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From the experimental data of the Compton Effect we know that
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- For the scattering angle θ = 135 o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0749nm.
- For the scattering angle θ = 90 o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0731nm.
- For the scattering angle θ = 45 o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0715nm.
- Nuclear Density
- Robert G. Ingersoll, The Works of Robert G. Ingersoll, Volume 1: Lectures Hawking Radiation
From the experimental data of the Compton Effect we know that: For the scattering angle θ = 135o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0749nm. For the scattering angle θ = 90o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0731nm. For the scattering angle θ = 45o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0715nm. For the scattering angle θ = 135o and the wavelength of the incident photon 0.0709nm, the wavelength of the scattered photon was found to be 0.0749nm. The energy of the incident photon E = hc/λ = (6.625 × 10 − 34 × 3 × 108) / 0.0709× 10 − 9 = 280.324 × 10 − 17 J. The energy of the incident photon Es = hc/λs = (6.625 × 10 − 34 × 3 × 108) / 0.0749× 10 − 9 = 265.353 × 10 − 17 J. From the law of conservation of energy, E + m0c2 = Es + mc2 mc2 – m0c2 = (E – Es) = 14.971 × 10 − 17 J. Which on rearranging we get: mc2 = m0c2 + 14.971 × 10 − 17 J. mc2 = (9.1 × 10 − 31 × 9 × 1016) J + 14.971 × 10 − 17 J = 82.049 × 10−15 J m = 82.049 × 10−15 / c2 = 9.1165 × 10 − 31kg … (1) From the law of conservation of momentum, pe2 = p2 + ps2 − 2p ps cosθ p = h /λ = 6.625 × 10 −34 / 0.0709× 10 − 9 = 93.441 × 10 − 25 Js/m ps= h /λs = 6.625 × 10 − 34 / 0.0749× 10 −9 = 88.451 × 10 − 25 Js/m θ = 135o pe2 = 28243.06 × 10 − 50 J 2s 2/m2 pe = 168.0567 × 10 − 25 Js /m In physics, we find out that momentum is mass multiplied by velocity. Special relativity (which overturned the understanding of space and time: space and time cannot be thought of as universal concepts experienced identically by everyone but they are malleable constructs whose form and appearance depends on one’s state of motion) has something to say about momentum. In particular, special relativity gets its (1− v2/c2) ½ factor into the momentum mix like this: pe = m0v / (1− v2/c2) ½. For non-relativistic case: v << c. Therefore, we have pe = m0v Suppose the particle is brought to rest, then (v = 0, pe = 0). Under this condition the rest mass of the particle becomes undefined i.e., m0 = pe/v = 0/0 There can be no bigger limitation than this because m0 cannot be undefined (it is always well defined). However, substituting m = 9.1165 × 10 − 31kg and pe =168.0567 × 10 − 25 Js /m in the equation pe = mv, we get: v = 18.434× 106 m/s Substituting this value in the equation m = m0 / (1 − v2/c2) ½, we get: m = 9.1172 × 10 − 31kg … (2) From (1) m = 9.1165 × 10 − 31kg From (2) m = 9.1172 × 10 − 31kg Difference = 7 × 10 − 4
The energy of the incident photon E = hc/λ = (6.625 × 10 − 34 × 3 × 108) / 0.0709× 10−9 = 280.324 × 10 − 17 J. The energy of the incident photon Es = hc/λs = (6.625 × 10 − 34 × 3 × 108) / 0.0731× 10 − 9 = 271.887 × 10 − 17 J. From the law of conservation of energy, E + m0c2 = Es+ mc2 mc2 – m0c2 = (E – Es) = 8.437 × 10 − 17 J. Which on rearranging we get: mc2 = m0c2 + 8.437 × 10 − 17 J mc2 = (9.1 × 10 − 31 × 9 × 1016) J + 8.437 × 10 − 17J mc2 = 81.984 × 10−15 J m = 81.984 × 10−15 / c2 = 9.10933 × 10 − 31kg … (1) From the law of conservation of momentum, pe2 = p2 + ps2 − 2ppscosθ p = h /λ = 6.625 × 10 − 34 / 0.0709 × 10 − 9 = 93.441 × 10 − 25 Js/m ps= h /λs= 6.625 × 10 − 34 / 0.0731 × 10 − 9 = 90.629 × 10 − 25 Js/m θ = 90o pe2 = 16944.83 × 10 − 50 J 2s 2/m2 pe = 130.172 × 10 −25 Js /m Substituting m = 9.10933 × 10 − 31kg and p =130.172 × 10 − 25 Js /m in the equation pe = mv, we get: v = 14.2899 × 106 m/s Substituting this value in the equation m = m0 / (1 − v2/c2) ½, we get: m = 9.11033 × 10 − 31kg … (2) From (1) m = 9.10933 × 10 − 31kg From (2) m = 9.11033 × 10 − 31kg Difference = 1 × 10 − 3
The energy of the incident photon E = hc/λ = (6.625 × 10 − 34 × 3 × 108) / 0.0709× 10 − 9 = 280.324 × 10 − 17 J. The energy of the incident photon Es = hc/λs = (6.625 × 10 − 34 × 3 × 108) / 0.0715× 10 − 9 = 277.972 × 10 − 17 J. From the law of conservation of energy, E + m0c2 = Es + mc2 mc2 – m0c2 = (E – Es) = 2.352 × 10 − 17 J. Which on rearranging we get: mc2 = m0c2 + 2.352 × 10 −17 J. mc2 = (9.1 × 10 − 31 × 9 × 1016) J + 2.352 × 10 −17 J mc2 = 81.923 × 10−15 J m = 81.923 × 10−15 / c2 = 9.10255 × 10 − 31kg … (1) From the law of conservation of momentum, pe2 = p2 + ps2 − 2ppscosθ p = h /λ = 6.625 × 10 − 34 / 0.0709 × 10 − 9 = 93.441 × 10 − 25 Js/m ps = h /λs= 6.625 × 10 − 34 / 0.0715 × 10 − 9 = 92.657× 10 − 25 Js/m θ = 45o pe2 = 5072.386 × 10 − 50 J 2s 2/m2 pe = 71.220 × 10 − 25 Js /m Substituting m = 9.10255 × 10 − 31kg and p = 71.220 × 10 − 25 Js /m in the equation pe = mv, we get: v = 7.824 × 106 m/s Substituting this value in the equation m = m0 / (1 − v2/c2) ½, we get: m = 9.10034 × 10 − 31kg … (2) From (1) m = 9.10255 × 10 − 31kg From (2) m = 9.10034 × 10 − 31kg Difference = 2.21 × 10 −3 CONCLUSION: For the scattering angle θ = 135o: m = 9.1165 × 10 − 31kg … (1) m = 9.1172 × 10 − 31kg … (2) m = 9.1165 × 10 − 31kg … (1) is less than m = 9.1172 × 10 − 31kg … (2) For the scattering angle θ = 90o: m = 9.10933 × 10 − 31kg … (1) m = 9.11033 × 10 − 31kg … (2) m = 9.10933 × 10 − 31kg … (1) is less than m = 9.11033 × 10 − 31kg … (2) However, For the scattering angle θ = 45o: m = 9.10255 × 10 − 31kg … (1) m = 9.10034 × 10 − 31kg … (2) m = 9.10255× 10 − 31kg … (1) is greater than m = 9.10034 × 10 − 31kg … (2) But WHY? The question lingers, unanswered until now.
As we know that: KE = mv2 But KE = E − Es. Therefore: mv2 = E − Es or pe2v2 = (E − Es) 2 or pe2v2 = E2 + Es2 − 2 E Es But pe2 = p2 + ps2 − 2ppscosθ or pe2c2 = E2 + Es2 − 2 E Escosθ Therefore: pe2v2/ pe2c2 = (E2 + Es2 − 2 E Es) / (E2 + Es2 − 2 E Escosθ) or v2/ c2 = (E2 + Es2 − 2 E Es) / (E2 + Es2 − 2 E Escosθ) From the above equation it is clear that if θ = 0o then v = c (which is a wrong and meaningless result because when θ = 0o there is no change in frequency / wavelength of the incident photon i.e., absorption of photon energy does not take place then how can the electron be accelerated to the velocity v = c). “Science is uncertain. Theories are subject to revision; observations are open to a variety of interpretations, and scientists quarrel amongst themselves. This is disillusioning for those untrained in the scientific method, who thus turn to the rigid certainty of the Bible instead. There is something comfortable about a view that allows for no deviation and that spares you the painful necessity of having to think.” — Isaac Asimov We know that virtual photon is to electromagnetism; why not to gravity? Nuclear Density Mass of the neutron, mneutron = 1.6750 × 10 −27 kg Mass of the proton, mProton = 1.6726 × 10 −27 kg Mneutron / mProton = 1.00143 Nuclear density = mass of the nucleus / its volume ρNucleus = M/V But M = (ZmProton + Nmneutron) V = (4/3) πr03A (where: Z = number of protons in the nucleus, N = number of neutrons in the nucleus, R0 = 1.2 × 10−15m, A = Z +N) Therefore: ρNucleus = 3mProton (Z + 1.00143N) / 4πr03A Which on rearranging: A = (3mProton / 4π R03ρNucleus) Z + (3.00429mProton / 4π R03ρNucleus) N Since A = (Z + N): (Z + N) = (3mProton / 4πr03ρNucleus) Z + (3.00429mProton / 4πr03ρNucleus) N Any equation is valid only if LHS = RHS. Hence the above equation is valid only if Z + N = Z +N. Z + N = Z +N is achieved only if ρNucleus attains 2 values i.e., ρNucleus = 3mProton / 4π R03 and ρNucleus = 3.00429mProton / 4π R03 at the same time. But how ρNucleus can attain 2 values at the same time? It’s highly impossible. “An actual perfection cannot exist.” “Nothing is perfect. To start with, perfection is ideal.” “A perfect thing cannot exist.” “Like the most of you, I was raised among people who knew - who were certain. They did not reason or investigate. They had no doubts. They knew that they had the truth. In their creed there was no guess — no perhaps. They had a revelation from God. They knew the beginning of things. They knew that God commenced to create one Monday morning, four thousand and four years before Christ. They knew that in the eternity — back of that morning, he had done nothing. They knew that it took him six days to make the earth — all plants, all animals, all life, and all the globes that wheel in space. They knew exactly what he did each day and when he rested. They knew the origin, the cause of evil, of all crime, of all disease and death. At the same time they knew that God created man in his own image and was perfectly satisfied with his work... They knew all about the Flood -- knew that God, with the exception of eight, drowned all his children -- the old and young -- the bowed patriarch and the dimpled babe -- the young man and the merry maiden -- the loving mother and the laughing child -- because his mercy endureth forever. They knew too, that he drowned the beasts and birds -- everything that walked or crawled or flew -- because his loving kindness is over all his works. They knew that God, for the purpose of civilizing his children, had devoured some with earthquakes, destroyed some with storms of fire, killed some with his lightnings, millions with famine, with pestilence, and sacrificed countless thousands upon the fields of war. They knew that it was necessary to believe these things and to love God. They knew that there could be no salvation except by faith, and through the atoning blood of Jesus Christ. Then I asked myself the question: Is there a supernatural power -- an arbitrary mind -- an enthroned God -- a supreme will that sways the tides and currents of the world -- to which all causes bow? I do not deny. I do not know - but I do not believe. I believe that the natural is supreme - that from the infinite chain no link can be lost or broken — that there is no supernatural power that can answer prayer - no power that worship can persuade or change — no power that cares for man. Is there a God? I do not know. Is man immortal? I do not know. One thing I do know, and that is, that neither hope, nor fear, belief, nor denial, can change the fact. It is as it is, and it will be as it must be. We can be as honest as we are ignorant. If we are, when asked what is beyond the horizon of the known, we must say that we do not know. We can tell the truth, and we can enjoy the blessed freedom that the brave have won. We can destroy the monsters of superstition, the hissing snakes of ignorance and fear. We can drive from our minds the frightful things that tear and wound with beak and fang. We can civilize our fellow-men. We can fill our lives with generous deeds, with loving words, with art and song, and all the ecstasies of love. We can flood our years with sunshine — with the divine climate of kindness, and we can drain to the last drop the golden cup of joy.” ― Robert G. Ingersoll, The Works of Robert G. Ingersoll, Volume 1: Lectures
“The area formula for the entropy — or number of internal states — of a black hole suggests that information about what falls into a black hole may be stored like that on a record, and played back as the black hole evaporates.” “There is no escape from a black hole in classical theory, but quantum theory enables energy and information to escape.” When stars are born, they form from existing gas dust of large amount of gas (mostly hydrogen). This is called interstellar matter. When cloud of interstellar matter crosses the spiral arm of a galaxy, it begins to form clumps. The gravitational forces within the clumps cause them to contract, forming protostar. The center of a protostar may reach a temperature of a several million of degree Celsius. At this high temperature, a fusion reaction begins. The energy released by this reaction prevents the protostar to contract. Thus, a star has been formed. There are so many stages of a star from its birth to death. The black hole is the final stage of dying star having masses 5 times the solar mass – 20 times the solar mass i.e., the star shrink to a certain critical radius, the gravitational field at the surface becomes so strong that the light cones are bent inward so much that light can no longer escape to reach a distant observer. Thus if light cannot escape, neither can anything else; everything is dragged back by the gravitational field. However, slow leakage of radiation from a black hole is allowed by quantum field effects near the event horizon (the boundary of a black hole where gravity is just strong enough to drag light back, and prevent it escaping) which will carry away energy, which mean that the black hole will lose mass and get smaller. In turn, this will mean that its temperature will rise and the rate of emission of radiation will increase (giving off x-rays and gamma rays, at a rate of about 10 million Megawatts, enough to power the world’s electricity supply). It is named after the renowned English physicist Stephen Hawking, who provided a theoretical argument for its existence in 1974). The rate of loss of energy of a black hole in the form of Hawking radiation (which make black hole to glow like a piece of hot metal) is given by the equation: − dMc2/dt = ħc6/ 15360πG2M2 Since the black hole temperature T = (ħc3 / 8πGMkB). Therefore: dT/dt = (kB3Gπ2/30ħc5) T4 or dT/dt = bt4 where: b = (kB3Gπ2/30ħc5) = 1.629 × 10 – 65 Kelvin– 3 second– 1 On rearranging: dt T −4 = b × dt which on integration we get: − 1/ 3T3 = bt + constant T = T1 (initial temperature of the black hole) when t = 0 − 1/ 3T13 = b (0) + constant − 1/ 3T13 = constant Solving for constant we get: −1/ 3T3 = bt – 1/ 3T13 T = T2 when t = half of the evaporation time i.e., tev /2 (where tev = evaporation time of the black hole). −1/ 3T23 = btev /2 – 1/ 3T13 or 1/ 3T23 = 1/ 3T13− btev /2 For a black hole of initial mass = one solar mass (i.e., M = 2 × 1030kg): tev= 6.7396 × 10 74 s T1 = 6.156 × 10 – 8 K 1/ 3 T23 = 1/3 × (6.156 × 10 – 8)3 – (1.629 × 10 – 65 × 3.369 × 10 74) 1/ 3 T23 = 1.4288× 10 21 – 5.4894× 10 9
or T2 = 6.156 × 10 – 8 K From the above calculation it is clear that: T1 = T2 i.e., temperature of the black hole when t = 0 is equal to the temperature of the black hole when t= tev/2. This means: T remains constant throughout the evaporation process. If T remains constant throughout the evaporation process, then from the equation: T = ħc3/ 8πGMkB M must remain constant throughout the evaporation process. But how can M remain constant because M varies throughout the evaporation process because the black hole loses its mass throughout its evaporation process. A virtual-particle pair has a wave function that predicts that both particles will have opposite spins. But if one particle falls into the black hole, it is impossible to predict with certainty the spin of the remaining particle. − S. W. Hawking Black holes have no Hair, says no hair Theorem: Wait, What? Characterizing the black hole The answer is then simple. Mass, Charge and Angular momentum. As photon travel near the event horizon of a black hole they can still escape being pulled in by gravity of a black hole (which is created when particularly massive star use up all its fuel and collapse inwardly to form super-dense object, much smaller than the original star. Only very large star end up as black hole. Smaller star don’t collapse that far; it often end up as neutron star instead) by traveling at a vertical direction known as exit cone. A photon on the boundary of this cone will not completely escape the gravity of the black hole. Instead it orbits the black hole. For a photon of mass m orbiting the black hole, the necessary centripetal force mv2/r is provided by the force of gravitation between the black hole and the photon GMm/r2. Therefore: mv2/r = GMm/r2 where: m = mass of the photon orbiting the black hole of mass M in a circular orbit of radius r and G is the gravitational constant. Since photon always travels with a speed equal to c. Therefore: v = c
mc2/r = GMm/r2 or r = GM/c2 Since RG = 2GM/c2 (where RG = radius of the black hole). Therefore: r = RG /2 WHICH MEANS: r < RG i.e., photon orbit exist inside the black hole. The photon orbit of radius r always exists in the space surrounding an extremely compact object such as a black hole. Hence r should be > RG. Therefore, it is clear that the condition mv2/r = GMm/r2 not always holds well. However, the image we often see of photons as a tiny bit of light circling a black hole in well-defined circular orbit of radius r = 3GM/c2 (where G = Newton’s universal constant of gravitation, c = speed of light in vacuum and M = mass of the black hole) is actually quite interesting. The angular velocity of the photon orbiting the black hole is given by: ω = c/r. For circular motion the angular velocity is the same as the angular frequency. Thus ω = c/r = 2πc/λ or λ =2πr Since Einstein’s E=mc2 relates mass to energy and Planck’s E = hυ energy to the frequency of light waves, therefore, by combining the two, photon mass should have a wave-like incarnation as well (exhibit interference phenomena - the telltale sign of waves). The De Broglie wavelength λ associated with the photon of mass m orbiting the black hole is given by Planck’s constant divided by the photon’s momentum): λ= h/mc. Therefore: r = ħ/mc, where ħ is the reduced Planck constant (since ħ is so small, the resulting photon wavelength is similarly minuscule compared with everyday scales - that is why the wavelike character of photon is directly apparent only upon careful microscopic investigation). The photon must satisfy the condition r = ħ/mc much like an electron moving in a circular orbit. Since this condition forces the photon to orbit the hole in a circular orbit. r = 3GM/c2 = ħ/mc or 3GM/c2 = ħ/mc or 3mM = (Planck mass) 2 Because of this condition the photons orbiting the small black hole carry more mass than those orbiting the big black hole. For a black hole of one Planck mass (M = Planck mass), m = 1/3 × Planck mass Since a black hole possess a nonzero temperature (no matter how small) the most basic and well-established physical principles would require it to emit radiation, much like a glowing poker. Therefore: the maximum energy an emitted radiation photon can possess is given by the equation: Lmax = 2.821 kBT (where kB = Boltzmann constant and T = black hole temperature = ħc3 / 8πGM). Lmax = 2.821 kBT or Lmax = 2.821 (ħc3 / 8πGM) which on rearranging: GM / c2 = 2.821 (ħc / 8πLmax) Since 3GM/c2 = ħ/mc. Therefore: ħ/ 3mc= 2.821 (ħc / 8πLmax) or mc2 = 2.968Lmax which means: mc2 > Lmax If a photon with energy mc2 orbiting the black hole can’t slip out of its influence, and so how can a Hawking radiation photon with maximum energy Lmax < mc2 is emitted from the event horizon of the Schwarzschild black hole (the edge of a black hole; the boundary of the region from which it is not possible to escape to infinity)? FGravity = force of gravitation experienced by the radiation photon at the surface of the black hole and FPhoton = force which moves the radiation photon. FGravity = GMm/ RG2 and FPhoton = mc2 / λ (where G = Newton’s universal constant of gravitation, c = speed of light in vacuum and M = mass of the black hole, m and λ = mass and wavelength of the radiation photon, RG = 2GM/c2 (the radius of the black hole). FGravity / FPhoton = c2 λ/4GM In MOST PHYSICS literature the energy of an emitted radiation photon is given by the equation: L = kBT (where kB = Boltzmann constant and T = black hole temperature). L = kBT = (ħc3 / 8πGM) By Planck’s energy-frequency relationship: L = hc/λ Hence:
hc/λ = (ħc3 / 8πGM) which on rearranging: λ= 16π2GM/c2 Solving for λ in the equation (FGravity / FPhoton = c2 λ/4GM) we get: FGravity / FPhoton = 16π2/ 4 = 39.43 FGravity = 39.43 FPhoton Which means: FGravity > FPhoton If the photon wants to detach from the surface of the black hole − (which is called its horizon, because someone outside the horizon can’t see what happens inside. That’s because seeing involves light, and no light can get out of a black hole) − it should obey the condition: FGravity = FPhoton GMm/RG 2 = mc2/λ (where RG = radius of the black hole = 2GM/c2) i.e., λ = 2 RG (wavelength of the photon should be twice the radius of the black hole) or FPhoton > FGravity. Because FGravity is > FPhoton, it is hard to claim the emission of radiation photon from the Schwarzschild black hole. However, Hawking radiation (a quantum phenomenon that leads to the eventual evaporation of an isolated black hole) has not been observed after over two decades of searching. Despite its strong theoretical foundation (i.e., it is widely regarded as one of the first real steps toward a quantum theory of gravity and allows physicists to define the entropy of a black hole), the existence of this effect is still in question and we have indirect observational evidence for this effect, and that evidence comes from the early universe. And looking at the unusual nature of Hawking radiation; it may be natural to question if such radiation exists in nature or to suggest that it is merely a theoretical solution to the hidden world of quantum gravity. However, if Schwarzschild black hole (which is indeed black body, absorbing everything that falls on them) does not emit any radiation, then it will continue to grow by absorbing surrounding matter and radiation. This would mean that the black hole would gain energy (and therefore mass by E=Mc2). Because Mc2 = − 3.33U, the gravitational binding energy becomes more negative with the increase in energy Mc2 of the black hole to shrink the black hole in size. And if we regard the nature of gravitational force so developed is similar to inter-molecular force. The gravitational force is attractive up to some extent [i.e., it is attractive until the distance between the constituents of the black hole is greater than or equal to the optimum distance (x Aº)] and when distance between the constituents of the black hole becomes < than x Aº it turns to a strong repulsive force. As the gravitational binding energy of the black hole become more negative, the distance between the constituents of the black hole decreases. As long as the distance between the constituents of the black hole is optimum, there is no considerable repulsion between the constituents. When the distance between the constituents of the black hole is further decreased i.e., the distance between the constituents of the black hole becomes < than x Aº and then at this stage, the singularity of the black hole may explode with unimaginable force, propelling the compressed matter into space. This matter then may condense into the stars, planets, and satellites that make up solar systems like our own. But perhaps not very scientific since no observational evidence available but still a nice mind exercise. However, if this is confirmed by observation, it will be the successful conclusion of a search going back more than 3,000 years. We will have found the grand design that we hope we will feel cheated that we hadn’t known about them until now − which no longer leaves omnipotent God (who play a central role in the operations of the universe and in the lives of humans) pretty much on the bench of philosophers and theologians for a long, long time − no need to offer an explanation for questions like: "What was God doing before the divine creation? Was he preparing hell for people who asked such questions?" Would the tidal forces kill an astronaut? Since gravity weakens with distance, the earth pulls on your head with less force than it pulls on your feet, which are a meter or two closer to the earth’s center. The difference is so tiny we cannot feel it, but an astronaut near the surface of a black hole would be literally torn apart. Quantum field theory = {Group theory + quantum mechanics} Lim N → ∞ Quantum mechanics = Quantum field theory The entropy of the black hole is given by the equation: SBH = c3 kB A / 4ħG, where c = speed of light in vacuum, kB = Boltzmann constant, ħ = Planck’s constant, G = gravitational constant and A = area of the event horizon. Since A = 4πRg2 = 4π (2GM/c2) 2. Therefore: SBH = 4πkB GM2 / ħc Differentiating the above equation we get: dSBH = (8πkB GM / ħc) dM dSBH = (8πkB GM / ħc3) dMc2 But T = ħc3/8πkB GM. Therefore: T × dSBH = dMc2 The rate of increase of black hole energy due to the absorption of energy from the surroundings is given by the equation: R1 = dMc2 /dt = T × (dSBH /dt) Suppose black hole absorbs no energy from the surroundings, then R1 = 0 (dSBH /dt) which is the rate of increase of black hole entropy = 0 T = {R1 / (dSBH /dt)} = 0/0 i.e., in order to maintain a well-defined temperature black hole must absorb energy from the surroundings. As we know that: mass energy of the black hole is = the twice its entropic energy Mc2 = 2 T × SBH Differentiating the above equation we get: dMc2 = 2 (T × dSBH + dT × dSBH) Since T × dSBH = dMc2. Therefore: dMc2 = 2 (dT × SBH) + 2dMc2 − dMc2 = 2 (SBH × dT) The rate of decrease of black hole energy due to the emission of energy in the form of Hawking radiation is given by the equation: R2 = − dMc2 /dt = 2SBH × (dT /dt) Suppose black hole emits no radiation, then R2 = 0 (dT /dt) which is the rate of increase of black hole temperature = 0 SBH = {R2 / 2 (dT /dt)} = 0/0 i.e., in order to maintain a well-defined entropy black hole must emit energy in the form of Hawking radiation. Taking natural logarithm of the equation SBH = 4πkB GM2 / ħc we get: lnSBH = ln (4πkB G / ħc) + 2lnM Differentiating the above equation we get: dlnSBH = 2dlnM Since M is proportional to 1/ T. Therefore: dlnSBH = − 2dlnT dSBH/SBH = −2 (dT/T) On rearranging we get: T × dSBH = −2 (dT × SBH) T × (dSBH /dt) = −2 SBH × (dT /dt) which can also be rewritten as: R1= − R2 From above equation it clear that R1 is = R2. But, because of the negative sign the actual value of R1 is = 1/ R2. Are Neutrinos Massless? If not they could contribute significantly to the mass of the universe? Evidence of neutrino oscillations prove that neutrinos are not massless but instead have a mass less than one-hundred-thousandth that of an electron. Dear “Dr.Science,” I hear that scientists have now made antiprotons and antielectrons… My question is: if you mixed antiprotons with antielectrons, could you make anti-oxygen? If so, could it be used to put out combustion, rather than supporting it? Yours,
Curious Harris. Hmmmmmmmmmmmm?? Hummmmmm… ? Hummmmmm mmmmmmmmmmmmmmmmmmmmm Dear Curious Harris. Unfortunately, Dr. Science is currently unable to provide a response to your recent query…. I think your question might have blocked his brain. “Our quest for knowledge would have been much simpler if all the mathematical indeterminates like 0/0, 1/0, etc. would have been well-defined.”
m0 = 2KE/v2 = (2 × 0) /0 = 0/0 There can be no bigger limitation than this. Rest mass cannot be undefined because rest mass is a physical property of the body. Did you know that simulation of the map of the cosmic microwave background that is being obtained by NASA’s Microwave Anisotropy Probe (MAP) shows that the CMB is not perfectly smooth. But has Ripples in it. If we measure the change in temperature on the Kelvin scale, then the change in kinetic energy is given by a simple equation: ∆KE = 3/2 × kB ∆T, where kB is called Boltzmann’s constant (which is = 1.380 × 10 to the power of – 23 Joules per Kelvin) Suppose ∆T → 0, then ∆KE = 0 Under this condition the Boltzmann’s constant ‘kB’ becomes UNDEFINED i.e., kB = (2 × 0) / (3 × 0) = 0/0 There can be no bigger limitation than this. Boltzmann’s constant cannot be undefined because kB = 1.380 × 10 – 23 J/ K. Gαβ = (8πG/c4) Tαβ Gαβ → curvature of space Tαβ → distribution of mass/ energy (8πG/c4) → constant But WHY? Maybe because matter and energy warp time and cause the time dimension to mix with the space? If we subtract the total energy mc2 by the energy at rest m0c2 we get the kinetic energy: KE = mc2 – m0c2 Because m = m0 / (1– v2/c2) ½ Therefore: KE = m0c2 [(1– v2/c2) –1/2 – 1] By Taylor series [(1– v2/c2) –1/2 – 1] = (v2/2c2 + 3v4/8c4 + 5v6/16c6 +……) Hence:
KE = m0c2 (v2/2c2 + 3v4/8c4 + 5v6/16c6 +……) For very low speed (i.e., v << c) all the terms except the first one are very small and can be ignored: KE = m0c2 (v2/2c2) = m0v2 /2 KE = m0c2 [(1– v2/c2) –1/2 – 1] For very low speed (i.e., v << c), (1– v2/c2) –1/2 ≈ 1 KE = m0c2 [(1 – 1] = 0 (which is an illogical and invalid result because at v << c KE is = m0v2 /2 not zero). The quantity of electric charge flowing through the filament of an incandescent bulb is given by: q = current × time or q = I × t If N is the number of electrons passing through the filament in the same time then q = Ne
or I × t = Ne or e = {I / (N/t)} where: e is the electron charge = – 1.602 × 10 –19 Coulombs and (N / t) = rate of flow of electrons. Suppose no electrons flow through the filament of an incandescent bulb, then I = 0 and (N/t) = 0 Under this condition the electron charge becomes UNDEFINED i.e., e = 0/0
“Actually, everything that can be known has a Number; for it is impossible to grasp anything with the mind or to recognize it without this.”− PHILOLAUS, C. 470 – C. 385 BC. The change in energy ∆E is related to the change in mass ∆m by the Einstein famous equation (which has entered into one’s mental frameworks due to its large impact thus gaining the status of more than a mere equation): ∆E = ∆mc2 Suppose ∆m = 0, then ∆E = 0 Under this condition the speed of light squared i.e., c2 becomes UNDEFINED i.e., c2 = 0/0 There can be no bigger limitation than this. c2 cannot be undefined because c2 = 9 × 10 16 m2/ s2. The change in energy ∆E is related to the change in frequency (i.e., number of oscillations per second) ∆υ by the Planck’s energy frequency relationship (which is a wonderful formula, because it tells us what change in frequency really means: it’s just change in energy in a new guise): ∆E = h∆υ
Suppose ∆υ = 0, then ∆E = 0
Under this condition the Planck’s constant becomes UNDEFINED i.e., h = 0/0. There can be no bigger limitation than this. h cannot be undefined because h is = 6.625 × 10 to the power of –34 Js. “So far as we know. All the fundamental laws of physics, like Newton’s Equations, are reversible. Then where does irreversibility come from? It comes from order going to disorder. But we do not understand this until we know the origin of order.” --Richard Feynman
P = (e2 / 6πε0c3) a2 where e is the charge on the electron and a is the acceleration of the electron, ε0 is the absolute permittivity of free space; c is the speed of light in vacuum. If a = 0, then P = 0. Under this condition (e2 / 6πε0c3) becomes UNDEFINED i.e., (e2 / 6πε0c3) = 0/0 Did you know that: By analyzing the stellar spectrum, one can determine both the temperature of a star and the composition of its atmosphere. Electric and magnetic forces are far stronger than gravity, but remain unnoticeable because every macroscopic body contain almost equal numbers of positive and negative electrical charges (i.e., the electric and magnetic forces nearly cancel each other out). The gigantic instrument constructed by Raymond Davis Jr. and Masatoshi Koshiba to detect neutrinos from the Sun confirmed the prediction that the Sun is powered by nuclear fusion. The Unruh temperature, derived by William Unruh in 1976, is the effective temperature experienced by a uniformly accelerating observer in a vacuum field. It is given by: TUnruh = (ħa/2πckB), where a is the acceleration of the observer, kB is the Boltzmann constant, ħ is the reduced Planck constant, and c is the speed of light in vacuum. Suppose the acceleration of the observer is zero (a = 0), then TUnruh = 0 Under this condition (ħ/2πckB) becomes UNDEFINED i.e., (ħ/2πckB) = 0/0. The change in entropy of the photon gas ∆S is related to the change in number of photons ∆N by the equation: ∆S = 3.6 kB ∆N. Suppose there is no change in number of photons (i.e., ∆N = 0), then ∆S = 0
Under this condition the Boltzmann’s constant ‘kB’ (which is = 1.380 × 10 –23 J/K) becomes UNDEFINED i.e., kB = 0 / (3.6 × 0) = 0/0. The energy required to lift a body of weight ‘w’ up to a height of h meter is mgh i.e., E = w × h. If h = 0, then the energy required to lift a body of weight w will be zero (i.e., E = 0). Under this condition the weight of the body ‘w’ becomes UNDEFINED i.e., w = 0/0. There can be no bigger limitation than this. ‘w’ cannot be undefined because weight is a physical property of the body. “I believe in God. It makes no sense to me to assume that the universe and our existence is just a cosmic accident, that life emerged due to random physical processes in an environment which simply happened to have the right properties.” : Antony Hewish (1974 Nobel Prize in Physics for his discovery of pulsars) W = F × S × cosφ, where W = work, F = force, S = displacement and φ is angle between force and displacement. For an electron moving in a circular orbit, F = mv2/r and S = rθ W = mv2 × θ × cosφ For one complete revolution θ = 2π
W = 2π mv2cosφ For an electron moving in a circular orbit, force and displacement are perpendicular to each other (i.e., φ = 90o). Now under the condition (φ = 90o): W = 0 m = W / 2πv2cosφ = 0 / (2πv2 × 0) m = 0/0 i.e., mass becomes UNDEFINED. Yüklə 283,52 Kb. Dostları ilə paylaş:
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