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A Warning To All Oxygen Breathing Humans
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- Bu səhifədəki naviqasiya:
- S.N. Bose’s letter to Einstein
- Decoding The Universe Since 1905 Atom → nucleus → proton → quark So, particle physics finished…… Or is it not
A Warning To All Oxygen Breathing Humans “IF YOU MEET SOMEONE FROM ANOTHER PLANET AND HE HOLDS OUT HIS LEFT HAND, DON’T SHAKE IT. HE MIGHT BE MADE OF ANTIMATTER. YOU WOULD BOTH DISAPPEAR IN A TREMENDOUS FLASH OF LIGHT.” --STEPHEN HAWKING In 1923 French physicist Louis de Broglie suggested that the wave-particle duality applied not only to light but to matter as well (mid-1920s proof came from the work of Clinton Davisson and Lester Germer: electrons [were found to] exhibit interference phenomena – the telltale sign of waves). Since Einstein’s E = mc2 relates mass to energy, that [since] Planck and Einstein related energy to the frequency of waves i.e., E = hυ, [that] therefore, by combining the two, hυ = mc2 (this relation is applicable only for relativistic particle and for non-relativistic particle mv2/2 = hυ) A small change in the frequency of the wave (∆υ) is followed by a small change in the mass (∆m) i.e., hdυ = dmc2 If dυ = 0, then dm = 0 h /c2 = dm/dυ = 0/0 i.e., h /c2 becomes UNDEFINED. “Science is a game — but a game with reality, a game with sharpened knives … If a man cuts a picture carefully into 1000 pieces, you solve the puzzle when you reassemble the pieces into a picture; in the success or failure, both your intelligences compete. In the presentation of a scientific problem, the other player is the good Lord. He has not only set the problem but also has devised the rules of the game — but they are not completely known, half of them are left for you to discover or to deduce. The experiment is the tempered blade which you wield with success against the spirits of darkness — or which defeats you shamefully. The uncertainty is how many of the rules God himself has permanently ordained, and how many apparently are caused by your own mental inertia, while the solution generally becomes possible only through freedom from its limitations.”
If not for a force called gravity, we would all go zinging off into outer space. The change in number of moles dn is related to the change in number of molecules dN by the Avogadro constant L: dn = dN/L If dN = 0, then dn = 0 Under this condition the Avogadro’s constant (the number of particles in a mole, 6.022 × 10 23) becomes UNDEFINED i.e., L = 0/0. There can be no bigger limitation than this (because Avogadro’s constant is = 6.022 × 10 23 particles). The density of solute ρ is related to its concentration C by the equation: ρ = M × C, where M is a constant for a given solute and it is termed the molecular mass. Now under the condition (C = 0): ρ = 0
M = ρ /C = 0/0 i.e., the molecular mass of the solute becomes undefined. There can be no bigger limitation than this. M cannot be undefined because molecular mass is a physical property of the solute. Atom: Why can’t you possibly measure where I am and how fast I’m moving at the same time? Physicist: ∆x ∆p ≥ ħ prevents me from doing so. “Scientific views end in awe and mystery, lost at the edge in uncertainty, but they appear to be so deep and so impressive that the theory that it is all arranged as a stage for God to watch man’s struggle for good and evil seems inadequate.” --Richard Feynman
Respected Master, I have ventured to send you the accompanying article for your perusal and opinion. I am anxious to know what you think of it… I do not know sufficient German to translate the paper. If you think the paper worth publication I shall be grateful if you arrange for its publication in Zeitschrift fur Physik. Though a complete stranger to you, I do not feel any hesitation in making such a request. Because we are all your pupils though profiting only by your teachings through your writings. I do not know whether you still remember that somebody from Calcutta asked your permission to translate your papers on Relativity in English. You acceded to the request. The book has since been published. I was the one who translated your paper on Generalised Relativity. Yours faithfully S. N. Bose According to Faraday’s law (introduced by British physicist and chemist Michael Faraday), the amount of a substance deposited on an electrode in an electrolytic cell is directly proportional to the quantity of electricity that passes through the cell. Faraday’s law can be summarized by: n = q / ZF, where n is the number of moles of the substance deposited on an electrode in an electrolytic cell, q is the quantity of electricity that passes through the cell, F = 96485 C/ mol is the Faraday constant and z is the valency number of ions of the substance. Suppose no electricity passes through the cell (q = 0), the amount of the substance deposited on an electrode in an electrolytic cell is 0 (i.e., n= 0). Under this condition q = 0, n = 0 F = q / (z × n) = 0 / (z × 0) = 0/0 i.e., Faradays constant (which is = 96485 Coulombs per mole) becomes Undefined. Did you know that the static on your television is caused by radiation left over from the Big Bang? If a quantity of heat Q is added to a system of mass m, then the added heat will go to raise the temperature of the system by ΔT = Q/Cm where C is a constant called the specific heat capacity (A system’s heat capacity per kilogram – which is the measure of how much heat a system can hold). ΔT = Q/Cm which on rearranging: m = Q / (C × ΔT). Suppose no heat is added to the system (Q = 0), then ΔT = 0 m = 0/ (C × 0) = 0/0 i.e., the mass of a system becomes UNDEFINED. The faster you move, The shorter and the heavier you are. And that is the THEORY OF RELATIVITY. “In a scientific sense, earthquakes are unpredictable. But that does not mean that you can’t predict things about them.” —PETER SAMMONDS “To suppose that the eye… could have been formed by natural selection, seems, I freely confess, absurd in the highest possible degree.” − Charles Darwin Entropy (a thermodynamic quantity -- first introduced by the German physicist Rudolf Clausius (1822--1888) -- a measure of untidiness in a system and a measure of how much information a system contains) is defined as S = kB ln {number of states} which, for N particles of the same type, will be S = kB ln {(no of one-particle states) N} S = kBN ln {a not-too-big number} S = kBN This means: the more particles, the more disorder. If no particles (i.e., N = 0), then no disorder (i.e., S = 0). Now under this condition: kB = S / N = 0/0 i.e., Boltzmann’s constant ‘kB’ (which is = 1.380 × 10 –23 J/K) becomes UNDEFINED. Note: The universe is expanding because the energy of expansion which is (which is proportional to MH2R2 is greater than the gravitational binding energy of the universe (which is proportional to ‒ GM2/R). M = mass and R = radius of the universe. H = Hubble constant and G = Gravitational constant. If the energy of expansion is less than the gravitational binding energy of the universe, the universe will stop expanding and collapse and if the energy of expansion is equal to the gravitational binding energy of the universe, the universe will neither expand nor contracts. Λ The Cosmological Constant was My GREATEST Mistake? : Albert Einstein Cosmic Gall by John Updike Neutrinos, they are very small. They have no charge and have no mass and do not interact at all. The earth is just a silly ball to them, through which they simply pass, like dust maids down a drafty hall or photons through a sheet of glass. They snub the most exquisite gas, ignore the most substantial wall, cold-shoulder steel and sounding brass, insult the stallion in his stall, and, scorning barriers of class, infiltrate you and me! Like tall and painless guillotines, they fall down through our heads into the grass. At night, they enter at Nepal and pierce the lover and his lass from underneath the bed—you call it wonderful; I call it crass. (a2 – b2) = (a+ b) (a−b) Which on rearranging: (a2 – b2) / (a – b) = (a+ b) If a= b=1, then 0/0 = 2 (illogical and meaningless result). tanθ = sinθ / cosθ which on rearranging: cosθ = sinθ / tanθ If θ = 0o, then 1= 0/0 (illogical and meaningless result). Absorbance = − log (Transmittance) Absorbance = − 2.303 × ln (Transmittance) If Transmittance = 1 (i.e., no light passed through the solution is absorbed), then Absorbance = 0. Now under this condition: Absorbance / ln (Transmittance) = − 2.303 take the form 0/ln1 = − 2.303 0/0 = − 2.303 (illogical and meaningless result). For a particle in its rest frame, the momentum is zero, so the energy-momentum relation E2 = p2c2 + m02c4 simplifies to: E0 = m0c2 where m0 is the rest mass of the particle. For a massless particle like photon m0 = 0 Hence: E0 = 0 c2= 0/0 i.e., c2 → undefined. c2 cannot be undefined. Since c2 cannot be undefined Does it mean that photon possess rest mass? Because m0 = m (1− v2/c2) 1/2 If v = c, m0 = 0 i.e., only zero rest mass particles can travel at the speed of light. Hence, photon possesses zero rest mass… We can ask what happens when an electron jumps from one energy level to another. If the electron jumps down in energy, then it sheds the excess energy by emitting a photon. The photon’s energy is the difference between the electron’s energy before it jumped and after i.e., E photon = hυ = E2 – E1 But E1 = electron’s energy before it jumped = − (2π2me e4 / n12h2) and E2 = electron’s energy after it jumped = − (2π2me e4 / n22h2) Therefore: hυ = (2π2me e4 / h2) [1/n12 – 1/n22] Suppose hυ = 0, then 0 = (2π2me e4 / h2) [1/n12 – 1/n22] From this it follows that n1= n2 Now under the condition (hυ = 0, n1= n2): (2π2me e4 / h2) = hυ / [1/n12 – 1/n22] = 0/0 i.e., (2π2me e4 / h2) becomes UNDEFINED. What is our physical place in the universe? Present 13.8 billion years after the Big Bang We can only see the surface of the sky where light was scattered. “Science itself, no matter whether it is the search for truth or merely the need to gain control over the external world, to alleviate suffering, or to prolong life, is ultimately a matter of feeling, or rather, of despite—the desire to know or the desire to realize.” --Louis Victor de Broglie Is the density of the Black Hole: 0.1253c6/ πG3M2 or 0.00585c6/ πG3M2? The density of the black hole is given by the expression: ρ = 3M/ 4πRG3, where M is the mass and RG is the radius of the black hole. Since RG = 2GM/c2. Therefore: ρ = 3c6/ 24πG3M2 or ρ = 0.1253c6/ πG3M2 According to Stefan – Boltzmann-Schwarzschild – Hawking black hole radiation power law, the rate of change in a black hole’s energy is: P = є × σ × T4 × (4π RG2) or P = 1 × (π2 kB4 /60ħ3c2) × (ħc3/8πGM) 4 × (16πG2M2/c4) or P = ħc6/ 15360πG2M2 Mario Rabinowitz discovered the simplest possible representation for the rate of change in a black hole’s energy in terms of black hole density ρ: P = Gρħ/90 or P = ħc6/ 15360πG2M2 = Gρħ/90 or ρ = 90c6/ 15360πG3M2 or ρ = 0.00585c6/ πG3M2 Conclusion: Two results for the density of the black hole: ρ = 0.1253c6/ πG3M2 ρ = 0.00585c6/ πG3M2 Is the Life time of our power house the sun: 2.63 × 10 18 or 3.98 × 1020 seconds?
Let’s calculate age of the sun according to nuclear considerations. Inside the sun, we have NProtons (say), which can be calculated as follows NProtons = M / mProton = 2 × 1030 / 1.672 × 10 −27 = 1.196 × 10 57, where M = mass of the sun and mProton = mass of the proton. Hence, the number of fusion reactions inside the sun is N Reactions = 1.196 × 10 57 / 4 = 2.99 × 10 56 So, star has the capacity of releasing 0.196 × 10 56 × 40 × 10 − 13 = 1.19 × 10 45 J The rate of loss of energy of the sun in the form of radiation i.e., power radiated by the sun, P = 4.52 × 10 26 J/s, the sun has the capacity to shine for t = 1.19 × 10 45 /4.52 × 10 26 = 2.63 × 10 18 seconds.
NProtons = M / mProton or M = NProtons × mProton Differentiating this with respect to time, we get (dM/dt) = mProton × (dNProtons /dt) This can also be written as: − (dMc2/dt) = mProtonc2 × − (dNProtons /dt) Since − (dMc2/dt) = P = 4.52 × 10 26 J/s and mProtonc2 = 15.04 × 10 − 11 J. Therefore: − (dN Protons /dt) = (4.52 × 10 26 / 15.04 × 10 − 11) or − (dN Protons /dt) = 3.005 × 10 36 protons per second 0.196 × 10 36 protons are utilized per second to release energy in the form of radiation. 0.196 × 10 36 protons → one second 1.196 × 10 57 protons → t seconds t = 1.196 × 10 57/3.005 × 10 36 = 3.98 × 1020 seconds. 1.196 × 10 57 protons are utilized per 3.980× 1020 seconds to release energy in the form of radiation. Therefore, the sun has the capacity to shine for 3.98 × 1020 seconds. Conclusion: Two results for the LIFE TIME of the sun: t = 2.63 × 10 18 seconds t = 3.98 × 1020 seconds Did you know that: Niels Bohr imagined the atom as consisting of electron waves of wavelength λ = h/mv endlessly circling atomic nuclei. In his picture, only orbits with circumferences corresponding to an integral multiple of electron wavelengths could survive without destructive interference (i.e., r = nħ/mv could survive without destructive interference). As mercury repeatedly orbits the sun, the long axis of its elliptical path slowly rotates, coming full circle roughly every 360,000 years. Because the square of the time it takes for the planet to complete one revolution around the sun is proportional to the cube of its average distance from the sun, the mercury move rapidly in its orbit and Venus, Earth and Mars move progressively less rapidly about the sun and the outer planets such as Jupiter, Saturn, Uranus, Neptune and Pluto move stately and slow. Newton rings is a phenomenon in which an interference pattern is created by the reflection of light between two surfaces — a spherical surface and an adjacent flat surface. It is named after Isaac Newton, who first studied them in 1717. Quantum mechanics says that the position of a particle is uncertain, and therefore that there is some possibility that a particle will be within an energy barrier rather than outside of it. The process of moving from outside to inside without traversing the distance between is known as quantum tunneling, and it is very important for the fusion reactions in stars like the Sun. The three kinematic equations that describe an object's motion are: d = ut + ½ at2 v2 = u2 + 2ad v = u + at There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for final velocity of the object, u stands for the initial velocity of the object. Assuming the initial velocity of the object is zero (u = 0): d = ½ at 2 v2 =2ad v = at
Since velocity is equal to displacement divided by time (i.e., v =d / t): a = 2d /t2 a =d / 2t2 a = d / t2 Conclusion: 3 different results for a.
The rest masses of proton and neutron are regarded as fundamental physical constants in existing physics and it is believed that they are invariant. Rest mass of proton plus neutron = 1.007825 + 1.008665 = 2.01649 u. But inside the deuteron nucleus, it is experimentally confirmed that rest mass of proton plus neutron = 2.01410 u i.e., rest mass of proton plus neutron inside the nucleus has decreased from 2.01649 u to 2.01410 u. The rest masses of neutrons and protons are fundamental constants only if they remain same universally (inside and outside the nucleus). Failure to meet universal equality proves that the rest masses of neutrons and protons are Variant. Decoding the quantum mechanics to find the solution to the Schrödinger equation for the hydrogen atom in arbitrary electric and magnetic fields-- If we can, we can know everything about the system? As per Albert Einstein’s special theory of relativity: m = m0 / [1 – v2 /c2] 1/2 L0 = L / [1 – v2 /c2] 1/2 ∆t = ∆t0 / [1 – v2 /c2] 1/2 If v = c, m → ∞
L0 → ∞ ∆t → ∞
Which means: if v=c then Albert Einstein’s special theory of relativity breaks down. Violation of the foundation of the fundamental theory of the twentieth century. Nevertheless, it is now completely accepted by the scientific community, and its predictions have been verified in countless applications. If a PART mc2 of the photon energy is absorbed by the electron at rest, then the absorbed energy mc2 manifests as the Kinetic energy KE of the electron and the momentum mc of the absorbed photon manifests as the momentum p of the electron. Therefore, the equation KE = ∆p × v where ∆p = p2 – p1, p2 = final momentum of the electron = p and p1 = initial momentum of the electron = 0 (since the electron was initially at rest) Becomes:
mc2 = mc × v From this it follows that v = c The idea which states that nothing with mass can travel at the speed of light is a cornerstone of Albert Einstein’s special theory of relativity, which claims that observers in relative motion will have different perceptions of distance and of time (and gives explanations for the behavior of objects near the speed of light, such as time dilation and length contraction) which itself forms the fundamental precept of modern physics. If the electron recoils with a velocity v=c, then the basic laws of physics have to be rewritten. Note: 6 × 0 = 0 2 × 0 = 0 0 = 0
6 × 0 = 2 × 0 6 / 2 = 0/0 i.e., 6 / 2 → UNDEFINED. There can be no bigger limitation than this because 6/2 is 3 not 0/0. For a source moving at angle θ = 0o towards the stationary observer, the relativistic Doppler effect equation is given by: υ observed = υ emitted × {(1 + v/c) / (1 − v/c)}½ From this it follows that (υ observed /υemitted )−1 = {(1 + v/c) / (1 − v/c)} ½ − 1 (υ observed − υ emitted) / υ emitted = {(1 + v/c) / (1 − v/c)} ½ − 1 Since redshift z = (υ emitted − υ observed) / υ emitted. Therefore: −z = {(1 + v/c) / (1 − v/c)} ½ − 1 (1− z) = {(1 + v/c) / (1 − v/c)} ½ On squaring we get: (1− z) 2 = (1 + v/c) / (1 − v/c) (1− z) 2 (1 − v/c) = (1 + v/c) (1− z) 2 − v/c (1− z) 2 = 1 + v/c On rearranging: (1− z) 2 – 1 = v/c {(1− z) 2 + 1} If v = c (some quasars or other heavenly bodies may attain the velocity v = c due to the Hubble expansion of space), then (1− z) 2 – 1 = (1− z) 2 + 1 i.e., LHS ≠ RHS, which is never justified. “Get your facts first, and then you can distort them as you please.” − MARK TWAIN
“For the first half of geological time our ancestors were bacteria. Most creatures still are bacteria, and each one of our trillions of cells is a colony of bacteria.” —RICHARD DAWKINS For a source moving at angle θ = 0o away from the stationary observer, the relativistic Doppler Effect equation is given by: υ observed = υ emitted × {(1 − v/c) / (1 + v/c)} ½ Since the force which moves the photon is given by: F = hυ2/ c, where h is the Planck’s constant, υ is the frequency of the photon. Therefore: F observed = F emitted × {(1 − v/c) 2 / (1 − v2/c2)} If v = c (some quasars or other heavenly bodies may attain the velocity v = c), then F observed = 0/0. The equation F observed = F source × {(1 − v/c) 2 / (1 − v2/c2)} can also be written as: F observed = F emitted × {(1 − v/c) / (1 + v/c)} If v = c, then F observed = 0. CONCLUSION: The same equation (in unsolved and solved forms) under similar conditions (v → c) gives different results i.e. (F observed →0/0 and F observed → 0), which is never justified. One of the key signature of quantum gravity would be the observation of a small quantum black hole?
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