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I do not co-locate them at the termination. I use the SR clock simultaneity convention which is logically equivalent to co-location. If they are co-located, then you can look at their clocks. If they are in the same frame at the endpoint, then you use the SR clock sync method to look at the clocks and establish ordinality. The ordinality can be established same as with the eyeballs.

Alternative Twins Paradox The traditional twins paradox is solved such that all agree the traveling twin is younger because of non-symmetric acceleration. However, the uniform acceleration equations and Lorentzian transformations of special relativity are all that are needed. So, this twins paradox will produce symmetric acceleration and force SR to exclusively explain reciprocal time dilation. O and O' are physicists testing special relativity. These physicists are going to test the aging of two twins. O, O', twin1 and twin2 are in the same frame and sync their clocks with Einstein's clock synchronization method. They all agree to burn for BT and agree on the number. Next, all set their clocks to 0 and reside at the same origin for a common start point. Then, twin1 take off from the frame in a spaceship at proper acceleration a and for burn time BT in the proper time of the accelerating frame such that they attain a relative motion v compared to the prior frame. Then, for a long period of time, they remain in relative motion to O' and twin2. So, O and twin1 see O' and twin2 moving away at relative motion v and O' and twin2 see O and twin1 moving away at relative motion v. After some long period of t, O' and twin2 take off in an identical spaceship in exactly the same direction as O and twin1 did at the same acceleration a and for the same burn time BT. After the burn is complete, all are back in the same frame but at a distance between them and O' immediately sends out a light pulse. O records the time it receives the light pulse. Then, O performs the round trip speed of light calculation to decide the distance between the two ships. Once the distance D between the two ships is known, O subtracts D/c from the time it received the light pulse from O'. This gives O the exact time in its own proper time when O' stopped accelerating and entered the frame of O. In addition, this gives a correct endpoint to the experiment that matches the endpoint of O'. Thus, O and O' share a common start point and end point to the experiment. Let BT be the burn time in the accelerating frame. Let a = acceleration and v = the relative speed attained by the acceleration. O and O' calculate the proper time of the twins as follows. Calculations of O for the twins Elapsed proper time calculation for twin 1 twin1's acceleration phase BT twin1's relative motion phase t' twin2's acceleration phase c/a sinh(aBT/c) Total elapsed proper time calculation of O for twin1 BT + t' + c/a sinh(aBT/c) Please note above, O marked an endpoint to the experiment after receiving a light pulse from O', call that time, Te. Then, Te = BT + t' + c/a sinh(aBT/c) + D/c where D is the distance between the two ships in the resulting frame as derived above. Thus, t' = Te - BT - c/a sinh(aBT/c) - D/c. Elapsed proper time calculation for twin 2 twin1's acceleration phase c/a sinh(aBT/c) twin2's relative motion phase t'/ λ twin2's acceleration phase BT Total elapsed proper time calculation of O for twin2 c/a sinh(aBT/c) + t'/λ + BT Conclusion of O, twin1 is older. Calculations of O' for the twins Elapsed proper time calculation for twin 1 twin1's acceleration phase BT twin1's relative motion phase t/λ twin2's acceleration phase c/a sinh(aBT/c) Total elapsed proper time calculation of O' for twin1 BT + t/λ + c/a sinh(aBT/c) Elapsed proper time calculation for twin 2 twin1's acceleration phase c/a sinh(aBT/c) twin2's relative motion phase t twin2's acceleration phase BT Total elapsed proper time calculation of O' for twin2 c/a sinh(aBT/c) + t + BT Conclusion of O', twin1 is younger. Since all are now in the same frame, they again use Einstein's clock synchronization method for the actual hard data of the experiment to test the clocks of the twins for age ordinality. This answer of the clock sync is to be accepted by all those in the frame because as Einstein said, "We assume that this definition of synchronism is free from contradictions". This clock sync can also establish the time ordinality of the clocks. This is how it is done. Twin1 sends its time ttwin1 to twin2. Twin2 immediately sends its time ttwin2 back to the clock of twin1. Twin1 receives this light at t'twin1*** Einstein concludes the two clocks are in sync if ttwin2 = ½ ( t'twin1 - ttwin1). Thus, the age ordinality is established as ttwin2 - ½ ( t'twin1 - ttwin1) http://www.fourmilab.ch/etexts/einstein/specrel/www/ SR is supposed to be a reliable tool for the calculation of the proper time of one frame to another frame. However, both O' and O cannot be correct with their SR conclusions when presented with the factual data of the clock sync. References http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html http://en.wikipedia.org/wiki/Twins_paradox

OK, you believe the satellites are not in sync from the view of the ground clocks. You believe also the emission are not in sync from M' and thus M' will believe B was emitted before A in order to preserve SR. This differential is caused by the distance to M' upon light emission. You believe they are not in sync based on distance to the clock. This is not supported by GPS but that is OK. If you read here, you will find the triangulation depends on the satellites have the same common time and earth time and then use distance to determine position. If the satellites were not in sync, this would not work. So, you are wrong in your thesis. http://www.beaglesoft.com/gpstechnology.htm Now, going back to the original R of S thought experiment from SR, both clocks are equidistant to M' upon light emission. M believes they are simultaneous Since they are equidistant to M' upon emission and they are simultaneous emissions in M, then they must also be simultaneous emissions in M' since there is no other basis to claim they are not simultaneous emissions for M' since they are equidistant at emission. Since they are equidistant and simultaneous in M', then they must reach M' simultaneously since the motion of the light sources has no bearing on the problem. But, they are equidistant and simultaneous in M also. Thus, both M and M' will be strick by the light beams simultaneously at different positions, a contradiction. Thus, either road you take you run into a contradiction.

Well, my argument agrees that GPS works. Each satellite remains synched within 100 nano-seconds of the ground clocks. GPS Specification States +/- 100 Nanoseconds to UTC http://www.endruntechnologies.com/timingspec.htm So, if GPS1 emits a laser sat 12:00 UTC and GPS2 emits at 12:00 UTC, then the ground clocks all agree the laser emits ~ at the same time. They are not out of sync up to the limits of the technology. Now, that means in my thought experiment, the M frame A and B emits lasers simultaneously, thus the M' frame will conclude they are simultaneous and thus will measure a different value c for each direction. This contradicts SR.

Does not work. These adjustments remove the effects. So, your view is simply not working. When you do an experiment, you want to exclude other factors, no? Try again. Time dilation refutes the universal t=d/c.

Well, you are intelligent. We are not communicating. I never said time dilation did not depend on v. That indeed would be absurd. I said it did not depend on position. A moving clock at position A1 moving to B1 will experience the same time interval as a clock at a different position A2 moving to B2 givemn the same moving frame. Einstein: It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be second slow. Thence we conclude that a balance-clock7 at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions. Now, he said time dilation applies to the time interval for a clock to move from point A to to B. Since the position A is arbitrary, then all clocks in a moving frame beat in sync at a time dilated interval from the view of the stationary frame since A can be anwhere in the frame. Let's be specific. From the earth frame, are the satellite clocks in sync or not. Uh, they are the same clocks. The same clocks only means they are able to maintain the same frequency and about the same accuracy. That is correct and I think you are getting it. When using clocks as moving objects, you use time dilation. When looking at light moving, you use full LT. Thus, if clocks in a moving frame are synched at some agreed upon time, then they remain in sync except they will all be time dilated for the interval of time of the stationary frame. This contradicts t=d/c as universal. No, I am OK with the GPS.

Well, I gave both arguments and you are asserting the one that refutes time dilation and has been experimentally verified. No, time dilation does not require this and this contradicts time dilation. Further this does not show up in GPS. In addition I am backed up by the scientific evidence of time dilation that does not depend on position and is simply consistent within the frame. Tests of Time Dilation and Transverse Doppler Effect http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Tests_of_time_dilation same frame. You are comparing two different situations, and it's not valid. False, they are corrected for time dilation. That whole "nonsense" was to try to help you understand that the shift term does not apply for clocks. You disagreed with it then but now agree with it. You must refute time dilation to maintain your position or redefine it. Let's see your proof against mine that time dilation is different for each clock in a moving frame. I am in the mainstream with experiments to back up my logic.

I mean, ignoring GR and Sagnac since they are not part of this problem, the GPS system preprograms the satellite clock according to time dilation prior to launch. All satellites in one orbit frame have one time. This is also true from the view of the ground clocks. So, if two GPS satellites emitted lasers at UTC 12:00, then the ground would conclude the emissions were simultaneous. R of S needs emissions not to be simultaneous to work. Now, I am not talking reception of the light. That is very clear. If a stationary frame sees two equal light path lengths, then a moving frame will record two different light path lengths. We are not talking light paths though. We are talking emission times. There is nothing in here to address this problem. One would need to perform one way speed of light calculations using clocks. None have been done. What predictions have I made that are not accurate? Merged post follows: Consecutive posts merged This may or may not be true. M contends the emission are simultaneous. If they are also simultaneous emissions in M', which is consistent with time dilation, then M' is struck simultaneously while co-located with M and therefore, M' must conclude two different light speeds. You must bring timing into this experiment also. That is the nature of R of S. My contentions is that clocks frame to frame only beat time dilated. R of S requires them to beat differently at different positions. They are identical except they are corrected for GR, SR and sagnac. Leave in GR and sagnac corrections because they are not part of this problem and therefore that removes their effects as if they did not exist, they all you have is the orbital frame beating time dilation and synchronous. This is consistent with my reasoing but not with R of S. That is the incorrect way to derive time dilation. Here is the correct way. x = vt and t' = ( t - vx/c² )λ substitute x=vt because we are looking at a moving clock along the x-axis at any time t. t' = ( t - v(vt)/c² )λ t' = t( 1 - v²/c² )λ t' = t( (c² - v²) /c² )λ t' = t( (c² - v²) /c² )( √[c²/(c² - v²)] ) t' = t( √[(c² - v²) /c²] ) t' = t/λ. Thus, the moving t' clock beats slower than the stationary clock. You can find this analysis by Einstein in the middle of sections 4. http://www.fourmilab.ch/etexts/einstein/specrel/www/ ------------------------------------------------------------------------------------------------------------------- OK, here is the bottom line on what this thought experiment exposes. If you assume time dilation is true, then all clocks in a moving frame beat the same time dilated value compared to a corresponding clock in the stationary frame. So, if O is a line of synchronized clocks that is stationary, and O' is a line of synchronized clocks that is moving, any query of a stationary clock to a moving clock for an interval t, will find that moving clock to read t/λ. If all points of O query simultaneously, then all will conclude the same result. This is time dilation. Proof Given the description of O above, let cl be any clock in O' and place an O origin at that clock location. That clock will have a position x at after time t according to x = vt after the start time. Also, t' = ( t - vx/c² )λ. Substitute x = vt. t' = ( t - v(vt)/c² )λ. t' = t( 1 - v² /c² )λ t' = t( 1 - v² /c² )λ. t' = t( (c² - v²) /c² )λ. t' = t( (c² - v²) /c² )√[c²/(c² - v²)] t' = t √[(c² - v²)/c²] t' = t / λ. Since the point was chosen arbitrarily, this is true for all points in O'. Since t was used in the equation and this is the time of all synchronized clocks in the O frame, then the proof is complete. Now, if one assumes time dilation is true, then if a frame M emits light at 2 different points simultaneously, then a moving frame M' will conclude the emissions were simultaneous but at the same time dilated interval for both. Since the light paths are different lengths, M' is struck simultaneously for both light beams and both emitted simultaneously, then M' will conclude t=d/c for both paths is false. Other the other hand, if you assume t=d/c for both paths is true, then you must conclude point B emits before point A. But, this contradicts time dilation. Now, GPS and other experiments support time dilation. So that assumption has been validated. However, no one way time light timing experiments have been performed using clocks to support the other assumption that t=d/c is universal. Either way, both cannot be true.

Good argument. Actually, SR says that B will emit before A from the view of M' and M concludes they emit simultaneously. From this, SR claims the clock of B in M' is ahead of the clock of A. Thus, from the view of M', the "emission time t0" in M is reached at B first so emission occurs and then eventually that t0 is reached at A and emission occurs. Now, let M and M' agree on some start time. Let M' be an array of synched clocks. A corrsponding clock that is located with B and any time t, would claim the clock at B is beating faster than the M' clock. Also, handling the A clock the same way, the M' clock would record that the clock of A is beating slower than the M' time. Anyway, GPS does not beat at different rates based on satellite positions. So, it is my view that GPS is the experiment. What is your opinion?

Sure I can set up a correspondence. Let clocks of M' be synchronized. Each clock of M becomes a coordinate system. Set x = vt for the moving system, and t' = ( t - vx/c² )λ Then you get, t' = t/λ for any interval. Yea, I see crappy writing. I meant that because of GPS, the earth frame cannot conclude GPS satellites are not in sync based on position. This is supported by the GPS system. Thus, if two satellites emitted light at exactly 12:00 UTC, the earth frame would not translate that to two different times. Thus, the earth frame would conclude the emissions were simultaneous. This is from the view of M. Yes, there is and this is supported by GPS otherwise, the system would fail. That is because an earth clock would have to conclude a GPS1 and GPS2 did not have the same time and any time t in its frame if they are at different positions according to this branch of SR logic. This is not supported by the scientific evidence.

M' is co-located with M when both light beams meet at M. There is no choice that M' is also met by both light beams simultaneously also.

That is the reasoning that must be used to get it to work. But, it is wrong. Correct me if I am wrong, here is the reasoning. Both M' and M are hit by both the light of A and B simultaneously. Since they were equidistant, by design, they were also simultaneous by design, then this logic is in place for M without question. Now, for M' as stationary, M' is struck by both at the same time. Then, one says, well since the light paths are different, it must be the case that the lights emitted at different times in the frame of M'. Correct? Well, instead of saying is must be the case, I am looking if it is. Let A and B correspond to positions in frame M'. At each point of M' there is a synchronized clock. Also A and B are synched in M. Given any time interval, A and B are at some positions in M' and M' calculates that interval as ∆t and views that interval as ∆t/λ elapsed in M. M' is not concluding M has different times at different positions of M unless light receiving events are taking place. This is GPS. The earth frame does not conclude two satellites in the same orbit have different times based on positions. So, when the interval is at ∆t1/λ in M, both A and B emit. That corresponds to one interval ∆t1 in M'. Therefore, in spite of the fact that SR says they must emit at different times in M' they do not.

I cannot see how the times will be different for M'. When the strikes hit M, M and M' are co-located. I do realize LT claims they will be different. But, this method is testing the validity of the relationship t = d/c.

Got it. Yea, perhaps that was unclear which experiment ws being referenced. OK, we are down to whether the flashes were indeed simultaneous in M' if they are simultaneous in M. Your analysis is correct above, but we are looking at this problem with M' stationary. What is not questioned is that A and B were equidistant from M' when the flashes occured. Now if an M' observer is placed at A at emission and one is placed at B at emission and the clocks are synched in M', the issue is will A' and B' have the same time on their clocks. We know A and B do in M. This is similar to GPS. If GPS1 emits light at 12:00 and GPS2 emits light at 12:00, what is the time of the earth clocks? These emission events will translate to one time in the earth frame. Now, once light starts to travel, then we end up using full LT. I will post a simpler thought experiment. Merged post follows: Consecutive posts mergedExperiment 2 First, take M as stationary. This time, based on the motion of M', the frame of M calculates a simultaneous emission from A and B such that lights will strike M simultaneously when M' and M are coincident. Light emits M' ->v A--------------M--------------B Light received M' ->v A--------------M--------------B Now, take the position that M' is stationary. According to LT and SR, the light path for a frame is measured from the emission point in the frame to the receiver, not from the position a moving light source will reside when the receiver is struck by light. Clearly then, when the light is emitted from A and B, the light emission point in the M' frame at the position where A emits is closer to M' than is the light emission point corresponding to the light emission of B. But, M' will receive the light flashes simultaneously since M' will be coincident with M when the light flashes hit by design. Yet, M' calculates two different light path lengths with equal times and will not measure the same speed of light for both paths.

In the Einstein train scenario, M and M' are co-located when M sees the flashes, i.e. the flashes were emitted when M and M' are not co-located. No, the were co-located when the flashes occured in his stuff. Just when the flashes of lightning occur, this point M' naturally coincides with the point M http://www.bartleby.com/173/9.html They are co-located in mine as well when the flashes occur. Everyone agrees the flashes will hit M at the same time. Einstein said they will not hit M' at the same time. But, A and B were equidistant to M' at light emission for both A and B. These positions become the light emission points in the frame of M' and the light emission points are stationary to M'. Since they are equidistant to M', then M' must be hit by both at the same time. Relativity does not care about the positions of the moving light sources, from the frame of M', after light emission. But, if they hit M' at the same time and M at the same time, and M' and M are not co-located when hit, then we have a contradiction because we have two light beams meeting at two different points in space.

Logic insults? I accept your surrender. I do apologize that I did not fully understand your limitiations. Merged post follows: Consecutive posts merged OK, advice, if you back down with your theory, and do not stand up to me, then you are not confident in it. You should have sucked down my data and then used it to support your "true" theory. You ran instead.

I think you are confused with models and axioms. They are different. And if you have found others that support your "thinking" good. I am providing you the modern way of operating with theories and models. I am quite certain you can find many dark age thinkers that have yet to evolve to the modern methods of mathematical logic. Personally, I would never present anything new without my list of axioms and then a model to prove its logical consistency. Hey, that is just me. Then, if you are going to toss out names and some human concensus logic as some replacement for mathematical logic, go for it. There were many flat earthers at some time but that concensus is not a replacement for the rules and discipline of logic. Clearly, you are not in the mainstream thought for mathematical logic. If you want to debate concensus and democracy, I will take you to a political site and we can operate on that style there. If you want to debate mathematical theories (physics is a sub-branch) and the models that satisfy them, then we will talk logic. You choose.

OK, I read your stuff. You do not have a list of axioms such that your system can construct deductions based on your axioms. Here, theory (set of sentences) http://en.wikipedia.org/wiki/Model_theory This definition is primitive though but will start this discussion. In reality, a theory is a set of L formulas. Now, when you develop a theory, then you must find a model so you can prove your theory is consistent. "Gödel's completeness theorem (not to be confused with his incompleteness theorems) says that a theory has a model if and only if it is consistent," http://en.wikipedia.org/wiki/Model_theory If you cannot produce this model, it is on you, then your theory is useless. It is not on others to prove your theory is consistent. That is the deal.

Let me guess, you don't have any math to go with this.

I cannot wait to see this path/method. We must all be on it in some way or another since there is only your path. It is just how close we match your path is how correct we are. Specifically, what is this path?

The units of time are seconds. Does the poster use other units? And I had a specific question of the poster of where the distance was located in the posters equation. Since the poster could not answer this simple question, what is the point of having a discussion. I am quite sure this will now make your curious of the question I had of the poster, Naturally, I am sure you can understand my reasoning. The thought experiment in this thread is based on distances. The poster offered an equation without a distance. That's because you took some convincing to answer that one properly too. I answered it properly with my thought experiment as I saw it. However, once I saw the nature of your questions, I then realized what I needed to add to make it clearer. It was already quite clear to me. So, your assessment that I divert is completely false since I have presented iron tight math here and clearly presented thought experiments. Instead of making your false and baseless accusations, perhaps you can make statements like "your argument is not clear". So, the topic of this debate: Is it logical to pit LT up against R of S and arrive at a contradiction. Thus far, this is holding up as true. Merged post follows: Consecutive posts merged OK, I will use your stuff and show you where you are wrong. Here are the correct equations. d'/(λ(c+v)) and d'/(λ(c-v)) These translate to t1 = d'/c √((c-v)/(c+v)) t2 = d'/c √((c+v)/(c-v)) In your "lingo", √((c-v)/(c+v)) is 1/(λ(1+ β)) after you do the simplification that λ = 1/(√(1- β² ) and assume v is in terms of light units. Error 1, you forgot d' Error 2: You only did one of the times for M'. There are two different times in the argument for R of S. But, this in no way addresses the LT requirement that ( tA - τ ) = d'/c and ( tB - τ ) = d'/c Thus, tA = tB. By taking M' as the stationary frame, the light sources were both a distance d' at emission and this LT logic also applies.

The only way this could be true is if when light moves through space at the same speed c regardless of any circumstances, the frame moves through space also somehow, then I imagine a specific v could be true. As to absolute time, no I do not believe there is one specific time. I would think the only possibility of this concept, but this is not absolute time, is that clocks can be synchronized across frames similar to GPS. It would seem to me, absolute time would require you ro prove you know the absolute beginning of the universe and you can prove time flows at one beat only at every point of the universe. I am not sure how this could ever be proven. So, I guess I see time as sort of like the North star. You pick a standard frame and use equations for other frames to adjust clocks to beat at the same rate as that standard. Merged post follows: Consecutive posts merged This is a normal debate tactic of yours. You did this to me on the speciific problem I found pitting LT up against R of S. I answered that specifically. Now, I answered this time question already and supplied a link as support. If the poster disagrees, then let the poster supply a disproof since I have already presented a time argument that represents the R of S side of this debate. Here is the link again. Page 17 http://www.tc.umn.edu/~janss011/pdf%20files/appendix-SR.pdf If you disagree or the other poster disagrees with the results of this link, since I agree with the link, you are welcome to present a specific counter argument. But, that would be simple minded to stop at the point and call the problem resolved. There is also the LT side which specifies t=d'/c. I have dealt specifically with this issue as well.

Your opinions do not concern me since you offered an equation you cannot even explain. Further, the coordinate of A or B is not relevent as long as the two are equidistant to M' at light emission. You see, I defined the problem in terms of distances from the emitter to the receiver. So, at both light emissions, it is time t in M and time τ in M'. Since the light sources are equidistant to M' at emission, and both emit at time t in M and this corresponds to time τ in M', then we apply the logic of LT that (t'a - τ) = d'/c and (t'b - τ) = d'/c. Thus, t'a = t'b. But, this disagrees with the logic of R of S.

I have no idea what you are trying to do. You said that equation was for the following, I.e. the time the moving observer measures between passing through the origin (time coordinate equals zero) and hitting the light of the laser is

Well, what can that be? I would welcome comments as to the flaw if there is one. I did something to get around all that anyway. I used: Based on the motion of M', when both the light sources are equidistant to M', both emit light. So, I worked with distances. The frame M is able to use the motion of M' to predict when M and M' will be coincident. So, this part is viable. Next, A and B are equidistant to M as the setup. Therefore, A and B are equidistant to M' when M and M' are coincident. Since the two light sources are equidistant to M' at the time of emission, and the time of their emission is simultaneous in the frame of M, regardless of the motion of the light sources relative to M', M' must conclude d'/c = t for both directions.