A Model of Black Hole Evaporation and 4D Weyl Anomaly RIKEN-iTHES - - PowerPoint PPT Presentation

A Model of Black Hole Evaporation and 4D Weyl Anomaly

RIKEN-iTHES

Yuki Yokokura

with H. Kawai (kyoto university)

2017 August 7 @ Strings and Fields

[H. Kawai and Y. Y, Universe 3, 51, (2017)]

Motivation

What is the “black hole” in quantum mechanics? ⇒I will give a possibility that in spherically- symmetric systems

• No horizon appears in QM.
• The quantum BH is a dense star that looks like the

classical BH from the outside. ⇒I will solve time evolution of the whole spacetime in a self-consistent manner: 𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

• 𝜈𝜈

Basic idea: step1

spherical thin shell Hawking radiation

BH

What happens? ⇒The shell will never reach “the horizon”.

Consider a spherically-symmetric evaporating BH. Add a spherical thin shell (or a particle).

particle

The shell never crosses “the horizon”.

𝑠

𝑡(𝑢)

𝑏(𝑢) −𝑏 𝑢 𝑒𝑏(𝑢) 𝑒𝑢 = 2𝜏 𝑏(𝑢) Δ𝑢~𝑏 𝑢

𝑢 𝑠 𝑠

𝑡(𝑢)

𝑏(𝑢) Δ𝑢𝑚𝑚𝑚𝑚~𝑏3/𝜏 𝑒𝑡2 = − 𝑠 − 𝑏 𝑢 𝑠 𝑒𝑢2 + 𝑠 𝑠 − 𝑏(𝑢) 𝑒𝑠2 + 𝑠2𝑒Ω2.

𝑒 𝑒𝑢 𝑏 𝑢 = − 2𝜏 𝑏 𝑢 2

𝑏 ≡ 2𝐻𝐻

intensity: 𝜏 = 𝑃 1 ~ℏ𝐻𝐻 (𝐻=d.o.f. of fields)

e.o.m for 𝑠

𝑡~𝑏:

𝑒𝑠

𝑡(𝑢)

𝑒𝑢 = − 𝑠

𝑡 𝑢 − 𝑏 𝑢

𝑠

𝑡 𝑢

. ⇒ 𝑠

𝑡 𝑢 ≃ 𝑏 𝑢 − 𝑏 𝑢 𝑒𝑏 𝑢

𝑒𝑢 + 𝐷𝑏 𝑢 𝑓

− 𝑢 𝑏(𝑢)

→ 𝑏 𝑢 +

2𝜏 𝑏 𝑢

⇒The shell will approach 𝑏 𝑢 +

2𝜏 𝑏 𝑢 .

Outside ≈ time-dependent “Schwarzschild” metric

Hawking radiation

The shell never crosses “the horizon”.

𝑠

𝑡(𝑢)

𝑏(𝑢) −𝑏 𝑢 𝑒𝑏(𝑢) 𝑒𝑢 = 2𝜏 𝑏(𝑢) Δ𝑢~𝑏 𝑢

𝑢 𝑠 𝑠

𝑡(𝑢)

𝑏(𝑢) Δ𝑢𝑚𝑚𝑚𝑚~𝑏3/𝜏 𝑒𝑡2 = − 𝑠 − 𝑏 𝑢 𝑠 𝑒𝑢2 + 𝑠 𝑠 − 𝑏(𝑢) 𝑒𝑠2 + 𝑠2𝑒Ω2.

𝑒 𝑒𝑢 𝑏 𝑢 = − 2𝜏 𝑏 𝑢 2

𝑏 ≡ 2𝐻𝐻

intensity: 𝜏 = 𝑃 1 ~ℏ𝐻𝐻 (𝐻=d.o.f. of fields)

e.o.m for 𝑠

𝑡~𝑏:

𝑒𝑠

𝑡(𝑢)

𝑒𝑢 = − 𝑠

𝑡 𝑢 − 𝑏 𝑢

𝑠

𝑡 𝑢

. ⇒ 𝑠

𝑡 𝑢 ≃ 𝑏 𝑢 − 𝑏 𝑢 𝑒𝑏 𝑢

𝑒𝑢 + 𝐷𝑏 𝑢 𝑓

− 𝑢 𝑏(𝑢)

→ 𝑏 𝑢 +

2𝜏 𝑏 𝑢

⇒The shell will approach 𝑏 𝑢 +

2𝜏 𝑏 𝑢 .

Outside ≈ time-dependent “Schwarzschild” metric

Hawking radiation

・The proper length is non-trivial Δ𝑚 = 𝑕𝑠𝑠 2𝜏 𝑏 ≈ 𝜏 ∼ 𝐻𝑚𝑞 ≫ 𝑚𝑞 if there are many matter fields: 𝐻 ≫ 1. (Note: The (𝑢, 𝑠)-coordinates are complete to describe this motion.)

Indeed, we can check that, (radiation from the shell with Δ𝐻) +(small redshift factor) (radiation from the core BH with 𝐻) = (radiation from a BH with 𝐻 + Δ𝐻)

BH

𝐻 shell Δ𝐻

Basic idea: step2

BH 𝐻 + Δ𝐻

=

radiation

(radiation from the shell with Δ𝐻) +(redshift factor) (radiation from the core BH with 𝐻) = (radiation from a BH with 𝐻 + Δ𝐻) After the shell comes sufficiently close to 𝑠 = 𝑏 + 2𝜏

𝑏 ,

the total system composed of the BH and the shell behaves like an ordinary BH with mass 𝐻 + Δ𝐻.

Basic idea: step3

BH

=

Any object we recognize as a BH should be such an object. There is not a horizon but a surface. 𝑠 = 𝑏 + 2𝜏 𝑏 𝑢𝑢 𝑠 𝑠′(𝑢𝑢) 𝑏𝑢(𝑢𝑢)

−𝑏′ 𝑒𝑏′ 𝑒𝑢𝑢 = 2𝜏 𝑏𝑢

continuous collapsing matter (not BH) Apply the previous result to each shell recursively. Regard this as many shells.

Strong angular pressure is induced by 4D Weyl anomaly.

𝑒𝑞𝑠 𝑒𝑠 + 2 𝑠 𝑞𝑠 − 𝑞𝜄 + 𝑕𝑠 𝜍 + 𝑞𝑠 = 0

Balance eq for spherical star (TOV eq)

balance ⇒This object is stable!

𝑞𝑠 ≪ 𝑞𝜄 ⇒not a fluid

gravitational force

Our strategy: self-consistent eq.

𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

• 𝜈𝜈

𝑕𝜈𝜈= a classical field =collapsing matter + Hawking radiation 𝜚 = quantum matter fields 𝛼

𝑕 2𝜚

𝑚 = 0 the self-consistent eq.

How to obtain the solution?

We assume spherical symmetry.

𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

• 𝜈𝜈

Step1: Construct a candidate metric 𝑕𝜈𝜈 by a simple model. Step2: Evaluate 𝑈

𝜈𝜈 on 𝑕𝜈𝜈

by considering conformal matter. Step3: Put this and determine the self-consistent 𝑕𝜈𝜈.

the self-consistent eq.

See [H.Kawai and Y.Y, PRD 2016] for a more general case.

inside = flat: 𝑒𝑡2 = −𝑒𝑉2 − 2𝑒𝑉𝑒𝑠 + 𝑠2𝑒Ω2 𝑠

𝑣

𝑣

𝑠

𝑡(𝑣)

𝑉 𝑏(𝑣)

Assumption Radiation goes to infinity without reflection.

a collapsing null shell

Step1: Construction of a candidate 𝑕𝜈𝜈(1/4) Preliminary: A single-shell model

flat

𝑏

flat

Consider time evolution

• f a spherical null shell.

scattering (or mass)

𝑠 𝑣

Scattering of radiations

inside = flat: 𝑒𝑡2 = −𝑒𝑉2 − 2𝑒𝑉𝑒𝑠 + 𝑠2𝑒Ω2

• utside = Vaidya metric:

𝑒𝑡2 = − 1 −

𝑏 𝑣 𝑠

𝑒𝑣2 − 2𝑒𝑣𝑒𝑠 + 𝑠2𝑒Ω2

𝐻𝑣𝑣 = − 𝑏̇ 𝑣 𝑠2 , 𝐻𝑝𝑢𝑝𝑚𝑠𝑡 = 0

𝑏 𝑣 is not fixed yet.

𝑠

𝑣

𝑣

𝑠

𝑡(𝑣)

𝑉 𝑏(𝑣)

Assumption Radiation goes to infinity without reflection.

a collapsing null shell

Note: At this stage we don’t claim that this is the solution.

Step1: Construction of a candidate 𝑕𝜈𝜈(1/4) Preliminary: A single-shell model

flat

𝑏

flat

Consider time evolution

• f a spherical null shell.

flat 𝑠 𝑠𝑚 𝑏𝑚−1

⋯

𝑠𝑚−1 𝑏𝑚

𝑒𝑡𝑚

2 = − 𝑠 − 𝑏𝑚(𝑣𝑚)

𝑠 𝑒𝑣𝑚

2 − 2𝑒𝑣𝑚𝑒𝑠 + 𝑠2𝑒Ω2

𝑒𝑡𝑚−1

2

= − 𝑠 − 𝑏𝑚−1(𝑣𝑚−1) 𝑠 𝑒𝑣𝑚−1

2

− 2𝑒𝑣𝑚−1𝑒𝑠 + 𝑠2𝑒Ω2

𝑠

1

𝑠

𝑜

⋯

𝑒𝑡𝑝𝑣𝑢

2

= − 𝑠 − 𝑏(𝑣) 𝑠 𝑒𝑣2 − 2𝑒𝑣𝑒𝑠 + 𝑠2𝑒Ω2 𝑒𝑡0

2 = −𝑒𝑉2 − 2𝑒𝑉𝑒𝑠 + 𝑠2𝑒Ω2

𝑣

Step1: Construction of a candidate 𝑕𝜈𝜈(2/4) A multi-shell model

Consider a continuous spherical matter. ⇒Model this as many shells.

Ansatz: Each shell behaves like the ordinary evaporating BH: 𝑒𝑏𝑚 𝑒𝑣𝑚 = − 𝜏 𝑏𝑚

2 ,

and that each shell has already come close to 𝑠𝑚 = 𝑏𝑚 + 2𝜏 𝑏𝑚

𝑣𝑚 𝑠 𝑠𝑚(𝑣𝑚) 𝑏𝑚(𝑣𝑚)

−2𝑏𝑚 𝑒𝑏𝑚 𝑒𝑣𝑚 = 2𝜏 𝑏𝑚

Step1: Construction of a candidate 𝑕𝜈𝜈(3/4): Self-consistent ansatz

⇒After taking continuum limit (Δ𝑏𝑚 ≡ 𝑏𝑚 − 𝑏𝑚−1 → 0), we obtain….

𝑒𝑡2 = − 1 − 𝑏 𝑣 𝑠 𝑒𝑣2 − 2𝑒𝑣𝑒𝑠 + 𝑠2𝑒𝛻2 − 2𝜏 𝑠2 𝑓− 1

2𝜏 𝑆 𝑏 𝑣

2−𝑠2 𝑒𝑣2 − 2𝑓− 1

4𝜏 𝑆 𝑏 𝑣

2−𝑠2 𝑒𝑣𝑒𝑠 + 𝑠2𝑒𝛻2

Here 𝑒𝑏 𝑣 𝑒𝑣 = − 𝜏 𝑏 𝑣 2 , 𝑆 𝑏 ≡ 𝑏 + 2𝜏 𝑏 𝑣 𝑠

𝑠 = 𝑆(𝑏(𝑣))

Δ𝑣𝑚𝑚𝑚𝑚~𝑏3/𝜏

Note: At this stage, 𝜏 is not determined. Note: At this stage we don’t claim yet that this is the solution.

Step1: Construction of a candidate 𝑕𝜈𝜈(4/4): the metric

Consider the interior region. The background metric is static:

𝑒𝑡2 = −

2𝜏 𝑠2 𝑓

𝑠2 2𝜏𝑒𝑉2 − 2𝑓 𝑠2 4𝜏𝑒𝑉𝑒𝑠 + 𝑠2𝑒𝛻2

= −𝑓𝜒 𝑠(𝑉,𝑊) 𝑒𝑉𝑒𝑒 + 𝑠 𝑉, 𝑒 2𝑒Ω2, ⇒ 𝑈

𝜈𝜈 also should be static:

𝑈

𝜈𝜈 = 𝑈 𝜈𝜈(𝑠) ,

𝑈

𝑉𝑉 = 𝑈 𝑊𝑊

⇒The use of Vaidya metric means 𝑈

𝑉𝑊 = 0,

we have to determine only 𝑈

𝑉𝑉 ,

𝑈

𝜄 𝜄 .

Step2: Evaluation of 𝑈

𝜈𝜈 (1/3):

Setup

(⇒We can remove this artificial assumption and generalize it to 〈𝑈𝑉𝑊〉 ≠ 0.) scattering (or mass)

𝑠 𝑉

Physical origin of 𝑈𝑉𝑊 ≠ 0 𝑈𝑉𝑉 𝑈𝑉𝑊 =0

・1st eq. 𝑈

𝜈 𝜈 = 2𝑕𝑉𝑊 𝑈 𝑉𝑊 + 2 𝑈 𝜄 𝜄 leads to

𝑈

𝜄 𝜄 = 1

2 𝑈

𝜈 𝜈

・2nd eq. 𝛼𝜈 𝑈

𝜈𝑉 = 0 provides

𝑠2 𝑈

𝑉𝑉 = 1

2 𝑒𝑠′𝑠′𝑓𝜒(𝑠′) 𝑈

𝜄 𝜄(𝑠′) 𝑠

+ 𝑠2 𝑈

𝑉𝑉 𝑠=0 ,

𝑉

𝑠

𝑒 =const. surface at 𝑠 = 𝑆(𝑏(𝑣)) 𝑉 =const.

flat

Boundary condition: 𝑈

𝜈𝜈(𝑠 = 0) = 0

Step2: Evaluation of 𝑈

𝜈𝜈 (2/3):

The relations of 𝑈

𝜈𝜈

All components are expressed in terms of 𝑈

𝜈 𝜈 .

Step2: Evaluation of 𝑈

𝜈𝜈 (3/3):

4D Weyl anomaly

For simplicity, consider conformal matters. ⇒ 𝑈

𝜈 𝜈 is determined by the 4D Weyl anomaly:

𝑈

𝜈 𝜈 = ℏ𝑑𝑥ℱ − ℏ𝑏𝑥ℊ

where ℱ ≡ 𝐷𝜈𝜈𝜈𝜈𝐷𝜈𝜈𝜈𝜈, ℊ ≡ 𝑆𝜈𝜈𝜈𝜈𝑆𝜈𝜈𝜈𝜈 − 4𝑆𝜈𝜈𝑆𝜈𝜈 + 𝑆2 ⇒The metric determines 𝑈

𝜈 𝜈 = ℏ𝑑𝑥

3𝜏2

←state-independent

Step3: Check of 𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

𝜈𝜈 (1/2):

The 𝑕𝜈𝜈 is the self-consistent solution. Thus , we have obtained 𝑈

𝜄 𝜄 = ℏ𝑑𝑋

6𝜏2 , 𝑈𝑉𝑉 = 𝑈𝑊𝑊 = ℏ𝑑𝑋 3𝑠4 𝑓

1 2𝜏𝑠2,

𝑈𝑉𝑊 = 0 On the other hand, the metric gives 𝐻𝜄

𝜄 = 1

2𝜏 , 𝐻𝑉𝑉 = 𝐻𝑊𝑊 = 𝜏 𝑠4 𝑓

1 2𝜏𝑠2,

𝐻𝑉𝑊 = 0 ⇒𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

𝜈𝜈 is satisfied by 𝑕𝜈𝜈 with

𝜏 = 8𝜌𝐻ℏ 3 𝑑𝑋.

 Hawking radiation ∝ 𝑑𝑋

𝑒𝑏 𝑣 𝑒𝑣 = − 𝜏 𝑏 𝑣 2

In the macroscopic region (𝑠 > 𝑚𝑄), 𝑆, 𝑆𝜈𝜈𝑆𝜈𝜈, 𝑆𝜈𝜈𝜈𝜈𝑆𝜈𝜈𝜈𝜈~ 1 𝑑𝑥𝑚𝑞

2 ≪ 1

𝑚𝑞

2

⇒No singularity Validity of the classical gravity by 𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

• 𝜈𝜈

as long as 𝑏 > 𝑚𝑞.

Step3: Check of 𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

𝜈𝜈 (2/2):

Validity of the classical gravity if c𝑥 ≫ 1

How to obtain Hawking radiation?

• So far, we have used the ansatz 𝑒𝑏𝑗

𝑒𝑣𝑗 = − 𝜏 𝑏𝑗

2 and obtained

the self-consistent solution with 𝜏 = 8𝜌𝜌ℏ

3 𝑑𝑋.

⇒ Hawking radiation appears self-consistently.

• Indeed, we can show in the metric

0 𝐻 𝜕 0 = 1 𝑓ℏ𝜕/𝑈 − 1 , 𝑈 = ℏ 4𝜌𝑏(𝑣)

𝑠 𝑣

flat:𝑉

|0 >

Quantum BHs are described by the field theory 𝐻𝜈𝜈 = 8𝜌𝐻 𝑈

𝜈𝜈 with 𝑑𝑋 ≫ 1.

Conclusion

“Hawking radiation” collapsing matter Δ𝑢~ 𝑏3 𝜏 Quantum BH without horizon or singularity

Δ𝑠~𝑏 𝑒𝑡2 = − 1 − 𝑏 𝑣 𝑠 𝑒𝑣2 − 2𝑒𝑣𝑒𝑠 + 𝑠2𝑒𝛻2 − 2𝜏 𝑠2 𝑓

− 1 2𝜏(1+𝑚) 𝑆 𝑏 𝑣

2−𝑠2

𝑒𝑣2 − 2𝑓

− 1 4𝜏(1+𝑚) 𝑆 𝑏 𝑣

2−𝑠2

𝑒𝑣𝑒𝑠 + 𝑠2𝑒𝛻2

𝜏 ≡ 8𝜌𝐻ℏ𝑑𝑋 3 1 + 𝑔 2

𝑒𝑏 𝑣 𝑒𝑣 = − 𝜏 𝑏 𝑣 2

The surface exists at 𝑆 𝑏 ≡ 𝑏 + 2𝜏 𝑏

𝑈𝑉𝑊 = 𝑔〈𝑈𝑉𝑉〉

What is the self-consistent state |𝜔〉? How will the matter 𝜚 come out? (work in progress) ⇒Information recovery “Hawking radiation” in |𝜔〉 Collapsing matter in |𝜔〉 What happens to baryons inside the BH?

Final stage? ⇒Beyond the semi-classical approximation

Discussions

How can we understand the BH entropy? Is this solution stable to perturbations? Can we generalize this picture to rotating BHs? Can we obtain some observable signal showing this picture?

Thank you very much!

Appendix

When any particle gets close to 𝑏(𝑢) , it becomes ultra relativistic. ⇒The position 𝑠

𝑡(𝑢) of a particle is determined by setting 𝑒𝑡2 = 0 :

𝑒𝑠

𝑡 𝑢

𝑒𝑢 = − 𝑠

𝑡 𝑢 − 𝑏(𝑢)

𝑠

𝑡 𝑢

⇒ 𝑒𝛦𝑠 𝑢 𝑒𝑢 = − 𝛦𝑠 𝑢 𝑠 𝑢 − 𝑒𝑏 𝑢 𝑒𝑢 ≃ − 𝛦𝑠 𝑢 𝑏 𝑢 − 𝑒𝑏 𝑢 𝑒𝑢 . The general solution is given by Δ𝑠 𝑢 = 𝑑0 𝑓

− ∫ 𝑒𝑢′ 𝑏 𝑢′

𝑢 𝑢0

+ 𝑒𝑢′{−

𝑢 𝑢0

(𝑒𝑏 𝑒𝑢 𝑢′ ) 𝑓

− ∫ 𝑒𝑢′′

𝑢 𝑢′

1 𝑏 𝑢′′ }.

Motion of the shell near the evaporating BH(1/2)

Δ𝑠 𝑢 ≡ 𝑠

𝑡 𝑢 − 𝑏 𝑢

Δ𝑠 𝑢 ≪ 𝑏(𝑢)

𝑏 𝑢 ≈const.

𝑒𝑏 𝑢 𝑒𝑢

≈const. in the time scale ∼ 𝑏 ≃ − 𝑒𝑏 𝑢 𝑒𝑢 𝑒𝑢′

𝑢 𝑢0

𝑓

− 𝑢−𝑢′ 𝑏 𝑢

= − 𝑒𝑏(𝑢) 𝑒𝑢 𝑏(𝑢)(1 − 𝑓

− 𝑢−𝑢0 𝑏 𝑢 )

Δ𝑢𝑚𝑚𝑚𝑚~𝑏3 𝜏

Motion of the shell near the evaporating BH(2/2)

𝑢 𝑠 𝑠

𝑡(𝑢)

𝑏(𝑢)

−𝑏 𝑢 𝑒𝑏(𝑢) 𝑒𝑢 = 2𝜏 𝑏(𝑢)

Δ𝑢~𝑏 𝑢

𝑢 𝑠 𝑠

𝑡(𝑢)

𝑏(𝑢)

Δ𝑢𝑚𝑚𝑚𝑚~𝑏3/𝜏

Therefore we have Δ𝑠 𝑢 ≃ 𝑑0 𝑓

−𝑢−𝑢0 𝑏(𝑢) − 𝑒𝑏 𝑢

𝑒𝑢 𝑏 𝑢 [1 − 𝑓

− 𝑢−𝑢0 𝑏 𝑢 ]

which leads to 𝑠

𝑡 𝑢 ≃ 𝑏 𝑢 − 𝑒𝑏 𝑢

𝑒𝑢 𝑏 𝑢 = 𝑏 𝑢 + 2𝜏 𝑏 𝑢 ≡ 𝑆(𝑏 𝑢 ) when 𝑢 − 𝑢0 ≫ 𝑏 𝑢 .

𝑒 𝑒𝑢 𝑏 𝑢 = − 2𝜏 𝑏 𝑢 2

Note: The (𝑢, 𝑠)-coordinates are not singular.

⇒Any particle will never reach “the horizon”.

𝑢

𝑏 𝑢 𝑠 𝑢 − 𝑏 𝑢

Note : A numerical solution of 𝑠

𝑡 𝑢

We solve

𝑒𝑠𝑡 𝑢 𝑒𝑢

= −

𝑠𝑡 𝑢 −𝑏(𝑢) 𝑠 𝑢

numerically under

𝑒𝑏 𝑒𝑢 = − 2𝜏 𝑏 𝑢 2 ⇒ 𝑏 𝑢 = 𝑏 0 3 − 6𝜏𝑢

1 3 .

⇒This shows that 𝑠 𝑢 → 𝑏 𝑢 + 2𝜏

𝑏(𝑢) ≡ 𝑆(𝑏(𝑢)). ← 2𝜏 = 2