Entropy inside
Evaporating Black Holes
RIKEN iTHEMS

Yuki Yokokurawith H.Kawai (Kyoto)
2018/12/17@ QFT and string theory

BHs evaporate by nature.
BH

BH

ℏ
Stefan-Boltzmann law of 𝑇𝑇𝐻𝐻 =:8𝜋𝜋𝜋𝜋𝜋𝜋

𝑑𝑑𝑑𝑑
1
ℏ𝑁𝑁
4
= − 3 × 𝑁𝑁: d. o. f. of fields × (𝑂𝑂 1 : graybody fact𝑜𝑜𝑜𝑜) × Area × 𝑇𝑇𝐻𝐻 ∝ − 2
𝑑𝑑𝑑𝑑
ℏ
𝑀𝑀

⇒We parametrize this as𝑎𝑎 = 2𝐺𝐺𝐺𝐺

𝑑𝑑𝑑𝑑(𝑡𝑡)
2𝜎𝜎
=−
𝑑𝑑𝑑𝑑
𝑎𝑎(𝑡𝑡)2

𝑎𝑎3⇒BHs evaporate in Δ𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ~𝜎𝜎Intensity of Hawking radiation
𝜎𝜎 = 𝑂𝑂 1 ∝ 𝑁𝑁𝑁𝑁𝑁 = 𝑁𝑁 𝑙𝑙𝑝𝑝2

BH can be in equilibrium with a heat bath.

BHℏ
𝑇𝑇𝐻𝐻 =
4𝜋𝜋𝜋𝜋

BH has entropy.
• Suppose that we form a BH adiabatically in the heatbath by controlling the temperature and size properly.

High temperature

𝑀𝑀𝑀

𝑀𝑀

Law temperature• The entropy of radiation absorbed during this process is
𝑀𝑀
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑𝑑
8𝜋𝜋𝜋𝜋 𝑀𝑀2ℏ==�
𝑆𝑆𝑟𝑟𝑟𝑟𝑟𝑟 = �
𝑇𝑇𝐻𝐻 (𝑀𝑀) =8𝜋𝜋𝜋𝜋𝜋𝜋
(𝑀𝑀𝑀)
2
𝑇𝑇
𝑇𝑇
ℏ
𝐻𝐻
0
⇒BH has entropy:𝑎𝑎 ≡ 2𝐺𝐺𝐺𝐺
𝐴𝐴𝐴𝐴 ≡ 4𝜋𝜋𝑎𝑎2
𝑆𝑆𝐵𝐵𝐵𝐵 = 24𝑙𝑙𝑝𝑝
𝑙𝑙𝑝𝑝 ≡ ℏ𝐺𝐺

Where is the entropy?
• Entropy should be inside the BH.
• In classical mechanics, BH is a vacuum region with thehorizon, but the BH evaporates in a finite time.
radiation in |𝜓𝜓⟩
horizon|𝜓𝜓⟩×vacuum(nothing)|𝜓𝜓⟩ is trapped by the horizon.

×

evaporation⇒Information seems to be lost after evaporation.
⇒The existence of horizon is not consistent with QM?

What is BH in QM?

BH?

Kawai, Matsuo and Y. Y., Int. J. Mod. Phys. A 28, 1350050 (2013),
Kawai and Y. Y., Int. J. Mod. Phys. A 30, 1550091 (2015),
Kawai and Y. Y., Phys.Rev.D.93.044011 (2016),
Kawai and Y. Y., Universe 3, 51, (2017)

Our approach• Reconsider the time evolution of a 4D sphericallysymmetric collapsing matter (radiation from the bath).
What happens?
Horizon is formed or not?

Our approach• Reconsider the time evolution of a 4D sphericallysymmetric collapsing matter (radiation from the bath).
What happens?
Horizon is formed or not?

Generically, particle creation occurs in a time-dependent metric (even without horizon).
⇒ We need to include the back reaction of evaporation in the formation process.
Some people might wonder that such effect is negligible because
𝑎𝑎3
time scale of evaporation~ 2 ≫ 𝑎𝑎~time scale of collapse.
𝑙𝑙𝑝𝑝
𝑎𝑎3⇒No! because 2 is measured at infinity and 𝑎𝑎 by a comoving observer.
𝑙𝑙𝑝𝑝⇒Consider dynamics of collapse and evaporation in a common time coordinate!

Our approach• Reconsider the time evolution of a 4D sphericallysymmetric collapsing matter (radiation from the bath).
What happens?
Horizon is formed or not?

Generically, particle creation occurs in a time-dependent metric (even without horizon).
⇒ We need to include the back reaction of evaporation in the formation process.

⇒ We will solve the self-consistent eq:

𝑔𝑔𝜇𝜇𝜇𝜇 = classical metric

𝐺𝐺𝜇𝜇𝜇𝜇 = 8𝜋𝜋𝐺𝐺 𝑇𝑇�𝜇𝜇𝜇𝜇

� quantum matter fields
𝜙𝜙=□𝜙𝜙� = 0=collapsing matter+ Hawking radiation

Basic idea: step1
Imagine that a spherically-symmetric BH is evaporating in the vacuum.
Add a spherical thin shell (or a particle) to it.
Hawking radiation
particle

spherical thin shell

BH

What happens if dynamics of the matter and spacetime is considered?
⇒The shell will never reach the Schwarzschild radius.

The shell never crosses “the horizon”.
𝑟𝑟

no gray-body factor

𝑟𝑟𝑠𝑠 (𝑢𝑢)Hawking radiation

Background ≈ outgoing Vaidya metric𝑑𝑑𝑠𝑠 2 = −𝑎𝑎 𝑢𝑢 = 2𝐺𝐺𝐺𝐺(𝑢𝑢)Δ𝑢𝑢𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ~𝑎𝑎3 /𝜎𝜎

𝑢𝑢

𝑎𝑎 𝑢𝑢
1−
𝑟𝑟

𝑑𝑑𝑢𝑢2 − 2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑𝛺𝛺2

𝑑𝑑
𝜎𝜎
𝑎𝑎 𝑢𝑢 = −←Stephan-Boltzmann law
𝑑𝑑𝑢𝑢
𝑎𝑎 𝑢𝑢 2ℏof 𝑇𝑇 =𝑎𝑎 ≡ 2𝐺𝐺𝐺𝐺

𝐻𝐻

4𝜋𝜋𝜋𝜋

intensity: 𝜎𝜎 = 𝑂𝑂 1 ~ℏ𝐺𝐺𝐺𝐺(𝑁𝑁=d.o.f. of fields)

For 𝑟𝑟𝑠𝑠 ~𝑎𝑎, a particle with any (𝑙𝑙, 𝑚𝑚) behaves lightlike:
𝑑𝑑𝑟𝑟𝑠𝑠 (𝑢𝑢)𝑟𝑟𝑠𝑠 𝑢𝑢 − 𝑎𝑎 𝑢𝑢=−.
𝑑𝑑𝑑𝑑(𝑢𝑢)
2𝜎𝜎
𝑑𝑑𝑢𝑢
2𝑟𝑟
𝑢𝑢
𝑠𝑠
−2𝑎𝑎 𝑢𝑢=𝑢𝑢
𝑑𝑑𝑢𝑢
𝑎𝑎(𝑢𝑢)
𝑑𝑑𝑑𝑑 𝑢𝑢−+ 𝐶𝐶𝐶𝐶 𝑢𝑢 𝑒𝑒 2𝑎𝑎(𝑢𝑢)⇒ 𝑟𝑟𝑠𝑠 𝑢𝑢 ≃ 𝑎𝑎 𝑢𝑢 − 2𝑎𝑎 𝑢𝑢
𝑑𝑑𝑢𝑢
2𝜎𝜎→ 𝑎𝑎 𝑢𝑢 +Δ𝑢𝑢~2𝑎𝑎 𝑢𝑢

𝑎𝑎 𝑢𝑢⇒Universally, particles will approach 𝑎𝑎 𝑢𝑢 +

2𝜎𝜎
.
𝑎𝑎 𝑢𝑢

The shell never crosses “the horizon”.
𝑟𝑟

no gray-body factor

𝑟𝑟𝑠𝑠 (𝑢𝑢)Hawking radiation

Background ≈ outgoing Vaidya metric𝑑𝑑𝑠𝑠 2 = −𝑎𝑎 𝑢𝑢 = 2𝐺𝐺𝐺𝐺(𝑢𝑢)Δ𝑢𝑢𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ~𝑎𝑎3 /𝜎𝜎

𝑢𝑢

𝑎𝑎 𝑢𝑢
1−
𝑟𝑟

𝑑𝑑𝑢𝑢2 − 2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑𝛺𝛺2

𝑑𝑑
𝜎𝜎
𝑎𝑎 𝑢𝑢 = −←Stephan-Boltzmann law
𝑑𝑑𝑢𝑢
𝑎𝑎 𝑢𝑢 2ℏof 𝑇𝑇 =
𝐻𝐻𝑎𝑎 ≡ 2𝐺𝐺𝐺𝐺

4𝜋𝜋𝜋𝜋

intensity: 𝜎𝜎 = 𝑂𝑂 1 ~ℏ𝐺𝐺𝐺𝐺(𝑁𝑁=d.o.f. of fields)

For・The𝑟𝑟𝑠𝑠 ~𝑎𝑎,propera particle
withofany
lengthΔ𝑟𝑟 (𝑙𝑙,= 𝑚𝑚)isbehaves
given bylightlike:𝑎𝑎
𝑑𝑑𝑟𝑟𝑠𝑠 (𝑢𝑢)𝑟𝑟𝑠𝑠 𝑢𝑢 − 𝑎𝑎 𝑢𝑢
𝜎𝜎=∼ − 𝑁𝑁𝑙𝑙𝑝𝑝 ≫ 𝑙𝑙𝑝𝑝.
𝑑𝑑𝑑𝑑(𝑢𝑢)
2𝜎𝜎
𝑑𝑑𝑢𝑢
2𝑟𝑟
𝑢𝑢
𝑠𝑠
⇒If there are many d.o.f. of fields,−2𝑎𝑎 𝑢𝑢=𝑢𝑢
𝑑𝑑𝑢𝑢
𝑎𝑎(𝑢𝑢)
𝑑𝑑𝑑𝑑
𝑢𝑢−+ 𝐶𝐶𝐶𝐶 𝑢𝑢 𝑒𝑒 2𝑎𝑎(𝑢𝑢)⇒ 𝑟𝑟𝑠𝑠 𝑢𝑢 ≃ 𝑎𝑎 𝑢𝑢 − 2𝑎𝑎𝑁𝑁 𝑢𝑢≫ 1,
𝑑𝑑𝑢𝑢
the shell is physically
2𝜎𝜎 outside the “BH”.
→ 𝑎𝑎 𝑢𝑢 +is complete, as we will see later.)(This coordinate𝑎𝑎 𝑢𝑢

Δ𝑢𝑢~2𝑎𝑎 𝑢𝑢

2𝜎𝜎⇒Universally, particles will approach 𝑎𝑎 𝑢𝑢 +

2𝜎𝜎
.
𝑎𝑎 𝑢𝑢

Basic idea: step2
Radiation is emitted from themetric around the shell.

shell
Δ𝑀𝑀

BH

𝑀𝑀

On the shell, a large tangential pressure
occurs, which decelerates the shell.
⇒Dominant energy condition is broken.

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

Basic idea: step2

BH

𝑀𝑀

shell
Δ𝑀𝑀

=
radiation

BH
𝑀𝑀 + Δ𝑀𝑀

Indeed, we can check that,(radiation from the shell with Δ𝑀𝑀)+(redshift
factor)(radiation from theshell with
Δ𝑀𝑀)(radiation from the core BH with 𝑀𝑀)= (radiationfrom redshift
a BH with
𝑀𝑀 + (radiation
Δ𝑀𝑀)+(smallfactor)from the core BH with 𝑀𝑀)= (radiation
BH with 𝑀𝑀+ Δ𝑀𝑀)⇒This
compositefrom
systema(=shell+BH)
behaves
like a larger BH.

Basic idea: step3

Each shell is evaporating and
2𝜎𝜎approaching 𝑟𝑟 = 𝑎𝑎′ + ′ .

𝑟𝑟 ′𝑟𝑟𝑠𝑠 (𝑢𝑢′)continuous
collapsing matter (not BH)

Regard thisas many shells.

The shells pile up andform a dense object.
𝑀𝑀 =

𝑎𝑎
2𝐺𝐺

𝑅𝑅(𝑎𝑎)=The matter is filled inside
2𝜎𝜎≡ 𝑅𝑅 𝑎𝑎 ,𝑟𝑟 = 𝑎𝑎 +𝑎𝑎which is the boundary of the object (surface).
⇒ No horizon exists.

𝑎𝑎𝑑𝑑𝑎𝑎′ 2𝜎𝜎
=
−2𝑎𝑎
𝑎𝑎𝑎
𝑑𝑑𝑢𝑢′
′

𝑎𝑎𝑎(𝑢𝑢′)𝑢𝑢′

Apply the previous result
to each shell recursively.

BH

But this looks like an ordinary BH
from the outside.
⇒Any object we recognize as a
BH should be such an object.

Talk Plan: How to obtain the solution?
⇒A self-consistent argument
We assume spherical symmetry.

(mean-field approximation in field theory)

Step2: Assume conformal maters and evaluate 𝑇𝑇𝜇𝜇𝜇𝜇 on 𝑔𝑔𝜇𝜇𝜇𝜇
𝜇𝜇by using ∇𝜇𝜇 𝑇𝑇𝜈𝜈 = 0 and 4D Weyl anomaly formula.

𝐺𝐺𝜇𝜇𝜇𝜇 = 8𝜋𝜋𝐺𝐺 𝑇𝑇�𝜇𝜇𝜇𝜇

Step1: Construct a candidate
metric 𝑔𝑔𝜇𝜇𝜇𝜇 by a phenomenological discussion
in the adiabatic formation.

Step3: Put this anddetermine the self-consistent 𝑔𝑔𝜇𝜇𝜇𝜇 .

Step4: Consistency check
Reproduce Stephan-Boltzmann law, Planck-like
distribution, and entropy area law.

Step1: Construct a candidate 𝑔𝑔𝜇𝜇𝜇𝜇
𝑀𝑀𝑀 =

′

𝑅𝑅(𝑎𝑎𝑎)

𝑎𝑎
2𝐺𝐺

𝑀𝑀 =

𝑎𝑎
2𝐺𝐺

𝑅𝑅(𝑎𝑎)

Kawai and Y. Y. IJMPA (2015)

No 𝑎𝑎-dependence of 𝐴𝐴 𝑟𝑟 , 𝐵𝐵 𝑟𝑟

Equilibrium and spherical symmetry at each step
⇒We can write the interior metric as𝑑𝑑𝑠𝑠 2 = −1 𝐴𝐴 𝑟𝑟 2𝑒𝑒𝑑𝑑𝑡𝑡 + 𝐵𝐵 𝑟𝑟 𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2 .
𝐵𝐵 𝑟𝑟

𝑀𝑀 =

𝑎𝑎
2𝐺𝐺

𝑅𝑅(𝑎𝑎)• In general, 𝐴𝐴 𝑟𝑟 , 𝐵𝐵(𝑟𝑟) depend on the size of the BH, 𝑎𝑎.
• But we can show that
their 𝑎𝑎-dependence is small:𝐴𝐴 𝑟𝑟; 𝑎𝑎 , 𝐵𝐵 𝑟𝑟; 𝑎𝑎

for 𝑟𝑟 < 𝑅𝑅(𝑎𝑎)Why? 𝐴𝐴 𝑟𝑟; 𝑎𝑎 , 𝐵𝐵 𝑟𝑟; 𝑎𝑎 (1/2)• Postulate 1: The redshift inside the BH is exponentially⇒We will check it later.
large.
The time inside is almost frozen.

• Postulate 2 : The time scale of the adiabatic change of the⇒We will check it later.
BH is of order 𝑂𝑂(𝑎𝑎3 /𝑙𝑙𝑝𝑝2 ).
𝑎𝑎3
Δ𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ~ 2
𝑙𝑙𝑝𝑝

𝑎𝑎3
(Note: In the vacuum, a BH with 𝑎𝑎 evaporates in Δ𝑡𝑡𝑒𝑒𝑒𝑒𝑒𝑒 ~ 2 .
𝑙𝑙𝑝𝑝⇒The relaxation time scale should be at most order of 𝑂𝑂(𝑎𝑎3 /𝑙𝑙𝑝𝑝2 ).)Why? 𝐴𝐴 𝑟𝑟; 𝑎𝑎 , 𝐵𝐵 𝑟𝑟; 𝑎𝑎 (2/2)• Suppose a process in which a BH with radius 𝑎𝑎 isadiabatically shrunk to radius 𝑎𝑎𝑎:
𝑎𝑎

Δ𝑡𝑡~𝐶𝐶𝑎𝑎3

𝑎𝑎′

The inside is almost frozen due to the exponentially redshift.

⇒The metric at 𝑟𝑟 < 𝑎𝑎′ hardly changes:
𝑔𝑔𝜇𝜇𝜇𝜇 𝑟𝑟 < 𝑎𝑎′ of BH with 𝑎𝑎 = 𝑔𝑔𝜇𝜇𝜇𝜇 𝑟𝑟 < 𝑎𝑎′ of BH with 𝑎𝑎𝑎⇒As a first trial, we assume that 𝐴𝐴 𝑟𝑟 and 𝐵𝐵(𝑟𝑟) areuniversal functions independent of the size of the BH.

Interior metric of evaporating BHs
The inside of the evaporating BH which has adiabatically grown has
the same metric as that of the stationary BH in the heat bath.

in vacuum

𝑒𝑒𝑒𝑒𝑒𝑒𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑔𝑔𝜇𝜇𝜇𝜇𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑔𝑔𝜇𝜇𝜇𝜇(𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)

in heat bath

• Why?

When the size of the BH becomes 𝑎𝑎,ℏ
we put it into the bath of 𝑇𝑇 =.
4𝜋𝜋𝜋𝜋

Δ𝑡𝑡~𝐶𝐶𝑎𝑎3

It becomes stationary in Δ𝑡𝑡~𝐶𝐶𝑎𝑎3 , inwhich the interior does not change
due to the time freezing.

Our ansatz
Thus, we can use universal function 𝐴𝐴 𝑟𝑟 , 𝐵𝐵(𝑟𝑟) and set
Stationary BH in the heat bath
2
𝑎𝑎
𝑑𝑑𝑟𝑟
2 𝑑𝑑Ω2 ,− 1 − 𝑑𝑑𝑡𝑡 2 +
+
𝑟𝑟
𝑎𝑎
𝑟𝑟
1
−
𝑟𝑟
𝑑𝑑𝑠𝑠 2 =1 𝐴𝐴 𝑟𝑟 2𝑒𝑒𝑑𝑑𝑡𝑡 + 𝐵𝐵 𝑟𝑟 𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2 ,−𝐵𝐵 𝑟𝑟
𝑅𝑅(𝑎𝑎)

𝑟𝑟 > 𝑅𝑅(𝑎𝑎)𝑟𝑟 < 𝑅𝑅(𝑎𝑎)

𝑅𝑅 𝑎𝑎 = 𝑎𝑎 +2𝜎𝜎
𝑎𝑎⇒Let’s determine 𝐴𝐴 𝑟𝑟 , 𝐵𝐵(𝑟𝑟) from now.

(Note: This is just a first approximation. In future, we want to consider the back reaction
from the dilute radiation around the BH and the surface fluctuation with Δ𝑟𝑟~

2
𝑙𝑙𝑝𝑝

𝑎𝑎

.)

Determination of 𝐵𝐵(𝑟𝑟)𝑑𝑑𝑠𝑠 2 =

− 1−

−

𝑎𝑎
1
𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2
𝑑𝑑𝑡𝑡 2 +
𝑎𝑎
𝑟𝑟
1−
𝑟𝑟

1 𝐴𝐴 𝑟𝑟 2
𝑑𝑑𝑡𝑡 + 𝐵𝐵 𝑟𝑟 𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2
𝑒𝑒
𝐵𝐵 𝑟𝑟

On the surface of BH, we have
𝑅𝑅(𝑎𝑎)
𝑎𝑎𝑎𝑎(𝑎𝑎) 𝑅𝑅 𝑎𝑎 2
𝑔𝑔𝑟𝑟𝑟𝑟 �
=
,
=
≈
2𝜎𝜎
𝑅𝑅(𝑎𝑎) − 𝑎𝑎2𝜎𝜎𝑟𝑟=𝑅𝑅(𝑎𝑎)Radial coordinate 𝑟𝑟 is uniquely fixed
because of spherical symmetry.

𝑅𝑅 𝑎𝑎 = 𝑎𝑎 +2𝜎𝜎𝑎𝑎

𝑅𝑅 𝑎𝑎 2𝐵𝐵 𝑅𝑅 𝑎𝑎 =2𝜎𝜎
Because this holds for any 𝑎𝑎 and 𝐵𝐵(𝑟𝑟) is independent of 𝑎𝑎,
we have𝑟𝑟 2
𝐵𝐵 𝑟𝑟 =.
2𝜎𝜎Determination of 𝐴𝐴(𝑟𝑟) (1/3)

The system is in equilibrium.
⇒The energy-momentum flow inside the BH is time
reversal invariant, which can be expressed covariantly as− 𝑇𝑇𝜇𝜇𝜇𝜇 𝑘𝑘𝜈𝜈 = 𝜂𝜂 𝑙𝑙 𝜇𝜇 + 𝑓𝑓𝑘𝑘 𝜇𝜇− 𝑇𝑇𝜇𝜇𝜇𝜇 𝑙𝑙𝜈𝜈 = 𝜂𝜂 𝑘𝑘 𝜇𝜇 + 𝑓𝑓𝑙𝑙 𝜇𝜇

𝑟𝑟

1 𝐴𝐴
𝒌𝒌 = 𝐵𝐵 𝑒𝑒 2 𝑑𝑑𝑑𝑑 −1 𝐴𝐴
𝒍𝒍 = 𝐵𝐵 𝑒𝑒 2 𝑑𝑑𝑑𝑑 +𝐵𝐵𝑑𝑑𝑑𝑑: radially ingoing null vector → −𝒍𝒍

Exchangein 𝑡𝑡 → −𝑡𝑡

𝐵𝐵𝑑𝑑𝑑𝑑: radially outgoing null vector (→ −𝒌𝒌)

− 𝑇𝑇𝜇𝜇𝜇𝜇 𝑘𝑘𝜈𝜈
𝒌𝒌𝑡𝑡

Energy-momentum flow througha falling null spherical shellDetermination of 𝐴𝐴(𝑟𝑟) (2/3)• 𝑓𝑓(𝑟𝑟) gives the ratio of outgoing and ingoing flow:(1 > 𝑓𝑓(𝑟𝑟) ≥ 0)
𝑇𝑇 𝒌𝒌𝒌𝒌 : 𝑇𝑇 𝒌𝒌𝒍𝒍 = 1 ∶ 𝑓𝑓
⇒Where does 𝑓𝑓 ≠ 0 come from?
⇒massive particles and reflection of the emitted particles𝑟𝑟

𝑇𝑇 𝒌𝒌𝒌𝒌

𝑇𝑇 𝒌𝒌𝒌𝒌𝑡𝑡

Δ𝑡𝑡~𝑎𝑎𝑟𝑟

𝑇𝑇 𝒌𝒌𝒌𝒌

𝑇𝑇 𝒌𝒌𝒌𝒌

𝑓𝑓 also appears in the ratio of energy density − 𝑇𝑇𝑡𝑡𝑡𝑡
and radial pressure 𝑇𝑇𝑟𝑟𝑟𝑟 :𝑇𝑇𝑟𝑟𝑟𝑟1 − 𝑓𝑓𝑡𝑡 = 1 + 𝑓𝑓
− 𝑇𝑇𝑡𝑡𝑡𝑡

Determination of 𝐴𝐴(𝑟𝑟) (3/3)
𝑑𝑑𝑑𝑑 2 = −1 𝐴𝐴 𝑟𝑟 2𝑒𝑒𝑑𝑑𝑡𝑡 + 𝐵𝐵 𝑟𝑟 𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2
𝐵𝐵 𝑟𝑟

• If 𝑓𝑓(𝑟𝑟) is given, we can determine 𝐴𝐴(𝑟𝑟).
From Einstein eq., we have
𝐺𝐺 𝒌𝒌𝒌𝒌 : 𝐺𝐺 𝒌𝒌𝒌𝒌 = 𝑇𝑇 𝒌𝒌𝒌𝒌 : 𝑇𝑇 𝒌𝒌𝒌𝒌 = 1: 𝑓𝑓.
𝑑𝑑𝑑𝑑
2𝐵𝐵
𝑟𝑟 2
𝐵𝐵 𝑟𝑟 ==⇒
2𝜎𝜎
1 + 𝑓𝑓 𝑟𝑟
𝑑𝑑𝑑𝑑
Here, for simplicity, we assume
In general,
𝑓𝑓 𝑟𝑟 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐.
𝑓𝑓 𝑟𝑟 should vary slowly:
𝑟𝑟 2⇒ 𝐴𝐴 𝑟𝑟 =2 1 + 𝑓𝑓 𝜎𝜎𝑙𝑙𝑝𝑝

𝑑𝑑𝑑𝑑(𝑟𝑟)≪ 𝑓𝑓(𝑟𝑟)
𝑑𝑑𝑑𝑑

A candidate 𝑔𝑔𝜇𝜇𝜇𝜇 of the stationary BH
2
𝑎𝑎
𝑑𝑑𝑟𝑟
2 𝑑𝑑Ω2 ,− 1 − 𝑑𝑑𝑡𝑡 2 +𝑟𝑟 > 𝑅𝑅(𝑎𝑎)
+
𝑟𝑟
𝑎𝑎
𝑟𝑟
1−
2
𝑟𝑟
𝑑𝑑𝑠𝑠 =
2 −𝑟𝑟 22
2𝜎𝜎 −𝑅𝑅2𝜎𝜎𝑎𝑎 1+𝑓𝑓
𝑟𝑟
− 2 𝑒𝑒𝑑𝑑𝑡𝑡 2 +𝑟𝑟 < 𝑅𝑅(𝑎𝑎)𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2 ,𝑟𝑟 Exponentially large redshift 2𝜎𝜎→The interior is frozen.

These are connected at 𝑟𝑟 = 𝑅𝑅 𝑎𝑎balanceradiationfrom the bath

Hawking radiation

2𝜎𝜎= 𝑎𝑎 + =const.
𝑎𝑎𝑟𝑟

𝑅𝑅(𝑎𝑎)

No trapped region inside!

𝑡𝑡

𝜎𝜎Time runs only in Δ𝑟𝑟~ 𝑎𝑎

A candidate 𝑔𝑔𝜇𝜇𝜇𝜇 of the evaporating BH
𝑑𝑑𝑠𝑠 2=𝑟𝑟 − 𝑎𝑎 𝑢𝑢−
𝑑𝑑𝑢𝑢2 − 2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑Ω2 ,𝑟𝑟
2𝑟𝑟 > 𝑅𝑅(𝑎𝑎(𝑢𝑢))2

𝑅𝑅 𝑎𝑎 𝑢𝑢 −𝑟𝑟 2
𝑅𝑅 𝑎𝑎 𝑢𝑢 −𝑟𝑟 2
2𝜎𝜎 − 2𝜎𝜎 1+𝑓𝑓−
2
− 2 𝑒𝑒𝑑𝑑𝑢𝑢 − 2𝑒𝑒 4𝜎𝜎 1+𝑓𝑓 𝑑𝑑𝑑𝑑𝑑𝑑𝑟𝑟 + 𝑟𝑟 2 𝑑𝑑Ω2
𝑟𝑟

These are connected at𝑟𝑟

2𝜎𝜎𝑟𝑟 = 𝑅𝑅 𝑎𝑎(𝑢𝑢) = 𝑎𝑎(𝑢𝑢) +𝑎𝑎(𝑢𝑢)

stationary𝑢𝑢

𝑟𝑟 < 𝑅𝑅(𝑎𝑎(𝑢𝑢))𝑑𝑑𝜎𝜎𝑎𝑎 𝑢𝑢 = −𝑑𝑑𝑑𝑑
𝑎𝑎 𝑢𝑢 2 ⇒We will show
this later.
𝑅𝑅(𝑎𝑎(𝑢𝑢))Note: A multi-shell model• We can use a multi-shell model and obtain thesame interior metric with 𝑓𝑓 = 0.
𝑟𝑟𝑛𝑛

⋯

𝑟𝑟

Kawai, Matsuo and Y. Y. IJMPA (2013)
Kawai and Y. Y. Universe (2017)

𝑎𝑎𝑖𝑖
𝑎𝑎𝑖𝑖−1

𝑟𝑟1
flat

Without assumption
about the stationarity,
we reach the stationary
interior metric!

𝑟𝑟 − 𝑎𝑎 2
𝑑𝑑𝑢𝑢 − 2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑Ω2
𝑟𝑟

𝑟𝑟 − 𝑎𝑎𝑖𝑖 2
𝑑𝑑𝑢𝑢𝑖𝑖 − 2𝑑𝑑𝑢𝑢𝑖𝑖 𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑Ω2
𝑟𝑟
𝑑𝑑𝑎𝑎𝑖𝑖𝜎𝜎=− 2𝑑𝑑𝑢𝑢𝑖𝑖𝑎𝑎𝑖𝑖
𝑑𝑑𝑠𝑠𝑖𝑖2 = −⋯𝑟𝑟𝑖𝑖𝑟𝑟𝑖𝑖−1

2
𝑑𝑑𝑠𝑠𝑜𝑜𝑜𝑜𝑜𝑜=−

This approaches the same stationary metric.

𝑢𝑢

𝑟𝑟

Each shell keeps falling.
⇒This object is shrinking.
in the local time 𝒖𝒖𝒖𝑢𝑢Step2: Evaluation of ⟨𝑇𝑇𝜇𝜇𝜇𝜇 ⟩Setup𝑟𝑟

Consider the interior region.
The background metric is static:
2
2
𝑟𝑟
𝑟𝑟
2𝜎𝜎
2
𝑑𝑑𝑠𝑠 = − 2 𝑒𝑒 2𝜎𝜎(1+𝑓𝑓) 𝑑𝑑𝑈𝑈 2 − 2𝑒𝑒 4𝜎𝜎(1+𝑓𝑓) 𝑑𝑑𝑈𝑈𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑𝛺𝛺2
𝑟𝑟 𝑟𝑟(𝑈𝑈,𝑉𝑉)
𝜑𝜑= −𝑒𝑒
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑟𝑟 𝑈𝑈, 𝑉𝑉 2 𝑑𝑑Ω2 ,⇒ 𝑇𝑇𝜇𝜇𝜇𝜇 also should be static:
𝑇𝑇𝜇𝜇𝜇𝜇 = 𝑇𝑇𝜇𝜇𝜇𝜇 (𝑟𝑟) ,⇒We have the relation𝑢𝑢

𝑇𝑇𝑈𝑈𝑈𝑈 = 𝑇𝑇𝑉𝑉𝑉𝑉𝑇𝑇𝑈𝑈𝑈𝑈 = 𝑓𝑓⟨𝑇𝑇𝑈𝑈𝑈𝑈 ⟩,we have to determine only
𝑇𝑇𝑈𝑈𝑈𝑈 ,𝑇𝑇𝜃𝜃𝜃𝜃 .

𝑟𝑟

Physical origin of 𝑇𝑇𝑈𝑈𝑈𝑈 ≠ 0
𝑇𝑇𝑈𝑈𝑈𝑈scattering

𝑇𝑇𝑈𝑈𝑈𝑈 ≠ 0

𝑈𝑈・1st eq.

The relations of 𝑇𝑇𝜇𝜇𝜇𝜇

𝑇𝑇𝜇𝜇 = 2𝑔𝑔𝑈𝑈𝑈𝑈 𝑇𝑇𝑈𝑈𝑈𝑈 + 2 𝑇𝑇𝜃𝜃𝜃𝜃 leads to1 𝜇𝜇= 𝑓𝑓⟨𝑇𝑇𝑈𝑈𝑈𝑈 ⟩𝜃𝜃𝑇𝑇𝜃𝜃 = 𝑇𝑇𝜇𝜇 + 2e−𝜑𝜑 𝑓𝑓⟨𝑇𝑇𝑈𝑈𝑈𝑈 ⟩2・2nd eq.
𝜇𝜇

All components are
expressed in terms
𝜇𝜇of 𝑇𝑇𝜇𝜇 .

Integrating 𝛻𝛻𝜇𝜇 𝑇𝑇𝜇𝜇𝑈𝑈 = 0 from 𝑟𝑟 = 0 to 𝑟𝑟, we have
1 + 𝑓𝑓 𝜑𝜑 𝜇𝜇
𝑇𝑇𝑈𝑈𝑈𝑈 =𝜎𝜎𝑒𝑒 𝑇𝑇𝜇𝜇 ,22𝑟𝑟
where we have used the boundary condition
𝑇𝑇𝜇𝜇𝜇𝜇 (𝑟𝑟 = 0) = 0 (⇔initial condition that the system𝑟𝑟

𝑉𝑉 =const.

flat

surface at
𝑟𝑟 = 𝑅𝑅(𝑎𝑎(𝑢𝑢))

𝑈𝑈 =const.

𝑈𝑈

started from the collapsing matter.)

4D Weyl anomaly
For simplicity, consider conformal matters.
𝜇𝜇⇒ 𝑇𝑇𝜇𝜇is determined by the 4D Weyl anomaly:𝜇𝜇
𝑇𝑇𝜇𝜇 = ℏ𝑐𝑐𝑤𝑤 ℱ − ℏ𝑎𝑎𝑤𝑤 ℊ←state-independentwhereℱ ≡ 𝐶𝐶𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝐶𝐶𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 ,ℊ ≡ 𝑅𝑅𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝑅𝑅𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 − 4𝑅𝑅𝜇𝜇𝜇𝜇 𝑅𝑅𝜇𝜇𝜇𝜇 + 𝑅𝑅2⇒The candidate metric determinesℏ𝑐𝑐𝑤𝑤
𝜇𝜇
𝑇𝑇𝜇𝜇 =
3 1 + 𝑓𝑓 4 𝜎𝜎 2

⇒We will discuss thecase of non-conformal
matters later.

step3: Determine the self-consistent 𝑔𝑔𝜇𝜇𝜇𝜇

The 𝑔𝑔𝜇𝜇𝜇𝜇 is the self-consistent solution.

Thus , we have obtained
2
𝑟𝑟
ℏ𝑐𝑐𝑊𝑊
ℏ𝑐𝑐𝑊𝑊
𝜃𝜃
2(1+𝑓𝑓)𝜎𝜎 ,𝑇𝑇𝜃𝜃 =
,
𝑇𝑇
=
𝑇𝑇
=
𝑒𝑒
𝑈𝑈𝑈𝑈
𝑉𝑉𝑉𝑉
6 1 + 𝑓𝑓 4 𝜎𝜎 2
3 1 + 𝑓𝑓 3 𝑟𝑟 4
𝑇𝑇𝑈𝑈𝑈𝑈 = 𝑓𝑓 𝑇𝑇𝑈𝑈𝑈𝑈
On the other hand, the metric gives
2
𝑟𝑟
1
𝜎𝜎
𝜃𝜃
2(1+𝑓𝑓)𝜎𝜎 ,𝐺𝐺𝜃𝜃 =
=
𝐺𝐺
=
𝑒𝑒
,
𝐺𝐺
𝑈𝑈𝑈𝑈
𝑉𝑉𝑉𝑉
2 1 + 𝑓𝑓 2 𝜎𝜎
(1 + 𝑓𝑓)𝑟𝑟 4
𝐺𝐺𝑈𝑈𝑈𝑈 = 𝑓𝑓𝐺𝐺𝑈𝑈𝑈𝑈⇒𝐺𝐺𝜇𝜇𝜇𝜇 = 8𝜋𝜋𝐺𝐺 𝑇𝑇𝜇𝜇𝜇𝜇 is satisfied if we identify
8𝜋𝜋𝑙𝑙𝑝𝑝2 𝑐𝑐𝑊𝑊𝜎𝜎 =.
2
3 1 + 𝑓𝑓

(We can do the same program for the exterior and check the self-consistency.)

Self-consistent solution
𝑑𝑑𝑠𝑠 2 =

𝑎𝑎 𝑢𝑢
− 1−
𝑟𝑟

𝑑𝑑𝑢𝑢2 − 2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑𝛺𝛺222112𝜎𝜎 −2𝜎𝜎(1+𝑓𝑓)
𝑅𝑅 𝑎𝑎 𝑢𝑢 −𝑟𝑟 2
−
𝑅𝑅
𝑎𝑎
𝑢𝑢
−𝑟𝑟 2
2
− 2 𝑒𝑒𝑑𝑑𝑢𝑢 − 2𝑒𝑒 4𝜎𝜎(1+𝑓𝑓)
𝑑𝑑𝑢𝑢𝑑𝑑𝑑𝑑 + 𝑟𝑟 2 𝑑𝑑𝛺𝛺2
𝑟𝑟←Non-perturbative solution w.r.t. ℏ

The surface exists at
𝑅𝑅(𝑎𝑎 𝑢𝑢 ) ≡ 𝑎𝑎 𝑢𝑢 +

2𝜎𝜎
𝑎𝑎 𝑢𝑢

𝑟𝑟

8𝜋𝜋𝜋𝜋𝜋𝑐𝑐𝑊𝑊
𝜎𝜎 ≡3 1 + 𝑓𝑓 2

⇒Stefan-Boltzmann law isreproduced self-consistently:
Hawking radiation ∝ 𝑐𝑐𝑊𝑊
𝑑𝑑𝑑𝑑 𝑢𝑢
𝜎𝜎=−
𝑑𝑑𝑑𝑑
𝑎𝑎 𝑢𝑢 2𝜇𝜇

𝑇𝑇𝜇𝜇 = ℏ𝑐𝑐𝑤𝑤 𝐶𝐶𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝐶𝐶𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 − ℏ𝑎𝑎𝑊𝑊 ℊ

No trapped region
Δ𝑢𝑢𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ~𝑎𝑎3 /𝜎𝜎𝑢𝑢

Note: This configuration is typical in a thermodynamical sense.

Consistency check (1/3)
Validity of the classical gravity
In the macroscopic region (𝑟𝑟 > 𝑙𝑙𝑃𝑃 ),𝑅𝑅, 𝑅𝑅𝛼𝛼𝛼𝛼

𝑅𝑅𝛼𝛼𝛼𝛼 ,𝑅𝑅𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇

𝑅𝑅𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 ~if c𝑤𝑤 ≫ 1(𝑐𝑐𝑤𝑤 plays a role of 𝑁𝑁 in the introduction.)⇒No singularity:
𝐺𝐺𝜇𝜇𝜇𝜇 = 8𝜋𝜋𝐺𝐺 𝑇𝑇�𝜇𝜇𝜇𝜇 is valid
as long as 𝑎𝑎 > 𝑙𝑙𝑝𝑝 .

1
1
2 ≪ 2
𝑐𝑐𝑤𝑤 𝑙𝑙𝑝𝑝 𝑙𝑙𝑝𝑝
Hawking radiation

𝑎𝑎3
Δ𝑡𝑡~
𝜎𝜎

Δ𝑟𝑟~𝑎𝑎BH

collapsing matter

Note: This is consistent with the singularity theorem because there is no trapped region
and the dominant energy-condition is broken.

Consistency check (2/3):
Hawking radiation• Hawking radiation appears self-consistently:
By a similar manner to Hawking’s derivation,
we can show1ℏ�𝜔𝜔 0 =,𝑇𝑇 =0 𝑁𝑁
ℏ𝜔𝜔/𝑇𝑇
4𝜋𝜋𝜋𝜋(𝑢𝑢)
𝑒𝑒
−1
𝑟𝑟

|0 >

(for s-wave)

𝑑𝑑𝑑𝑑 𝑢𝑢
𝜎𝜎=−
𝑑𝑑𝑑𝑑
𝑎𝑎 𝑢𝑢 2

flat:𝑈𝑈

𝑢𝑢

Consistency check (3/3)
Energy condition of 𝑇𝑇𝜇𝜇𝜇𝜇

4𝜋𝜋

𝑅𝑅 𝑎𝑎

� 𝑑𝑑𝑑𝑑 𝑟𝑟 2 − 𝑇𝑇𝑡𝑡𝑡𝑡~ 𝑐𝑐𝑊𝑊 𝑙𝑙𝑝𝑝

− 𝑇𝑇𝑡𝑡𝑡𝑡

≈

𝑎𝑎
= 𝑀𝑀
2𝐺𝐺

1 1
1 1 − 𝑓𝑓 1
𝑟𝑟=
, 𝑇𝑇𝑟𝑟 =28𝜋𝜋𝜋𝜋 𝑟𝑟
8𝜋𝜋𝜋𝜋 1 + 𝑓𝑓 𝑟𝑟 2≪

The dominant energy
condition (𝜌𝜌 ≥ 𝑝𝑝𝑖𝑖 > 0)
is broken.

𝑇𝑇𝜃𝜃𝜃𝜃

anisotropic ⇒ Not a fluid

13=8𝜋𝜋𝜋𝜋 16𝜋𝜋𝑐𝑐𝑊𝑊 𝑙𝑙𝑝𝑝2

This large angular pressure, which is consistent
with 4D Weyl anomaly, supports the object.

Why is𝜃𝜃𝑇𝑇𝜃𝜃

so large?

• To study it, we want to evaluate ⟨𝜓𝜓 𝑇𝑇𝜇𝜇𝜇𝜇 𝜓𝜓⟩, where𝜓𝜓 = equilibrium at each stage .
𝑅𝑅(𝑎𝑎)• As a first trial, we are now calculating ⟨0 𝑇𝑇𝜇𝜇𝜇𝜇 0⟩ direcly.
⇒Result: For 𝑁𝑁 massless scalar fields 𝜙𝜙𝑖𝑖 ,1𝜇𝜇𝜃𝜃⇒consistent!
0 𝑇𝑇𝜃𝜃 0 ~ 0 𝑇𝑇𝜇𝜇 0 = 𝑂𝑂
2 ,𝑁𝑁𝑙𝑙𝑝𝑝which appears after integrating angular modes ∑∞𝑙𝑙=0 .

[Kawai and Y.Y.,
work in progress]

Step4:
Information problem and BH entropy[Kawai and Y.Y. PRD 2016]

Information recovery by interaction• Hawking radiation is created inside the collapsing matter.
⇒The collapsing matter and Hawking radiation interact.
1d scattering model

𝑟𝑟

|𝜓𝜓⟩

𝑡𝑡

matterradiation|𝜓𝜓⟩

𝜆𝜆: coupling constant⇒We can estimate the scattering time scale𝑎𝑎Δ𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∼ 𝑎𝑎 log
𝜆𝜆𝜆𝜆𝑙𝑙𝑝𝑝～scrambling time
⇒information flow = energy flow
⇒information recovery?

(|𝜓𝜓⟩, 𝑀𝑀)(|𝜓𝜓⟩, 𝑀𝑀)Entropy from the interior
𝑀𝑀𝑀 =𝑅𝑅(𝑎𝑎𝑎)
′

𝑎𝑎
2𝐺𝐺

𝑀𝑀 =

𝑅𝑅(𝑎𝑎)

𝑎𝑎
2𝐺𝐺

8𝜋𝜋𝑙𝑙𝑝𝑝2 𝑐𝑐𝑊𝑊𝜎𝜎 =3 1 + 𝑓𝑓 2

• The interior metric of the stationary BH is given by
2
1
2 −𝑟𝑟 2
2𝜎𝜎
𝑟𝑟
−
𝑅𝑅
𝑎𝑎
𝑑𝑑𝑟𝑟 2 + 𝑟𝑟 2 𝑑𝑑Ω2 .
𝑑𝑑𝑠𝑠 2 = − 2 𝑒𝑒 2(1+𝑓𝑓)𝜎𝜎
𝑑𝑑𝑡𝑡 2 +𝑟𝑟
2𝜎𝜎• We can evaluate the entropy density as
1 2𝜋𝜋 2𝜎𝜎
⇒ 4𝜋𝜋𝑟𝑟 2 𝑠𝑠~ 𝑁𝑁/𝑙𝑙𝑝𝑝
𝑠𝑠 =.
4𝜋𝜋𝑟𝑟 2 𝑙𝑙𝑝𝑝2• Summing up it over the interior volume, we obtain
𝑅𝑅(𝑎𝑎)
𝑎𝑎
2
𝑟𝑟
2𝜋𝜋
2𝜎𝜎
𝜋𝜋𝑎𝑎
𝐴𝐴
2
= 2 = 2𝑆𝑆 = �𝑑𝑑𝑑𝑑 𝑠𝑠 = � 𝑑𝑑𝑑𝑑 𝑔𝑔𝑟𝑟𝑟𝑟 4𝜋𝜋𝑟𝑟 𝑠𝑠 ≈ � 𝑑𝑑𝑑𝑑
2
𝑙𝑙𝑝𝑝
4𝑙𝑙𝑝𝑝
2𝜎𝜎 𝑙𝑙𝑝𝑝
00⇒The information should be stored inside the BH.

Conclusions

Conclusion: The new picture of BH
BH evaporates by nature.
⇒Quantum BH is described by field theory 𝐺𝐺𝜇𝜇𝜇𝜇 = 8𝜋𝜋𝐺𝐺 𝑇𝑇𝜇𝜇𝜇𝜇
with 𝑐𝑐𝑊𝑊 ≫ 1 (as long as 𝑎𝑎 ≫ 𝑙𝑙𝑝𝑝 ).
The surface exists at←non-perturbative solution w.r.t. ℏ

2𝜎𝜎
𝑅𝑅(𝑎𝑎 𝑢𝑢 ) ≡ 𝑎𝑎 𝑢𝑢 +𝑎𝑎 𝑢𝑢Information may comeback through interactions.
|𝜓𝜓⟩

|𝜓𝜓⟩

𝑎𝑎
𝑀𝑀 =
2𝐺𝐺

𝑅𝑅(𝑎𝑎)

Large pressure 𝑇𝑇𝜃𝜃𝜃𝜃No horizon or singularity
BH

Each shell keeps falling,𝑎𝑎3
but the interior is stationary. Δ𝑡𝑡~

𝑟𝑟𝑢𝑢

in the local time 𝒖𝒖𝒖

𝜎𝜎Δ𝑟𝑟~𝑎𝑎No horizon but surface.
This is the BH.

Thank you very much!