2k

2π

∫ ∞0

dωω2e−2πωeiω(x−y)e−iµρ(sinhx−sinh y)

=

∫ ∫dxdy

∫d2k

2π

2

[2π − i(x− y)]3e−iµρ(sinhx−sinh y)

=

∫ ∫dudv

∫d2k

2π

2e−iµρv

[2π − i(u− arcsinh(sinhu− v))]3√

1 + (sinhu− v)2

(5.20)

in the last step we made a change of variables as follows

x → u

sinhx− sinh y → v (5.21)

23

and the square root in the denominator comes about as the Jacobi determi-nant. Next we can perform the integration over k in the following way

∫d2k

2πe−iµρv =

∫ ∞0

dkke−iµρv

=

∫ ∞m

dµµe−iµρv

=

∫ ∞0

dk(k +m)e−ikρve−imρv

=e−imρv

2[i

ρ2δ′(v) +

m

ρδ(v)] (5.22)

where we used the following formula:∫ ∞0

dkkne−ikv =in

2

dn

dvn

∫ ∞−∞

dke−ikv =in

2

dn

dvnδ(v)

with n an non-negative integer.

Inserting Eq.(5.22) into Eq.(5.20) we get

< ∂τφ∂τφ >s =

∫ ∫dudv

e−imρv[ iρ2δ′(v) + m

ρδ(v)]

[2π − i(u− arcsinh(sinhu− v))]3√

1 + (sinhu− v)2

=

∫ ∞−∞

du[− i

ρ2A′(0) +

m

ρA(0)]

=3

8π4ρ2(5.23)

where

A(v) =e−imρv

[2π − i(u− arcsinh(sinhu− v))]3√

1 + (sinhu− v)2

In precisely the same manner we find exact values for the following quantities

< ∂ρφ∂ρφ >s =1

8π4ρ4+

1

3π2ρ4(5.24)

< ∂Xφ∂Xφ >s =1

8π4ρ4+

1

12π2ρ4− m2

8π2ρ2(5.25)

< ∂Y φ∂Y φ >s =1

8π4ρ4+

1

12π2ρ4− m2

8π2ρ2(5.26)

< m2φ2 >s =m2

4π2ρ2(5.27)

< Lm >s = − 1

4π2ρ4(5.28)

24

Therefore one can work out for the Energy-momentum tensor

< Tµν >s=

26666664

38π4ρ2

+ 14π2ρ2

0 0 0

0 18π4ρ4

+ 112π2ρ4

0 0

0 0 18π4ρ4

− 16π2ρ4

− m2

8π2ρ20

0 0 0 18π4ρ4

− 16π2ρ4

− m2

8π2ρ2

37777775(5.29)

We have got a neat result. To check the correctness let’s verify it meets theconservation law of the Energy-momentum, i.e.

∇µTµν = 0 (5.30)

For the metric in Eq.(4.7), the non-vannishing components of the connectionsare as following:

Γρττ = −ρ (5.31)

Γτρτ = Γττρ =1

ρ(5.32)

Then a bit algebra shows that

∇µTµν = gµαgνβ∇µTαβ (5.33)

= gµαgνβ(∂µTαβ + ΓθµαTθβ + ΓθµβTθα) (5.34)

= gνν(gνν∂νTνν +1

ρTρν −

1

ρ2Γττ νTττ ) (5.35)

where ν indicates that summing on the index is not taken. From this formulait is evident that when ν 6= ρ

∇µ < T µν >s= 0 (5.36)

For the case ν = ρ we have

∇µ < T µρ >s= ∂ρ < Tρρ >s +1

ρ< Tρρ >s −

1

ρ3< Tττ >s= 0 (5.37)

Therefore the conservation is checked. Of course this is garanteed by the defi-nition of the Energy-momentum tensor and the Einstein equation. Our checkjust ensures us that the caculation is most likely correct. At this stage we canmake some comment on the result. Hawking radiation in Schwartschild met-ric is also expected to be spherically symetric. Hence the Energy-momentumtensor should be spherically symmetric,too. The tensor we find is diagonaland depends only on ρ means that it agrees with this consideration. More-over, the component < Tττ >s goes as 1/ρ2 agrees to the feature of theenergy flux. However, there seems an evident drawback of this result, it istime-independent! Because the black hole is lossing mass through emitting

25

particles, this seems impossible. One may attribute this to that at the be-ginning we assumed a static metric. Since the rediation takes place in aspherically symmetric way, the metric remains the same form up to a rescalevia the dependence of the ρ on the black hole Mass at every moment. If weassume a time-depending mass of the black hole, i.e. a function M(τ), theresult for the stress tensor will also be dynamic.

Another thing to notice is, the tensor diverges at the horizon, no mattterhow much mass the black hole possesses. Since we can always go to thehorizon by taking the limit ρ→ 0 which lead to divergence in the tensor. Weremember that in Hawking’s derivation of the Black Hole radiation, a normalmode of particle with advanced time close to the last moment for escapingthe horizon will gain an effectively infinite frequency when approaching thehorizon, hence causing effectively infinite energy there. This is the origin ofthe divergence of the tensor at the horizon and suggests a universal featurenear the horizon for black holes. One tough question about this result is howto explain the conflict betweeen the divergent stress tensor and the flatnessof spacetime at the horzion? To this question the following explanation isproposed: the divergence is due to that the observer is sitting in Rindlerspacetime and undergoing an accerleration. For an freely falling observer hesees no radiation and no dramatic change in spacetime. We can understandthis difference in the way that the Rindler observer sees only part of the’truth’ while the other part is hidden behind the horizon.

Then what’s the significance of this calculation except it looks a bit nicerthan a divergent one? On one hand, it might serve to convince us thatthe Rindler particles are also normal particles that can be measured with afinite result. On the other hand, we can use this result to calculate the backreaction of the radiation to the metric of the spacetime.

Appendix 1

The equation to solve reads:

4g − g2

g= l

where l = l(ε) is a function of ε. Since the derivative is with respect to τ , wecan view l as a constant. Then we replace the second order derivative with:

g =dg

dτ=gdg

dg=

1

2

dg2

dg

Then the equation becomes:

2d(g2 − l ∗ g)

dg− g2 − l ∗ g

g= 0

This is a first-order linear equation for g2 − l ∗ g and can be solved by:

g2 = k√g + l ∗ g

or,

g = ±√k√g + l ∗ g

where k = k(ε) is an arbitrary function. Now we make a substitution:

√g = y

hence,2yy = ±

√k ∗ y + l ∗ y2.

We can complete the square and get:

2yy = ±√l[(y +

k

2l)2 − (

k

2l)2]

One more substitution:

y +k

2l=k

2lcoshφ

26

27

eventually reduces the equation into the form:

sinhφ− φ = ± l√l

kτ + n

where n = n(ε) is again an arbitrary function due to the integration. Sincek(ε) is arbitrary, we can rescale and finally write down the following solutions:

g =k2

4(coshφ− 1)2

and

sinhφ− φ =

√l

kτ + n

In the last equation the sign ± is dropped because we have chosen a specificparametrization.

Appendix 2

We are going to prove

< ∂τφ∂ρφ+ ∂ρφ∂τφ >s=

∫ ∞0

dω(iω)e−2πω

∫d2k

2π[∂ρKK

∗ −K∂ρK∗] = 0

(5.38)We see that the A ≡ ∂ρKK

∗ −K∂ρK∗ = 0 is sufficient. We write down thespecific expression

A = −iµ2

∫ ∞0

∫ ∞0

dsdtsiω−1t−iω−1(s− 1

s+ t− 1

t)e

iµρ2

(t− 1t−s+ 1

s) (5.39)

Now we make a change of variables

s→ 1

u

t→ 1

v

Then the above integral turns into

A = −iµ2

∫ ∞0

∫ ∞0

dudvu−iω−1viω−1(1

u− u+

1

v− v)e

iµρ2

( 1v−v+u+ 1

u) (5.40)

Now we rename the variables

u→ t

v → s

and the above integral turns to

A = −iµ2

∫ ∞0

∫ ∞0

dsdtsiω−1t−iω−1(1

s− s+

1

t− t)e

iµρ2

(t− 1t−s+ 1

s)

=iµ

2

∫ ∞0

∫ ∞0

dsdtsiω−1t−iω−1(s− 1

s+ t− 1

t)e

iµρ2

(t− 1t−s+ 1

s)

= −A (5.41)

Therefore,A = 0 (5.42)

28

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[13] P. Kraus, F. Wilczek:Self-interaction correction to black hole radi-anceNuclear Physics B 433403-420(1995)

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30 BIBLIOGRAPHY

[14] G. ’t Hooft: The Black Hole Interpretation of String Theory NuclearPhysics B335138-154(1990)