Iggy Posted May 11, 2013 Share Posted May 11, 2013 It is very somple , really. If you reset the C clock to the value of the B clock, there is a jump in elapsed time because at the event BC , when the clocks met, they have accumulated different amounts of proper time. This is what post 177 demonstrates mathematically. Post 73 showed the same calculations (not in Latex). C starts his stopwatch at event BC. He measures no proper time before that event according to Md's thought experiment. At event BC, C merely writes down the reading he notices on B's stopwatch and starts his own. If you have a different scenario where C does something different, or earlier, then I am very interested in that, and I'd enjoy reading whatever thread you might decide to create on that topic. 1 Link to comment Share on other sites More sharing options...
xyzt Posted May 11, 2013 Share Posted May 11, 2013 (edited) C starts his stopwatch at event BC. He measures no proper time before that event according to Md's thought experiment. At event BC, C merely writes down the reading he notices on B's stopwatch and starts his own. If you have a different scenario where C does something different, or earlier, then I am very interested in that, and I'd enjoy reading whatever thread you might decide to create on that topic. This is false: "Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C." Clock C accumulated a proper time [math]\tau_C[/math] between start and BC. At BC, clock C was reset to mathch clock B value, [math]\tau_B[/math] where [math]\tau_B \ne \tau_C[/math]. Even if you claimed that C is simply set to [math]\tau_B[/math] at BC ignoring its history, the above is STILL equivalent to executing a step in speed from [math]v_B[/math] to [math]v_C[/math], i.e. accelerating from [math]v_B[/math] to [math]v_C[/math]. This was explained twice, once in post 73 and a second time in post 177. Edited May 11, 2013 by xyzt Link to comment Share on other sites More sharing options...
Iggy Posted May 11, 2013 Share Posted May 11, 2013 (edited) This is false: "Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C." Clock C accumulated a proper time [math]\tau_C[/math] between start and BC. No. You are incorrect and I am now correcting you. The history of clock C before event BC is irrelevant to the thought experiment and it is irrelevant to the claim you quoted above and called false. The quoted claim is true, and it is true regardless of, and independently from, anything you want to say about C before BC. At BC, clock C was reset to mathch clock B value, [math]\tau_B[/math] where [math]\tau_B \ne \tau_C[/math]. The measurement of [math]\tau_C[/math] does not start until BC. Whatever arbitrary length of time one wishes to dream about C measuring his proper time before that event, it has no bearing on the results of the thought experiment proposed and described in detail in this thread. I couldn't reasonably talk about such a thing at any more length without knowing that it serves no purpose but hijacking Md's thread away from any possible relevant considerations, and I do not want to do that.. Even if you claimed that C is simply set to [math]\tau_B[/math] at BC ignoring its history, the above is STILL equivalent to executing a step in speed from [math]v_B[/math] to [math]v_C[/math], i.e. accelerating from [math]v_B[/math] to [math]v_C[/math]. This was explained twice, once in post 73 and a second time in post 177. I'll now say for the fourth time: if one considers a coordinate transformation to be part of the definition of acceleration then acceleration is key to the twin paradox results, and if not then not. It is a question of semantics. No other honest and real disagreements persist between you and Md. I won't comment on the dishonest ones that serve no purpose but obfuscating and arguing over nothing endlessly. I'm also to understand that the results of the experiment can be completed and calculated without any additional needed information when ignoring the history of clock C before event BC completely, and those results will be entirely consistent, and indeed quantitatively identical to, the traditional twin paradox? You admit that all we could do with C's proper time before BC is to measure it for any length of arbitrary time, then at the end of the experiment throw the results away because they are irrelevant to, and a compete distraction from, any relevant information that affects the results of the thought experiment? The sentence I quoted directly above does admit precisely as much, yes? Edited May 11, 2013 by Iggy Link to comment Share on other sites More sharing options...
xyzt Posted May 11, 2013 Share Posted May 11, 2013 (edited) No. You are incorrect and I am now correcting you. The history of clock C before event BC is irrelevant to the thought experiment and it is irrelevant to the claim you quoted above and called false. The quoted claim is true, and it is true regardless of, and independently from, anything you want to say about C before BC. The measurement of [math]\tau_C[/math] does not start until BC. Not according to MD. From post 74: One way to describe the discrepancy in MEASUREMENT while observing the same thing, is to consider the differing distances that the light signal travels from AB to BC. Remember each observer measures light traveling at a speed of c, and if you accept that, then relativity of simultaneity has these unintuitive effects. Using gamma=2, v=.866, proper time T from AB to BC = 1 year (measured by B)... For A, the rest distance between itself and BC (imagine a marker in space at that location, or consider that events occur in all frames, they don't have an inertial frame and can be considered at rest relative to anyone) is the length-contracted distance traveled by B, converted to A's frame: v*T*gamma = 1.732 LY. Observer A ages 2 years while B ages the one, and t = 2-1.732 = 0.268, the age A is when it must send a signal to reach event BC. For B: If A sent a signal at time t, the time at B is t*gamma = 0.536, and the light spends the rest of proper time T incoming, ie. T-t*gamma = 0.464 years. For B, the signal travels 0.464 LY, during which A ages 0.464/gamma = 0.232. Add t, the age when A sent the signal, and you get that A aged 0.5 years when B reaches C, in agreement with the Lorentz transformation. For C: Observer C is closing on B extremely quickly. Using composition of velocities, its relative velocity is 0.990c, Lorentz factor gammaB=7. While B ages 1 year, C ages 7 before they pass. Taking 7 years to reach BC, it measures that A ages 7/gamma (it is also moving at -v relative to A) = 3.5 years. If A sends its signal at 0.268 years old, it will age 3.5-0.268 = 3.232 years while the light signal approaches BC, during which C ages 3.232*gamma = 6.464 years. So to recap: Due to relativity of simultaneity, and light traveling the same relative speed according to all the observers, A observes that the signal from AB to BC travels 1.732 LY, B measures 0.464 LY, and C measures 6.464 years. You admit that all we could do with C's proper time before BC is to measure it for any length of arbitrary time, You need to keep on reading, IF you do that , this is equivalent to accelerating from [math]v_B[/math] to [math]v_C[/math]. And no, this is not a coordinate transformation. It is very simple really. Edited May 11, 2013 by xyzt Link to comment Share on other sites More sharing options...
Delta1212 Posted May 11, 2013 Share Posted May 11, 2013 (edited) The entire point of this thread is that a coordinate transformation provides equivalent results to acceleration. You are trying to disprove this by stating that the coordinate transformation present in the experiment is... equivalent in effect to acceleration. Edit to elaborate: The thought experiment states that transferring B's time to C at BC will end the experiment with the C reading a time at the end of the experiment as if a single clock had started the experiment at B's velocity and accelerated to C's velocity at point BC. None of the clocks need actually undergo acceleration to accomplish this and the time at the end will not reflect the proper time experienced by either B or C over the course of the experiment. Edited May 11, 2013 by Delta1212 Link to comment Share on other sites More sharing options...
Iggy Posted May 11, 2013 Share Posted May 11, 2013 (edited) You need to keep on reading, IF you do that , this is equivalent to accelerating from [math]v_B[/math] to [math]v_C[/math]. And no, this is not a coordinate transformation. It is very simple really. That it is quantitatively the same as having one outbound and inbound observer whom accelerates (the traditional twin paradox) is exactly the claim of the OP. Nevertheless, no clock actually accelerates in the thought experiment. The only question remaining is if one semantically cares to consider if the passing of information from one frame to another via light is considered "acceleration". If so, acceleration is key. If not then not. You've now agreed with the OP and myself as confrontationaly as possible. One wonders if you knew just now that you were fiercely supporting our contentions with that comment. (-1)ing all my posts while I've tried to be helpful, drawn diagrams and did math that took considerable time only because you've asked for them, discussed in good faith, and not once neg repped you in this thread, only serves to demonstrate the breadth of your desire to argue, and the shallow nature of the evidence you bring to honestly work out in unbiased inquiry in this thread. Edited May 11, 2013 by Iggy Link to comment Share on other sites More sharing options...
xyzt Posted May 11, 2013 Share Posted May 11, 2013 (edited) The entire point of this thread is that a coordinate transformation provides equivalent results to acceleration. You are trying to disprove this by stating that the coordinate transformation present in the experiment is... equivalent in effect to acceleration. Edit to elaborate: The thought experiment states that transferring B's time to C at BC will end the experiment with the C reading a time at the end of the experiment as if a single clock had started the experiment at B's velocity and accelerated to C's velocity at point BC. None of the clocks need actually undergo acceleration to accomplish this and the time at the end will not reflect the proper time experienced by either B or C over the course of the experiment. There is no coordinate transformation in the explanation. So, no. That it is quantitatively the same as having one outbound and inbound observer whom accelerates (the traditional twin paradox) is exactly the claim of the OP. Nevertheless, no clock actually accelerates in the thought experiment. The only question remaining is if one semantically cares to consider the passing of information from one frame to another via light is considered "acceleration". If so, acceleration is key. If not then not. You've now agreed with the OP and myself as confrontationaly as possible. One wonders if you knew just now that you were fiercely supporting our contentions with that comment. Like I said, it is very simple: you either - dump C's time and reset the clock to B's time (see citation from MD's claim in the previous post, just in case you missed it), meaning that you are generating a discontinuity in C's clock reading - make up the journey from two DIFFERENT parts, one whereby B moves at [math]v_B[/math] and accumulates proper time [math]\tau_B[/math] and the second leg whereby C moves at [math]v_C[/math] and accumulates proper time [math]\tau_C[/math]. This means that at event BC the speed jumped from [math]v_B[/math] to [math]v_C[/math] (-1)ing all my posts while I've tried to be helpful, drawn diagrams and did math that took considerable time only because you've asked for them, discussed in good faith, and not once neg repped you in this thread, only serves to demonstrate the breadth of your desire to argue, and the shallow nature of the evidence you bring to honestly work out in unbiased inquiry. It reflects your tone. Edit to elaborate: The thought experiment states that transferring B's time to C at BC will end the experiment with the C reading a time at the end of the experiment as if a single clock had started the experiment at B's velocity and accelerated to C's velocity at point BC. Correct. None of the clocks need actually undergo acceleration to accomplish this and the time at the end will not reflect the proper time experienced by either B or C over the course of the experiment. Correct, the acceleration is "hidden" through this trick that has been exposed since post 73. In other words, the acceleration has always been there, it just takes an experienced debunker to expose it. Edited May 11, 2013 by xyzt -1 Link to comment Share on other sites More sharing options...
md65536 Posted May 11, 2013 Author Share Posted May 11, 2013 Not according to MD. From post 74: [...] You need to keep on reading, IF you do that , this is equivalent to accelerating from [math]v_B[/math] to [math]v_C[/math]. And no, this is not a coordinate transformation. It is very simple really. You've misunderstood what I said in 74. I agree with Iggy's assessment. In any post of mine that refers to C, it is treated as an inertial observer and there is no acceleration from v_B to v_C nor any equivalent. If you speak of C's history, I am assuming that C has remained inertial for all relevant history. Link to comment Share on other sites More sharing options...
Delta1212 Posted May 11, 2013 Share Posted May 11, 2013 (edited) There is no coordinate transformation in the explanation. So, no. Like I said, it is very simple: you either - dump C's time and reset the clock to B's time (see citation from MD's claim in the previous post, just in case you missed it), meaning that you are generating a discontinuity in C's clock reading- make up the journey from two DIFFERENT parts, one whereby B moves at [math]v_B[/math] and accumulates proper time [math]\tau_B[/math] and the second leg whereby C moves at [math]v_C[/math] and accumulates proper time [math]\tau_C[/math]. This means that at event BC the speed jumped from [math]v_B[/math] to [math]v_C[/math]It reflects your tone. Correct.Correct, the acceleration is "hidden" through this trick that has been exposed since post 73. In other words, the acceleration has always been there, it just takes an experienced debunker to expose it. Except you're not exposing anything. Demonstrating this equivalence was md's entire point, and was presumably understood by everyone arguing in his favor. I don't believe anyone here believes you can measure the time of an accelerated frame on a single inertial clock. The point wasn't that the twin experiment doesn't require undergoing a change in direction in order to create time discrepancy. It's that this discrepancy is entirely dependent upon the use of two (or more) different inertial frames. If you define switching frames as acceleration, then absolutely, the twin paradox is impossible without acceleration. A better understanding of the point of the thread is that there is nothing about experiencing acceleration that creates the twin paradox other than the fact that it causes the observer to change frames, and it is this use of two different frames which results in the discrepancy. The reason that "Acceleration is unimportant" is being claimed is solely because the effect can be measured using two clocks that, themselves, do not have to actually accelerate. The time they measure is understood to be the time in an accelerated frame and not reflective of either clock's individual proper time over the course of the experiment. Edited May 11, 2013 by Delta1212 Link to comment Share on other sites More sharing options...
xyzt Posted May 12, 2013 Share Posted May 12, 2013 (edited) You've misunderstood what I said in 74. I agree with Iggy's assessment. In any post of mine that refers to C, it is treated as an inertial observer and there is no acceleration from v_B to v_C nor any equivalent. If you speak of C's history, I am assuming that C has remained inertial for all relevant history. The acceleration is hidden through a trick. You jump from observer B to observer C at event BC. If you add up the proper times of B (for the outboard leg) with the time of C (for the return leg) it is as if you accelearted. You may not admit it but this is what you did. The reason that "Acceleration is unimportant" is being claimed is solely because the effect can be measured using two clocks that, themselves, do not have to actually accelerate. Yet, the acceleration is there, right under your nose. Edited May 12, 2013 by xyzt Link to comment Share on other sites More sharing options...
Delta1212 Posted May 12, 2013 Share Posted May 12, 2013 xyzt, if you're defining acceleration as changing the frame in which a measurement is being conducted, then yes, there is acceleration present in this experiment. Is that what you're looking for? I feel like you think md is trying to claim that, because clock B is inertial, and clock C is inertial, and clock C's reading (which was set to B's at BC partway through the experiment) is equivalent to that of the accelerated twin in the twin paradox, that he must have then produced an example of the twin paradox that shows a discrepancy in elapsed time between inertial frames. You are then stating that this is inaccurate because upon switching the measurement from B's clock to C's clock, you are changing frames and, in effect, accelerating in the same way that the accelerating twin does in the typical version of the experiment. Is that accurate or am I mistaking your view of this thread? Link to comment Share on other sites More sharing options...
xyzt Posted May 12, 2013 Share Posted May 12, 2013 (edited) xyzt, if you're defining acceleration as changing the frame in which a measurement is being conducted, then yes, there is acceleration present in this experiment. Is that what you're looking for? I feel like you think md is trying to claim that, because clock B is inertial, and clock C is inertial, and clock C's reading (which was set to B's at BC partway through the experiment) is equivalent to that of the accelerated twin in the twin paradox, that he must have then produced an example of the twin paradox that shows a discrepancy in elapsed time between inertial frames. You are then stating that this is inaccurate because upon switching the measurement from B's clock to C's clock, you are changing frames and, in effect, accelerating in the same way that the accelerating twin does in the typical version of the experiment. Is that accurate or am I mistaking your view of this thread? Yes, for the past 190 posts. Edited May 12, 2013 by xyzt Link to comment Share on other sites More sharing options...
Delta1212 Posted May 12, 2013 Share Posted May 12, 2013 (edited) Yes, for the past 190 posts.Ok, so here's where the disconnect is happening. I didn't interpret md's claim that way. I'm assuming Iggy and md didn't either. I was working under the assumption that the equivalence of transferring B's reading to C's clock with an acceleration of one of the clocks was understood, and that md wasn't attempting to demonstrate a discrepancy between the elapsed times of inertial frames, but rather demonstrate that you could measure the elapsed time of an accelerated frame using observers in two different inertial frames making up different legs of the trip, without having to accelerate the actual observers. The final measurement would then not correspond to either inertial observer, but to the accelerated path that would be made up of their component paths. For that to be the case, the equivalence with an accelerated observer is, in fact, integral to md's entire premise. The acceleration of the measurement isn't hidden, it's the entire point. The acceleration that isn't important is that of the individual clocks, none of which measure an elapsed time in agreement with the final accelerated time measurement that is read by C at the end of the experiment. Edit to be extra clear: No one (I believe) thinks that the time on C at the end of the experiment is reflective of C's elapsed proper time since the beginning of the experiment. Edited May 12, 2013 by Delta1212 Link to comment Share on other sites More sharing options...
xyzt Posted May 12, 2013 Share Posted May 12, 2013 (edited) I was working under the assumption that the equivalence of transferring B's reading to C's clock with an acceleration of one of the clocks was understood, No, it wasn't understood. MD has admited it at times only to recant later, followed by admitting it just a few posts later. Edited May 12, 2013 by xyzt Link to comment Share on other sites More sharing options...
Delta1212 Posted May 12, 2013 Share Posted May 12, 2013 I'm pretty sure at this point that this whole thread has been an argument over semantic differences and ambiguous language, but I guess I'll wait to see how md and Iggy respond to my view of the thought experiment. I've been assuming they both basically agree with what I said above, but I suppose it's possible they don't and there really is more meat to the disagreement. Link to comment Share on other sites More sharing options...
xyzt Posted May 12, 2013 Share Posted May 12, 2013 (edited) I'm pretty sure at this point that this whole thread has been an argument over semantic differences and ambiguous language, but I guess I'll wait to see how md and Iggy respond to my view of the thought experiment. I've been assuming they both basically agree with what I said above, but I suppose it's possible they don't and there really is more meat to the disagreement. The whole thread is about pretending that "acceleration is not important in the twin paradox" while trying to hide it all along. Edited May 12, 2013 by xyzt Link to comment Share on other sites More sharing options...
DimaMazin Posted May 12, 2013 Share Posted May 12, 2013 In system of observer the observer has no acceleration.Observer can see only motions of masses relatively of him,the motions of the masses slow down his time. Link to comment Share on other sites More sharing options...
Markus Hanke Posted May 12, 2013 Share Posted May 12, 2013 Clocks in uniform relative motion do disagree, so let's try this with a different approach to see if we can clear it up. If an alien ship were passing though our solar system inertially at close to c relative to earth would you agree that it could cover the distance between the earth and the sun in one second (according to the clocks on the ship)? Of course it could, but that is because, in the frame of reference of the alien ship, the distance between earth and sun is length contracted. The pilot of your UFO will not notice anything special happening on his own clock. The point is that the pilot of your alien vessel could consider himself at rest, and argue that it is in fact the sun which is rushing towards him, or the earth receding from him. Because his frame is inertial, in the absence of another outside point of reference he has no way to decide whether it is himself moving, or whether it is the sun/earth that is moving, or some combination of the two. If he was to observe a clock co-moving with the sun or earth, he would also read 1 second on it since it is dilated by exactly the same factor, just like on his own clock is from the other frame. The two frames of reference are freely interchangeable without affecting the physical outcome, so long as all coordinate effects are taken into account; that is what I mean when I say that the clocks cannot disagree, and neither can rulers. Not so in the twin paradox - because the travelling twin measures a local gravitational field in his vessel, his frame is physically distinguishable from the one of the stationary stay-behind twin. They are no longer symmetric, so if you swap frames you will find that the clocks are dilated by different factors. They are not symmetric. Link to comment Share on other sites More sharing options...
Iggy Posted May 12, 2013 Share Posted May 12, 2013 Of course it could, but that is because, in the frame of reference of the alien ship, the distance between earth and sun is length contracted. The pilot of your UFO will not notice anything special happening on his own clock. The point is that the pilot of your alien vessel could consider himself at rest, and argue that it is in fact the sun which is rushing towards him, or the earth receding from him. Because his frame is inertial, in the absence of another outside point of reference he has no way to decide whether it is himself moving, or whether it is the sun/earth that is moving, or some combination of the two. If he was to observe a clock co-moving with the sun or earth, he would also read 1 second on it since it is dilated by exactly the same factor, just like on his own clock is from the other frame. The two frames of reference are freely interchangeable without affecting the physical outcome, so long as all coordinate effects are taken into account; that is what I mean when I say that the clocks cannot disagree, and neither can rulers. Not so in the twin paradox - because the travelling twin measures a local gravitational field in his vessel, his frame is physically distinguishable from the one of the stationary stay-behind twin. They are no longer symmetric, so if you swap frames you will find that the clocks are dilated by different factors. They are not symmetric. I know what you're saying, Markus. The velocity between the earthling and the alien is relative and symmetric. The situation, however, is not symmetric. The alien intersected the world line of the earth and sun, and the earthling did not. This doesn't mean that one is moving and the other is stationary. It doesn't establish a preferred frame. It simply acknowledges that one of them is present for an event on the sun and the other is not. Can we further agree that a different alien could be heading inertially through the solar system the opposite direction -- that it could pass within meters of the first alien ship as the first alien ship passes the sun -- then traverse the distance between the sun and earth in one second (according to the clocks on the second alien ship)? Assuming your answer is yes, let's compare it to your statement: Given that, I think it is quite clear that in the scenario of three clocks, all of which are in uniform relative motion with respect to one another, you will never find a "twin paradox". That is really all I was trying to point out. For me, it is most clearly obvious in the proper time integral for each observer. The path length of the earth observer is roughly 17 minutes between the first alien flyby and the second. The path length of the first alien is one second, and the path length of the second alien is one second. My interpretation of the OP is that two inertial observers can measure the path, earth -> sun -> earth, just as well as a single twin undergoing instant acceleration can do. The cause of the resolution of the twin paradox is therefore not the proper acceleration felt by any particular clock. Rather, the cause of the resolution of the paradox is the fact that the path (earth event 1 -> sun -> earth event 2) through spacetime is shorter than the path (earth event 1 -> earth event 2). Ok, so here's where the disconnect is happening. I didn't interpret md's claim that way. I'm assuming Iggy and md didn't either. I was working under the assumption that the equivalence of transferring B's reading to C's clock with an acceleration of one of the clocks was understood, and that md wasn't attempting to demonstrate a discrepancy between the elapsed times of inertial frames, but rather demonstrate that you could measure the elapsed time of an accelerated frame using observers in two different inertial frames making up different legs of the trip, without having to accelerate the actual observers. The final measurement would then not correspond to either inertial observer, but to the accelerated path that would be made up of their component paths. Yes, that is my understanding. By the way, you just stated the clock postulate very precisely. It postulates that an accelerated path can be broken up into "momentarily comoving inertial frames"... (what you just called "inertial frames making up different legs of the trip"). I believe Md independently discovered the clock postulate, and its claims are essentially equivalent to his. 1 Link to comment Share on other sites More sharing options...
md65536 Posted May 12, 2013 Author Share Posted May 12, 2013 (edited) The acceleration is hidden through a trick. You jump from observer B to observer C at event BC. If you add up the proper times of B (for the outboard leg) with the time of C (for the return leg) it is as if you accelearted. You may not admit it but this is what you did.I won't admit it but let's say that it's true. It's still not important because it doesn't have a measurable (detectable) effect. If it does, what is the difference between measurements when the "hidden" acceleration is neglected, vs. when it is included? I have not seen anyone include acceleration in any equation in this thread. In the typical twin paradox with the traveler using one clock, change of velocity is real and important because there is a difference between measurements when the acceleration is applied (the twin returns home) vs. not (the twin continues out into space). What difference occurs if your "hidden" effects are not considered? Is it possible that you're inventing a hidden effect to explain something that you don't understand? I believe Md independently discovered the clock postulateOnly partly, and I made the mistake described in your link (http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html) of thinking that the equations of SR prove the postulate is true, when really applying the equations of SR, when acceleration is involved, assumes that the postulate is true. If the postulate was not true, the accepted calculations of the twin paradox effect according to SR would be wrong, while SR could still correctly predict the outcome of the experiment in post #1, which would then not be equivalent to a twin paradox setup. Edited May 12, 2013 by md65536 1 Link to comment Share on other sites More sharing options...
Markus Hanke Posted May 12, 2013 Share Posted May 12, 2013 (edited) I know what you're saying, Markus. The velocity between the earthling and the alien is relative and symmetric. The situation, however, is not symmetric. The alien intersected the world line of the earth and sun, and the earthling did not. This doesn't mean that one is moving and the other is stationary. It doesn't establish a preferred frame. It simply acknowledges that one of them is present for an event on the sun and the other is not. I'm afraid I don't follow you. I never said anything about "preferred frames". In fact I explicitly stated that the frames are interchangeable without affecting the physical outcome, thereby ruling out that one of them is "preferred". That automatically makes them symmetric. I do not know why you state that on the one hand the motion is relative and symmetric, whereas the "situation" is not. That makes no sense. My interpretation of the OP is that two inertial observers can measure the path, earth -> sun -> earth, just as well as a single twin undergoing instant acceleration can do. There is no such thing as "instant acceleration". I must admit that I am starting to wonder what the actual purpose of this thread really is. Of course it is possible to make up the elapsed proper time of an accelerating observer by combining the times of two purely inertial observers. So what ? That does not allow us to state that "acceleration is not important in the twin paradox". The two scenarios are not physically equivalent, in the sense that if you conduct them right next to each other, the participants will always be able to tell which experiment they take part in. So could somebody just state in plain text what the premise of the OP actually is ? I am getting more and more confused. So far as I am concerned acceleration is an integral part of the "twin paradox" scenario as it appears in relativity textbooks, because the point of the exercise is to teach students the physical difference between symmetric and non-symmetric frames. If you eliminate the asymmetry, you eliminate the discrepancy between the two twins after they are brought back together in the same frame at relative rest, defeating the purpose. Likewise, if you eliminate the "bring back together at rest in the same frame" bit, you depart from the original twin paradox scenario. In the typical twin paradox with the traveler using one clock, change of velocity is real and important because there is a difference between measurements when the acceleration is applied (the twin returns home) vs. not (the twin continues out into space). Yes, precisely my point. So, the premise of the thread that "acceleration is not important in the twin paradox" is clearly false. You can consider alternatives to the typical twin paradox scenario to achieve the same outcome, but these are then different scenarios, which are not physically equivalent. Edited May 12, 2013 by Markus Hanke 1 Link to comment Share on other sites More sharing options...
Iggy Posted May 12, 2013 Share Posted May 12, 2013 Only partly, and I made the mistake described in your link (http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html) of thinking that the equations of SR prove the postulate is true, when really applying the equations of SR, when acceleration is involved, assumes that the postulate is true. If the postulate was not true, the accepted calculations of the twin paradox effect according to SR would be wrong, while SR could still correctly predict the outcome of the experiment in post #1, which would then not be equivalent to a twin paradox setup. I agree. Important distinction. I'm afraid I don't follow you. I never said anything about "preferred frames". In fact I explicitly stated that the frames are interchangeable without affecting the physical outcome, thereby ruling out that one of them is "preferred". That automatically makes them symmetric. I do not know why you state that on the one hand the motion is relative and symmetric, whereas the "situation" is not. That makes no sense. I meant to agree with you when saying there is no preferred frame. The velocity is symmetric. The situation is not symmetric because the earthling is at rest relative to the earth and sun, and the alien is not at rest relative to the earth and sun. The first half of the thought experiment involves 4 objects. Three of them are in one frame of reference (the earthling, the earth, and the sun) while the alien is the only thing at rest in his frame. If we label one event on the earth and another on the sun it is inevitable that the two observers will measure different times between the two events. If you label one event "alien passes earth" and another "alien passes sun" then the earthling and the alien will measure different amounts of proper time between those events. That aspect is not symmetrical. There is no such thing as "instant acceleration". I agree. It is a useful simplification is all I must admit that I am starting to wonder what the actual purpose of this thread really is. Of course it is possible to make up the elapsed proper time of an accelerating observer by combining the times of two purely inertial observers. So what ? The conclusion is that the twin paradox is conceptually consistent with the clock postulate. The clock hypothesis is an assumption in special relativity. It states that the rate of a clock doesn't depend on its acceleration but only on its instantaneous velocity. This is equivalent to stating, that a clock moving along a path [math]P[/math] measures the proper time, defined by: [math]d \tau = \int_P \sqrt {dt^2 - dx^2/c^2 - dy^2/c^2 - dz^2/c^2}[/math]. The clock hypothesis was implicitly (but not explicitly) included in Einstein's original 1905 formulation of special relativity. Since then, it has become a standard assumption and is usually included in the axioms of special relativity, especially in the light of experimental verification up to very high accelerations in particle accelerators. Clock hypothesis When you earlier said, Given that, I think it is quite clear that in the scenario of three clocks, all of which are in uniform relative motion with respect to one another, you will never find a "twin paradox". That is really all I was trying to point out. For me, it is most clearly obvious in the proper time integral for each observer. I wasn't sure if you thought the results of the traditional twin paradox were quantitatively equivalent to Md's thought experiment. It looked (probably mistakenly on my part) like you thought they would not be. Link to comment Share on other sites More sharing options...
Delta1212 Posted May 12, 2013 Share Posted May 12, 2013 I'm afraid I don't follow you. I never said anything about "preferred frames". In fact I explicitly stated that the frames are interchangeable without affecting the physical outcome, thereby ruling out that one of them is "preferred". That automatically makes them symmetric. I do not know why you state that on the one hand the motion is relative and symmetric, whereas the "situation" is not. That makes no sense. There is no such thing as "instant acceleration". I must admit that I am starting to wonder what the actual purpose of this thread really is. Of course it is possible to make up the elapsed proper time of an accelerating observer by combining the times of two purely inertial observers. So what ? That does not allow us to state that "acceleration is not important in the twin paradox". The two scenarios are not physically equivalent, in the sense that if you conduct them right next to each other, the participants will always be able to tell which experiment they take part in. So could somebody just state in plain text what the premise of the OP actually is ? I am getting more and more confused. So far as I am concerned acceleration is an integral part of the "twin paradox" scenario as it appears in relativity textbooks, because the point of the exercise is to teach students the physical difference between symmetric and non-symmetric frames. If you eliminate the asymmetry, you eliminate the discrepancy between the two twins after they are brought back together in the same frame at relative rest, defeating the purpose. Likewise, if you eliminate the "bring back together at rest in the same frame" bit, you depart from the original twin paradox scenario. Yes, precisely my point. So, the premise of the thread that "acceleration is not important in the twin paradox" is clearly false. You can consider alternatives to the typical twin paradox scenario to achieve the same outcome, but these are then different scenarios, which are not physically equivalent. The sentence "Acceleration is not important in the twin paradox" is not here being used to mean "acceleration is not a necessary aspect of the classic twin paradox" because clearly you can't have an observer leave and then return without accelerating. It is rather being used to mean "acceleration is only necessary to switch frames for one observer but itself is otherwise incidental to the time dilation effect, which can be demonstrated by the fact that the same path through spacetime can be accurately measured by two inertial observers." Or stated another way, if disagreement of proper elapsed time is the demonstrated effect of the twin paradox, then the cause of the twin paradox is the path through spacetime through two inertial frames, while the acceleration has no direct contribution to the effect except that a single observer cannot switch frames without first accelerating. It is necessary to, but not the cause of, the time dilation. In other words, the point of the thread is, as Iggy said, to illustrate the clock postulate by setting up a thought experiment that will measure an equivalent time over an equivalent path to the twin paradox sans acceleration. It is not to set up a physically identical experiment to the twin paradox, but without acceleration, as this is clearly impossible. Md appears to be learning about relativity at the moment because I've noticed that he frequently makes topics about it as he discovers/learns new aspects of it, apparently in an effort to both work through the information himself as we'll as to receive input and help demonstrate the information to others. I don't think it has much objective beyond that, but this is an assumption about md that I have made. If I am wrong, I'm certain that I'll be corrected. Link to comment Share on other sites More sharing options...
Markus Hanke Posted May 12, 2013 Share Posted May 12, 2013 In other words, the point of the thread is, as Iggy said, to illustrate the clock postulate by setting up a thought experiment that will measure an equivalent time over an equivalent path to the twin paradox sans acceleration. It is not to set up a physically identical experiment to the twin paradox, but without acceleration, as this is clearly impossible. Excellent, if we can all agree on this then we're good I was just starting to get confused as to what the actual argument was, since participants here appeared to be going in different directions, talking about different things. Link to comment Share on other sites More sharing options...
Delta1212 Posted May 12, 2013 Share Posted May 12, 2013 Excellent, if we can all agree on this then we're good I was just starting to get confused as to what the actual argument was, since participants here appeared to be going in different directions, talking about different things. I suspect that confusion was common to most of the participants, whether we all realized it or not. Link to comment Share on other sites More sharing options...
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