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Natural Science Forum / Physics / General Physics / February 2008LEE SMOLIN AGAINST ISAAC NEWTON
http://philipball.blogspot.com/2007/09/arthur-eddington-was-innocent-this-is.html "With the technology then available, measuring the bending of starlight was very challenging. And contrary to popular belief, Newtonian physics did not predict that light would remain undeflected - Einstein himself pointed out in 1911 that Newtonian gravity should cause some deviation too. So the matter was not that of an all-or- nothing shift in stars' positions, but hinged on the exact numbers." However Budding Young Einsteins are taught in a different way: http://pirsa.org/speaker/Lee_Smolin Lee Smolin - ISSYP Keynote Session Speaker(s): Lee Smolin Abstract: Date: 01/08/2007 - 1:00 pm http://streamer.perimeterinstitute.ca/mediasite/viewer/?peid=5f32739a-624d-4ec8- 9ecc-4d44d3d16fe9 Lee Smolin: "Newton's theory predicts that light goes in straight lines and therefore if the star passes behind the sun, we can't see it. Einstein's theory predicts that light is bent...." Pentcho Valev pvalev@yahoo.com > http://philipball.blogspot.com/2007/09/arthur-eddington-was-innocent-... > "With the technology then available, measuring the bending of [quoted text clipped - 3 lines] > cause some deviation too. So the matter was not that of an all-or- > nothing shift in stars' positions, but hinged on the exact numbers." Right. Classical Newtonian mechanics predicts that light will bend with twice the angle observed. Relativistic Newtonian mechanics however predicts the correct angle since it is now understood that only half a photon's energy is sensitive to transverse interaction. > However Budding Young Einsteins are taught in a different way: > [quoted text clipped - 11 lines] > Pentcho Valev > pva...@yahoo.com Agreed. Outright falsehood. Probably plain ignorance on the part of this author. André Michaud On Feb 6, 5:33 pm, srp2...@gmail.com wrote: > >http://philipball.blogspot.com/2007/09/arthur-eddington-was-innocent-... > > "With the technology then available, measuring the bending of [quoted text clipped - 12 lines] > angle since it is now understood that only half a photon's energy > is sensitive to transverse interaction. If you wish your analysis to be complete, you should discuss Einstein's 1911 equation c'=c(1+V/c^2) showing how the speed of light varies with the gravitational potential V. Note that c'=c(1+V/c^2) is consistent with the gravitational redshift factor 1+V/c^2 confirmed experimentally: http://www.blazelabs.com/f-g-gcont.asp "The first confirmation of a long range variation in the speed of light travelling in space came in 1964. Irwin Shapiro, it seems, was the first to make use of a previously forgotten facet of general relativity theory -- that the speed of light is reduced when it passes through a gravitational field....Faced with this evidence, Einstein stated:"In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position."......Today we find that since the Special Theory of Relativity unfortunately became part of the so called mainstream science, it is considered a sacrilege to even suggest that the speed of light be anything other than a constant. This is somewhat surprising since even Einstein himself suggested in a paper "On the Influence of Gravitation on the Propagation of Light," Annalen der Physik, 35, 1911, that the speed of light might vary with the gravitational potential. Indeed, the variation of the speed of light in a vacuum or space is explicitly shown in Einstein's calculation for the angle at which light should bend upon the influence of gravity. One can find his calculation in his paper. The result is c'=c(1+V/c^2) where V is the gravitational potential relative to the point where the measurement is taken. 1+V/c^2 is also known as the GRAVITATIONAL REDSHIFT FACTOR." Note also that, by applying Einstein's equivalence principle, you can prove that c'=c(1+V/c^2) is equivalent to c'=c+v, an equation given by Newton's emission theory and showing how the speed of light varies with v, the relative speed of the light source and the observer, in the absence of a gravitational field. Pentcho Valev pvalev@yahoo.com > > However Budding Young Einsteins are taught in a different way: > [quoted text clipped - 16 lines] > > André Michaud > On Feb 6, 5:33 pm, srp2...@gmail.com wrote: > [quoted text clipped - 27 lines] > speed of light is reduced when it passes through a gravitational > field.... This simply is impossible. From electromagnetic theory and confirming observation, when em radiation moves deeper into a gravity field, its energy increases but its velocity doesn't change, and the opposite when radiation moves away from a gravity field, its energy diminishes but its velocity remains constant. Doppler radars would not work even here on earth if this was not the case, and they do work perfectly, based on the underlying principle that the speed of light is constant. > Faced with this evidence, Einstein stated:"In the second > place our result shows that, according to the general theory of [quoted text clipped - 23 lines] > with v, the relative speed of the light source and the observer, in > the absence of a gravitational field. My analysis is complete. The verified invariance of the speed of light has nothing to do with GR or even SR. It is an electromagnetic property of em radiation. Whatever Einstein or anyone else could have said to the contrary reveals only partial understanding of electromagnetism at the time of statement. André Michaud On Feb 6, 7:50 pm, srp2...@gmail.com wrote: > > On Feb 6, 5:33 pm, srp2...@gmail.com wrote: > [quoted text clipped - 75 lines] > reveals only partial understanding of electromagnetism at the time > of statement. No you cannot just avoid the discussion of the following five points: (1) Frequency shift in a gravitational field (confirmed experimentally by Pound and Rebka): f'=f(1+V/c^2) (2) Speed of light shift in a gravitational field (Einstein's 1911 equation): c'=c(1+V/c^2) (3) Textbook equation demonstrating consistency between (1) and (2): (frequency)=(speed of light)/(wavelength) (4) Newton's emission theory's equation (absence of a gravitational field): c'=c+v (5) Einstein's equivalence principle allowing one to PROVE equivalence between (2) and (4). If you do not agree with (2) and insist on the speed of light being constant in a gravitational field, then your should (re)interprete (1), (3), (4) and (5) accordingly. Pentcho Valev pvalev@yahoo.com > On Feb 6, 9:34 am, Pentcho Valev <pva...@yahoo.com> wrote: > [quoted text clipped - 91 lines] > simple and familiar. He introduced his second postulate, more or less > evident as one thinks in terms of waves in aether." Yes criminal Einsteinians are mysterious individuals. They are either extremely silly or extremely dishonest or both but one can never be sure: http://edge.org/3rd_culture/smolin03/smolin03_index.html Lee Smolin: "Now, here is the really interesting part: Some of the effects predicted by the theory appear to be in conflict with one of the principles of Einstein's special theory of relativity, the theory that says that the speed of light is a universal constant. It's the same for all photons, and it is independent of the motion of the sender or observer. How is this possible, if that theory is itself based on the principles of relativity? The principle of the constancy of the speed of light is part of special relativity, but we quantized Einstein's general theory of relativity. Because Einstein's special theory is only a kind of approximation to his general theory, we can implement the principles of the latter but find modifications to the former. And this is what seems to be happening! So Gambini, Pullin, and others calculated how light travels in a quantum geometry and found that the theory predicts that the speed of light has a small dependence on energy. Photons of higher energy travel slightly slower than low-energy photons....A very exciting question we are now wrestling with is, How drastically shall we be forced to modify Einstein's special theory of relativity if the predicted effect is observed? The most severe possibility is that the principle of relativity simply fails....But there is another possibility. This is that the principle of relativity is preserved, but Einstein's special theory of relativity requires modification so as to allow photons to have a speed that depends on energy. The most shocking thing I have learned in the last year is that this is a real possibility. A photon can have an energy-dependent speed without violating the principle of relativity!" Pentcho Valev pvalev@yahoo.com On 6 fév, 02:34, Pentcho Valev <pva...@yahoo.com> wrote: > http://philipball.blogspot.com/2007/09/arthur-eddington-was-innocent-... > "With the technology then available, measuring the bending of [quoted text clipped - 3 lines] > cause some deviation too. So the matter was not that of an all-or- > nothing shift in stars' positions, but hinged on the exact numbers." | Right. | Classical Newtonian mechanics predicts that light will bend with | twice the angle observed. How? > <srp2...@gmail.com> wrote in message > [quoted text clipped - 15 lines] > > How? Lapsus. Its the reverse. I meant to write half the angle observed. and for corrected relativistic Newtonian, the angle is twice that of classical Newtonian, as is observed. André Michaud On 6 fév, 11:48, "Androcles" <Headmas...@Hogwarts.physics> wrote: > <srp2...@gmail.com> wrote in message > [quoted text clipped - 15 lines] > > How? | Lapsus. Its the reverse. | I meant to write half the angle observed. and for corrected | relativistic | Newtonian, the angle is twice that of classical Newtonian, as is | observed. Ok, but my real question is how does classical Newtonian mechanics prophesy that light will bend at all? It does, I'm wondering if you, Eddington or Einstein knows how. > <srp2...@gmail.com> wrote in message > [quoted text clipped - 31 lines] > prophesy that light will bend at all? It does, I'm wondering if you, > Eddington or Einstein knows how. They certainly did since Einstein did the calculation himself. And I do too. Explained summarily, to proceed, for the requirements of calculation he converted the energy of a photon to its equivalent mass (m=E/c^2) and proceeded to calculate the deflection of theoretical "masses" for visible light photons as they grazed the Sun mass. This gave half the deflection angle that was later observed. His calculations are in a paper from 1911 titled "Über den Einfluß der Schwerkraft auf die Ausbreitung des Lichter" He later 1915 corrected the figures and obtained the right deflection (twice that of classical Newtonian). However, the same correct deflection angle can easily be obtained from upgraded relativistic Newtonian, something that has never been documented, since the 1919 Eddington et al. observation was specifically meant to prove the superiority of GR over classical mechanics. Just like the community never considered confirming atomic clock speeding up with altitude with parallel mechanical clocks experiments. They were not looking for the truth, but to prove a point. André Michaud On 6 fév, 12:15, "Androcles" <Headmas...@Hogwarts.physics> wrote: > <srp2...@gmail.com> wrote in message > [quoted text clipped - 30 lines] > prophesy that light will bend at all? It does, I'm wondering if you, > Eddington or Einstein knows how. | They certainly did since Einstein did the calculation himself. | And I do too. | Explained summarily, to proceed, for the requirements of | calculation he converted the energy of a photon to its | equivalent mass (m=E/c^2) and proceeded to calculate | the deflection of theoretical "masses" for visible light | photons as they grazed the Sun mass. | This gave half the deflection angle that was later observed. | His calculations are in a paper from 1911 titled | "Über den Einfluß der Schwerkraft auf die Ausbreitung | des Lichter" | He later 1915 corrected the figures and obtained the | right deflection (twice that of classical Newtonian). | However, the same correct deflection angle can easily | be obtained from upgraded relativistic Newtonian, | something that has never been documented, since the | 1919 Eddington et al. observation was specifically | meant to prove the superiority of GR over classical | mechanics. | Just like the community never considered confirming | atomic clock speeding up with altitude with parallel | mechanical clocks experiments. | They were not looking for the truth, but to prove | a point. Androcles' third law: For every photon there is an equal and opposite rephoton. http://www.androcles01.pwp.blueyonder.co.uk/rephoton.gif (from Newton's third law and Huygens's wave superposition, photons have direction and come in pairs) The equivalent mass of a photon is m = 2E/c^2, from which one might surmise the gravitational deflection, should one know the energy. However, http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov Riding the Earth, you, Eddington and Einstein (who didn't bother to look anyway, the only one that did is Eddington who obtained ONE suitable photograph) are observers on the roundabout. The light has to curve. From simple trigonometry, the deflection has to be ~v/c. A star in the direction of the Earth's travel will show no aberration, a star at right angles will show maximum aberration. The observed star close to the sun has to be identified in daylight and there is further curvature as a result of refraction in Earth's atmosphere. > On 6 fév, 12:15, "Androcles" <Headmas...@Hogwarts.physics> wrote: >> <srp2...@gmail.com> wrote in message > The equivalent mass of a photon is m = 2E/c^2 Yes. E=m*c^2 ----> m=E/c^2 multiplié par deux. Parce qu'un photon passe par les deux fentes..... dans l'exprérience de Young. et non par une seule fente. C'est bon? Fripounette. On 6 fév, 15:12, "Fripounette" <clementine.nectar...@mandarine.fr> wrote: > > <srp2...@gmail.com> wrote in message > >news:6ba03bd0-12cf-4b68-a7e7-0b84a3e036c6@i12g2000prf.googlegroups.com... > > On 6 fév, 12:15, "Androcles" <Headmas...@Hogwarts.physics> wrote: > >> <srp2...@gmail.com> wrote in message > > The equivalent mass of a photon is m = 2E/c^2 Possible erreur d'attribution. Cette ligne n'a pas été écrite par moi, mais par Androcles André Michaud > Yes. > E=m*c^2 ----> m=E/c^2 multiplié par deux. > Parce qu'un photon passe par les deux fentes..... dans l'exprérience > de Young. et non par une seule fente. C'est bon? > Fripounette. > <srp2...@gmail.com> wrote in message > [quoted text clipped - 73 lines] > (from Newton's third law and Huygens's wave superposition, > photons have direction and come in pairs) My view is de Broglie's on photons. I see them as complex standing harmonic oscillators in motion, with only half their energy oscillating electromagnetically. The other half is is oriented in the direction of motion and is thus impervious to transverse force interaction. > The equivalent mass of a photon is m = 2E/c^2, Yes. Obvious and perfectly explained by relativistic Newtonian. > from which one might surmise the gravitational deflection, > should one know the energy. Since it invariantly amounts to half the full quantum, we always know the energy that maintains the velocity. > However, > http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov > Riding the Earth, you, Eddington and Einstein (who didn't bother > to look anyway, the only one that did is Eddington who obtained ONE > suitable photograph) are observers on the roundabout. The light has > to curve. From simple trigonometry, the deflection has to be ~v/c. This does not take into account that the permanent half-half equilibrium of the photons energy (directed half plus oscillating inert half) continuously takes up the slack so that v can equal no other velocity than c. Any energy added adds to both halves and any energy expended subtracts from both halves. This is what the de Broglie photon structure reveals. > A star in the direction of the Earth's travel will show no aberration, > a star at right angles will show maximum aberration. The observed > star close to the sun has to be identified in daylight and there is > further curvature as a result of refraction in Earth's atmosphere. Agreed. André Michaud On 6 fév, 15:04, "Androcles" <Headmas...@Hogwarts.physics> wrote: > <srp2...@gmail.com> wrote in message > [quoted text clipped - 74 lines] > (from Newton's third law and Huygens's wave superposition, > photons have direction and come in pairs) | My view is de Broglie's on photons. I see them as complex | standing harmonic oscillators in motion, with only half | their energy oscillating electromagnetically. The other | half is is oriented in the direction of motion and is | thus impervious to transverse force interaction. Whilst I concur with "harmonic oscillators in motion" http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.gif http://www.androcles01.pwp.blueyonder.co.uk/AC/Photon.gif http://www.androcles01.pwp.blueyonder.co.uk/photon.gif inasmuch as a spinning bullet has energy in the spin and energy of forward motion, I cannot agree to the ratio 1/2. If one travels alongside the spinning bullet then it has no energy of forward motion (which is necessarily relative) but retains its energy of spin. > The equivalent mass of a photon is m = 2E/c^2, Yes. Obvious and perfectly explained by relativistic Newtonian. > from which one might surmise the gravitational deflection, > should one know the energy. | Since it invariantly amounts to half the full quantum, | we always know the energy that maintains the velocity. BUT!.... An object falling in a gravitational field accelerates alongside an object with the same mass or twice that mass. A photon of mass 2 will curve exactly the same as a photon of mass 1. Therefore the mass or (mass equivalent) is irrelevant. > However, > http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov > Riding the Earth, you, Eddington and Einstein (who didn't bother > to look anyway, the only one that did is Eddington who obtained ONE > suitable photograph) are observers on the roundabout. The light has > to curve. From simple trigonometry, the deflection has to be ~v/c. | This does not take into account that the permanent half-half | equilibrium of the photons energy (directed half plus oscillating | inert half) continuously takes up the slack so that v can equal no | other velocity than c. You've misunderstood, v is the velocity of the Earth at right-angles to the direction of the light and is approximately 0.0001c. sin(v/c) ~= v/c for small angles. This has nothing to do with the intrinsic oscillation, it is strictly the Coriolis effect. | Any energy added adds to both halves and any energy expended | subtracts from both halves. | This is what the de Broglie photon structure reveals. > A star in the direction of the Earth's travel will show no aberration, > a star at right angles will show maximum aberration. The observed > star close to the sun has to be identified in daylight and there is > further curvature as a result of refraction in Earth's atmosphere. | Agreed. André Michaud > <srp2...@gmail.com> wrote in message > [quoted text clipped - 90 lines] > http://www.androcles01.pwp.blueyonder.co.uk/AC/Photon.gif > http://www.androcles01.pwp.blueyonder.co.uk/photon.gif I see. Standing dephased pi/2. Does not accomodate real physical quantum localization. In the 3-spaces model, the dephasing is 180 and 2 orthogonal spaces allow the electromagnetic oscillation while a 3rd orthogonal space (normal space) harbor the directed energy sustaining the motion of the oscillating half. > inasmuch as a spinning bullet has energy in the spin and > energy of forward motion, I cannot agree to the ratio 1/2. > If one travels alongside the spinning bullet then it has no energy > of forward motion (which is necessarily relative) but retains > its energy of spin. This is where we differ. In relativistic Newtonian, if you travel alongside the bullet, it can only mean that you have sufficient energy in the forward motion yourself to be able to travel alongside the moving bullet. In the underlying 3-spaces model, the energy is not relative but has physical existence. The energy of the speeding bullet does not diminish as your own increases to catch up with it. > > The equivalent mass of a photon is m = 2E/c^2, > [quoted text clipped - 11 lines] > A photon of mass 2 will curve exactly the same as a photon > of mass 1 In the 3-spaces model, this is because its velocity sustaining energy also falls to 1 For photons, acceleration only increases the total quantum of energy. Does not affect the equilibrium velocity, which is c. > Therefore the mass or (mass equivalent) is irrelevant. Yes. > > However, > > http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov [quoted text clipped - 12 lines] > sin(v/c) ~= v/c for small angles. This has nothing to do with the > intrinsic oscillation, it is strictly the Coriolis effect. Right. My bad. André Michaud > | Any energy added adds to both halves and any energy expended > | subtracts from both halves. [quoted text clipped - 7 lines] > > | Agreed. On 6 fév, 16:36, "Androcles" <Headmas...@Hogwarts.physics> wrote: > <srp2...@gmail.com> wrote in message > [quoted text clipped - 91 lines] > http://www.androcles01.pwp.blueyonder.co.uk/AC/Photon.gif > http://www.androcles01.pwp.blueyonder.co.uk/photon.gif | I see. Standing dephased pi/2. Of course... the magnetic field cannot be zero at the same time as the electric field or energy would not be conserved. E = -dB/dt | Does not accomodate real physical quantum localization. "If the facts don't fit the theory, change the facts." - Einstein. I don't care about your real physical quantum localization, that means nothing. Energy is conserved. | In the 3-spaces model, the dephasing is 180 and 2 orthogonal | spaces allow the electromagnetic oscillation while a 3rd | orthogonal space (normal space) harbor the directed energy | sustaining the motion of the oscillating half. Then your model is wrong. > inasmuch as a spinning bullet has energy in the spin and > energy of forward motion, I cannot agree to the ratio 1/2. > If one travels alongside the spinning bullet then it has no energy > of forward motion (which is necessarily relative) but retains > its energy of spin. | This is where we differ. In relativistic Newtonian, if you | travel alongside the bullet, it can only mean that you have | sufficient energy in the forward motion yourself to be | able to travel alongside the moving bullet. We certainly do differ, I have no idea what "relativistic Newtonian" is. In the real world I can spin a top while sitting in my chair, but I certainly have enough energy to travel alongside it as it goes around the Sun. | In the underlying 3-spaces model, the energy is not relative | but has physical existence. The energy of the speeding bullet | does not diminish as your own increases to catch up with it. A brick on a shelf has enough energy to break your toe, but not if you happen to be in the attic with the shelf below you. Energy is relative. > > The equivalent mass of a photon is m = 2E/c^2, > [quoted text clipped - 11 lines] > A photon of mass 2 will curve exactly the same as a photon > of mass 1 | In the 3-spaces model, this is because its velocity sustaining | energy also falls to 1 KE = 1/2mv^2, v is relative, hence energy is relative. More correctly, KE = 1/2mv^2.cos(phi). A bullet traveling away from you can do no work on your body and is therefore harmless. cos(pi) = -1 To reverse the bullet, first you must stop it, absorbing its energy in (say) compressing a spring, and then give it back. The energy of the cordite is converted to the KE of the bullet, converted again to compress the spring and then converted back into the KE of the bullet which is then absorbed as heat tearing in hole in you. http://www.youtube.com/watch?v=HjYJTs4I5uY | For photons, acceleration only increases the total quantum | of energy. Does not affect the equilibrium velocity, which | is c. Acceleration is rate of change of velocity by definition. Only in crackpot relativity does light travel at 2AB/(t'A-tA) = c, both ways at once. http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img7.gif > Therefore the mass or (mass equivalent) is irrelevant. | Yes. > > However, > > http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov [quoted text clipped - 12 lines] > sin(v/c) ~= v/c for small angles. This has nothing to do with the > intrinsic oscillation, it is strictly the Coriolis effect. Right. My bad. André Michaud > | Any energy added adds to both halves and any energy expended > | subtracts from both halves. [quoted text clipped - 7 lines] > > | Agreed. > <srp2...@gmail.com> wrote in message > [quoted text clipped - 102 lines] > time as the electric field or energy would not be conserved. > E = -dB/dt In 3D + time classical wave theory yes. But with 9D + time it is perfectly possible. In fact, the only possibility turns out to be pi dephasing to accomodate the full em oscillation for localized em events. > | Does not accomodate real physical quantum localization. > > "If the facts don't fit the theory, change the facts." - Einstein. > I don't care about your real physical quantum localization, that > means nothing. Energy is conserved. Of course energy is conserved. > | In the 3-spaces model, the dephasing is 180 and 2 orthogonal > | spaces allow the electromagnetic oscillation while a 3rd > | orthogonal space (normal space) harbor the directed energy > | sustaining the motion of the oscillating half. > > Then your model is wrong. Fine. Your loss. André Michaud > > inasmuch as a spinning bullet has energy in the spin and > > energy of forward motion, I cannot agree to the ratio 1/2. [quoted text clipped - 97 lines] > > > | Agreed. On 6 fév, 18:03, "Androcles" <Headmas...@Hogwarts.physics> wrote: > <srp2...@gmail.com> wrote in message > [quoted text clipped - 103 lines] > time as the electric field or energy would not be conserved. > E = -dB/dt | In 3D + time classical wave theory yes. Fine, that's the world I live in. | But with 9D + time Go away, crank, go live in your 9D universe. > <srp2...@gmail.com> wrote in message > [quoted text clipped - 15 lines] > > How? I'll tell you how. In order to see how newtonian gravity can bend the trajectory of a photon, relativists fake it. They magically tranform a photon with energy E into a particle with mass m = E/2c^2, it is saying half the mass in E = mc^2. Then, relativists assume that particle passes by the massive body from infinity travelling locally at c, and then they can apply newtonian gravity to see how its trajectory is deflected into a hyperbolic one. As a result there is a deflection angle twice the observed one. A particle with mass m = E/c^2, travelling locally at c from infinity, would be deflected in the correct angle, under newtonian gravity. On Feb 6, 9:39 am, AlbertShito <albertito1...@gmail.com> > I'll tell you how. In order to see how newtonian gravity can > bend the trajectory of a photon, relativists fake it. They [quoted text clipped - 4 lines] > apply newtonian gravity to see how its trajectory is deflected > into a hyperbolic one. No, little sh.t, the relativists don't do any of the above idiocies you attribute to them. Right now, you are just a little sh.t, when you grow up you will be a big sh.t, like Kobee-Wublee for example. > On Feb 6, 9:39 am, AlbertShito <albertito1...@gmail.com> > [quoted text clipped - 11 lines] > Right now, you are just a little sh.t, when you grow up you will be a > big sh.t, like Kobee-Wublee for example. dunnoshito dunnoshito still the same shito, you punky http://www.upload-images.net/imagen/58a516b007.jpg On Feb 6, 10:41 am, AlbertShito <albertito1...@gmail.com> > > No, little sh.t, the relativists don't do any of the above idiocies > > you attribute to them. > > Right now, you are just a little sh.t, when you grow up you will be a > > big sh.t, like Kobee-Wublee for example. I see, you are ALREADY a BIG sh.t. Sorry for downgrading you to little sh.t. :-) > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 27 lines] > travelling locally at c from infinity, would be deflected in > the correct angle, under newtonian gravity. Why don't you do the Newtonian calculation before making this claim? - Randy > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 32 lines] > > - Randy The calculations have been already done by you relativists, http://www.theory.caltech.edu/people/patricia/obsertop.html I'm only witness of that, and I see that if we double the mass assumed for a photon we attain the correct answer. On Feb 6, 11:00 am, AlbertShito <albertito1...@gmail.com> wrote: > The calculations have been already done by you relativists,http://www.theory.caltech.edu/people/patricia/obsertop.html > I'm only witness of that, and I see that if we double the mass > assumed for a photon we attain the correct answer.> Dumbshit, In the link you cited, do you see the naive Newtonian approach? Do you see the relativitic approach? Do you see the difference, you despicable pile of dung? Supertroll Dono trolled: Dumbshit, Do you see the difference, you despicable pile of dung?  SignatureDono is concubine Lady Chacha http://en.wikipedia.org/wiki/Yodo-Dono > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 34 lines] > > The calculations have been already done by you relativists,http://www.theory.caltech.edu/people/patricia/obsertop.html Correct. > I'm only witness of that, and I see that if we double the mass > assumed for a photon we attain the correct answer. There are a great many assumptions in those equations which you are not aware of. You are trying to change the meaning of the symbols midstream. Here's what those orbital mechanics equations look like for planetary orbits: http://www.go.ednet.ns.ca/~larry/orbits/kepler.html Note that E is the sum of kinetic energy and potential energy. So what are you going to use for this E? Do you really want to use mc^2 and claim it's a Newtonian model? If you were doing Newtonian gravitation the way Newton did it to calculate the planetary orbits, you'd say the kinetic energy is (1/2)*mc^2, and the potential energy is -GMm/r. Then you'd find out that the mass m drops out, that is the parameter E/m = [c^2/2 - GM/r] does not depend on m. Let's start from the basics: What is the initial velocity, what is the acceleration, what is the result? Start out with a velocity c, mass m, distance R as it passes near mass M. The acceleration experienced by the mass m is GM/R^2. That is what you would apply to the velocity vector to get the change in velocity. It does not depend on m. Trajectories of small bodies through solar systems do not depend on mass. - Randy > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 73 lines] > > - Randy It's simpler than that. Under newtonian gravity you never can say that a photon is predicted to be deflected by twice the correct angle, because under newtonian gravity photons can't be addressed at all, only bodies with mass. You never can fake a photon to behave as a particle with non-zero mass, that's unphysical. If you do fake a photon to behave so, then you attain a particle with mass travelling at c, which is impossible under your assumptions (i.e nothing with mass can travel at c). > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 76 lines] > the correct angle, because under newtonian gravity photons > can't be addressed at all, only bodies with mass. Right. Unless you assume it has mass, as people did long before Einstein. Once you do, the path is independent of whatever mass you assume, just as any Newtonian path is. > You never > can fake a photon to behave as a particle with non-zero mass, > that's unphysical. If you do fake a photon to behave so, then > you attain a particle with mass travelling at c, which is impossible > under your assumptions (i.e nothing with mass can travel at c). What point are you trying to make? If the photon truly has no mass (as we believe) then the Newtonian model says it won't be deflected. Therefore gravitational lensing won't happen. Yet you seemed to be making a claim that there is a Newtonian model that would predict light bending. Presumably that would involve assuming a nonzero mass. So which model are you proposing? Is it (a) the photon has no mass, so Newton says it doesn't bend, or (b) the photon has a mass, so Newton says it bends (at half the GR prediction) independent of mass? You seemed to be proposing a third model which was (c) the photon has a mass, and unlike every other calculation with Newtonian gravity, the trajectory depends on mass. Is that your claim? Stick to one model and we can follow the consequences. Does the photon have mass or not? Do you claim Newton predicts bending or not? - Randy > > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 93 lines] > model says it won't be deflected. Therefore gravitational > lensing won't happen. To claim that a theory predicts a photon will not be deflected and to claim photons are outside the domain of applicability of a theory are not the same claim. Photons are outside the domain of applicability of newtonian gravity. Therefore, newtonian gravity gives no answers to questions concerning photons. Newtonian gravity is concerned with neutral bulk matter, not with bosons. Even if it were found out that photons have mass, newtonian gravity would still remain without answers for them, because they would still be bosons. If it were found out that photons can decay into neutrinos, then newtonian gravity would account for the gravitational behavior of those neutrinos. Finally, I can conclude saying that newtonian gravity deals only with fermionic particles (half-integer spin), not with bosonic particles (integer spin). [snipped already answered redundant questions, above] > > > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 99 lines] > the same claim. Photons are outside the domain of > applicability of newtonian gravity. If massless. This is part of the standard model and would certainly upset a lot of modern physics if false, but it's still a caveat in all your statements, an assumption you're making. Thus the only theory which predicts deflection of photons in gravitational fields (to date) is GR. Are we agreeing on this? If so, I can't figure out what anti-relativistic stance you're trying to make. > Therefore, > newtonian gravity gives no answers to questions > concerning photons. Unless you assume gravitational mass which is nonzero. Once you do, you get a deflection which is half the GR prediction. What YOU stated was that the deflection depends on the assumed mass. It does not. YOU are the one claiming that a Newtonian calculation is possible which will show deflection, if you assume the right mass. Are you backing off of that? > Newtonian gravity is concerned > with neutral bulk matter, not with bosons. Newtonian gravity is concerned with anything with mass. > Even if > it were found out that photons have mass, newtonian > gravity would still remain without answers for them, > because they would still be bosons. ??? And where are you getting this from? So what's your point? Are you backing off your claim that the evil relativists goofed on the Newtonian calculation and if they did it right, Newtonian gravity would predict the correct deflection? Because that seems to contradict this post. - Randy > > > > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 108 lines] > in gravitational fields (to date) is GR. Are we agreeing > on this? Yes, GR tries to predict deflection of photons, and it works nearly fine :-) > If so, I can't figure out what anti-relativistic stance > you're trying to make. Are you seeing in my stance any anti-relativist fashion?. I only say that newtonian gravity can't address bosons, and that it is not correct claiming that newtonian gravity predicts half the correct deflection angle. Even for W(+), W(-) and Z bosons, which have mass, newtonian gravity can't predict their respective deflection angles. Bosons are outside the domain of applicability of newtonian gravity. > > Therefore, > > newtonian gravity gives no answers to questions [quoted text clipped - 4 lines] > Once you do, you get a deflection which is half the > GR prediction. No. It is clear that gravity behaves differently for fermions that for bosons (included bosons with nonzero mass). > What YOU stated was that the deflection depends on the > assumed mass. It does not. > > YOU are the one claiming that a Newtonian calculation > is possible which will show deflection, if you assume > the right mass. Are you backing off of that? No, I only said that if you can say that newtonian gravity predicts half the correct angle, then I could also say that newtonian gravity can predict the correct angle if you assume the right mass for a fake photon. > > Newtonian gravity is concerned > > with neutral bulk matter, not with bosons. > > Newtonian gravity is concerned with anything with > mass. No. Newtonian gravity is not concerned with bosons, regardless they are massless or not. > > Even if > > it were found out that photons have mass, newtonian [quoted text clipped - 4 lines] > > And where are you getting this from? This should need a new dedicated thread to explain you where :-) . But, I'll give you a clue. Pauli exclusion principle. > So what's your point? Are you backing off your claim > that the evil relativists goofed on the Newtonian > calculation and if they did it right, Newtonian > gravity would predict the correct deflection? > > Because that seems to contradict this post. No, see above > > Thus the only theory which predicts deflection of photons > > in gravitational fields (to date) is GR. Are we agreeing > > on this? > > Yes, GR tries to predict deflection of photons, and it works > nearly fine :-) Where does it fail in predicting deflection of photons? > > If so, I can't figure out what anti-relativistic stance > > you're trying to make. > > Are you seeing in my stance any anti-relativist fashion?. Yes. Since you've been on the newsgroup. In practically every post. > I only say that newtonian gravity can't address bosons, > and that it is not correct claiming that newtonian gravity > predicts half the correct deflection angle. Even for W(+), > W(-) and Z bosons, which have mass, newtonian gravity > can't predict their respective deflection angles. Bosons > are outside the domain of applicability of newtonian gravity. Why do you declare this? Newtonian gravity predicts the interaction of masses. What is special about these masses that makes them exempt? > > > Therefore, > > > newtonian gravity gives no answers to questions [quoted text clipped - 7 lines] > No. It is clear that gravity behaves differently for fermions > that for bosons (included bosons with nonzero mass). It is clear from what? Show your reasoning. By "Newtonian gravity" you still seem to have something other than F = GMm/r^2 in mind, since that is silent on fermions, bosons, or any property other than mass. > > What YOU stated was that the deflection depends on the > > assumed mass. It does not. [quoted text clipped - 7 lines] > also say that newtonian gravity can predict the correct > angle if you assume the right mass for a fake photon. Ah... finally we get to an assertion about Newtonian gravity. So the answer is "Yes", not "No". You claim that you could do a Newtonian prediction with a different mass and get a different deflection. Or by saying "can predict the correct angle" you don't really mean getting a different answer from "half the correct angle". Once again, you are saying that if you change the mass, you'll get a different angle. Yes? Fine. The answer is No. Not in standard Newtonian mechanics. There is nothing about the trajectory of small masses near large ones which depends on the small mass. What you are talking about is the parameter E/m, the ratio of total energy (kinetic plus gravitational potential) to mass. In Newtonian mechanics, KE = 0.5*mv^2. So there is no possible ratio E/m except [0.5*v^2 - GM/r], a quantity which you will note does not depend on m. Now, what you are REALLY claiming is that if we use E/m = c^2, then the "Newtonian" prediction would be a different angle. Look at the web page you cited again. On that I certainly agree. But that's not Newtonian mechanics. You are using a non-Newtonian expression for KE. If you use Newtonian mechanics, then E/m = c^2 for any mass moving at c. And if you don't change E/m, then what happens to those formulas you were looking at? I tried to get you to see this by talking about planetary orbits. The orbital parameters of a satellite around the earth, or of a planet around the sun, do NOT depend on the mass of the satellite/planet. You won't get a different trajectory by changing mass. - Randy > > > Even if > > > it were found out that photons have mass, newtonian [quoted text clipped - 8 lines] > explain you where :-) . But, I'll give you a clue. > Pauli exclusion principle. Your hint fails. Newtonian gravity is silent on the Pauli exclusion principle, as Pauli is silent on gravitational potential between two masses. There is no relation between these things. If you think otherwise, state why. In a new thread if you like. But one small sanity check: Do you really think it should be called "Newtonian mechanics" if it includes the Pauli exclusion principle? - Randy > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 75 lines] > > - Show quoted text - Well, you are also using the concept of absolute time if you say this. But according to scientists, they have proven that a clock in S' will run slower than a clock in S, so my question is, What does that do to Newton's equations? If a clock in a satellite is running slower than a clock on earth, then an astronaut in the satellite will compute his velocity to be faster in an orbit around the earth than a scientist on earth would compute it if they are both using Newton's equations and the concept of absolute time, believing the two clocks agree with each other. As far as I can see, the concept of absolute time is a correct principle, but you cannot keep track of it with clocks without making adjustments for relative velocities. In other words, a moving clock has to have its rate adjusted in order to agree with a clock that is not moving, so Newton's equations, as he used them, describe a system with clocks that automatically adjust according to velocity within a gravitational system to agree with a clock at the center of gravity. But since scientists say they have proven that a moving clock is slower than a stationary one, that has to be figured into the equations. Robert B. Winn > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 78 lines] > Well, you are also using the concept of absolute time if you say > this. If I'm doing a Newtonian calculation, I do it with Newtonian physics. > But according to scientists, they have proven that a clock in > S' will run slower than a clock in S, so my question is, What does > that do to Newton's equations? Whatever it does (and I have no interest in your nonsense), the result is not Newtonian physics. Call it something else. If you want to ask, "what does pure Newtonian physics predict" then the answer is done with Newtonian physics, not some hybrid. - Randy > > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 95 lines] > > - Show quoted text - Well, OK, Randy, let's use pure Newtonian physics. Newton is sitting under a tree trying to figure out the orbit of the moon, and an apple falls and hits him on the head. So Newton remembers Galileo's experiment with the two different size rocks on the leaning tower or Pisa and decides that the moon is falling toward the earth at a rate that keeps it always at the same altitude. So he uses Kepler's equation for the radius of an orbit and figures the force required. All I asked was how a slower clock on the moon would apply to the equations. You people were the ones who said that a clock on the moon would be running slower. Now you have all of these GPS satellites, etc. How do the times on the clocks in those satellites relate to their orbits? You keep claiming that the equations that Newton used are scripture of some kind. Kepler's number was just an approximation. It did not exactly match the orbit of anything from the start. It was just an average taken from the orbits that could be observed. Maybe if clock time was used instead of absolute time, Kepler's calculation of the orbits would be closer, maybe it would be less accurate. I have no way of determining any of this. I have no time to run an experiment of that kind. I have to work every day welding. So I ask a scientist. No, Newton's equations are scripture. You cannot use them any way except the way Newton did. OK, I do not want to get into an argument with you over your religion. Just go ahead and use Newton's equations the same way he did. He used absolute time. According to him, a clock in a satellite would read the same as a clock on earth. I don't want to get you all irrational over something like this. Robert B. Winn > > > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 103 lines] > that keeps it always at the same altitude. So he uses Kepler's > equation for the radius of an orbit and figures the force required. Garbled. But at any rate, in Newtonian physics we can write the equation GMm/r^2 = d^2/dt^2 [m*x ] and calculate the behavior of mass m in the presence of mass M. > All I asked was how a slower clock on the moon would apply to the > equations. Didn't you start this post with this statement: "let's use pure Newtonian physics."? There's no "slower clock on the moon" in pure Newtonian physics. > You people were the ones who said that a clock on the moon > would be running slower. Faster, actually. In the theory called GR. Not in "pure Newtonian physics". > Now you have all of these GPS satellites, > etc. How do the times on the clocks in those satellites relate to > their orbits? Well, that's a different question. Did you want to ask a question about satellite orbits in GR or in Newtonian mechanics? Don't say "Newtonian" and then ask a GR question. I'm going to try to make an analogy with a subject you might understand: welding. If I weld the left half of a Mercedes-Benz to the right half of a Volkswagon Beetle, and find this creates all sorts of problems, does this tell me that the Benz is poorly designed? OK, OK, let's just talk about the pure Benz and how poorly it handles. I took my welded Benz-Beetle out for a drive and found... what's that? That's not a pure Benz anymore? What's wrong with you? I'm just trying to ask you a question about handling of the Mercedes Benz! > You keep claiming that the equations that Newton used are > scripture of some kind. No, I'm claiming that if you want to claim you're doing Newtonian mechanics, you should be using the Newtonian physics. If you want to throw something else in there, you can certainly do it, there's no law against it. But it's no longer Newtonian physics. What, is the Mercedes Benz some holy grail that I'm not allowed to weld onto? Was it created by St. Peter or something? What's wrong with my welding job? Just answer the question about why the handling of the Mercedes is so bad, based on my experiments with my welded vehicle! > Kepler's number was just an approximation. What number? > It did not exactly match the orbit of anything from the start. But Newton's equations matched the behavior astonishingly well, and could even be extended to model the interaction of multiple bodies with high accuracy. To the extent that NASA still uses them, and so does every agency which tracks the thousands of bits of spacecraft and rubble traveling around the earth. > It was > just an average taken from the orbits that could be observed. That's nice, but we were talking about Newton's theory, not Kepler's. Newton explained Kepler's Laws, and also in what way they are an approximation. > Maybe if clock time was used instead of absolute time, Kepler's > calculation of the orbits would be closer, maybe it would be less > accurate. I have no way of determining any of this. I have no time > to run an experiment of that kind. I have to work every day welding. > So I ask a scientist. No, Newton's equations are scripture. If you ask me about Newton's equations, then those are the equations I'm going to use. You've dragged both pre-Newton and post-Newton science into this discussion. Are you asking about Newton's equations or not? What the hell is wrong with this Mercedes Benz anyway? It's just a terrible car. > You cannot use them any way except the way Newton did. I can use them any way I want. And I can even change them. But then I have to say they aren't Newton's equations any more. What's wrong with saying that? Why do you want to change them and then pretend they're unchanged? You want to explore the welded theory, let's do it. - Randy > > > > > > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 209 lines] > > - Show quoted text - OK, so you do not believe Kepler had a number. Kepler thought that he did. It was R^3/T^2 = 3.353, more or less. Newton used Kepler's number in his equation for the centripetal force on a planet. F=4(pi)^2 Km/R^2 Now I don't know what you are saying about welding. So is this equation used according to time on a clock on earth a Mercedes Benz and this equation used according to time on a clock in a satellite a Volkswagen? Well, all right. But the fact remains that you get two different values for the centripetal force on the satellite. That means you will get two different values for a gravitational constant, depending on which clock you use. Now here is something you may not realize about welding. Just because you have a Mercedes Benz and a Volkswagen, it does not mean you have to cut both cars in half and weld them together. You could leave both cars in their original condition. I do not know if you ever considered two cars this way. Anyway, it is something you might want to consider, just leaving the Mercedex Benz as a Mercedes Benz and the Volkswagen as a Volkswagen. But, then, if your religious beliefs require you to cut the cars in half, who am I to criticize? Robert B. Winn > OK, so you do not believe Kepler had a number. I didn't say whether I believed Kepler had a number of not. I asked you what number you were referring to. > Kepler thought that he > did. It was R^3/T^2 = 3.353, more or less. OK, that's a number. What about it? > Newton used Kepler's number in his equation for the centripetal force > on a planet. No, Newton didn't "use Kepler's number". Newton proved why R and T have this relation from more fundamental considerations. > Now I don't know what you are saying about welding. I'm asking you what theory you want to talk about. (1) Newton's theory, which would be the theory that, you know, was Newton's. I will only insist on sticking to Newton physics if you say that we are having a "pure Newtonian" discussion. That would be what is implied by the words "Newtonian" and "pure". (2) GR, where time is affected by gravitational potential and orbits are geodesics, or (3) Some theory where you have welded GR results with Newtonian equations, which is neither GR nor Newton but a hybrid monster, and possibly not drivable. I'm happy to talk about one, but you have to pick. Don't say you're talking about (1) and then immediately switch to (3). You have some question about the GR effect of time and gravitational fields. So we're in either (2) or (3). Which is it, and what's your question? - Randy > > OK, so you do not believe Kepler had a number. > [quoted text clipped - 39 lines] > > � � � � - Randy I don't really have any questions any more. I had a question about 20 years ago which I posted in this newsgroup which was about the two little equations Einstein extracted from the Lorentz equations, x=ct, x'=ct'. After years and years of nonsense, I finally figured out the answer to my question. Einstein's two little equations were wrong. The Lorentz equations were not the correct equations for what Einstein said he was doing. What I was discussing with you was the difference that according to scientists exists in Newton's equations if you use time from a clock in a satellite as compared with time from a clock on earth. Newton knew of no such time difference, so he never considered it. According to you, since Newton never considered it, it does not apply to his equations, and it would be "impure" in some way to compare this difference. I say that if Newton's equations are applied using clock times from two different frames of reference, and two different gravitational constants are found, then one or the other or something in between must be the actual constant, since two different answers are not constant. As for whether or not Newton used Kepler's ratio, I was basing what I said on what is said in a physics book, "Newton explained many years later that he was led to the correct force law by working backwards from Kepler's third law." So the fact of the matter is that either Newton did use Kepler's ratio R^3/T^2 or else the writers of that physics book lied about it. It does not matter to me. Kepler's ratio definitely fits into Newton's equation, whether the writers of that physics book lied about Newton using it or not. Scientists lie all the time. Possibly the writers of that book did lie as you are claiming, or maybe they just came to a wrong conclusion because it seemed logical to them. Perhaps we will never know. Why was it so important to you that Newton got his equation from somewhere else than Kepler's third law? Robert B. Winn > > > OK, so you do not believe Kepler had a number. > [quoted text clipped - 47 lines] > The Lorentz equations were not the correct equations for what Einstein > said he was doing. 20 years and you still don't understand? Nice. [...] > > > > OK, so you do not believe Kepler had a number. > [quoted text clipped - 49 lines] > > 20 years and you still don't understand? Nice. Well, yes, I do understand, Eric. If you had read the entire post, you would have read the sentence that said, Scientists lie all the time. As I recall, you were going to prove that these equations are wrong. Just take a few minutes and show all of these scientists how well you understand. x'=x-vt y'=y z'=z t'=t w=velocity of light x=wt x'=wn' wn'=wt-vt n'=t(1-v/w) Robert B. Winn > > > > > OK, so you do not believe Kepler had a number. > [quoted text clipped - 55 lines] > wrong. Just take a few minutes and show all of these scientists how > well you understand. What chance do I have of making you understand anything through 20+ years of failure? Go back to whatever it was you were doing before you decided to pollute USENET with your shitposting. > x'=x-vt > y'=y [quoted text clipped - 8 lines] > > Robert B. Winn > > > OK, so you do not believe Kepler had a number. > [quoted text clipped - 65 lines] > So the fact of the matter is that either Newton did use Kepler's ratio > R^3/T^2 or else the writers of that physics book lied about it. Your quote is correct. Georges Gamow even explained how Newton mathematically proceeded in his small book "Gravity", Science Study Series, Doubleday, 1962, chapters 2, 3 and 4. André Michaud > It > does not matter to me. Kepler's ratio definitely fits into Newton's [quoted text clipped - 5 lines] > equation from somewhere else than Kepler's third law? > Robert B. Winn On Feb 10, 4:01 pm, srp2...@gmail.com wrote: > > > > OK, so you do not believe Kepler had a number. > [quoted text clipped - 85 lines] > > - Show quoted text - Thank you for clearing that up, Andre. Robert B. Winn > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 30 lines] > Why don't you do the Newtonian calculation before > making this claim? I'll give you a hint: When you have a mass moving under the influence of a much heavier mass (say, a planet moving under the influence of the sun, or a moon moving around the planet), does anything in the orbit depend on the mass of the lighter object? That is, if the moon's mass were half what it is, would it have a different path around the earth? If you started a satellite in orbit around the earth, and another satellite with half the mass in the same starting orbit, would they continue in the same orbit? If you fire a projectile from a gun, and then another with the same initial velocity but half the mass, does it follow a different path (neglecting air resistance)? Once you've pondered that, decide whether you really believe that two particles with different mass but the same starting trajectory as they approach the sun follow different paths. - Randy > > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 58 lines] > > - Show quoted text - Randy, Good to see that you decided to look up the principle of equivalence. Robert B. Winn > > > <srp2...@gmail.com> wrote in message > [quoted text clipped - 34 lines] > > - Show quoted text - Well, going back to the astronaut with his slower clock, it appears to me that he would come up with a different gravitational constant than the observer on earth because the velocity of the satellite relative to earth would be faster according to his clock. Robert B. Winn On 6 feb, 16:48, "Androcles" <Headmas...@Hogwarts.physics> wrote: > <srp2...@gmail.com> wrote in message > [quoted text clipped - 15 lines] > > How? | I'll tell you how. In order to see how newtonian gravity can | bend the trajectory of a photon, relativists fake it. They [quoted text clipped - 7 lines] | travelling locally at c from infinity, would be deflected in | the correct angle, under newtonian gravity. See my reply to Andre. > | Classical Newtonian mechanics predicts that light will bend with > | twice the angle observed. > > How? Good question. Actually, I think it's 1/2 the angle, not twice. To come up with refined tests for Newtonian vs. Relativistic gravity; it's common for people to embody the former in the framework of Newton- Cartan geometry. This minimizes the differences between the classical and relativistic theory (a necessity if you want to do experimental tests), down to their minimal core differences. Newton-Cartan gravity is a curved spacetime theory compatible with Newtonian physics and the Galilean symmetry group that presides over Newtonian physics. If you go with straight classical theory, treating Newtonian gravity as a field on top of a flat spacetime, then I think Smolin is right -- no bending of light. On the other hand, if you use the classical version of the Equivalence Principle, then that would imply light undergoes the same acceleration near a gravitating body as does everything else. Then, in effect, you're back to the curved spacetime form of Newtonian gravity -- that is, the same results follows ... a bending in Newtonian gravity at 1/2 the angle predicted by General Relativity. On Feb 6, 6:33 am, srp2...@gmail.com wrote: > >http://philipball.blogspot.com/2007/09/arthur-eddington-was-innocent-... > > "With the technology then available, measuring the bending of [quoted text clipped - 12 lines] > angle since it is now understood that only half a photon's energy > is sensitive to transverse interaction. Uhhhhh....no. General relativity [is NOT "relativistic Newtonian mechanics] makes the correct prediction without assuming _anything_ abut the photon other than it traveling along a null path. [...] > On Feb 6, 6:33 am, srp2...@gmail.com wrote: > [quoted text clipped - 22 lines] > > - Show quoted text - Light falls. The null path is the space curvature itself. Mitch Raemsch Twice Nobel Laureate 2008 > On Feb 6, 6:33 am, srp2...@gmail.com wrote: > [quoted text clipped - 17 lines] > Uhhhhh....no. General relativity [is NOT "relativistic Newtonian > mechanics] makes the correct prediction So does relativistic Newtonian mechanics, in perfect harmony with Maxwell's theory. André Michaud On Feb 6, 2:10 pm, srp2...@gmail.com wrote: > > On Feb 6, 6:33 am, srp2...@gmail.com wrote: > [quoted text clipped - 20 lines] > So does relativistic Newtonian mechanics, in perfect harmony > with Maxwell's theory. A relativistic Newtonian mechanics is a contradiction in terms. The fact is that general relativity makes the predictions about light bending /without/ reference to Maxwell's equations or anything about light /other/ than it traveling on a null path. Is there a particular reason you can't open up a textbook, see the derivation, repeat the derivation, and compare with observation? > André Michaud > On Feb 6, 2:10 pm, srp2...@gmail.com wrote: > [quoted text clipped - 36 lines] > > - Show quoted text - Light falls. Mitch Raemsch Twice Nobel Laureate 2008 > On Feb 6, 2:10 pm, srp2...@gmail.com wrote: > [quoted text clipped - 28 lines] > bending /without/ reference to Maxwell's equations or anything about > light /other/ than it traveling on a null path. Any theory that predicts anything about light without reference to electromagnetism and Maxwell can only be a mathematical patchwork doomed to end up in file 13. > Is there a particular > reason you can't open up a textbook, see the derivation, repeat the > derivation, and compare with observation? I have and have rejected the method. I have a much better integrated one that can even explain the stuff that GR can't deal with. André Michaud On Feb 7, 4:35 am, srp2...@gmail.com wrote: > > On Feb 6, 2:10 pm, srp2...@gmail.com wrote: > [quoted text clipped - 32 lines] > to electromagnetism and Maxwell can only be a mathematical patchwork > doomed to end up in file 13. Where do you think the massless assumption came from? You could, in theory, solve the field equations for a electromagnetic stress-energy tensor with a plane wave solution. Good luck solving that. > > Is there a particular > > reason you can't open up a textbook, see the derivation, repeat the > > derivation, and compare with observation? > > I have and have rejected the method. I have a much better integrated > one that can even explain the stuff that GR can't deal with. If you still don't understand how GR can make the predictions it does and that it doesn't assume the things you don't think it does, no, you haven't. Work on the mundane before you attack the subtle. If your theory can't explain prosaic effects like perihelion precession, why would it accurately explain whatever fringe effect you are focusing on this time? > André Michaud > On Feb 7, 4:35 am, srp2...@gmail.com wrote: > [quoted text clipped - 36 lines] > > Where do you think the massless assumption came from? From ignorance of the nature of electromagnetic energy. > You could, in theory, solve the field equations for a electromagnetic > stress-energy tensor with a plane wave solution. Good luck solving > that. The Copenhagen community's moto: Why make things simple when you can make them complicated ? > > > Is there a particular > > > reason you can't open up a textbook, see the derivation, repeat the [quoted text clipped - 11 lines] > accurately explain whatever fringe effect you are focusing on this > time? You are so disconnected Eric that it edges on unconsciousness. Why don't you wake up. André Michaud On Feb 7, 10:51 am, srp2...@gmail.com wrote: [...} > > > Any theory that predicts anything about light without reference > > > to electromagnetism and Maxwell can only be a mathematical patchwork [quoted text clipped - 3 lines] > > From ignorance of the nature of electromagnetic energy. Wrong, try again. 1) Solid research places error bars on photon mass at the 10^-17 eV level. Consistent with zero. 2) Maxwell's equations have no mass coupling. 3) Proca's equations - Maxwell's equations for a massive photon - are distinctly different from Maxwell's equations. > > You could, in theory, solve the field equations for a electromagnetic > > stress-energy tensor with a plane wave solution. Good luck solving [quoted text clipped - 3 lines] > > Why make things simple when you can make them complicated ? You were the one bitching about me giving the simple answer. In the optics limit, Maxwell's equations do not matter so long as waves travel along null paths. If you want the exact answer, yea it's gonna be complicated. Pretending it isn't is dishonest. > > > > Is there a particular > > > > reason you can't open up a textbook, see the derivation, repeat the [quoted text clipped - 15 lines] > > Why don't you wake up. Why don't you prove your ability before making sweeping claims about the physics community? > André Michaud > On Feb 7, 10:51 am, srp2...@gmail.com wrote: > [...} [quoted text clipped - 52 lines] > Why don't you prove your ability before making sweeping claims about > the physics community? What conceit! I have nothing to prove to you. You are meaningless. André Michaud >> On Feb 6, 6:33 am, srp2...@gmail.com wrote: >> [quoted text clipped - 10 lines] > So does relativistic Newtonian mechanics, in perfect harmony > with Maxwell's theory. OK, I'll bite. What is "relativistic Newtonian mechanics"? How is it in perfect harmony with Maxwell's theory? To clarify, do you mean _Maxwell's_ theory, or do you mean modern classical electrodynamics? Ordinary Newtonian mechanics is in perfect harmony with Maxwell's theory, although not with modern classical EM. SignatureTimo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html > >> On Feb 6, 6:33 am, srp2...@gmail.com wrote: > [quoted text clipped - 14 lines] > perfect harmony with Maxwell's theory? To clarify, do you mean _Maxwell's_ > theory, or do you mean modern classical electrodynamics? I mean Maxwell's wave theory. The point is that Maxwell's theory can be seamlessly expanded, contrary to belief, from a disregarded de Broglie hypothesis, to directly describe localized photons and by linking it with classical Newton, to also describe localized massive particles, moving or not. Now as to what it is, this media couldn't possibly allow full explanation. but as an example, if you correct Newton's classical kinetic energy equation to correctly include the energy he could know nothing about and that goes into relativistic mass increase and if you convert it to its correct exponential form, you then end up with an equation that allows calculating the complete range of possible velocities, from zero for massive particles at rest to c for free photons not associated with massive particles. Here is a generalized form of this equation f(x)=c [sqrt(4ax + x^2)) / (2a - x)] where f(x |
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