Celeritas Posted September 20, 2016 Posted September 20, 2016 Tim88, Then we disagree. Best regards, Celeritas
Tim88 Posted September 20, 2016 Posted September 20, 2016 Thank you Tim88. I think the issue here is a wording one. When I talk about an apparent time loss (or apparent time gain), I am not excluding the presence of a real time loss (or a real time gain if that is appropriate). So if I understand you correctly, I agree with your point. A further conclusion (which please discuss if you disagree), is that it is not necessary to include an apparent time loss in order to deduce the real time loss - they are separate things. When a real time loss occurs, the presence of the real time loss cannot be 'denied' by any particular observer's point of view. Whereas for an apparent time loss, this can be dependent on a particular observer's point of view. So to use the example of two objects in motion, moving at different speeds. This means that the distance between them is changing and thus the change in distance is a real effect. But as to how much that distance is changing, or which one (or both) are moving, or even what their apparent closing speed (or separation speed) is, can be dependent on an observer's point of view. Because of the wrong statement on which you based your self contradiction, I have an issue with the part that I put in bold. Expanding on my illustration, would you agree with the following: "A further conclusion, is that it is not necessary to include an apparent trajectory in order to deduce the real trajectory - they are separate things. When a real trajectory is followed to the point of the ball hitting the net, the presence of the real trajectory cannot be 'denied' by any particular observer's point of view. Whereas for an apparent trajectory, this can be dependent on a particular observer's point of view." It sounds as if you tend to confound "apparent trajectory" (or "apparent clock retardation") with the statement "it may be denied that the ball moved at all" (or "it may be denied that something happened to the clock"). I say this because your earlier conclusion would be, restated in the words of my example: "If these two trajectories were apparent effects – such a scenario would be possible. But the the effect of the trajectory - the ball hitting the net - is real and so these two conditions cannot occur together – hence a logical contradiction has occurred". But there only would be a logical contradiction if one would conclude that no trajectories occurred, based on the deduction that it is not necessary to include an apparent trajectory in order to deduce the real trajectory. By the way, the old literature is full of this kind of discussions, because SR removed the physical model that originally was at the basis of the Lorentz transformation, as "unnecessary". To fill up the conceptual black hole, people next made up their own interpretations to make sense of it; but not all of those interpretations and formulations really match SR, leading to much unnecessary confusion.
robinpike Posted September 20, 2016 Author Posted September 20, 2016 Because of the wrong statement on which you based your self contradiction, I have an issue with the part that I put in bold. Expanding on my illustration, would you agree with the following: "A further conclusion, is that it is not necessary to include an apparent trajectory in order to deduce that there is a real trajectory - they are separate things. When a real trajectory is followed to the point of the ball hitting the net, the presence of the real trajectory cannot be 'denied' by any particular observer's point of view. Whereas for an apparent trajectory, this can be dependent on a particular observer's point of view." My bit in bold - there is a difference between knowing that there is a real trajectory - and knowing what direction, or what speed etc that trajectory might be. It sounds as if you tend to confound "apparent trajectory" (or "apparent clock retardation") with the statement "it may be denied that the ball moved at all" (or "it may be denied that something happened to the clock"). I say this because your earlier conclusion would be, restated in the words of my example: "If these two trajectories were apparent effects – such a scenario would be possible. But the the effect of the trajectory - the ball hitting the net - is real and so these two conditions cannot occur together – hence a logical contradiction has occurred". But there only would be a logical contradiction if one would conclude that no trajectories occurred, based on the deduction that it is not necessary to include an apparent trajectory in order to deduce the real trajectory. My bit in bold - it may not follow that my argument for the travelling clocks one and two causing a logical contradiction, fits the above example? By the way, the old literature is full of this kind of discussions, because SR removed the physical model that originally was at the basis of the Lorentz transformation, as "unnecessary". To fill up the conceptual black hole, people next made up their own interpretations to make sense of it; but not all of those interpretations and formulations really match SR, leading to much unnecessary confusion. Tim88, I am at work at the moment and may not be able to reply for a while. I need to read your post carefully before commenting further. This is important to me to sort out, as I want to remove discussions that are at cross purposes.
swansont Posted September 20, 2016 Posted September 20, 2016 Describing the whole argument in one post is making the subsequent discussions too difficult for me to ascertain which step(s) in the logic are being refuted. Here are the first steps in the logic. Even at this stage of the discussion I do not know if any of these deductions are being disputed, so continuing in discrete, sequential steps can only help. First statements... When the travelling clock completes its round trip journey, it loses time compared to the stay at home clock. This is a loss in time for the travelling clock (as opposed to a gain in time by the stay at home clock). This loss in time is not an apparent loss - it is a real loss in time. To reiterate what Tim88 stated, the measurements are relative, so you can only state that one clock ran slow as compared to another, which means that the other ran fast in comparison. There is no absolute time that you can use as a basis for the analysis. I wish you would dispense with this "apparent" nonsense. It will not be an issue if you do measurements properly and consistently. Your example of an "apparent" measurement is synonymous with an incorrect measurement. From the above statements the following deductions can be made... At some point in the travelling clock's round trip, the travelling clock loses real time with respect to the stay at home clock. It is not necessary to identify at which parts of the travelling clock's journey that real time is lost, or the mechanism as to how that real time is lost, for deduction 1 to be valid. It is not necessary to identify by how much that real time is lost, for deduction 1 to be valid. It is not necessary to include any apparent loss in time (or any apparent gain in time) during the travelling clock's journey for deductions 1, 2 and 3 to be valid. What cannot be deduced from the statements 1, 2 and 3, is whether the travelling clock exclusively loses real time, or whether at any point in its round trip, it gains real time with respect to the stay at home clock. All that can be deduced with those statements is that overall the travelling clock loses real time - and therefore at some point during its journey, it loses real time with respect to the stay at home clock, i.e.deduction 1. 5 can't be deduced from your logic, but it can be determined by measurement and by application of relativity. But when the travelling clock de-accelerated, it effectively accelerated away from the second travelling clock (i.e. it increased its relative speed with respect to the second travelling clock). But such an acceleration is just a repeat of the scenario when the travelling clocks first accelerated away from the stay at home clock (although now with respect to different reference frames). This means that the de-accelerated clock’s time has to run slow with respect to the second travelling clock’s rate of time, which already is running slow with respect to the stay at home clock’s rate of time – AND yet the de-accelerated clock’s rate of time has to be at the same rate as the stay at home clock’s rate of time (because they are stationary with respect to each other). There is no contradiction here. Again, the issue is that clock A runs slow relative to clock B in clock B's frame, and B runs slow relative to A is A's frame. Your error is ignoring that clock rates are not invariant values. Your invalid assumption in your logic is that they should be. A simpler example: In A's frame it is is rest and has no kinetic energy. B is moving, and has KEB. KEB > KEA (=0) In B's frame it is is rest and has no kinetic energy. A is moving, and has KEA. (0=) KEB < KEA Is there a contradiction in those statements? You can argue that they can't be true simultaneously, but you can't be in two frames at once, so that's not a problem. 1
VandD Posted September 20, 2016 Posted September 20, 2016 (edited) .. Tim88, I already addressed this post as well, wrt the twin B. I'll briefly restate it here more generally ... There is no harm whatever, that one models the heavens circling himself as the result of earthly proper-rotation. That, is kinematics. One would have to be a masochist to want to do that, but it is not improper. However, if one is to address energy-considerations, well then energy is absolute. And so if the earth bound observer chooses to make himself the zero reference for all relative motion, he cannot also say that "all of the heavens expended energy to cause the relative motion". He must assign the energy expenditure (that caused the relative motion) to that which in fact experienced the applied force. This is more important in relativity than in classical mechanics, for otherwise the twin B could never accurately predict twin A's clock wrt return. Navigation systems always assign itself the zero reference for motion. For a period of proper-acceleration, a spaceship nav system computes the final location and speed wrt the initial parameters as reference. However, for relativistic rates ... Just before B's virtually-instant turnabout, an inertial phase (v=0.866c so gamma=2) is coming to an end. The location of twin A just before B's proper acceleration is (say) 1 light hr. At the moment halfway thru his turnabout, he is momentarily at rest with earth and twin A, and so twin A "must then be" 2 light hr distant. That happens "in virtually no time at all", because of B's virtually instant turnabout. 2 lt-hr being the proper separation between earth and the turnabout point. Not only that, but with A's wild shift in B-space, comes a wild shift in A-time (in only the B system). Twin A's worldline rapidly rotates in its angular orientation wrt the B worldline, because B's line-of-simultaneity remains always-fixed (since he assumes himself the zero-ref for motion). That rotation also requires twin A to advance wildly along his own worldline (not something that A experiences), but only in the B system, and so (per B) the twin A clock ticks wildly fast during his turnabout. Given such, twin B may accurately predict the twin A clock on return, without any consideration of how A himself experiences it all. Again, this is what the MCCIRF method does. It allows twin B to assign himself the zero reference for all motion, but dynamically adjusts worldlines within B's spacetime system knowing that it was "B himself" who underwent the applied force (causing the change in relative motion). In this way, his spacetime predictions always match that of any inertial observer, who we know all concur with each other using an accepted theory. Twin B's nav system accounts for its own proper acceleration, and adjusts all worldlines in B's own system accordingly. B does not ignore "the consideration of energy-expenditure", but rather accounts for it as necessary. So I"ve already addressed this. If you have a different point that I did not address, maybe just say it, cut to the chase. My point here is that energy considerations are not ignored by twin B using the MCCIRF method. He accounts for it. Twin A knows it was twin B who changed in his own state of motion, because twin A is always inertial and never feels his own inertia. However, the spacetime predictions made by twin A are indifferent wrt who properly accelerated". It matters not, to twin A, because he's always inertial. Twin A knows that for higher relative velocity, the B clock slows down, and that's all A needs. Twin B knows that for higher relative velocity, the A clock slows down, and that's all B needs during his inertial phases of his roundtrip. However during non-inertial phases, twin B "cannot and does not" ignore energy considerations, for otherwise his spacetime predictions of twin A fail. Best regards, Celeritas Good job, Celeritas. One could draw a kind if diagram (not exactly a Minkowski diagram although it shows the same information) representing the traveler's path by a straight line (see right diagram on sketch). At traveler's proper time 4 he feels decelertion/acceleration, which makes his 3D reference frame (red on the sketch) rotate in 4D spacetime. It's a matter of relativity of simultaneity. Many people battle to understand this, even by visualizing it with spacetime diagrams! Edited September 20, 2016 by VandD
Tim88 Posted September 20, 2016 Posted September 20, 2016 (edited) Bold face mine: Good job, Celeritas. One could draw a kind if diagram (not exactly a Minkowski diagram although it shows the same information) representing the traveler's path by a straight line (see right diagram on sketch). At traveler's proper time 4 he feels decelertion/acceleration, which makes his 3D reference frame (red on the sketch) rotate in 4D spacetime. It's a matter of relativity of simultaneity. Many people battle to understand this, even by visualizing it with spacetime diagrams! Apparently Celeritas finally understood my criticism on his phrasing, and we agreed to disagree. The three of us teach here relativity of simultaneity. As I pointed out earlier, clocks in physical reference systems (by means of which simultaneity is indicated) do not automatically adjust to the velocity change as sketched. Therefore your statement that acceleration makes the traveler's reference frame rotate in spacetime -illustrated by your non-Minkowski spacetime diagram- suggests to me a misunderstanding of relativity of simultaneity (despite your great ability of drawing all kinds of instructive diagrams). In any case, I find it unhelpful for showing robinpike that the Lorentz transformations are self consistent. Edited September 20, 2016 by Tim88
robinpike Posted September 20, 2016 Author Posted September 20, 2016 ... I wish you would dispense with this "apparent" nonsense. It will not be an issue if you do measurements properly and consistently. Your example of an "apparent" measurement is synonymous with an incorrect measurement. ... Point taken, I can see that using the terms 'apparent' and 'real' are bad choices and confusing for this discussion (and probably for any discussion). I will re-phrase the initial statements and deductions in a different way. Let's see if I can do it based on measurements and point of views according to observers.
robinpike Posted September 20, 2016 Author Posted September 20, 2016 Here are the first steps in the logic, now rewritten using better phrasing. First statements... When the travelling clock completes its round trip journey, it has less time on its clock than the stay at home clock. This is caused by a loss in time for the travelling clock (as opposed to a gain in time by the stay at home clock). That the travelling clock lost time is agreed by all observers. All observers agree the amount of time lost with respect to the stay at home clock, but it is possible for an observer to measure a different amount of lost time when measured by their own clock. From the above statements the following deductions can be made... At some point in the travelling clock's round trip, the travelling clock lost time compared to the stay at home clock. It is not necessary to identify at which parts of the travelling clock's journey that the time was lost, or the mechanism as to how that time was lost, for deduction 1 to be valid. It is not necessary to state how much time was lost, for deduction 1 to be valid. What cannot be deduced from these first statements (i.e. in the absence of any further statements) is whether the travelling clock at all points on its journey lost time with respect to the stay at home clock, or whether there were some parts of the journey when the travelling clock was gaining time with respect to the travelling clock. I will continue with the next steps in due course.
J.C.MacSwell Posted September 21, 2016 Posted September 21, 2016 Here are the first steps in the logic, now rewritten using better phrasing. First statements... When the travelling clock completes its round trip journey, it has less time on its clock than the stay at home clock. This is caused by a loss in time for the travelling clock (as opposed to a gain in time by the stay at home clock). That the travelling clock lost time is agreed by all observers. All observers agree the amount of time lost with respect to the stay at home clock, but it is possible for an observer to measure a different amount of lost time when measured by their own clock. From the above statements the following deductions can be made... At some point in the travelling clock's round trip, the travelling clock lost time compared to the stay at home clock. It is not necessary to identify at which parts of the travelling clock's journey that the time was lost, or the mechanism as to how that time was lost, for deduction 1 to be valid. It is not necessary to state how much time was lost, for deduction 1 to be valid. What cannot be deduced from these first statements (i.e. in the absence of any further statements) is whether the travelling clock at all points on its journey lost time with respect to the stay at home clock, or whether there were some parts of the journey when the travelling clock was gaining time with respect to the travelling clock. I will continue with the next steps in due course. I think this seems right. It is not standard Minkowski space /SR interpretation (where I think you can deduce where time was lost or gained with respect to any well defined frame) but if I am reading it correctly it is consistent with your statements (and also consistent with experiment as far as I can tell)
VandD Posted September 21, 2016 Posted September 21, 2016 (edited) At some point in the travelling clock's round trip, the travelling clock lost time compared to the stay at home clock. Special relativity tells you exactly where, why and how -relative to the home clock- the travelling clock ticks slower. What else do you want to know? - ... the MCCIRF method ... Wiki quote: <<Spacetime diagram of an accelerating observer in special relativity. The momentarily co-moving inertial frames along the world line of a rapidly accelerating observer (center). The vertical direction indicates time, while the horizontal indicates distance, the dashed line is the spacetime trajectory ("world line") of the observer. The small dots are specific events in spacetime. If one imagines these events to be the flashing of a light, then the events that pass the two diagonal lines in the bottom half of the image (the past light cone of the observer in the origin) are the events visible to the observer. The slope of the world line (deviation from being vertical) gives the relative velocity to the observer. Note how the momentarily co-moving inertial frame changes when the observer accelerates.>> https://en.wikipedia.org/wiki/Minkowski_diagram Edited September 21, 2016 by VandD
robinpike Posted September 21, 2016 Author Posted September 21, 2016 (edited) Special relativity tells you exactly where, why and how -relative to the home clock- the travelling clock ticks slower. What else do you want to know? - Wiki quote: <<Spacetime diagram of an accelerating observer in special relativity. The momentarily co-moving inertial frames along the world line of a rapidly accelerating observer (center). The vertical direction indicates time, while the horizontal indicates distance, the dashed line is the spacetime trajectory ("world line") of the observer. The small dots are specific events in spacetime. If one imagines these events to be the flashing of a light, then the events that pass the two diagonal lines in the bottom half of the image (the past light cone of the observer in the origin) are the events visible to the observer. The slope of the world line (deviation from being vertical) gives the relative velocity to the observer. Note how the momentarily co-moving inertial frame changes when the observer accelerates.>> https://en.wikipedia.org/wiki/Minkowski_diagram True, special relativity does give us that information. However, the starting statements do not mention (special) relativity because the initial deductions need to be independent of any assumptions connected to relativity. Any assumptions / premises that relativity requires / predicts will be included in the upcoming steps in logic, as that is when I think the logical contradiction occurs - and then it can be discussed as to why that logical contradiction is flawed / not flawed etc. Before I put together the details of the argument (and as Swansont suggests, I'll try to think of a way to keep it as simple as possible), To summarize, the focus will be on when the travelling clock changes its progress through space-time at the start of the round trip, and when the travelling clock changes its progress through space-time at the end of the round trip. I think those two steps present a logical contradiction for relativity. But please wait for me to post the logical steps first before explaining why there is no logical issue. =============================================================== To help clarify how it is even possible to have a logical contradiction when relativity agrees absolutely with experiment, here is a pretend example of a logical argument... In the above graphic, the mass of the earth causes space-time to bend. If someone were to explain gravity as an object following the curved lines of space-time, then that explanation has a potential logical contradiction. For if gravity is a result of curved space-time, then what is the force that is distorting the space-time in the first place? Before anyone replies - that is just a made up example to help show what a logical argument is. Edited September 21, 2016 by robinpike
Mordred Posted September 21, 2016 Posted September 21, 2016 (edited) Don't feel bad Robin. Lots of people have difficulty with what is termed Spacetime curvature. In GR space is your x,z,z coordinates. Nothing more... time is just an added coordinate. It is simply geometric relations between events that is curved. Though if you add a multiparticle distribution the density distribution will also be curved in density values. At no point does GR require some spacetime "fabric" to curve. The curvature is geometric relations. The stress tensor in GR tells space how to curve. That stress tensor includes changes in density/pressure,temperature etc. Edited September 21, 2016 by Mordred
swansont Posted September 21, 2016 Posted September 21, 2016 True, special relativity does give us that information. However, the starting statements do not mention (special) relativity because the initial deductions need to be independent of any assumptions connected to relativity. Any assumptions / premises that relativity requires / predicts will be included in the upcoming steps in logic, as that is when I think the logical contradiction occurs - and then it can be discussed as to why that logical contradiction is flawed / not flawed etc. I think there's a chance to go massively astray here. Your second declaration has to be part of the first. You must be sure you haven't assumed anything contrary to relativity. To help clarify how it is even possible to have a logical contradiction when relativity agrees absolutely with experiment, here is a pretend example of a logical argument... In the above graphic, the mass of the earth causes space-time to bend. If someone were to explain gravity as an object following the curved lines of space-time, then that explanation has a potential logical contradiction. For if gravity is a result of curved space-time, then what is the force that is distorting the space-time in the first place? Before anyone replies - that is just a made up example to help show what a logical argument is. You should find a better example; this one has a flawed assumption. Gravity and curved spacetime are the same thing. Before you go to too much trouble restating your argument, I wish you would address this There is no contradiction here. Again, the issue is that clock A runs slow relative to clock B in clock B's frame, and B runs slow relative to A is A's frame. Your error is ignoring that clock rates are not invariant values. Your invalid assumption in your logic is that they should be. A simpler example: In A's frame it is is rest and has no kinetic energy. B is moving, and has KEB. KEB > KEA (=0) In B's frame it is is rest and has no kinetic energy. A is moving, and has KEA. (0=) KEB < KEA Is there a contradiction in those statements? You can argue that they can't be true simultaneously, but you can't be in two frames at once, so that's not a problem. 2
Tim88 Posted September 21, 2016 Posted September 21, 2016 (edited) With a little correction by me: Here are the first steps in the logic, now rewritten using better phrasing. First statements... When the travelling clock completes its round trip journey, it has less time on its clock than the stay at home clock. This is caused by a loss in time for the travelling clock (as opposed to a gain in time by the stay at home clock). That the travelling clock lost time is agreed by all observers. All observers agree the amount of time lost with respect to the stay at home clock, but it is possible for an observer to measure a different amount of lost time when measured by their own clock. From the above statements the following deductions can be made... At some point in the travelling clock's round trip, the travelling clock lost time compared to the stay at home clock. It is not necessary to identify at which parts of the travelling clock's journey that the time was lost, or the mechanism as to how that time was lost, for deduction 1 to be valid. It is not necessary to state how much time was lost, for deduction 1 to be valid. What cannot be deduced from these first statements (i.e. in the absence of any further statements) is whether the travelling clock at all points on its journey lost time with respect to the stay at home clock, or whether there were some parts of the journey when the travelling clock was gaining time with respect to the travelling stay at home clock. I will continue with the next steps in due course. At a first glance that's quite correct I would say (not nitpicking on "loosing time" but making the shown correction). In particular your last statement shows improved understanding in accordance with SR, as it's easy to identify a group of reference frames according to which in a part of the journey the traveling clock was advancing on the stay-at-home clock. Edited September 21, 2016 by Tim88
Celeritas Posted September 22, 2016 Posted September 22, 2016 (edited) Special relativity tells you exactly where, why and how -relative to the home clock- the travelling clock ticks slower. What else do you want to know? - Wiki quote: ... The momentarily co-moving inertial frames ... https://en.wikipedia.org/wiki/Minkowski_diagram Thanx. That wiki animation was better before someone went and removed the twin B horizontal line of simultaneity from it. Best Regards, Celeritas Edited September 22, 2016 by Celeritas
robinpike Posted September 22, 2016 Author Posted September 22, 2016 I think there's a chance to go massively astray here. Your second declaration has to be part of the first. You must be sure you haven't assumed anything contrary to relativity. When the travelling clock completes its round trip journey, it has less time on its clock than the stay at home clock. This is caused by a loss in time for the travelling clock (as opposed to a gain in time by the stay at home clock). Yes, thanks. What I've written isn't right. The second declaration isn't a declaration at all - it is a deduction based on this declaration below that is missing... 2. The travelling clock is the clock that undergoes acceleration to cause the two clocks to move apart and return back together. So then it becomes an additional deduction... 1. The time difference between the two clocks at the end of the journey is caused by a loss in time for the travelling clock (as opposed to a gain in time by the stay at home clock). Before you go to too much trouble restating your argument, I wish you would address this There is no contradiction here. Again, the issue is that clock A runs slow relative to clock B in clock B's frame, and B runs slow relative to A is A's frame. Your error is ignoring that clock rates are not invariant values. Your invalid assumption in your logic is that they should be. A simpler example: In A's frame it is is rest and has no kinetic energy. B is moving, and has KEB. KEB > KEA (=0) In B's frame it is is rest and has no kinetic energy. A is moving, and has KEA. (0=) KEB < KEA Is there a contradiction in those statements? You can argue that they can't be true simultaneously, but you can't be in two frames at once, so that's not a problem. Yes thanks, I am going to have to be careful about things like that so as to avoid false contradictions. The next steps are going to take a bit of thought. The reason for those first steps though, is that I wanted to identify deductions that are valid on their own merit - which means that they should remain valid even when the next steps include deductions provided by relativity. The part that I am currently thinking about, is when the travelling clock undergoes acceleration. When the travelling clock first moves away from the stay at home clock, relativity doesn't care about which direction the travelling clock moves off in. And yet the direction of acceleration does seem to be relevant when the travelling clock applies further acceleration to turn around and finally at the journey's end, when the travelling clock comes to a stop next to the stay at home clock. Someone did already mention there is no contention with that, because it is like moving to the left and then moving right to get back to where you started. Or alternatively moving to the right first, and then to the left. Nothing special about positions in space. But my line of thinking is a bit more subtle than just how the acceleration affects the travelling clock's position in space. The travelling clock ends up on a shorter route through space-time after the first acceleration. And yet the final piece of acceleration at the end of the journey causes the travelling clock to return back to the stay at home clock's (longer) route through space-time. So I am just having a think about whether there is any logical inconsistency in the effects of those two accelerations? Of course the expected outcome is that there is no logical inconsistency - but that is what I am going to think about. With a little correction in bold by me: 4. What cannot be deduced from these first statements (i.e. in the absence of any further statements) is whether the travelling clock at all points on its journey lost time with respect to the stay at home clock, or whether there were some parts of the journey when the travelling clock was gaining time with respect to the travelling stay at home clock. At a first glance that's quite correct I would say (not nitpicking on "losing time" but making the shown correction). In particular your last statement shows improved understanding in accordance with SR, as it's easy to identify a group of reference frames according to which in a part of the journey the traveling clock was advancing on the stay-at-home clock. Yes thank you Tim for noticing that slip up - your correction is what I meant to write.
swansont Posted September 22, 2016 Posted September 22, 2016 When the travelling clock first moves away from the stay at home clock, relativity doesn't care about which direction the travelling clock moves off in. And yet the direction of acceleration does seem to be relevant when the travelling clock applies further acceleration to turn around and finally at the journey's end, when the travelling clock comes to a stop next to the stay at home clock. Someone did already mention there is no contention with that, because it is like moving to the left and then moving right to get back to where you started. Or alternatively moving to the right first, and then to the left. Nothing special about positions in space. But my line of thinking is a bit more subtle than just how the acceleration affects the travelling clock's position in space. The travelling clock ends up on a shorter route through space-time after the first acceleration. And yet the final piece of acceleration at the end of the journey causes the travelling clock to return back to the stay at home clock's (longer) route through space-time. So I am just having a think about whether there is any logical inconsistency in the effects of those two accelerations? Of course the expected outcome is that there is no logical inconsistency - but that is what I am going to think about. The only reason the direction of the second acceleration matters is that you have placed a constraint on the problem, and want the clock to end up at a particular position. That's a boundary condition of your problem that has nothing to do with relativity. And still there are options. You could reverse course, or travel along some other path (e.g. you could travel in a circle through 180 degrees and then retrace the path) and the answer is still the same. The details of the acceleration are irrelevant under the assumptions of the twins paradox.
robinpike Posted September 22, 2016 Author Posted September 22, 2016 The only reason the direction of the second acceleration matters is that you have placed a constraint on the problem, and want the clock to end up at a particular position. That's a boundary condition of your problem that has nothing to do with relativity. And still there are options. You could reverse course, or travel along some other path (e.g. you could travel in a circle through 180 degrees and then retrace the path) and the answer is still the same. The details of the acceleration are irrelevant under the assumptions of the twins paradox. Yes mentioning the direction of the acceleration was a bit muddled - probably because my thoughts went on to mention the comparison about moving from left to right, and right to left. Of course, the travelling clock can choose to arrive back at the stay at home clock from any direction. The bit that I am thinking about is the consequence of the initial acceleration away from the stay at home clock, and the final acceleration that arrives back, stationary again next to the stay at home clock. The first acceleration puts the travelling clock onto a progress through space-time that is shorter than the stay at home clock's progress through space-time, whereas the final acceleration puts it back onto the same progress through space-time as the stay at home clock. Is it logically consistent that those two accelerations can do that? That is where my thoughts are at the moment.
swansont Posted September 22, 2016 Posted September 22, 2016 Yes mentioning the direction of the acceleration was a bit muddled - probably because my thoughts went on to mention the comparison about moving from left to right, and right to left. Of course, the travelling clock can choose to arrive back at the stay at home clock from any direction. The bit that I am thinking about is the consequence of the initial acceleration away from the stay at home clock, and the final acceleration that arrives back, stationary again next to the stay at home clock. The first acceleration puts the travelling clock onto a progress through space-time that is shorter than the stay at home clock's progress through space-time, whereas the final acceleration puts it back onto the same progress through space-time as the stay at home clock. Is it logically consistent that those two accelerations can do that? That is where my thoughts are at the moment. The clock's rate, measured from any inertial reference frame, depends on its speed relative to that frame. Math is at least as rigorous as logic, so if this is mathematically consistent, it should be logically consistent. (And it agrees with observation) Your logic must incorporate relativity or there is no point in discussing this. If you assume something contrary to relativity you can make a logical argument which comes to a conclusion that's inconsistent with relativity. The logic will be fine, but the conclusion has no validity when compared to the real world. By framing this as a logic issue, you're asking the wrong question. If you assume a Galilean world, then the answers may not be logically consistent, because Galileo's relativity and Einstein's relativity don't agree.
Tim88 Posted September 22, 2016 Posted September 22, 2016 bold emphasis mine: [..]But my line of thinking is a bit more subtle than just how the acceleration affects the travelling clock's position in space. The travelling clock ends up on a shorter route through space-time after the first acceleration. And yet the final piece of acceleration at the end of the journey causes the travelling clock to return back to the stay at home clock's (longer) route through space-time. So I am just having a think about whether there is any logical inconsistency in the effects of those two accelerations? Of course the expected outcome is that there is no logical inconsistency - but that is what I am going to think about. [..] Are you not complicating things unnecessarily? I'm just asking, as I don't really know what you have in mind, but the simplest case is a traveling clock that is passing the Earth's clock at high speed and later turns around and passes the Earth's clock again. Only one acceleration step involved, and roughly the same prediction about the difference in clock times between the two events.
robinpike Posted September 22, 2016 Author Posted September 22, 2016 (edited) bold emphasis mine: Are you not complicating things unnecessarily? I'm just asking, as I don't really know what you have in mind, but the simplest case is a traveling clock that is passing the Earth's clock at high speed and later turns around and passes the Earth's clock again. Only one acceleration step involved, and roughly the same prediction about the difference in clock times between the two events. Perhaps, but I would prefer not to use a scenario where different observers may measure the travelling clock in different ways. If I keep to when the travelling clock begins next to the stay at home clock, and ends its round trip, again next to the stay at home clock, then at least all observers can agree that the travelling clock lost time compared to the stay at home clock. That may make it easier to find if a logical contradiction has occurred? Also it is the acceleration of those two events that results in the travelling clock making different progress through space-time (shorter / as opposed to back to not shorter). Edited September 22, 2016 by robinpike
michel123456 Posted September 22, 2016 Posted September 22, 2016 (...), time dilation is symmetric. Each clock sees the other as running slow. (...)This is the corner stone of all misunderstandings. i use to compare it to the optical effect caused by distance. When A see B at a distance, B looks smaller. At the same time, when B looks at A, then A is smaller. So it happens that A and B "are" smaller than each other. But it is not a contradiction. This is to be parallelized to standard comparison of heights: If A is bigger than B, then B is smaller than A. These are 2 totally different concepts.
VandD Posted September 23, 2016 Posted September 23, 2016 This is the corner stone of all misunderstandings. There is no misunderstanding. Only for time intervals including acceleration/deceleration the symmetry is disrupted.
michel123456 Posted September 23, 2016 Posted September 23, 2016 There is no misunderstanding. Only for time intervals including acceleration/deceleration the symmetry is disrupted. Sure there is. Because in my examples nobody is arguing that B becomes "really" shorter because there is a distance between A & B. In Relativity some argue that B is "really" shorter and that time is "really" dilated because these are not optical effects. You must understand that it is really troubling. Especially when it is linked to the symmetrical effect. IOW any layman will accept that if B is "really" shorter than A, the A is "really" bigger than B. And a layman will not accept easily that they are "both really shorter than the other".
swansont Posted September 23, 2016 Posted September 23, 2016 Sure there is. Because in my examples nobody is arguing that B becomes "really" shorter because there is a distance between A & B. In Relativity some argue that B is "really" shorter and that time is "really" dilated because these are not optical effects. You must understand that it is really troubling. Especially when it is linked to the symmetrical effect. IOW any layman will accept that if B is "really" shorter than A, the A is "really" bigger than B. And a layman will not accept easily that they are "both really shorter than the other". I will ask you what I asked robinpike elsewhere in the thread. Two objects, A and B, are in relative motion In A's frame it is is rest and has no kinetic energy. B is moving, and has KEB. KEB > KEA (=0) In B's frame it is is rest and has no kinetic energy. A is moving, and has KEA. (0=) KEB < KEA Can you accept this? Or put another way, that they each see themselves as moving slower than the other one, since they see themselves as at rest? 1
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