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Home , Hawking radiation, Superradiance

Hawking is probably a type of superradiation

Yi-Xiao Zhang South China Normal University, Guangzhou 510006,China

Wen-Xiang Chen∗ Department of Astronomy, School of and Materials , GuangZhou University

In this article, it mainly discusses that when the scalar field equation presets boundary condi- tions, the effective action form of is consistent with the effective action form of superradiation. From this I conclude that Hawking radiation may be a form of superradiation. Keywords: Hawking radiation, , effective action

I. INTRODUCTION

Hawking radiation is a kind of emitted by black holes, which is speculated by effect theory. This theory was put forward by physicist in 1974. With the Hawking radiation theory, we can explain how to reduce the of black holes to cause evapotranspiration. And because Hawking radiation can cause black holes to lose mass, when black holes lose more mass than they increase, they will shrink and eventually disappear. The divergence of a relatively small black hole is usually larger than that of a normal black hole, so the former shrinks and disappears faster than the latter. Hawking’s analysis quickly became the first convincing theory of quantum , although the existence of Hawking radiation has not yet been actually observed. In June 2008, NASA launched the GLAST satellite, which can search for flashes of gamma rays in evaporating black holes. In the theory of extra dimensions, collisions of high- particles may also create black holes that disappear by themselves. In September 2010, the results of a simulated gravity study were considered by some scientists to demonstrate the possible existence and possible of Hawking radiation for the first time. However, Hawking radiation has not yet been actually observed. A black hole is a place with great gravitation, and the around it will be pulled in by gravity. In classical mechanics, its gravity is so strong that even electromagnetic radiation waves cannot escape. Although it is not yet known how to unify gravity and , the gravitational effect far away from the black hole is so weak that the calculation results can still conform to the quantum field theory framework of curved space-time. Hawking said that quantum effects allow black holes to emit precise radiation. This electromagnetic radiation seems to be emitted by a black body whose is inversely proportional to the mass of the black hole. For example, the temperature of a solar-mass black hole is only 60nK; in fact, a black hole absorbs much more

∗Electronic address: [email protected] 2 cosmic microwave background radiation than it emits. A black hole with a mass of 4.5×1022 kg (similar to the mass of the ) will maintain its temperature at 2.7K and absorb the same amount of radiation as it emits. A smaller primary black hole emits more radiation than it absorbs, and therefore gradually loses mass. Before the concept of Hawking radiation, there was a problem in the physics world. If you throw things with a lot of into a black hole, will that entropy be eliminated, but entropy will never decrease in the universe, so this represents a black hole. There should also be a lot of entropy, and anything with entropy will release black body radiation. Whether black holes will also release black body radiation, but what is the mechanism of release? Hawking radiation explains the mechanism of black body radiation. According to the Heisenberg Uncertainty Principle, many particle- () pairs are generated in a instantly and naturally out of thin air, and they are annihilated in pairs in a very short time, and there is no mass production in the macroscopic view. Physicists such as Yakov Borisovich Zeldovich, , and Stephen Hawking combined quantum me- chanics and , and the results showed that the temperature of the horizon is not zero, but also Glow, although extremely weak. This kind of light is the so-called ”Hawking radiation”; when two pairs of particles-such as and positrons, or a pair of -are created in a strong gravitational field, one of the particles will fall into the black hole, and the other One will flee, thus generating this radiation. If a pair of particles is formed near a black hole, due to the strong gravitational field of the black hole, the paired positive and negative particles are torn apart. It is possible that one of them will fall into the , and the other will not, thereby being lifted to reality by the black hole’s gravity. particle. But this violates the law of , so the mass of another particle must come from the mass of the black hole itself-this is a simplified explanation of the radiation emitted by the black hole. Basically, massive black holes can survive longer. Generally, black holes produced by the death of stars can live for 1066 years, while supermassive black holes can live for 1090 years. Hawking radiation can also explain why we cannot observe the micro-black holes produced when the universe was born because they have evaporated. Absolute vacuum violates the uncertainty principle of quantum mechanics, so it does not exist. When the space moves towards an absolute vacuum, a pair of virtual particles will be produced, and the two particles will disappear after colliding, so that neither quantum mechanics nor the conservation of matter will be violated. When this quantum phenomenon occurs at the edge of the black hole’s horizon, the virtual particles outside the horizon can be observed because they are outside the horizon, and thus become real particles, while the virtual particles within the horizon are within the horizon, so Will be swallowed by black holes and will not be observed. Because the particles outside the horizon are real particles with mass, according to the law of conservation of mass and energy, the particles swallowed by the black hole within the horizon have negative mass, so the mass of the black hole will be reduced due to this effect. From the outside, it seems that the black hole is slowly evaporating. The smaller the black hole, the faster the evaporation rate, until the black hole is completely evaporated. In 1972, Press and Teukolsky[18] proposed that it is possible to create a by adding a mirror to the outside of the black hole (a scattering process that, according to current interpretations, involves classical and quantum mechanics)[1,3, 11, 12, 15, 17, 19]). When a bosonic wave is impinging upon a , the wave reflected by the event horizon will be 3 amplified if the wave ω lies in the following superradiant regime[13, 14, 18, 20, 21]

a 0 < ω < mΩH , ΩH = 2 2 , (1) r+ + a where m is azimuthal number of the bosonic wave mode, ΩH is the angular velocity of black hole horizon.This amplification is called superradiant scattering.Therefore, the rotational energy of the black hole can be extracted by the superradiation process.If there is a mirror between the event horizon of the black hole and the infinite space, the amplified wave will scatter back and forth and grow exponentially, which will cause the superradiation of the black hole to become unstable. Associate Professor Hasegawa Yuji of the Vienna University of Technology and Professor Masaaki Ozawa of Nagoya University and other scholars published empirical results against Heisenberg’s uncertainty principle on January 15, 2012[10]. They measured the angle of neutrons with two instruments, and calculated the neutrons with a smaller error than the Heisenberg uncertainty principle, which proved that the limit of the measurement method created by the Heisenberg uncertainty principle is wrong. But due to the inherent quantum nature of particles. In that paper[8], the method of my research on superradiation is adopted to relate the uncertainty principle to the superradiation effect. It is found that under the superradiation effect, the measurement range of the uncertainty principle can be reduced. In the paper[5],when ∆x∆p < ~/2 happens at the same time when the entropy reaches its maximum value, the boson will condense, and if there is a potential well but it does not explode, then the boson will gain high energy (more than normal).This article is to illustrate the possibility of a kind of Bose particle to obtain high energy. There is an inference there, quantum radiation cannot produce thermal effects. In the article[5], under super- radiation, first preset the boundary conditions of the boson, it is possible to get a larger energy than the conventional quantum effect, and that extra energy belongs to the classical domain, that is, heat. In this article, it mainly discusses that when the scalar field equation presets boundary conditions, the effective action form of Hawking radiation is consistent with the effective action form of superradiation. From this I conclude that Hawking radiation may be a form of superradiation.

II. THE SUPERRADIATION EFFECT OF BOSON SCATTERING

We find the Klein-Gordon equation[2]

;µ Φ;µ = 0 , (2)

µ where we defined Φ;µ ≡ (∂µ − ieAµ)Φ and e is the of the scalar field.We get A = {A0(x), 0},and eA0(x)can be equal to µ(where µ is the mass).   0 as x → −∞ A0 → . (3)  V as x → +∞

With Φ = e−iωtf(x), which is determined by the ordinary differential equation

d2f + (ω − eA )2 f = 0 . (4) dx2 0 4

We see that particles coming from −∞ and scattering off the potential with reflection and transmission amplitudes R and T respectively. With these boundary conditions, the solution to behaves asymptotically as

iωx −iωx fin(x) = Ie + Re , x → −∞, (5)

ikx fin(x) = T e , x → +∞ (6) where k = ±(ω − eV ). To define the sign of ω and k we must look at the wave’s group velocity. We require ∂ω/∂k > 0, so that they travel from the left to the right in the x–direction and we take ω > 0.

The reflection coefficient and transmission coefficient depend on the specific shape of the potential A0. However one can easily show that the Wronskian

df˜ df˜ W = f˜ 2 − f˜ 1 , (7) 1 dx 2 dx ˜ ˜ between two independent solutions, f1 and f2, of is conserved. From the equation on the other hand, if f is a solution ∗ 2 2 ω−eV 2 then its complex conjugate f is another linearly independent solution. We find|R| = |I| − ω |T | .Thus,for 0 < ω < eV ,it is possible to have superradiant amplification of the reflected current, i.e, |R| > |I|.

III. CLASSICAL SUPERRADIATION EFFECT IN THE SPACE-TIME OF A STEADY BLACK HOLE

2 2 dr∗ r We know that ψ ∼ exp(−iωt+imφ),and the ordinary differential equation(We use tortoise coordinate r∗ , dr2 = ∆ )

d2ψ 2 + V ψ = 0. (8) dr∗ There are other potentials that can be completely resolved, which can also show superradiation explicitly.We see that particles coming from −∞ and scattering off the potential with reflection and transmission amplitudes C and D respectively. With these boundary conditions, the solution to behaves asymptotically as  iωH r∗ −iωH r∗  Ae + Be , r → r+ ψ = (9)  Ceiω∞r∗ + De−iω∞r∗ , r → ∞ . The reflection coefficient and transmission coefficient depend on the specific shape of the potential V.We show that the Wronskian

dψ¯ dψ W ≡ ψ − ψ¯ , (10) dr∗ dr∗

2 2 2 2 2 2 W (r → r+) = 2iωH |A| − |B| ,W (r → ∞) = 2iω∞ |C| − |D| is conserved. We find|C| − |D| =

ωH 2 2 |A| − |B| .Thus,for ωH /ω∞ < 0,it is possible to have superradiant amplification of the reflected current, i.e,if ω∞ |A| = 0, |C| > |D|. 5

IV. THAT RESULT CONTAINED SOME CONCLUSIONS THAT VIOLATED UNCERTAINTY PRINCIPLE

The joint uncertainty principle shows that joint measurement of position and is impossible, that is, simultaneous measurement of position and momentum can only be approximate joint measurement, and the error 2 2 ω−eV 2 2 follows the inequality ∆x∆p ≥ 1/2(in natural unit system).We find|R| = |I| − ω |T | ,and we know that|R| ≥ ω−eV 2 − ω |T | is a necessary condition for the inequality ∆x∆p ≥ 1/2 to be established[2, 10].We can pre-set the boundary conditions eA0(x) = yω(which can be µ = yω)[4][6][7][9], and we see that when y is relatively large(according 2 ω−eV 2 to the properties of the boson, y can be very large),|R| ≥ − ω |T | may not hold.In the end,we can get ∆x∆p ≥ 1/2 may not hold.If the boundary conditions of the incident boson are set in advance, the flow density equations on both sides of the probability are not equal due to the boundary conditions.At that time the particle enters the classical mode (including the quantum mode plus the classical mode). In the quantum mode of the component, the uncertainty principle is established (take the extreme value of the uncertainty principle 1/2), and in the classical mode of the other component, the Bose particles under super-radiation continue to condense and gain energy.

V. HAWKING RADIATION FROM KERR BLACK HOLES

We will prove that the Hawking radiation of the Kerr black hole can be understood as the flux that offsets the . The key is that near the horizon, the scalar field theory in the of a 4-dimensional Kerr black hole can be simplified to a 2-dimensional field theory. Since space-time is not spherically symmetric, this is an unexpected result. In Boyer-Linquist coordinates, reads[16]

2 ∆ − a2 sin2 θ r2 + a2 − ∆ r2 + a2 − ∆a2 sin2 θ Σ ds2 = − dt2 − 2a sin2 θ dtdφ + sin2 θdφ2 + dr2 + Σdθ2 (11) Σ Σ Σ ∆

2 2 2 2 2 Σ = r + a cos θ, ∆ = r − 2Mr + a = (r − r+)(r − r−) . (12)

The action for the scalar field in the Kerr spacetime is 1 Z √ S[ϕ] = d4x −gϕ∇2ϕ 2 " 2 ! 1 Z √ 1 r2 + a2 = d4x −gϕ − − a2 sin2 θ ∂2 2 Σ ∆ t (13) 2 2   2  2a r + a − ∆ 1 a 2 − ∂t∂φ + − ∂ ∆ sin2 θ ∆ φ 1  +∂ ∆∂ + ∂ sin θ∂ ϕ r r sin θ θ θ

Taking the limit r → r+ and leaving the dominant terms, we have " 2 1 Z r2 + a2 S[ϕ] = d4x sin θϕ − + ∂2 2 ∆ t # (14) 2a r2 + a2 a2 − + ∂ ∂ − ∂2 + ∂ ∆∂ ϕ ∆ t φ ∆ φ r r 6

Now we transform the coordinates to the locally non-rotating coordinate system by   ψ = φ − ΩH t (15)  ξ = t where

a ΩH ≡ 2 2 . (16) r+ + a We can rewrite the action Z   a 4 1 2 S[ϕ] = d x sin θϕ − ∂ξ + ∂rf(r)∂r ϕ (17) 2ΩH f(r) We know that when sin θ = 0, the pull equation for action can conform to the above form, but the boundary becomes 0. However, if the boundary conditions are preset, the boundary conditionsµ = yω(y takes a larger number) act as sin θ, and then the effective action form of superradiation satisfies the effective action form of Hawking radiation, and it is not necessarily on the boundary of the horizon.

VI. SUMMARY

In the paper [5], in the case of superradiation, the boson boundary condition is presupposed first, which is possible to obtain higher energy than the traditional quantum effect, while the extra energy belongs to the classical domain, namely heat.In this paper, it is discussed that the effective form of Hawking radiation is consistent with the effective form of superradiation when the scalar field equation assumes the boundary conditions.From this I conclude that Hawking radiation is probably a type of superradiation.

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