SciForums.com > On the Fringe > Alternative Theories > Master Theory (edition 2) PDA View Full Version : Master Theory (edition 2) Post ReplyCreate New ThreadPages : [1] 2 Masterov08-16-11, 12:32 AM1. Master Theory - relativistic theory, because it is the same problem ("Electrodynamics of moving material bodies"), which solved the Einstein. 2. Einstein gave the absoluteness to a cross-scale, but not to the time. In Master Theory time is absolutely (absent "Twin Paradox"), than it differs from Einstein's Special Relativity Theory (SRT) - a cross-scale is absolutely. But this difference has profound implications. (For example: in Master Theory are absent "Twin Paradox" and "Ehrenfest's Paradox".) 3. Author of Master Theory (Alexander V. Masterov) has proved that the problem (that are solved by Einstein) has an infinite number of solutions. SRT - only one of this infinite number of solutions. Where did these solutions? The longitudinal scales of SRT (along the direction of motion) are relative and depends on the relative velocity, whereas (by default) are the absolute transverse (to the direction of motion) scales. The author of Master Theory are assume a regarding of the transverse scope, that let has at its disposal a free parameter. For each value this parameter to can build a individual Theory of Relativity. This theory will be equal footing with Einstein's theory. Among this infinite number of theories discovered one in which time is absolute. The author named it - "Master Theory". 4. As Master Theory is a solution of the same problem (that Einstein's theory), then all the experimental results that confirm SRT - confirmed by Master Theory. The exception is fact of relativistic dilation time, which today has not been experimentally proved. 5. In Master Theory to exist absoluteness of: a. light's speed; b. acceleration (which can be measured indirectly through a mass attached to a spring, for example); c. number Pi (absent "Ehrenfest's Paradox"); 6. Master Theory (as against SRT) is valid in all reference frames (not only in inertial). 7. In Master Theory identified two types of coordinates: real and visual: a. real-coordinates obey the Galilean Transformations, and can be calculated by integrating an acceleration; b. visual-coordinates determined by the properties of EMF and can be calculated from a real-coordinates; c. in Master Theory defined the inverse transformations coordinates (from a visual-coordinates to a real-coordinates). Further exposition of the theory is impossible because of the strange limitations of this forum. Read full text: http : //masterov.qptova.ru/MasterTheory/ Download: http : //masterov.qptova.ru/MasterTheory/MasterTheory.zip Me-Ki-Gal08-16-11, 12:43 AMWe don't got no stinking limitations . We don't need no Limitations . Sock it to us cosmictraveler08-16-11, 01:08 AMThis theory will be equal footing with Einstein's theory. http://4.bp.blogspot.com/-TV7ovGDvaWw/Tht9Vz88emI/AAAAAAAAAJ8/DoZh2XMgyxU/s1600/bullshit+detector+1.gif rpenner08-16-11, 01:45 AM1) Professional mathematicians and scientists don't name theorems and theories respectively after themselves. Their peers award them this recognition by citing their published work. Thus it is a huge abuse of professionalism to break with this convention and it speaks both to the hubris-blinded judgement of the author and the poor communication skills of the author who spends all his energy making claims and no time demonstrating them. Indeed, this breach has been codified as a hallmark of the physics crackpot: Item 25 (http://math.ucr.edu/home/baez/crackpot.html) 2) The author persists in spamming forums with his tripe, and ignores corrections: http://www.physforum.com/index.php?showtopic=29575&st=0 For example: [waitedavid137] In terms of the proper length, L_0 we have L \; = \; L_0 \sqrt{1 - \frac{v^2}{c^2}} \; = \; \gamma^{-1} L_0 and so if \tau \; = \; 2 \frac{L_0}{c}, then t \; = \; \frac{L}{c - v} - \frac{L}{c + v} \; = \; \frac{2 c L}{(c - v)(c + v)} \; = \; 2 \frac{L}{c} \frac{1}{1 - \frac{v^2}{c^2}} \; = \; 2 \gamma^2 \frac{L}{c} \; = \; 2 \gamma \frac{L_0}{c} \; = \; \gamma \tau 3) What special relativity isn't is a magic formula that talks about length contraction or time dilation. Special Relativity is the principle that my laws of physics are the same as your laws of physics even when we are in relative (inertial) motion and the physical observation since 1859 that those laws of physics are non-Euclidean giving rise to a hyperbolic geometry which preserves between any two events the invariant measure: c^2(\Delta t)^2 - ( \Delta x)^2 - ( \Delta y)^2 - ( \Delta z)^2. By showing ignorance of what is physics and what is Special Relativity, the author fails to actual address the subject matter of physics -- precise and useful communications about the behavior of reality. Masterov08-16-11, 03:26 AMYou propose to discuss a of the particular task solution, which found Einstein (for alpha = 0). http : //masterov.qptova.ru/MasterTheory/Formuls/at.gif But such solutions (which Einstein found) infinite much. Master Theory is the subject of our speculation (alpha = 0.5). _________________________ PS You have an error. (In gamma + change to -) Masterov08-16-11, 05:08 AMEinstein did not have a compelling reason to attach a absoluteness to the cross-scales. I set free the cross-scales, that provided to me by a free parameter (alpha). For each value of alpha can build a separate theory of relativity. Such a theory will be equal to Einstein's theory. In Master Theory a time is absolute. It is a characteristic property of Master Theory. In Einstein's theory a cross-scales is absolute. It is a characteristic property of Einstein's theory. rpenner08-16-11, 11:46 AMThe Lorentz transformation is a hyperbolic rotation which uniquely preserves the measure between two events. \begin{pmatrix} c\Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda \begin{pmatrix} c\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix}\cosh \phi & - \sinh \phi & 0 & 0 \\ - \sinh \phi & \cosh \phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \gamma & - \gamma \frac{v}{c} & 0 & 0 \\ - \gamma \frac{v}{c} & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} We can see that this preserves the measure: (c \Delta t')^2 - (\Delta x')^2 = (c \Delta t \; \cosh \phi - \Delta x \; \sinh \phi )^2 - (\Delta x \; \cosh \phi - c \Delta t \; \sinh \phi)^2 = (\cosh^2 \phi - \sinh^2 \phi ) \left( (c \Delta t)^2 - (\Delta x)^2 \right) = (c \Delta t)^2 - (\Delta x)^2 because \cosh^2 \phi - \sinh^2 \phi = 1 = \gamma^2 - \gamma^2 \frac{v^2}{c^2} Further, the product of two Lorentz transforms also preserves this measure. And this measure says the speed of light is constant -- whenever one observer agrees that a hypothetical particle may pass from one event to another at less than, equal to, or faster than the speed of light all other observers agree. Your transform does not preserve the geometry of the universe and therefore doesn't say anything physically interesting about it. Masterov08-16-11, 12:37 PMNo need to slow down time in order to make it all happen. Me-Ki-Gal08-16-11, 12:39 PMThe Lorentz transformation is a hyperbolic rotation which uniquely preserves the measure between two events. \begin{pmatrix} c\Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda \begin{pmatrix} c\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix}\cosh \phi & - \sinh \phi & 0 & 0 \\ - \sinh \phi & \cosh \phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \gamma & - \gamma \frac{v}{c} & 0 & 0 \\ - \gamma \frac{v}{c} & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} We can see that this preserves the measure: (c \Delta t')^2 - (\Delta x')^2 = (c \Delta t \; \cosh \phi - \Delta x \; \sinh \phi )^2 - (\Delta x \; \cosh \phi - c \Delta t \; \sinh \phi)^2 = (\cosh^2 \phi - \sinh^2 \phi ) \left( (c \Delta t)^2 - (\Delta x)^2 \right) = (c \Delta t)^2 - (\Delta x)^2 because \cosh^2 \phi - \sinh^2 \phi = 1 = \gamma^2 - \gamma^2 \frac{v^2}{c^2} Further, the product of two Lorentz transforms also preserves this measure. And this measure says the speed of light is constant -- whenever one observer agrees that a hypothetical particle may pass from one event to another at less than, equal to, or faster than the speed of light all other observers agree. Your transform does not preserve the geometry of the universe and therefore doesn't say anything physically interesting about it. O.K. there is more than one way to skin a cat . The thing that is fascinating to me is incremental time slotting by humans is a human construct where we put things in incremental time slots . The division of the lapsing is set in specific increments. It can be like water you know . That might sound stupid to a mathematician but to a time keeper ? Fucking the death of Me ( Slave to the beat can be quite harder than you think , Me life struggle) Catch the wave is all I can say RJBeery08-16-11, 03:32 PMMasterov, I like the idea of your work but how can you say that time dilation has not been proven? What about atmospheric muon decay? What about atomic clocks flown around the planet compared to those on the ground? Masterov08-17-11, 12:15 AMNo one measures the real speed and transit time. Each measure only the distance and divide this distance by the speed of light. Aircraft's speed is negligible compared to the speed of light. (Especially - a square of their output of divide.) Relativistic effects are negligible. The vibration or other factors could affect the results of the experiments. RJBeery08-17-11, 12:35 AMThe point is that the atomic clocks in flight and the atomic clocks on the ground became out of synchronization by an amount that was predicted by Einstein's explanation. It would be extraordinarily odd (albeit possible I suppose) that vibration of the aircraft happened to affect the clocks in precisely this manner. Masterov08-17-11, 05:34 AMI propose to discuss one of these 'paradoxes'. Rocket's time slow down when it goes from one (A) an inertial reference frame to another (B). Einstein's theory contend that it. If the rocket return back into the original reference frame, rocket's time slow down in relation to B. As a result: we have two a times, wich have different speeds. How this can be possible? For any other theory this paradox would be a death sentence, but not for Einstein's theory. (For some reason.) rpenner08-17-11, 10:50 AMIt's hyperbolic geometry. In Euclidean geometry, the straight line is the shortest path between two points in space. Thus in a triangle, any side is shorter than the sum of the lengths of the other two sides. In hyperbolic geometry, the inertial path is the path of longest duration between two events in space-time that a material body can traverse at a speed slower than the speed of light. Let A, B, C be any three distinct events in space time such that it is possible to get from A to B or C by moving slower than the speed of light and it is also possible to get from C to B slower than the speed of light. In arbitrary inertial coordinates this is: t_A \; < \; t_B \; < \; t_C \\ 0 \; \le \; \sqrt{ ( x_C - x_A )^2 + ( y_C - y_A )^2 + ( z_C - z_A )^2 } \; < \; c(t_C - t_A) \\ 0 \; \le \; \sqrt{ ( x_B - x_A )^2 + ( y_B - y_A )^2 + ( z_B - z_A )^2 } \; < \; c(t_B - t_A) \\ 0 \; \le \; \sqrt{ ( x_C - x_B )^2 + ( y_C - y_B )^2 + ( z_C - z_B )^2 } \; < \; c(t_C - t_B) Since these numbers are all non-negative we may legally square them and preserve the inequalities. 0 \; \le \; ( x_C - x_A )^2 + ( y_C - y_A )^2 + ( z_C - z_A )^2 \; < \; c^2(t_C - t_A)^2 \\ 0 \; \le \; ( x_B - x_A )^2 + ( y_B - y_A )^2 + ( z_B - z_A )^2 \; < \; c^2(t_B - t_A)^2 \\ 0 \; \le \; ( x_C - x_B )^2 + ( y_C - y_B )^2 + ( z_C - z_B )^2 \; < \; c^2(t_C - t_B)^2 Likewise, we may subtract equals from both sides, so we have a statement about measures: 0 \; < \; c^2(t_C - t_A)^2 - ( x_C - x_A )^2 - ( y_C - y_A )^2 - ( z_C - z_A )^2 \; \equiv \; I_{CA} \; \equiv \; c^2 (\Delta \tau_{CA})^2 \\ 0 \; < \; c^2(t_B - t_A)^2 - ( x_B - x_A )^2 - ( y_B - y_A )^2 - ( z_B - z_A )^2 \; \equiv \; I_{BA} \; \equiv \; c^2 (\Delta \tau_{BA})^2 \\ 0 \; < \; c^2(t_C - t_B)^2 - ( x_C - x_B )^2 - ( y_C - y_B )^2 - ( z_C - z_B )^2 \; \equiv \; I_{CB} \; \equiv \; c^2 (\Delta \tau_{CB})^2 So my claim is that a) \Delta \tau_{CA} \ge \Delta \tau_{BA} + \Delta \tau_{CB} , b) that the equality only holds when all three segments are colinear in space-time, c) that \Delta \tau is the elapsed time between any two events in the inertial coordinate system where there is no change of spatial coordinates, thus justifying the common name of "proper time." It is instructive to compute: \left( c^2 (\Delta \tau_{CA})^2 - c^2 (\Delta \tau_{BA})^2 - c^2 (\Delta \tau_{CB})^2 \right)^2 - \left( c^2 (\Delta \tau_{BA} + \Delta \tau_{CB})^2 - c^2 (\Delta \tau_{BA})^2- c^2 (\Delta \tau_{CB})^2 \right)^2 But I don't have the time to typeset the calculation at this time. Tach08-17-11, 11:33 AMBut I don't have the time to typeset the calculation at this time. In 1+1 spacetime, using your notation: c^2(t_C-t_A)^2-(x_C-x_A)^2=c^2(t_C-t_B)^2-(x_C-x_B)^2+c^2(t_B-t_A)^2-(x_B-x_A)^2+2[c(t_C-t_B) c(t_B-t_A)-(x_C-x_B)(x_B-x_A)] Thus: c^2(t_C-t_A)^2-(x_C-x_A)^2>c^2(t_C-t_B)^2-(x_C-x_B)^2+c^2(t_B-t_A)^2-(x_B-x_A)^2 This generalizes trivially to 3+1 dimensions. In other words: 1. Euclidian spacetime ||x+y||<||x||+||y|| 2. Minkowski spacetime ||x+y||>||x||+||y|| OnlyMe08-17-11, 12:40 PMWhile I have some difficulty understanding the specifics of the theory presented in the OP, I assume that it is at least in part a language barrier and translation issue. That said, i don't find what I think I understand of the general intent credible. As to the time dilation issue... Masterov, ... What about atmospheric muon decay? What about atomic clocks flown around the planet compared to those on the ground? No one measures the real speed and transit time. Each measure only the distance and divide this distance by the speed of light. Aircraft's speed is negligible compared to the speed of light. (Especially - a square of their output of divide.) Relativistic effects are negligible. The vibration or other factors could affect the results of the experiments. The point is that the atomic clocks in flight and the atomic clocks on the ground became out of synchronization by an amount that was predicted by Einstein's explanation. It would be extraordinarily odd (albeit possible I suppose) that vibration of the aircraft happened to affect the clocks in precisely this manner. Masterov, does present a valid point of view above, though not a generally accepted one. (Generally accepted is not always the same as accurate!) The muon decay proof is the clumsiest, as it cannot be directly observed. It is by it's very nature dependent upon the validity of the principles it "proves". The reasoning is somewhat circular in that if time dilation occurs as described in SR and to some extent in GR, muon decay maybe a proof. If not muon decay under such conditions may be due to some as yet to be understood phenomena/interaction. The atomic clock in motion issue is also not a proof. Even though, the assumed time dilation observed in atomic clocks in motion both in aircraft and GPS satellites, appears to be consistent with Einstein's predictions, here predominantly an affect of SR, as opposed to a counter affect predicted by GR (which in either case would be negligible), the Lorentz transformations Einstein applies were repurposed theirselves from earlier work of Lorentz and Fitzgerald, originally developed for other purposes. While it is true that we observe a variation in the rate of electron transitions, which are the basis of atomic clocks, in such cases, we do not really know enough about space itself and even inertia, to know with certainty that the difference is the result of the mechanisms suggested by either SR and/or GR. In the case of GPS satellites, clocks are synchronized not by an application of relativistic mathematical means. Instead they are just routinely corrected to match time as define by similar ground based clocks. This is most likely because it is just far easier. However, until and unless we carry out a mathematically based relativistic synchronization that holds up over time, we can only assume the origin of the change in transition rate, to be consistent with not the product of relativistic effects. One further note. The basis for time dilation in both situations is derived from the Lorentz transformations, which also predicts length contractions. A real length contraction has never been observed, or measured by experiment. That does not mean it does not exist, as our means to measure lengths, is not as fine as our measurement of time relative to electron transitions. It may just be that we await the technology to better examine and test both, in a more objective manner and in finer detail. Tach08-17-11, 01:59 PMIn the case of GPS satellites, clocks are synchronized not by an application of relativistic mathematical means. Instead they are just routinely corrected to match time as define by similar ground based clocks. Nonsense, you do not understand the basic principles of GPS, see here (http://relativity.livingreviews.org/Articles/lrr-2003-1/). The periodic adjustments are due to ephemerides, the frequency calculation is done using GR effects (for a list, check the reference). The rest of the fringe stuff you posted about time dilation is just as bad. Anyways, time dilation is tested directly thought the experiments on transverse Doppler effect. Masterov08-17-11, 02:03 PM2OnlyMe I translated your post in two different programs, and readed twice. I caught the general sense of what you have said, but not sure of the details. (My English is poor.) I got the main thing: give cause for a specialists to think. I can spent a lot of time on the development of Master Theory, without having to have the information, having scant experience of relativity and electrodynamics. And most importantly: I have no desire that to do so. I know that some of my former colleagues are making efforts to rewrite Maxwell's equations in the newly discovered facts. But I do not know the condition of the their work. AlexG08-17-11, 02:14 PMI can spent a lot of time on the development of Master Theory, without having to have the information, having scant experience of relativity and electrodynamics That your 'theory' is built on ignorance is obvious from the first post. Masterov08-17-11, 02:39 PMThese words I constantly hear from people who are uncapable demonstrate their skills. Whether you're an exception to this rule? OnlyMe08-17-11, 03:42 PMNonsense, you do not understand the basic principles of GPS, see here (http://relativity.livingreviews.org/Articles/lrr-2003-1/). The periodic adjustments are due to ephemerides, the frequency calculation is done using GR effects (for a list, check the reference). The rest of the fringe stuff you posted about time dilation is just as bad. Anyways, time dilation is tested directly thought the experiments on transverse Doppler effect. Tach, I think you missed the point of my post. A quote from the abstract you linked pretty much says it. This paper discusses the conceptual basis, I would add to that the theoretical basis. I did not read the whole paper but in searching through I found no reference to the actual method used to synchronize satellite clocks with ground based clocks. The theory is absolutely based on relativistic effects. The mechanics of actually maintaining some measure of synchronization? I have never found any reference that put the actual process as a computation of the relativistic divergence. It could not be done on a satellite or a GPS receiver as that calculation would require a significant time on a super computer. Those kinds of resources are not committed to the everyday operation of the GPS system. I don't see where experiments involving the transverse Doppler effect involves time dilation. Even so, as it, as far as I know, involves astronomical observations, it is subject to the same issues that the muon decay reference would be. They are both dependent on assumptions that the underlying theory and models are accurate. It any case, I never said at any point that time dilation and even length contraction do not occur. All I said is that the evidence we currently have supporting either or both is not such that it excludes all other possible known or unknown causes. And that for that reason I began with, Masterov, does present a valid point of view above, though not a generally accepted one. (Generally accepted is not always the same as accurate!) Sometimes it is better to read a post within its larger context than to respond to isolated statements. Incidently, I have not posted what I believe... Tach08-17-11, 05:47 PMTach, I think you missed the point of my post.. No, I didn't, you posted fringe BS and tried to pass it as science. I don't see where experiments involving the transverse Doppler effect involves time dilation. Then you need to get educated. Stop posting BS and get an education. rpenner08-17-11, 06:02 PMI propose to discuss one of these 'paradoxes'. Please don't get distracted. Please continue the conversation on this topic. Rocket's time slow down when it goes from one (A) an inertial reference frame to another (B). Einstein's theory contend that it. If the rocket return back into the original reference frame, rocket's time slow down in relation to B. As a result: we have two a times, wich have different speeds. How this can be possible? For any other theory this paradox would be a death sentence, but not for Einstein's theory. (For some reason.) It's hyperbolic geometry. ... So my claim is that a) \Delta \tau_{CA} \ge \Delta \tau_{BA} + \Delta \tau_{CB} , Since I am emphasizing hyperbolic geometry, I choose to parametrize the 12 coordinates thus: \begin{eqnarray} t_C & = & t_B \; + \; \xi_{CB} \, \cosh \eta_{CB} \\ x_C & = & x_B \; + \; c \xi_{CB} \, \sinh \eta_{CB} \, \cos \phi_{CB} \, \cos \theta_{CB}\\ y_C & = & y_B \; + \; c \xi_{CB} \, \sinh \eta_{CB} \, \cos \phi_{CB} \, \sin \theta_{CB}\\ z_C & = & z_B \; + \; c \xi_{CB} \, \sinh \eta_{CB} \, \sin \phi_{CB} \\ t_B & = & t_A \; + \; \xi_{BA} \, \cosh \eta_{BA} \\ x_B & = & x_A \; + \; c \xi_{BA} \, \sinh \eta_{BA} \, \cos \phi_{BA} \, \cos \theta_{BA}\\ y_B & = & y_A \; + \; c \xi_{BA} \, \sinh \eta_{BA} \, \cos \phi_{BA} \, \sin \theta_{BA}\\ z_B & = & z_A \; + \; c \xi_{BA} \, \sinh \eta_{BA} \, \sin \phi_{BA} \end{eqnarray} Since I_{CB} = c^2 \xi_{CB}^2 \, \left( \cosh^2 \eta_{CB} - \sinh^2 \eta_{CB} \, \left( \cos^2 \phi_{CB} \, \left( \cos^2 \theta_{CB} + \sin^2 \theta_{CB} \right) + \sin^2 \phi_{CB} \right) \right) = c^2 \xi_{CB}^2 and similarly I_{BA} = c^2 \xi_{BA}^2, we automatically satisfy all of the inequalities related to traveling slower than the speed of light. And we guarantee t_A \; < \; t_B \; < \; t_C by requiring \xi_{BA} > 0 \; , \quad \xi_{CB} > 0 . So \begin{eqnarray} I_{CA} & = & c^2 \left( \left( \xi_{CB} \cosh \eta_{CB} + \xi_{BA} \cosh \eta_{BA} \right)^2 \right. \\ & & \quad - \left( \xi_{CB} \sinh \eta_{CB} \cos \phi_{CB} \cos \theta_{CB} + \xi_{BA} \sinh \eta_{BA} \cos \phi_{BA} \cos \theta_{BA} \right)^2 \\ & & \quad - \left( \xi_{CB} \sinh \eta_{CB} \cos \phi_{CB} \sin \theta_{CB} + \xi_{BA} \sinh \eta_{BA} \cos \phi_{BA} \sin \theta_{BA} \right)^2 \\ & & \quad \left. - \left( \xi_{CB} \sinh \eta_{CB} \sin \phi_{CB}+ \xi_{BA} \sinh \eta_{BA} \sin \phi_{BA} \right)^2 \right) \\ & = & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \left( \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \cos \phi_{CB} \cos \theta_{CB} \sinh \eta_{BA} \cos \phi_{BA} \cos \theta_{BA} \right. \\ & & \quad \quad \left. - \sinh \eta_{CB} \cos \phi_{CB} \sin \theta_{CB} \sinh \eta_{BA} \cos \phi_{BA} \sin \theta_{BA} - \sinh \eta_{CB} \sin \phi_{CB} \sinh \eta_{BA} \sin \phi_{BA} \right) \\ & = & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \left( \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \left( \cos \phi_{CB} \cos \phi_{BA} \left( \cos \theta_{CB} \cos \theta_{BA} + \sin \theta_{CB} \sin \theta_{BA} \right) + \sin \phi_{CB} \sin \phi_{BA} \right) \right) \\ & = & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \left( \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \left( \cos \phi_{CB} \cos \phi_{BA} \cos ( \theta_{CB} - \theta_{BA} ) + \sin \phi_{CB} \sin \phi_{BA} \right) \right) \\ & \ge & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \left( \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \left( \cos \phi_{CB} \cos \phi_{BA} + \sin \phi_{CB} \sin \phi_{BA} \right) \right) \\ & = & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \left( \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \cos ( \phi_{CB} - \phi_{BA} ) \right) \\ & \ge & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \left( \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \right) \\ & = & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \cosh ( \eta_{CB} - \eta_{BA} ) \\ & \ge & I_{CB} + I_{BA} + 2 c^2 \xi_{CB} \xi_{BA} \\ & = & c^2 \left( \xi_{CB} + \xi_{BA} \right) ^2 \\ & = & c^2 \left( \tau_{CB} + \tau_{BA} \right) ^2 \end{eqnarray} b) that the equality only holds when all three segments are colinear in space-time, , Note that for equality to hold across all of the previous statement, we need the following three conditions to hold: \eta_{CB} = \eta_{BA} \\ \cos( \phi_{CB} - \phi_{BA} ) = 1 \\ \cos ( \theta_{CB} - \theta_{BA} ) = 1 which is equivalent to saying that the velocity vector corresponding to both intervals is the same. c) that \Delta \tau is the elapsed time between any two events in the inertial coordinate system where there is no change of spatial coordinates, thus justifying the common name of "proper time.", Rotate by -\theta, Rotate by -\phi then use a Lorentz transform corresponding to hyperbolic angle -\eta. Done. It is instructive to compute: \left( c^2 (\Delta \tau_{CA})^2 - c^2 (\Delta \tau_{BA})^2 - c^2 (\Delta \tau_{CB})^2 \right)^2 - \left( c^2 (\Delta \tau_{BA} + \Delta \tau_{CB})^2 - c^2 (\Delta \tau_{BA})^2- c^2 (\Delta \tau_{CB})^2 \right)^2 It's also always greater than or equal to zero. AlexG08-17-11, 06:19 PMThese words I constantly hear from people who are uncapable demonstrate their skills. Whether you're an exception to this rule? You constantly hear that because you're an idiot. OnlyMe08-17-11, 06:53 PMNo, I didn't, you posted fringe BS and tried to pass it as science. Then you need to get educated. Stop posting BS and get an education. This sounds like educated discussion, to you? AlexG08-17-11, 06:55 PMThis sounds like educated discussion, to you? You can't have an educated discussion unless both sides are educated. Otherwise, it's arguing with ignorance. OnlyMe08-17-11, 07:20 PMYou can't have an educated discussion unless both sides are educated. Otherwise, it's arguing with ignorance. True, and yet! It often seems to me that what is passed off as discussion even educated discussion, often represents no more than the lowest form of argument, in which the "debate" becomes a personal exchange with no intent or possibility for the exchange of ideas, perspective or information. But this now begins to diverge from the topic of the thread and adds nothing to anyone's understanding of practically anything. rpenner08-17-11, 08:20 PMNote that for equality to hold across all of the previous statement, we need the following three conditions to hold: \eta_{CB} = \eta_{BA} \\ \cos( \phi_{CB} - \phi_{BA} ) = 1 \\ \cos ( \theta_{CB} - \theta_{BA} ) = 1 which is equivalent to saying that the velocity vector corresponding to both intervals is the same. Actually there are some other cases, but correspond to some sign flipping, such as anti-parallel directions and flipping the sign of eta. Yet another case is associated with the coordinate singularity of spherical coordinates (i.e. East and West don't matter at the North Pole -- there is only South). Tach08-17-11, 09:22 PMSince I am emphasizing hyperbolic geometry, I choose to parametrize the 12 coordinates thus: You are making the solution more complicated than it needs (http://www.sciforums.com/showpost.php?p=2798914&postcount=15) to be. Me-Ki-Gal08-17-11, 09:28 PMActually there are some other cases, but correspond to some sign flipping, such as anti-parallel directions and flipping the sign of eta. Yet another case is associated with the coordinate singularity of spherical coordinates (i.e. East and West don't matter at the North Pole -- there is only South). Oh that is good !! I am going to use that . Please don't sue Me rpenner08-18-11, 02:58 PMYou are making the solution more complicated than it needs (http://www.sciforums.com/showpost.php?p=2798914&postcount=15) to be. Perhaps you would care to expand on that by displaying how your calculation results in the requested inequality in \tau not \tau^2. I can rewrite the section in parentheses in terms of just cosh and cos. \frac{1}{4} \left( \cosh ( \eta_{CB} - \eta_{BA} ) \left( \cos( \phi_{CB} - \phi_{BA} ) \left( \cos( \theta_{CB} - \theta_{BA} ) + 1 \right) + \cos( \phi_{CB} + \phi_{BA} ) \left( \cos( \theta_{CB} - \theta_{BA} ) - 1 \right) + 2 \right) \right. \\ \quad \quad \quad \left. - \cosh ( \eta_{CB} + \eta_{BA} ) \left( \cos( \phi_{CB} - \phi_{BA} ) \left( \cos( \theta_{CB} - \theta_{BA} ) + 1 \right) + \cos( \phi_{CB} + \phi_{BA} ) \left( \cos( \theta_{CB} - \theta_{BA} ) - 1 \right) - 2 \right) \right) \\ = \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \left( \cos \phi_{CB} \cos \phi_{BA} \left( \cos \theta_{CB} \cos \theta_{BA} + \sin \theta_{CB} \sin \theta_{BA} \right) + \sin \phi_{CB} \sin \phi_{BA} But we still have a ugly breakdown of this into cases to prove it is always >= 1. Tach08-18-11, 07:12 PMPerhaps you would care to expand on that by displaying how your calculation results in the requested inequality in \tau not \tau^2. Easy, using your notation: c^2(t_C-t_A)^2-(x_C-x_A)^2=c^2(t_C-t_B)^2-(x_C-x_B)^2+c^2(t_B-t_A)^2-(x_B-x_A)^2+2[c(t_C-t_B) c(t_B-t_A)-(x_C-x_B)(x_B-x_A)] Thus: (c d \tau_{CA})^2=(c d \tau_{CB})^2+(c d \tau_{BA})^2+2c^2d \tau_{CB}c d \tau_{BA}-2c^2d \tau_{CB}c d \tau_{BA}+2c^2(t_C-t_B)(t_B-t_A)-2(x_C-x_B)(x_B-x_A) You are left with proving that: -2c^2d \tau_{CB}c d \tau_{BA}+2c^2(t_C-t_B)(t_B-t_A)-2(x_C-x_B)(x_B-x_A)>0 Rearranging: c^2(t_C-t_B)(t_B-t_A)-(x_C-x_B)(x_B-x_A)>\sqrt{c^2(t_C-t_B)^2-(x_C-x_B)^2} \sqrt{c^2(t_B-t_A)^2-(x_B-x_A)^2} .....which is done by squaring both sides (since they are both positive). But we still have a ugly breakdown of this into cases to prove it is always >= 1. I don't think so, see the above solution. rpenner08-19-11, 12:10 PMAh .. a refresher on the law of cosines in hyperbolic and spherical geometry and we are done. \begin{eqnarray} & &\cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \left( \cos \phi_{CB} \cos \phi_{BA} \left( \cos \theta_{CB} \cos \theta_{BA} + \sin \theta_{CB} \sin \theta_{BA} \right) + \sin \phi_{CB} \sin \phi_{BA} \right) \\ & = & \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \left( \sin (\frac{\pi}{2} - \phi_{CB} ) \sin (\frac{\pi}{2} - \phi_{BA}) \cos (\theta_{CB}-\theta_{BA}) + \cos (\frac{\pi}{2} - \phi_{CB} ) \cos (\frac{\pi}{2} -\phi_{BA} ) \right) \\ & = & \cosh \eta_{CB} \cosh \eta_{BA} - \sinh \eta_{CB} \sinh \eta_{BA} \cos \Phi_{\textrm{spherical}} \\ & = & \cosh H_{\textrm{hyperbolic}} \\ & \ge & 1 \end{eqnarray} Masterov08-22-11, 08:17 AMRelativistic Maxwell Since any differential operator is local (in close proximity), and (consequently) a finiteness of a propagation speed of the fields (in infinitely small distance) can not make adjustments to the equation. Therefore, Maxwell's equations in differential form in the context of the Master Theory would not be changed in comparison with the classics. Namely: http: //masterov.qptova.ru/MasterTheory/Formuls/MaxvellDiff.gif But integral Maxwell (because of a nonlocality of a integral operators) will have a fundamentally different (non-classical, but - relativistic) form. This is explained by the fact that Stokes' and Gauss' theorems: http: //masterov.qptova.ru/MasterTheory/Formuls/Stox.gif and http: //masterov.qptova.ru/MasterTheory/Formuls/Gaus.gif valid only for stationary fields, or fields, the speed of which is infinite. Application of these theorems (for the output of the integral form of Maxwell's equations) can not be correct in the relativistic case. Masterov08-22-11, 09:21 AMOnlyMe, please, do not respond to AlexG's messages. His are not worth one's salt. OnlyMe08-22-11, 09:43 AMOnlyMe, please, do not respond to AlexG's messages. His are not worth one's salt. I make a choice to treat everyone's opinions with the same respect and consideration. Even when I do not agree with either the facts or sentiment of their post. I would hope that others extend the same latitude to myself. Alex and I do not always see things the same way, but generally I find his contributions to be thought provoking, which is why I come to these discussions in the first place. That said, our last exchange (between Alex and myself) and this post are off topic. Though I often disagree I have no problems with Alex's post. I don't think I agree with much of what you have presented.., still I find the discussion again thought provoking... Anything that makes me think it over, is a good thing. Masterov08-22-11, 10:10 AMAnything that makes me think it over, is a good thing.Your words are right, but it can not be attributed to the offensive remarks, in which are absent arguments. rpenner08-22-11, 08:19 PMRelativistic Maxwell Since any differential operator is local (in close proximity), and (consequently) a finiteness of a propagation speed of the fields (in infinitely small distance) can not make adjustments to the equation. Therefore, Maxwell's equations in differential form in the context of the Master Theory would not be changed in comparison with the classics. Namely: \vec{\nabla} \vec{E} = \frac{\rho}{\epsilon \epsilon_0} \, ; \; \vec{\nabla} \vec{B} = 0 \, ; \; \vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{ \partial t } \, ; \; \vec{\nabla} \times \vec{B} = \mu \mu_0 \left( \vec{j} + \epsilon \epsilon_0 \frac{ \partial \vec{E} }{ \partial t } \right) \, ; \; \vec{E} = \sigma \vec{j} But integral Maxwell (because of a nonlocality of a integral operators) will have a fundamentally different (non-classical, but - relativistic) form. This is explained by the fact that Stokes' and Gauss' theorems: \oint _L \vec{A} d\vec{l} = \int _S \vec{\nabla} \times \vec{A} \; d\vec{s} and \int _S \vec{A} d\vec{s} = \int _V \vec{\nabla} \vec{A} dV valid only for stationary fields, or fields, the speed of which is infinite. Application of these theorems (for the output of the integral form of Maxwell's equations) can not be correct in the relativistic case. Corrections: Your conventions for dot product and integrals are a little unclear. Your Maxwell equations are for linear, isotropic matter where \epsilon = 1 + \chi_e is the relative permittivity and \mu = 1 + \chi_m is the relative permeability. This differs from the common treatment where absolute permittivities and permeabilities are uses and differs from the microscopic fundamental theory is that a linear theory for matter is used rather than modeling the matter in EM theory itself. For example, there is no expectation of isotropism in moving matter. So, to be clear, the currents and charges are free charges and not the charges inside of atoms. So \vec{\nabla} \cdot \vec{D} = \vec{\nabla} \cdot ( \epsilon_{\textrm{rel}} \epsilon_0 \vec{E} ) = \rho_{\textrm{free}}, \vec{\nabla} \cdot \vec{B} = 0, \vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{ \partial t }, \vec{\nabla} \times \vec{H} = \vec{\nabla} \times ( \frac{1}{ \mu_{\textrm{rel}} \mu_0} \vec{B} ) = \vec{j}_{\textrm{free}} + \epsilon_{\textrm{rel}} \epsilon_0 \frac{ \partial \vec{E} }{ \partial t } are not fundamental, but approximations to Maxwell's equations in isotropic, linear matter. Likewise, Ohm's law, is for isotropic, linear matter where time-varying currents are unimportant and is written as \vec{j}_{\textrm{free}} = \sigma \vec{E} when \sigma is conductivity. When there are no charges bound to matter, we have Maxwell's equations in this form: \vec{\nabla} \cdot ( \epsilon_0 \vec{E} ) = \rho \, ; \; \vec{\nabla} \cdot \vec{B} = 0 \, ; \; \vec{\nabla} \times \vec{E} = - \frac{ \partial \vec{B} }{ \partial t } \, ; \; \vec{\nabla} \times ( \frac{1}{ \mu_0} \vec{B} ) = \vec{j} + \epsilon_0 \frac{ \partial \vec{E} }{ \partial t } Instead of tracking 6 parameters throughout space and time, we can track 4 parameters if we write the above in terms of the scalar potential \varphi and the vector potential, \vec{A}. Then we define \vec{E} = - \vec{\nabla} \varphi - \frac{ \partial \vec{A} }{ \partial t } and \vec{B} = \vec{\nabla} \times \vec{A}. Thus by definition, \vec{\nabla} \cdot \vec{B} = \vec{\nabla} \cdot \vec{\nabla} \times \vec{A} = 0 and \vec{\nabla} \times \vec{E} = - \vec{\nabla} \times \vec{\nabla} \varphi - \vec{\nabla} \times \frac{ \partial \vec{A} }{ \partial t } = 0 - \frac{ \partial \vec{\nabla} \times \vec{A} }{ \partial t } = - \frac{ \partial \vec{B} }{ \partial t }. So the physics content of Maxwell's equations reduce to these two: \vec{\nabla} \cdot \vec{E} = - \vec{\nabla} \cdot \vec{\nabla} \varphi - \vec{\nabla} \cdot \frac{ \partial \vec{A} }{ \partial t } = - \nabla^2 \varphi - \frac{ \partial \vec{\nabla} \cdot \vec{A} }{ \partial t }= \frac{\rho}{\epsilon_0} and \vec{\nabla} \times \vec{B} - \epsilon_0 \mu_0 \frac{ \partial \vec{E} }{ \partial t } = \vec{\nabla} \times \vec{\nabla} \times \vec{A} + \frac{1}{c^2} \frac{ \partial }{ \partial t } \left( \vec{\nabla} \varphi + \frac{ \partial \vec{A} }{ \partial t } \right) = \vec{\nabla} \vec{\nabla} \cdot \vec{A} - \nabla^2 \vec{A} +\vec{\nabla} \frac{1}{c^2} \frac{ \partial \varphi }{ \partial t } + \frac{1}{c^2} \frac{ \partial^2 \vec{A} }{ \partial t^2 } = \vec{\nabla} \left( \vec{\nabla} \cdot \vec{A} + \frac{1}{c^2} \frac{ \partial \varphi }{ \partial t } \right) + \frac{1}{c^2} \frac{ \partial^2 \vec{A} }{ \partial t^2 } - \nabla^2 \vec{A} = \mu_0 \vec{j}. Although we have less variables, by using the potentials we introduce a choice of gauge which does not affect the physics. So I may freely choose the Lorentz Gauge: \vec{\nabla} \cdot \vec{A} + \frac{1}{c^2} \frac{ \partial \varphi }{ \partial t } = 0. So we have, \frac{1}{c^2} \frac{ \partial^2 \vec{A} }{ \partial t^2 } - \nabla^2 \vec{A} = \mu_0 \vec{j} and - \nabla^2 \varphi - \frac{ \partial \vec{\nabla} \cdot \vec{A} }{ \partial t } = - \nabla^2 \varphi + \frac{1}{c^2} \frac{ \partial^2 \varphi }{ \partial t^2 } = \frac{\rho}{\epsilon_0} Finally, we can go a step further if we work in four dimensions, not three. Then we have a single equation in four-vectors: \partial^2 A^{k} = \left( \frac{1}{c^2} \frac{ \partial^2 }{ \partial t^2 } - \nabla^2 \right) \begin{pmatrix} \frac{1}{c} \varphi \\ \vec{A} \end{pmatrix} = \mu_0 \begin{pmatrix} c \rho \\ \vec{j} \end{pmatrix} = \mu_0 J^{k} Furthermore, this last expression begins to suggest the concept of Lorentz covariance rendering the laws of physics invariant. All of this is carried in a standard electromagnetism course and all of it is available in Wikipedia pages. I have labored to make the connections explicit. Masterov08-23-11, 12:49 AMСorrected: http: //masterov.qptova.ru/MasterTheory/Formuls/MaxvellDiff.gif Masterov08-25-11, 12:30 AMSuppose we have a closed-loop surface S, which have a characteristic size L. Then the integral expressions of the classical Maxwell's Equations: http: //masterov.qptova.ru/MasterTheory/Formuls/Maxvell1.gif valid only in cases where a characteristic time of variation of the fields (and electric charge) in these equations are much smaller than L/c. A similar can be said about the other pair of expressions. http: //masterov.qptova.ru/MasterTheory/Formuls/Maxvell2.gif rpenner08-25-11, 02:13 AMSuppose we have a closed-loop surface S, which have a characteristic size L. Then the integral expressions of the classical Maxwell's Equations: \oint _S \vec{D} d\vec{s} = Q \, ; \; \oint _S \vec{B} d\vec{s} = 0 valid only in cases where a characteristic time of variation of the fields (and electric charge) in these equations are much smaller than L/c. A similar can be said about the other pair of expressions. \oint _L \vec{E} d\vec{l} = - \frac{d }{dt} \int _S \vec{B} d\vec{s} \oint _L \vec{H} d\vec{l} = I + \frac{d }{dt} \int _S \vec{D} d\vec{s} That's just not the way it works. \begin{eqnarray} \oint _S \vec{D} \cdot d\vec{s} & = & \varepsilon_0 \iint _{\partial V} \vec{E} \cdot d\vec{s} \\ & = & \varepsilon_0 \iint _{\partial V} \left( - \vec{\nabla} \varphi - \frac{ \partial \vec{A} }{ \partial t } \right) \cdot d\vec{s} \\ & = & \varepsilon_0 \iiint _V \vec{\nabla} \cdot \left( - \vec{\nabla} \varphi - \frac{ \partial \vec{A} }{ \partial t } \right) dv \\ & = & \varepsilon_0 \iiint _V \left( - \nabla^2 \varphi - \frac{ \partial }{ \partial t } ( \vec{\nabla} \cdot \vec{A} ) \right) dv \\ & = & \varepsilon_0 \iiint _V \left( \frac{1}{c^2} \frac{ \partial^2 }{ \partial t^2 } \varphi - \nabla^2 \varphi \right) dv \\ & = & \varepsilon_0 \iiint _V \frac{\rho}{\varepsilon_0 } dv \\ & = & \iiint _V \rho dv = Q \end{eqnarray} The differential laws work in integral form for the same reason that the continuity equation for current is related to the time integral of charge. \frac{\partial \rho}{\partial t} = - \vec{\nabla} \cdot \vec{j} Nothing changes for the volume integration. What changes is when you have two comoving coordinate systems, the limits of integration also change. But we didn't begin to address this and how Maxwell's equations hold up in this form, because you didn't argue for your claim. You merely asserted it. http://ru.wikipedia.org/wiki/Лоренц-ковариантность Masterov08-25-11, 01:30 PMThis is true for quasi-static fields only. Gauss's and Stokes's theorems are correctly applies to quasi-static fields only. These theorems do not take into account the time delay from the V-volume to the S-surface. The result of S-integration will depend on the shape of the surface in the general case. rpenner08-25-11, 03:12 PMThis is true for quasi-static fields only. Where is the fact-based argument to support this? You have asserted it, but not given a reason to believe it. Gauss's and Stokes's theorems are correctly applies to quasi-static fields only.Where is the fact-based argument to support this? You have asserted it, but not given a reason to believe it. These theorems do not take into account the time delay from the V-volume to the S-surface.Where is the fact-based argument to support this? You have asserted it, but not given a reason to believe it. How can there be a "time delay" from the V-volume to the S-surface when the S-surface is the limits of the V-volume, S = \partial V, and both are arbitrary and imaginary volumes and surfaces conjured by the mind of the physicist, not the laws of nature? The result of S-integration will depend on the shape of the surface in the general case. Absolutely I agree if we were talking about the general case, but we aren't talking about the general case; we are talking about the surface integral of a flux and the volume integral of a divergence of that flux. Where is the fact-based argument to support your viewpoint? You have asserted it, but not given a reason to believe it. Masterov08-25-11, 11:54 PMIt is necessary to recall the proofs of (Gauss's and Stokes's) theorems. They use a static field or fields, a speed of which is infinite. Spend a thought experiment: 1. Electric current in a conductor changed. 2. Do L-contour need time to learn about this event, if L-contour located at some distance from the conductor? Masterov08-26-11, 11:45 AMFor sample If: Real coordinates: http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple1.png Then: Visual coordinates: http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple2.png Next: Visual speed (with Dopler's effect): http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple3.png http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple4.png Masterov08-26-11, 12:07 PMCommon case Non inertial reference frame Real coordinates: http: //masterov.qptova.ru/MasterTheory/Formuls/comn1.png Real speed: http: //masterov.qptova.ru/MasterTheory/Formuls/comn2.png http: //masterov.qptova.ru/MasterTheory/Formuls/comn3.png Visual speed: http: //masterov.qptova.ru/MasterTheory/Formuls/comn4.png Visual coordinates: http: //masterov.qptova.ru/MasterTheory/Formuls/comn5.png Visual time: http: //masterov.qptova.ru/MasterTheory/Formuls/comn6.png Inertial reference frame Real coordinates: http: //masterov.qptova.ru/MasterTheory/Formuls/comn7.png Visual coordinates: http: //masterov.qptova.ru/MasterTheory/Formuls/comn8.png http: //masterov.qptova.ru/MasterTheory/Formuls/comn9.png - Dopler's effect. http: //masterov.qptova.ru/MasterTheory/Formuls/comna.png http: //masterov.qptova.ru/MasterTheory/Formuls/comnb.png OnlyMe08-26-11, 12:16 PMmasterov Your are not posting the URLs as links and the server continues to come up as not found. The last two post are useless if their content is unavailable. Masterov08-26-11, 01:10 PMRemove(delite) a spacer after "http:" http:?//masterov.qptova.ru/MasterTheory/Formuls/comna.png OnlyMe08-26-11, 01:36 PMRemove(delite) a spacer after "http:" http:?//masterov.qptova.ru/MasterTheory/Formuls/comna.png Ah! An obvious over site on my part. funkstar08-26-11, 06:05 PMThere's a reason that you're not allowed to post links and images before you've accrued a certain number of posts on the site. I'll leave you to speculate on the particulars... OnlyMe08-26-11, 08:22 PMThere's a reason that you're not allowed to post links and images before you've accrued a certain number of posts on the site. I'll leave you to speculate on the particulars... Yea, I knew that I had just missed the dead space in his URL until he pointed out the ? In my cut and paste. It would be very difficult to follow the math from a series of images like that. Too much work for me. I have to work to keep up with the stuff these days when it is properly formatted .... and fully explained. rpenner08-27-11, 12:30 AMIt is necessary to recall the proofs of (Gauss's and Stokes's) theorems. They use a static field or fields, a speed of which is infinite. That's an issue if you use only part of Maxwell's equations. With all of Maxwell's equations, a time-varying charge requires the existance of currents which are the source of magnetic fields. And the net effect of electromagnetism on physical measurements is the sum of the effects of electric and magnetic fields. Another way to see this is that electromagnetism has never predicted that signals travel with inifinite velocity, so your claim that E and B respond instantaneously to the charge and current distribution ignores that the charge and current distribution are physically constrained to vary in limited manner. A third way to see this is to use the 4-dimensional version of the divergence theorem, which gives us: \iiint _{\partial \Omega} \partial_k A^m d\omega^k = \iiint \int_{\Omega} \partial^k \partial_k A^m dV dt = \iiint \int_{\Omega} \partial^k \partial_k A^m dV dt = \iiint \int_{\Omega} \partial^2 A^m dV dt = \mu_0 \iiint \int _{\Omega} J^m dV dt But a college level education in electromagnetism is usually only gained by people going on to actually get a degree in physics or an advanced degree in electrical engineering. So your ignorance is understandable. Your motivations to claim expertise in this field is not understood. Spend a thought experiment: 1. Electric current in a conductor changed. 2. Do L-contour need time to learn about this event, if L-contour located at some distance from the conductor? Electric current flows in circuits. So lets make this concrete. Imagine an eternal loop of charge of radius R, which is large, and linear charge density of \ell = \frac{Q}{2 \pi R}. Before time t=0, the loop is motionless. After time t=0, the loop rigidly rotates with linear velocity v, which gives a current of I = v \ell = \frac{Q v}{2 \pi R}. Or using the step function, I(t) = \frac{Q v}{2 \pi R} \theta(t) Thus Ampère's circuital law suggests that for a loop of constant radius r just outside the charges wire (r << R), that the magnetic field is B = \frac{\mu_0}{2 \pi r} I(t) = \frac{\mu_0 Q v}{4 \pi^2 R r} \theta(t) But this is not Maxwell's equation, but an equation of magnetostatics misapplied to a case where the current changes. Maxwell realized that the Ampère's circuital law was wrong, and added a term he called the displacement current to it. The displacement current is not really a current, but it's the standard name for this term. So since we have a lot of heavy lifting to do, lets review: \vec{E}(\vec{r},t) = -\vec{\nabla}\varphi(\vec{r},t) - \frac{\partial}{\partial t}\vec{A}(\vec{r},t) \, ; \; \vec{B}(\vec{r},t) = \vec{\nabla} \times \vec{A}(\vec{r},t) \, ; \; \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \varphi(\vec{r},t) - \nabla^2 \varphi(\vec{r},t) \varphi(\vec{r},t) = \frac{\rho(\vec{r},t)}{\varepsilon_0} \, ; \; \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec{A}(\vec{r},t) - \nabla^2 \varphi(\vec{r},t) \vec{A}(\vec{r},t) = \mu_0 \vec{j}(\vec{r},t) Our plan to solve \left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \varphi(\vec{r},t) \right) Y(\vec{r},t) = X(\vec{r},t) is to express X in terms of sums of point functions and then have Y in terms of Green's functions. So we need to solve: \left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \varphi(\vec{r},t) \right) G(\vec{r},t,\vec{r}',t') = \delta(\vec{r}-\vec{r}') \delta(t - t'). Please verify that G(\vec{r},t,\vec{r}',t') = \frac{\delta \left( t - t' - \frac{\left| \vec{r}-\vec{r}' \right| }{c} \right) }{4 \pi \left| \vec{r}-\vec{r}' \right|} is a solution. So for the neighborhood near the wire at r'=0, we have a source term which is \mu_0 \vec{j}(\vec{r},t) = \frac{\mu_0 Q v}{2 \pi R} \theta(t) \delta(r_x)\delta(r_y) (in the z-direction) which gives rise to vector potential \vec{A}(\vec{r},t) = \frac{\mu_0 Q v}{4 \pi^2 R} \frac{\theta \left(t - \left| \vec{r} \right| /c \right)}{\left| \vec{r} \right|} (in the direction of the current) which gives us \vec{B}(\vec{r},t) = \vec{\nabla} \times \vec{A}(\vec{r},t) = \frac{\mu_0 Q v}{4 \pi^2 R \left| r \right|} \theta \left(ct - \left| \vec{r} \right| \right) (in the direction expected ). So the expectation is that the B field "turns on" at a later time depending on how far away you are from the current. For further reading, I direct you to any college textbook for a year-long course in electromagnetism or http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html (especially the section on the Time-Dependent Maxwell's equations). Masterov08-27-11, 01:49 AMComment Assume that the relativistic mass-point gives a flash of light at regular intervals (in its reference frame). Then, while visual and visual coordinates can be measured by measuring (by optical instruments) coordinates the flash and time intervals between outbreaks. Real coordinates are calculated by double integration of acceleration, which may be measuring (in mass-point reference frame). Masterov08-27-11, 02:10 AMRpenner, imagine a conductor with alternating current, which passes through the two contours of different radii. And (if speed of propagation of EMF is finite) whether left integral of expression: http://masterov.qptova.ru/MasterTheory/Formuls/Maxvell3.gif is the same for both circuits in a t the same time? rpenner08-27-11, 01:21 PMWhat I (and everyone who studies electromagnetism since Maxwell) am saying is that if the current, I, which is at the center of the circular surface changes, there is a pulse of electric field flux that originates with the change in time and space and propagates out with the speed of light. So after of time T from the time of the change, the magnetic field B reflects the old value of I if cT < r and the new value of B if cT > r. This comes from the part of E which comes from the - \frac{\partial \vec{A}}{\partial t} term. This is not obvious when a person with limited math education just looks at Maxwell's four equations. But physics education is about actually solving equations, and a changing current results in fields that vary over space and over time. So physics education also strengthens one's math intuition for some types of problems. If you study the link I provided, it gives a guide on how Maxwell's four equations can be turned into solutions for the electric and magnetic fields now in terms of the charges and currents of the past and their time derivatives. equations 541 and 542 of http://farside.ph.utexas.edu/teaching/em/lectures/node52.html \vec{E}(\vec{r},t) = \frac{1}{4\pi \varepsilon_0} \iiint \left( \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right| ^3} \rho \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) + \frac{\vec{r} - \vec{r}'}{c \left| \vec{r} - \vec{r}' \right| ^2} \dot{\rho} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) - \frac{1}{c^2 \left| \vec{r} - \vec{r}' \right|} \dot{\vec{j}} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) \right) d V' \vec{B}(\vec{r},t) = - \frac{\mu_0}{4\pi} \iiint \left( \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right| ^3} \times \vec{j} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) + \frac{\vec{r} - \vec{r}'}{c \left| \vec{r} - \vec{r}' \right| ^2} \times \dot{\vec{j}} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) \right) d V' I'm recovering from surgery to remove a tumor, so please excuse that I probably made some math mistakes in the prior post. But you seem particularly ill-equipped to criticize either special relativity or Maxwell's electromagnetism. Masterov08-27-11, 11:56 PMI did not understand your text, but your formula look like to true. And it (your formula) is not look like to the integral Maxwell equations. (This is what I mentioned above.) I repeat: my specialty - nonlinear dynamics (attractors, bifurcations, auto-wave, etc.) Master Theory was born out of a different theory. This (other) theory describe a model of a matter-generator. This generator is a Hamiltonian generator of solitons. My attempts to reconcile this theory with Einstein's theory were futile, but nothing has given me. I had to delve into Einstein's theory. Einstein gave the absolute cross-scale, but not for time. I'm looking for a specialist in Einstein's theory, which justified this choice of Einstein. Such a person absent on this forum, unfortunately. I wish you a speedy recovery and hope to see Maxwell's integral equations (of Master Theory context) in your performance. Masterov09-02-11, 03:38 AMYour formula: \vec{E}(\vec{r},t) = \frac{1}{4\pi \varepsilon_0} \iiint \left( \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right| ^3} \rho \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) + \frac{\vec{r} - \vec{r}'}{c \left| \vec{r} - \vec{r}' \right| ^2} \dot{\rho} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) - \frac{1}{c^2 \left| \vec{r} - \vec{r}' \right|} \dot{\vec{j}} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) \right) d V' \vec{B}(\vec{r},t) = - \frac{\mu_0}{4\pi} \iiint \left( \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right| ^3} \times \vec{j} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) + \frac{\vec{r} - \vec{r}'}{c \left| \vec{r} - \vec{r}' \right| ^2} \times \dot{\vec{j}} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) \right) d V'is an approximation. It can not be used for derivation of Maxwell's integral equations. Masterov09-08-11, 01:22 PMI came to the forum not in order to my a knowledges demonstrate, but in order to get answers to questions. I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Einstein had entitled (bestowed the absoluteness to time), but no did it. Why? rpenner09-08-11, 05:07 PMYour formula:is an approximation. Incorrect -- it is not an approximation to Maxwell's equations; rather it is a general solution to Maxwell's equations, as it tells you how to calculate B and E. This has been in textbooks since the 1960's. http://en.wikipedia.org/wiki/Jefimenko%27s_equations It can not be used for derivation of Maxwell's integral equations. Incorrect, with the continuity equation for electric charge, one may prove Maxwell's time-invariant equations follow as consequences. http://rediscoveries.blogspot.com/2011/07/in-2007-statement-and-proof-of.html http://arxiv.org/abs/0812.4785v1 As for the assertion that the differential and integral forms of Maxwell's equations are inequivalent, the world is still waiting for your physically valid demonstration of this claim of pure mathematics. I came to the forum not in order to my a knowledges demonstrate, but in order to get answers to questions. I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Translation: A Lorentz transformation corresponding to relative motion does not affect the projection of distances measured perpendicularly to the direction of relative motion. Or, shorter: a boost in the X direction leaves ΔY and ΔZ unchanged. Why? Answer 1: Physics tells us c(Δt)²−(Δx)²−(Δy)²−(Δz)² is physically relevant to the extent that observers agree on this value in a boosted environment. c(Δt)²−(Δx)²−(Δy)²−(Δz)² = c(Δt')²−(Δx')²−(Δy')²−(Δz')². The corresponding Euclidean relationship for rotation of axes: (Δx)²+(Δy)²+(Δz)²=(Δx')²+(Δy')²+(Δz')² factors into (Δx)²+(Δy)²=(Δx')²+(Δy')² and (Δz)²=(Δz')² when the rotation is confined the the x-y plane. So it is natural to assume that motion in the t-x plane factors the relativistic invariant as c(Δt)²−(Δx)² = c(Δt')²−(Δx')² and (Δy)²=(Δy')² and (Δz)²=(Δz')² and the only way to continuously maintain (Δz)²=(Δz')² across all possible boost speeds and still identify v=0 with the identity transform is Δz=Δz'. Similarly for Δy=Δy'. Einstein had entitled (bestowed the absoluteness to time), but no did it. Why? Translation 1 (low-confidence guess): Why is Δt = Δt' not the physics adopted by Einstein. Answer: Because Einstein understood Maxwell's equations better than you understand Maxwell's equations, and Einstein trusted that they accurately reflected the physics of the universe. Galileo and Newton thought Δt = Δt' because they had no experience with objects moving faster than 0.01% the speed of light, and with their clocks of 300+ years ago didn't have the precision to measure the time effects on the order of 0.00001% or smaller. Since 1859, as experiments and observations accumulated, more experimental evidence points at Δt ≠ Δt' for high-speed objects. http://sciforums.com/showpost.php?p=2039656&postcount=28 Masterov09-09-11, 04:44 AMRetarded potentials and Jefimenko's equations studied in other specialization. I agree to: retarded potentials or Jefimenko's equations can be used for create Maxwell's equations for Master Theory. But it sum no for me. (I not want to deprive somebody of a livelihood and my scientific career had completed eight years ago.) As for the assertion that the differential and integral forms of Maxwell's equations are inequivalent, the world is still waiting for your physically valid demonstration of this claim of pure mathematics.Gaus and Stokes use infinite velocity of light (or static fields). Used retarded potentials or Jefimenko's equations. A Lorentz transformation corresponding to relative motion does not affect the projection of distances measured perpendicularly to the direction of relative motion. Or, shorter: a boost in the X direction leaves ΔY and ΔZ unchanged. Why?A Lorentz transformation is absonant (is irrational). Two observers can not see the slowdown of each other simultaneously. If one sees a slowdown, then the second sees the acceleration. (Otherwise causality principle come to incorrect.) But such asymmetry violates the equality of inertial reference frames. Time can not slow down. Translation 1 (low-confidence guess): Why is Δt = Δt' not the physics adopted by Einstein. Answer: Because Einstein understood Maxwell's equations better than you understand Maxwell's equations, and Einstein trusted that they accurately reflected the physics of the universe. Galileo and Newton thought Δt = Δt' because they had no experience with objects moving faster than 0.01% the speed of light, and with their clocks of 300+ years ago didn't have the precision to measure the time effects on the order of 0.00001% or smaller. Since 1859, as experiments and observations accumulated, more experimental evidence points at Δt ≠ Δt' for high-speed objects.It is not a scientific argument. __________________________________________________ _____________ Do I understand your questions: you (just as me) do not see any reason for the absoluteness of the cross-scale (transverse dimensions)? Tach09-09-11, 09:49 AMTwo observers can not see the slowdown of each other simultaneously. If one sees a slowdown, then the second sees the acceleration. (Otherwise causality principle come to incorrect.) But such asymmetry violates the equality of inertial reference frames. Time can not slow down.It is not a scientific argument. Ah, Dingle. Classical. rpenner09-09-11, 11:21 AMI not want to deprive somebody of a livelihood and my scientific career had completed eight years Then you admit you are anti-science today? You can not deprive anyone of a livelihood when you are just spreading untrue and useless ideas based on your personal ignorance of math and physics. Every time you make an untrue math claim you ignore requests to supply a mathematical proof and every time you make a useless physics claim you ignore requests to provide fact-based arguments for the claim. You are anti-science and anti-truth. As for the assertion that the differential and integral forms of Maxwell's equations are inequivalent, the world is still waiting for your physically valid demonstration of this claim of pure mathematics. Gaus and Stokes use infinite velocity of light (or static fields). Infinite light propagation speed is not necessary for Gauss's and Stokes's theorems to be correct. They also apply in other fields of physics like fluid dynamics and do not imply that the speed of flowing water is infinite. It is obviously untrue that they require static fields since there are time derivatives in the integral representation. Where is the fact-based argument to support this ridiculous claim? You have asserted it, but not given a reason to believe it. Used retarded potentials or Jefimenko's equations.Yes, they used Jefimenko's equations and the continuity equation of charge to prove that Maxwell's equations result. Nowhere have you attempted to actually calculate the left and right sides of any equation to demonstrate that Maxwell's equation is untrue. Translation [of Masterov's question]: A Lorentz transformation corresponding to relative motion does not affect the projection of distances measured perpendicularly to the direction of relative motion. Or, shorter: a boost in the X direction leaves ΔY and ΔZ unchanged. Why? A Lorentz transformation is absonant (is irrational). You are confused. I was merely translating your question into better English. Lorentz transformations are the subject of your question. You have failed to provide a reason for your claim that: \begin{pmatrix} \cosh \varphi & \quad & \sinh \varphi \\ \sinh \varphi & \quad & \cosh \varphi \end{pmatrix} is any more or less rational than the well-known Euclidean rotation transform: \begin{pmatrix} \cos \theta & \quad & - \sin \theta \\ \sin \theta & \quad & \cos \theta \end{pmatrix} Answer 1: Physics tells us c(Δt)²−(Δx)²−(Δy)²−(Δz)² is physically relevant to the extent that observers agree on this value in a boosted environment. c(Δt)²−(Δx)²−(Δy)²−(Δz)² = c(Δt')²−(Δx')²−(Δy')²−(Δz')². The corresponding Euclidean relationship for rotation of axes: (Δx)²+(Δy)²+(Δz)²=(Δx')²+(Δy')²+(Δz')² factors into (Δx)²+(Δy)²=(Δx')²+(Δy')² and (Δz)²=(Δz')² when the rotation is confined the the x-y plane. So it is natural to assume that motion in the t-x plane factors the relativistic invariant as c(Δt)²−(Δx)² = c(Δt')²−(Δx')² and (Δy)²=(Δy')² and (Δz)²=(Δz')² and the only way to continuously maintain (Δz)²=(Δz')² across all possible boost speeds and still identify v=0 with the identity transform is Δz=Δz'. Similarly for Δy=Δy'. Two observers can not see the slowdown of each other simultaneously. If one sees a slowdown, then the second sees the acceleration. (Otherwise causality principle come to incorrect.) But such asymmetry violates the equality of inertial reference frames. This has nothing to do with your question or my answer. Further, it is another paragraph filled with unsupported claims. Are you the pope? No? Then you are not allowed to make unsupported claims based on your personal authority and expect people to believe as you do. Further, Tach seems to recognize this a plagiarized claim. I wish he would expand on that cryptic reference to "Dingle". Time can not slow down This belongs after my translation and answer to your following question. Translation 1 (low-confidence guess): Why is Δt = Δt' not the physics adopted by Einstein. Time can not slow down Where is the fact-based argument to support this claim that time can not slow down? You have asserted it, but not given a reason to believe it. Further, as I show below we do have many reasons to believe time can slow down. Answer: Because Einstein understood Maxwell's equations better than you understand Maxwell's equations, and Einstein trusted that they accurately reflected the physics of the universe. Galileo and Newton thought Δt = Δt' because they had no experience with objects moving faster than 0.01% the speed of light, and with their clocks of 300+ years ago didn't have the precision to measure the time effects on the order of 0.00001% or smaller. Since 1859, as experiments and observations accumulated, more experimental evidence points at Δt ≠ Δt' for high-speed objects. http://sciforums.com/showpost.php?p=2039656&postcount=28 It is not a scientific argument. How, when it is based in facts and logic and math, is it not scientific? Where is the fact-based argument to support this claim that evidence that time can slow down is not a scientific argument for the proposition that time can slow down? You have asserted it, but not given a reason to believe it. Do I understand your questions: you (just as me) do not see any reason for the absoluteness of the cross-scale (transverse dimensions)? Untrue. Physics supports the Lorentz transformation. Physics does not support your transformation. In fact, I do not see that if you use your transformation in 1 direction and then a transformation in the opposite direction that you can ever get the identity transformation, therefore your transformation cannot describe a group. Galileo said v_3 = v_1 \oplus_{G} v_2 = v_1 + v_2 so v \oplus_{G} (-v) = 0 and so \frac{c}{2} \oplus_{G} \frac{c}{2} = c Einstein said v_3 = v_1 \oplus_{E} v_2 = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} so v \oplus_{E} (-v) = 0 and so \frac{c}{2} \oplus_{E} \frac{c}{2} = \frac{4}{5} c Masterov says v \oplus_{M} (-v) = \cdots ? Masterov says \frac{c}{2} \oplus_{M} \frac{c}{2} = \cdots ? In general, there is no velocity parameter which makes the Masterov transformation on the right side equal to the product of the two Masterov transformations on the left side. Tach09-09-11, 11:36 AMFurther, Tach seems to recognize this a plagiarized claim. I wish he would expand on that cryptic reference to "Dingle". Herbert Dingle (http://www.mathpages.com/home/kmath024/kmath024.htm) spent a lot of his life, time and effort fighting the reciprocity of time dilation. Masterov shares the same misconceptions with Dingle (among many other misconceptions). AlphaNumeric09-09-11, 03:34 PMGaus and Stokes use infinite velocity of light (or static fields). No, they don't. They are not statements about some physically varying system where the (for instance) material contained within a volume must communicate with the surface so that they are all updated on what bit of stuff is doing what, it is a statement about mathematical functions in non-trivial (or sometimes trivial) spaces. Consider something simpler, Cauchy's residue theorem. Given a holomorphic function then the integral along a simple closed curve is proportional to the sum of residues of the function at the poles surrounded by the curve. This is in fact a form of Stoke's theorem (or vice versa). There's no communication, no physical properties, no signals, no propogation, no light or sound, nothing other than a function with a very specific property, it is holomorphic. Consider the function f(z) = \frac{1}{z} and a contour integral on |z| = R for some huge R. It is what it is, regardless of how large R is made. If you say "Make R 20 times bigger!" there's no change in the integral, no need to wait for signals to propagate out that far. It is a statement about how holomorphicity constraints certain properties of functions. Stokes' theorem is likewise, a statement about how certain properties, which are stated in the definition of the theorem, lead to certain other properties. It is a relationship between mathematical structures, nothing to do with physical properties. If you don't grasp this when you failed to understand a critical conceptual point in mathematics and its application to physics. Just like you obviously don't understand how Lorentz transforms work in physics or their mathematical structure. To someone living in \mathbb{C}^{2} rather than Minkowski or Euclidean space-time there is no distinction between standard rotations and Lorentzian boosts, they are just special cases of a more general set of transformations. Ask Penrose, he loves \mathbb{C}^{2} twistors and generalised rotations. quantum_wave09-09-11, 05:24 PMI came to the forum not in order to my a knowledges demonstrate, but in order to get answers to questions. ... Einstein had entitled (bestowed the absoluteness to time), but no did it. Why? Translation 1 (low-confidence guess): Why is Δt = Δt' not the physics adopted by Einstein. I'm a pea brain and simply reading along, but I can't tell if this low-confidence guess as to what Masterov was asking in his second question has been confirmed to be what he meant to ask, given the language barrier. Answer: Because Einstein understood Maxwell's equations better than you understand Maxwell's equations, and Einstein trusted that they accurately reflected the physics of the universe. Galileo and Newton thought Δt = Δt' because they had no experience with objects moving faster than 0.01% the speed of light, and with their clocks of 300+ years ago didn't have the precision to measure the time effects on the order of 0.00001% or smaller. Since 1859, as experiments and observations accumulated, more experimental evidence points at Δt ≠ Δt' for high-speed objects. http://sciforums.com/showpost.php?p=2039656&postcount=28Given that Masterov responded: ... It is not a scientific argument. How, when it is based in facts and logic and math, is it not scientific? Where is the fact-based argument to support this claim that evidence that time can slow down is not a scientific argument for the proposition that time can slow down? You have asserted it, but not given a reason to believe it. Are you referencing the earlier arguments about atmospheric muon decay time dilation and the atomic clocks getting out of sync when one is flown at high altitude and speed (this post) (http://www.sciforums.com/showpost.php?p=2798443&postcount=10)as the scientific evidence that time can slow down? I'm just asking if that is the evidence or if I missed other evidence that you are referring to? rpenner09-09-11, 07:03 PMGalileo said v_3 = v_1 \oplus_{G} v_2 = v_1 + v_2 so v \oplus_{G} (-v) = 0 and so \frac{c}{2} \oplus_{G} \frac{c}{2} = c Define G_x(v) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} then the Galiean transform for a space-time interval \begin{pmatrix} c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = G_x(v)\begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} c \Delta t \\ \Delta x - v \Delta t \\ \Delta y \\ \Delta z \end{pmatrix} . Note that G_x(0) = I which is the identity transform. So we wish to solve G_x(v_2) G_x(v_1) \begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = G_x(v_3) \begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} for all space-time intervals. Obviously G_x(v_2) G_x(v_1) = G_x(v_3) \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_1}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_2}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_3}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} Expanding the left half, we have : \left( I - \frac{v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \right) \left( I - \frac{v_1}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \right) = I - \frac{v_1 + v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \frac{v_1 v_2}{c^2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}^2 = I - \frac{v_1 + v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = G_x(v_1 + v_2) = G_x(v_3). So v_3 = v_1 + v_2 as Galileo and Newton would expect. Einstein said v_3 = v_1 \oplus_{E} v_2 = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} so v \oplus_{E} (-v) = 0 and so \frac{c}{2} \oplus_{E} \frac{c}{2} = \frac{4}{5} c Define \Lambda_x(v) = \begin{pmatrix} \cosh \, \tanh^{-1} \frac{v}{c} & \quad & - \sinh \, \tanh^{-1} \frac{v}{c} & \quad & 0 & 0 \\ - \sinh \, \tanh^{-1} \frac{v}{c} & \quad & \cosh \, \tanh^{-1} \frac{v}{c} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix} Note that \Lambda_x(0) = I. We wish to solve \Lambda_x(v_2)\Lambda_x(v_1) = \Lambda_x(v_3) and since \begin{pmatrix}\cosh a & \quad & \sinh a \\ \sinh a & \quad & \cosh a \end{pmatrix} \begin{pmatrix}\cosh b & \quad & \sinh b \\ \sinh b & \quad & \cosh b \end{pmatrix} = \begin{pmatrix}\cosh a \cosh b + \sinh a \sinh b & \quad & \cosh a \sinh b + \sinh a \cosh b \\ \cosh a \sinh b + \sinh a \cosh b & \quad & \cosh a \cosh b + \sinh a \sinh b \end{pmatrix} = \begin{pmatrix}\cosh (a + b) & \quad & \sinh (a + b) \\ \sinh (a + b) & \quad & \cosh (a + b) \end{pmatrix} , we have the solution v_3 = c \tanh \left( \tanh^{-1} \frac{v_1}{c} \; + \; \tanh^{-1} \frac{v_2}{c} \right) = c \frac{ \frac{v_1}{c} + \frac{v_2}{c} }{1 + \frac{v_1}{c} \frac{v_2}{c}} = \frac{ v_1 + v_2 }{1 + \frac{v_1 v_2}{c^2}}. Masterov says v \oplus_{M} (-v) = \cdots ? Masterov says \frac{c}{2} \oplus_{M} \frac{c}{2} = \cdots ? In general, there is no velocity parameter which makes the Masterov transformation on the right side equal to the product of the two Masterov transformations on the left side. We don't know what Masterov's equivalent of \Lambda_x(v) because he only defines: L' = L ( 1 - v^2/c^2) \\ H' = H \sqrt{ 1 - v^2/c^2} \\ T' = T so we can define: \mathcal{L}_x(v) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -\frac{v A(v)}{c} & 1 - \frac{v^2}{c^2} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} such that \begin{pmatrix} 0 \\ \Delta x' \\ 0 \\ 0 \end{pmatrix} = \mathcal{L}_x(v)\begin{pmatrix} 0 \\ \Delta x \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ (1 - \frac{v^2}{c^2}) \Delta x \\ 0 \\ 0 \end{pmatrix} This suggests that 1 - \frac{v_3^2}{c^2} = (1 - \frac{v_2^2}{c^2}) (1 - \frac{v_1^2}{c^2}) or v_3^2 = v_1^2 + v_2^2 - \frac{v_1^2 + v_2^2}{c^2} But to solve v_3 = 0 we have the relation v_2 = \pm \frac{v_1}{\sqrt{\frac{v_1^2}{c^2} -1}} which if 0 < \left| v_1 \right| < c requires v_2 to be imaginary. Likewise we can define: \mathcal{H}_x(v) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \cos B(v) & -\sqrt{1 - \frac{v_2^2}{c^2}} \sin B(v) \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \sin B(v) & \sqrt{1 - \frac{v_2^2}{c^2}} \cos B(v) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} & 0 \\ 0 & 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos B(v) & - \sin B(v) \\ 0 & 0 &\sin B(v) & \cos B(v) \end{pmatrix} such that \begin{pmatrix} c \Delta t' \\ 0 \\ \Delta y' \\ \Delta z' \end{pmatrix} = \mathcal{H}_x(v)\begin{pmatrix} c \Delta t \\ 0 \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} c \Delta t \\ 0 \\ \sqrt{1 - \frac{v_2^2}{c^2}} \left( \Delta y \cos B(v) - \Delta z \sin B(v) \right) \\ \sqrt{1 - \frac{v_2^2}{c^2}} \left( \Delta z \cos B(v) + \Delta y \sin B(v) \right) \end{pmatrix} Similar problems exist, and the physical requirement that v be a observed velocity indicates this cannot describe the physics of this universe. rpenner09-09-11, 07:16 PMAre you referencing the earlier arguments about atmospheric muon decay time dilation and the atomic clocks getting out of sync when one is flown at high altitude and speed (this post) (http://www.sciforums.com/showpost.php?p=2798443&postcount=10)as the scientific evidence that time can slow down? I'm just asking if that is the evidence or if I missed other evidence that you are referring to? I'm referencing the entire experimental record, which since 1859 has included more precision and faster experimental speeds than in the pre-1859 record and so highlights that Newtonian and Galilean low-precision, low-velocity experiments do not describe nature as well as Special Relativity, which works at all speeds. http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html But I linked to a description of one particular 1859 observation. quantum_wave09-09-11, 09:12 PMI'm referencing the entire experimental record, which since 1859 has included more precision and faster experimental speeds than in the pre-1859 record and so highlights that Newtonian and Galilean low-precision, low-velocity experiments do not describe nature as well as Special Relativity, which works at all speeds. http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html But I linked to a description of one particular 1859 observation.Thank you, great link too. You made a point early in the thread about SR being about the fact that the physics are the same in all reference frames. To me that and the fact that the math works are the keys. I follow the threads on SR to some extent and the "controversy" about evidence seems to focus on the spherically expanding light wave front being observed as spherical in the rest frame and spherical to observers in all inertial frames as opposed to being seen as an oblate spheroid in the non-rest frame. Also, to see the light wave front as spherical the point of origin of the light must move with the observer in his frame. I guess that all frames can be considered the rest frame and so all frames must see the spherical light wave front. It is just that someone said that when you do the transformations the wave front doesn't come out spherical in the transformed frame. Can you address that for me too? Tach09-09-11, 10:27 PMThank you, great link too. You made a point early in the thread about SR being about the fact that the physics are the same in all reference frames. To me that and the fact that the math works are the keys. I follow the threads on SR to some extent and the "controversy" about evidence seems to focus on the spherically expanding light wave front being observed as spherical in the rest frame and spherical to observers in all inertial frames as opposed to being seen as an oblate spheroid in the non-rest frame. Also, to see the light wave front as spherical the point of origin of the light must move with the observer in his frame. I guess that all frames can be considered the rest frame and so all frames must see the spherical light wave front. It is just that someone said that when you do the transformations the wave front doesn't come out spherical in the transformed frame. Can you address that for me too? A family of spherical light fronts is emitted at times t_i, i=1,n in frame S. Their equation is (in one dimension, for simplicity): x^2-(ct_i)^2=0 Now, consider frame S', moving away from S at speed v along the x axis. Applying the Lorentz transforms between S and S': x'=\gamma(x-vt) t'=\gamma(t-vx/c^2) you can easily prove that : x'^2-(ct'_i)^2=0 That is, the spherical fronts are also concentric in S' and they are centered around O', the origin of the system of axes in S'. Repeat the calculations in a frame S" moving towards S: x"=\gamma(x+vt) t"=\gamma(t+vx/c^2) If you do them right, you should get: x"^2-(ct"_i)^2=0 As an added bonus, if you consider that the time separation between two spherical fronts is t_{i+1}-t_i in S, you can easily deduce the relativistic Doppler effect in S' and in S". rpenner09-09-11, 11:32 PMPoints in space need three numbers (coordinates) to describe them. A popular choice are Cartesian coordinates x, y and z since each of them by themselves is a real number line just like any other, and each of the number lines are at right angles to the other two. The definition of distance in space between any two points, distance(A,B), is given by the square root of the sums of the squares of each difference of coordinates, \textrm{distance}(A,B) = \sqrt{ (x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2 }. This follows by using the famous theorem named for Pythagoras. The definition of a sphere is that all points, P, have the same distance, R, from the center, O = (x_O,y_O,z_O) so R = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }. The definition of something spherically expanding is that at any instant in time, t, the points at that instant in time, P, form a sphere about the center, O, and the radius of that sphere is a function that increase as time goes on, \frac{\partial R(t)}{\partial t} > 0. The simplest motion is a constant increase of R at the same rate and so we can describe the radius growing with the speed of light as: R(t) = c(t - t_0) where the radius is zero at time = t_0. Putting the two sides together, we have c(t-t_0) = R(t) = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }. Since the distance is a non-negative number and we don't care about times prior to t_0, we can square both sides. c^2(t-t_0)^2 = (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 So for all times we care about, we have (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 - c^2(t-t_0)^2. Or concisely, (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 = 0 Relativity says that every coordinate system where the laws of physics are the same are equally valid. In Special Relativity, we are talking about inertial coordinate systems. They don't rotate, they don't accelerate, in them Newton's law of inertia is valid. So two different inertial coordinate systems can have a velocity difference between them. In what follows, lets assume that x,y,and z of one coordinate system was instantly aligned with the x', y' and z' of another at some instant and that the only motion is along the x axis. Then the simplest version of the Lorentz transformation applies. \begin{pmatrix}c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ 0 & \quad & 0 & \quad & 1 & \quad & 0 \\ 0 & \quad & 0 & \quad & 0 & \quad & 1 \end{pmatrix} \begin{pmatrix}c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} This allow us to calculate: \begin{eqnarray} c \Delta t' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( c \Delta t - \frac{v}{c} \Delta x \right) \\ \Delta x' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( \Delta x - \frac{v}{c} c \Delta t \right) \\ \Delta y' \quad & = & \Delta y \\ \Delta z' \quad & = & \Delta z \end{eqnarray} So (c \Delta t')^2 = \frac{(c \Delta t)^2 + \frac{v^2}{c^2}(\Delta x)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ (\Delta x' )^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ ( \Delta y')^2 = ( \Delta y)^2 \\ ( \Delta z')^2 = ( \Delta z)^2 And therefore (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t) - (c \Delta t)^2 - \frac{v^2}{c^2}(\Delta x)^2 + \frac{v}{c} (\Delta x)(c \Delta t) }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 \\ = \frac{(\Delta x)^2 - \frac{v^2}{c^2}(\Delta x)^2 }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 - \frac{ (c \Delta t)^2 - \frac{v^2}{c^2}(c \Delta t)^2 }{1 - \frac{v^2}{c^2}} \\ = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 This is the most important property of the Lorentz transform: (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 Because we are dealing with a sphere expanding at the speed of light, the right side equals zero, and therefore the left side equals zero. Therefore both coordinate systems agree that the light is expanding spherically at the speed of light. Masterov09-10-11, 02:43 AMThen you admit you are anti-science today? You can not deprive anyone of a livelihood when you are just spreading untrue and useless ideas based on your personal ignorance of math and physics. Every time you make an untrue math claim you ignore requests to supply a mathematical proof and every time you make a useless physics claim you ignore requests to provide fact-based arguments for the claim. You are anti-science and anti-truth.Such an arguments are worthy of a medieval Inquisition, but not a scientist. I asked the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Einstein had entitled (bestowed the absoluteness to time), but no did it. Why? I do asked a question only. Why do you accuse me of blasphemy and heresy? Masterov09-10-11, 03:30 AMInfinite light propagation speed is not necessary for Gauss's and Stokes's theorems to be correct. They also apply in other fields of physics like fluid dynamics and do not imply that the speed of flowing water is infinite. It is obviously untrue that they require static fields since there are time derivatives in the integral representation. Where is the fact-based argument to support this ridiculous claim? You have asserted it, but not given a reason to believe it. Yes, they used Jefimenko's equations and the continuity equation of charge to prove that Maxwell's equations result. Nowhere have you attempted to actually calculate the left and right sides of any equation to demonstrate that Maxwell's equation is untrue.Check Stokes's theorem by a simple example: Two loops of different radii and an electric conductor of alternating current pierces through which. Stokes's theorem return equal results for both loops. Is it true? Yes, they used Jefimenko's equations and the continuity equation of charge to prove that Maxwell's equations result. Nowhere have you attempted to actually calculate the left and right sides of any equation to demonstrate that Maxwell's equation is untrue.I do not get it. Lorentz transformations are the subject of your question.No. My question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Einstein had entitled (bestowed the absoluteness to time), but no did it. Why? This has nothing to do with your question or my answer. Further, it is another paragraph filled with unsupported claims. Are you the pope? No? Then you are not allowed to make unsupported claims based on your personal authority and expect people to believe as you do. Further, Tach seems to recognize this a plagiarized claim. I wish he would expand on that cryptic reference to "Dingle".I do not get it. Where is the fact-based argument to support this claim that time can not slow down? You have asserted it, but not given a reason to believe it.It to do causality principle to incorrect or violates the equality of inertial reference frames. Masterov09-10-11, 07:16 AMHow the expression created? x^2-(ct)^2=0 The absoluteness of this expression is sequent what? AlphaNumeric09-10-11, 07:54 AMHow the expression created? x^2-(ct)^2=0 The absoluteness of this expression is sequent what?Seriously? You're asking where that comes from? It's the integrated (for constant motion) space-time interval for something moving through Minkowski space-time at the speed of light. Have you even read a book on special relativity? Tach09-10-11, 08:59 AMHow the expression created? x^2-(ct)^2=0 The absoluteness of this expression is sequent what? OK, let's try differently, what do you think that the following expression means: x^2+y^2+z^2-(ct)^2=0 Hint: a spherical wavefront propagating at light speed. quantum_wave09-10-11, 09:35 AMA family of spherical light fronts is emitted at times t_i, i=1,n in frame S. Their equation is (in one dimension, for simplicity): x^2-(ct_i)^2=0 Now, consider frame S', moving away from S at speed v along the x axis. Applying the Lorentz transforms between S and S': x'=\gamma(x-vt) t'=\gamma(t-vx/c^2) you can easily prove that : x'^2-(ct'_i)^2=0 That is, the spherical fronts are also concentric in S' and they are centered around O', the origin of the system of axes in S'. Repeat the calculations in a frame S" moving towards S: x"=\gamma(x+vt) t"=\gamma(t+vx/c^2) If you do them right, you should get: x"^2-(ct"_i)^2=0 As an added bonus, if you consider that the time separation between two spherical fronts is t_{i+1}-t_i in S, you can easily deduce the relativistic Doppler effect in S' and in S".I see. So you could deduce the same in S'' and S'''? Or is that taking it too far? Points in space need three numbers (coordinates) to describe them. A popular choice are Cartesian coordinates x, y and z since each of them by themselves is a real number line just like any other, and each of the number lines are at right angles to the other two. The definition of distance in space between any two points, distance(A,B), is given by the square root of the sums of the squares of each difference of coordinates, \textrm{distance}(A,B) = \sqrt{ (x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2 }. This follows by using the famous theorem named for Pythagoras. The definition of a sphere is that all points, P, have the same distance, R, from the center, O = (x_O,y_O,z_O) so R = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }. The definition of something spherically expanding is that at any instant in time, t, the points at that instant in time, P, form a sphere about the center, O, and the radius of that sphere is a function that increase as time goes on, \frac{\partial R(t)}{\partial t} > 0. The simplest motion is a constant increase of R at the same rate and so we can describe the radius growing with the speed of light as: R(t) = c(t - t_0) where the radius is zero at time = t_0. Putting the two sides together, we have c(t-t_0) = R(t) = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }. Since the distance is a non-negative number and we don't care about times prior to t_0, we can square both sides. c^2(t-t_0) = (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 So for all times we care about, we have (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 - c^2(t-t_0). Or concisely, (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 = 0 Relativity says that every coordinate system where the laws of physics are the same are equally valid. In Special Relativity, we are talking about inertial coordinate systems. They don't rotate, they don't accelerate, in them Newton's law of inertia is valid. So two different inertial coordinate systems can have a velocity difference between them. In what follows, lets assume that x,y,and z of one coordinate system was instantly aligned with the x', y' and z' of another at some instant and that the only motion is along the x axis. Then the simplest version of the Lorentz transformation applies. \begin{pmatrix}c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ 0 & \quad & 0 & \quad & 1 & \quad & 0 \\ 0 & \quad & 0 & \quad & 0 & \quad & 1 \end{pmatrix} \begin{pmatrix}c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} This allow us to calculate: \begin{eqnarray} c \Delta t' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( c \Delta t - \frac{v}{c} \Delta x \right) \\ \Delta x' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( \Delta x - \frac{v}{c} c \Delta t \right) \\ \Delta y' \quad & = & \Delta y \\ \Delta z' \quad & = & \Delta z \end{eqnarray} So (c \Delta t') = \frac{(c \Delta t)^2 + \frac{v^2}{c^2}(\Delta x)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ (\Delta x' )^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ ( \Delta y')^2 = ( \Delta y)^2 \\ ( \Delta z')^2 = ( \Delta z)^2 And therefore (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t) - (c \Delta t)^2 - \frac{v^2}{c^2}(\Delta x)^2 + \frac{v}{c} (\Delta x)(c \Delta t) }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 \\ = \frac{(\Delta x)^2 - \frac{v^2}{c^2}(\Delta x)^2 }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 - \frac{ (c \Delta t)^2 - \frac{v^2}{c^2}(c \Delta t)^2 }{1 - \frac{v^2}{c^2}} \\ = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 This is the most important property of the Lorentz transform: (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 Because we are dealing with a sphere expanding at the speed of light, the right si[d]e equals zero, and therefore the left side equals zero. Therefore both coordinate systems agree that the light is expanding spherically at the speed of light.And therefore the math works perfectly as I knew it did :). And 0' moves relative 0 along one axis, or is that my misconception? Apologies to Masterov for jumping in but I wanted to clarify that rpenner was correct as to what you were asking. No responses needed to this post. Thanks for the help. Tach09-10-11, 10:04 AMI see. So you could deduce the same in S'' and S'''? Or is that taking it too far? works in any inertial frame Masterov09-10-11, 10:24 AMOK, let's try differently, what do you think that the following expression means: x^2+y^2+z^2-(ct)^2=0 Hint: a spherical wavefront propagating at light speed.It is boundary between events that may be a cause and consequence of each other of Einsteinian theory, that is also a consequence of no-proven assumption that nothing can travel faster than light. x^2-(ct)^2=0 - special case (y,z=const) AlphaNumeric09-10-11, 11:02 AMthat is also a consequence of no-proven assumption that nothing can travel faster than light.No, it isn't. It is a statement about trajectories. If v^{\mu}v_{\mu} = -v^{0}v^{0} + \mathbf{v}\cdot \mathbf{v} = 0 then it means v = v^{\mu}\partial_{\mu} is a null vector. A vector which corresponds to something moving faster than light locally has (for this metric signature) v^{\mu}v_{\mu} > 0. A consequence of the Lorentzian signature of space-time is that no object with mass can go faster than the local speed of light, not the reverse. x^2-(ct)^2=0 - special case (y,z=const)For any constant null velocity vector v = v^{\mu}\partial_{\mu} there is always a frame where v^{\mu}v_{\mu} = 0 takes the form -v^{0}v^{0} + v^{x}v^{x} = 0. Simply use an SO(3) rotation (which is contained within the SO(3,1) symmetry group of flat Lorentzian space-time metrics) to align the x axis with the velocity vector and you're done. Seriously, learn some special relativity. Masterov09-10-11, 11:15 AMFor example: from target of elementary particle accelerator generate unstable particles emitted from the known lifetime (tau leptons - 5 10^-13 seconds) sometimes. During life, they manage to overcome the distance that can not be overcome if the move at the speed of light. Tach09-10-11, 11:54 AMFor example: from target of elementary particle accelerator generate unstable particles emitted from the known lifetime (tau leptons - 5 10^-13 seconds) sometimes. During life, they manage to overcome the distance that can not be overcome if the move at the speed of light. Have you learned about why muons formed in the upper atmosphere reach the ground? For the same reason, leptons manage to cross distances that exceed your stupid misconceptions. How stupid are you? Really. AlphaNumeric09-10-11, 12:05 PMFor example: from target of elementary particle accelerator generate unstable particles emitted from the known lifetime (tau leptons - 5 10^-13 seconds) sometimes. During life, they manage to overcome the distance that can not be overcome if the move at the speed of light.Because they experience time dilation relative to the accelerator. They experience about 5 \times 10^{-13} seconds but to us it seems like 10^{-10} seconds and thus during that time they move further than 5 \times 10^{-13} \times c. The muons are seen to move at something like 0.9999c and they live for around \gamma(0.9999c) \times 5 \times 10^{-13} seconds, during which time they travel a distance \gamma(0.9999c) \times 5 \times 10^{-13} s \times 0.9999c. See how it works? If a particle lives for a time T from its point of view then if it moves at a speed v relative to the accelerator then it will be seen to live for a time \gamma(v)T, where \gamma(v) = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}. Since distance is speed times velocity it'll be seen to move a distance v\gamma(v)T. No one, either the particle or accelerator, sees anyone move faster than light, it is all consistent. Seriously, learn some special relativity. Masterov09-10-11, 11:21 PMBecause they experience time dilation relative to the accelerator.It is not possible. Time dilation (of Einstein's theory) is a consequence of the presence of velocity, but does not the acceleration. Einstein speek: \frac{dt'}{dt}=\gamma. (Transverse Doppler effect.) t' - time of under review IRF. Masterov09-11-11, 07:59 AMHow stupid are you? Really.You'll pay for this your words (if you happen to be wrong). In this case, you (if you're an honest person) do to finish his scientist career. (I am sure that your colleagues will help you with this if you show indecision.) Tach09-11-11, 09:24 AMIt is not possible. Einstein speek: \frac{dt'}{dt}=\gamma. (Transverse Doppler effect.) t' - time of under review IRF. Why isn't it possible? Because you don't understand basic stuff? d \tau= dt \sqrt {1-(v/c)^2} so dt=\gamma d \tau Using Alfanumeric's notation, if a particle lives a time T in its proper frame , it will live a time \gamma T in a frame where the particle moves at speed v. This is basic stuff, you are totally ignorant. Masterov09-11-11, 10:17 AMRepeat: I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? You not has knowledge to give a qualified answer to this question, but repeated a phrases: you are totally ignorant. How stupid are you? Really.You hear these phrase to your address from your colleagues every day and to do cackle like parrots. (I suspect). I see no reason to continue to communicate with you. Tach09-11-11, 11:07 AMEinstein speek: \frac{dt'}{dt}=\gamma. (Transverse Doppler effect.) t' - time of under review IRF. You mean \frac{dt}{d \tau}=\gamma where \tau is the proper time? You don't know the basics yet you pretend to discuss relativity. OnlyMe09-11-11, 11:50 AMRepeat: I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Masterov, No one answers the question you have posed several times now likely for two reasons. First is we, or at least "I" am not entirely sure what you mean. And second, I am not sure Einstein ever addressed the question himself and again we, or "I" have no real idea what his thoughts were or might have been. I can give you my thoughts on the issue, with the disclaimer that though I have read some of the older literature I have not had access to all of the literature that might have some influence on this topic and anything I say is only my opinion. (And yes I am fully aware of what some folks think about personal opinions. It really does not concern me. I remain opinionated anyway!) In special relativity Einstein was dealing with a modernization of Relativity, which predates his efforts by several hundred years at least. His effort seems to have been largely to address Relativity from a context that incorporated Maxwell's work on electrodynamics. Hence, the title of his 1905 paper, "On the electrodynamics of moving bodies" and a flat Minkowski space-time, in which time moves in only one direction, consistent with experience. The last three words there are perhaps significant, "consistent with experience". Einstein seemed at least in the early stages of his work to base his theoretical models on the results of experiment and experience. And time as a mater of experience moves in only one direction. After publishing his 1905 paper on Relativity he began to explore its impact on gravity and at least two known and unresolved observations that Newton's vision of gravity could not explain, the perihelion advancement of Mercury's orbit and the apparent instantaneous action at a distance required, for Newton's field equation to be functionally accurate, which it was and remains to be. I am not sure that anyone can truly say or know what it was that initially lead Einstein to the concept of curved space, as an explanation of the known problems of gravity at the time. It may well have been nothing more than recognizing the impact that the Lorentz Transformations would have on flat Minkowski space-time. However, he arrived at the conclusions he did, and space-time as defined from the perspective of GR became curved... And while we have experience in the world such that spacial coordinate systems can physically have meaning in all directions from a starting point, our experience of time was still limited to change progressing in only one direction, forward and into the future. However, in associating the Lorentz Transformations with a curvature of space also came implied changes to time. Time when viewed from the perspective of a curved and dynamic space-time and the Lorentz Transformations, lost any ridged structure, and become to some extent flexible, in its rate of change. The flow of time could a can be imagined to be subject to both the curvature of space and relativistic affects, while consistent with experience it still remains constrained to motion or change in only one direction, forward and into the future. From the perspective of GR while time cannot be imagined to change directions it can be imagined to move at differing rates depending upon the conditions within which it is viewed..., whether that is the now curved space of GR or the relativistic effects of velocity as described within SR. Since our experience of time involves motion in only one direction, representing time as moving in two directions from a starting point or zero reference point is not consistent with experience. We can at least at present only imagine time as past, present and future and that it only moves from the present to the future, the past being as far as experience is concerned an artifact of memory. Though I am sure that the past would remain, should our memories fail, experience continues to move only into the future. AlphaNumeric09-11-11, 12:08 PMIt is not possible. Time dilation (of Einstein's theory) is a consequence of the presence of velocity, but does not the acceleration.Accelerators accelerate particles so that they have, relative to the Earth, high velocity. Apply an acceleration and you change the velocity, by definition! The accelerators get the particles to speeds like 0.9999c. Lots of relative speed. You'll pay for this your words (if you happen to be wrong). In this case, you (if you're an honest person) do to finish his scientist career. (I am sure that your colleagues will help you with this if you show indecision.)What on Earth are you talking about? If you were an honest person you'd learn some basic physics. Repeat: I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis?Rpenner has provided a ton of formal mathematics, haven't you realised he understands it? If you want people with various letters after their name then I've got a PhD in physics. I see no reason to continue to communicate with you.You have no understanding of relativity and you don't seem to even realise it. The question is why should anyone continue to communicate with you? Masterov09-11-11, 12:13 PMYour English is good. Even though you were writing a lot, I understood almost everything. But some sentences i do not understand. And time as a mater of experience moves in only one direction.What is the meaning "mater"? (matter?) in flat Minkowski space-time time moves in only one direction, consistent with experience.In flat Minkowski space-time time moves in only one direction: s^2=(ct)^2-x^2-y^2-z^2=(-ct)^2-x^2-y^2-z^2 Direction of time is not definite. (Is it?) It does not matter now. NietzscheHimself09-11-11, 12:38 PMTime doesn't move. It is a concept used to explain motion. Masterov09-11-11, 01:14 PMTime doesn't move. It is a concept used to explain motion.Yes! I'm agree. I tried to realize the model of matter (hamiltonian) generator in Minkowski space, but I could not. (It was eight years ago.) This forced me to try to understand Einstein's theory. So originated Master Theory. Time of my generator is static also. OnlyMe09-11-11, 01:21 PMYour English is good. Even though you were writing a lot, I understood almost everything. But some sentences i do not understand.What is the meaning "mater"? (matter?) This was a bad phrasing. Perhaps... Time as a result of experience moves only toward the future.., or Time is experienced only to move in one direction... In flat Minkowski space-time time moves in only one direction: s^2=(ct)^2-x^2-y^2-z^2=(-ct)^2-x^2-y^2-z^2 Direction of time is not definite. (Is it?) It does not matter now. Mathematically the direction of time is not definite perhaps. Mathematics and experience are not always equivalent and remember I was attempting to "imagine" some basis for Einstein's treatment of time and space. And yes I tend to over think and over explain things. One of my former bosses used to routinely complain that I was providing too much unnecessary information. This is as it it is. There is no help for it. Just as there is no help for the fact that though I try to see the perspectives of others, I remain opinionated myself. Masterov09-11-11, 01:22 PMWhy should me slow down time? I've shown that we can restrict the deformation of space. Why? OnlyMe09-11-11, 01:23 PMTime doesn't move. It is a concept used to explain motion. This is true and yet, the direction of time remains a subject of great discussion, both in the past and present. Masterov09-11-11, 01:33 PMOnlyMe, Matter generator must be hamiltonian. Ie time must be reversible. This means that: if exist Matter, then exist anti-Matter (time flowing in the opposite direction). The electron and positron are distinguished direction of time. I understand it. AlphaNumeric09-11-11, 01:50 PMYour English is good. Even though you were writing a lot, I understood almost everything.Firstly, English is my native language. Secondly, I think it's important to be coherent. And thirdly, writing technical physics is a coherent way is literally my job. But some sentences i do not understand.If you are unfamiliar with the concepts I have referred to then it is a sign you need to familiarise yourself with basic relativity. This stuff is covered during undergraduate physics courses. I've taught it to 1st years. What is the meaning "mater"? (matter?)Matter, the substance of materials and objects. [QUOTE=Masterov;2814170]In flat Minkowski space-time time moves in only one direction: s^2=(ct)^2-x^2-y^2-z^2=(-ct)^2-x^2-y^2-z^2[/tex]No, only one of those expressions is even possibly right. There is a choice in how you write down the metric of Minkowski space-time, it is known as the signature. The space-time metric is either \eta = \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & +1 & 0 & 0 \\ 0 & 0 & +1 & 0 \\ 0 & 0 & 0 & +1 \end{array}\right) OR it is \eta = \left( \begin{array}{cccc} +1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right). It is not both. The first is often written as (-+++), while the second is written as (+---), referring to the signs on the diagonal. For the (-+++) signature the space-time interval is ds^{2} = \eta_{ab}dx^{a}dx^{b} = -dt^{2} + dx^{2}+dy^{2}+dz^{2}. This has nothing to do with the direction of motion, it is a statement about lengths of curves through 4 dimensional space-time. The length of a curve is independent of the direction you move along it. OnlyMe09-11-11, 01:54 PMOnlyMe, Matter generator must be hamiltonian. Ie time must be reversible. This means that: if exist Matter, then exist anti-Matter (time flowing in the opposite direction). The electron and positron are distinguished direction of time. I understand it. I must admit that, I do not. It does not seem to be consistent with experience and I remain to some extent still influenced by a classical understanding of experience. Though I do remember on one occasion when experimenting with a particular mushroom that events did seem to take on an order not entirely consistence with my previous experiences. Sadly the experiment was not repeatable in a consistent manner.., though I did try on numerous occasions. AlphaNumeric09-11-11, 02:17 PMOnlyMe, Matter generator must be hamiltonian. Ie time must be reversible. This means that: if exist Matter, then exist anti-Matter (time flowing in the opposite direction).Hamiltonian flows are something very specific. A system can have a well defined Hamiltonian but that doesn't mean all things which follow from that related to Hamiltonian flows. You mention antimatter, but that isn't due to a Hamiltonian flow, it is due to conjugacy properties of fields. For instance, in QED the Lagrangian involves both \psi and \bar{\psi}, specifically \mathcal{L} = \bar{\psi}(i\gamma^{a}\partial_{a}-m + ie\gamma^{a}A_{a})\psi. The fields are \psi and A_{a}, the electron and photon respectively. The field \bar{\psi} is obtained from \psi but represents positrons. A Hamiltonian flow describes how the dynamics/properties of a system can be modelled by trajectories through the system's phase space. The Hamiltonian flow basically refers to the motion a system moves through in phase space as time passes. For instance, if you want to know how some quantity A varies as the system evolves then you compute \dot{A} \propto \{A,H\}, where { , } is a Poisson bracket (http://en.wikipedia.org/wiki/Poisson_bracket). This is seen in quantum mechanics (after quantisation obviously) in the form of Heisenberg's Equations (http://en.wikipedia.org/wiki/Heisenberg_picture), where the rate of change of some quantity is proportional to the commutation with the Hamiltonian, \dot{A} \propto [A,H]. funkstar09-11-11, 03:25 PMYou not has knowledge to give a qualified answer to this question, but repeated a phrases:You hear these phrase to your address from your colleagues every day and to do cackle like parrots. (I suspect). There's a wonderful insult here, I suspect: "to do cackle like parrots" sound particularly good... Tach09-11-11, 10:58 PMThere's a wonderful insult here, I suspect: "to do cackle like parrots" sound particularly good... He must be using an automatic translator program from Russian to English, he doesn't know much English. As a mater of fact, he doesn't know much physics either. Masterov09-12-11, 02:47 AMIf you are unfamiliar with the concepts I have referred to then it is a sign you need to familiarise yourself with basic relativity. This stuff is covered during undergraduate physics courses. I've taught it to 1st years.I have a problem with translation into Russian. My dictionaries do not always cope with the scientific terminology. I can not adequately understand the English text is not always a case. For example: mater === "brain-tunic" or "mother". This has nothing to do with the direction of motion, it is a statement about lengths of curves through 4 dimensional space-time. The length of a curve is independent of the direction you move along it.This is not true. The space-time isotropy setted limits for a types of matter-generator. This generator must be hamiltonian. This means that a solution to his equations will always give a paired results. These decisions will differ a time direction. (For AlphaNumeric also.) OnlyMe09-12-11, 09:49 AMYour English is good... But some sentences i do not understand.What is the meaning "mater"? (matter?) I can not adequately understand the English text is not always a case. For example: mater === "brain-tunic" or "mother".This is not true. Masterov, this is and was my fault. You were being generous when you said my English was good. A complement my English teachers would have had some disagreement with, though it is my first language. I used the word "mater" incorrectly and it does indeed translate to "mother". I should have been using the word "matter" all along. Spelling was never a strong suit for me and with the advent of spell checkers it has only become worse. My apologies, for the confusion. I have since obtained a dictionary of my own. Digital of course. Now the problem will be remembering to use it.... Masterov09-12-11, 01:47 PMSpelling was never a strong suit for me and with the advent of spell checkers it has only become worse.Don't mention apologize. We are not linguists. (Both.) Masterov09-13-11, 02:15 AMEinstein allege: time can slow down (by acceleration), but time can not accelerate (by acceleration). If time slow down, then - it can not accelerate, can not return back. Traveler will have its own (slow) time on Earth. How can this be? arfa brane09-13-11, 04:49 AMEinstein allege: time can slow down (by acceleration), but time can not accelerate (by acceleration).No he didn't. What he alleged was the same thing that Lorentz alleged: that an accelerated frame has a slowed time rate relative to a non-accelerated frame. Which means a non-accelerated frame has a speeded up time rate relative to an accelerated frame. By "time rate" is meant the time recorded by otherwise identical clocks. I read about this when I was 10, and I thought I understood it then (but perhaps not). At least I thought I understood what the twin paradox meant--if you move through space faster than someone else you age less, relative to the slower traveller, who appears to age more quickly. Both travellers see themselves aging at a "normal" rate, and their clocks record time at the same rate locally. Masterov09-13-11, 05:23 AMthat an accelerated frame has a slowed time rate relative to a non-accelerated frame.That's impossible! Return home will require the acceleration, which again slows down time. Subsequent braking will slow down again. rpenner09-13-11, 10:29 AMThat's impossible! Return home will require the acceleration, which again slows down time. Subsequent braking will slow down again. That's why it was claimed you were repeating one of the mistake of Dingle, because you were focusing on the term \frac{\partial \Delta t'}{\partial \Delta t} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} when the Lorentz transform is more than that, it's \Delta t' = \frac{\Delta t - \frac{v}{c^2}\Delta x}{\sqrt{1 - \frac{v^2}{c^2}}}. This is the same mistake as claiming a rotation left by 30 degrees isn't canceled by a rotation right by 30 degrees because \cos \, 30^{\circ} = \cos \, -30^{\circ} = \frac{\sqrt{3}}{2} when the relevant expression is \begin{pmatrix} \cos \, 30 ^{\circ} & \quad & - \sin \, 30^{\circ} \\ \sin \, 30^{\circ} & \quad & \cos \, 30^{\circ} \end{pmatrix} \begin{pmatrix} \cos \, -30 ^{\circ} & \quad & - \sin \, -30^{\circ} \\ \sin \, -30^{\circ} & \quad & \cos \, -30^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & \quad & - \frac{1}{2} \\ \frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{3}}{2} & \quad & \frac{1}{2} \\ -\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2}\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \frac{1}{2} - \frac{\sqrt{3}}{2}\frac{1}{2} \\ \frac{\sqrt{3}}{2} \frac{1}{2} - \frac{\sqrt{3}}{2}\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2}\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} So you criticize relativity because you don't understand relativity. And you don't understand relativity because, at a minimum, you don't understand the mathematics of the Lorentz transform. Specifically, you don't understand that a Lorentz transform of v is canceled by a Lorentz transform of -v (in the same direction). Masterov09-13-11, 01:11 PMMoving across a view axis (the distance does not change): \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta x = 0 rpenner09-13-11, 03:30 PMMoving across a view axis (the distance does not change): \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta x = 0 Attempted translation: "But if the velocity and the x direction are perpendicular, the time-dilation only accumulates." Attempted translation: "But if original change in x is zero, the time-dilation only accumulates." Response: Untrue. Please consult my previous post. Restrict: 0 < |v| < c Define: \Lambda_x (v) = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ 0 & \quad & 0 & 1 & 0 \\ 0 & \quad & 0 & 0 & 1\end{pmatrix} Define: \Lambda_y (v) = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 \\ 0 & \quad & 1 & \quad & 0 & 0 \\ \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix} So \Lambda_x (-v) \Lambda_x (v) = \Lambda_x (v) \Lambda_x (-v) = I Likewise \Lambda_y (-v) \Lambda_y (v) = \Lambda_y (v) \Lambda_y (-v) = I Likewise for \Lambda_z. Also: \Lambda_x (-v) \Lambda_y (-v) \Lambda_y (v) \Lambda_x (v) = \Lambda_x (-v) \Lambda_y (v) \Lambda_y (-v) \Lambda_x (v) = \Lambda_x (v) \Lambda_y (-v) \Lambda_y (v) \Lambda_x (-v) = \Lambda_x (v) \Lambda_y (v) \Lambda_y (-v) \Lambda_x (-v) = I This means \begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x (-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} and \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x (v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} together mean \begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} Proof: \begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x (-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x (-v) \left( \Lambda_x (v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} \right) = \left( \Lambda_x (-v) \Lambda_x (v) \right) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = I \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}. Even if \Delta x = 0 or \Delta x' = 0 we always have \Delta t'' = \Delta t. Alternate proof: \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x (v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t - \frac{v}{c^2} \Delta x}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{\Delta x - v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta y \\ \Delta z \end{pmatrix}, so \begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x (-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{\Delta t' - \frac{-v}{c^2} \Delta x'}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{\Delta x' - (-v) \Delta t'}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{\left( \Delta t - \frac{v}{c^2} \Delta x \right) - \frac{-v}{c^2} \left(\Delta x - v \Delta t \right)}{\sqrt{1 - \frac{v^2}{c^2}}\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{\left(\Delta x - v \Delta t \right) - (-v) \left( \Delta t - \frac{v}{c^2} \Delta x \right)}{\sqrt{1 - \frac{v^2}{c^2}}\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t - \frac{v}{c^2} \Delta x + \frac{v}{c^2} \Delta x - \frac{v^2}{c^2} \Delta t}{1 - \frac{v^2}{c^2}} \\ \frac{\Delta x - v \Delta t + v \Delta t - \frac{v^2}{c^2} \Delta x}{1 - \frac{v^2}{c^2}} \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t - \frac{v^2}{c^2} \Delta t}{1 - \frac{v^2}{c^2}} \\ \frac{\Delta x - \frac{v^2}{c^2} \Delta x}{1 - \frac{v^2}{c^2}} \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} AlphaNumeric09-13-11, 03:31 PMMoving across a view axis (the distance does not change): \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta x = 0That is not true. Such a transformation wouldn't leave the space-time interval unchanged. A transform (\Delta t,\Delta x) \to (\Delta t',\Delta x') must be such that -(\Delta t)^{2} + (\Delta x)^{2} = -(\Delta t')^{2} + (\Delta x')^{2} , assuming \Delta y and \Delta z are unchanged. This is the central tenant of Lorentz transformations. Masterov09-14-11, 02:07 AMI did mean sideways movement: If: \Delta x = 0 \Delta y'/\Delta t' = v then: \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} Longitudinal Doppler effect is absent. Tach09-14-11, 09:16 AMI did mean sideways movement: If: \Delta x = 0 \Delta y'/\Delta t' = v then: \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} Longitudinal Doppler effect is absent. So, is your problem that you : A. Do not understand Transverse Doppler Effect? or B. Do not understand the reciprocity of time dilation? rpenner09-14-11, 10:32 AMI did mean sideways movement: If: \Delta x = 0 \Delta y'/\Delta t' = v then: \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} Longitudinal Doppler effect is absent. You still aren't being clear. Did you mean: \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_y(-v) \begin{pmatrix} \Delta t \\ 0 \\ 0 \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 \\ \frac{v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta z \end{pmatrix} ? Well, \Delta x = 0, \; \frac{\Delta y'}{\Delta t'} = v, \; \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} but that doesn't stop me from computing: \begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_y(v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_y(v) \begin{pmatrix} \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 \\ \frac{v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{1 - \frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} \Delta t \\ 0 \\ \frac{v \Delta t - v \Delta t}{1 - \frac{v^2}{c^2}} \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ 0 \\ 0 \\ \Delta z \end{pmatrix} And so the Lorentz transform in the opposite direction undoes the coordinate change, including any time dilation effects. ------------------- Or did you mean: \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x(v) \begin{pmatrix} \Delta t \\ 0 \\ \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{- v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta z \end{pmatrix} ? Well, \Delta x = 0, \; \frac{\Delta y'}{\Delta t'} = v, \; \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} but that doesn't stop me from computing: \begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x(-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x(-v) \Lambda_x(v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ 0 \\ \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \\ \Delta z \end{pmatrix} . Once again, the Lorentz transform in the opposite direction undoes the coordinate change, including any time dilation effects. It still appears you are repeating the mistake of Dingle and so your criticisms of Special Relativity reveal only your lack of familiarity with Special Relativity and therefore miss the target. Masterov09-15-11, 04:32 AMI was referring to the case: d|\vec r|/dt = 0 d\phi/dt = const Only transverse Doppler effect. Longitudinal Doppler effect is absent completely. Tach09-15-11, 08:37 AMI was referring to the case: d|\vec r|/dt = 0 d\phi/dt = const Only transverse Doppler effect. Longitudinal Doppler effect is absent completely. But you don't understand the TDE, it is a direct confirmation of time dilation. Masterov09-15-11, 08:41 AMCompare both teories: If the observer is stationary. http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L.gif then a flight time is independent of direction: T_1=T_2=\frac{L}{c} If the observer moves: http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif then the flight time in different directions will be different for all theories (Einstein's theory included). T'_1=\frac{L'}{c+v} T'_2=\frac{L'}{c-v} For both theories: T'_1\neq T'_2 In Master Theory: T'_1+T'_2=T_1+T_2 \frac{L'}{c+v}+\frac{L'}{c-v}=\frac{2L}{c} ______________________________________________ Einstein theory: (T'_1+T'_2)\sqrt{1-v^2/c^2} =T_1+T_2 (\frac{L'}{c+v}+\frac{L'}{c-v})\sqrt{1-v^2/c^2} =\frac{2L}{c} ______________________________________________ In Master Theory: \frac{T'}{T}=1 \frac{L'}{L}=1-\frac{v^2}{c^2} Einstein theory: \frac{T'}{T}=\frac{1}{\sqrt{1-v^2/c^2}} \frac{L'}{L}=\sqrt{1-v^2/c^2} ______________________________________________ x'^2 - (ct')^2 = 0 - is not correct. Tach09-15-11, 08:57 AMCompare both teories: If the observer is stationary. http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L.gif then a flight time is independent of direction: [/tex]T_1=T_2=\frac{L}{c} [/tex] If the observer moves: http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif then the flight time in different directions will be different for all theories (Einstein's theory included). T'_1=\frac{L}{c+v} T'_2=\frac{L}{c-v} Both cases: T'_1\neq T'_2 In Master Theory: T'_1+T'_2=T_1+T_2 This is because "Master" theory is so stupid that it has no notion of time dilation/length contraction. This, in turn, is a consequence of the fact that the "Master" theory is so stupid that it still uses the Galilei transforms (absolute time) instead of the Lorentz transforms. The above idiocies mean, amongst other things that : 1. you cannot explain the null result of the Michelson-Morley experiment 2. your theory fails the invariance of Maxwell's equations (we have already seen ample proof that you don't understand this issue). Masterov09-15-11, 09:09 AMTach, I wrote to you, that talk to you I have no intentions. Do not you understand me? (Apparently, my English is not good enough.) Tach09-15-11, 09:19 AMTach, I wrote to you, that talk to you I have no intentions. Do not you understand me? (Apparently, my English is not good enough.) This doesn't stop me from pointing out the errors in your theory. Masterov09-15-11, 10:14 AMThis doesn't stop me from pointing out the errors in your theory.You do not indicate an error. You (in rough form) are allege about my error, but not able to substantiate his claim. So behave religious fanatics. You are not a scientist. You - a religious fanatic who professes "scientific" theories such as Einstein's theory. Scientist not only indicates an error opponent, but proves it. The fanatic knows the correct answer, but can not prove its correctness. Scientists theorize that knows the correct answer, and in the controversy is trying to prove it. You did not interest me. Tach09-15-11, 10:44 AMYou do not indicate an error. You have some very strong delusions. You (in rough form) are allege about my error, but not able to substantiate his claim. Really?Then how come you are unable to explain the null result of Michelson-Morley with your "theory"? How come that you can't show the invariance of the Maxwell equations? How come you don't understand the transverse Doppler effect? How come you can't explain why the muons formed high in the atmosphere reach the Earth? Eh? So behave religious fanatics. You are not a scientist. You - a religious fanatic who professes "scientific" theories such as Einstein's theory. LOL. Scientist not only indicates an error opponent, but proves it. I did, and so did rpenner, you are just too stupid to understand the proofs. Masterov09-15-11, 11:11 AMYou have some very strong delusions.1. You are not proved it. 2. You importuned me (as a gypsy on market-place). Tach09-15-11, 11:17 AM1. You are not proved it. Of course not, this is why you are a crackpot, insensitive to all disproofs :-) Masterov09-15-11, 12:34 PMOf course not, this is why you are a crackpot, insensitive to all disproofs :-)From this point on, I'm going to ignore you. Tach09-15-11, 01:32 PMFrom this point on, I'm going to ignore you. That's ok, the mainstream physicists ignore your crackpot theories. rpenner09-15-11, 03:12 PMThat's impossible! Return home will require the acceleration, which again slows down time. Subsequent braking will slow down again. That's why it was claimed you were repeating one of the mistake of Dingle, because you were focusing on the term \frac{\partial \Delta t'}{\partial \Delta t} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} when the Lorentz transform is more than that, it's \Delta t' = \frac{\Delta t - \frac{v}{c^2}\Delta x}{\sqrt{1 - \frac{v^2}{c^2}}}. This is the same mistake as claiming a rotation left by 30 degrees isn't canceled by a rotation right by 30 degrees because \cos \, 30^{\circ} = \cos \, -30^{\circ} = \frac{\sqrt{3}}{2} when the relevant expression is \begin{pmatrix} \cos \, 30 ^{\circ} & \quad & - \sin \, 30^{\circ} \\ \sin \, 30^{\circ} & \quad & \cos \, 30^{\circ} \end{pmatrix} \begin{pmatrix} \cos \, -30 ^{\circ} & \quad & - \sin \, -30^{\circ} \\ \sin \, -30^{\circ} & \quad & \cos \, -30^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & \quad & - \frac{1}{2} \\ \frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{3}}{2} & \quad & \frac{1}{2} \\ -\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2}\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \frac{1}{2} - \frac{\sqrt{3}}{2}\frac{1}{2} \\ \frac{\sqrt{3}}{2} \frac{1}{2} - \frac{\sqrt{3}}{2}\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2}\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} So you criticize relativity because you don't understand relativity. And you don't understand relativity because, at a minimum, you don't understand the mathematics of the Lorentz transform. Specifically, you don't understand that a Lorentz transform of v is canceled by a Lorentz transform of -v (in the same direction). Moving across a view axis (the distance does not change): \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta x = 0 I was referring to the case: d|\vec r|/dt = 0 d\phi/dt = const Only transverse Doppler effect. Longitudinal Doppler effect is absent completely. How is that a rebuttal of my disproof of your claim that "Return home will require the acceleration, which again slows down time. Subsequent braking will slow down again" ? (emphasis added) You have either lost track of what you were claiming or dishonestly changed the subject. Masterov09-16-11, 12:55 AMAcceleration is the acceleration, even if you call it braking. So: any inclusion of jet engines will result in a time dilation. You have either lost track of what you were claiming or dishonestly changed the subject.I did not change the subject, but tried to return to it. Masterov09-16-11, 05:46 AMEssence of Master Theory are reduced to a simple idea: the absoluteness of a cross-scales of SRT is not justified. Therefore: we obtain opportunity deprive the cross-scale of this quality, that is, cross-scale can be relative. Thus get a free parameter for each value that you can build a separate theory of relativity, each of which will have as much right exist, as special relativity Einstein's theory has. Then I prove rigorously that time must be absolute (otherwise violates the principle of causality). So (of the entire infinite set of alternatives SRT) remain in force only one a theory which I called Master Theory. Tach09-16-11, 09:09 AM. Then I prove rigorously that time must be absolute (otherwise violates the principle of causality). So (of the entire infinite set of alternatives SRT) remain in force only one a theory which I called Master Theory. ...which is a piece of garbage no one gives a sh^t about Masterov09-23-11, 06:10 AMBBC news 22 September 2011 Last updated at 17:28 GMT Speed-of-light experiments give baffling result at Cern (http://www.bbc.co.uk/news/science-environment-15017484) hyperz, my thanks to you for this news. rpenner09-23-11, 06:14 AMWrong thread. All discussion of the OPERA neutrino experiment and evaluations of it's strength belong here: http://sciforums.com/showthread.php?t=110051 Masterov09-23-11, 07:03 AMWrong thread. All discussion of the OPERA neutrino experiment and evaluations of it's strength belong here: http://sciforums.com/showthread.php?t=110051It's new for me too. Thank you. Master Teory meet with experimental approval. Einstein theory meet with experimental disproof. It cause joy to me. Tach09-23-11, 08:57 AMIt's new for me too. Thank you. Master Teory meet with experimental approval. Einstein theory meet with experimental disproof. It cause joy to me. You have no proof that the experiment confirms your garbage. Besides, what will you do when the experiment is proven wrong? Masterov09-24-11, 02:34 AMEinstein bestowed the absoluteness to a cross-scale by x^2 - (ct)^2 =x'^2 - (ct')^2=0 , but this expression is not correct.(T_1'\neq T_2') x^2 - (ct)^2 =0 - the observer stood motionless. x_1=ct x_2=-ct (x' - (c+v)t_1')(x' - (c-v)t_2')=0 - the observer moved. x_1'=(c+v)t_1' x_2'=(-c+v)t_2' x_1'=-x_2'=x' t_1'+t_2'=2t t_2'-t_1'=\frac{x'}{c-v}-\frac{x'}{c+v}=\frac{2x'v}{c^2-v^2} t_1'=t-\frac{x'v}{c^2-v^2}; t_2'=t+\frac{x'v}{c^2-v^2} (x' - (c+v)(t-\frac{x'v}{c^2-v^2}))(x' - (c-v)(t+\frac{x'v}{c^2-v^2}))=0 (x' - (c+v)t+\frac{x'v}{c-v})(x' - (c-v)t-\frac{x'v}{c+v})=0 (x'(c-v) - (c^2-v^2)t+x'v)(x'(c+v) - (c^2-v^2)t-x'v)=0 (x'c - (c^2-v^2)t)(x'c - (c^2-v^2)t)=0 (x'c - (c^2-v^2)t)^2=0 \frac{x'}{1-v^2/c^2}-ct=0 x'=x(1-v^2/c^2) For SRT: t_1'+t_2'=2t/\sqrt{1-v^2/c^2} x'=x\sqrt{1-v^2/c^2} x'^2 - (ct')^2\neq 0 (x' - (c-v)t_1')(x' - (c+v)t_2')=0 Einstein was wrong. ______________________________________ Tach, I do not read your messages, because I have no translation of it. But your cackling do weary people who can (and therefore - are forced to) read your crow. You are importunate to an obscene. You must roll into a tube your mustachioed uncle and his wacky theories and shove it up your ass. Place of it there. I hope that the moderator will not to do summons to me (to force me to do courtesy for this boor). Masterov09-24-11, 03:29 AMEasy mathematical problem: Two travelers going same speed (v) L - distance one another. A dog runs Between them (c - speed of it). How long does dog to run forth and how long - back? A childs capable of solve this problem, but Sirs: Einstein, Lorentz, Minkowski, and all those who professed SRT - no capable. Tach09-24-11, 09:13 AMFor SRT: t_1'+t_2'=2t/\sqrt{1-v^2/c^2} x'=x\sqrt{1-v^2/c^2} x'^2 - (ct')^2\neq 0 (x' - (c-v)t_1')(x' - (c+v)t_2')=0 Einstein was wrong. . You don't have a clue what SRT is saying. AlphaNumeric09-24-11, 09:21 AMx^2 - (ct)^2 =0 - the observer stood motionless.No, that means the object is moving at light speed. If v = \frac{dx}{dt} and x = ct then v=c. Simple. That invalidates everything afterwards because you put in v incorrectly. Einstein was wrong.No, you just can't do simple algebra. You must roll into a tube your mustachioed uncle and his wacky theories and shove it up your ass. Place of it there.Is that some non-English phrase which doesn't translate properly? I hope that the moderator will not to do summons to me (to force me to do courtesy for this boor).No, I think they will 'do summons to you' for telling someone to shove it up their ass. Masterov09-24-11, 12:39 PMNo, that means the object is moving at light speed. If v = \frac{dx}{dt} and x = ct then v=c. Simple. That invalidates everything afterwards because you put in v incorrectly. No, you just can't do simple algebra. Give prove (in SRT context): T_1'=T_2' http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif ______________________________________ Trailer record of previous my post was addressed to Tach (not to you). Tach09-24-11, 12:41 PMGive prove (in SRT context): T_1'=T_2' http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif In the frame of the train, T_1=T_2. This was known since Galilei. Are you THAT ignorant? In any other frame, moving with respect to the train, T'_1 \ne T'_2. This was also known since Galilei. Are you THAT ignorant? Masterov09-29-11, 05:37 AMThe mathematical expression: x^2-(ct)^2=(x')^2-(ct')^2=0 is not a consequence of the universality of the speed of light. So the result is the following expression: x^2-(ct)^2=(x'-vt')^2-(ct')^2=0 Look at the following animation to see this: http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif You should see that the transit time of red spots are different in different directions, despite the fact that the rate of red spots is equal(same) in both directions. The first expression allege that these times are identical. Let solve one a school's puzzle: 1. Two travelers do walking on the road with equal speed (v) in one direction at a distance (L) from each other. 2. Between them runs a dog (speed of it c). QUESTION: How much time a dog runs forward, and how many - back? ANSWER: T_1=L/(c+v) and T_2=L/(c-v) But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression. Masterov09-29-11, 08:17 AMLet talk about an electron-neutrino and an electron-antineutrino. The latter occurs in a pair with a positron. (A first - in a pair with an electron.) But those neutrinos (and antineutrinos) are not some separate particles. When paired with an electron positron born. And nothing else. What is called the electron neutrino is a positron, whose rate is greater than the speed of light. Well, the electronic antineutrino is - an electron whose velocity is greater than the speed of light. The mass of both are 0.5 MeV (0.28 eV obtained by SRT-formulas, when the mass of the speed dependent). But mass no depend on a speed today. These neutrinos and antineutrinos are produced as a result of common (cascading) decay, when an particle decays into a pair of electron + positron. In some cases, the electron flies in the same direction, which gave birth to it. In this case, the electron (speed of which > c) is called the electron antineutrino. If the positron flew forward (the electron - back) then the positron called by electron neutrino. In those cases where the electron and positron fly away (to broadside direction), and result of their speed does not exceed the speed of light, we are witnessing the birth of an pair of electron+positron. It would be strange to call the bus, which moves in some other way (not, as we call the bus, which stoped). But physics have it. We called neutrino all elementary particles (whose velocity is greater than the speed of light). Tach09-29-11, 09:17 AM1. Two travelers do walking on the road with equal speed (v) in one direction at a distance (L) from each other. 2. Between them runs a dog (speed of it c). QUESTION: How much time a dog runs forward, and how many - back? ANSWER: T_1=L/(c+v) and T_2=L/(c-v) But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression. No, none of them says such an idiocy, only YOU claim that they do. Big difference. Masterov10-03-11, 01:08 AMMy elucidations: The observer at rest: x_1 - it's path of light from right to left = distance between the mirrors. x_2 - it's path of return. -x_2 - the distance between the mirrors. ( x_2=-x_1 ) path_1 = -path_2 = x_1 The observer move ( v>0 ) for Master Theory: x_1' - distance between the mirrors for moving observer. ( x_1'=x_1(1-v^2/c^2) ) path_1' = x_1'/(1-v/c) - it's path of light from right to left for moving observer. path_2' = -x_1'/(1+v/c) - it's path of return. Tese pathes is roots of this equation: (path'-vt)^2-(ct)^2=0 and path_1' \neq -path_2' SRT pathes is roots of this equation: (path')^2-(ct')^2=0 and path_1' = -path_2' . (It's not correct.) AlphaNumeric10-03-11, 02:15 AMQUESTION: How much time a dog runs forward, and how many - back? ANSWER: T_1=L/(c+v) and T_2=L/(c-v) But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression.Einstein et al don't use the transformations you do. You use the following : x' = x-vt and t' = t They use the following : x' = \gamma(x-vt) and t' = \gamma(t-vx) If you put their expressions into x^{2}-t^{2} you indeed get (x')^{2}-(t')^{2}. The transformation rules you use, the Galilean ones, are demonstrably not how the universe works. Your entire argument is "Relativity is wrong because it isn't what I claim". Masterov10-03-11, 02:52 AMSlowing down time can not change the fact: T_1\neq T_2. AlphaNumeric10-03-11, 03:09 AMIt depends on your frame. In the frame where the box is moving, no the times are different. In the frame where the box is stationary the times are the same, since in that frame v=0. Masterov10-03-11, 03:34 AM... Masterov10-03-11, 03:34 AMThis is gibberish. Look at the following animation to see this: http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif You should see that the transit time of red spots are different in different directions, despite the fact that the rate of red spots is equal(same) in both directions. Slowing down time can not change the fact: T_1\neq T_2. AlphaNumeric10-03-11, 07:26 AMYes, in the frame where the box moves the transition times are different. But if someone is inside the box, moving along with it, then they are in the box's rest frame and in that frame the transition times are equal. Do the Lorentz transform and you'll see. Different frames will not necessarily agree on transit times. Some will see T_{1}=T_{2} and some will see T_{1} \neq T_{2}. Masterov10-03-11, 07:45 AMI think that a neutrino detector can be constructed from a balloon with compressed hydrogen. (The mass of the hydrogen atom is 28 times lighter than the nitrogen molecule and 32 times lighter than oxygen, which means that hydrogen is more than five faster than any of them.) This balloon should be lowered to great depths in the ocean. (Near Mariana Trench.) At 11 km depth, under pressure of 1,100 atmospheres, the density of hydrogen would be 1,000 times more. (100kg/m3) In this case, hydrogen can be obtained (by electrolysis of distillate) directly into the cylinder, while immersing the container into the depths. (This - as an option.) Can be in a different way: using a cascade of pumps at different depths. And it is possible: instead of pumps can be used with a balloon+plunger. Hanging on the ropes these balloons (in the form of two garlands), dropping and picking up by turns - you compel them inject hydrogen into the depths. And you can still way: Balon Dewar (with liquid hydrogen) set cylinder. The hydrogen to do evaporation - balon go down. (This option seems most appropriate.) Let me explain: Neutrinos with matter should not have to interact with, and (if their rate significantly greater than the speed of light) - so be it. Methods which try to detect neutrinos today, you can catch those neutrinos, whose rate is only slightly greater than the speed of light (more on the rate of Brownian motion). Of the neutrino flux filtered out by those whose speed exceeds the speed of light is greater than the rate of Brownian motion of granite. Therefore, the number of neutrinos passing through the Earth increases sharply if their speed exceeds the speed of light is greater than the rate of Brownian motion in the interior of the earth. Water molecules are lighter molecules granite. Therefore, the velocity of the Brownian motion of water molecules in two - two and a half times (the square root of mass ratio) of greater. Therefore, water molecules move twice (at least) faster than the molecules of granite. This means that there is a chance to catch the water in those neutrinos which are not extinguished in the interior of the earth, but the speed is (still) not much greater than the speed of light. If used as a medium for hydrogen, the sensitivity will increase many times, since the rate of Brownian motion of the atoms of hydrogen is ten times higher than that of granite. Hence: the hydrogen is able to catch those neutrinos, which are almost not able to catch the granite (and water). Should be a lot of neutrinos (for which the thickness of the Earth is transparent). And in Baykal lake to catch a cosmic neutrinos is even more difficult, because Brownian speed of motion of atmosphere much more than in water. Atmosphere do swallow up those neutrinos which could be fixed in the lake. Masterov10-03-11, 07:48 AMYes, in the frame where the box moves the transition times are different. But if someone is inside the box, moving along with it, then they are in the box's rest frame and in that frame the transition times are equal. Do the Lorentz transform and you'll see. Different frames will not necessarily agree on transit times. Some will see T_{1}=T_{2} and some will see T_{1} \neq T_{2}.T_{1}=T_{2} for v=0 only. If v \neq 0 then T_{1}\neq T_{2}. Masterov10-03-11, 10:54 AMElectron neutrino is a positron, whose rate is higher than the speed of light. And an electron antineutrino is an electron whose velocity is greater than the speed of light. We can calculate the velocity of neutrinos, since the mass of the electron and positron are known. origin10-03-11, 11:30 AMElectron neutrino is a positron, whose rate is higher than the speed of light. And an electron antineutrino is an electron whose velocity is greater than the speed of light. Really? What do you base this on. One would think that if a neutrino had a charge that it would readily interact with matter. Tach10-03-11, 11:37 AMT_{1}=T_{2} for v=0 only. If v \neq 0 then T_{1}\neq T_{2}. You have been TOLD the above about 50 times already. Now you come back by telling us the same thing we've been telling you? What sort of twisted brain do you have? You aren't only ignorant, you are also utterly dishonest. OnlyMe10-03-11, 11:45 AMYou have been TOLD the above about 50 times already. Now you come back by telling us the same thing we've been telling you? What sort of twisted brain do you have? Maybe he has learned.., without recognizing he has! Masterov10-03-11, 12:30 PMReally? What do you base this on. One would think that if a neutrino had a charge that it would readily interact with matter.No. 1. Electron is born in a pair with a positron. 2. Mass on the velocity-independent. 3. Neutrinos move faster than the speed of light (otherwise it would not neutrinos). origin10-03-11, 12:38 PMNo. 1. Electron is born in a pair with a positron. 2. Mass on the velocity-independent. 3. Neutrinos move faster than the speed of light (otherwise it would not neutrinos). Huh? You said: Electron neutrino is a positron So if the neutrino is a positron (has a charge), shouldn't it interact rather easily with matter? As far as #3, you seem to be very skeptical of scientific data - but you are accepting this preliminary finding as a fact. Seems rather ironic to me.;) Masterov10-03-11, 12:55 PMSo if the neutrino is a positron (has a charge), shouldn't it interact rather easily with matter?Matter, whose speed is greater than the speed of light does not interact with EMF, if the source field is static. More than that, the matter (whose speed is greater than the speed of light) does not interact with a static matter. Positron seems neutral, since his speed than the speed of light. Positron does not interact with static EMF and a static matter, since has speed greater than the speed of light. As far as #3, you seem to be very skeptical of scientific data - but you are accepting this preliminary finding as a fact. Seems rather ironic to me.;)I suggest you discuss these facts. origin10-03-11, 02:24 PMMatter, whose speed is greater than the speed of light does not interact with EMF, if the source field is static. More than that, the matter (whose speed is greater than the speed of light) does not interact with a static matter. Positron seems neutral, since his speed than the speed of light. Positron does not interact with static EMF and a static matter, since has speed greater than the speed of light.I suggest you discuss these facts. For my degree I only needed 3 semesters of physics so I a may not have all of the facts, but I am fairly certain I recall that in physics you cannot just make shit up to support your position. RJBeery10-03-11, 03:01 PMfor my degree i only needed 3 semesters of physics so i a may not have all of the facts, but i am fairly certain i recall that in physics you cannot just make shit up to support your position. lol! AlphaNumeric10-03-11, 03:02 PMMatter, whose speed is greater than the speed of light does not interact with EMF, if the source field is static. More than that, the matter (whose speed is greater than the speed of light) does not interact with a static matter. Positron seems neutral, since his speed than the speed of light. Positron does not interact with static EMF and a static matter, since has speed greater than the speed of light.Except that we have instances where charged particles move faster than light in a medium and not only do the charged particles interact with things but they give off a very distinct signature. It's known as Cherenkov radiation. Your argument is like saying things which move faster than the speed of sound should be silent. Trust me, you'd know it it an F22 went past you. 3. Neutrinos move faster than the speed of light (otherwise it would not neutrinos).That is an assertion, not evidence. Calling it a fact doesn't make it so. Masterov10-04-11, 02:06 AMYour argument is like saying things which move faster than the speed of sound should be silent. Trust me, you'd know it it an F22 went past you.This noise creates a jet of gas of engine backwards. This jet velocity (relative to the still air) is less than the speed of sound. That is an assertion, not evidence. Calling it a fact doesn't make it so.An argument is not always proof. Argument justifies the assumption of certainty, but evidence proves a hypothesis. AlphaNumeric10-04-11, 02:47 AMThis noise creates a jet of gas of engine backwards. This jet velocity (relative to the still air) is less than the speed of sound.Yes, but it still interacts and produces noise. Despite moving faster than sound it still has an effect to produce sound in some obvious way. You're claiming the neutrino is a faster than light charged particle. The neutrino doesn't interact with electromagnetism at all, while charged particles which move faster than light (in the local medium) still produce large electromagnetic effects. The electromagnetic version of a sonic boom is Cherenkov radiation. If the neutrino were a faster than light charged particle it would produce such an effect. It doesn't. Your claim does not square with reality. An argument is not always proof. Argument justifies the assumption of certainty, but evidence proves a hypothesis.And the evidence is against you. Masterov10-04-11, 06:44 AMYes, but it still interacts and produces noise. Despite moving faster than sound it still has an effect to produce sound in some obvious way. You're claiming the neutrino is a faster than light charged particle. The neutrino doesn't interact with electromagnetism at all, while charged particles which move faster than light (in the local medium) still produce large electromagnetic effects. The electromagnetic version of a sonic boom is Cherenkov radiation. If the neutrino were a faster than light charged particle it would produce such an effect. It doesn't. Your claim does not square with reality.Any accelerator can not disperse a particles to speeds greater than the speed of light, but not because the mass or energy of a particles increases without limit, so that charged particles with velocities approaching the speed of light, cease to interact with the field, whose source is stationary. Once the particle velocity was greater than the speed of light, it ceases to interact with static fields and static matter. Such a particle is converted into neutrinos. An argument is not always proof. Argument justifies the assumption of certainty, but evidence proves a hypothesis.And the evidence is against you.Show me it. AlphaNumeric10-04-11, 07:51 AMCherenkov radiation, look it up. I never said charged particles move faster than light in a vacuum, but they can be faster than light in a medium like water. Masterov10-04-11, 09:25 AMCherenkov radiation, look it up. I never said charged particles move faster than light in a vacuum, but they can be faster than light in a medium like water.Once any particle overcomes the speed of light, this particle is no longer interact with the environment. So do not expect Cherenkov radiation. origin10-04-11, 09:52 AMso that charged particles with velocities approaching the speed of light, cease to interact with the field, whose source is stationary. Absolutely not true. As AlphaNumeric pointed out there are electrons (beta radiation) that eminate from reactors that is moving faster than the speed of light in water and these electrons emit Cherenkov radiation. As someone who worked in radcon around a nuclear reactor believe me when I say that beta radiation composed of electrons traveling close to the speed of light will readily interact with matter causing a large number of ionizing event before they stop. A 2.3 MeV beta particle will only make it 10 mm or less through most materials, heck it will only make it about 8 meters through the air. origin10-04-11, 09:53 AMOnce any particle overcomes the speed of light, this particle is no longer interact with the environment. Why? OnlyMe10-04-11, 10:20 AMWhy? Because! (This was not intended as a serious response.) Masterov10-04-11, 11:09 AMso that charged particles with velocities approaching the speed of light, cease to interact with the field, whose source is stationary.Absolutely not true. As AlphaNumeric pointed out there are electrons (beta radiation) that eminate from reactors that is moving faster than the speed of light in water and these electrons emit Cherenkov radiation. As someone who worked in radcon around a nuclear reactor believe me when I say that beta radiation composed of electrons traveling close to the speed of light will readily interact with matter causing a large number of ionizing event before they stop. A 2.3 MeV beta particle will only make it 10 mm or less through most materials, heck it will only make it about 8 meters through the air.I know that the penetration of relativistic particles if their speed approaching the speed of light. Am I right, or this information of fast neutrons only. But in any case, our dispute may only allow direct measurement of the energy of relativistic particles in the calorimeter. If it is proved experimentally that the energy of relativistic particles actually grows indefinitely (if the velocity of the particles approaches the speed of light) - any reason for the dispute will remain. If you have any experimental results of these experiments, please give a link to where you can see it. Masterov10-04-11, 11:12 AMOnce any particle overcomes the speed of light, this particle is no longer interact with the environment.Why?Doppler effect prevents. t'=t-\frac{x(t')}{c} V(t')=\frac{dx(t')}{dt'} \frac{dt'}{dt}=1-\frac{V}{c}\frac{dt'}{dt} \frac{dt'}{dt}=\frac{1}{1+V/c} if V=-c then \frac{dt'}{dt}=\infty if V<-c then \frac{dt'}{dt}<0 - observer will see the reverse sequence of events, and this is impossible. OnlyMe10-04-11, 11:13 AMDoppler effect prevents. How does the Doppler effect prevent this? origin10-04-11, 12:06 PMI know that the penetration of relativistic particles if their speed approaching the speed of light. Am I right, or this information of fast neutrons only. I am not sure what you are trying to say - sorry. Maybe this will help. The shielding of nuclear reactors is based on 1/10 thickness of material. Essentially, one tenth thickness will attenuate 90 of the ionizing radiation from the reactor (or radiactive source). The next tenth thickness will attenuate 90% of the remaining radiation and so on. Neutrons are not ionizing since they have no charge but by using a material such as water or a large chain organic like polypropylene the neutrons will 'hit' a proton transfering some portion of their KE to the proton causeing secondary ionizations. At any rate the energy of a radioactive decay is relatively easy to determine so that we know what the energy and hence the speed of the ejected particles. The makers of shielding test the ability of there materials to attenuate different types of radiaton by simple and effective experiments. Simply measure the radiation at the source and on the otherside of the shielding material, while taking into account the 1/r^2 distance. I have proved the effectiveness of shielding during my radcon training. But in any case, our dispute may only allow direct measurement of the energy of relativistic particles in the calorimeter. If it is proved experimentally that the energy of relativistic particles actually grows indefinitely (if the velocity of the particles approaches the speed of light) - any reason for the dispute will remain. Not sure what you are getting at here. Look at this (http://en.wikipedia.org/wiki/Decay_energy) (yeah I know it is wiki) If you have any experimental results of these experiments, please give a link to where you can see it. Don't know what you are looking for. origin10-04-11, 12:10 PMBecause! (This was not intended as a serious response.) It is however, pretty close to the actual response.:D Masterov10-05-11, 02:30 AMI am not sure what you are trying to say - sorry.Neutrino has an unlimited ability to penetrate into matter. Relativistic neutrons seek to do so if their rate tends to the speed of light. Cherenkov's effect does not interfere for a neutrons. The charged particles would behave the same way as the neutron, if Cherenkov effect were not interfere. However, all particles (electrons and positrons - too) get unlimited ability to penetrate into matter, if their speed more than the speed of light. Cherenkov's effect does not operate at speeds exceeding the speed of light in a vacuum. Any (hyper-light-speed) matter has unlimited ability to penetrate into any statical matter. All converted into neutrinos, if moving faster than light. origin10-05-11, 12:49 PMNeutrino has an unlimited ability to penetrate into matter. If that was true we would NEVER detect them. Relativistic neutrons seek to do so if their rate tends to the speed of light. Untrue Cherenkov's effect does not interfere for a neutrons. Correct, Cherenkov Radiation is limited to charged particles. The charged particles would behave the same way as the neutron, if Cherenkov effect were not interfere. No they would not. A charged particle slows by causing ionizations of the surrounding material even with out Cherenkov radiation. Neutrons do not cause ionization, they slow by collisions with other particles. However, all particles (electrons and positrons - too) get unlimited ability to penetrate into matter, if their speed more than the speed of light. Completely unevidenced crap that you just made up. You can't do that - it is cheating. Cherenkov's effect does not operate at speeds exceeding the speed of light in a vacuum. Why? How do you know? Any (hyper-light-speed) matter has unlimited ability to penetrate into any statical matter. Completely unevidenced crap that you just made up. You can't do that - it is cheating. All converted into neutrinos, if moving faster than light.[/QUOTE] Completely unevidenced crap that you just made up. You can't do that - it is cheating. Masterov10-07-11, 02:16 AMNeutrino has an unlimited ability to penetrate into matter. ” If that was true we would NEVER detect them.It's true. We can register the slowest neutrinos only, whose velocity exceeds the speed of light on the rate of Brownian motion in the environment of the registrar. We can not register faster neutrinos today. In order to build a more sensitive detector of neutrinos, must be used such matter, molecules of which is lightweight . For that can used hydrogen or helium under tremendous pressure (in a deep ocean). Propose to use a balloon, such as: http://masterov.qptova.ru/Images/Ballon.gif As the evaporation of the hydrogen / helium will increase the pressure inside the balloon. The higher internal pressure, the deeper into the ocean, omit the balloon. At a depth of 10 kilometers pressure exceeds 1,000 kilograms per square centimeter. The density of hydrogen will be exceed 90 kilograms per cubic meter. Relativistic neutrons seek to do so if their rate tends to the speed of light.UntrueTrue, true. It is well known: the higher the energy of the neutrons, the smaller scattering cross-section of it, the deeper they penetrate into matter. This fact has long been known and has long been used in nuclear power plants. In nuclear power plants are used graphitic shanks (in order to slow the neutrons and increase the likelihood of their interactions with uranium). Cherenkov's effect does not interfere for a neutrons.Correct, Cherenkov Radiation is limited to charged particles.It's true, as long as the speed electron not exceeds the speed of light. When the electron velocity exceeds the speed of light, the electrons turn into a neutrino. The charged particles would behave the same way as the neutron, if Cherenkov effect were not interfere.No they would not. A charged particle slows by causing ionizations of the surrounding material even with out Cherenkov radiation. Neutrons do not cause ionization, they slow by collisions with other particles.Cherenkov effect are actual for speed < c. However, all particles (electrons and positrons - too) get unlimited ability to penetrate into matter, if their speed more than the speed of light.Completely unevidenced crap that you just made up. You can't do that - it is cheating.Experiments argue that the SRT is Completely unevidenced crap and cheating. Cherenkov's effect does not operate at speeds exceeding the speed of light in a vacuum.Why? How do you know?Because the electron neutrinos and electron antineutrinos === positron and electron (respectively). Calculate the mass of the neutrino and antineutrino, to be sure. (Mass of it does not depend on the speed.) On 4th December 1930, Pauli wrote his famous letter http://www.library.ethz.ch/exhibit/pauli/images/M23-2.GIF Pauli wrote: neutrinos has Spin 1/2 and comparable electron mass. But mass neutrinos = 0.28Ev and mass neutrinos = 500 000Ev Is it comparable? Yes? origin10-07-11, 10:04 PM.Because the electron neutrinos and electron antineutrinos === positron and electron (respectively). The neutrino was first theorized becasue energy and momentum were not conserved in certain radioactive decays. An example of a radiactive decay that produces a neutrino is: http://content.answcdn.com/main/content/img/oxford/Oxford_Chemistry/0192801015.beta-decay.1.jpg If a neutrino has the charge of a electron or a positron how do you explain the above radioactive decay? Why is the charge not conserved? Masterov10-07-11, 11:34 PMThe neutrino was first theorized becasue energy and momentum were not conserved in certain radioactive decays. An example of a radiactive decay that produces a neutrino is: http://content.answcdn.com/main/content/img/oxford/Oxford_Chemistry/0192801015.beta-decay.1.jpg If a neutrino has the charge of a electron or a positron how do you explain the above radioactive decay? Why is the charge not conserved?This question is complicated. I can only assume. Positron annihilation is possible, which turn a neutron into a proton. Electron leaves the nucleus of an atom. Neutrino rest-mass transform into tiny in consequence of an assumption that the mass depends on speed. This assumption is erroneous. Mass is an absolutely. (Time, \pi, speed of light - too.) funkstar10-08-11, 03:00 AMAn example of a radiactive decay that produces a neutrino is: http://content.answcdn.com/main/content/img/oxford/Oxford_Chemistry/0192801015.beta-decay.1.jpg LaTeX is excellent for typesetting stuff like that: {}^{14}_{\;6}\text{C} \rightarrow {}^{14}_{\;7}\text{N} + \text{e}^- + \overline{\nu} Then you don't have to rely on external image hosts that may decay over time... origin10-08-11, 08:19 AMThis question is complicated. I can only assume. Positron annihilation is possible, which turn a neutron into a proton. Electron leaves the nucleus of an atom. What the hell is that suppose to mean. Positron annihilation? A positron is annihilated with an electron, where did the electron come from? If a positron was emitted it could be detected and if an annihilated event occured the gama rays sure as hell would be detected. Why not go with the more reasonable answer - you are wrong that a neutrino is an electron or positron.;) origin10-08-11, 08:21 AMLaTeX is excellent for typesetting stuff like that: {}^{14}_{\;6}\text{C} \rightarrow {}^{14}_{\;7}\text{N} + \text{e}^- + \overline{\nu} Then you don't have to rely on external image hosts that may decay over time... Are there some instructions on how to use this? Thanks prometheus10-08-11, 08:44 AMCheck out this thread (http://www.sciforums.com/showthread.php?t=61223) to start with. I know it feels wrong to google for something like "help with latex" but you will honestly get results that are safe for work. :) origin10-08-11, 09:21 PMThanks for the info on LaTex. Masterov10-09-11, 02:00 AMWhat the hell is that suppose to mean. Positron annihilation? A positron is annihilated with an electron, where did the electron come from? If a positron was emitted it could be detected and if an annihilated event occured the gama rays sure as hell would be detected. Why not go with the more reasonable answer - you are wrong that a neutrino is an electron or positron.;)If neutrino are known to us particles (whose velocity exceeds the speed of light), then candidates for only two: neutrinos are electron and positron. By the way: taon can also be neutrinos (very short period of time). AlphaNumeric10-09-11, 04:52 AMIf neutrino are known to us particles (whose velocity exceeds the speed of light), then candidates for only two: neutrinos are electron and positron.Provide evidence or reasoning. By the way: taon can also be neutrinos (very short period of time).Firstly, what evidence/reason do you have for that. Secondly, what about muons? Thirdly, if neutrinos are electrons and positrons and taus are neutrinos doesn't that make taus electrons? There are ways of distinguishing between electrons, muons and taus. They are all leptons but are in different families, with different rest masses. If the muon were an electron with different energy then there wouldn't be this discrete jump in scattering processes when the energies reach what we consider to be the muon rest mass. Why wouldn't there be a smooth change in cross sections for electron processes as we up the energy? The reason we can measure the muon and tau rest masses is that when the centre of mass energy for those processes each 2 particular energies there is a sudden change in the cross section amplitudes, because now the particles can be created on shell. It's seen directly in e^{-}+e^{+} \to e^{-}+e^{+} processes. For 2m_{e}< E < 2m_{\mu} the amplitude of this process has one profile. When E = 2m_{\mu} the amplitude changes significantly, as now the channel e^{-}+e^{+} \to \mu^{-}+\mu^{+} is open. Ramp it up to E= 2m_{\tau} and the channel e^{-}+e^{+} \to \tau^{-}+\tau^{+} opens. We can observe these things. If you have a copy of Peskin & Schroder it's covered in the section on e^{-}+e^{+} processes. prometheus10-09-11, 04:59 AMIf neutrino are known to us particles (whose velocity exceeds the speed of light), then candidates for only two: neutrinos are electron and positron. By the way: taon can also be neutrinos (very short period of time). You are obviously not aware that neutrinos are produced via weak processes, like d \to u + W^- \to u + e^- + \overline{\nu}_e. Neutrinos exist in three generations associated with electrons, muons and tau particles and each generation contains a neutrino and an antineutrino, making a total of six different types. Masterov10-10-11, 02:09 AMIf neutrino are known to us particles (whose velocity exceeds the speed of light), then candidates for only two: neutrinos are electron and positron.Provide evidence or reasoning.Provide candidate (light and stable, as the electron and positron) of fermions, which could compete with this pair of well known particles. Firstly, what evidence/reason do you have for that. Secondly, what about muons?In the neutrinos can transform any matter if its speed will be faster than the speed of light. None of the decay of quasi-stationary elementary particle can not produce neutrinos. Neutrinos are born as a result of cascading decays: 1. Quasi-stationary material (by decay) creates a moving (no faster of light) matter. 2. This agile (mobile) matter (by decay) creates a moving matter, part of which move faster of light. As a result of the cascade decay of heavier unstable particles are transformed into two (or - more) light particles, one of which moves in the same direction as the moving parent particle. This descendant could be a neutrino, since speed of it added at the speed of the parent particle. Brothers and sisters of the particles are moving in the opposite direction, so parent particle speed subtracted from their speeds. 1. Light particles are more easily converted into neutrinos, because neutrinos are produced in cascade decay. 2. Who among the descendants of the parent will be neutrino - will determine the case. For example, if the decay of a heavy mobile particle to do birth a couple of electron + positron, then we can observe three types of decay: 1. electron + positron, if the parent particle has been slow and / or both of her offspring scattered in the transverse direction. 2. electronic neutrino + electron, if the positron scattered in the same direction as the parent particle. 3. pozitron + electronic antineutrinos, if the electron scattered in the same direction as the parent particle. Thirdly, if neutrinos are electrons and positrons and taus are neutrinos doesn't that make taus electrons? There are ways of distinguishing between electrons, muons and taus. They are all leptons but are in different families, with different rest masses.Any matter are neutrinos, if moving faster than light. Heavy a matter is hard to compel moving faster of light. At any point of space may exist an infinite number of matter, each of which moves faster than light relative to the other. If the muon were an electron with different energy then there wouldn't be this discrete jump in scattering processes when the energies reach what we consider to be the muon rest mass. Why wouldn't there be a smooth change in cross sections for electron processes as we up the energy? The reason we can measure the muon and tau rest masses is that when the centre of mass energy for those processes each 2 particular energies there is a sudden change in the cross section amplitudes, because now the particles can be created on shell. It's seen directly in e^{-}+e^{+} \to e^{-}+e^{+} processes. For 2m_{e}< E < 2m_{\mu} the amplitude of this process has one profile. When E = 2m_{\mu} the amplitude changes significantly, as now the channel e^{-}+e^{+} \to \mu^{-}+\mu^{+} is open. Ramp it up to E= 2m_{\tau} and the channel e^{-}+e^{+} \to \tau^{-}+\tau^{+} opens. We can observe these things. If you have a copy of Peskin & Schroder it's covered in the section on e^{-}+e^{+} processes.Mass is absoluteness (no depend on speed). Light speed, acceleration, \pi and time is absoluteness too. Real speed and real positions are a result of integrating by time of acceleration and do not depend on any properties of electromagnetic wave. All physical laws are the same for any inertial reference frames. Masterov10-10-11, 03:57 AMYou are obviously not aware that neutrinos are produced via weak processes, like d \to u + W^- \to u + e^- + \overline{\nu}_e. Neutrinos exist in three generations associated with electrons, muons and tau particles and each generation contains a neutrino and an antineutrino, making a total of six different types.Any matter are neutrinos, if moving faster of light. Heavy matter is hard to compel moving faster of light. prometheus10-10-11, 04:04 AMMod note: This thread has been moved to a forum that is more appropriate for it's content. origin10-10-11, 08:16 AMIt must be kinda fun making up your own 'master' theory. Everytime some new info comes along you just make up more junk to add to your "theory". Prelimary data seems to show that neutrinos can exceed c. masterov: Of course and neutrinos are, uh, electrons! If they were electrons they would interact with matter. masterov: No because, uh, uh, if you move faster than light you don't interact with matter. Where is your proof? masterov: make a series of wild unsubstantiated claims without evidence, even thought there is plenty of real evidence refuting these claims. Continue to make up additional claims everytime your ideas are refute.... Masterov10-10-11, 11:05 AMNo because, uh, uh, if you move faster than light you don't interact with matter. Where is your proof?My proof lies on your bookshelf: 1. Einstein says that mass and energy increases without limit, if the particle velocity approaches the speed of light. It impede to disperse the particles in an accelerator, to velocities greater than the speed of light. Einstein's lying, and you know it: the mass and energy increases indefinitely on paper only. This monstrous energy give ridiculous zilch in a calorimeter . (Speed up by accelerator a particles faster of light is impossible because the accelerating fields propagate at the speed of light: relativistic particle ceases to interact with these fields (due to the Doppler effect), because source of these fields is stationary.) 2. Time is absolute. Otherwise violates the principle of causality is violated or equality of inertial reference systems. But Einstein says (not having reason for it) that the cross-scales (as opposed to a longitudinal scale) is absolute. 3. Matter can travel faster than light. It demonstrate OPERA experiment. Earlier experimental data were obtained in which taon overcame an incredible distance. This distance Taon could overcome (during Taon lifetime = 5\times 10^{-13} seconds) by moving faster of light only. 4. The mathematical expression: x^2-(ct)^2=(x')^2-(ct')^2=0 is not a consequence of the universality of the speed of light. So the result is the following expression: x^2-(ct)^2=(x'-vt')^2-(ct')^2=0 Look at the following animation to see this: http://masterov.qptova.ru/MasterTheory/Clocks/Clock_L_move.gif You should see that the transit time of red spots are different in different directions, despite the fact that the rate of red spots is equal(same) in both directions. The first expression allege that these times are identical. Let solve one a school's puzzle: 1. Two travelers do walking on the road with equal speed (v) in one direction at a distance (L) from each other. 2. Between them runs a dog (speed of it c). QUESTION: How much time a dog runs forward, and how many - back? ANSWER: T_1=L/(c+v) and T_2=L/(c-v) But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression. origin10-10-11, 01:39 PM3. Matter can travel faster than light. It demonstrate OPERA experiment. You're a hoot. You poo poo all scientific data that shows your conjectures wrong, but give you one piece of preliminary data that you like and all the sudden we have proof. You are just so transparent and disingenuous ... Earlier experimental data were obtained in which taon overcame an incredible distance. This distance Taon could overcome (during Taon lifetime = 5\times 10^{-13} seconds) by moving faster of light only. What is Taon? I googled it and the only hits were this Master Theory something or other? origin10-10-11, 01:44 PM1. Two travelers do walking on the road with equal speed (v) in one direction at a distance (L) from each other. 2. Between them runs a dog (speed of it c). QUESTION: How much time a dog runs forward, and how many - back? ANSWER: T_1=L/(c+v) and T_2=L/(c-v) But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression. If by c you mean the speed of light then I would prefere that fido is a photon. Not only do the physicist say it will be the same measurements prove that it will be the same. Masterov10-11-11, 01:55 AMMatter can travel faster than light. It demonstrate OPERA experiment.You're a hoot. You poo poo all scientific data that shows your conjectures wrong, but give you one piece of preliminary data that you like and all the sudden we have proof. You are just so transparent and disingenuous ...Well, why? There is other evidence that had been received earlier: 1. Were recorded neutrinos, which passed ahead of light on a few hours before. 2. About taone (tau-lepton), I've already said: this particle has flown a longer distance than she could over a period of life by light speed. Taon (obviously) moved faster than light. Masterov10-11-11, 02:05 AMIf by c you mean the speed of light then I would prefere that fido is a photon. Not only do the physicist say it will be the same measurements prove that it will be the same. This text is very difficult to understand. (What is fido?) ____________________________________________ PS You have not answered my first two paragraphs: No because, uh, uh, if you move faster than light you don't interact with matter. Where is your proof?My proof lies on your bookshelf: 1. Einstein says that mass and energy increases without limit, if the particle velocity approaches the speed of light. It impede to disperse the particles in an accelerator, to velocities greater than the speed of light. Einstein's lying, and you know it: the mass and energy increases indefinitely on paper only. This monstrous energy give ridiculous zilch in a calorimeter . (Speed up by accelerator a particles faster of light is impossible because the accelerating fields propagate at the speed of light: relativistic particle ceases to interact with these fields (due to the Doppler effect), because source of these fields is stationary.) 2. Time is absolute. Otherwise violates the principle of causality is violated or equality of inertial reference systems. But Einstein says (not having reason for it) that the cross-scales (as opposed to a longitudinal scale) is absolute. origin10-11-11, 09:21 AMEinstein's lying, and you know it. You really are a freaking genius! Not only do you know Einstien's motivation you know what I think! Maybe this should be in parapsychology. You are starting to rant and rave that does not help your (indefensible) position.;) OnlyMe10-11-11, 10:03 AMThis text is very difficult to understand. (What is fido?) ____________________________________________ PS You have not answered my first two paragraphs: My proof lies on your bookshelf: 1. Einstein says that mass and energy increases without limit, if the particle velocity approaches the speed of light. It impede to disperse the particles in an accelerator, to velocities greater than the speed of light. Einstein's lying, and you know it: the mass and energy increases indefinitely on paper only. This monstrous energy give ridiculous zilch in a calorimeter . (Speed up by accelerator a particles faster of light is impossible because the accelerating fields propagate at the speed of light: relativistic particle ceases to interact with these fields (due to the Doppler effect), because source of these fields is stationary.) 2. Time is absolute. Otherwise violates the principle of causality is violated or equality of inertial reference systems. But Einstein says (not having reason for it) that the cross-scales (as opposed to a longitudinal scale) is absolute. That portion in bold implies that Einstein had access to information that was not available u til long after the theory was published.., and it misrepresents his intent and meaning. It is obvious that at the time of publication his theory on this was, on paper only, as QM and our experience with accelerators came later. However, Einstein was referring to what at the time was referred to as relativistic mass (that termonology is now loosing favor as it results in some confussion) and not an increase in rest mass. Experiments in particle physics demonstrate with some significant accuracy that as you increase the velocity of a particle with mass it requires an exponentially increasing energy to further accelerate it? Calling Eimstein a liar on such flimsy reasoning and more than 50 years after his death is almost as bad as attempting to invoke his name as proof of some knowledge, experiment or idea that he had no knowledge of... Masterov10-11-11, 10:27 AMNot only do you know Einstien's motivation you know what I think!1. Einstein understood that he make mistakes. But he could not to tell World about it because his tribesmen made ​​him a great advertising and turn Einstein into a symbol of genius. Einstein was forced to lie. 2. My questions you understand, but refuse to give an answer. You to evade a question. This proves that you see that Einstein was wrong, but You refuse openly admit it. Hence: you have the motivation to not tell the truth. I think you have the same motivation to lie, what was Einstein. Masterov10-11-11, 10:41 AMExperiments in particle physics demonstrate with some significant accuracy that as you increase the velocity of a particle with mass it requires an exponentially increasing energy to further accelerate it?That does not prove that the energy of relativistic particles well increase exponentially. Moreover, you know that the momentum and energy of relativistic particles do not grow indefinitely, if the particle velocity approaches the speed of light. You understand that this proves the fallacy of Einstein's theory. You have the motivation to not tell the truth too. Calling Eimstein a liar on such flimsy reasoning and more than 50 years after his death is almost as bad as attempting to invoke his name as proof of some knowledge, experiment or idea that he had no knowledge of...1. This forum has not honored courtesy. (Have you noticed it?) 2. To protect SRT used inadmissible methods. Medieval Inquisition applied these methods to protect the religious doctrines. Einstein's theory is not a scientific theory. Einstein's theory is a religious doctrine. This religious doctrine imposed to us. This forum demonstrates it. Knowledge does not make you a scientific. All scientists are motivated to learn the truth. You have another motivation. Einstein did not had the motivation of a scientist too. You have not science interested. You are interested in your position of Science. I would not be fair to you if you answered the questions, but refuse to give an answer. origin10-11-11, 12:03 PM1. This forum has not honored courtesy. (Have you noticed it?) Yes, I noticed it when you said, "I think you have the same motivation to lie, what was Einstein." I do not think you are lying. I think you are too ignorant to know that what you are saying sounds like the ramblings of a fool. origin10-11-11, 12:16 PM1. Einstein says that mass and energy increases without limit, if the particle velocity approaches the speed of light. It impede to disperse the particles in an accelerator, to velocities greater than the speed of light. Einstein's lying, and you know it: the mass and energy increases indefinitely on paper only. The concept that mass cannot travel at c as proposed by SR has been shown experimentally to be accurate. Yes, the amount of energy needed to accelerate a particle increases at an increasing rate the closer you get to c. Again, this has been shown to be true over and over. That you would accuse Einstein of lying about this when it has been experimentally proven time after time shows your willful ignorance. This monstrous energy give ridiculous zilch in a calorimeter . (Speed up by accelerator a particles faster of light is impossible because the accelerating fields propagate at the speed of light: relativistic particle ceases to interact with these fields (due to the Doppler effect), because source of these fields is stationary.) Pretty much meaningless word salad.:rolleyes: 2. Time is absolute. Otherwise violates the principle of causality is violated or equality of inertial reference systems. But Einstein says (not having reason for it) that the cross-scales (as opposed to a longitudinal scale) is absolute. Again, the time dilation affect has been shown experiemntally to be true time and time again. It is an affect that reveals itself in every GPS system in use. The fact that you ignore the obvious is an act of willful ignorance and frankly embarrassingly foolish. Masterov10-12-11, 01:10 AMI do not think you are lying. I think you are too ignorant to know that what you are saying sounds like the ramblings of a fool.Question can not be stupid, but an answer can be stupid. I ask question to you. You have no clever answer. Who is stupid? Masterov10-12-11, 04:35 AMThe concept that mass cannot travel at c as proposed by SR has been shown experimentally to be accurate. Yes, the amount of energy needed to accelerate a particle increases at an increasing rate the closer you get to c. Again, this has been shown to be true over and over. That you would accuse Einstein of lying about this when it has been experimentally proven time after time shows your willful ignorance.Give a (modern and reliable) experimental results of direct measurements of energy and momentum of relativistic particles. Einstein says that energy and momentum infinitely rise, when the particle velocity approaches the speed of light. You know that this is false. But you can not substantiate a negative result. It forces you to lie. You claim that you were able to speed up an electron to monstrous energy, as measured TEv. But this is false. You know it. Pretty much meaningless word salad.:rolleyes: Again, the time dilation affect has been shown experiemntally to be true time and time again. It is an affect that reveals itself in every GPS system in use. The fact that you ignore the obvious is an act of willful ignorance and frankly embarrassingly foolish.The clocks of GPS-satellite differently make mistakes. One clocks outstrip. Other clocks behind time. Patterns are observed. A clocks synchronize when the GPS-satellite passes perihelion. origin10-12-11, 04:51 AMgive a (modern and reliable) experimental results of direct measurements of energy and momentum of relativistic particles. How about the LHC? origin10-12-11, 05:04 AMEinstein says that energy and momentum infinitely rise, when the particle velocity approaches the speed of light. You know that this is false. But you can not substantiate a negative result. It forces you to lie. You have no theory. Experimental evidence shows that SR is correct for the amount of energy needed to raise the speed of a particle near the speed of light. You willful ignorance on this points forces you to be an idiot. You claim that you were able to speed up an electron to monstrous energy, as measured TEv. But this is false. You know it. This is not false. You are too stupid to know it. Masterov10-12-11, 05:16 AMHow about the LHC?Large Hadron Collider? Ok. Give several options to choose from. (If possible.) All should be a description of the experiment to measure the energy of relativistic particles in the calorimeter. Masterov10-12-11, 05:24 AMYou have no theory. Experimental evidence shows that SR is correct for the amount of energy needed to raise the speed of a particle near the speed of light.The energy of relativistic particles is important. An energy that was spent on heating of the experimental setup is unimportant. You claim that you were able to speed up an electron to monstrous energy, as measured TEv. But this is false. You know it.This is not false. You are too stupid to know it.This is an argument? No. I do not think so. Think again. origin10-12-11, 07:18 AMLarge Hadron Collider? Ok. Give several options to choose from. (If possible.) The entire experiment demonstrates that the closer you approach the speed of light you have to add more and more energy and it is asymptotic as you approach the speed of light. That is one of the fundemental aspects of accelerators, if you are going to willfully ignore that fact I really can't help you.:shrug: All should be a description of the experiment to measure the energy of relativistic particles in the calorimeter. This is your method of ignoring the reality? Really? If you can't measure the energy of the particles in a calorimeter then it isn't real? Your ability to delude yourself knows no bounds. Rather pathetic. Well at this point I don't know if this is an exercize in self delusion or just trolling.:shrug: origin10-12-11, 10:54 AMThe clocks of GPS-satellite differently make mistakes. One clocks outstrip. Other clocks behind time. Patterns are observed. A clocks synchronize when the GPS-satellite passes perihelion. Could you rephrase this? I don't know what point you are trying to make. Masterov10-12-11, 11:15 AMGive several options to choose from. (If possible.)The entire experiment demonstrates that the closer you approach the speed of light you have to add more and more energy and it is asymptotic as you approach the speed of light. That is one of the fundemental aspects of accelerators, if you are going to willfully ignore that fact I really can't help you.:shrug:With that I will not argue. No questions and no answers. My questions: 1. Whereon expend heavy expenses power (that consumes the accelerator)? 2. How many has realy energy of relativistic particles? 3. Is the energy of relativistic particles increases indefinitely, as maintained by Einstein's theory? Master Theory says: that Doppler effect impede acceleration of particles to superluminal velocities (but not the unlimited growth of the mass and energy of the particles, as maintained by Einstein's theory). Mass is absolute and not depends on speed. All should be a description of the experiment to measure the energy of relativistic particles in the calorimeter.This is your method of ignoring the reality? Really? If you can't measure the energy of the particles in a calorimeter then it isn't real? Your ability to delude yourself knows no bounds. Rather pathetic.As you prove that energy is transferred to the accelerated particles, and not only warms the accelerator. Master Theory argues that an increase in energy expended (nearly whole) on heating of the accelerator, but not for an particle acceleration. The clocks of GPS-satellite differently make mistakes. One clocks outstrip. Other clocks behind time. Patterns are observed. A clocks synchronize when the GPS-satellite passes perihelion.Could you rephrase this? I don't know what point you are trying to make.Clocks have fortuitous errors. origin10-12-11, 01:09 PMIt is really difficult to argue with someone who is so adebt at deluding themselves with willful ignorance. So, I say live your fantasy that one day the world will see your genius and adopt the Master Theory as the ultimate physics theory.;) Masterov10-12-11, 02:58 PMIt is really difficult to argue with someone who is so adebt at deluding themselves with willful ignorance. So, I say live your fantasy that one day the world will see your genius and adopt the Master Theory as the ultimate physics theory.;)You are not able to answer questions. You give up? You capitulated? origin10-12-11, 03:51 PMYou are not able to answer questions. I and many others have answered your questions you are unwilling or unable to understand the answers. You give up? I am bored to tears trying to get through your delusions. You capitulated Of course not. I said I was bored not insane...:rolleyes: Masterov10-13-11, 01:57 AMI and many others have answered your questions you are unwilling or unable to understand the answers.I offered you to publish the experimental results of the direct (of a calorimeter) measurement of actual energy of the relativistic particles. You are telling me that the accelerator consumes enormous energy. Your answer: Yes, the amount of energy needed to accelerate a particle increases at an increasing rate the closer you get to c. Again, this has been shown to be true over and over.What should I understand your answer? Master Theory demonstrate that speed up particles in the accelerator faster than the speed of light (through the fields, which propagate at the speed of light) it is impossibly (Doppler effect impedes for it). The increase in power consumption of the accelerator does not practically change the energy of relativistic particles. Direct (in the calorimeter) measurements of the relativistic energy is enough to verify this. I offered to publish any experimental results of the direct (in the calorimeter) measurement of actual energy of the relativistic particles. You are telling me that the accelerator consumes enormous energy. You report that requires a lot of energy accelerator instead of answering a question. Bottom line: you know, that energy and mass of relativistic particles does not grow indefinitely (if the velocity of the particles tends to the speed of light), as claimed by Einstein's theory. But you are hiding this information. Why? Pincho Paxton10-13-11, 02:36 AMPinchoism (I state a reply as Pinchoism if it is not scientific, but based on my own theory. It can either be ignored, or thought about.)... Propagation of light is like a Newtons Cradle, it is bump propagated. Imagine bulbs bumping together, and each time they bump they switch on, and the bump in rows like a Newton's Cradle. Now, in each bulb you drill a hole so that you can travel through the bulbs themselves. So light still bumps to switch on the bulbs, but mass has travel through the bulbs as a sort of liquid flow. Now mass has to push the other water along with it, but light only has to bump. You are correct in one way, if you can get the flow moving with you, you can catch up with light, and you can overtake light, but you can't use the flow of light to help you. You have to use the flow of Aether to help you (the liquid in the bulbs). There are natural flows in the Universe, and they can be used to accelerate mass. The problem is that Aether is invisible, and so finding a flow to help you naturally will be difficult. OnlyMe10-13-11, 08:40 AMBottom line: you know, that energy and mass of relativistic particles does not grow indefinitely (if the velocity of the particles tends to the speed of light), as claimed by Einstein's theory. But you are hiding this information. Why? You have to be misunderstanding something here... First Einstein never said that a particle's intrinsic or rest mass increases with velocity. In other words he never claimed that a moving mass has an increased gravitational force compared to the same mass/particle as rest. He and most physicists are referring to inertial mass or "relativistic mass" and most often don't even use those terms any longer as they create exactly the confusions you seem to be enjoying. Second it would seem obvious that as long as we are using the terms inertial mass and relativistic mass they are most certainly limited, to less than infinite, since before they reach c we run out of enough energy to further accelerate them. Masterov10-13-11, 10:25 AMhe never claimed that a moving mass has an increased gravitational force compared to the same mass/particle as rest. He and most physicists are referring to inertial mass or "relativistic mass" and most often don't even use those terms any longer as they create exactly the confusions you seem to be enjoying.I speak about relativistic mass too. m=\frac{m_0}{\sqrt{1-v^2/c^2}} prometheus10-13-11, 02:13 PMEveryone knows that mass is invariant in special relativity. tsch! OnlyMe10-13-11, 07:39 PMEveryone knows that mass is invariant in special relativity. tsch! Apparently not everyone. origin10-13-11, 07:49 PMYou guys are really confusing Masterov!:D OnlyMe10-13-11, 08:52 PMYou guys are really confusing Masterov!:D Yea, actually after my last post I remembered we were talking "Master Theory" not SR. Just a little off topic.... Masterov10-14-11, 12:20 AMThe formula: m=\frac{m_0}{\sqrt{1-v^2/c^2}} is no valid? When did this happen? Find for Google picture by "relativistic mass" phrase. You will see this formula in the galore (en masse). This formula was valid 30 years ago. This (supposedly unlimited) increase was cause for growth of energy and momentum of a relativistic particles. That mass increase was cause to the growth of the radius of the trajectory of relativistic particles in cyclotrons. R=\frac{p}{qB}=\frac{m_0v}{qB\sqrt{1-v^2/c^2}} This cyclotron radius no grow today? How did this happen? Well, well. Ok! p=mv=\frac{m_0}{\sqrt{1-v^2/c^2}}v - valid? E = mc^2=\frac{m_0}{\sqrt{1-v^2/c^2}}c^2 - valid? prometheus10-14-11, 03:09 AMBasically, m has to be invariant because it is the norm of the energy momentum 4 vector: m^2 = \eta^{a b} p_a p_b = E^2 - p^2. That means m is invariant under Lorentz transformations and therefore cannot depend on v. What you are calling m = \gamma m_0 is a funny and not very useful measurement of the energy. Masterov10-14-11, 03:19 AMBasically, m has to be invariant because it is the norm of the energy momentum 4 vector: m^2 = \eta^{a b} p_a p_b = E^2 - p^2. That means m is invariant under Lorentz transformations and therefore cannot depend on v. What you are calling m = \gamma m_0 is a funny and not very useful measurement of the energy.This is not any answer to my questions. I expect answers. ___________________________ Master Theory prove un-valid of E = mc^2=\frac{m_0}{\sqrt{1-v^2/c^2}}c^2. I'm looking for an experimental confirmation of this expression a few years. I need reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. Help me find them, please. OnlyMe10-14-11, 08:46 AMThis is not any answer to my questions. I expect answers. ___________________________ Master Theory prove un-valid of E = mc^2=\frac{m_0}{\sqrt{1-v^2/c^2}}c^2. I'm looking for an experimental confirmation of this expression a few years. I need reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. Help me find them, please. Masterov, when the terms "relativistic and inertial mass" were first used there was far less information available about particle physics. They are not used so much any longer. Yes you will find them on Google and even Wiki but both will also discuss many other things that are no longer considered accurate definitions. I am not sure it is really the math that is tripping you up in this instance. It is the concept behind those old definitions or terms. Try to think of relativistic mass as increasing inertia, as in increasing resistance to further acceleration. The faster you go the more resistance there is to going even faster. So at some point it begins to take an "exponentially" increasing amount of energy to go faster yet. (here exponentially is something of an exaggeration, but the point is it takes more than it did at a slower velocity.) Once an object, a particle is moving near c, it essentially holds all that extra energy as momentum, kind of, while its, invariant rest mass remains constant. This is not an exact description of what is happening but it is difficult to talk about the conceptual aspects without the math.., and yet without the conceptual basis the math does not always seem to say the same thing. Masterov10-14-11, 11:34 AMTry to think of relativistic mass as increasing inertia, as in increasing resistance to further acceleration.I so think already. I always thought so. Today - too. My meditations is not important. The important thing is what gives the experiment. The experiment let measure the energy and momentum of a relativistic particle. The experiment let determine truth of the expression: p=mv=\frac{m_0}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_0}{\sqrt{1-v^2/c^2}}c^2 Master Theory indicates the falsity of these expressions. If the experiment demonstrate truth of these expressions, then these experimental data have been everywhere published. (Pornographic magazines - inclusive.) You stubbornly refuse to provide experimental confirmation of these expressions. Thus you confirm that the experiment shows the falsity of these expressions. Your misdealing is the basis to suspect and accuse you of falsehood. Any scientist has no right to lie in science. You are not scientists. You're lying. AlphaNumeric10-16-11, 03:52 AMIf the experiment demonstrate truth of these expressions, then these experimental data have been everywhere published. (Pornographic magazines - inclusive.)Exaggerate much? You stubbornly refuse to provide experimental confirmation of these expressions.You stubbornly refuse to look for them. http://physics.dickinson.edu/~dept_web/activities/papers/relativity.pdf First hit when you Google for 'relativistic mass, experiment'. It's an experiment students do, because it's so basic. It works on much the same principle as modern accelerators, bending particle paths via magnetic fields. If the relativistic formulae weren't very close to right we wouldn't be able to build working accelerators based on relativity. Thus you confirm that the experiment shows the falsity of these expressions. Your misdealing is the basis to suspect and accuse you of falsehood. Any scientist has no right to lie in science. You are not scientists. You're lying.Your logic is terrible. You have a conclusion and you're not waiting for or finding evidence, you're just leaping to the conclusion you want. It took 4 seconds for me to find that example of experiments about relativistic mass. Why couldn't you find that yourself? Why do you demand other people do things you're either too lazy or too dishonest to do? Why haven't you got yourself some data about accelerators and analysed it using your own model? There's plenty of material available online about how accelerators work, such as their magnetic fields, their energies, their timings, their radiation braking issues. All of it can be found on websites or in journals or books. Yes, you may have to pay a bit of money for some stuff but that's life. You accuse us of lying but all you've done is presuppose your own conclusions. That is lying, that is a cardinal sin in science. To lead the evidence, rather than to be lead by the evidence, is the sort of things which gets people fired from research jobs. In medicine you can be stripped of your license if you did such a thing in a clinical trial. You are the one being dishonest here. You have access to Google, you simply haven't used it and to compound your dishonest you're just making up things. The person with the least right to refer to themselves as a scientist here appears to be you. Masterov10-16-11, 12:06 PMExaggerate much?Nuu ... Yes - exaggerating. I'm right in fact. You stubbornly refuse to look for them. http://physics.dickinson.edu/~dept_web/activities/papers/relativity.pdfYou (stubbornly) give me not what is necessary. bending particle paths via magnetic fields.The results of this experiment can be interpreted otherwise: the radius of the trajectory increases due to slow down increase of Lorentz force. This experiment does not prove fact of increase of relativist mass. If the relativistic formulae weren't very close to right we wouldn't be able to build working accelerators based on relativity.Einstein's theory poorly approximates the experiment. But this theory approximates. This allows you to build accelerators. The quality of these accelerators match the quality of approximation. Your logic is terrible. You have a conclusion and you're not waiting for or finding evidence, you're just leaping to the conclusion you want. It took 4 seconds for me to find that example of experiments about relativistic mass. Why couldn't you find that yourself? Why do you demand other people do things you're either too lazy or too dishonest to do?You give me not what is necessary. I need reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. Why haven't you got yourself some data about accelerators and analysed it using your own model? There's plenty of material available online about how accelerators work, such as their magnetic fields, their energies, their timings, their radiation braking issues. All of it can be found on websites or in journals or books. Yes, you may have to pay a bit of money for some stuff but that's life.These experiments were performed many times, but the results of these experiments can not get into print, because they are contrary to Einstein's theory. I suggest you find reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. (I failed to do so.) You accuse us of lying but all you've done is presuppose your own conclusions. That is lying, that is a cardinal sin in science. To lead the evidence, rather than to be lead by the evidence, is the sort of things which gets people fired from research jobs. In medicine you can be stripped of your license if you did such a thing in a clinical trial. You are the one being dishonest here. You have access to Google, you simply haven't used it and to compound your dishonest you're just making up things.I agree with you. But your word is no excuse for fraud in science. EMW have light speed. EMW can not overclock mass faster than light. The experiment: p=mv=\frac{m_0}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_0}{\sqrt{1-v^2/c^2}}c^2 is false and you know it. If you are an honest man, why do not you talk about it. The person with the least right to refer to themselves as a scientist here appears to be you.An accountant has the right to call themselves an accountant? Is it presumptuous? If an accountant steals money, it is still considered to be an accountant? Maybe it properly be called her a thief? Is a scientist an honest man, if money is spent on the experiment and to do downright lie for public? AlphaNumeric10-16-11, 01:25 PMNuu ... Yes - exaggerating. I'm right in fact.So you honestly think pornographic magazines would publish the results? The results of this experiment can be interpreted otherwise: the radius of the trajectory increases due to slow down increase of Lorentz force.Sure. Any experiment can be interpreted differently. The question is whether a single model can provide accurate interpretations for all relevant experiments. That's the point of unifying models, so you end up with less explaining more. This experiment does not prove fact of increase of relativist mass.Experiments never prove models. However, it does demonstrate that such a conclusion about the behaviour of mass at relativistic speeds is a justifiable one. Einstein's theory poorly approximates the experiment. But this theory approximates. This allows you to build accelerators. The quality of these accelerators match the quality of approximation.Using accelerators we have measured the accuracy of quantum electrodynamics, which includes both quantum mechanics and special relativity, to parts per trillion. It is the most accurate and precisely tested model in human history. As such I'd say that's a pretty good approximation. You give me not what is necessary.What is necessary is you getting 4~10 years of physics education. I need reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles.These experiments were performed many times, but the results of these experiments can not get into print, because they are contrary to Einstein's theory.Prove your assertion they are contrary to special relativity. I've worked with people who do analysis of experimental data from accelerators, the data does not contradict relativity. Do you think there's a massive conspiracy to suppress information or something? It would have to include every single particle physicist in the world. For example, I didn't do experimental particle physics but I have worked along side people who have, I've seen their data, their computer programs, their calculations, their publications. As such if you think they are involved in a conspiracy then you must think I am too. I suggest you find reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. (I failed to do so.)You suggest I find it? Sorry Chuckles but you're the one making assertions about all this. I don't have to do anything. I paid attention in school, I learnt some physics. If you're either too lazy or too incompetent to find the information I'm not going to help you. Get a journal subscription. Or better yet, get a physics degree, a particle physics PhD and then a research position at an accelerator facility doing collider physics and then you can see the data for yourself! I agree with you. But your word is no excuse for fraud in science.There isn't the conspiracy of silence you're implying. Your evidence-less claims about fraud are themselves dishonest. You have an axe to grind and you're fabricating justifications. You suck at physics, I get it. Unfortunately you need to learn to accept the reason you're getting nowhere is because you suck at physics and you lie, not because there's some elaborate conspiracy against you. The experiment: p=mv=\frac{m_0}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_0}{\sqrt{1-v^2/c^2}}c^2 is false and you know it. If you are an honest man, why do not you talk about it.Because I'm paid $500,000 a year to suppress the truth, along with the tens, even hundreds of thousands of particle physics PhDs like me. Opps! I shouldn't have said that, if only I could edit this post! There isn't a conspiracy of silence. There isn't suppression of evidence. You're just too incompetent to find it yourself. Rather than being honest and saying "I couldn't find it therefore I don't know what the evidence says" you're saying "I couldn't find it therefore it's exactly what I claim it is and it's being suppressed". Here (http://arxiv.org/archive/hep-ex) is an index of pretty much every experimental particle physics paper from the last 17 years. There's tons of data there. If you're unable to understand it that doesn't mean it's wrong or a lie, it's means you're insufficiently educated. An accountant has the right to call themselves an accountant? Is it presumptuous?By what criteria would you consider yourself to be a scientist? If an accountant steals money, it is still considered to be an accountant? Maybe it properly be called her a thief?If someone accuses an honest accountant of stealing does that make the accountant a thief? Is a scientist an honest man, if money is spent on the experiment and to do downright lie for public?What do you call someone who rants about conspiracies against him for which he has no evidence? OnlyMe10-16-11, 02:25 PMBecause I'm paid$500,000 a year to suppress the truth, along with the tens, even hundreds of thousands of particle physics PhDs like me. Opps! I shouldn't have said that, if only I could edit this post! Really! \$500,000 a year... And how much then for your day job? I did pretty well in business and there is a forty year difference in the time frames, starting out.., but that's a chunk of change Alpha. AlphaNumeric10-16-11, 04:51 PMI don't mean my day job pays that (that's easily more than a decade's worth of salary to me!), my involvement in a massive global conspiracy to suppress the truth of Einstein being incorrect pays it. After all, when you're part of a conspiracy spanning countries, continents, governments, religions and ideologies it's easy to lay your hands on massive of money. Masterov's fantasy world pays pretty well! Masterov10-17-11, 03:08 AMThe results of this experiment can be interpreted otherwise: the radius of the trajectory increases due to slow down increase of Lorentz force.Sure. Any experiment can be interpreted differently. The question is whether a single model can provide accurate interpretations for all relevant experiments. That's the point of unifying models, so you end up with less explaining more. This experiment does not prove fact of increase of relativist mass.Experiments never prove models. However, it does demonstrate that such a conclusion about the behaviour of mass at relativistic speeds is a justifiable one.The validity of expressions: p=mv=\frac{m_o}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 can clearly demonstrate a direct measurement of the energy and momentum of the relativistic particles. The experiments were conducted such, but the results of these experiments is not public print. The only justification for it: are a contradiction of these experiments to Einstein's theory. Do you think there's a massive conspiracy to suppress information or something? It would have to include every single particle physicist in the world. For example, I didn't do experimental particle physics but I have worked along side people who have, I've seen their data, their computer programs, their calculations, their publications. As such if you think they are involved in a conspiracy then you must think I am too. There isn't the conspiracy of silence you're implying. Your evidence-less claims about fraud are themselves dishonest. You have an axe to grind and you're fabricating justifications. You suck at physics, I get it. Unfortunately you need to learn to accept the reason you're getting nowhere is because you suck at physics and you lie, not because there's some elaborate conspiracy against you. Russian Academy of Sciences three times makes a decision that prohibits the publication of scientific papers, if they contradict Einstein's theory. I suspect that such bans exist not only in Russia. Europe spent a lot of money for the construction of LHC. (Other countries are spending money to build a large accelerators, too.) If I will be right, this would mean that the LHC (and others) was vain built. This may be the cause of big scandal. Exist are people who are not interested in such scandal. These people have an interest to hide the truth. Destroy my suspicions. It's done simply. It's enough to publish experiments that directly (of calorimeter) measured the energy and momentum of the relativistic particles. Einstein's theory poorly approximates the experiment. But this theory approximates. This allows you to build accelerators. The quality of these accelerators match the quality of approximation.Using accelerators we have measured the accuracy of quantum electrodynamics, which includes both quantum mechanics and special relativity, to parts per trillion. It is the most accurate and precisely tested model in human history. As such I'd say that's a pretty good approximation.If you look closely at the derivation of Master Theory, you will notice that the Master Theory comes from the same initial postulates. Master Theory gives absoluteness to time, but Einstein's gives absoluteness to cross-scale. In cases where time dilation in the coordinate frame of a relativistic particle does not apparent, the equations of Master Theory and Einstein's theory of differences have not. Both theories will provide a good approximation in these cases. Einstein's expressions: p=mv=\frac{m_o}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 for Master Theory has classical form: p=m_oV E = \frac{m_oV^2}{2}+Const Experiment (of direct measurement of the energy and momentum of the relativistic particles) let determine truth of these expressions. You give me not what is necessary.What is necessary is you getting 4~10 years of physics education.I'm not interested in knowledge that is built on lies and I am quite familiar with this knowledge, to see lie of them. Einstein's false was causal for that my big difficulty and I could not finish the construction of another theory ten years ago. I managed to build a model of "Generator of Matter". Attempts to generalize this model in Minkowski space engendered nonsense. Today I understand why I am had big difficulty ten years ago. I suggest you find reliable experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. (I failed to do so.)You suggest I find it? Sorry Chuckles but you're the one making assertions about all this. I don't have to do anything. I paid attention in school, I learnt some physics. If you're either too lazy or too incompetent to find the information I'm not going to help you. Get a journal subscription. Or better yet, get a physics degree, a particle physics PhD and then a research position at an accelerator facility doing collider physics and then you can see the data for yourself!If the data (which I'm looking for) have been published (someplace), then I would have found them. I would do not invited you do search for them. I wanted to you see for yourself that the experimental results of direct measurements of energy and momentum of relativistic particles in the press there be absent. There isn't a conspiracy of silence. There isn't suppression of evidence. You're just too incompetent to find it yourself. Rather than being honest and saying "I couldn't find it therefore I don't know what the evidence says" you're saying "I couldn't find it therefore it's exactly what I claim it is and it's being suppressed". Here is an index of pretty much every experimental particle physics paper from the last 17 years. There's tons of data there. If you're unable to understand it that doesn't mean it's wrong or a lie, it's means you're insufficiently educated.Accusations of incompetence in his address, I hear not the first time of some years. But no one (of accusers) has yet proved these (unsupported by evidence) allegations. Despite the fact that it's easy to do. It's enough to publish experiments that directly (of calorimeter) measured the energy and momentum of the relativistic particles. An accountant has the right to call themselves an accountant? Is it presumptuous?By what criteria would you consider yourself to be a scientist?1. I studied to be a scientist. 2. I've been doing this job since 1986. 3. I want to know the truth. 4. I do not lie. If an accountant steals money, it is still considered to be an accountant? Maybe it properly be called her a thief?If someone accuses an honest accountant of stealing does that make the accountant a thief?An honest man does not lie. If a man lies, he can not be an honest man. It's obvious. You are participating in concealment and forgery. You are not an honest man. I have proof of these accusations: you suppress publish experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles. You are not a scientist. You suck at physics only. Is a scientist an honest man, if money is spent on the experiment and to do downright lie for public?What do you call someone who rants about conspiracies against him for which he has no evidence?1. I would call that person crazy. 2. I'm not accusing anyone in conspiracies against personally me here. I don't mean my day job pays that (that's easily more than a decade's worth of salary to me!), my involvement in a massive global conspiracy to suppress the truth of Einstein being incorrect pays it. After all, when you're part of a conspiracy spanning countries, continents, governments, religions and ideologies it's easy to lay your hands on massive of money. Masterov's fantasy world pays pretty well!It's simply prove. It's enough to publish experiments that directly (of calorimeter) measured the energy of the relativistic particles. What prevents you from doing so? origin10-17-11, 07:43 AMThis is trending more toward a Conspiracy thread instead of a Silly Unsubstantiated Theory thread. AlphaNumeric10-17-11, 08:12 AMTThe experiments were conducted such, but the results of these experiments is not public print. The only justification for it: are a contradiction of these experiments to Einstein's theory.Or rather the information is in journals and available to researchers. Since you won't pay for journals and you're too incompetent to be a physicist you don't have access to the data. The data from places like the LHC rubs into the thousands of petabytes. Dedicated supercomputing networks have been built, linking locations across the world, to handle the data. Even if you had access to it you wouldn't know the first thing to do with it. Russian Academy of Sciences three times makes a decision that prohibits the publication of scientific papers, if they contradict Einstein's theory.Provide evidence, otherwise you sound like a paranoid nut. I suspect that such bans exist not only in Russia.You can 'suspect' all you like, but you have no evidence. In fact recently there's been a lot of talk about how neutrinos might have been observed contradicting Einstein. It wasn't suppressed, it became one of the most talked about things in physics for years! Europe spent a lot of money for the construction of LHC. (Other countries are spending money to build a large accelerators, too.) If I will be right, this would mean that the LHC (and others) was vain built. This may be the cause of big scandal. Exist are people who are not interested in such scandal. These people have an interest to hide the truth.Paranoid delusions without evidence. In fact it contradicts evidence. Already the LHC has produced data which has been of use to physics. If the LHC demonstrated relativity is wrong then it would be a bigger contribution to physics than anything it was deliberately built for. Disproving relativity would make someone's career. It would make them the most famous physicist since Einstein himself. If I could prove relativity wrong I'd not suppress it, quite the opposite. Knocking over well established models is the fastest way to become famous, get a job for life, get tons of awards and be remembered forever in science. And yet you claim that everyone in the particle physics community is involved in a conspiracy where they give up all of that? If you think we're all greedy and looking out for ourselves it would be in my interest to expose the conspiracy. It's basic game theory, like the prisoners dilemma, there would be no reason for me not to defect. Destroy my suspicions. It's done simply.I just did. Anyone involved in such a conspiracy would have more to gain by revealing it. Your claims imply there's tens, even hundreds, of thousands of people involved. No one has 'defected', despite there being no reason not to and plenty of reasons to do so. Your claims fail in the face of basic human nature. It's enough to publish experiments that directly (of calorimeter) measured the energy and momentum of the relativistic particles.Which would involve petabytes of data. The ATLAS detector produces 100GB of data A SECOND. In cases where time dilation in the coordinate frame of a relativistic particle does not apparent, the equations of Master Theory and Einstein's theory of differences have not. Both theories will provide a good approximation in these cases. Einstein's expressions: p=mv=\frac{m_o}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 for Master Theory has classical form: p=m_oV E = \frac{m_oV^2}{2}+ConstIn other words you take the Newtonian approximation of relativity. I'm not interested in knowledge that is built on lies and I am quite familiar with this knowledge, to see lie of them.You aren't interested in knowledge full stop. Today I understand why I am had big difficulty ten years ago.Because you don't know any physics? I wanted to you see for yourself that the experimental results of direct measurements of energy and momentum of relativistic particles in the press there be absent.The hep-ex section of ArXiv illustrates just how much experimental data there is. The raw data isn't given, as it is simply too large to publish in print, but its available to researchers in universities. I know because I've worked along side a number of them. But no one (of accusers) has yet proved these (unsupported by evidence) allegations. You haven't proven any of your accusations about suppression, conspiracies, lies, frauds. You're just hurling out fabrications you have no evidence for. Despite the fact that it's easy to do. It's enough to publish experiments that directly (of calorimeter) measured the energy and momentum of the relativistic particles.You clearly don't know the amount of data involved. As I said, we're talking thousands of petabytes of raw data. The raw data is not literally printed in journals, it's kept on servers and is available to those who are doing actual research into this stuff. If you could get yourself a particle physics job then you'd be able to get access to it. 1. I studied to be a scientist.Did you actually get to become a scientist? And what do you mean 'studied'? Do you mean you have a science related degree? Masters? PhD? 2. I've been doing this job since 1986.What job? 3. I want to know the truth.And yet you have no interest in being honest and listening to evidence, instead you make suppositions and lead the evidence. 4. I do not lie.An honest man does not lie. If a man lies, he can not be an honest man. It's obvious.Then you are not an honest man. You are participating in concealment and forgery. You are not an honest man. I have proof of these accusations: you suppress publish experimental results of the direct (from calorimeter) measurement of actual energy of the relativistic particles.Your proof is you don't have access to it? That isn't proof of concealment, it's proof you have not advanced far enough in physics research to be allowed access to raw experimental data. If you had gotten a physics degree, PhD and then a research position you'd be allowed access to it. Presently you lack the knowledge and ability to understand the data. You are not a scientist. You suck at physics only.I have a physics PhD and a research job. I'm more of a scientist than you. 1. I would call that person crazy.Then you admit you're crazy. Masterov10-17-11, 08:17 AMThis is trending more toward a Conspiracy thread instead of a Silly Unsubstantiated Theory thread.I answered a questions only. And I expect that will be published data of experiments that directly measured (of calorimeter) the energy of a relativistic particles. origin10-17-11, 09:28 AMI answered a questions only. And I expect that will be published data of experiments that directly measured (of calorimeter) the energy of a relativistic particles. You are the one with the Alternative Theory! You supply the data that supports YOUR theory. If you cannot do that then why should we even consider your theory? OnlyMe10-17-11, 11:10 AMAnd I expect that will be published data of experiments that directly measured (of calorimeter) the energy of a relativistic particles. You keep asking for evidence based on calorimeter measurements of relativistic particles. You should understand that as a result of the scales involved in the technologies that is not currently possible. Relativistic subatomic particles are difficult and in most cases impossible to measure directly. Calorimeters measure change directly, as heat or some other direct measure of change. If you have any reference to a Calorimeter that functions directly at a subatomic scale please provide a link. AlphaNumeric10-17-11, 12:46 PMModern detectors can follow individual particles, because they produce significant numbers of ionised particles as they blast through detectors. Different materials 'block' different particles in different amounts so we can tell a muon from an electron. Masterov, I told you, the raw data is petabytes of information. You're not going to find that on some website. The hep-ex section of www.arxiv.org has plenty of summaries of experiments from all the detectors in the world, you seem to be refusing to even look there. The fact is you have no evidence for your deranged conspiracy delusions, you're just looking for an excuse why you've failed to accomplish anything in physics. I don't find it surprising in the slightest you refused to answer my question about what you actually have done in regards to being a scientist. You obviously know you've backed yourself into a corner, else you'd try to answer my questions. Instead you ignore them, showing to everyone you don't want to admit the answers. Masterov10-17-11, 01:35 PMYou are the one with the Alternative Theory! You supply the data that supports YOUR theory. If you cannot do that then why should we even consider your theory?You made me smile. These expressions: p=m_oV E = \frac{m_oV^2}{2}+Const are repeatedly experimental proven and beyond doubt. These expressions: p=mv=\frac{m_o}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 are experimental not proven. These expressions require experimental confirmation. You I was accused of incompetence. But you refuse to demonstrate their competence. Honest people do not do that. Please present directly (of calorimeter) experimental confirmation of expression: E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 martillo10-17-11, 01:57 PMJust curious. You said: “ bending particle paths via magnetic fields. ” The results of this experiment can be interpreted otherwise: the radius of the trajectory increases due to slow down increase of Lorentz force. This experiment does not prove fact of increase of relativist mass. How do you explain by "slow down the Lorentz Force" the appearence of the term root(1-v2/c2) in the results of those experiments? Do you redefine the Lorentz Force in your theory? OnlyMe10-17-11, 02:52 PMModern detectors can follow individual particles, because they produce significant numbers of ionised particles as they blast through detectors. Different materials 'block' different particles in different amounts so we can tell a muon from an electron. Alpha, aren't these still indirect observations of photon emissions by the secondary particles? origin10-17-11, 03:32 PMYou I was accused of incompetence. But you refuse to demonstrate their competence. Honest people do not do that. You are accused of incompetence because you demonstrate incompetence. Honesty never enters into this discussion, just your lack of physics knowledge. Please present directly (of calorimeter) experimental confirmation of expression: E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 Lets try this again. YOU disagree with the accepted physics. YOU are presenting an alternate theory. It is up to YOU to prove that your theory is better, it is NOT up to me to prove that your theory is garbage. Get it. No one is going to waste their time doing YOUR work for you, OK? AlphaNumeric10-17-11, 03:52 PMYou made me smile. These expressions: p=m_oV E = \frac{m_oV^2}{2}+Const are repeatedly experimental proven and beyond doubt.No, those expressions have been shown to be close to the truth for small values of V. That isn't the same as being 'proven'. You can never form a formula like that, only provide more and more evidence for it being valid within particular domains. And obviously it isn't 'beyond doubt', relativity calls it into doubt and demonstrates how they can be low V approximations to more general results. These expressions: p=mv=\frac{m_o}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 are experimental not proven. These expressions require experimental confirmation.The fact you haven't seen the evidence (and even if you had you obviously wouldn't understand it) doesn't mean the evidence doesn't exist. I've never seen Japan but I don't demand people bring Tokyo to me (there's a terrible analogy....) You I was accused of incompetence. You we all accuse of ignorance and dishonesty. Aren't you going to answer my questions about what precisely you've been working as since the mid 1980s? What are your qualifications as a scientist? But you refuse to demonstrate their competence. Honest people do not do that.Honest people don't claim there's a global conspiracy for which they have no evidence and assert people they are talking to are involved in said conspiracy. You said it yourself, such a person is crazy. You are that person. OnlyMe, yes what is seen are things like photons given off by decelerating electrons or muons produced. But then that's how you're seeing the monitor in front of you now, by the photons the matter is giving off. For things like neutrinos we see the light given off by electrons moving faster than light in the medium (like water in massive underground detectors) which are produced by neutrinos producing a W boson and knocking the electron out of an atom in the water. It's convoluted by if something doesn't directly couple to electromagnetism we cannot detect its photon radiation directly as it has none! Suffice to say we can see the tracks produced by individual particles so we can literally count precisely how many particles made it out of the fire ball produced by a collision. Masterov10-17-11, 11:52 PMThe results of this experiment can be interpreted otherwise: the radius of the trajectory increases due to slow down increase of Lorentz force. This experiment does not prove fact of increase of relativist mass.How do you explain by "slow down the Lorentz Force" the appearence of the term root(1-v2/c2) in the results of those experiments? Do you redefine the Lorentz Force in your theory? Yes. EMW have light speed. EMW can not overclock mass faster than light. Therefore, an interaction of a relativistic particle with EM-field (source of which is stationary) becomes weaker if a particle speed tends to light speed. Therefore, I search (several years) experimental proven for expressions: p=mv=\frac{m_o}{\sqrt{1-v^2/c^2}}v E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 My opponents (who profess to Einstein's theory) constantly accuse me of incompetent, of unqualified, of imbecility and of weak-mindedness, but no one wants to provide proof of their qualified and experimental proof of this pair of expressions. Masterov10-18-11, 12:40 AMTo prove of expression: E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2 is enough publish experiments that directly (of calorimeter) measured the energy and momentum of the relativistic particles.Which would involve petabytes of data. The ATLAS detector produces 100GB of data A SECOND. The data from places like the LHC rubs into the thousands of petabytes. Dedicated supercomputing networks have been built, linking locations across the world, to handle the data. Even if you had access to it you wouldn't know the first thing to do with it.It's a lie. Such experiments were conducted Bertozzi more than forty years ago in a lab for students. Unfortunately Bertozzi had set layout, and not a complete experimental setup. His report for dean and not a full scientific report. Bertozzi to do Experiments long ago. For these reasons, experiments Bertozzi not be taken as an argument in the proof of this expression. But Bertozzi showed that such an experiment is fairly easy to implement. Sufficient to establish the fact of the growth temperature of the calorimeter is practically no increase in the rate of relativistic particles in order to prove this expression. I have a physics PhD and a research job. I'm more of a scientist than you.The experimental results of direct measurements of the energy of relativistic particles will be published shortly. Whereupon your thesis will cost no more than the paper on which is printed your thesis. (Not your only.) martillo10-18-11, 01:50 AMYes. EMW have light speed. EMW can not overclock mass faster than light. Therefore, an interaction of a relativistic particle with EM-field (source of which is stationary) becomes weaker if a particle speed tends to light speed. Then you have to redefine the EM-field... Masterov10-18-11, 02:24 AMThen you have to redefine the EM-field...This is the least that will do. Many textbooks and reference books on physics will be reissued. (It's least too.) Statements that accelerators could accelerate particles to MeV and TeV is a lie. It will be a huge scandal. Such has happened in the history of mankind. This happened when completely unexpected result was that the Earth is a sphere. This led to the fact that human knowledge is divided between science and religion. Today, science is not unlike the medieval church. Methods that protect the Einstein theory, borrowed from the medieval Inquisition. The church defended its religious doctrine by these methods. The main question of the day will be this: why science become an inveterate liar so long. Who is to blame, and - what to do? Who lobbied for the SRT, and - why they did it?... AlphaNumeric10-18-11, 02:36 AMStatements that accelerators could accelerate particles to MeV and TeV is a lie. It will be a huge scandal. Such has happened in the history of mankind. This happened when the Earth was a sphere. This led to the fact that human knowledge is divided between science and religion. Today, science is not unlike the medieval church. Methods that protect the Einstein theory, borrowed from the medieval Inquisition. The church defended its religious doctrine by these methods. http://math.ucr.edu/home/baez/crackpot.html You've hit at least 3 of the last 4 points on that list with that post alone. You've a conspiracy nut. You have no evidence for your own claims, you have no evidence for anything you say about mainstream science, you have no right to call yourself a scientist while you simultaneously accuse others of the same, you are dishonest, you are ignorant and you're paranoid. Such experiments were conducted Bertozzi more than forty years ago in a lab for students. So why don't you contact someone at the relevant research lab/university and enquire about where to find the data then? Have you done this? Or has the extent of your 'searching' been to use Google and whine on forums? But Bertozzi showed that such an experiment is fairly easy to implement. Then why don't you do the experiments yourself, if it's so easy? Sufficient to establish the fact of the growth temperature of the calorimeter is practically no increase in the rate of relativistic particles in order to prove this expression.So you can't find his work but you know what it says? Do you realise how dishonest saying stuff like that makes you? It's like me saying that although I have no idea who you are in real life I'm sure you've been convicted of arson, theft and assault. Would it be honest of me to fabricate such claims simply because I haven't got any information about you? Of course not! Yet you seem happy to do the physics version of that, calling into question the character and honesty of everyone in the particle physics community for the last century. The experimental results of direct measurements of the energy of relativistic particles will be published shortly.So you do know where to find it then? Do you or don't you? Whereupon your thesis will cost no more than the paper on which is printed your thesis. (Not your only.)What does the cost of my thesis have to do with accelerator data being made available? How does that retort my comment that I'm more a scientist than you? You continue to ignore my questions. You said you have learnt something for your job. What is your education in physics? You said you'd been doing your job since the mid 80s. What is your job? From your silence I suspect it's not related to physics. Post ReplyCreate New Thread