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| Tags: asymptotical, bars, error, incompleteness, its, prove |
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#31
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On May 13, 2:04�am, PD wrote:
On May 13, 3:31�am, Albertito wrote: On May 12, 11:21 pm, PD wrote: On May 12, 6:56 am, Albertito wrote: On May 11, 10:07 pm, PD wrote: On May 9, 6:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c �below the third-order term approximation. The power series expansion of r = arctanh(v/c) is � � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't �still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors � � � � � � � �w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities � � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler � � � � � � � �f' = Exp(-r) f , � � � � � � � �where r = arctanh(v/c), can't still be experimentally distinguished from this one � � � � � � � �f' = Exp(-v/c) f There is actually an abundance of these tests. I'll mention one. Muon beamlines are created by allowing charged pions to decay in flight. The pions have momenta such that their speed is close to that of light. Since it is an exoenergetic decay, the muon has extra kinetic energy from the decay. Since the pion's decay mechanism doesn't give a whit whether the pion is in flight when it decays (and in fact, the principle of relativity says the physics of the pion decay has to be the same for pions at rest vs. pions in free flight), we can guess what that kinetic energy profile is for the moving pions by using the distribution for decaying pions at rest. Or, put another way, we can take the velocity distribution of muons in the pion rest frame and boost them to the frame where the pions are moving close to the speed of light. Using rapidities to perform that boost results in a velocity distribution that agrees with direct measurement. The Galilean transformations, on the other hand, predict that the muon velocity would have a peak in the forward direction that exceeds c.. Muon time of flight can be measured directly via a triplet of scintillator paddles, and in fact this is routinely done in muon beamlines. The muon speed distribution never exceeds c. PD Interpretation under my model and that provided by SR can't still be experimentally distinguished. Consider this sum of betas, � � v/c = v_p/c + v_mu'/c , � � with � � v_p, pion speed at the moment of its decay wrt muon � � � � � � detector.. � � v_mu', �the relative speed of a muon with respect � � � � � � � � �to the point where the pion has decayed Consider this sum of rapidities, �arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c) In both sums, neither v_p nor v_m_u' exceed c. SR assumes the final speed of the muon in the detector must be � � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)), � � � � but it needs also the assumption there is time dilation � � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2), � � � � in order to fit the prediction to the observation. In my model, the final speed of the muon in the detector is � � � � � � �v �= v_p + v_mu', � � � � � � without the assumption of any time dilation at all. If it still is not clear to you, I can repeat, � � � � � � �v/c =(v_p + v_mu')/c And, if you do the math, you find out the left-hand-side of this formula should be greater than 1, given the amount of kinetic energy released in pions decaying to muons. Go ahead, plug in the numbers. What's the problem with finding out the left-hand-side of this formula is greater than 1? Well, for one thing, it doesn't happen. As I mentioned to you, the time of flight of muons is *measured*. We know what the speed is. It's not greater than 1. Well, the way I figure it with the Galilean transformation equations and the slower clock in the frame of reference of the muon, from the frame of reference of the muon, it reaches the speed of light at .5c as measured according to a clock at rest. So if the muon is measured at traveling near the speed of light by a clock at rest, a clock in the frame of reference of the muon would show the clock at rest to be traveling at about nine times the speed of light. What really happens is that the rate of decay of the muon is slowed down because of the energy of its motion. Really, there is no clock in the frame of reference of the muon because a muon is a sub- atomic particle, not a force moving the earth and clocks on the earth around. So all you can really say is that the rate of decay of a muon slows down when its speed increases. Robert B. Winn |
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#32
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On May 13, 3:01�am, Eric Gisse wrote:
On May 13, 1:53�am, Albertito wrote: On May 13, 10:04 am, PD wrote: On May 13, 3:31 am, Albertito wrote: On May 12, 11:21 pm, PD wrote: On May 12, 6:56 am, Albertito wrote: On May 11, 10:07 pm, PD wrote: On May 9, 6:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c �below the third-order term approximation. The power series expansion of r = arctanh(v/c) is � � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't �still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors � � � � � � � �w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities � � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler � � � � � � � �f' = Exp(-r) f , � � � � � � � �where r = arctanh(v/c), can't still be experimentally distinguished from this one � � � � � � � �f' = Exp(-v/c) f There is actually an abundance of these tests. I'll mention one. Muon beamlines are created by allowing charged pions to decay in flight. The pions have momenta such that their speed is close to that of light. Since it is an exoenergetic decay, the muon has extra kinetic energy from the decay. Since the pion's decay mechanism doesn't give a whit whether the pion is in flight when it decays (and in fact, the principle of relativity says the physics of the pion decay has to be the same for pions at rest vs. pions in free flight), we can guess what that kinetic energy profile is for the moving pions by using the distribution for decaying pions at rest. Or, put another way, we can take the velocity distribution of muons in the pion rest frame and boost them to the frame where the pions are moving close to the speed of light. Using rapidities to perform that boost results in a velocity distribution that agrees with direct measurement. The Galilean transformations, on the other hand, predict that the muon velocity would have a peak in the forward direction that exceeds c. Muon time of flight can be measured directly via a triplet of scintillator paddles, and in fact this is routinely done in muon beamlines. The muon speed distribution never exceeds c. PD Interpretation under my model and that provided by SR can't still be experimentally distinguished. Consider this sum of betas, � � v/c = v_p/c + v_mu'/c , � � with � � v_p, pion speed at the moment of its decay wrt muon � � � � � � detector. � � v_mu', �the relative speed of a muon with respect � � � � � � � � �to the point where the pion has decayed Consider this sum of rapidities, �arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c) In both sums, neither v_p nor v_m_u' exceed c. SR assumes the final speed of the muon in the detector must be � � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)), � � � � but it needs also the assumption there is time dilation � � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2), � � � � in order to fit the prediction to the observation. In my model, the final speed of the muon in the detector is � � � � � � �v �= v_p + v_mu', � � � � � � without the assumption of any time dilation at all. If it still is not clear to you, I can repeat, � � � � � � �v/c =(v_p + v_mu')/c And, if you do the math, you find out the left-hand-side of this formula should be greater than 1, given the amount of kinetic energy released in pions decaying to muons. Go ahead, plug in the numbers.. What's the problem with finding out the left-hand-side of this formula is greater than 1? Well, for one thing, it doesn't happen. As I mentioned to you, the time of flight of muons is *measured*. We know what the speed is. It's not greater than 1. That's only a problem if you use the relativistic kinetic energy. No, sir. It's a problem with measurement. Of course, v/c = (v_p + v_mu')/c �is greater than 1, because v_p/c is very close to 1, and v_mu'/c can be greater than (1 - v_p/c). You can't keep track of single muons to measure their 'time of flight'. All you can do is to perform statistics and extrapolate to single muons. You can't say a muon called A was emitted by a pion called P at time t=0, and then A was detected at distance R at time t'. The extrapolation for a single muon you perfom is pure nonsense. That's the reason why we find out in particle physics there are three odd generations of charged leptons, and nobody can guess why there are actually those three generations and why they exhibit the rest masses they exhibit. SR is unable to account for the mass of muons and tau leptons with respect to the mass of electrons. Should we call that impossibility a fraud?. If you avoid SR, and try to explain why muon and tau lepton have the rest masses they have wrt that of an electron, you will be able to find out �those three generations of charged leptons, are just exact ratios derived from multiples of c. I'd like to take a moment to point out that you are arguing about particle physics with an actual published particle physicist.- Hide quoted text - - Show quoted text - Ptolemy was an actual published astronomer when Galileo decided that the earth was rotating instead of the sun orbiting the earth. Special relativity has just as many epicycles as Ptolemaic astronomy. Robert B. Winn |
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#33
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On May 11, 2:26*pm, Albertito wrote:
On May 11, 7:10 pm, Igor wrote: On May 9, 7:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c *below the third-order term approximation. The power series expansion of r = arctanh(v/c) is * * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't *still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors * * * * * * * *w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities * * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler * * * * * * * *f' = Exp(-r) f , * * * * * * * *where r = arctanh(v/c), can't still be experimentally distinguished from this one * * * * * * * *f' = Exp(-v/c) f Uhh, you do understand that v/c = tanh r don't you? *I didn't think so. Really?, didn't you? Didn't *you realize that both v/c and its rapidity, r = arctanh(v/c), currently are experimentally indistinguishable? You must understand that if I know one of them, I know both of them. So what is your point? |
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#34
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On May 13, 6:26 pm, Igor wrote:
On May 11, 2:26 pm, Albertito wrote: On May 11, 7:10 pm, Igor wrote: On May 9, 7:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c below the third-order term approximation. The power series expansion of r = arctanh(v/c) is r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler f' = Exp(-r) f , where r = arctanh(v/c), can't still be experimentally distinguished from this one f' = Exp(-v/c) f Uhh, you do understand that v/c = tanh r don't you? I didn't think so. Really?, didn't you? Didn't you realize that both v/c and its rapidity, r = arctanh(v/c), currently are experimentally indistinguishable? You must understand that if I know one of them, I know both of them. So what is your point? The point is that you can't know both of them. How do you know that v/c is the beta you have measured for a moving body, but it is not its rapidity r = arctanh(v/c). Error bars on (v/c ± d) do not allow you to discriminate v/c from its rapidity beyond its second-order term. |
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#35
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On May 13, 1:52*pm, Albertito wrote:
On May 13, 6:26 pm, Igor wrote: On May 11, 2:26 pm, Albertito wrote: On May 11, 7:10 pm, Igor wrote: On May 9, 7:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c *below the third-order term approximation. The power series expansion of r = arctanh(v/c) is * * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + .... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't *still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors * * * * * * * *w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities * * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler * * * * * * * *f' = Exp(-r) f , * * * * * * * *where r = arctanh(v/c), can't still be experimentally distinguished from this one * * * * * * * *f' = Exp(-v/c) f Uhh, you do understand that v/c = tanh r don't you? *I didn't think so. Really?, didn't you? Didn't *you realize that both v/c and its rapidity, r = arctanh(v/c), currently are experimentally indistinguishable? You must understand that if I know one of them, I know both of them. So what is your point? The point is that you can't know both of them. How do you know that v/c is the beta you have measured for a moving body, but it is not its rapidity r = arctanh(v/c). Error bars on (v/c ± d) do not allow you to discriminate v/c from its rapidity beyond its second-order term. Now you're just being either silly, stupid, or both. I don't need to measure both of them. Or maybe you haven't noticed that they are functions of each other? |
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#36
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On May 13, 6:59 pm, Igor wrote:
On May 13, 1:52 pm, Albertito wrote: On May 13, 6:26 pm, Igor wrote: On May 11, 2:26 pm, Albertito wrote: On May 11, 7:10 pm, Igor wrote: On May 9, 7:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c below the third-order term approximation. The power series expansion of r = arctanh(v/c) is r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler f' = Exp(-r) f , where r = arctanh(v/c), can't still be experimentally distinguished from this one f' = Exp(-v/c) f Uhh, you do understand that v/c = tanh r don't you? I didn't think so. Really?, didn't you? Didn't you realize that both v/c and its rapidity, r = arctanh(v/c), currently are experimentally indistinguishable? You must understand that if I know one of them, I know both of them. So what is your point? The point is that you can't know both of them. How do you know that v/c is the beta you have measured for a moving body, but it is not its rapidity r = arctanh(v/c). Error bars on (v/c ± d) do not allow you to discriminate v/c from its rapidity beyond its second-order term. Now you're just being either silly, stupid, or both. I don't need to measure both of them. Or maybe you haven't noticed that they are functions of each other? It's unnecessary to blurt out curses. I have not said you need to measure both of them, I've said that if you measure v/c, its error bars make its rapidity useless for relativistic corrections, for example relativistic addition of velocities. |
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#37
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On May 13, 6:19Â*am, rbwinn wrote:
On May 13, 3:01�am, Eric Gisse wrote: On May 13, 1:53�am, Albertito wrote: On May 13, 10:04 am, PD wrote: On May 13, 3:31 am, Albertito wrote: On May 12, 11:21 pm, PD wrote: On May 12, 6:56 am, Albertito wrote: On May 11, 10:07 pm, PD wrote: On May 9, 6:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c �below the third-order term approximation. The power series expansion of r = arctanh(v/c) is � � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't �still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors � � � � � � � �w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities � � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler � � � � � � � �f' = Exp(-r) f , � � � � � � � �where r = arctanh(v/c), can't still be experimentally distinguished from this one � � � � � � � �f' = Exp(-v/c) f There is actually an abundance of these tests. I'll mention one. Muon beamlines are created by allowing charged pions to decay in flight. The pions have momenta such that their speed is close to that of light. Since it is an exoenergetic decay, the muon has extra kinetic energy from the decay. Since the pion's decay mechanism doesn't give a whit whether the pion is in flight when it decays (and in fact, the principle of relativity says the physics of the pion decay has to be the same for pions at rest vs. pions in free flight), we can guess what that kinetic energy profile is for the moving pions by using the distribution for decaying pions at rest. Or, put another way, we can take the velocity distribution of muons in the pion rest frame and boost them to the frame where the pions are moving close to the speed of light. Using rapidities to perform that boost results in a velocity distribution that agrees with direct measurement. The Galilean transformations, on the other hand, predict that the muon velocity would have a peak in the forward direction that exceeds c. Muon time of flight can be measured directly via a triplet of scintillator paddles, and in fact this is routinely done in muon beamlines. The muon speed distribution never exceeds c. PD Interpretation under my model and that provided by SR can't still be experimentally distinguished. Consider this sum of betas, � � v/c = v_p/c + v_mu'/c , � � with � � v_p, pion speed at the moment of its decay wrt muon � � � � � � detector. � � v_mu', �the relative speed of a muon with respect � � � � � � � � �to the point where the pion has decayed Consider this sum of rapidities, �arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c) In both sums, neither v_p nor v_m_u' exceed c. SR assumes the final speed of the muon in the detector must be � � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)), � � � � but it needs also the assumption there is time dilation � � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2), � � � � in order to fit the prediction to the observation. In my model, the final speed of the muon in the detector is � � � � � � �v �= v_p + v_mu', � � � � � � without the assumption of any time dilation at all. If it still is not clear to you, I can repeat, � � � � � � �v/c =(v_p + v_mu')/c And, if you do the math, you find out the left-hand-side of this formula should be greater than 1, given the amount of kinetic energy released in pions decaying to muons. Go ahead, plug in the numbers. What's the problem with finding out the left-hand-side of this formula is greater than 1? Well, for one thing, it doesn't happen. As I mentioned to you, the time of flight of muons is *measured*. We know what the speed is. It's not greater than 1. That's only a problem if you use the relativistic kinetic energy. No, sir. It's a problem with measurement. Of course, v/c = (v_p + v_mu')/c �is greater than 1, because v_p/c is very close to 1, and v_mu'/c can be greater than (1 - v_p/c). You can't keep track of single muons to measure their 'time of flight'. All you can do is to perform statistics and extrapolate to single muons. You can't say a muon called A was emitted by a pion called P at time t=0, and then A was detected at distance R at time t'. The extrapolation for a single muon you perfom is pure nonsense. That's the reason why we find out in particle physics there are three odd generations of charged leptons, and nobody can guess why there are actually those three generations and why they exhibit the rest masses they exhibit. SR is unable to account for the mass of muons and tau leptons with respect to the mass of electrons. Should we call that impossibility a fraud?. If you avoid SR, and try to explain why muon and tau lepton have the rest masses they have wrt that of an electron, you will be able to find out �those three generations of charged leptons, are just exact ratios derived from multiples of c. I'd like to take a moment to point out that you are arguing about particle physics with an actual published particle physicist.- Hide quoted text - - Show quoted text - Ptolemy was an actual published astronomer when Galileo decided that the earth was rotating instead of the sun orbiting the earth. Special relativity has just as many epicycles as Ptolemaic astronomy. Robert B. Winn Where are the epicycles, Robert? |
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#38
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On May 13, 8:23 pm, Eric Gisse wrote:
On May 13, 6:19 am, rbwinn wrote: On May 13, 3:01�am, Eric Gisse wrote: On May 13, 1:53�am, Albertito wrote: On May 13, 10:04 am, PD wrote: On May 13, 3:31 am, Albertito wrote: On May 12, 11:21 pm, PD wrote: On May 12, 6:56 am, Albertito wrote: On May 11, 10:07 pm, PD wrote: On May 9, 6:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c �below the third-order term approximation. The power series expansion of r = arctanh(v/c) is � � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't �still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors � � � � � � � �w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities � � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler � � � � � � � �f' = Exp(-r) f , � � � � � � � �where r = arctanh(v/c), can't still be experimentally distinguished from this one � � � � � � � �f' = Exp(-v/c) f There is actually an abundance of these tests. I'll mention one. Muon beamlines are created by allowing charged pions to decay in flight. The pions have momenta such that their speed is close to that of light. Since it is an exoenergetic decay, the muon has extra kinetic energy from the decay. Since the pion's decay mechanism doesn't give a whit whether the pion is in flight when it decays (and in fact, the principle of relativity says the physics of the pion decay has to be the same for pions at rest vs. pions in free flight), we can guess what that kinetic energy profile is for the moving pions by using the distribution for decaying pions at rest. Or, put another way, we can take the velocity distribution of muons in the pion rest frame and boost them to the frame where the pions are moving close to the speed of light. Using rapidities to perform that boost results in a velocity distribution that agrees with direct measurement. The Galilean transformations, on the other hand, predict that the muon velocity would have a peak in the forward direction that exceeds c. Muon time of flight can be measured directly via a triplet of scintillator paddles, and in fact this is routinely done in muon beamlines. The muon speed distribution never exceeds c. PD Interpretation under my model and that provided by SR can't still be experimentally distinguished. Consider this sum of betas, � � v/c = v_p/c + v_mu'/c , � � with � � v_p, pion speed at the moment of its decay wrt muon � � � � � � detector. � � v_mu', �the relative speed of a muon with respect � � � � � � � � �to the point where the pion has decayed Consider this sum of rapidities, �arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c) In both sums, neither v_p nor v_m_u' exceed c. SR assumes the final speed of the muon in the detector must be � � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)), � � � � but it needs also the assumption there is time dilation � � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2), � � � � in order to fit the prediction to the observation. In my model, the final speed of the muon in the detector is � � � � � � �v �= v_p + v_mu', � � � � � � without the assumption of any time dilation at all. If it still is not clear to you, I can repeat, � � � � � � �v/c =(v_p + v_mu')/c And, if you do the math, you find out the left-hand-side of this formula should be greater than 1, given the amount of kinetic energy released in pions decaying to muons. Go ahead, plug in the numbers. What's the problem with finding out the left-hand-side of this formula is greater than 1? Well, for one thing, it doesn't happen. As I mentioned to you, the time of flight of muons is *measured*. We know what the speed is. It's not greater than 1. That's only a problem if you use the relativistic kinetic energy. No, sir. It's a problem with measurement. Of course, v/c = (v_p + v_mu')/c �is greater than 1, because v_p/c is very close to 1, and v_mu'/c can be greater than (1 - v_p/c). You can't keep track of single muons to measure their 'time of flight'. All you can do is to perform statistics and extrapolate to single muons. You can't say a muon called A was emitted by a pion called P at time t=0, and then A was detected at distance R at time t'. The extrapolation for a single muon you perfom is pure nonsense. That's the reason why we find out in particle physics there are three odd generations of charged leptons, and nobody can guess why there are actually those three generations and why they exhibit the rest masses they exhibit. SR is unable to account for the mass of muons and tau leptons with respect to the mass of electrons. Should we call that impossibility a fraud?. If you avoid SR, and try to explain why muon and tau lepton have the rest masses they have wrt that of an electron, you will be able to find out �those three generations of charged leptons, are just exact ratios derived from multiples of c. I'd like to take a moment to point out that you are arguing about particle physics with an actual published particle physicist.- Hide quoted text - - Show quoted text - Ptolemy was an actual published astronomer when Galileo decided that the earth was rotating instead of the sun orbiting the earth. Special relativity has just as many epicycles as Ptolemaic astronomy. Robert B. Winn Where are the epicycles, Robert? Dark matter and dark energy are epicycles. |
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On May 12, 3:01 pm, Albertito wrote:
On May 12, 1:13 pm, Curt Jurgens wrote: On May 12, 1:56 pm, Albertito wrote: On May 11, 10:07 pm, PD wrote: On May 9, 6:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c below the third-order term approximation. The power series expansion of r = arctanh(v/c) is r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler f' = Exp(-r) f , where r = arctanh(v/c), can't still be experimentally distinguished from this one f' = Exp(-v/c) f There is actually an abundance of these tests. I'll mention one. Muon beamlines are created by allowing charged pions to decay in flight. The pions have momenta such that their speed is close to that of light. Since it is an exoenergetic decay, the muon has extra kinetic energy from the decay. Since the pion's decay mechanism doesn't give a whit whether the pion is in flight when it decays (and in fact, the principle of relativity says the physics of the pion decay has to be the same for pions at rest vs. pions in free flight), we can guess what that kinetic energy profile is for the moving pions by using the distribution for decaying pions at rest. Or, put another way, we can take the velocity distribution of muons in the pion rest frame and boost them to the frame where the pions are moving close to the speed of light. Using rapidities to perform that boost results in a velocity distribution that agrees with direct measurement. The Galilean transformations, on the other hand, predict that the muon velocity would have a peak in the forward direction that exceeds c. Muon time of flight can be measured directly via a triplet of scintillator paddles, and in fact this is routinely done in muon beamlines. The muon speed distribution never exceeds c. PD Interpretation under my model and that provided by SR can't still be experimentally distinguished. Consider this sum of betas, v/c = v_p/c + v_mu'/c , with v_p, pion speed at the moment of its decay wrt muon detector. v_mu', the relative speed of a muon with respect to the point where the pion has decayed Consider this sum of rapidities, arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c) I spot a big mistake here, is atanh distributive? tanh seems to be not that tanh(x+y) = {tanh x + tanh y} / {1 + tanh x tanh y} arctanh(v_p/c) = x - v_p/c = tanh(x), arctanh(v_mu'/c) = y - v_mu'/c = tanh(y), arctanh(w/c) = x + y, w/c = tanh(x + y) = (tanh x + tanh y)/(1 + tanh x tanh y) where is the great mistake? You said this **** v/c = v_p/c + v_mu'/c Then you take arctanh of both sides arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c) Which you may not, but arctanh(w/c) = arctanh(v_p/c + v_mu'/c) Which gives something completely else |
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On May 13, 11:34Â*am, Albertito wrote:
On May 13, 8:23 pm, Eric Gisse wrote: On May 13, 6:19 am, rbwinn wrote: On May 13, 3:01�am, Eric Gisse wrote: On May 13, 1:53�am, Albertito wrote: On May 13, 10:04 am, PD wrote: On May 13, 3:31 am, Albertito wrote: On May 12, 11:21 pm, PD wrote: On May 12, 6:56 am, Albertito wrote: On May 11, 10:07 pm, PD wrote: On May 9, 6:44 am, Albertito wrote: Let us consider the simple case of addition of velocities along a straight line. The incompleteness of SR resides in the fact that a speed v can't currently be experimentally distinguished from its rapidity r = arctanh(v/c), for values of that beta = v/c �below the third-order term approximation. The power series expansion of r = arctanh(v/c) is � � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ... Provide references of any experimental test of SR, showing that the rapidity r can be distinguished from its beta v/c, beyond its second-order approximation. Prove at least that the third-order term v*3/3c^3 lives outside the error bars. Since, we can't �still perform such accurate experimental tests, we must conclude the addition of velocities still remains within the euclidean sum of vectors � � � � � � � �w = u + v. That sum can't still be experimentally distinguished from the sum of rapidities � � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c). In addition, we must also conclude that the relativistic Doppler � � � � � � � �f' = Exp(-r) f , � � � � � � � �where r = arctanh(v/c), can't still be experimentally distinguished from this one � � � � � � � �f' = Exp(-v/c) f There is actually an abundance of these tests. I'll mention one. Muon beamlines are created by allowing charged pions to decay in flight. The pions have momenta such that their speed is close to that of light. Since it is an exoenergetic decay, the muon has extra kinetic energy from the decay. Since the pion's decay mechanism doesn't give a whit whether the pion is in flight when it decays (and in fact, the principle of relativity says the physics of the pion decay has to be the same for pions at rest vs. pions in free flight), we can guess what that kinetic energy profile is for the moving pions by using the distribution for decaying pions at rest. Or, put another way, we can take the velocity distribution of muons in the pion rest frame and boost them to the frame where the pions are moving close to the speed of light. Using rapidities to perform that boost results in a velocity distribution that agrees with direct measurement. The Galilean transformations, on the other hand, predict that the muon velocity would have a peak in the forward direction that exceeds c. Muon time of flight can be measured directly via a triplet of scintillator paddles, and in fact this is routinely done in muon beamlines. The muon speed distribution never exceeds c. PD Interpretation under my model and that provided by SR can't still be experimentally distinguished. Consider this sum of betas, � � v/c = v_p/c + v_mu'/c , � � with � � v_p, pion speed at the moment of its decay wrt muon � � � � � � detector. � � v_mu', �the relative speed of a muon with respect � � � � � � � � �to the point where the pion has decayed Consider this sum of rapidities, �arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c) In both sums, neither v_p nor v_m_u' exceed c. SR assumes the final speed of the muon in the detector must be � � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)), � � � � but it needs also the assumption there is time dilation � � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2), � � � � in order to fit the prediction to the observation. In my model, the final speed of the muon in the detector is � � � � � � �v �= v_p + v_mu', � � � � � � without the assumption of any time dilation at all. If it still is not clear to you, I can repeat, � � � � � � �v/c =(v_p + v_mu')/c And, if you do the math, you find out the left-hand-side of this formula should be greater than 1, given the amount of kinetic energy released in pions decaying to muons. Go ahead, plug in the numbers. What's the problem with finding out the left-hand-side of this formula is greater than 1? Well, for one thing, it doesn't happen. As I mentioned to you, the time of flight of muons is *measured*. We know what the speed is.. It's not greater than 1. That's only a problem if you use the relativistic kinetic energy. No, sir. It's a problem with measurement. Of course, v/c = (v_p + v_mu')/c �is greater than 1, because v_p/c is very close to 1, and v_mu'/c can be greater than (1 - v_p/c). You can't keep track of single muons to measure their 'time of flight'. All you can do is to perform statistics and extrapolate to single muons. You can't say a muon called A was emitted by a pion called P at time t=0, and then A was detected at distance R at time t'. The extrapolation for a single muon you perfom is pure nonsense. That's the reason why we find out in particle physics there are three odd generations of charged leptons, and nobody can guess why there are actually those three generations and why they exhibit the rest masses they exhibit. SR is unable to account for the mass of muons and tau leptons with respect to the mass of electrons. Should we call that impossibility a fraud?. If you avoid SR, and try to explain why muon and tau lepton have the rest masses they have wrt that of an electron, you will be able to find out �those three generations of charged leptons, are just exact ratios derived from multiples of c. I'd like to take a moment to point out that you are arguing about particle physics with an actual published particle physicist.- Hide quoted text - - Show quoted text - Ptolemy was an actual published astronomer when Galileo decided that the earth was rotating instead of the sun orbiting the earth. Special relativity has just as many epicycles as Ptolemaic astronomy. Robert B. Winn Where are the epicycles, Robert? Dark matter and dark energy are epicycles. Bullet cluster? Casimir effect? Just because _you do not understand_ does not make them "epicycles". |