SciForums.com > sciforums.com > Archives > Pseudoscience Archive > Time Dilation PDA View Full Version : Time Dilation Post ReplyCreate New ThreadPages : [1] 2 chinglu12-14-10, 08:01 PMHave coordinate at x' = -5/6ls. What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0. t' = ( t - vx/c² )γ t' = ( t - 0 )γ = tγ. So, moving clock coming toward rest origin beat not time dilated. Is this wrong? Tach12-14-10, 09:46 PMHave coordinate at x' = -5/6ls. What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0. t' = ( t - vx/c² )γ t' = ( t - 0 )γ = tγ. So, moving clock coming toward rest origin beat not time dilated. Is this wrong? Yes, it is wrong but you are too stupid to figure out why. arfa brane12-14-10, 11:00 PMDon't worry about Tach, he doesn't actually know anything. Well, he does know how to type. James R12-14-10, 11:39 PMHave coordinate at x' = -5/6ls. That's half a set of spacetime coordinates. What is the t' value? What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0. I don't understand this. Which two events are you talking about? You have given one coordinate of one event, but then you start talking as if there are two events. So, moving clock coming toward rest origin beat not time dilated. I don't understand this either. Perhaps your English isn't so good. rpenner12-15-10, 04:19 AMI think I have mastered the translation process. Please allow me to assist. Have coordinate at x' = -5/6ls. In a 1+1 space time, let us have two inertial frames S, with unprimed coordinates x and t, and S', with primed coordinates x' and t'. Let A be the time-like world-line containing all events in this spacetime such that x' = - \frac{5}{6} \, \textrm{light-seconds}. What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0. Assuming that the frames S and S' agree on event O as being the origin of their respective space and time coordinate systems, we call this standard configuration such that event O has the following coordinates: x_O = 0, \, t_O = 0, \, x'_O = 0, \, t'_O = 0 Let B be the time-like world-line containing all events in this spacetime such that x = 0. Obviously, unless the frame S and S' are not in relative motion, world-lines A and B share only one event in common. Following convention (established below) we call this event Q. Let us call the motion of S' relative to S as v, so now we may compute the coordinates of Q in both frames: x_Q = 0, \, t_Q = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}}, \, x'_Q = - \frac{5}{6} \, \textrm{light-seconds}, \, t'_Q = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} t' = ( t - vx/c² )γ t' = ( t - 0 )γ = tγ. Applying the Lorentz transform from S to S' for Q, we see that \begin{pmatrix} t'_Q \\ x'_Q \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} & \frac{-v}{c^2 \sqrt{1- \frac{v^2}{c^2}}} \\ \frac{-v}{\sqrt{1- \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} t_Q \\ x_Q \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} \\ \frac{-v}{\sqrt{1- \frac{v^2}{c^2}}} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} \end{pmatrix} =\begin{pmatrix} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \\ - \frac{5}{6} \, \textrm{light-seconds} \end{pmatrix} So, moving clock coming toward rest origin beat not time dilated. The poster then asserts that a clock on world-line A, which is moving at speed v according to S, is not time-dilated. Is this wrong? It is wrong for a number of reasons. Just because at event O (which is on world-line B, and not on world-line A) t_O = t'_O = 0 does not imply that on world-line A (the world-line of the moving clock) that t = t'. Specifically, on world-line A, t=0 and t'=0 specify two different events. To see this parameterize the events on A with the parameter a (with units of length): x_{A_a} = \frac{a}{\sqrt{1-\frac{v^2}{c^2}}}, \, t_{A_a} = \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{c^2-v^2}{v c^2}) + \frac{a}{v}}{\sqrt{1-\frac{v^2}{c^2}}}, \, x'_{A_a} = - \frac{5}{6} \, \textrm{light-seconds}, \, t'_{A_a} = \frac{(\frac{5}{6} \, \textrm{light-seconds}) + a}{v} So if you wanted to answer a question about time dilation of a clock moving (according to S) on world-line A and ending at event Q, you need to choose a starting event P on A. (Once again, O is not on A.) P_{t'=0} \, = \, P_{a = - \frac{5}{6} \, \textrm{light-seconds}} with coordinates: x = \frac{- \frac{5}{6} \, \textrm{light-seconds}}{\sqrt{1-\frac{v^2}{c^2}}}, \, t = \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{-v}{c^2})}{\sqrt{1-\frac{v^2}{c^2}}}, \, x' = - \frac{5}{6} \, \textrm{light-seconds}, \, t' = 0 But P_{t=0} \, = \, P_{a = - (1 - \frac{v^2}{c^2}) (\frac{5}{6} \, \textrm{light-seconds} )} has different coordinates: x = - \sqrt{1 - \frac{v^2}{c^2}} (\frac{5}{6} \, \textrm{light-seconds} ) , \, t = 0, \, x' = - \frac{5}{6} \, \textrm{light-seconds}, \, t' = \frac{v (\frac{5}{6} \, \textrm{light-seconds})}{c^2} So because P_{t'=0} \neq P_{t=0} you have to be extra careful when talking about time dilation as a clock moves from P to Q, since t_Q - t_P depends on the choice of P. FYI, this is likely chinglu1998 who recently appeared on various forums with contrived physical situations which he seeks to leverage into arguing that the Lorentz transform predicts absolute simultaneity or absence of time dilation for moving clocks. PhysForum: OP Edited, thread title changes, 10 day suspension (http://www.physforum.com/index.php?showtopic=28676) BAUT: Thread closed (http://www.bautforum.com/showthread.php/110308-Time-Dilation-and-Negative-Coordinates-in-Moving-Frame?p=1828259) Physics Forums: Backed into a corner (http://www.physicsforums.com/showthread.php?t=455613) Watch out for ambiguous problem statements that leave one or more endpoints of an interval ambiguously defined so that an apples and oranges comparison results or arguments that a Lorentz transformation of a uniformly moving particle may result in a frame where the particle has the same time dilation factor as the original speed. (This happens in 1+1 space time when you switch to a frame with twice the rapidity of the particle.) chinglu12-15-10, 10:35 AMThat's half a set of spacetime coordinates. What is the t' value? I don't understand this. Which two events are you talking about? You have given one coordinate of one event, but then you start talking as if there are two events. I don't understand this either. Perhaps your English isn't so good. This picture. |---> v ---------x'---------|------------------ | Origins of frames common here Interval starts when origins are common. Interval ends when x' and unprime origin common. Unprime, apply length contraction dx = ( 0 - x'/γ ) dx = vdt -x'/γ = vdt dt = (-x'/γ)/v prime The unprime origin moves to x' coordinate in this frame. dx' = -x' dx' = vdt' = -x' dt' = -x'/v. dt = dt'/γ or dtγ = dt'. For x' to move to unprime origin prime clock beats faster than origin clock. This my calculation. chinglu12-15-10, 10:44 AMWatch out for ambiguous problem statements that leave one or more endpoints of an interval ambiguously defined so that an apples and oranges comparison results or arguments that a Lorentz transformation of a uniformly moving particle may result in a frame where the particle has the same time dilation factor as the original speed. (This happens in 1+1 space time when you switch to a frame with twice the rapidity of the particle.) Start of time interval is origins same. End interval is when x' at same place as unprime origin. This is well defined interval. Next, when origins same, do not set t'=t=0. It is impossible to sync clocks in frames to 0 during motion. This only makes sense for light pulse. So When origins same only record time on clocks as t0 and t0'. Now simple question in unprimed frame how long it takes for x' to reach origin. dx = vdt. It is that simple. dx = -x'/γ. This applys lengths contraction. This is simplest motion possible in a frame. So very simple dt = (-x'/γ)/v. Now simple question in primed frame how long it takes for unprime origin to reach x'. No length contraction. dx' = vdt'. dt' = dx'/v = -x'/v. Check ratio dt'/dt. chinglu12-15-10, 11:07 AMDon't worry about Tach, he doesn't actually know anything. Well, he does know how to type. Many thanks. Since his post had no math I simply dismissed it out of hand. Tach12-15-10, 11:46 AMMany thanks. Since his post had no math I simply dismissed it out of hand. If you want to refute relativity , you will need to learn it first. So far, despite of posting the same mistakes all over the internet, you haven't. chinglu12-15-10, 12:04 PMIf you want to refute relativity , you will need to learn it first. So far, despite of posting the same mistakes all over the internet, you haven't. Here is thinking in a simple way. Both frames can understand when origins same. Both frames can understand when unprimed origin and coordinate (x',0,0) same. dt = dx/v = (-x'/γ)/v. dt' = dx'/v = -x'/v. This does not refute relativity. Tach12-15-10, 12:12 PMHere is thinking in a simple way. Both frames can understand when origins same. Both frames can understand when unprimed origin and coordinate (x',0,0) same. dt = dx/v = (-x'/γ)/v. dt' = dx'/v = -x'/v. This does not refute relativity. No, it only shows that after spamming multiple forums on the internet you still don't understand any. chinglu12-15-10, 12:46 PMNo, it only shows that after spamming multiple forums on the internet you still don't understand any. Here is general problem. Let (x',y',z' ) coordinate in primed frame. Let (x,y,z ) coordinate in unprimed frame. Assume standard configuration. this means motion along x-axis and time starts when origins same. Question to answer how long it takes for (x',y,z ) to be at same location as (x,y,z ) from when origins same. Answer in unprimed frame. dx = vdt. dx = x - x'/γ because of length contraction of x'. x - x'/γ = vdt x'/γ = x - vdt x' = (x - vdt)γ Have you seen this equation before? Answer in primed frame. dx' = vdt'. dx' = x/γ - x' because of length contraction of x. x/γ - x' = vdt' x/γ = x' + vdt' x = (x' + vdt')γ Have you seen this equation before? Tach12-15-10, 12:53 PMHere is general problem. The "general problem" is that you don't know SR. You TRY to apply it, without knowing what you are doing. You have been refuted countless times on countless forums, yet , you persist. Why? chinglu12-15-10, 01:36 PMThe "general problem" is that you don't know SR. You TRY to apply it, without knowing what you are doing. You have been refuted countless times on countless forums, yet , you persist. Why? As any person can see in above post all I have been doing is a different way of rewriting Lorentz Transforms. This mean if your thoughts are true, then Lorentz Transforms have been refuted. I have read some your posts. I have see you use math when you have actual argument. I see also you do not use math when responding to my posts. Tach12-15-10, 02:11 PMAs any person can see in above post all I have been doing is a different way of rewriting Lorentz Transforms. ....while showing a total lack of understanding of what they represent. This mean if your thoughts are true, then Lorentz Transforms have been refuted. No, it means you don't understand basic relativity. As proven by your threads being locked and you being banned for trolling. I see also you do not use math when responding to my posts. There is no point , you would not understand it anyways. chinglu12-15-10, 02:49 PM....while showing a total lack of understanding of what they represent. No, it means you don't understand basic relativity. As proven by your threads being locked and you being banned for trolling. There is no point , you would not understand it anyways. Your writings are false. This is respected poster in this forum. He agrees with my calculations. physicsforums.com/showpost.php?p=3032483&postcount=23 physicsforums.com/showpost.php?p=3032528&postcount=24 Tach12-15-10, 03:08 PMYour writings are false. This is respected poster in this forum. He agrees with my calculations. physicsforums.com/showpost.php?p=3032483&postcount=23 physicsforums.com/showpost.php?p=3032528&postcount=24 No one agrees with your nonsense. rpenner made an effort to chronicle all your mistakes and all your bans. rpenner12-15-10, 03:11 PMRecap: World-line B has events O and Q. World-line A has events P_{t'=0}, P_{t=0} and Q. In frame S, world-line A moves at speed v and world-line B moves at speed 0. In frame S', world-line A moves at speed 0 and world-line B moves at speed -v. Every event has a x and t coordinate in S, and a x' and t' coordinate in S'. The Lorentz transformation connects the coordinates in S with the coordinates in S' for the same event. \begin{tabular}{l|cccc} \textrm{Event} & x & t & x' & t' \\ \hline \\ P_{t'=0} & \frac{- \frac{5}{6} \, \textrm{light-seconds}}{\sqrt{1-\frac{v^2}{c^2}}} & \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{-v}{c^2})}{\sqrt{1-\frac{v^2}{c^2}}} & - \frac{5}{6} \, \textrm{light-seconds} & 0 \\ O & 0 & 0 & 0 & 0 \\ P_{t=0} & - \sqrt{1 - \frac{v^2}{c^2}} (\frac{5}{6} \, \textrm{light-seconds} ) & 0 & - \frac{5}{6} \, \textrm{light-seconds} & \frac{v (\frac{5}{6} \, \textrm{light-seconds})}{c^2} \\ Q & 0 & \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} & - \frac{5}{6} \, \textrm{light-seconds} & \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \end{tabular} There are therefore 3 time-like intervals in the forward direction that terminate at Q Which clock is considered moving depends on what world-line the time-like intervals describe. In all cases the time dilation appears based on v or -v, depending on in which frame the interval describes movement. So it is important both when and where your clock starts and when and where your clock ends to determine who sees it moving and who sees time-dilation \begin{tabular}{l|ccc|cccc|c} \textrm{Interval} & \textrm{World-line} & \textrm{Moving} & \textrm{Unmoving} & \Delta x_{Moving} & \Delta t_{Moving} & \Delta x_{Unmoving} & \Delta t_{Unmoving} & \frac{\Delta t_{Moving}}{\Delta t_{Unmoving}} \\ \hline \\ Q - P_{t'=0} & A & S & S' & \Delta x = \frac{5}{6} \, \textrm{light-seconds} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \Delta t = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}& \Delta x' = 0 & \Delta t' = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} & \frac{\Delta t}{\Delta t'} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ Q - O & B & S' & S & \Delta x' = - \frac{5}{6} \, \textrm{light-seconds} & \Delta t' = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} & \Delta x = 0 & \Delta t = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} & \frac{\Delta t'}{\Delta t} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ Q - P_{t=0} & A & S & S' & \Delta x = \frac{5}{6} \, \textrm{light-seconds} \sqrt{1 - \frac{v^2}{c^2}} & \Delta t = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} & \Delta x' = 0 & \Delta t' = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \left(1 - \frac{v^2}{c^2} \right) & \frac{\Delta t}{\Delta t'} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ \hline \end{tabular} Importantly, Special Relativity has a geometric interpretation, so that you can graph these events as points and intervals as line segments for either S or S' coordinates. Watch out for ambiguous problem statements that leave one or more endpoints of an interval ambiguously defined so that an apples and oranges comparison results ... This picture. |---> v ---------x'---------|------------------ | Origins of frames common here Interval starts when origins are common.We are talking about a space-time interval, so you have to say when and where with clarity. Only at x'=0 does t=t'=0. At x' = -5/6 light-seconds t'=0 and t=0 are different events with different coordinates. See the tables above. Interval ends when x' and unprime origin common. Event Q is the unique event in space-time when x' = -5/6 light-seconds and x = 0. Unprime, apply length contraction dx = ( 0 - x'/γ ) dx = vdt -x'/γ = vdt dt = (-x'/γ)/v This is correct for t_Q - t_{P_{t=0}}. prime The unprime origin moves to x' coordinate in this frame. dx' = -x' dx' = vdt' = -x' dt' = -x'/v. dt = dt'/γ or dtγ = dt'. This is correct for t'_Q - t'_{P_{t'=0}} but importantly P_{t'=0} \neq P_{t=0}. For x' to move to unprime origin prime clock beats faster than origin clock. This my calculation. Your calculation mixes apples and oranges, and therefore in physically invalid and appears based on assuming absolute simultaneity in that you do not recognize that P_{t'=0} \neq P_{t=0} since x' \neq 0 and that you have not considered the geometrical content of what is world-line A and B, and what is the the locus of all events where t = 0 and t' = 0. Alternately for the interval Q-O (which is on world-line B) your calculations are correct, but world-line B is not moving in the primed coordinates. So the physical meaning of t and t' swap places and time dilation of the moving clock is still exactly as predicted. Start of time interval is origins same.Actually, we have shown that at x' = -5/6 light-seconds, t' = 0 is not the same time as t = 0. You need a time and a place for the start or end of an interval. End interval is when x' at same place as unprime origin.The event Q has both a time and a place and has x' = -5/6 light-seconds and x = 0. This is well defined interval.Not until you pick t' = 0 or t = 0, because you can't have both when x' = -5/6 light-seconds. Next, when origins same, do not set t'=t=0. It is impossible to sync clocks in frames to 0 during motion. This only makes sense for light pulse. t'=t=0 only at the event where x'=x=0 i.e. at event O. You were the one who insisted on "standard configuration". t and t' are coordinate time not synchronized clock time -- we don't care about the synchronized time in this example, we only care about the time elapsed between two events in space time, so synchronization is not an issue. The Lorentz transformation is homogeneous, so it works perfectly fine with coordinate differences, provided you are correctly working with well-defined intervals. So When origins same only record time on clocks as t0 and t0'.That's fine for a clock on world-line B, but the clock you seem to want to talk about is on world-line A. So you have to choose, t = 0 or t' = 0. Now simple question in unprimed frame how long it takes for x' to reach origin. That's starting at P_{t=0} and ending at Q. dx = vdt. It is that simple. dx = -x'/γ. This applys lengths contraction. This is simplest motion possible in a frame. So very simple dt = (-x'/γ)/v.The length contraction formula is only valid when the t coordinate of the ends is the same, and since you are having problems with keeping your events straight, you should probably stick with the full Lorentz transforms. Now simple question in primed frame how long it takes for unprime origin to reach x'. No length contraction. dx' = vdt'. dt' = dx'/v = -x'/v. But this is starting from P_{t'=0} and ending at Q. Check ratio dt'/dt. There's no pedagogically valid reason to, since you have impermissible mixed apples and oranges and due to your error of assuming absolute simultaneity. Many thanks. Since his post had no math I simply dismissed it out of hand. Your posts have bad and misapplied math, which we do not dismiss out of hand, but attempt to engage you and determine where your geometrical and physical reasoning has gone awry so that you might learn how to apply these concepts correctly. Here is thinking in a simple way. Both frames can understand when origins same. Both frames can understand when unprimed origin and coordinate (x',0,0) same. O and Q are well-defined events in space-time. They have a well-defined place and a well-defined when. But for Q, x_Q \neq x'_Q and t_Q \neq t'_Q so obviously someone should be careful when talking about t = 0 or t' = 0 since they only both happen at O. dt = dx/v = (-x'/γ)/v. dt' = dx'/v = -x'/v. This does not refute relativity. It badly misapplies relativity, and your thinking is too simple. chinglu12-15-10, 03:38 PMYour posts have bad and misapplied math, which we do not dismiss out of hand, but attempt to engage you and determine where your geometrical and physical reasoning has gone awry so that you might learn how to apply these concepts correctly. O and Q are well-defined events in space-time. They have a well-defined place and a well-defined when. But for Q, x_Q \neq x'_Q and t_Q \neq t'_Q so obviously someone should be careful when talking about t = 0 or t' = 0 since they only both happen at O. It badly misapplies relativity, and your thinking is too simple. First, the coordinate ('x,0,0) is not an event. It is a space coordinate. It translates to ('x/γ,0,0) when origins are same. There are two events in this problem. 1) The event of both origins being same. 2) The event of x' and x being same. Next you cannot set t'=t=0 when time-like intervals. It is impossible to Einstein synchronize the clocks during the motion and get meaningful results. So all you can deal with is time intervals as relativity produces. So you can only record the frame times when origins same and work with time intervals. This is simple question and if you refute it you refute lorentz transforms. Is below true or false. If false please give specific reasons on these simple equations. Here is general problem. Let (x',y',z' ) coordinate in primed frame. Let (x,y,z ) coordinate in unprimed frame. Assume standard configuration. this means motion along x-axis and time starts when origins same. Question to answer how long it takes for (x',y,z ) to be at same location as (x,y,z ) from when origins same. Answer in unprimed frame. dx = vdt. dx = x - x'/γ because of length contraction of x'. x - x'/γ = vdt x'/γ = x - vdt x' = (x - vdt)γ Have you seen this equation before? Answer in primed frame. dx' = vdt'. dx' = x/γ - x' because of length contraction of x. x/γ - x' = vdt' x/γ = x' + vdt' x = (x' + vdt')γ Have you seen this equation before? Maybe you should look at Galilean transformations and how they function. They answer two questions in a frame if time start when origins are same. 1) If t in a unprimed frame elapsed, what x' will at same place as x. 2) If given x' and x what is elapsed time these coordinates be at same place. Lorentz no different except must use length contraction of the other frames coordinates to get relativity answer. chinglu12-15-10, 03:53 PMRecap: It badly misapplies relativity, and your thinking is too simple. We also went through this in other forum. We assumed your answer was correct that all time intervals for moving frame time dilated. We then had clock at negative coordinate x'. We had clock at origin of moving frame and rest frame. We then start time intervals when origins same. Record time t0 and t0'. All clocks in moving frame have t0' when origins same and all clock in rest frame have t0 in rest frame. These apply only to frames themselves. We then assume when x' clock reach rest origin and rest origin elapsed t and rest clock has t0 + t as its time. We said clock at x' elapsed t/γ and has t0' + t/γ to agree with your calculations. But clock at moving origin elapsed t/γ also and has t0' + t/γ on its clock. This mean rest frame concludes these clock remain synchronized which contradict relativity of simultaneity. James R12-16-10, 03:09 AMIf this conversation is happening on two forums at once, I suggest you keep it in one forum. Please choose one. chinglu12-16-10, 05:21 PMIf this conversation is happening on two forums at once, I suggest you keep it in one forum. Please choose one. We are not have this conversation anywhere else. We had been but discontinued it. chinglu12-16-10, 07:12 PMI can not know why posters in other forums are typing about me and are afraid to come here. This is where I am going to have this discussion and no other forums. If they do not come here that means they are afraid. Beer w/Straw12-16-10, 07:54 PMI can not know why posters in other forums are typing about me and are afraid to come here. This is where I am going to have this discussion and no other forums. If they do not come here that means they are afraid. OK This I couldn't help. I have giggles :o http://www.physforum.com/index.php?showtopic=28676&st=75 chinglu12-16-10, 08:16 PMOK This I couldn't help. I have giggles :o Did you have math or statement of fear? Beer w/Straw12-16-10, 08:28 PMI don't know your aim for one. Two, I hope fear is not the word you really want to use. To say "statement of fear" is absurd. I'm not going over everything that's been said. I have not even gone over rpenner's retort once, forget thrice! So, if I can be forgiven for my laziness, what is it exactly you are saying? rpenner12-17-10, 02:37 AMI don't know what he's saying. But in this thread alone, he has demonstrated the absolute necessity to not just talk about intervals of space or intervals of time when doing relativity because you need to quantify the exact same interval of space and time. These are the points I wish to convey: Special Relativity is a description of space and time as an inseparable whole -- the concepts of "place" and "when" don't mean much without each other. Length contraction you only fairly derive between co-moving time-like world-lines of endpoints because it is an effect caused by relativity of simultaneity. http://www.sciforums.com/showpost.php?p=2518556&postcount=33 Time dilation only happens between two events on the world-line of the clock under consideration. Relativity of simultaneity means that even though clocks might be synchronized at one event, those clocks need not be synchronized at another event. And the full Lorentz transformations are homogeneous, meaning they work just as well on an interval of coordinate differences as they do on sets of coordinates. but I have not seen that the poster managed to engage any of them. //Edit -- Fun fact -- e^{\begin{pmatrix} 0 & \ln \sqrt{\frac{c+v}{c-v}} \\ \ln \sqrt{\frac{c+v}{c-v}} & 0 \end{pmatrix} } = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\end{pmatrix} chinglu12-17-10, 06:37 PMI don't know your aim for one. Two, I hope fear is not the word you really want to use. To say "statement of fear" is absurd. I'm not going over everything that's been said. I have not even gone over rpenner's retort once, forget thrice! So, if I can be forgiven for my laziness, what is it exactly you are saying? Yes. Assume clock is located at x' < 0 in the primed coordinates. Assume standard configuration and time interval begins for both frame when origins same. Assume time interval ends when x' and unprimed origin same. What is this time interval in each frame. Unprimed/rest frame Apply length contraction. -x'/γ = vt. t = -x'/(vγ) Primed frame -x' = vt'. t' = -x'/v. Note in unprimed frame/rest frame clock elapsed more time in moving frame than rest frame. This verified by Lorentz Transforms. x = 0, origin at rest. t' = ( t - vx/c² )γ = tγ chinglu12-17-10, 06:53 PMI don't know what he's saying. But in this thread alone, he has demonstrated the absolute necessity to not just talk about intervals of space or intervals of time when doing relativity because you need to quantify the exact same interval of space and time. These are the points I wish to convey: Special Relativity is a description of space and time as an inseparable whole -- the concepts of "place" and "when" don't mean much without each other. Length contraction you only fairly derive between co-moving time-like world-lines of endpoints because it is an effect caused by relativity of simultaneity. Time dilation only happens between two events on the world-line of the clock under consideration. Relativity of simultaneity means that even though clocks might be synchronized at one event, those clocks need not be synchronized at another event. And the full Lorentz transformations are homogeneous, meaning they work just as well on an interval of coordinate differences as they do on sets of coordinates. but I have not seen that the poster managed to engage any of them. //Edit -- Fun fact -- e^{\begin{pmatrix} 0 & \ln \sqrt{\frac{c+v}{c-v}} \\ \ln \sqrt{\frac{c+v}{c-v}} & 0 \end{pmatrix} } = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\end{pmatrix} I quantified the intervals precisely. How long in rest frame does it take a clock located at x' to reach the rest origin when everything starts when origins are same? Start event: origins same End event: x' same with rest origin. Beer w/Straw12-17-10, 07:17 PMI was actually looking at what you replied to me with. It wasn't hard to restrain myself from posting cause it was actually enjoyable to have no idea what you typed. Hence, a mystery was afoot and tedious deciphering to come - in between whatever else of more importance I was doing. It could have been fun, but then you had to go and respond to rpenner and suddenly I didn't feel so special. :mad: :bawl: rpenner12-18-10, 04:29 AMI quantified the intervals precisely. How long in rest frame does it take a clock located at x' to reach the rest origin when everything starts when origins are same? Start event: origins same End event: x' same with rest origin. Event's have a place and a time. Your start event is O, when you need to pick exactly one of P_{t'=0} or P_{t=0} if you want to answer the full question. When you start at O and go to Q, you are only talking about a clock which is not moving in S (what you call the rest frame) and is moving in S'. James R12-18-10, 05:17 AMAssume standard configuration and time interval begins for both frame when origins same. In other words, assume the origins of the x and x' frames are co-located at t=t'=0. Assume clock is located at x' < 0 in the primed coordinates. i.e. Assume the clock doesn't move in primed frame, so it always has x'=-c, for c a positive constant. So at t'=t=0, the clock is at x'=-c. At this time, the clock's location in the x frame is: x=\gamma (x' + vt') = -c\gamma Assume time interval ends when x' and unprimed origin same. What is this time interval in each frame. If x' frame moves in positive x direction with speed v, then the time interval ends when the x coordinate of the clock is zero. The clock moves at speed v in the x frame (along with x' frame). In the x frame, we have distance covered = speed multiplied by time taken, so: c\gamma = v\Delta t which gives the time interval in the x frame. i.e. \Delta t = \frac{c\gamma}{v} The corresponding time in the t' frame is for the event at x=0 is: \Delta t' = \frac{c\gamma^2}{v} Beer w/Straw12-18-10, 11:13 AMI was actually looking at what you replied to me with. It wasn't hard to restrain myself from posting cause it was actually enjoyable to have no idea what you typed. Hence, a mystery was afoot and tedious deciphering to come - in between whatever else of more importance I was doing. It could have been fun, but then you had to go and respond to rpenner and suddenly I didn't feel so special. :mad: :bawl: On further evaluation, however, your question appears a little bit too textbook and given the extraordinary measures to explain mathematically you have done yourself a tremendous disservice by demanding only the math. It could be so simple as explaining your dissension in mere words. Just starting a post with saying "This is what my textbook says and this is what I don't understand." Rather than jump in with any incorrect mathematical conclusion and leaving others' bewildered in how you reached it. chinglu12-18-10, 11:18 AMEvent's have a place and a time. Your start event is O, when you need to pick exactly one of P_{t'=0} or P_{t=0} if you want to answer the full question. When you start at O and go to Q, you are only talking about a clock which is not moving in S (what you call the rest frame) and is moving in S'. Is it not a well defined and agreed upon event when origins are the same. An event has a place and elapsed time since origins same. It is not a well defined event when x' < 0 and the unprime origin are at same place? I am not understand problem. chinglu12-18-10, 11:24 AMIn other words, assume the origins of the x and x' frames are co-located at t=t'=0. i.e. Assume the clock doesn't move in primed frame, so it always has x'=-c, for c a positive constant. So at t'=t=0, the clock is at x'=-c. At this time, the clock's location in the x frame is: x=\gamma (x' + vt') = -c\gamma No clock does not move. But, unprime origin move to x'. That is the end event for primed frame. Since unprime origin moves at -vt, t = ( t' + vx'/c²)γ = t'γ If x' frame moves in positive x direction with speed v, then the time interval ends when the x coordinate of the clock is zero. The clock moves at speed v in the x frame (along with x' frame). In the x frame, we have distance covered = speed multiplied by time taken, so: This I can agree. c\gamma = v\Delta t which gives the time interval in the x frame. i.e. \Delta t = \frac{c\gamma}{v} The corresponding time in the t' frame is for the event at x=0 is: \Delta t' = \frac{c\gamma^2}{v} This I can not agree. You forgot length contraction. -c/\gamma = v\Delta t chinglu12-18-10, 11:29 AMOn further evaluation, however, your question appears a little bit too textbook and given the extraordinary measures to explain mathematically you have done yourself a tremendous disservice by demanding only the math. It could be so simple as explaining your dissension in mere words. Just starting a post with saying "This is what my textbook says and this is what I don't understand." Rather than jump in with any incorrect mathematical conclusion and leaving others' bewildered in how you reached it. Assume start event origins same. What is distance in unprimed frame x' must travel to reach unprimed origin. What is distance in primed frame unprime origin must travel to reach x'. Then, simple (x' < 0) Δx' = ( 0 - x') = v Δt' Δx = ( 0 - x'/γ) = v Δt Δt'/Δt = γ Beer w/Straw12-18-10, 12:04 PMRead my post again, please. chinglu12-18-10, 12:50 PMRead my post again, please. Of course I will write your way. I am positive you know how to show time dilation for primed origin clock starting when origins same. Just set x=vt. Plug this into Lorenntz Transforms of t' = ( t - vx/c² )γ and get t'=t/γ. Likewise for unprimed frame clock at origin primed frame calculates x' = -vt', t = ( t' + vx'/c² )γ = t=t'/γ. Many then conclude reciprocal time dilation. Just look t=t'/γ in prime frame for unprimed origin clock and t'=t/γ in unprime frame for prime origin clock. Sure looks like reciprocal time dilation. But, these are not the same events for the moving clocks since they are at different locations when motion is complete. This is apples and oranges. So, in unprimed frame, it views a clock moving toward its origin whereas this same sequence in the primed frame looks like the unprime origin is move toward a coordinate that has a clock. If we now remember that when Lorenntz Transforms calculates time, it is always the case that both frame agree with both times. This mean for this situation, both frames agree the origin of the unprimed frame elapses t = t'/γ. But, also both frame agree the clock in primed frame elapses tγ = t'. There is no dispute for correctly defined agreed events. In short there is no such thing as reciprocal time dilation since it is a comparison of two completely different end events but same start event, common origins. So here is geometry. start event O' and O at same place. . end event x' and O origin at same place. x'<0 x'-------------------------O' x'/γ-----------------------O For O to reach x' in view of primed frame, it moves left, no length contraction longer distance, same v longer time on x' clock. For x'/γ to reach O, it moves right, length contraction, shorter distance, same v shorter time on O clock. So, this means the x' clock must elapse more time than the O origin clock from start of O and O' the same place to O and x' the same place. rpenner12-18-10, 02:39 PMIn other words, assume the origins of the x and x' frames are co-located at t=t'=0. i.e. Assume the clock doesn't move in primed frame, so it always has x'=-c, for c a positive constant. So at t'=t=0, the clock is at x'=-c. At this time, the clock's location in the x frame is: x=\gamma (x' + vt') = -c\gamma If x' frame moves in positive x direction with speed v, then the time interval ends when the x coordinate of the clock is zero. The clock moves at speed v in the x frame (along with x' frame). In the x frame, we have distance covered = speed multiplied by time taken, so: c\gamma = v\Delta t which gives the time interval in the x frame. i.e. \Delta t = \frac{c\gamma}{v} The corresponding time in the t' frame is for the event at x=0 is: \Delta t' = \frac{c\gamma^2}{v} (c \Delta t')^2 - (\Delta x')^2 = \left( \frac{c^4}{v^2 (c^2 - v^2)} \right) c^2 \gamma^2 but (c \Delta t)^2 - (\Delta x)^2 = \left(\frac{c^2}{v^2} - 1 \right) c^2 \gamma^2 So it looks like you are not talking about the same space-time interval throughout. That's because at x'=-c, and t'=0, t is not also 0. chinglu12-18-10, 04:28 PM(c \Delta t')^2 - (\Delta x')^2 = \left( \frac{c^4}{v^2 (c^2 - v^2)} \right) c^2 \gamma^2 but (c \Delta t)^2 - (\Delta x)^2 = \left(\frac{c^2}{v^2} - 1 \right) c^2 \gamma^2 Thanks, invariance of space-time interval gives alternative way for me to prove my case. (c \Delta t')^2 - (\Delta x')^2 = (c \Delta t)^2 - (\Delta x)^2 Assume x'=-x then \Delta x' = x/\gamma - x' \Delta x = x - x'/\gamma Since x'=-x then \Delta x' = x/\gamma - (-x) \Delta x = x - (-x/\gamma) So \Delta x = \Delta x' So, based on the invariance of the space-time interval (c \Delta t')^2 = (c \Delta t)^2 \Delta t' = \Delta t Beer w/Straw12-18-10, 05:06 PMThanks, invariance of space-time interval gives alternative way for me to prove my case. (c \Delta t')^2 - (\Delta x')^2 = (c \Delta t)^2 - (\Delta x)^2 Assume x'=-x then \Delta x' = x/\gamma - x' \Delta x = x - x'/\gamma Since x'=-x then \Delta x' = x/\gamma - (-x) \Delta x = x - (-x/\gamma) So \Delta x = \Delta x' So, based on the invariance of the space-time interval (c \Delta t')^2 = (c \Delta t)^2 You have to do better than the above. Saying 1=1 or two events are infact the same event measured in proper time is not an argument for anything. chinglu12-18-10, 05:21 PMYou have to do better than the above. Saying 1=1 or two events are infact the same event measured in proper time is not an argument for anything. I can not know what you mean. It is one event when x' and x are at same place time starts when origins same place. There is a space interval between them. Here is picture what unprime frame sees when origins same. x'/γ moves toward x coordinate. x'/γ-------------O-------------------x Here is picture what unprime frame sees when origins same. x/γ moves toward x' coordinate. x'----------------O'---------------x/γ Beer w/Straw12-18-10, 05:27 PMThen use an analogy. Or make a thought experiment. chinglu12-18-10, 06:35 PMThen use an analogy. Or make a thought experiment. I can not simplify it any further. It is one event when x' and x are at same place time starts when origins same place. There is a space interval between them. Here is picture what unprime frame sees when origins same. x'/γ moves toward x coordinate. x'/γ-------------O-------------------x Here is picture what unprime frame sees when origins same. x/γ moves toward x' coordinate. x'----------------O'---------------x/γ Tach12-18-10, 06:51 PMThanks, invariance of space-time interval gives alternative way for me to prove my case. (c \Delta t')^2 - (\Delta x')^2 = (c \Delta t)^2 - (\Delta x)^2 Assume x'=-x then Why would you assume the above? Physics is not numerology. What you have to assume is that you are measuring the time interval on a clock in one of the frames. The spatial separation is 0: \Delta x=0 So: (c \Delta t')^2 - (\Delta x')^2 = (c \Delta t)^2 -0 This means: \Delta t=\Delta t' \sqrt{1-(v'/c)^2} where v'=\frac{\Delta x'}{\Delta t'} I bet that you've been shown this before. I also bet that you will never understand it and that you will continue arguing to death. Beer w/Straw12-18-10, 06:54 PMI can not simplify it any further. No. I think you just don't want too for some reason. chinglu12-18-10, 07:12 PMWhy would you assume the above? Physics is not numerology. What you have to assume is that you are measuring the time interval on a clock in one of the frames. The spatial separation is 0: \Delta x=0 You are claiming the spatial separation between x' and x in the view of the unprimed frame is 0? How did you arrive at this conclusion? This is about an observer at x having the primed coordinate x' come toward it. The spatial separation between the two is x - x'/ γ when origins same. The start of the event is the origins being same. The end of the event is when x' and x are the same place. That is not a zero spatial separation. Why would you not then claim \Delta x'=0 in the view of the primed frame. Otherwise you make absolute claims. \Delta x=0 v\Delta t=\Delta x=0 So v=0 or \Delta t\ = 0 chinglu12-18-10, 07:13 PMNo. I think you just don't want too for some reason. What? Beer w/Straw12-18-10, 07:16 PMWhat? That's what everyone keeps asking you. Tach12-18-10, 07:35 PMYou are claiming the spatial separation between x' and x in the view of the unprimed frame is 0? No, this is not what I am claiming. You don't know the meaning of \Delta x. One day, in a few years, it will come to you. Give it time, you aren't ready yet. chinglu12-18-10, 07:42 PMNo, this is not what I am claiming. You don't know the meaning of \Delta x. One day, in a few years, it will come to you. Give it time, you aren't ready yet. Oh, I guess it is some deep thought. You have \Delta x=0. But, you also have \Delta x'<>0. How do you work this out given the relativity postulate? Also, \Delta x=0 mean no spatial separation between x' and x. How do you make this true? I would sure like to understand how the observer at x' moved to x in the unprimed frame and yet no distance was between them but time was between them. Tach12-18-10, 08:19 PMOh, I guess it is some deep thought. You have \Delta x=0. But, you also have \Delta x'<>0. How do you work this out given the relativity postulate? Simple, it has nothing to do with PoR. You can easily have \Delta x=0 while \Delta x'<>0. Also, \Delta x=0 mean no spatial separation See? You are starting to get it. It means that in the unprimed frame we have two events with pure temporal separation, like two clock ticks. between x' and x. Err, you are wrong again, \Delta x has NOTHING to do with the separation between x' and x. You must think some more, one day you'll get it. How do you make this true? I would sure like to understand how the observer at x' moved to x It didn't. chinglu12-18-10, 08:33 PMSimple, it has nothing to do with PoR. You can easily have \Delta x=0 while \Delta x'<>0. See? You are starting to get it. It means that in the unprimed frame we have two events with pure temporal separation, like two clock ticks. Err, you are wrong again, \Delta x has NOTHING to do with the separation between x' and x. You must think some more, one day you'll get it. It didn't. This not what lorentz transforms say. x' = ( x- vt )γ x' /γ = x - vt x' /γ - x = vt = Δx Sorry your thoughts are not the same as lorentz transforms. See? You are starting to get it. It means that in the unprimed frame we have two events with pure temporal separation, like two clock ticks. I must apply relativity postulate making your argument false. if Δx = 0 then Δx' = 0. Tach12-18-10, 08:39 PMThis not what lorentz transforms say. x' = ( x- vt )γ x' /γ = x - vt x' /γ - x = vt = Δx It is not my problem that you confuse physics with numerology. x' = ( x- vt )γ implies \Delta x' =\gamma ( \Delta x- v \Delta t ) I am quite sure others tried to teach you tha. Don't worry, in a few years, you'll get it. I must apply relativity postulate making your argument false. if Δx = 0 then Δx' = 0. One day you'll get it. It will take you many years but you are still young. chinglu12-18-10, 08:43 PMIt is not my problem that you confuse physics with numerology. x' = ( x- vt )γ implies \Delta x' =\gamma ( \Delta x- v \Delta t ) numerology Yes I am seeing your point. I have coordinate at x' and one at x. Can you explain \Delta x in unprimed frame and \Delta x' in primed frame. Many apologies to interrupt your view of self but x' and x are coordinates not deltas. Tach12-18-10, 08:47 PMnumerology Yes I am seeing your point. I have coordinate at x' and one at x. Can you explain \Delta x in unprimed frame and \Delta x' in primed frame. Because this is what basic calculus tells you. Do they teach calculus in China? Many apologies to interrupt your view of self but x' and x are coordinates not deltas. You are welcome. See what classes they teach in your province. Come back when you passed them. chinglu12-18-10, 08:51 PMBecause this is what basic calculus tells you. Do they teach calculus in China? You are welcome. See what classes they teach in your province. Come back when you passed them. Many sorries but I passed them. They are coordinate mappings not deltas. We learn to admit when we are wrong. That way we do not make same simple error over and over. Do you have this course? If so, you failed. Lorentz transforms map coordinates not deltas. Tach12-18-10, 08:57 PMMany sorries but I passed them. They are coordinate mappings not deltas. Then, you must have bribed your teachers. You are utterly hopeless. We learn to admit when we are wrong. That way we do not make same simple error over and over. Too bad that you are not applying this, you are making a fool of yourself everywhere on the physics forums. Lorentz transforms map coordinates not deltas. By basic differentiation, they also map coordinate differences. You say that you passed calculus? You lie. chinglu12-18-10, 09:04 PMThen, you must have bribed your teachers. You are utterly hopeless. Too bad that you are not applying this, you are making a fool of yourself everywhere on the physics forums. By basic differentiation, they also map coordinate differences. You say that you passed calculus? You lie. It seem you finally understand Lorentz map coordinates. Now you should admit you are wrong so as to not make that simple error again. You can do this with self as we are taught. And since we are dealing with coordintes, what is derivative of a coordinate? For example, x=0 is contant, origin. t' = ( t - xv/c² )γ = tγ. So moving clock at x' beat time expanded in view of rest frame. Tach12-18-10, 10:09 PMIt seem you finally understand Lorentz map coordinates. Now you should admit you are wrong so as to not make that simple error again. You can do this with self as we are taught. And since we are dealing with coordintes, what is derivative of a coordinate? You claimed that you took calculus. So, you lied. For example, x=0 is contant, origin. t' = ( t - xv/c² )γ = tγ. So moving clock at x' beat time expanded in view of rest frame. I am sure that you've been shown this before. Since you lied about taking calculus, I do not expect you to understand what it means. Here: \Delta t'=\gamma (\Delta t - \frac{v}{c^2} \Delta x) What happens if a clock at rest in the unprimed frame (\Delta x=0) is viewed from the primed frame? chinglu12-19-10, 03:27 PMYou claimed that you took calculus. So, you lied. I am sure that you've been shown this before. Since you lied about taking calculus, I do not expect you to understand what it means. Here: \Delta t'=\gamma (\Delta t - \frac{v}{c^2} \Delta x) What happens if a clock at rest in the unprimed frame (\Delta x=0) is viewed from the primed frame? You do not seem to understand, Lorentz transforms do not map intervals, they map coordinates. The interval is built in. x'=\gamma ( x - vt) x' and x are coordinates not deltas. you can call t a delta and that part is OK. Here is a basic fact. \Delta x = v\Delta t Now if have moving observer at x' and want to know what happens for it to move to x when all events begin when origins same, you calculate the distance between the two applying length contraction to x' since it is measured in the moving frame. \Delta x = x - x'/\gamma Substitute this result into \Delta x = v\Delta t x - x'/\gamma= v\Delta t x'/\gamma= x - v\Delta t x'= \gamma(x - v\Delta t) This is Lorentz transform. As you can understand the interval is built in from the coordinates x' and x. Now the point of this thread is the comparison of the time intervals between the frames for a moving clock at x'<0 to move to the stationary origin. All math thus far have proved the moving clock beats faster than the clock at the stationary origin. Tach12-19-10, 03:46 PMYou do not seem to understand, Lorentz transforms do not map intervals, they map coordinates. Find a different hobby, your trolling is getting boring. chinglu12-19-10, 06:19 PMFind a different hobby, your trolling is getting boring. You have the maths I typed. You did not address maths so I must conclude you give up. I can understand after reading your posts. Tach12-19-10, 06:32 PMYou do not seem to understand, Lorentz transforms do not map intervals, they map coordinates. The interval is built in. x'=\gamma ( x - vt) x' and x are coordinates not deltas. you can call t a delta and that part is OK. Here is a basic fact A basic fact is that you lied when you said you learned calculus. See the proof: x'=\gamma ( x - vt) means: x_1'=\gamma ( x_1 - vt_1) x_2'=\gamma ( x_2 - vt_2) Subtract and you get: x_2'-x_1'=\gamma ( (x_2 -x_1)- v(t_2-t_1)) You never took calculus. chinglu12-19-10, 06:58 PMA basic fact is that you lied when you said you learned calculus. See the proof: x'=\gamma ( x - vt) means: x_1'=\gamma ( x_1 - vt_1) x_2'=\gamma ( x_2 - vt_2) Subtract and you get: x_2'-x_1'=\gamma ( (x_2 -x_1)- v(t_2-t_1)) You never took calculus. You maths correct except should look like x_2'-x_1'=\gamma ( (x_2 -x_1)- v( \Delta t_2 - \Delta t_1 )) This is not Lorentz transform. Also, why not continue forward with this and prove unrestricted time dilation is false. Also, you set \Delta x' = 0 in prior post. Can you now understand how false that is. It implies exact motion forward and back, no motion at all or no time. What good is that? Next, you thought \Delta x' had something to do with coordinate transformations as this thread is doing. Can you now see all your posts were false? Sometimes it is good to write things down to see exactly what you mean and where you are wrong. So can you refute the conclusions of the OP with maths? That is point of this thread. Tach12-19-10, 07:33 PMyou maths correct except should look like x_2'-x_1'=\gamma ( (x_2 -x_1)- v( \Delta t_2 - \Delta t_1 )) lol this is not lorentz transform. lol. So can you refute the conclusions of the OP with maths? That is point of this thread. Yes, I already did. The problem is that you understand neither the "maths" nor the physics. And I can't fix that. James R12-19-10, 07:50 PM(c \Delta t')^2 - (\Delta x')^2 = \left( \frac{c^4}{v^2 (c^2 - v^2)} \right) c^2 \gamma^2 but (c \Delta t)^2 - (\Delta x)^2 = \left(\frac{c^2}{v^2} - 1 \right) c^2 \gamma^2 So it looks like you are not talking about the same space-time interval throughout. That's because at x'=-c, and t'=0, t is not also 0. Yeah. I was a bit hasty with that post. It's hard to work out what chinglu's scenario actually is. I might try again later if I have some spare time. chinglu12-19-10, 08:09 PMYes, I already did. The problem is that you understand neither the "maths" nor the physics. And I can't fix that. No, you set delta x = 0. I already proved this is no good and does nothing. Are you still wanting this? Otherwise you proved nothing as usual. chinglu12-19-10, 08:11 PMlol Can you show your equation in Einstein's paper? I can not find it. You claimed your equation is Lorentz transform. Tach12-19-10, 08:11 PMNo, you set delta x = 0. Yes, I can see that you still have trouble understanding this. Don't worry, you will NEVER understand it. Tach12-19-10, 08:13 PMYou claimed your equation is Lorentz transform. It is. It is also your problem that you can't recognize it. chinglu12-19-10, 08:16 PMYes, I can see that you still have trouble understanding this. Don't worry, you will NEVER understand it. Oh, so when delta x' = 0 and not motion resulted you have found some great idea? Why not give direct explanations instead of evasive. Readers know this mean you can not answer because you are wrong. chinglu12-19-10, 08:18 PMIt is. It is also your problem that you can't recognize it. If your equation is lorentz transform readers would like to see the link in Einstein's paper as you can find mine. That gives you no credability if you can not show that in Einstein's paper. Readers will think you crackpot without this link. Tach12-19-10, 08:19 PMOh, so when delta x' = 0 and not motion resulted you have found some great idea? \Delta x =0 means that we are measuring the tempotal interval on a clock at rest in the unprimed frame. You still don't understand this? Readers know this mean you can not answer because you are wrong. No, readers have figured out that you are retarded. chinglu12-19-10, 08:26 PM\Delta x =0 means that we are measuring the tempotal interval on a clock at rest in the unprimed frame. You still don't understand this? Given the relativity postulate, the lorentz transforms indicate the motion as seen by the stationary coordinates. There may be a temporal interval, but that must be associated with some \Delta x >0 motion. This elementary. Now, children know this. \Delta x = v\Delta t Since you have \Delta t>0, obviously you must have \Delta x>0 or \Delta x = v\Delta t=0. Simple mind can understand this. Tach12-19-10, 08:30 PMGiven the relativity postulate, the lorentz transforms indicate the motion as seen by the stationary coordinates. There may be a temporal interval, but that must be associated with some \Delta x >0 motion. No, it isn't. You are making up your own demented rules. Now, children know this. \Delta x = v\Delta t Nope, v is the relative speed between frames, it has nothing to do with \Delta x. You need to start studying and to stop spewing stupidities. chinglu12-19-10, 08:33 PMNo, it isn't. You are making up your own demented rules. Nope, v is the relative speed between frames, it has nothing to do with \Delta x. You need to start studying and to stop spewing stupidities. Oh, so \Delta x = v\Delta t is false for you? :roflmao: What was your reason again this false? v is the relative speed between frames, it has nothing to do with \Delta x Tach12-19-10, 08:36 PMOh, so \Delta x = v\Delta t is false for you? In the Lorentz transforms , v represents the speed between frames. You need to start learning and to stop spewing stupidities. chinglu12-19-10, 08:38 PMIn the Lorentz transforms , v represents the speed between frames. You need to start learning and to stop spewing stupidities. You did not answer question. Oh, so \Delta x = v\Delta t is false for you? Tach12-19-10, 08:42 PMYou did not answer question. Oh, so \Delta x = v\Delta t is false for you? It is false for anyone who understands physics. You are excluded . Forever. chinglu12-19-10, 09:10 PMOh, so \Delta x = v\Delta t is false for you? It is false for anyone who understands physics. You are excluded . Forever. Here is Einstein. Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt and http://www.fourmilab.ch/etexts/einstein/specrel/www/ Since time start when origins same, x coordinate is \Delta x = x - 0 Since t = 0 when origins same, \Delta t = t - 0 This means Einstein wrote \Delta x = v\Delta t. You are not very good at all. James R12-19-10, 09:45 PMOk. Let's try this again. Assume clock is located at x' < 0 in the primed coordinates. Assume standard configuration and time interval begins for both frame when origins same. Assume time interval ends when x' and unprimed origin same. What is this time interval in each frame. Let's assume the clock is initially located at x'=-k, with k a positive constant. The primed frame sees the x=0 origin moving in the negative direction with speed v. For x=0 to coincide with x'=-k, the x=0 origin must move distance k at speed v, which in the primed frame takes time \Delta t' = \frac{k}{v} In the unprimed frame, at t=0 the distance to the point x'=-k is contracted by the Lorentz factor. Therefore, in the unprimed frame: \Delta t = \frac{k}{\gamma v} I think this is what chinglu originally calculated. His answers are correct for this particular problem. So, chinglu, what I don't understand is why you believe this is a problem for relativity? Tach12-19-10, 09:45 PMSince time start when origins same, x coordinate is \Delta x = x - 0 Wrong. You claimed you learned calculus, you lied. Since t = 0 when origins same, \Delta t = t - 0 Wrong. You claimed you learned calculus, you lied. This means Einstein wrote \Delta x = v\Delta t. You are not very good at all. no , this only means you don't have a clue. You never will. Tach12-19-10, 09:49 PMOk. Let's try this again. Let's assume the clock is initially located at x'=-k, with k a positive constant. The primed frame sees the x=0 origin moving in the negative direction with speed v. For x=0 to coincide with x'=-k, the x=0 origin must move distance k at speed v, which in the primed frame takes time \Delta t' = \frac{k}{v} In the unprimed frame, at t=0 the distance to the point x'=-k is contracted by the Lorentz factor. Therefore, in the unprimed frame: \Delta t = \frac{k}{\gamma v} I think this is what chinglu originally calculated. His answers are correct for this particular problem. So, chinglu, what I don't understand is why you believe this is a problem for relativity? This is not what he calculated, you do not understand what he's doing. He is claiming that : t'=\gamma (t-\frac{vx}{c^2}) accepts a solution of the form t'=t (which is true but irrelevant) thus disproving time dilation (which is obviously false). As an alternative, he uses the other Lorentz transform: x'=\gamma(x-vt) and , by making x'=-x he concludes, just as incorrectly, that he has disproven time dilation. Chinglu does not even begin to understand the notion of \Delta as in \Delta x or \Delta t. He has lied about having taken calculus. James R12-19-10, 10:07 PMJust to add a little to my previous post, let's look at the Lorentz transformations. Note that we are considering THREE events here: 1. The origins of the primed and unprimed frames coincide. 2. The clock is located at position x'=-k in the primed frame at t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. In the primed frame, the spacetime coordinates of these three events are: 1. (x',t') = (0,0) 2. (x',t') = (-k,0) 3. (x',t') = (-k, k/v) Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame: 1. (x,t) = (0,0) 2. (x,t)=(-\gamma k, \frac{\gamma k v}{c^2}) 3. (x,t)=(0,\frac{k}{\gamma v}) Note that using events 1 and 3, the time intervals are as calculated in my previous post, which are also in agreement with chinglu's original post. But what is interesting is event 2. Note that events 1 and 2 are simultaneous in the primed frame, but NOT in the unprimed frame. The time interval between events 2 and 3 is k/v in the primed frame, but in the unprimed frame the time interval between those two events is: \Delta t = \frac{\gamma k}{v}\left(1-\frac{2v^2}{c^2}\right) James R12-19-10, 10:14 PMThis is not what he calculated, you do not understand what he's doing. I have reproduced the relevant parts of his post that I concentrated on in my posts #82 and #85. Do you assert that I have not applied his prescription correctly? I have no interest in what he has said or done since that post. I have merely concentrated on what he initially said. I do not particularly appreciate your telling me that I don't understand. If you wish to make such an assertion, then I suggest you point out exactly where I have gone wrong in posts #82 and #85. Tach12-19-10, 10:53 PMI have reproduced the relevant parts of his post that I concentrated on in my posts #82 and #85. Do you assert that I have not applied his prescription correctly? yes, you didn't I do not particularly appreciate your telling me that I don't understand. If you wish to make such an assertion, then I suggest you point out exactly where I have gone wrong in posts #82 and #85. i explained to you what is the mechanism of his idea. it is different from what you understand it to be. Tach12-19-10, 11:19 PMJust to add a little to my previous post, let's look at the Lorentz transformations. Note that we are considering THREE events here: 1. The origins of the primed and unprimed frames coincide. 2. The clock is located at position x'=-k in the primed frame at t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. In the primed frame, the spacetime coordinates of these three events are: 1. (x',t') = (0,0) 2. (x',t') = (-k,0) 3. (x',t') = (-k, k/v) you mean 3. (x',t') = (-k, -k/v). otherwise , you you couldn't possibly have 3. (x,t)=(0,\frac{k}{\gamma v}) The time interval between events 2 and 3 is k/v in the primed frame, but in the unprimed frame the time interval between those two events is: \Delta t = \frac{\gamma k}{v}\left(1-\frac{2v^2}{c^2}\right) This appears to be incorrect, you may want to check your calculations. James R12-19-10, 11:31 PMTach: I have reproduced the relevant parts of his post that I concentrated on in my posts #82 and #85. Do you assert that I have not applied his prescription correctly? yes, you didn't Then go back to posts #82 and #85 and point out exactly where I went wrong. i explained to you what is the mechanism of his idea. it is different from what you understand it to be. I'll wait for you to point out the errors in posts #82 and #85 before I discuss what his "mechanism" might be. you mean 3. (x',t') = (-k, -k/v). otherwise , you you couldn't possibly have 3. (x,t)=(0,\frac{k}{\gamma v}) No. I mean what I wrote. Why don't you apply the Lorentz transformation to the (x,t) coordinates given and tell me what you get for (x',t')? This appears to be incorrect, you may want to check your calculations. I think you need to check yours. Mine are correct, as far as I can tell. Tach12-19-10, 11:33 PMTach: Then go back to posts #82 and #85 and point out exactly where I went wrong. I'll wait for you to point out the errors in posts #82 and #85 before I discuss what his "mechanism" might be. No. I mean what I wrote. Why don't you apply the Lorentz transformation to the (x,t) coordinates given and tell me what you get for (x',t')? You don't write it but it is obvious that you are trying to use: t=\gamma(t'-vx'/c^2) In order to get what you wish to get you must have t'=-k/v which is consistent with your result 3. I think you need to check yours. Mine are correct, as far as I can tell. Well, you have mistakes. James R12-19-10, 11:35 PMTach: Yeah. You already made that claim. Assertions without evidence are worthless. Tach12-19-10, 11:38 PMTach: Yeah. You already made that claim. Assertions without evidence are worthless. I added the math. The second mistake is more serious, try showing your steps and you'll find your error. Events 2 and 3 have both a temporal and a spatial separation, you calculated only the temporal separation. James R12-19-10, 11:43 PMTach: You've made the error, not me. The correct Lorentz transformation is: t = \gamma (t' + vx'/c^2) You had the sign wrong. There's no error in the other result, either. Will you admit your mistake now, or will you keep insisting you are right, as I have observed you do on many occasions when you have been proved wrong? Let's see... Tach12-19-10, 11:49 PMTach: You've made the error, not me. The correct Lorentz transformation is: t = \gamma (t' + vx'/c^2) You had the sign wrong. If that is the case, then you got 2. (x,t)=(-\gamma k, \frac{\gamma k v}{c^2}) wrong, since according to you 2. (x',t') = (-k,0). It is difficult to know what you did since you don't show the steps. Either 2 or 3 is wrong. You decide. James R12-20-10, 12:03 AMTach: You're right in post #94. I did make a mistake. As for following the steps, all I did was to apply the Lorentz transformations. I didn't think it was necessary to include the algebra, since I assume everybody following this conversation knows the transformations and can do the algebra themselves. Are you willing to concede that you, like me, made a mistake with the sign in the transformation? James R12-20-10, 12:07 AMHere's a corrected repost of my post #82... -------- Let's look at the Lorentz transformations. Note that we are considering THREE events here: 1. The origins of the primed and unprimed frames coincide at t=t'=0. 2. The clock is located at position x'=-k in the primed frame at t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. In the primed frame, the spacetime coordinates of these three events are: 1. (x',t') = (0,0) 2. (x',t') = (-k,0) 3. (x',t') = (-k, k/v) Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame: 1. (x,t) = (0,0) 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) 3. (x,t)=(0,\frac{k}{\gamma v}) Note that using events 1 and 3, the time intervals are as calculated in my previous post, which are also in agreement with chinglu's original post. But what is interesting is event 2. Note that events 1 and 2 are simultaneous in the primed frame, but NOT in the unprimed frame. The time interval between events 2 and 3 is k/v in the primed frame, but in the unprimed frame the time interval between those two events is: \Delta t = \frac{\gamma k}{v} Note that this time interval between events 1 and 3 is SHORTER in the unprimed frame than it is in the primed frame, whereas the time interval between events 2 and 3 is LONGER in the unprimed frame than in the primed frame. Tach12-20-10, 12:09 AMTach: You're right in post #94. I did make a mistake. As for following the steps, all I did was to apply the Lorentz transformations. I didn't think it was necessary to include the algebra, since I assume everybody following this conversation knows the transformations and can do the algebra themselves. Are you willing to concede that you, like me, made a mistake with the sign in the transformation? I told you that you have EITHER event 2 or event 3 computed incorrectly. You need to learn how to own when you make a mistake and not try to shift the burden on the other person. I had no idea what transform you used , based on 2 I assumed that you used the minus sign. You claim that you used the plus sign, that makes your event 3 correct but your event 2 incorrect. Either way, one was incorrect. You have not acknowledged a much more serious mistake, in calculating \Delta t. Do you understand your mistake? Tach12-20-10, 12:13 AM. The time interval between events 2 and 3 is k/v in the primed frame, Yes. but in the unprimed frame the time interval between those two events is: \Delta t = \frac{\gamma k}{v} How did you get that? James R12-20-10, 12:14 AMI told you that you have EITHER 2 or 3 computed incorrectly. No you didn't. And please don't keep going back and editing your posts. You start to look dishonest when you do that. You need to learn how to own when you make a mistake and not try to shift the burden on the other person. I owned my mistake as soon as we had together worked out what it was. You have yet to own yours. In fact, I can't recall you ever owning any mistake you've made since you joined this forum. I had no idea what transform you used , based on 2 I assumed that you used the minus sign. You claim that you used the plus sign, that makes your 3 correct but your 2 incorrect. In post #82 I explicitly stated that I used the Lorentz transformations. What more is needed? Note: I was also explicit about which was the primed frame and which was the unprimed frame. You have not acknowledged a much more serious mistake, in calculating \Delta t. Do you understand your mistake? What mistake? Are you saying I've made another mistake in my corrected post above? James R12-20-10, 12:15 AMHow did you get that? I subtracted the time of event 2 from the time of event 3. Tach12-20-10, 12:17 AMWhat mistake? Are you saying I've made another mistake in my corrected post above? Yes, you still have it wrong. James R12-20-10, 12:17 AMThen show me the correct calculation, Tach, if you can do it. Tach12-20-10, 12:21 AMThen show me the correct calculation, Tach, if you can do it. Events 2 and 3 have both a temporal and a spatial separation. You can't use such events in order to demonstrate time dilation. In a few other posts, I showed how you demonstrate time dilation correctly. James R12-20-10, 12:27 AMEvents 2 and 3 have both a temporal and a spatial separation. In the unprimed frame, events 1 and 3 occur at the same spatial location. In the primed frame events 2 and 3 occur at the same spatial location. You can't use such events in order to demonstrate time dilation. Tell that to chinglu. I haven't attempted to use these events to demonstrate time dilation. In a few other posts, I showed how you demonstrate time dilation correctly. Well, thanks for your efforts, but I already know all about that. Tach12-20-10, 12:29 AMIn the unprimed frame, events 1 and 3 occur at the same spatial location. In the primed frame events 2 and 3 occur at the same spatial location. Tell that to chinglu. I haven't attempted to use these events to demonstrate time dilation. Then, the whole exercise is pointless. Well, thanks for your efforts, but I already know all about that. Well, I can show you how to do this correctly. It takes much fewer steps than your approach , making it immune to errors. Do you want to see? James R12-20-10, 12:36 AMIn fact, if we look at my post #96, we see a rather nice demonstration of time dilation. Consider the primed frame to be "stationary" and the unprimed frame to be "moving". (Note: this is the opposite of what is usual done in this kind of demonstration.) Now look at events 1 and 3. 1. The origins of the primed and unprimed frames coincide at t=t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. From the point of view of an observer in the primed frame, the unprimed (x) axis moves in the negative x direction with speed v. The time it takes to move distance k is k/v as measured by clocks in the primed frame. As post #96 shows, the time interval between these two events in the unprimed frame is SHORTER by the Lorentz factor. But in the unprimed frame these two events occur at the same spatial location. Hence, it is the unprimed frame here that measures the proper time. And we see that the proper time is shorter than the time in the other frame. In other words, this confirms that moving clocks run slow. Neat, eh? James R12-20-10, 12:38 AMThen, the whole exercise is pointless. It's not my fault that you can't see the point. I think I've rather neatly shown where chinglu went wrong - something you failed to do over the course of 80 posts or so. Well, I can show you how to do this correctly. It takes much fewer steps than your approach , making it immune to errors. Do you want to see? Sure. It would make a nice change for you to actually do something, rather than simply claim that other people are making mistakes. Tach12-20-10, 12:42 AMIt's not my fault that you can't see the point. I think I've rather neatly shown where chinglu went wrong - something you failed to do over the course of 80 posts or so. You can't prove chinglu wrong. No one can. You will not be able to do that, just wait for his answers. Sure. It would make a nice change for you to actually do something, rather than simply claim that other people are making mistakes. Well, you made quite a few mistakes in your post 85 (http://www.sciforums.com/showpost.php?p=2665368&postcount=85) and I showed you how to correct them. I think that this is pretty constructive. James R12-20-10, 12:53 AMTach: Let's see. I made a mistake in post #85. I admitted the mistake in post #95 and posted a revised version of post #85 in post #96. Between posts #86 and #94, you repeatedly told me I have made a mistake, although you were initially unsure where I had gone wrong. In fact, in post #90, you made your own mistake by getting the sign wrong in the Lorentz transformation. Nowhere between #86 and #94 do I see you "showing me" how to correct my mistake. All you did was to assert repeatedly that there was a mistake there somewhere. I found what it was myself. You still haven't admitted your own mistake in #90. Will you now do that? Tach12-20-10, 12:56 AMTach: Let's see. I made a mistake in post #85. I admitted the mistake in post #95 and posted a revised version of post #85 in post #96. Between posts #86 and #94, you repeatedly told me I have made a mistake, although you were initially unsure where I had gone wrong. In fact, in post #90, you made your own mistake by getting the sign wrong in the Lorentz transformation. Nowhere between #86 and #94 do I see you "showing me" how to correct my mistake. All you did was to assert repeatedly that there was a mistake there somewhere. I found what it was myself. You made two mistakes, you got either 2 or 3 wrong and you also got \Delta t wrong. You still haven't admitted your own mistake in #90. Will you now do that? I guessed you inverse Lorentz transform wrong. If you use one sign, then your 2 is wrong, if you use the other sign, then your 3 is wrong. Eiither way, one of the events is wrong, you take your pick. So is your \Delta t. Next time I will not try to guess your mistakes, I will ask you outright to show your steps. James R12-20-10, 12:56 AMAlso, Tach, it turns out you were also wrong in post #103, where you claimed that events 2 and 3 cannot be used to demonstrate time dilation. Since those two events both occur at the same spatial location in the primed frame, the time interval between them is a proper time in the primed frame. Therefore, they can be used to demonstrate time dilation. Do you agree? For comparison, see post #106. I'd say I've given quite a nice demonstration of time dilation here. Wouldn't you? James R12-20-10, 12:58 AMYou made two mistakes, you got either 2 or 3 wrong and you also got \Delta t wrong. Well, a single sign error in the time of event 2 led to a consequential error in the calculation of \Delta t = t_3 - t_2. You can call that 2 mistakes if you like, but if I was marking an exam and a student did that I wouldn't be penalising them twice. Tach12-20-10, 01:01 AMWell, a single sign error in the time of event 2 led to a consequential error in the calculation of \Delta t = t_3 - t_2. You can call that 2 mistakes if you like, but if I was marking an exam and a student did that I wouldn't be penalising them twice. Well, considering that you are a moderator that should be held to higher ethical and scientific standards and that you were so aggressive towards me in denying any wrongs, I would . James R12-20-10, 01:04 AMYou missed a couple of questions: You still haven't admitted your own mistake in #90. Will you now do that? ... Also, Tach, it turns out you were also wrong in post #103.... Do you agree? ... I'd say I've given quite a nice demonstration of time dilation here. Wouldn't you? Tach12-20-10, 01:04 AMIn fact, if we look at my post #96, we see a rather nice demonstration of time dilation. Consider the primed frame to be "stationary" and the unprimed frame to be "moving". (Note: this is the opposite of what is usual done in this kind of demonstration.) This is a bad start, there is no such thing as "stationary" frame in SR. Now look at events 1 and 3. 1. The origins of the primed and unprimed frames coincide at t=t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. From the point of view of an observer in the primed frame, the unprimed (x) axis moves in the negative x direction with speed v. The time it takes to move distance k is k/v as measured by clocks in the primed frame. This is hard to follow, it needs the math that goes with it. As post #96 shows, the time interval between these two events in the unprimed frame is SHORTER by the Lorentz factor. While correct, this is a bad conclusion, you want to demonstrate time dilation , not time contraction. You don't want to start justifying why the contraction is actually a dilation. But in the unprimed frame these two events occur at the same spatial location. Hence, it is the unprimed frame here that measures the proper time. And we see that the proper time is shorter than the time in the other frame. In other words, this confirms that moving clocks run slow. Neat, eh? Not very, try putting all this in a mathematical formulation, let's see how it looks. James R12-20-10, 01:08 AMTach: Do you need me to join the dots for you between post #96 and post #106? And between #96 and #111? I'll do it if you need me to. James R12-20-10, 01:27 AMOk. Here we go... We have a primed frame with event coordinates (x',t'), and an unprimed frame with coordinates (x,t). In the unprimed coordinates, the primed frame moves in the positive x direction with speed v. (Note: in what follows, v is always taken to be a positive constant.) In the primed coordinates, the unprimed frame moves in the negative x' direction with speed v. Let's look at the Lorentz transformations. Note that we are considering THREE events here: 1. The origins of the primed and unprimed frames coincide at t=t'=0. 2. Some irrelevant object (e.g. a clock) is initially located at position x'=-k in the primed frame at t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. In the primed frame, the spacetime coordinates of these three events are: 1. (x',t') = (0,0) 2. (x',t') = (-k,0) 3. (x',t') = (-k, k/v) The coordinates of event 3 follow from the fact that in the primed frame the unprimed x axis moves a distance k at speed v in time k/v, where the unprimed axis moves in the negative x' direction in the primed frame. Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame as follows: x = \gamma (x' + vt'); t = \gamma (t' + vx'/c^2); \gamma = \frac{1}{\sqrt{1-(v/c)^2} 1. (x,t) = (0,0) 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) 3. (x,t)=(0,\frac{k}{\gamma v}) In the unprimed frame the time intervals between pairs of events (calculated by subtracting the time coordinate of one event from the other) are: 1 and 3: \Delta t = \frac{k}{\gamma v} 2 and 3: \Delta t = \frac{\gamma k}{v} In the primed frame, the corresponding intervals are: 1 and 3: \Delta t' = \frac{k}{v} 2 and 3: \Delta t' = \frac{k}{v} Now, in the unprimed frame, the spatial coordinates of events 1 and 3 are the same, whereas in the primed frame they are different. So, the time measured between 1 and 3 in the unprimed frame is a proper time. Note that the time interval between events 1 and 3 in the primed frame is LONGER than in the unprimed frame by a factor of \gamma. Relativity tells us that the proper time is always the SHORTEST time interval between two events, so this is the result we expect. Similarly, in the primed frame, the spatial coordinates of events 2 and 3 are the same, so the time interval between those events in the primed frame is a proper time. Comparing the time interval between events 2 and 3 in the unprimed frame, we again see that it is LONGER by a factor of \gamma. So, again, we see that the proper time is the shorter time of the two. All of this is consistent with the relativistic result often expressed sloppily as "moving clocks run slow". Considering events 1 and 3, a clock has to move in the primed frame to record both events at different locations, so for those events the primed frame is "moving", while the unprimed frame is "stationary". Considering events 2 and 3, a clock has to move in the unprimed frame to record both events at different locations, so for those events the unprimed frame is "moving", while the primed frame is "stationary". The mathematics shows that in both cases the "moving" clocks run slow - just as Einstein said. James R12-20-10, 02:13 AMI should also add that the above post shows what's wrong with chinglu's picture from post #6 of this thread. This picture. |---> v ---------x'---------|------------------ | Origins of frames common here Interval starts when origins are common. Interval ends when x' and unprime origin common. Unprime, apply length contraction dx = ( 0 - x'/γ ) dx = vdt -x'/γ = vdt dt = (-x'/γ)/v prime The unprime origin moves to x' coordinate in this frame. dx' = -x' dx' = vdt' = -x' dt' = -x'/v. dt = dt'/γ or dtγ = dt'. For x' to move to unprime origin prime clock beats faster than origin clock. This my calculation. In this post, chinglu is comparing what I have called events 1 and 3. He concludes, correctly, that the time interval between events 1 and 3 in the primed frame is longer than between those two events in the unprimed frame. Since these kinds of problems mostly start with a proper time in the primed frame, the usual conclusion is that the primed clocks ran "slow" compared to the unprimed clocks. However, chinglu has managed to choose events such that the proper time is actually measured in the unprimed frame, in which case we expect the unprimed frame to run "slow". As you can see clearly from my calculations, there is no problem for relativity here. The results are entirely consistent with the special theory of relativity. The lesson chinglu needs to take away from this is to be careful about what is and is not a proper time interval. I'll be happy to accept the thanks of chinglu and Tach for producing such a comprehensive and clear explanation and solving the question of the thread. Let's see if either of those two posters have any sense of manners and good grace. chinglu12-20-10, 08:51 PMOk. Let's try this again. Let's assume the clock is initially located at x'=-k, with k a positive constant. The primed frame sees the x=0 origin moving in the negative direction with speed v. For x=0 to coincide with x'=-k, the x=0 origin must move distance k at speed v, which in the primed frame takes time \Delta t' = \frac{k}{v} In the unprimed frame, at t=0 the distance to the point x'=-k is contracted by the Lorentz factor. Therefore, in the unprimed frame: \Delta t = \frac{k}{\gamma v} I think this is what chinglu originally calculated. His answers are correct for this particular problem. So, chinglu, what I don't understand is why you believe this is a problem for relativity? Because the time interval from start to end is longer for the primed frame interval for the motion over the unprimed. This mean moving clock beat time expanded. chinglu12-20-10, 08:55 PMJust to add a little to my previous post, let's look at the Lorentz transformations. Note that we are considering THREE events here: 1. The origins of the primed and unprimed frames coincide. 2. The clock is located at position x'=-k in the primed frame at t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. Only 2 events. (2.) is a consequence of (1.) It is not an event. The rest of post is error. chinglu12-20-10, 08:57 PMThis is not what he calculated, you do not understand what he's doing. He is claiming that : t'=\gamma (t-\frac{vx}{c^2}) accepts a solution of the form t'=t (which is true but irrelevant) thus disproving time dilation (which is obviously false). As an alternative, he uses the other Lorentz transform: x'=\gamma(x-vt) and , by making x'=-x he concludes, just as incorrectly, that he has disproven time dilation. Chinglu does not even begin to understand the notion of \Delta as in \Delta x or \Delta t. He has lied about having taken calculus. I take this one step at time. I am not doing that yet. As a recovering Δx = 0 addict, should you first come clean that you are wrong that this describes the motion of x' moving to the origin of the rest frame? chinglu12-20-10, 09:04 PMEvents 2 and 3 have both a temporal and a spatial separation. You can't use such events in order to demonstrate time dilation. In a few other posts, I showed how you demonstrate time dilation correctly. You are wrong. You cannot prove time dilation without the corresponding motion that describes the event based on the relativity postulate. Each frame is watching some event of motion and each frame is recording the time required for that motion to beginning to end. That is how time dilation is calculated. It is the ratio of these time intervals for two agreed upon events of beginning to end. chinglu12-20-10, 09:12 PMI should also add that the above post shows what's wrong with chinglu's picture from post #6 of this thread. In this post, chinglu is comparing what I have called events 1 and 3. He concludes, correctly, that the time interval between events 1 and 3 in the primed frame is longer than between those two events in the unprimed frame. Since these kinds of problems mostly start with a proper time in the primed frame, the usual conclusion is that the primed clocks ran "slow" compared to the unprimed clocks. However, chinglu has managed to choose events such that the proper time is actually measured in the unprimed frame, in which case we expect the unprimed frame to run "slow". As you can see clearly from my calculations, there is no problem for relativity here. The results are entirely consistent with the special theory of relativity. The lesson chinglu needs to take away from this is to be careful about what is and is not a proper time interval. I'll be happy to accept the thanks of chinglu and Tach for producing such a comprehensive and clear explanation and solving the question of the thread. Let's see if either of those two posters have any sense of manners and good grace. Very good. Now, accept what you have figured out. I reversed the normal time dilation logic. I took the primed frame as the origin and stationary with x' = -vt'. This proves t'/γ = t. So, the primed origin claims time dilation as expected. This is where you are. Now must continue with all thoughts. In the view of the unprimed frame, x' moved to the unprimed origin. It was length contracted. So t' = tγ. When you now take the unprimed frame as stationary, the unprimed origin concludes t' is time expanded. That is the problem. The moving x' beat faster than the unprimed origin clock. James R12-20-10, 09:27 PMchinglu: Because the time interval from start to end is longer for the primed frame interval for the motion over the unprimed. I have clearly shown in post #117 why this is not a problem for relativity. Only 2 events. (2.) is a consequence of (1.) It is not an event. The rest of post is error. No. My analysis is correct. A spacetime event is anything that takes place at a specific location in space at a specific time. Your scenario involves three separate events, as I have listed in post #117. Very good. Now, accept what you have figured out. I reversed the normal time dilation logic. I took the primed frame as the origin and stationary with x' = -vt'. This proves t'/γ = t. So, the primed origin claims time dilation as expected. This is where you are. Now must continue with all thoughts. In the view of the unprimed frame, x' moved to the unprimed origin. It was length contracted. So t' = tγ. When you now take the unprimed frame as stationary, the unprimed origin concludes t' is time expanded. That is the problem. The moving x' beat faster than the unprimed origin clock. I have pointed out your error with the time intervals in post #117, above. Here, you are making the same kind of error with length contraction. Note that a proper length is a length measured between two events that occur at the same time in a particular reference frame. For events 1, 2 and 3, can you work out which pairs produce a proper length in the two reference frames? chinglu12-20-10, 09:28 PMThis is not what he calculated, you do not understand what he's doing. He is claiming that : t'=\gamma (t-\frac{vx}{c^2}) accepts a solution of the form t'=t (which is true but irrelevant) thus disproving time dilation (which is obviously false). As an alternative, he uses the other Lorentz transform: x'=\gamma(x-vt) and , by making x'=-x he concludes, just as incorrectly, that he has disproven time dilation. Chinglu does not even begin to understand the notion of \Delta as in \Delta x or \Delta t. He has lied about having taken calculus. What do you think t'=t mean? I would not answer, but you think you smart. Tach12-20-10, 09:29 PMWhat do you think t'=t mean? I would not answer, but you think you smart. It is your idea of refuting time dilation. Perfectly incorrect. chinglu12-20-10, 09:43 PMOk. Here we go... We have a primed frame with event coordinates (x',t'), and an unprimed frame with coordinates (x,t). In the unprimed coordinates, the primed frame moves in the positive x direction with speed v. (Note: in what follows, v is always taken to be a positive constant.) In the primed coordinates, the unprimed frame moves in the negative x' direction with speed v. Let's look at the Lorentz transformations. Note that we are considering THREE events here: 1. The origins of the primed and unprimed frames coincide at t=t'=0. 2. Some irrelevant object (e.g. a clock) is initially located at position x'=-k in the primed frame at t'=0. 3. The coordinate x'=-k coincides with the coordinate x=0 at some later time. In the primed frame, the spacetime coordinates of these three events are: 1. (x',t') = (0,0) 2. (x',t') = (-k,0) 3. (x',t') = (-k, k/v) The coordinates of event 3 follow from the fact that in the primed frame the unprimed x axis moves a distance k at speed v in time k/v, where the unprimed axis moves in the negative x' direction in the primed frame. Using the Lorentz transformations, we determine the equivalent coordinates in the unprimed frame as follows: x = \gamma (x' + vt'); t = \gamma (t' + vx'/c^2); \gamma = \frac{1}{\sqrt{1-(v/c)^2} 1. (x,t) = (0,0) 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) 3. (x,t)=(0,\frac{k}{\gamma v}) In the unprimed frame the time intervals between pairs of events (calculated by subtracting the time coordinate of one event from the other) are: 1 and 3: \Delta t = \frac{k}{\gamma v} 2 and 3: \Delta t = \frac{\gamma k}{v} In the primed frame, the corresponding intervals are: 1 and 3: \Delta t' = \frac{k}{v} 2 and 3: \Delta t' = \frac{k}{v} Now, in the unprimed frame, the spatial coordinates of events 1 and 3 are the same, whereas in the primed frame they are different. So, the time measured between 1 and 3 in the unprimed frame is a proper time. Note that the time interval between events 1 and 3 in the primed frame is LONGER than in the unprimed frame by a factor of \gamma. Relativity tells us that the proper time is always the SHORTEST time interval between two events, so this is the result we expect. Similarly, in the primed frame, the spatial coordinates of events 2 and 3 are the same, so the time interval between those events in the primed frame is a proper time. Comparing the time interval between events 2 and 3 in the unprimed frame, we again see that it is LONGER by a factor of \gamma. So, again, we see that the proper time is the shorter time of the two. All of this is consistent with the relativistic result often expressed sloppily as "moving clocks run slow". Considering events 1 and 3, a clock has to move in the primed frame to record both events at different locations, so for those events the primed frame is "moving", while the unprimed frame is "stationary". Considering events 2 and 3, a clock has to move in the unprimed frame to record both events at different locations, so for those events the unprimed frame is "moving", while the primed frame is "stationary". The mathematics shows that in both cases the "moving" clocks run slow - just as Einstein said. This is your error. 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) You have put a negative elapsed time on clock at x' from view of unprimed. So, unprimed frame views origin of primed frame as elapsed 0 and x' as elapsed -\frac{\gamma k v}{c^2} These are the times on the clocks so to speak. Who cares. We are not concerned with the times on the clock but the elapsed times since the origins were same. So what is clock had x. Who cares. What if clock had y. Who cares. You get this yet? Once the start of events has commenced, times on clock at that instant is irrelevant. It is the elapsed times on clocks after that instant. This is a common error. So your step 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) has nothing to do with this problem. chinglu12-20-10, 09:44 PMIt is your idea of refuting time dilation. Perfectly incorrect. So explain the meaning. James R12-20-10, 10:09 PMThis is your error. 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) You have put a negative elapsed time on clock at x' from view of unprimed. Two spatially-separated events in spacetime cannot be simultaneous in more than one reference frame. Any two spatially-separated events that are simultaneous in one reference frame will not be simultaneous in the other. In this case, you have set up a problem where the events I have labelled 1 and 2 occur simultaneously in the primed frame. Therefore, they do not occur simultaneously in the unprimed frame. In fact, in the unprimed frame, event 2 occurs before event 1. So, unprimed frame views origin of primed frame as elapsed 0 and x' as elapsed -\frac{\gamma k v}{c^2} Correct. These are the times on the clocks so to speak. Who cares. We are not concerned with the times on the clock but the elapsed times since the origins were same. These are not times on a clock. They are the actual times at which those events occur in the two reference frames. If you like, you could put a clock at the locations of the two events. One of those clocks would be moving with the primed frame, and the other with the unprimed frame. They would read the values I have shown in post #117, provided they were synchronised with clocks at x=0 and x'=0 in their respective rest frames. So what is clock had x. Who cares. What if clock had y. Who cares. You get this yet? Sure I get it. Do you? I've explained it very thoroughly to you. Once the start of events has commenced, times on clock at that instant is irrelevant. It is the elapsed times on clocks after that instant. This is a common error. I have calculated the elapsed times by subtracting the final time from the initial time in each case. So, you're right - the particular reading on the clock at the start of the interval is irrelevant. Only the elapsed time will matter in the end. I have calculated the elapsed time in each frame assuming that the initial values at x=0 and x'=0 were set to zero at t=t'=0. If you want to set those initial values to some other constant, go right ahead. The elapsed time calculation still gives the same answer that I calculated. You get this yet? So your step 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) has nothing to do with this problem. It's a simple application of the Lorentz transformation, as I wrote in post #117. Do you understand the Lorentz tranformation? Do you believe in the Lorentz transformation? If not, then what is your argument? Are you claiming that relativity is wrong, or that it is inconsistent? I have shown that it is not inconsistent. You get this yet? chinglu12-20-10, 10:23 PMTwo spatially-separated events in spacetime cannot be simultaneous in more than one reference frame. Any two spatially-separated events that are simultaneous in one reference frame will not be simultaneous in the other. This has nothing to do with problem. You are quoting typical quotes. None has claimed two spatially-separated events in spacetime are simultaneous. Let's agree the primed clocks are not synchronized. So, one has time t'1 and the other t'2 on their clocks. That is where you get error. We are concerned with elapsed times not initial times. You are claiming initial times on clocks affects the outcomes of events of the universe. If that is true, then if I am late for a train, I simply adjust the clock at the station to whatever I want and I am never late. See how initial clock values do not affect elapsed times in the universe? I am thinking this will take several pages. In short, to prove your theory, you are adjusting the clock at the train station to whatever you like and then claim you are not late for the train. Special relativity is about time intervals and elapsed times, not intial times. That is your error. Tach12-20-10, 10:52 PMSo explain the meaning. To you - not possible. James R12-20-10, 11:16 PMchinglu: This has nothing to do with problem. You are quoting typical quotes. I'm telling you what relativity says. Do you think I've made a mistake in what relativity says, or do you think relativity is wrong? None has claimed two spatially-separated events in spacetime are simultaneous. Good. So you agree that spatially-separated clocks, such as at position -x and 0 in the primed frame, if they are synchronised in the primed frame they cannot be synchronised in the unprimed frame. Correct? Let's agree the primed clocks are not synchronized. So, one has time t'1 and the other t'2 on their clocks. No, we can't assume that. The definition of a reference frame includes the fact that all clocks in that frame are synchronised with each other. If you have two clocks in a frame that are not synchronised in that frame, then either (a) you have a faulty clock or (b) you're not using the standard definition of a reference frame. We are concerned with elapsed times not initial times. You are claiming initial times on clocks affects the outcomes of events of the universe. No. Read my previous post where I explained this exact point to you. Here's the relevant part again: I have calculated the elapsed times by subtracting the final time from the initial time in each case. So, you're right - the particular reading on the clock at the start of the interval is irrelevant. Only the elapsed time will matter in the end. I have calculated the elapsed time in each frame assuming that the initial values at x=0 and x'=0 were set to zero at t=t'=0. If you want to set those initial values to some other constant, go right ahead. The elapsed time calculation still gives the same answer that I calculated. You get this yet? See? If that is true, then if I am late for a train, I simply adjust the clock at the station to whatever I want and I am never late. See how initial clock values do not affect elapsed times in the universe? I am thinking this will take several pages. You're right. So, we agree. In short, to prove your theory, you are adjusting the clock at the train station to whatever you like and then claim you are not late for the train. No. Re-read post #117 and tell me exactly where you think I made a mistake. Go through it line by line if you have to. Because it seems to me that either you didn't read that post, or you didn't understand it. --- On a side-note, I must comment that you are very ill-mannered. I have spent a considerable amount of time showing you exactly where you went wrong. And either you haven't read posts #117 and #118 or you are ignoring what I showed you there. You can't pretend that I haven't shown you your mistake. It is quite clearly set out in post #118. If you wish to dispute this matter further, you will need to go through posts #117 and #118 and show where I have made a mistake. If you will not argue your case in good faith and with honesty and integrity, as any scholar would, then I don't think I will waste any more time on you. Are you willing to look at posts #117 and #118 honestly? We'll see. You should be thanking me for helping you, not telling me that I have made mistakes. I am educating you, and for free! On the other hand, if you can't understand posts #117 and #118, please ask me questions and I will do my best to answer them for you. I understand that relativity can be difficult for beginners such as yourself. chinglu12-21-10, 06:45 PMchinglu: I'm telling you what relativity says. Do you think I've made a mistake in what relativity says, or do you think relativity is wrong? Good. So you agree that spatially-separated clocks, such as at position -x and 0 in the primed frame, if they are synchronised in the primed frame they cannot be synchronised in the unprimed frame. Correct? No, we can't assume that. The definition of a reference frame includes the fact that all clocks in that frame are synchronised with each other. If you have two clocks in a frame that are not synchronised in that frame, then either (a) you have a faulty clock or (b) you're not using the standard definition of a reference frame. No. Read my previous post where I explained this exact point to you. Here's the relevant part again: See? You're right. So, we agree. No. Re-read post #117 and tell me exactly where you think I made a mistake. Go through it line by line if you have to. Because it seems to me that either you didn't read that post, or you didn't understand it. --- On a side-note, I must comment that you are very ill-mannered. I have spent a considerable amount of time showing you exactly where you went wrong. And either you haven't read posts #117 and #118 or you are ignoring what I showed you there. You can't pretend that I haven't shown you your mistake. It is quite clearly set out in post #118. If you wish to dispute this matter further, you will need to go through posts #117 and #118 and show where I have made a mistake. If you will not argue your case in good faith and with honesty and integrity, as any scholar would, then I don't think I will waste any more time on you. Are you willing to look at posts #117 and #118 honestly? We'll see. You should be thanking me for helping you, not telling me that I have made mistakes. I am educating you, and for free! On the other hand, if you can't understand posts #117 and #118, please ask me questions and I will do my best to answer them for you. I understand that relativity can be difficult for beginners such as yourself. I understand that relativity can be difficult for beginners such as yourself Yea. You are doing way to much typing. I told you what I am doing an you just keep at it because it went over you. I will now reverse it into a form you know. O. O' v--->---------------------x=vt Have you seen this before. It is the proof from Einstein for time dilation. In the O frame, you can plug in x=vt and calculate t' = t/γ. Everyone know this. So, in absolute sense one conclude moving clock time dilated. But, what does O' calculate? You are trying to claim an adjustment factor must be used for the x coordinate same way you think falsely it must be used for x' coordinate. There is no correction to the clock. It is only elapsed times since origins same. So, what does O' calculate? t' = x/(vγ). Since t = x/v. As can see t' = t/γ. So, O' frame as stationary thinks moving x clock beats slower exactly as I have typing. Look how both frame agree t' = t/γ. That is the way it works under relativity. Here is something to help you understand relativity better. When you calculate t in frame and use that to calculate t', if you did your calculations correctly, then using t' in Lorentz Transforms will result in t. If not, then you failed. You can check your work on your own and find your own failure. James R12-21-10, 07:04 PMchinglu: I think we're in agreement. Both frames see clocks in the other frame as running slow. Correct? Ok, are we done? I assume we are finished, since you agree with me and you agree with what I wrote in posts #117 and #118. Thankyou for the conversation. I have had fun educating you. chinglu12-21-10, 07:16 PMchinglu: I think we're in agreement. Both frames see clocks in the other frame as running slow. Correct? Ok, are we done? I assume we are finished, since you agree with me and you agree with what I wrote in posts #117 and #118. Thankyou for the conversation. I have had fun educating you. No, typing mistake on my part. t' = x/(vγ). Since t = x/v. As can see t' = t/γ. So, O' frame as stationary thinks moving x clock beats faster exactly as I have typing. As we can see, in the primed frame for x moving, t' = t/γ. Note how t is faster for the moving x. chinglu12-21-10, 07:19 PMchinglu: I think we're in agreement. Both frames see clocks in the other frame as running slow. Correct? Ok, are we done? I assume we are finished, since you agree with me and you agree with what I wrote in posts #117 and #118. Thankyou for the conversation. I have had fun educating you. Also, Lets assume you correct. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) Let's see you translation using Lorentz assuming light emits from origins when same. Make sure you use your theory above. James R12-21-10, 08:07 PMchinglu: t' = x/(vγ). Since t = x/v. As can see t' = t/γ. So, O' frame as stationary thinks moving x clock beats faster exactly as I have typing. As we can see, in the primed frame for x moving, t' = t/γ. Note how t is faster for the moving x. Yes. I explained this observation in post #117. It is important to correctly identify which time interval is a proper time and which is not. See? You still haven't pointed out any error in posts #117 or #118. Also, Lets assume you correct. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) Let's see you translation using Lorentz assuming light emits from origins when same. Make sure you use your theory above. I'm happy to do this, but I don't understand exactly what it is that you're asking me to do. Can you please be specific? Light emits from the origin when? Towards what? Are you asking me to calculate the flight time for a light pulse in each frame? Once you say what you want me to do, I'll do it. You really should point out any mistake in posts #117 and #118 first, though. rpenner12-21-10, 08:14 PM(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}). x' = \gamma (x - vt) = \gamma \left( -\gamma k - v (- \frac{\gamma k v}{c^2}) \right) = - \gamma^2 k \left( 1 - \frac{v^2}{c^2} \right) = -k t' = \gamma (t - \frac{v}{c^2} x) = \gamma \left( -\frac{\gamma k v}{c^2} - \frac{v}{c^2} (-\gamma k) \right) = - \gamma^2 k \left( \frac{v}{c^2} - \frac{v}{c^2} \right) = 0 So saying (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) is the same thing as saying (x', t') = ( -k, 0) -- (x,t) and (x', t') are two different names for the same point in space and time, just like (r,θ) and (x,y) are different names for the same point on the Euclidean plane. x = r \cos \theta\\y = r \sin \theta This is JamesR's event 2 from post 117. When k = 5/6 light-seconds this is event P_{t'=0} in post 18 (http://sciforums.com/showpost.php?p=2663466&postcount=18). chinglu12-21-10, 08:37 PMx' = \gamma (x - vt) = \gamma \left( -\gamma k - v (- \frac{\gamma k v}{c^2}) \right) = - \gamma^2 k \left( 1 - \frac{v^2}{c^2} \right) = -k t' = \gamma (t - \frac{v}{c^2} x) = \gamma \left( -\frac{\gamma k v}{c^2} - \frac{v}{c^2} (-\gamma k) \right) = - \gamma^2 k \left( \frac{v}{c^2} - \frac{v}{c^2} \right) = 0 So saying (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) is the same thing as saying (x', t') = ( -k, 0) -- (x,t) and (x', t') are two different names for the same point in space and time, just like (r,θ) and (x,y) are different names for the same point on the Euclidean plane. x = r \cos \theta\\y = r \sin \theta This is JamesR's event 2 from post 117. When k = 5/6 light-seconds this is event P_{t'=0} in post 18 (http://sciforums.com/showpost.php?p=2663466&postcount=18). I thought you were resting peacefully in old folks home. Apply this argument to Einstein's proof of time dilation. chinglu12-21-10, 08:41 PMchinglu: Yes. I explained this observation in post #117. It is important to correctly identify which time interval is a proper time and which is not. See? You still haven't pointed out any error in posts #117 or #118. I'm happy to do this, but I don't understand exactly what it is that you're asking me to do. Can you please be specific? Light emits from the origin when? Towards what? Are you asking me to calculate the flight time for a light pulse in each frame? Once you say what you want me to do, I'll do it. You really should point out any mistake in posts #117 and #118 first, though. First, you need to apply your argument to Einstein's proof of time dilation You will then find the coordinate moving toward the origin is time expanded. Next, assume a light pulse is emitted from origins. That is a timing device. Prove time dilation negative x-axis point and positive x-axis point. Let me know when you have failed. Remember, you told the readers you are the teacher. brucep12-21-10, 08:47 PMAnd you're the uneducable fool. chinglu12-21-10, 08:58 PMAnd you're the uneducable fool. So, you can prove time dilation for light pulse emitted from origins when they are at same place. Prove time dilation for positive x-axis points and negative x-axis points of light pulse for any x and -x. Do not forget, time dilation is an absolute truth. James R12-21-10, 09:33 PMchinglu: I note that you have still pointed out no errors in posts #117 and #118. First, you need to apply your argument to Einstein's proof of time dilation I did that in post #117, which has no errors. You will then find the coordinate moving toward the origin is time expanded. Coordinates are points. They have no length. Only spatial intervals can expand or contract. Next, assume a light pulse is emitted from origins. That is a timing device. Prove time dilation negative x-axis point and positive x-axis point. You still haven't explained exactly what you want me to prove, or how it is different from what I showed you in posts #117 and #118. Let me know when you have failed. I can't start until you explain what you want me to do. Remember, you told the readers you are the teacher. You still haven't thanked me for teaching you. Were you not taught to say thankyou when somebody helps you? So, you can prove time dilation for light pulse emitted from origins when they are at same place. I don't know what you mean by this. Please explain in more detail. Prove time dilation for positive x-axis points and negative x-axis points of light pulse for any x and -x. You'll need to explain what you mean in more detail. I can't understand you. Do not forget, time dilation is an absolute truth. No. Time dilation is a derived consequence of the postulates of special relativity. If the postulates are correct, then time dilation is a mathematical consequence. Which of the two postulates of special relativity do you think is wrong? chinglu12-22-10, 05:23 PMchinglu: I note that you have still pointed out no errors in posts #117 and #118. I told you step 2 was wrong and explained why. I did that in post #117, which has no errors. You do not understand his proof. Set x = vt and plug this into Lorentz transform for t'. You will notice he does not use any of your false ideas. Check his 1905 paper and understand how he derives time dilation before you claim expert opinion. Coordinates are points. They have no length. Only spatial intervals can expand or contract. Coordinates are used to make intervals. No. Time dilation is a derived consequence of the postulates of special relativity. If the postulates are correct, then time dilation is a mathematical consequence. First, read Einstein's proof for time dilation. Then write your statement. Maybe we can find your error in thinking with answer simple question. Given (t,x,y,z) mapped by Lorentz Transform to (t',x',y,z) what exactly is this telling you. Say everything you know about this mapping. James R12-23-10, 04:45 AMchinglu: I note that you have still pointed out no errors in posts #117 and #118. I told you step 2 was wrong and explained why. Please quote the wrong part of post #117 or #118 (I don't know which post you're talking about) and remind me what is wrong with it. You do not understand his proof. Set x = vt and plug this into Lorentz transform for t'. You will notice he does not use any of your false ideas. Check his 1905 paper and understand how he derives time dilation before you claim expert opinion. Ok. I had a look at Einstein's 1905 paper. He agrees with me. If you dispute this, please show where my results in posts #117 and #118 differ from Einstein's. It's strange that you're quoting Einstein as an authority, though. I thought you said Einstein was wrong. Do you think Einstein was right or wrong? It's not clear. Are you now agreeing that Einstein was right? No. Time dilation is a derived consequence of the postulates of special relativity. If the postulates are correct, then time dilation is a mathematical consequence. First, read Einstein's proof for time dilation. Then write your statement. Ok. I've done that. My statements are all still correct and in agreement with Einstein. What point are you making? Einstein and I both know where you went wrong, and I explained it to you in posts #117 and #118. Maybe we can find your error in thinking with answer simple question. Given (t,x,y,z) mapped by Lorentz Transform to (t',x',y,z) what exactly is this telling you. Say everything you know about this mapping. Oh, it would take me years to tell you everything I know about this mapping. I'll write down what the mapping is for you, ok? That's a good starting point. t = \gamma (t' + \frac{vx'}{c^2}) x = \gamma (x' + vt') y = y' z = z' This is assuming motion of the primed frame as speed v in the positive x direction. Do you think there's an error in this? If so, please point it out. Also, don't forget to keep searching for any error you can find in posts #117 and #118. Let's not get side-tracked from examining your mistakes, which I pointed out in those posts. Do you see where you went wrong yet? chinglu12-23-10, 06:49 PMchinglu: Please quote the wrong part of post #117 or #118 (I don't know which post you're talking about) and remind me what is wrong with it. Ok. I had a look at Einstein's 1905 paper. He agrees with me. If you dispute this, please show where my results in posts #117 and #118 differ from Einstein's. It's strange that you're quoting Einstein as an authority, though. I thought you said Einstein was wrong. Do you think Einstein was right or wrong? It's not clear. Are you now agreeing that Einstein was right? Ok. I've done that. My statements are all still correct and in agreement with Einstein. What point are you making? Einstein and I both know where you went wrong, and I explained it to you in posts #117 and #118. Oh, it would take me years to tell you everything I know about this mapping. I'll write down what the mapping is for you, ok? That's a good starting point. t = \gamma (t' + \frac{vx'}{c^2}) x = \gamma (x' + vt') y = y' z = z' This is assuming motion of the primed frame as speed v in the positive x direction. Do you think there's an error in this? If so, please point it out. Also, don't forget to keep searching for any error you can find in posts #117 and #118. Let's not get side-tracked from examining your mistakes, which I pointed out in those posts. Do you see where you went wrong yet? Take the primed frame as stationary. The moving unprimed origin moves to x'. In the primed frame, how long does this take? It is simple and consistent with Einstein's time dilation proof. x' = vt' or x'/v = t'. Your calculations calculate a different t' and so you contradict simply physics in the frame. You calculate t' in your posts and come up with a different value. That mean you are in error. rpenner12-23-10, 08:20 PMTake the primed frame as stationary. The moving unprimed origin moves to x'. In the primed frame, how long does this take? It takes the time elapsed in the primed coordinate between the start time at some event and the end time when x'=-k and x=0 are in the same place. The end event (place and time) has primed coordinates x' = -k t' = k/v But unless observers in the primed and unprimed coordinate systems agree on what point in space time the start is, they are going disagree on which is moving and how far it moved. A reasonable start for the "moving unprimed origin" as seen by the primed frame is x = 0 and t' = 0. [1] A very little math later, and we see that is: x' = 0 t' = 0 So the interval (Δx', Δt') = end - start = (-k, k/v) - (0, 0) = (-k, k/v). This is the interval that describes a particle moving at a speed of w' = Δx'/Δt' = -v. So in the primed coordinate system, the start and end describe the "moving unprimed origin" is moving at speed -v, and to answer your question, the time this takes is Δt' = k/v. Since the question is about time dilation, let us ask about the same two events in the unprimed coordinate system. For any event the relationship: x' = γ(x - vt) t' = γ(t - vx/c²) means x = γ(x' + vt') t = γ(t' + vx'/c²) [2] So in the order presented: End event (place and time) in unprimed coordinates: x = γ(-k + vk/v) = 0 t = γ(k/v - vk/c²) = (γk/v)(1 - v²/c²) = k/(γv). Start event: x = γ(0 + v0) = 0 t = γ(0 + v0/c²) = 0 (Δx, Δt) = (0, k/(γv)). Δt = k/(γv) w = Δx/Δt = 0 So the conclusion is that the primed coordinate system measures k/v between the events which represent something moving at speed w' = -v, and the unprimed coordinate system, like a clock which is not moving, w=0, on the origin of place for the unprimed coordinate system measures a smaller number: k/(γv). We can get the same conclusion in a number of ways [3]. For two events on the world line of a uniformly moving clock, the elapsed time is shortest in the inertial frame where the clock is not moving. [4] Appendix [1] Given: x = 0 t' = 0 x' = γ(x - vt) t' = γ(t - vx/c²) What is x' ? From t' = 0 and x = 0 and t' = γ(t - vx/c²) and γ≥1 we conclude t = 0. From x = 0 and t=0 and x' = γ(x - vt) we conclude x' = 0. Appendix [2] Given: x' = γ(x - vt) t' = γ(t - vx/c²) x'' = γ(x' + vt') t'' = γ(t' + vx'/c²) 1 = γ²(1 - v²/c²) what is x'' and t'' in terms of x and t? x'' = γ(x' + vt') = γ(γ(x - vt) + vγ(t - vx/c²)) = γ²(x - vt + vt - v²x/c²) = γ²(x - v²x/c²) = xγ²(1 - v²/c²) = x. t'' = γ(t' + vx'/c²) = γ(γ(t - vx/c²) + vγ(x - vt)/c²) = γ²(t - vx/c² + vx/c² - v²t/c²) = γ²(t - v²t/c²) = tγ²(1 - v²/c²) = t So x' = γ(x - vt) t' = γ(t - vx/c²) means x = γ(x' + vt') t = γ(t' + vx'/c²) Appendix [3] Given two events (ℓ,n) and (o,p) and defining the interval (Δx, Δt) as (ℓ-o,n-p), does the difference of the Lorentz transforms of the events equal the Lorentz transform of the interval? Does (Δx', Δt') = (ℓ'-o',n'-p') ? Well, by the Lorentz transform we have: ℓ' = γ(ℓ - vn) n' = γ(n - vℓ/c²) o' = γ(o - vp) p' = γ(p - vo/c²) And so: Δx' = γ(Δx - vΔt) = γ((ℓ-o) - v(n-p)) = γ((ℓ-vn)-(o - vp)) = γ(ℓ-vn)-γ(o - vp) = ℓ' - o'. Δt' = γ(Δt - vΔx/c²) = γ((n-p) - v(ℓ-o)/c²) = γ((n - vℓ/c²) - (p - vo/c²)) = γ(n - vℓ/c²) - γ(p - vo/c²) = n' - p' So if the interval is the space-time difference of the events, then the Lorentz transform of the coordinate differences is the same as the space-time difference of the Lorentz transformed coordinates. Appendix [4] Given a particle which is moving at speed w for a certain time Δt, what is the Lorentz transform of that time in a frame where the particle is not moving. The speed of the particle in the unprimed coordinates is w = Δx/Δt so Δx = wΔt. if we set v = w, then Δx' = γ(Δx - vΔt) = γ(wΔt - wΔt) = 0 Δt' = γ(Δt - vΔx/c²) = γ(Δt - v²Δt/c²) = Δt/γ So in the special frame where the particle is not moving (w' =Δx'/Δt' = 0), we see that the measured time is shorter by a factor of √(1 - w²/c²) than in the frame where it moves with speed w. chinglu12-23-10, 08:39 PMIt takes the time elapsed in the primed coordinate between the start time at some event and the end time when x'=-k and x=0 are in the same place. The end event (place and time) has primed coordinates x' = -k t' = k/v But unless observers in the primed and unprimed coordinate systems agree on what point in space time the start is, they are going disagree on which is moving and how far it moved. A reasonable start for the "moving unprimed origin" as seen by the primed frame is x = 0 and t' = 0. [1] A very little math later, and we see that is: x' = 0 t' = 0 So the interval (Δx', Δt') = end - start = (-k, k/v) - (0, 0) = (-k, k/v). This is the interval that describes a particle moving at a speed of w' = Δx'/Δt' = -v. I stop your lack of understanding above. The standard configuration mandates both origins are the same for the start of the event. This is not up for discussion. So, you setting x'=0 is complete false because that is not the situation and indicates you do not understand what is going on with relativity. x' is some negative number for the unprimed origin to reach. Are you not able to follow Einsten's proof of time dilation? He set x = vt. You are so willing to accept this reasoning. I am setting x' = -vt', which is exactly the same thing and you and James R are totally sliced by this. We have x' = -vt' for some negative x'. We plug the result into the equation t = ( t' + vx'/c² )γ. The result is t = t'/γ. There is nothing you can do except refute Einstein's proof of time dilation. Since the LT matrix is invertible, then from the unprimed frame at rest, tγ = t' or the primed moving clock coming toward the unprimed origin beats time expanded. There is no choice. James R12-23-10, 09:06 PMTake the primed frame as stationary. The moving unprimed origin moves to x'. In the primed frame, how long does this take? Assuming the origins were co-located at t'=0, this takes time t'=x'/v in the primed frame, where v is the velocity of the unprimed frame. It is simple and consistent with Einstein's time dilation proof. x' = vt' or x'/v = t'. Your calculations calculate a different t' and so you contradict simply physics in the frame. No. My calculations agree with yours in this instance. So, we agree now. That was easy. Are we finished? You calculate t' in your posts and come up with a different value. That mean you are in error. You still haven't pointed out any mistakes in posts #117 and #118. If you think I've made an error, you'll have to go through those two posts and show where the error is. Do it line by line. All you've managed to do since I posted those posts is to try to divert the topic onto something else, avoid the issue, deny your mistakes and generally be evasive. And you still haven't thanked me for helping you see your mistakes. chinglu12-23-10, 09:32 PMAssuming the origins were co-located at t'=0, this takes time t'=x'/v in the primed frame, where v is the velocity of the unprimed frame. No. My calculations agree with yours in this instance. So, we agree now. That was easy. Are we finished? You still haven't pointed out any mistakes in posts #117 and #118. If you think I've made an error, you'll have to go through those two posts and show where the error is. Do it line by line. All you've managed to do since I posted those posts is to try to divert the topic onto something else, avoid the issue, deny your mistakes and generally be evasive. And you still haven't thanked me for helping you see your mistakes. Since your calculations agree with mine, then you understand the unprimed frame claims the moving clock at x' is time expanded and thus time dilation is false. I guess we are done then. You have learned a lot. rpenner12-24-10, 12:04 AMChinglu wrote this: Take the primed frame as stationary. The moving unprimed origin moves to x'. In the primed frame, how long does this take? You take the "place" defined in the primed frame x'=-k as stationary and ask "how long from now" the moving "place" defined in the moving unprimed frame as x=0. So you need to know what you mean by the time of "now" and the time of "when" the "moving unprimed origin moves to x'=-k". Obviously, the "place" where x=0 is seen as moving by primed frame -- the x' of this "place" is a non-constant function of t'. x(t) = 0 \quad \Rightarrow \quad x'(t') = -vt' Both are uniform motions and this is true in both Galilean and Special Relativity. In Special Relativity, we can write: x'(t') = -vt' \quad \Rightarrow \quad x(t') = \gamma \left( x'(t') + vt' \right) = 0 \\ x'(t') = -vt' \quad \Rightarrow \quad t(t') = \gamma \left( t' + vx'(t')/c^2 \right) = \frac{1}{\gamma} t' \\ x'(t') = -vt' \quad \Rightarrow \quad x(t) = 0 This is the core content of the principle of relativity. What is a "place" for one frame can "move" in another and both descriptions are equally valid. It takes the time elapsed in the primed coordinate between the start time at some event and the end time when x'=-k and x=0 are in the same place. ... unless observers in the primed and unprimed coordinate systems agree on what point in space time the start is, they are going disagree on which is moving and how far it moved. A reasonable start for the "moving unprimed origin" as seen by the primed frame is x = 0 and t' = 0. [1] A very little math later, and we see that is: x' = 0 t' = 0 So the interval (Δx', Δt') = end - start = (-k, k/v) - (0, 0) = (-k, k/v). This is the interval that describes a particle moving at a speed of w' = Δx'/Δt' = -v .... Appendix [1] Given: x = 0 t' = 0 x' = γ(x - vt) t' = γ(t - vx/c²) What is x' ? From t' = 0 and x = 0 and t' = γ(t - vx/c²) and γ≥1 we conclude t = 0. From x = 0 and t=0 and x' = γ(x - vt) we conclude x' = 0. Note that there is math connecting the claim that t'=0 means t=0 when x=0. See below for why that math is important. I stop your lack of understanding above.The only thing I don't understand is why you think you understand this material better than Einstein and all who followed in the 105 years. The standard configuration mandates both origins are the same for the start of the event. This is not up for discussion. The standard configuration says when the space-time origin of the unprimed frame (x,t) = (0,0) is the same point (event) in space-time as the space-time origin of the (x',t') = (0,0). Since you started a new question unrelated to the questions of post #1, it was not clearly communicated that you assumed that this event was meant as the start, that you know that every event has both a place and time, or that you have a thesis statement in mind. What I have done is showing assuming we are talking about the moving "place" which is the spatial origin of the unprimed frame x=0, the choice of t' = 0 implies that the event we are talking about is the common space-time origin of both coordinate systems. This is also (more obviously) true if you pick x=0 and t=0. But my proof that t'=0 means t=0 depends on x=0, and is not true if x ≠ 0. This is not up for discussion.Please demonstrate superiority before asserting authority; you will still be wrong but at least you will have a reasoned argument for your authoritarian claim. So we are agreed that (x,t) = (0,0) is the same point (event) in space-time as (x',t') = (0,0) and is the start of movement of the "unprimed origin [place]" that we wish to follow. So, you setting x'=0 is complete false because that is not the situation and indicates you do not understand what is going on with relativity.You claim this, but have not demonstrated this. Just because you use horribly confusing notation like x' to mean the spatial coordinate in the primed coordinate frame of the end of the journey of the "unprimed origin [place]" is equal to some constant, does not forbid me from using the much more standard notation x' = -k to say the same thing. More than 100 posts ago, I would be even more explicit by writing x'_Q = -k, or even \left. x' = -k \right| _{End}. Since we are talking about the start of the journey, where the "place" picked out by the world-line of x=0 has not yet met the "place" picked by the "not-moving" world-line of x'=-k, the expectation is at the start x' ≠ -k, otherwise there has been no movement. That's why the math of Appendix [1] demonstrates that for the world-line of x = 0, the choice of t' = 0 uniquely picks the start event that forces x' = 0. More than 100 posts ago, I would be even more explicit by writing x_O = 0 \, \textrm{and} \, t'_O = 0 \quad \Rightarrow \quad (x_O,t_O) = (0,0) \, \textrm{and} \, (x'_O,t'_O) = (0,0), or even \left. x = 0 \, \textrm{and} \, t' = 0 \quad \Rightarrow \quad (x,t) = (0,0) \, \textrm{and} \, (x',t') = (0,0) \right| _{Start}. x' is some negative number for the unprimed origin to reach.Exactly why I said that was unnecessarily confusing. x' and x and t' and t are values which vary throughout space-time depending on what event we are talking about. Since you are talking about something moving to a "place" in the primed frame, at the start the place of that something cannot be the same place. So it is better to use some other symbol to represent that place, which is the world-line which consists of all events where the primed spatial coordinate is equal to that negative constant number. x' = -k is a valid way to describe that world-line in the language of algebra, just like x = 0 is a valid way to describe the world-line of the "unprimed origin [of space]". Are you not able to follow Einsten's proof of time dilation? He set x = vt. You are so willing to accept this reasoning.Well I also have exposure to the notation of Cartesian coordinates and the 300+ year history of physics to know that we are talking about the algebraic geometry of straight lines in space-time. In which case the most general expression of that concept is \left. \vec{x}_A(t) = \vec{x}_{0,A} + \vec{v}_{A} t \right|_S.where A selects the world-line and S selects the frame. The Lorentz transformation converts this expression into an expression of the same form for a different frame, say S'. What is at issue here is not Einstein's thesis, but what your thesis is and why you see unable to communicate it or a reason to trust it. I am setting x' = -vt', which is exactly the same thing and you and James R are totally sliced by this.I doubt we are "sliced." But here you are confusing apples and oranges, since you are criticizing me for talking about the start of the movement and here you are talking about the end of the movement. In my notation, your "x' = -v t'" would mean that for all points on the world-line of the "origin" are described by (x',t') = (-vt',t'), but in your notation you are just talking about the end of the journey, which in my notion is x' = -vt' = -k. We have x' = -vt' for some negative x'.It would be more useful to name this as x' = -k like James R did or x' = -5/6 light seconds as your post #1 did. We plug the result into the equation t = ( t' + vx'/c² )γ. The result is t = t'/γ.Here, based a situation such that t measures the elasped time in the frame where there is no movement of the relevant object Chinglu, Einstein, James R, and myself conclude that t = t'/γ < t' . There is nothing you can do except refute Einstein's proof of time dilation. None of the above refutes Einstein. The frame that sees the object as moving says it takes more time (t') to get from O to Q than the frame that sees the same object as not moving (t). Since the LT matrix is invertible, then from the unprimed frame at rest, tγ = t' or the primed moving clock coming toward the unprimed origin beats time expanded.ERROR. You can invert the Lorentz transform all you want, but you have lost the thread of your physical-geometrical reasoning. (tγ = t' is just another way of saying t = t'/γ, and is only valid for describing events on the world-line that is moving relative to the the primed spatial origin and stationary in the unprimed coordinates.) But if you want to talk now about (like you did in post #1) events on a different world line that is moving relative to the unprimed spatial origin and stationary in the primed coordinates, then the correct expression is Δt' = Δt/γ. If you flip-flop which coordinates see the object as stationary, then you have to flip-flop which coordinate measure the shortest elapsed time. See Appendix [4] where I already demonstrated this concept in anticipation of your continued (at least four forums now) failure to undertand place, time, space-time, event, and coordinate, which are prerequisites to having a working understanding of Special Relativity. James R12-24-10, 03:30 AMSince your calculations agree with mine, then you understand the unprimed frame claims the moving clock at x' is time expanded and thus time dilation is false. No. I have only agreed with one particular point - your primed frame calculation that I quoted above. I note once again that you have still not addressed the content of posts #117 and #118 in any way. I must assume at this point that you have no reply to those posts and are therefore avoiding responding to them because you can't admit you were wrong. Is that right? Just to be repeat myself: your error is that you have wrongly assumed that the time interval you measured in the unprimed frame is a proper time. In your example, it is the primed frame that measures the proper time between the two events you have used. Therefore, we expect the time interval for the same two events to be LONGER in the unprimed frame than it is in the primed frame, and this is exactly what we see in post #117. Your problem is that you think you are calculating the time interval between events 1 and 3 in post #117, when in fact you have calculated the interval between events 2 and 3. But all this is already clear from posts #117 and #118. You have not shown any error in either of those posts and you're avoiding the issue because you know there is no error. At this point you're just stringing us along like a troll. Your credibility has sunk to nothing. BenTheMan12-24-10, 12:29 PMI think this thread has FAR outstayed it's welcome. rpenner12-26-10, 05:00 AMI don't want it. Chinglu tried to take it back to my forum for Christmas. chinglu12-26-10, 11:22 AMNo. I have only agreed with one particular point - your primed frame calculation that I quoted above. I note once again that you have still not addressed the content of posts #117 and #118 in any way. I must assume at this point that you have no reply to those posts and are therefore avoiding responding to them because you can't admit you were wrong. Is that right? Just to be repeat myself: your error is that you have wrongly assumed that the time interval you measured in the unprimed frame is a proper time. In your example, it is the primed frame that measures the proper time between the two events you have used. Therefore, we expect the time interval for the same two events to be LONGER in the unprimed frame than it is in the primed frame, and this is exactly what we see in post #117. Your problem is that you think you are calculating the time interval between events 1 and 3 in post #117, when in fact you have calculated the interval between events 2 and 3. But all this is already clear from posts #117 and #118. You have not shown any error in either of those posts and you're avoiding the issue because you know there is no error. At this point you're just stringing us along like a troll. Your credibility has sunk to nothing. I thought I did address #117. 1. (x,t) = (0,0) 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) 3. (x,t)=(0,\frac{k}{\gamma v}) 1 and 3: \Delta t = \frac{k}{\gamma v} 2 and 3: \Delta t = \frac{\gamma k}{v} In the context of the unprimed frame, if it happened to be the case that the observer had been watching som object moving to the origin based on its own coordinate system that began at -k when the origins were the same, then we would calculate \Delta t = \frac{k}{v} I am not sure it you agree. Now, if on the other hand, this coordinate happened to be in the coordinates of the moving frame, then when the origins are the same, its distance to travel to the origin is \Delta t = \frac{k}{\gamma v} instead because of length contraction. I am not considering the clock at -k at all. The only thing being considered is the view of the unprimed origin observer. Do we agree on this? chinglu12-26-10, 11:38 AMChinglu wrote this: You take the "place" defined in the primed frame x'=-k as stationary and ask "how long from now" the moving "place" defined in the moving unprimed frame as x=0. So you need to know what you mean by the time of "now" and the time of "when" the "moving unprimed origin moves to x'=-k". Obviously, the "place" where x=0 is seen as moving by primed frame -- the x' of this "place" is a non-constant function of t'. x(t) = 0 \quad \Rightarrow \quad x'(t') = -vt' Both are uniform motions and this is true in both Galilean and Special Relativity. In Special Relativity, we can write: x'(t') = -vt' \quad \Rightarrow \quad x(t') = \gamma \left( x'(t') + vt' \right) = 0 \\ x'(t') = -vt' \quad \Rightarrow \quad t(t') = \gamma \left( t' + vx'(t')/c^2 \right) = \frac{1}{\gamma} t' \\ x'(t') = -vt' \quad \Rightarrow \quad x(t) = 0 This is the core content of the principle of relativity. What is a "place" for one frame can "move" in another and both descriptions are equally valid. Note that there is math connecting the claim that t'=0 means t=0 when x=0. See below for why that math is important. The only thing I don't understand is why you think you understand this material better than Einstein and all who followed in the 105 years. The standard configuration says when the space-time origin of the unprimed frame (x,t) = (0,0) is the same point (event) in space-time as the space-time origin of the (x',t') = (0,0). Since you started a new question unrelated to the questions of post #1, it was not clearly communicated that you assumed that this event was meant as the start, that you know that every event has both a place and time, or that you have a thesis statement in mind. What I have done is showing assuming we are talking about the moving "place" which is the spatial origin of the unprimed frame x=0, the choice of t' = 0 implies that the event we are talking about is the common space-time origin of both coordinate systems. This is also (more obviously) true if you pick x=0 and t=0. But my proof that t'=0 means t=0 depends on x=0, and is not true if x ≠ 0.Please demonstrate superiority before asserting authority; you will still be wrong but at least you will have a reasoned argument for your authoritarian claim. So we are agreed that (x,t) = (0,0) is the same point (event) in space-time as (x',t') = (0,0) and is the start of movement of the "unprimed origin [place]" that we wish to follow. You claim this, but have not demonstrated this. Just because you use horribly confusing notation like x' to mean the spatial coordinate in the primed coordinate frame of the end of the journey of the "unprimed origin [place]" is equal to some constant, does not forbid me from using the much more standard notation x' = -k to say the same thing. More than 100 posts ago, I would be even more explicit by writing x'_Q = -k, or even \left. x' = -k \right| _{End}. Since we are talking about the start of the journey, where the "place" picked out by the world-line of x=0 has not yet met the "place" picked by the "not-moving" world-line of x'=-k, the expectation is at the start x' ≠ -k, otherwise there has been no movement. That's why the math of Appendix [1] demonstrates that for the world-line of x = 0, the choice of t' = 0 uniquely picks the start event that forces x' = 0. More than 100 posts ago, I would be even more explicit by writing x_O = 0 \, \textrm{and} \, t'_O = 0 \quad \Rightarrow \quad (x_O,t_O) = (0,0) \, \textrm{and} \, (x'_O,t'_O) = (0,0), or even \left. x = 0 \, \textrm{and} \, t' = 0 \quad \Rightarrow \quad (x,t) = (0,0) \, \textrm{and} \, (x',t') = (0,0) \right| _{Start}. Exactly why I said that was unnecessarily confusing. x' and x and t' and t are values which vary throughout space-time depending on what event we are talking about. Since you are talking about something moving to a "place" in the primed frame, at the start the place of that something cannot be the same place. So it is better to use some other symbol to represent that place, which is the world-line which consists of all events where the primed spatial coordinate is equal to that negative constant number. x' = -k is a valid way to describe that world-line in the language of algebra, just like x = 0 is a valid way to describe the world-line of the "unprimed origin [of space]". Well I also have exposure to the notation of Cartesian coordinates and the 300+ year history of physics to know that we are talking about the algebraic geometry of straight lines in space-time. In which case the most general expression of that concept is \left. \vec{x}_A(t) = \vec{x}_{0,A} + \vec{v}_{A} t \right|_S.where A selects the world-line and S selects the frame. The Lorentz transformation converts this expression into an expression of the same form for a different frame, say S'. What is at issue here is not Einstein's thesis, but what your thesis is and why you see unable to communicate it or a reason to trust it. I doubt we are "sliced." But here you are confusing apples and oranges, since you are criticizing me for talking about the start of the movement and here you are talking about the end of the movement. In my notation, your "x' = -v t'" would mean that for all points on the world-line of the "origin" are described by (x',t') = (-vt',t'), but in your notation you are just talking about the end of the journey, which in my notion is x' = -vt' = -k. It would be more useful to name this as x' = -k like James R did or x' = -5/6 light seconds as your post #1 did. Here, based a situation such that t measures the elasped time in the frame where there is no movement of the relevant object Chinglu, Einstein, James R, and myself conclude that t = t'/γ < t' . None of the above refutes Einstein. The frame that sees the object as moving says it takes more time (t') to get from O to Q than the frame that sees the same object as not moving (t). ERROR. You can invert the Lorentz transform all you want, but you have lost the thread of your physical-geometrical reasoning. (tγ = t' is just another way of saying t = t'/γ, and is only valid for describing events on the world-line that is moving relative to the the primed spatial origin and stationary in the unprimed coordinates.) But if you want to talk now about (like you did in post #1) events on a different world line that is moving relative to the unprimed spatial origin and stationary in the primed coordinates, then the correct expression is Δt' = Δt/γ. If you flip-flop which coordinates see the object as stationary, then you have to flip-flop which coordinate measure the shortest elapsed time. See Appendix [4] where I already demonstrated this concept in anticipation of your continued (at least four forums now) failure to undertand place, time, space-time, event, and coordinate, which are prerequisites to having a working understanding of Special Relativity. ERROR. You can invert the Lorentz transform all you want, but you have lost the thread of your physical-geometrical reasoning. (tγ = t' is just another way of saying t = t'/γ, and is only valid for describing events on the world-line that is moving relative to the the primed spatial origin and stationary in the unprimed coordinates.) But if you want to talk now about (like you did in post #1) events on a different world line that is moving relative to the unprimed spatial origin and stationary in the primed coordinates, then the correct expression is Δt' = Δt/γ. If you flip-flop which coordinates see the object as stationary, then you have to flip-flop which coordinate measure the shortest elapsed time. See Appendix [4] where I already demonstrated this concept in anticipation of your continued (at least four forums now) failure to undertand place, time, space-time, event, and coordinate, which are prerequisites to having a working understanding of Special Relativity. This is not the case. In the view of the primed frame, all can agree in the view of the primed frame, the moving clock at the unprimed origin beats time dilated, that is the part all agree, the origin clock. So, in the primed frame, it calculates Δt' = γΔt or that the moving unprimed origin clock is time dilated. So, the origin moves to the primed coordinate -k a distance k at v or Δt' = k/v. Using substitution, Δt' = γΔt = k/v or Δt = k/(γ v). In fact, when unprimed is take stationary, based on length contraction, this is exactly waht the unprimed concludes for the primed coordinate -k to move to it origin such aht both frame agree on the start and end events. This also is maintains the invertibility of the Lorentz Transform matrix and so this method is consistent with that. I assume you conclude from the view of the unprimed frame, it must conclude Δt = γΔt' for this same start event of the origins same and primed -k and unprimed origin same. Since, Δt = k/(γ v) = γΔt'. Then, in your calculation Δt' = k/(γ² v) for the calculation of the primed -k clock to reach the unrpimed origin. This violates the calculation in the primed frame for the same events and also violates the invertibility of Lorentz transform matrix. My calculation violates nothing. arfa brane12-26-10, 07:26 PMI don't want it. Chinglu tried to take it back to my forum for Christmas. Just advertise it on ebay like thousands of others do with unwanted presents. . . (make sure it's still in the original wrapper, though). James R12-27-10, 04:42 AMchinglu: In the context of the unprimed frame, if it happened to be the case that the observer had been watching som object moving to the origin based on its own coordinate system that began at -k when the origins were the same, then we would calculate \Delta t = \frac{k}{v} I am not sure it you agree. If the object began at x=-k and moved to x=0 at speed v, I agree the time taken would be as you say. Now, if on the other hand, this coordinate happened to be in the coordinates of the moving frame, then when the origins are the same, its distance to travel to the origin is \Delta t = \frac{k}{\gamma v} instead because of length contraction. Well, that's a complicated statement, isn't it? The problem is the part where you say "when the origins are the same". The origins are assumed to be co-located at t=t'=0. But events that occur at x=-k/\gamma or x'=-k, for example, are not located at x=x'=0, and so cannot be simultaneous with events at both x=0 and x'=0 at t=t'=0. So, you need to carefully check when, in EACH frame, the event at, say x=-k/\gamma occurs. This was the source of your original error, and I quite clearly explained the problem to you in posts #117 and #118. It seems to me that you are still making the same mistake. The solution to your problem is this: two spatially-separated events that are simultaneous in one reference frame cannot be simultaneous in any other reference frame. chinglu12-27-10, 06:03 PMchinglu: If the object began at x=-k and moved to x=0 at speed v, I agree the time taken would be as you say. Well, that's a complicated statement, isn't it? The problem is the part where you say "when the origins are the same". The origins are assumed to be co-located at t=t'=0. But events that occur at x=-k/\gamma or x'=-k, for example, are not located at x=x'=0, and so cannot be simultaneous with events at both x=0 and x'=0 at t=t'=0. So, you need to carefully check when, in EACH frame, the event at, say x=-k/\gamma occurs. This was the source of your original error, and I quite clearly explained the problem to you in posts #117 and #118. It seems to me that you are still making the same mistake. The solution to your problem is this: two spatially-separated events that are simultaneous in one reference frame cannot be simultaneous in any other reference frame. If the object began at x=-k and moved to x=0 at speed v, I agree the time taken would be as you say. This part is resolved. We are not writing about simultaneity as I do not care what the clock at -k in the moving system says. If I cared about its time, you have point and our debate would move on different tract. So, I am only calculating how long that coordinate in moving system takes to reach origin and nothing more. I assume you believe in length contraction. So when origins same, that coordinate is a distance k/γ from the unprimed origin. Again I do not care what the time is on that clock and do not even care if clocks are synchronized in that frame. I only care how long the unprimed frame claims it takes to move to origin and nothing more. So if length contraction true, then when origins same the distance to the unprimed origin is k/γ. Do you agree or disagree. chinglu12-27-10, 07:44 PMI don't want it. Chinglu tried to take it back to my forum for Christmas. You are less than honest. In your forum you wrote Physics Forums: Backed into a corner Let's be honest. I posed this question in your forum. You then banned me and I moved to http://www.physicsforums.com/showthread.php?t=455613&page=2 to intellectually discuss the issue further. As an internet stalker, you follow me there. Here you droned on and on with false mathematics and even DaleSpam a respected moderator in that forum supported and agreed with my conclusions as anyone can see below. http://www.physicsforums.com/showpost.php?p=3032436&postcount=20 I demonstrated you falsified my post in your forum here. You are less than honest. http://www.physicsforums.com/showpost.php?p=3032669&postcount=26 Since that thread concluded with my conclusions as validated, I came here. Again, as an internet stalker, here you came again posting your failed math that I easily refute. I never chased you. Then, I went back to your forum in which your bullies on that forum attacked me. Then, as all can see here, http://www.physforum.com/index.php?showtopic=28676&st=120 I corrected the bully with pure math and back him down based only on math. So, your bullies can dish it out but can't take it. Then, you banned me because it was impossible for your home bullies to push me around. Again, DaleSpam validates my math and refutes yours. http://www.physicsforums.com/showpost.php?p=3032436&postcount=20 AlexG12-27-10, 08:46 PMThen, I went back to your forum in which your bullies on that forum attacked me. Would you like some cheese with your whine? chinglu12-27-10, 08:57 PMWould you like some cheese with your whine? I just love originality. Did you make this up or did you copy it from someone else? James R12-28-10, 05:01 AMchinglu: We are not writing about simultaneity as I do not care what the clock at -k in the moving system says. If I cared about its time, you have point and our debate would move on different tract. So, I am only calculating how long that coordinate in moving system takes to reach origin and nothing more. To know how long something takes to move from one place to another, you need to know the time it started moving and the time it finished moving. Those two times are two separate events in spacetime. I assume you believe in length contraction. So when origins same, that coordinate is a distance k/γ from the unprimed origin. Well, let's see. You're saying that when (x,t)=(0,0) and (x',t')=(0,0) (so that the origins are the same) then another event occurs at (x,t)=(k/\gamma,0). That's in the unprimed frame. But that other event does not occur at t'=0! Check it for yourself by calculating the Lorentz transformed coordinates of the event. Do you know how to do that? If not, see posts #117 and #118 where I explicitly gave you all the maths you need. Again I do not care what the time is on that clock and do not even care if clocks are synchronized in that frame. I only care how long the unprimed frame claims it takes to move to origin and nothing more. But ultimately you want to compare the time taken in the primed frame and in the unprimed frame for the same events, don't you? If you don't use the same events, but think you are using the same events, then you'll make a mistake and reach an incorrect conclusion, just like you did before. I explained where and why you made the error in posts #117 and #118, as you will recall. chinglu12-28-10, 05:32 PMchinglu: To know how long something takes to move from one place to another, you need to know the time it started moving and the time it finished moving. Those two times are two separate events in spacetime. Well, let's see. You're saying that when (x,t)=(0,0) and (x',t')=(0,0) (so that the origins are the same) then another event occurs at (x,t)=(k/\gamma,0). That's in the unprimed frame. But that other event does not occur at t'=0! Check it for yourself by calculating the Lorentz transformed coordinates of the event. Do you know how to do that? If not, see posts #117 and #118 where I explicitly gave you all the maths you need. But ultimately you want to compare the time taken in the primed frame and in the unprimed frame for the same events, don't you? If you don't use the same events, but think you are using the same events, then you'll make a mistake and reach an incorrect conclusion, just like you did before. I explained where and why you made the error in posts #117 and #118, as you will recall. James R All I am trying to get you to do here is tell the the distance -k in the moving coordinates is from the origin when the origins are the same. We are not trying to determine the anything about the clock time at that coordinate in the moving system. We are only calculating the distance to the origin in the view of unprimed frame when origins are same. So, what is answer? I calculate the distance is k/γ based on length contraction. What do you calculate for the distance? James R12-28-10, 10:10 PMchinglu: James R All I am trying to get you to do here is tell the the distance -k in the moving coordinates is from the origin when the origins are the same. In the moving frame, you're talking about two events here. The moving frame origin at (x',t')=(0,0) and the event (x',t')=(-k,0). Now, if the primed origin is at x=0 at time t=0 in the unprimed coordinates, then the event (x',t')=(-k,0) is at (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2}). Note that this event occurs at a time in the unprimed frame when the origins are not co-located, even though the same event occurs simultaenously with the origins being co-located in the primed frame. This is because events that spatially separated and simultaneous in one frame cannot be simultaenous in any other frame. I explained this to you in detail in posts #117 and #118, above. We are only calculating the distance to the origin in the view of unprimed frame when origins are same. So, what is answer? I calculate the distance is k/γ based on length contraction. What do you calculate for the distance? Now you're talking about the event (x,t)=(-\frac{k}{\gamma},0), which as you can see from above, is NOT the same event as the one at (x',t')=(-k,0). The event (x,t)=(-\frac{k}{\gamma},0) is the same as (x',t')=(-k,\frac{kv}{c^2}) Your mistake is that you are confusing two different events and thinking they are the same. To summarise, here are two events, with their spacetime coordinates in both frames: Event 2: (x',t')=(-k,0); (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2}). Event 3: (x',t')=(-k,\frac{kv}{c^2}); (x,t)=(-\frac{k}{\gamma},0) Note that event 2 occurs when the origins are co-located in the primed frame, and event 3 occurs when the origins are co-located in the unprimed frame. chinglu12-28-10, 10:29 PMchinglu: In the moving frame, you're talking about two events here. The moving frame origin at (x',t')=(0,0) and the event (x',t')=(-k,0). Now, if the primed origin is at x=0 at time t=0 in the unprimed coordinates, then the event (x',t')=(-k,0) is at (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2}). Note that this event occurs at a time in the unprimed frame when the origins are not co-located, even though the same event occurs simultaenously with the origins being co-located in the primed frame. This is because events that spatially separated and simultaneous in one frame cannot be simultaenous in any other frame. I explained this to you in detail in posts #117 and #118, above. Now you're talking about the event (x,t)=(-\frac{k}{\gamma},0), which as you can see from above, is NOT the same event as the one at (x',t')=(-k,0). The event (x,t)=(-\frac{k}{\gamma},0) is the same as (x',t')=(-k,\frac{kv}{c^2}) Your mistake is that you are confusing two different events and thinking they are the same. To summarise, here are two events, with their spacetime coordinates in both frames: Event 2: (x',t')=(-k,0); (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2}). Event 3: (x',t')=(-k,\frac{kv}{c^2}); (x,t)=(-\frac{k}{\gamma},0) Note that event 2 occurs when the origins are co-located in the primed frame, and event 3 occurs when the origins are co-located in the unprimed frame. We are not writing about the moving frame. Are you going to give the distance of moving k in the unprimed frame? If not, you will have to conceed to my calculations. In the unprimed frame, that distance is k/γ when the origins are same. You can not disagree since that mean you refute length contraction. I guess you can not know this value. It might be too hard. James R12-28-10, 10:36 PMWe are not writing about the moving frame. Yes we are. You are assuming the primed frame moves at speed v in the unprimed frame. Your original question asked when the x'=-k coordinate was in the same place as the x=0 coordinate, assuming the x'=0 coordinate coincides with x=0 at t=t'=0. Are you going to give the distance of moving k in the unprimed frame? I already did that in post #117, and repeated myself in my post immediately before this one. In the unprimed frame, that distance is k/γ when the origins are same. Yes, when the origins are the same in the unprimed frame. Of course, at that particular time the origins are NOT the same in the primed frame. You can not disagree since that mean you refute length contraction. I don't refute length contraction. As I explained to you in post #118, you have to be careful to apply length contraction the "right way" round. You need to identify which length is the proper length and which is not. Your mistake is that you got it wrong. You assumed something is a proper length when it is not. I guess you can not know this value. It might be too hard. Your job is to refute posts #117 and #118, if you can. Your insults won't work. chinglu12-28-10, 11:03 PMYes we are. You are assuming the primed frame moves at speed v in the unprimed frame. Your original question asked when the x'=-k coordinate was in the same place as the x=0 coordinate, assuming the x'=0 coordinate coincides with x=0 at t=t'=0. I already did that in post #117, and repeated myself in my post immediately before this one. Yes, when the origins are the same in the unprimed frame. Of course, at that particular time the origins are NOT the same in the primed frame. I don't refute length contraction. As I explained to you in post #118, you have to be careful to apply length contraction the "right way" round. You need to identify which length is the proper length and which is not. Your mistake is that you got it wrong. You assumed something is a proper length when it is not. Your job is to refute posts #117 and #118, if you can. Your insults won't work. Maybe we can make this simpler for both of us. If the moving frame has a rod of length k, how long is that rod in the view of the rest frame? Does its position in the moving frame change its length in the view of the rest frame? Now, what if I place one end of the rod at the origin of the moving frame and the otther along the negative x-axis, does this matter? Can you answer these questions? If you are unable to answer these simple questions, let's just go our separate ways. James R12-29-10, 12:14 AMchinglu: Maybe we can make this simpler for both of us. If the moving frame has a rod of length k, how long is that rod in the view of the rest frame? The rod has length \gamma k in the rest frame of the rod if the frame that sees the rod moving measures length k. This result is derived by considering spacetime events at the two ends of the rod in each frame. Notice that, once again, three events are involved, not two, because events at opposite ends of the rod that are simultaneous in one frame are NOT simultaneous in the other, and to measure a length we need two simultaneous ends of the rod. Does its position in the moving frame change its length in the view of the rest frame? No. Now, what if I place one end of the rod at the origin of the moving frame and the otther along the negative x-axis, does this matter? No, it doesn't matter. Can you answer these questions? If you are unable to answer these simple questions, let's just go our separate ways. I've answered all your questions. I showed you exactly where you went wrong, first in posts #117 and #118, and then again a few posts above this one. Are you unable to comprehend the explanations I have given to you? If you are unable to understand how to measure lengths, then let's just go our separate ways. chinglu12-29-10, 12:31 PMchinglu: The rod has length \gamma k in the rest frame of the rod if the frame that sees the rod moving measures length k. This result is derived by considering spacetime events at the two ends of the rod in each frame. Notice that, once again, three events are involved, not two, because events at opposite ends of the rod that are simultaneous in one frame are NOT simultaneous in the other, and to measure a length we need two simultaneous ends of the rod. No. No, it doesn't matter. I've answered all your questions. I showed you exactly where you went wrong, first in posts #117 and #118, and then again a few posts above this one. Are you unable to comprehend the explanations I have given to you? If you are unable to understand how to measure lengths, then let's just go our separate ways. James R: Yes, I do understand your posts. That is not what I am doing. We have not resolved the simple matter of length contraction yet. If the coordinate is -k in the moving system, then it is length contracted in the view of the rest coordinates, hence its length is k/γ in the view of the rest system. You wrote k γ is the length in the rest system. Did you mean this or what it a typing mistake? This is not length contraction of a moving rod in the view of the rest frame., James R12-29-10, 10:07 PMchinglu: We have not resolved the simple matter of length contraction yet. If the coordinate is -k in the moving system, then it is length contracted in the view of the rest coordinates, hence its length is k/γ in the view of the rest system. Coordinates are not length contracted. Spatial intervals are length contracted. Every length has two ends to it. Each end is one event in spacetime. A length is the spatial interval between two events. A proper length is a spatial interval between two events that occur at the same time. Therefore, a proper length in one reference frame CANNOT be a proper length in any other frame due to the relativity of simultaneity. You wrote k γ is the length in the rest system. Did you mean this or what it a typing mistake? This is not length contraction of a moving rod in the view of the rest frame., I was very specific. I used the spacetime events that you specified. The length k that you gave was not a proper length, but a contracted length. Changing frames in that case gives the proper length, which is longer by a factor of \gamma. Your mistake, as I pointed out in post #117 was to confuse which length is a proper length and which is not. You're making the same mistake here. chinglu12-30-10, 08:57 AMchinglu: Coordinates are not length contracted. Spatial intervals are length contracted. Every length has two ends to it. Each end is one event in spacetime. A length is the spatial interval between two events. A proper length is a spatial interval between two events that occur at the same time. Therefore, a proper length in one reference frame CANNOT be a proper length in any other frame due to the relativity of simultaneity. I was very specific. I used the spacetime events that you specified. The length k that you gave was not a proper length, but a contracted length. Changing frames in that case gives the proper length, which is longer by a factor of \gamma. Your mistake, as I pointed out in post #117 was to confuse which length is a proper length and which is not. You're making the same mistake here. You can read here if L0 = x2' - x1' is measured as a distance in the moving system, then L0/γ is the measured distance in the stationary frame. http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html Since we have L0 = 0 - (-k) = k as the diatance in the moving system, then k/γ is the distance in the stationary system. Here is another link http://physics.tamuk.edu/~hewett/ModPhy1/Unit1/SpecialRelativity/RelativeView/Space/Length/Length.html James R12-30-10, 06:59 PMchinglu: The problem is that you need to decide which frame is "stationary". For example, suppose I watch a plane flying past as I stand on the ground. As far as I'm concerned, I am stationary and the plane is length contracted. But a person on the plane thinks he is stationary and I am length contracted. The thing is: any frame can be regarded as stationary, so working out what is a contracted length and what is an uncontracted length isn't always obvious. That's why you need to work out who is measuring the proper length or rest length, and who is not, whenever you deal with a length. chinglu12-30-10, 07:13 PMchinglu: The problem is that you need to decide which frame is "stationary". For example, suppose I watch a plane flying past as I stand on the ground. As far as I'm concerned, I am stationary and the plane is length contracted. But a person on the plane thinks he is stationary and I am length contracted. The thing is: any frame can be regarded as stationary, so working out what is a contracted length and what is an uncontracted length isn't always obvious. That's why you need to work out who is measuring the proper length or rest length, and who is not, whenever you deal with a length. James R: I thought we already decided which frame we are taking as stationary for right now, it is the unprimed frame. So, k < 0 is in the measurements of the primed frame and it is moving. Now that we have resolved this issue, then I must assume you agree Δt1 = (k/γ)/v is the time it takes for k to move to the unprimed origin based on simple length contraction. James R12-30-10, 07:22 PMJames R: I thought we already decided which frame we are taking as stationary for right now, it is the unprimed frame. So, k < 0 is in the measurements of the primed frame and it is moving. Now that we have resolved this issue, then I must assume you agree Δt1 = (k/γ)/v is the time it takes for k to move to the unprimed origin based on simple length contraction. If an object starts at x=-\frac{k}{\gamma} at time t=0, then I agree with you that if it travels in the +x direction at speed v then it reaches x=0 at time t=\frac{k}{\gamma v}. This is different from your initial scenario, where the coordinate was not at x=-\frac{k}{\gamma} at t=0. If you're not sure about that, please read posts #117 and #118 again, where I explained it all to you in detail. Alternatively, look at post #165: To summarise, here are two events, with their spacetime coordinates in both frames: Event 2: (x',t')=(-k,0); (x,t)=(-\gamma k, -\frac{\gamma kv}{c^2}). Event 3: (x',t')=(-k,\frac{kv}{c^2}); (x,t)=(-\frac{k}{\gamma},0) Note that event 2 occurs when the origins are co-located in the primed frame, and event 3 occurs when the origins are co-located in the unprimed frame. chinglu12-30-10, 07:46 PMIf an object starts at x=-\frac{k}{\gamma} at time t=0, then I agree with you that if it travels in the +x direction at speed v then it reaches x=0 at time t=\frac{k}{\gamma v}. This is different from your initial scenario, where the coordinate was not at x=-\frac{k}{\gamma} at t=0. If you're not sure about that, please read posts #117 and #118 again, where I explained it all to you in detail. Alternatively, look at post #165: Now, that we have that straight that the unprimed frame will conclude Δt = (k/γ)/v let's calculate the view of the primed frame now. It sees the origin frame move to the coordinate -k. It calculates Δt' = k/v. Unprimed Frame So, these from the time the origins are same to the time k reaches the unprimed origin, Δt = (k/γ)/v Primed Frame So, these from the time the origins are same to the time unprimed origin reaches k, Δt' = k/v There is agreement on the stop and start of these two events between the frames. We can understand Δt = Δt'/γ. So, in view of unprimed frame at rest, Δt = Δt'/γ for the coordinate k in the moving system to reach the unprimed origin. Thus, the moving clock beats time expanded. Let's use Lorentz Transforms to check out work. We have x=0 because at origin in unprimed frame. t' = ( t - vx/c² )γ = t γ, so t = t'/γ, just as we calculated. No time dilation for clock moving to the rest origin when in coordinates of moving frame. We just proved it. James R12-30-10, 09:49 PMchinglu: Now, that we have that straight that the unprimed frame will conclude Δt = (k/γ)/v let's calculate the view of the primed frame now. Ok. But first let's make sure we know which events we're dealing with. The time interval you give here starts at t=0, when the origins of the primed and unprimed frames are in the same place. And at t=0 you're interested in the other point located at (x,t)=(-\frac{k}{\gamma},0). Now, let's set the clock at the origin in the primed frame to read t'=0 when the origins of the two frames are in the same place. Now, we calculate the coordinates of the other event: (x,t)=(-\frac{k}{\gamma},0); (x',t') = (-k, \frac{kv}{c^2}) Notice that the event at x=-k/\gamma does NOT occur at t'=0 in the primed frame. [The primed frame] sees the origin [of the unprimed] frame move to the coordinate -k.[/tex] Yes. It calculates Δt' = k/v. Yes. This is the time interval for the origin of the unprimed frame to move to the primed coordinate x'=-k. Unprimed Frame So, these from the time the origins are same to the time k reaches the unprimed origin, Δt = (k/γ)/v Now you're talking about different events. If you want to know where the primed coordinate x'=-k is at t=0 in the unprimed frame you need to calculate it: x=\gamma (x' + vt') = \gamma x' =-\gamma k Primed Frame So, these from the time the origins are same to the time unprimed origin reaches k, Δt' = k/v No. In the primed frame, the distance travelled by unprimed origin is \gamma k, which takes time \Delta t' = \frac{\gamma k}{v} There is agreement on the stop and start of these two events between the frames. Which two events? You've used three events here. So, in view of unprimed frame at rest, Δt = Δt'/γ for the coordinate k in the moving system to reach the unprimed origin. Thus, the moving clock beats time expanded. Which is the "moving" clock? It's better not to make assumptions about who is moving and who is not. Better to make sure you clearly specify which spacetime events you're talking about. Let's use Lorentz Transforms to check out work. We have x=0 because at origin in unprimed frame. t' = ( t - vx/c² )γ = t γ, so t = t'/γ, just as we calculated. So the event you're transforming here is located at x=0 at some time t=t. Note that you're not transforming a time interval here, but a coordinate. No time dilation for clock moving to the rest origin when in coordinates of moving frame. We just proved it. It's not clear what you proved. You mixed up two different events, for a start. chinglu12-30-10, 10:35 PMchinglu: Ok. But first let's make sure we know which events we're dealing with. The time interval you give here starts at t=0, when the origins of the primed and unprimed frames are in the same place. And at t=0 you're interested in the other point located at (x,t)=(-\frac{k}{\gamma},0). Now, let's set the clock at the origin in the primed frame to read t'=0 when the origins of the two frames are in the same place. Now, we calculate the coordinates of the other event: (x,t)=(-\frac{k}{\gamma},0); (x',t') = (-k, \frac{kv}{c^2}) Notice that the event at x=-k/\gamma does NOT occur at t'=0 in the primed frame. Yes. Yes. This is the time interval for the origin of the unprimed frame to move to the primed coordinate x'=-k. Now you're talking about different events. If you want to know where the primed coordinate x'=-k is at t=0 in the unprimed frame you need to calculate it: x=\gamma (x' + vt') = \gamma x' =-\gamma k No. In the primed frame, the distance travelled by unprimed origin is \gamma k, which takes time \Delta t' = \frac{\gamma k}{v} Which two events? You've used three events here. Which is the "moving" clock? It's better not to make assumptions about who is moving and who is not. Better to make sure you clearly specify which spacetime events you're talking about. So the event you're transforming here is located at x=0 at some time t=t. Note that you're not transforming a time interval here, but a coordinate. It's not clear what you proved. You mixed up two different events, for a start. No, I did not mix up anything. Let's keep it simple. Note that you're not transforming a time interval here, but a coordinate. Are you talking about a time coordinate? Be specific is it a GMT time or an elapsed time? What do you think time means under Lorentz transforms? Unprimed Frame So, these from the time the origins are same to the time k reaches the unprimed origin, Δt = (k/γ)/v Primed Frame So, these from the time the origins are same to the time unprimed origin reaches k, Δt' = k/v I wrote above. True or false. James R12-30-10, 10:44 PMchinglu: No, I did not mix up anything. Yes you did. I have now explained it to you in posts #117, #118, #165 and #177. It seems you're a slow learner, because you keep repeating the same mistake over and over again. Let's keep it simple. Let's not dumb it down so much that we start to make mistakes. Note that you're not transforming a time interval here, but a coordinate. Are you talking about a time coordinate? Be specific is it a GMT time or an elapsed time? Every event in spacetime has coordinates. To specify the coordinates of an event, you need a location and the time the event occurs. For example, (x,t) or (x',t'). Coordinates for the same event are generally different in different frames. An interval is the difference between two coordinates. For example, a time interval is \Delta t = t_1 - t_2 where t_1, t_2 are the times of two different events. A length (space interval) is \Delta x = x_1 - x_2 where x_1, x_2 are the two locations in space where two different events occur. You can also have a spacetime interval, which takes both the x and t coordinates into account. What do you think time means under Lorentz transforms? Time is a coordinate. The Lorentz transformations tell us how to transform a coordinate from one frame to another. Unprimed Frame So, these from the time the origins are same to the time k reaches the unprimed origin, Δt = (k/γ)/v Primed Frame So, these from the time the origins are same to the time unprimed origin reaches k, Δt' = k/v I wrote above. True or false. Read post #177! I went over it all in detail for you. chinglu12-30-10, 11:05 PMchinglu: Yes you did. I have now explained it to you in posts #117, #118, #165 and #177. It seems you're a slow learner, because you keep repeating the same mistake over and over again. Let's not dumb it down so much that we start to make mistakes. Every event in spacetime has coordinates. To specify the coordinates of an event, you need a location and the time the event occurs. For example, (x,t) or (x',t'). Coordinates for the same event are generally different in different frames. An interval is the difference between two coordinates. For example, a time interval is \Delta t = t_1 - t_2 where t_1, t_2 are the times of two different events. A length (space interval) is ]Delta x = x_1 - x_2 where x_1, x_2 are the two locations in space where two different events occur. You can also have a spacetime interval, which takes both the x and t coordinates into account. Time is a coordinate. The Lorentz transformations tell us how to transform a coordinate from one frame to another. Read post #177! I went over it all in detail for you. Let's try to remember what you said. You wrote here in the primed frame ∆t' = k/v. http://www.sciforums.com/showpost.php?p=2668002&postcount=158 You wrote here in the unprimed frame ∆t = (k/γ)/v. http://www.sciforums.com/showpost.php?p=2669157&postcount=175 I assume the frames agree on the start event when the frames are the same. We also have your previous agreements above. Now, there is only the question as to whether when k and origin of unprimed are same in primed frame and when k and unprimed origin are same in unprimed frame. If you disagree, perhaps you can explain why these events are not the same. If you try to use time, then you are confused since time does not work between the frames. Note that your calculations try to cross frames and predict some time on the clock in the primed frame when calculating in the unprimed frame. We do not need this. We simply look at each frame's calculations individually. This method you use above is called frame mixing. James R01-01-11, 12:42 AMchinglu: I assume the frames agree on the start event when the frames are the same. An event at the origin is one event. An event at x=-k is a different event. Those two spatially-separated events can only be simultaneous in one frame. They cannot be simultaneous in two frames. Note that your calculations try to cross frames and predict some time on the clock in the primed frame when calculating in the unprimed frame. We do not need this. We simply look at each frame's calculations individually. If you're only doing calculations in one frame, then you never need to worry about Lorentz transformations etc. But from the start of this thread you have been trying to compare time intervals in TWO frames. That requires translating events from one frame to another. And that means you need to make sure you're actually using the same event in both frames. Your constant mistake has been to assume that two different events are actually the same event. This method you use above is called frame mixing. The only method I have used is to apply the Lorentz transformation correctly to the events you have specified. You, on the other hand, have continually tried to bypass the hard work, which has meant that you have made the same error over and over again. Complete explanations of where you went wrong can be found in posts #117, #118, #165 and #177. rpenner01-01-11, 02:50 AMYou are less than honest. Suitable for this debate mutatis mutandis. http://atheismresource.com/wp-content/uploads/Debate-Flow-Chart.jpg chinglu01-01-11, 10:38 AMchinglu: An event at the origin is one event. An event at x=-k is a different event. Those two spatially-separated events can only be simultaneous in one frame. They cannot be simultaneous in two frames. If you're only doing calculations in one frame, then you never need to worry about Lorentz transformations etc. But from the start of this thread you have been trying to compare time intervals in TWO frames. That requires translating events from one frame to another. And that means you need to make sure you're actually using the same event in both frames. Your constant mistake has been to assume that two different events are actually the same event. The only method I have used is to apply the Lorentz transformation correctly to the events you have specified. You, on the other hand, have continually tried to bypass the hard work, which has meant that you have made the same error over and over again. Complete explanations of where you went wrong can be found in posts #117, #118, #165 and #177. What we have thus far is we both agree in the unrprimed frame, Δt = (-k/γ)/v What we have thus far is we both agree in the unrprimed frame, Δt' = (-k)/v Now, we both agree the origns same are same event. Next, we both agree when k and unprimed origin same, same event. So, we are both in agreement both frames can agree on the start event and the end event. Your calculations in #117, #118 do not equal to your conclusions of Δt = (-k/γ)/v Δt' = (-k)/v So, your own calculations are contradicting your own statements. As we can see, we compare the time intervals of the start and end events. We have no choice but to conclude Δt'/γ = Δt This mean, when primed frame is stationary, moving origin is time dilated or Δt' = Δtγ When unprimed frame is stationary, moving k is time expanded or Δt'/γ = Δt. This is also consistent with the invertibility of the LT matrix. Last, If coordinate k, then apply Lorentz Transforms t = ( t' + vx/c² )γ t = ( (-k)/v + v(-k)/c²)γ t = -k/v( 1 + v²/c²)γ = -(k/γ)/v Now, go other way. t' = ( t - vx/c² )γ t' = ( (-k/γ)/v - v(0)/c² )γ = -k/v We can see my calculations are the same as Lorentz transform and the results are correct with invertible LT matrix. Your calculations do not calculate Δt = (-k/γ)/v Δt' = (-k)/v You calculations do not consistent with invertibility of the LT matrix. chinglu01-01-11, 10:40 AMSuitable for this debate mutatis mutandis. http://atheismresource.com/wp-content/uploads/Debate-Flow-Chart.jpg Not sure what this has to with math. You calculations are not Lorentz Matrix invertible. So your math wrong. rpenner01-01-11, 11:13 AMNot sure what this has to with math. You calculations are not Lorentz Matrix invertible. So your math wrong. It's remarkable that you would make this claim in light of Post #18 (http://sciforums.com/showthread.php?p=2663466#post2663466) where the Lorentz transform takes S to S' and the inverse Lorentz transform takes S' to S for both events and intervals. Indeed, in Post 147 (http://sciforums.com/showthread.php?p=2667181#post2667181), Appendix [2], it was demonstrated that the inverse Lorentz transform composed with the Lorentz transform is always the identity transform, and you do not begin to explain how this could not be the case. chinglu01-01-11, 04:20 PMIt's remarkable that you would make this claim in light of Post #18 (http://sciforums.com/showthread.php?p=2663466#post2663466) where the Lorentz transform takes S to S' and the inverse Lorentz transform takes S' to S for both events and intervals. Indeed, in Post 147 (http://sciforums.com/showthread.php?p=2667181#post2667181), Appendix [2], it was demonstrated that the inverse Lorentz transform composed with the Lorentz transform is always the identity transform, and you do not begin to explain how this could not be the case. Here is normal Lorentz Transform inversion. Let x' and t' be given and time-like interval. x = ( x' + vt' ) γ t = ( t' + vx'/c² ) γ Now, given t and x have been calculated as above, x' = ( x - vt ) γ Plug in above x and t as calculated. x' = ( [ ( x' + vt' ) γ] - v[( t' + vx'/c² ) γ] ) γ x' = γ²( ( x' + vt' ) - v( t' + vx'/c² ) ) x' = γ²( x' - v²x'/c² ) ) x' = γ²( x' ( 1 - v²/c² ) ) x' = x' . This is invertability. But, your claim, if unprimed rest frame, always t = t' γ since moving frame interval is always time dilated. Let's try your theory. x' = ( x - vt ) γ Plug in x = ( x' + vt' ) γ but, according to your theory, must plug in t = t' γ always. x' = ( ( x' + vt' ) γ - v(t' γ)) γ x' = γ²( ( x' + vt' ) - vt') x' = γ² x' 1 = γ² Only if v = 0. So, your theory fails Lorentz Transform inversion test just like I said. James R01-01-11, 10:08 PMchinglu: There's little point in my repeating myself. You need to go back and read posts #117 and #118. If you like, you can also look at #165 and #177. Get back to me if you have any questions about those. What we have thus far is we both agree in the unrprimed frame, Δt = (-k/γ)/v What we have thus far is we both agree in the unrprimed frame, Δt' = (-k)/v You haven't specified which events you used to calculate these two time intervals. I did, in post #117. We can't "agree" about this unless you are specific about which events you used. Next, we both agree when k and unprimed origin same, same event. No. An event at x'=-k MUST be different to an event at the origin x'=0. Two different spatial locations = two different events. So, we are both in agreement both frames can agree on the start event and the end event. Yes. They can agree if you specify which events you're talking about. Your calculations in #117, #118 do not equal to your conclusions of Δt = (-k/γ)/v Δt' = (-k)/v So, your own calculations are contradicting your own statements. No. There's no contradiction. Nor have you shown any. You calculations do not consistent with invertibility of the LT matrix. Then go through post #117 and show exactly where I made a mistake. Go through it line by line. We've had almost 100 posts since then, and you've produced nothing new. rpenner01-02-11, 04:27 AMBut, your claim, if unprimed rest frame, always t = t' γ since moving frame interval is always time dilated. Let's try your theory. x' = ( x - vt ) γ Plug in x = ( x' + vt' ) γ but, according to your theory, must plug in t = t' γ always. x' = ( ( x' + vt' ) γ - v(t' γ)) γ x' = γ²( ( x' + vt' ) - vt') x' = γ² x' 1 = γ² Only if v = 0. So, your theory fails Lorentz Transform inversion test just like I said. My claim (which the second table of post 18 (http://sciforums.com/showthread.php?p=2663466#post2663466) makes explicit) is that if clock is at rest in the unprimed frame, then: Δx = 0, and each tick of the clock has a certain amount of time Δt > 0, while in the primed coordinates the same object is seen by the moving observer as moving at speed -v, or Δx' = −vΔt', and the moving observer sees the time between ticks as Δt' = γΔt > Δt. This is entirely consistent with the Lorentz transforms, which simplify with Δx = 0 as: Δx' = γ(Δx − vΔt) = −vγΔt = −vΔt' Δt' = γ(Δt − vΔx/c²) = γΔt Likewise entirely consistent with the inverse Lorentz transforms of Δx' = −vΔt' Δx = γ(Δx' + vΔt') = γ(−vΔt' + vΔt') = 0 Δt = γ(Δt' + vΔx'/c²) = γ(Δt' −v²Δt'/c²) = Δt'/γ chinglu01-02-11, 06:41 PMMy claim (which the second table of post 18 (http://sciforums.com/showthread.php?p=2663466#post2663466) makes explicit) is that if clock is at rest in the unprimed frame, then: Δx = 0, and each tick of the clock has a certain amount of time Δt > 0, while in the primed coordinates the same object is seen by the moving observer as moving at speed -v, or Δx' = −vΔt', and the moving observer sees the time between ticks as Δt' = γΔt > Δt. This is entirely consistent with the Lorentz transforms, which simplify with Δx = 0 as: Δx' = γ(Δx − vΔt) = −vγΔt = −vΔt' Δt' = γ(Δt − vΔx/c²) = γΔt Likewise entirely consistent with the inverse Lorentz transforms of Δx' = −vΔt' Δx = γ(Δx' + vΔt') = γ(−vΔt' + vΔt') = 0 Δt = γ(Δt' + vΔx'/c²) = γ(Δt' −v²Δt'/c²) = Δt'/γ You are now consistent with Lorentz Matrix invertibility. Let's take unprimed as stationary and use your calculations. that mean primed frame moving and use information in the rest frame to calculate the moving frame. You have Δt' = γ(Δt − vΔx/c²) = γΔt, thus Δt'/γ = Δt. By your own post, the moving frame is time expanded or beats faster than the at rest unprimed frame. quod erat demonstrandum chinglu01-02-11, 06:50 PMchinglu: There's little point in my repeating myself. You need to go back and read posts #117 and #118. If you like, you can also look at #165 and #177. Get back to me if you have any questions about those. You haven't specified which events you used to calculate these two time intervals. I did, in post #117. We can't "agree" about this unless you are specific about which events you used. No. An event at x'=-k MUST be different to an event at the origin x'=0. Two different spatial locations = two different events. Yes. They can agree if you specify which events you're talking about. No. There's no contradiction. Nor have you shown any. Then go through post #117 and show exactly where I made a mistake. Go through it line by line. We've had almost 100 posts since then, and you've produced nothing new. There is little point in repeating myself. I told you step2 is an error for the problem. You assigned a negative elapsed time to the k coordinate from the view of the unprimed frame. The unprimed frame does not care the time on that clock. It has nothing to do with anything. the unprimed frame is only timing the arrival of k to the unprimed origin in which you already confessed that time is Δt = (-k/γ)/v You also confessed Δt' = -k/v The ratio Δt'/Δt = γ which means the moving clock k beats time expanded. Do not forget, the ratio Δt'/Δt = γ is based on confessions you made. James R01-02-11, 07:43 PMThere is little point in repeating myself. I told you step2 is an error for the problem. You assigned a negative elapsed time to the k coordinate from the view of the unprimed frame. The unprimed frame does not care the time on that clock. It has nothing to do with anything. If you wish to compare time intervals for two events in two different frames, you need to make sure you're dealing with the same events in both frames. You did not do that. I have explained your errors quite clearly in posts #117, #118, #165 and #177. You have not directly addressed any of those posts. There's no point in talking to you further. You are unteachable. Goodbye. chinglu01-02-11, 07:56 PMIf you wish to compare time intervals for two events in two different frames, you need to make sure you're dealing with the same events in both frames. You did not do that. I have explained your errors quite clearly in posts #117, #118, #165 and #177. You have not directly addressed any of those posts. There's no point in talking to you further. You are unteachable. Goodbye. Wrong. I did that over and over. The start is origin same. The end is k same with unprimed origin. I asked you over and over if this end event could not be agreed upon. You never answered. The fact is it can. I clearly set up intervals that you also agreed upon. That proved dt'/dt = γ. I explained to you over and over your statement 2 in your posts has not issue on the problem. rpenner01-03-11, 01:35 AMBut, your claim, if unprimed rest frame, always t = t' γ since moving frame interval is always time dilated. Let's try your theory. x' = ( x - vt ) γ Plug in x = ( x' + vt' ) γ but, according to your theory, must plug in t = t' γ always. x' = ( ( x' + vt' ) γ - v(t' γ)) γ x' = γ²( ( x' + vt' ) - vt') x' = γ² x' 1 = γ² Only if v = 0. So, your theory fails Lorentz Transform inversion test just like I said. My claim (which the second table of post 18 (http://sciforums.com/showthread.php?p=2663466#post2663466) makes explicit) is that if clock is at rest in the unprimed frame, then: Δx = 0, and each tick of the clock has a certain amount of time Δt > 0, while in the primed coordinates the same object is seen by the moving observer as moving at speed -v, or Δx' = −vΔt', and the moving observer sees the time between ticks as Δt' = γΔt > Δt. This is entirely consistent with the Lorentz transforms, which simplify with Δx = 0 as: Δx' = γ(Δx − vΔt) = −vγΔt = −vΔt' Δt' = γ(Δt − vΔx/c²) = γΔt Likewise entirely consistent with the inverse Lorentz transforms of Δx' = −vΔt' Δx = γ(Δx' + vΔt') = γ(−vΔt' + vΔt') = 0 Δt = γ(Δt' + vΔx'/c²) = γ(Δt' −v²Δt'/c²) = Δt'/γ You are now consistent with Lorentz Matrix invertibility.I think you mean that I have always been consistent with the use of Lorentz transforms and inverse Lorentz transforms, since I quote you results from post 18 (http://sciforums.com/showthread.php?p=2663466#post2663466). But thank you for realizing this a mere 170 posts later. Let's take unprimed as stationary and use your calculations. that mean primed frame moving and use information in the rest frame to calculate the moving frame.It is a mistake to call the frames or clocks or objects as moving or stationary except in relationship to a labelled frame. Every distinct event has a unique set of space-time coordinates in every inertial frame -- these sets of coordinates are like the names of objects and every frame is a different language. Information doesn't flow between the frames on an event-by-event basis. The Lorentz transform is a dictionary which for a given pair of frames lets us look up the name (coordinates) of and event in the other frame. You have Δt' = γ(Δt − vΔx/c²) = γΔt, thus Δt'/γ = Δt. I have Δx between two events equal to zero in the unprimed frame and therefore in a different frame where Δx' is not equal to zero, then Δt' > Δt. In the most important case for this, the two events are a clock measuring 11:59:59 and the same clock measuring 12:00:00, so that Δx = 0 and Δt = 1 second means the clock is not moving for the unprimed observer and it measures one second in one second. Since γ ≠ 0, Δt' = γΔt and Δt'/γ = Δt represent the same algebraic statement. Importantly since v ≠ 0 means γ > 1, both Δt' = γΔt and Δt'/γ = Δt mean Δt' > Δt. By your own post, the moving frame is time expanded or beats faster than the at rest unprimed frame. The frame is not time-expanded, nor does it beat. The frame is a system of space-time coordinates assigned to events. If the unprimed frame is the coordinate system where we agree that a certain clock is stationary, then the primed frame is the frame where we have agreed that the clock is moving. The moving clock is time dilated, which means each tick of the moving clock needs a longer time than the unmoving clock. Δt' > Δt. But this means the beats slower since faster and slower refer to how many ticks pass per amount of time. So if the start and end events are ticks of the the clock, then Δt' > Δt means that the primed frame measures the lower tick frequency for the moving clock than then unprimed frame measures for the stationary clock, 1/Δt' < 1/Δt. For the same two events, the primed coordinates see the clock move, since Δx' ≠ 0, and since Δt' > 1 second, the "moving clock" is seen in the primed coordinates to need more than 1 second to move from 11:59:59 to 12:00:00. Therefore, you have gotten it backwards. If Δx = 0 and v ≠ 0, then γ > 1 and Δt' = γΔt means Δt' > Δt, so that if Δt represents the time between two ticks of a stationary clock, then Δt' represents the time between two ticks of a moving clock and Δt' > Δt means that the rate of the ticks is slower for the moving clock 1/Δt' < 1/Δt. quod erat demonstrandum Arrogant and wrong when remedial concepts like what "beats faster" or "frame" or "event" or "interval" have yet to have been mastered. James R01-03-11, 02:31 AMchinglu: If you can find a mistake in posts #117, #118, #165 and #177, then get back to me. Don't write one or two lines saying I have said something I have not said. You need to go through post #117 line by line and show me the point where I made an error, if any such error exists. Otherwise, don't bother writing to me again. I have wasted enough time on you. After four separate explanations of the same point, plus innumerable posts in between breaking it down into bite-sized chunks for you, you seem incapable of understanding a simple point. Either you're stupid, or your a troll. Either way, you're a waste of my time. chinglu01-03-11, 08:07 PMI think you mean that I have always been consistent with the use of Lorentz transforms and inverse Lorentz transforms, since I quote you results from post 18 (http://sciforums.com/showthread.php?p=2663466#post2663466). But thank you for realizing this a mere 170 posts later. Let's look at your post #18 all the way to the right with time. You said, for this same exercise, t' = t/γ for unprimed frame stationary and t' = tγ for primed frame take stationary. Afrer all these pages, you still do not understand this is not LT invertible. That is the proof I wrote for you. This mean you cannot even tell when you step into a contradiction. [QUOTE]It is a mistake to call the frames or clocks or objects as moving or stationary except in relationship to a labelled frame. Every distinct event has a unique set of space-time coordinates in every inertial frame -- these sets of coordinates are like the names of objects and every frame is a different language. Information doesn't flow between the frames on an event-by-event basis. The Lorentz transform is a dictionary which for a given pair of frames lets us look up the name (coordinates) of and event in the other frame. So, what is the semantics of this dictionary? When you have my understanding, you will not step into contradiction. Anyway, show readers you know the semantics. I have Δx between two events equal to zero in the unprimed frame and therefore in a different frame where Δx' is not equal to zero, then Δt' > Δt. In the most important case for this, the two events are a clock measuring 11:59:59 and the same clock measuring 12:00:00, so that Δx = 0 and Δt = 1 second means the clock is not moving for the unprimed observer and it measures one second in one second. Please explain if Δx = 0, how are you judging the start and stop of events? In other words, if hold clock with Δx = 0, how do you know when to stop time? Do you get a message from God? You need to specify this. Since γ ≠ 0, Δt' = γΔt and Δt'/γ = Δt represent the same algebraic statement. Importantly since v ≠ 0 means γ > 1, both Δt' = γΔt and Δt'/γ = Δt mean Δt' > Δt. Where exactly do you have reciprocal time dialation, since that is not SR? You seem to be non-mainsteam here which may make you a crackpot. Now, if you try to prove reciprocal time dilation, I will run you into a contradiction using the invertibility of the LT matrix. The frame is not time-expanded, nor does it beat. The frame is a system of space-time coordinates assigned to events. If the unprimed frame is the coordinate system where we agree that a certain clock is stationary, then the primed frame is the frame where we have agreed that the clock is moving. I can say what I want. I say the frame beats a time beat. Einstein said the frame has but one time. Are you refuiting this also? It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it the time of the stationary system.'' http://www.fourmilab.ch/etexts/einstein/specrel/www/ The moving clock is time dilated, which means each tick of the moving clock needs a longer time than the unmoving clock. Δt' > Δt. But this means the beats slower since faster and slower refer to how many ticks pass per amount of time. So if the start and end events are ticks of the the clock, then Δt' > Δt means that the primed frame measures the lower tick frequency for the moving clock than then unprimed frame measures for the stationary clock, 1/Δt' < 1/Δt. Can you please show this is consistent with the invertibility of the LT matrix given the same events between the frames? No you can not. Arrogant and wrong when remedial concepts like what "beats faster" or "frame" or "event" or "interval" have yet to have been mastered. Too bad. You continue to contradict yourself after all these pages where I showed you reciprocal time dilation is inconsistent with the invertibility of the LT matrix. I just keep hammering you over and over on this and you never learn. chinglu01-03-11, 08:11 PMchinglu: If you can find a mistake in posts #117, #118, #165 and #177, then get back to me. Don't write one or two lines saying I have said something I have not said. You need to go through post #117 line by line and show me the point where I made an error, if any such error exists. Otherwise, don't bother writing to me again. I have wasted enough time on you. After four separate explanations of the same point, plus innumerable posts in between breaking it down into bite-sized chunks for you, you seem incapable of understanding a simple point. Either you're stupid, or your a troll. Either way, you're a waste of my time. You have agreed that t' = k/v. and t = (k/γ)/v for the same events for the 2 frames. Is there any way possible way you can at least admit what you agreed to? James R01-03-11, 08:27 PMchinglu: You give me the spacetime coordinates for the two events I have supposedly agreed about. Make sure you give the coordinates of both events in both the unprimed and primed frame. Then we can talk further. Please express the coordinates in the form (x,t)= whatever and (x',t')= whatever. chinglu01-05-11, 06:38 PMchinglu: You give me the spacetime coordinates for the two events I have supposedly agreed about. Make sure you give the coordinates of both events in both the unprimed and primed frame. Then we can talk further. Please express the coordinates in the form (x,t)= whatever and (x',t')= whatever. Yes, for better use x' = -k where k > 0 Event 1 (x,t)=( 0,0 ) (x',t')=( 0,0) Event 2 (x,t)=( 0,(-k/γ)/v ) (x',t')=( k,-k/v ) Event 1 is always true for standard configuration. Event 2 Calculate from x frame. (x,t)=( 0,(-k/γ)/v ) x' = ( x - vt )γ x' = ( 0 - v((-k/γ)/v )γ x' = k t' = ( t - vx/c² )γ t' = ( (-k/γ)/v - v(0)/c² )γ t' = -k/v Calculate from x' frame. (x',t')=( k,-k/v ) x = ( x' + vt' )γ x = ( k + v(-k/v) )γ x = 0 t = ( t' + vx'/c² )γ t = ( -k/v + vk/c² )γ t = k/v( -1 + v²/c² )γ t = (-k/v)/γ rpenner01-05-11, 07:44 PMIf 0 < v < c, event 2 happens now before event 1. Also, event 2 happens at x' = k, not x' = -k. chinglu01-05-11, 08:42 PMIf 0 < v < c, event 2 happens now before event 1. Also, event 2 happens at x' = k, not x' = -k. Also, event 2 happens at x' = k, not x' = -k I may have made a mistake, I have before. I have above, Event 2 (x,t)=( 0,(-k/γ)/v ) (x',t')=( k,-k/v ) x' = ( 0 - v((-k/γ)/v )γ x' = k No event 2 does not happend before event 1. Event 1 is the originss being same. Event 2 is the unprimed origin and k being same. rpenner01-05-11, 09:55 PMYes, for better use x' = -k where k > 0 Event 1 (x,t)=( 0,0 ) (x',t')=( 0,0) Event 2 (x,t)=( 0,(-k/γ)/v ) (x',t')=( k,-k/v ) If 0 < v < c, event 2 happens now before event 1. Also, event 2 happens at x' = k, not x' = -k. Also, event 2 happens at x' = k, not x' = -k I may have made a mistake, I have before. I have above, Event 2 (x,t)=( 0,(-k/γ)/v ) (x',t')=( k,-k/v ) x' = ( 0 - v((-k/γ)/v )γ x' = k No event 2 does not happend before event 1. Event 1 is the originss being same. Event 2 is the unprimed origin and k being same. If 0 < v < c (like I say) and if k > 0 (like you say) Event 2 has t < 0 and t' < 0, so it happened before Event 1 (where, t = t' = 0) , exactly like 1989 < 2011 so 1989 came before 2011. Why would you even try to say the opposite? James R01-06-11, 11:29 PMEvent 1 (x,t)=( 0,0 ) (x',t')=( 0,0) Event 2 (x,t)=( 0,(-k/γ)/v ) (x',t')=( k,-k/v ) Event 1 is always true for standard configuration. Event 2 Calculate from x frame. (x,t)=( 0,(-k/γ)/v ) x' = ( x - vt )γ x' = ( 0 - v((-k/γ)/v )γ x' = k t' = ( t - vx/c² )γ t' = ( (-k/γ)/v - v(0)/c² )γ t' = -k/v Calculate from x' frame. (x',t')=( k,-k/v ) x = ( x' + vt' )γ x = ( k + v(-k/v) )γ x = 0 t = ( t' + vx'/c² )γ t = ( -k/v + vk/c² )γ t = k/v( -1 + v²/c² )γ t = (-k/v)/γ Ok. That all looks right. Since the position coordinate does not change in the x frame between events 1 and 2, the time interval in the x frame is a proper time for these two events. The time interval in the x' frame is not a proper time, so we should expect it to be longer than the x time for the same two events. And it is! It's longer by a factor of \gamma. It looks like you've got these calculations correct. Any questions? Are you sorted now? chinglu01-07-11, 06:53 PMIf 0 < v < c (like I say) and if k > 0 (like you say) Event 2 has t < 0 and t' < 0, so it happened before Event 1 (where, t = t' = 0) , exactly like 1989 < 2011 so 1989 came before 2011. Why would you even try to say the opposite? Yes, k < 0 for x'. I typed it wrong. The rest falls into place under that k < 0 assumption. chinglu01-07-11, 06:56 PMOk. That all looks right. Since the position coordinate does not change in the x frame between events 1 and 2, the time interval in the x frame is a proper time for these two events. The time interval in the x' frame is not a proper time, so we should expect it to be longer than the x time for the same two events. And it is! It's longer by a factor of \gamma. It looks like you've got these calculations correct. Any questions? Are you sorted now? Thanks, that was the point of this thread. If the primed frame is the moving frame, then it elapses more time then the stationary unprimed frame origin. That is not time dilation. arfa brane01-07-11, 07:12 PMLength contraction in a moving frame, as seen by an observer (that is, motion relative to their frame of reference) is a consequence of measuring say, the ends of a moving rod simultaneously. The length is contracted as a consequence of the Lorentz transformation. Time is dilated for the same reason the length is contracted, when you invert the transformation. This is entirely consistent with transforming a frequency into a wavelength when the source is moving relative to an observer. If the primed frame is the moving frame, then it elapses more time then the stationary unprimed frame origin. That is not time dilation. If the primed frame is the moving frame, events appear to take longer in that frame than the "stationary" frame. That IS time dilation, and it's relative to the observer of the moving frame. What would you call it? James R01-07-11, 07:33 PMThanks, that was the point of this thread. If the primed frame is the moving frame, then it elapses more time then the stationary unprimed frame origin. That is not time dilation. Nothing in the above calculation says which frame is "moving" and which is not. chinglu01-07-11, 08:42 PMLength contraction in a moving frame, as seen by an observer (that is, motion relative to their frame of reference) is a consequence of measuring say, the ends of a moving rod simultaneously. The length is contracted as a consequence of the Lorentz transformation. Time is dilated for the same reason the length is contracted, when you invert the transformation. This is entirely consistent with transforming a frequency into a wavelength when the source is moving relative to an observer. If the primed frame is the moving frame, events appear to take longer in that frame than the "stationary" frame. That IS time dilation, and it's relative to the observer of the moving frame. What would you call it? The math is above. Feel free. chinglu01-07-11, 08:42 PMNothing in the above calculation says which frame is "moving" and which is not. I have said over and over unprimed stationary. Does not matter. arfa brane01-07-11, 08:53 PMIf the primed frame is the moving frame, then it elapses more time then the stationary unprimed frame origin. That is not time dilation. If that last sentence is what you think, then what do YOU THINK it should be called? Feel free (I know I do). chinglu01-07-11, 09:04 PMIf that last sentence is what you think, then what do YOU THINK it should be called? Feel free (I know I do). t = (-k/v)/γ and t' = (-k/v) It does not matter which frame is stationary. t' is always larger. I am not sure what to call this mathematical fact. arfa brane01-07-11, 09:09 PMI am not sure what to call this mathematical fact. Well if you aren't sure, it's probably acceptable to call it "Lorentz contraction". Seeing that it's a consequence of the Lorentz transformation of moving coordinates relative to a stationary observer. Hell, you could get carried away and call it "a principle of relativity of simultaneity", but that's harder to type. chinglu01-07-11, 09:25 PMWell if you aren't sure, it's probably acceptable to call it "Lorentz contraction". Seeing that it's a consequence of the Lorentz transformation of moving coordinates relative to a stationary observer. Hell, you could get carried away and call it "a principle of relativity of simultaneity", but that's harder to type. I suppose you could call it R of S. In fact, this is a consequence of it. I am not sure about the "Lorentz contraction", because that is something completely different. arfa brane01-07-11, 10:23 PMI am not sure about the "Lorentz contraction", because that is something completely different. To what? The contraction of length and the dilation of time (together) is a consequence of the transformation. Ergo, distance and time in moving coordinate systems are both transformed (in parallel) relative to local coordinate systems chinglu01-08-11, 05:07 PMTo what? The contraction of length and the dilation of time (together) is a consequence of the transformation. Ergo, distance and time in moving coordinate systems are both transformed (in parallel) relative to local coordinate systems I doubt we communicate. The point of this thread is that a clock moving toward the origin between the start event of both origins being same to the end event of moving clock at same place as origin will beat faster than the clock of the stationary origin. That is what alll the valid math in this thread has proven. arfa brane01-08-11, 05:47 PMSo because an observer sees a clock beat faster, that can't be time dilation? If another observer is with the 'faster' clock, won't the other clock seem to be slower than their one? I think so. Any other possibility just doesn't seem logical, Jim. James R01-08-11, 08:52 PMI have said over and over unprimed stationary. Does not matter. You're right. It doesn't matter which frame you call "stationary", because an observer at rest in either frame can consider himself stationary. t = (-k/v)/γ and t' = (-k/v) It does not matter which frame is stationary. t' is always larger. I am not sure what to call this mathematical fact. t' is not always larger. It depends on which events you're looking at, obviously. chinglu02-03-11, 08:27 PMThe point of this thread is that a clock moving toward the origin between the start event of both origins being same to the end event of moving clock at same place as origin will beat faster than the clock of the stationary origin. That is what all the valid math in this thread has proven. James R02-03-11, 09:14 PMComplete explanations of where you went wrong can be found in posts #117, #118, #165 and #177 of the current thread. chinglu02-04-11, 07:54 PMComplete explanations of where you went wrong can be found in posts #117, #118, #165 and #177 of the current thread. In step 3 #117, you claimed (x',t') = (-k, k/v) when t'=t=0 with k > 0. Assume a clock is located at -k in the primed system. Now, let's shoot a light beam at that clock when the origins are the same. Hence, while light moves c from the origin, x' moves toward the light at v cutting the distance light must travel. ct = vt + k/γ t = k/(γ(c-v)) x = -k/γ According to LT, t' = ( t - vx/c² )γ Plug in the numbers. t' = ( k/(γ(c-v)) - v(-k/γ)/c² )γ t' = k/(c-v) + vk/c² But, this assume the clock at (x',t') = (-k, k/v) had t'= 0. We add k/v to the result above and have t'failed = k/(c-v) + vk/c² + k/v. Thus, you contradict LT. arfa brane02-04-11, 09:40 PMa clock moving toward the origin between the start event of both origins being same to the end event of moving clock at same place as origin will beat faster than the clock of the stationary origin. This is a bit impenetrable. The stationary origin doesn't make sense unless there is a clock there. But you say "the start event of both origins being same", but there is a moving clock, which can't be at the origin (or it would be stationary), then the moving clock is "at the same place", eventually. If there are two clocks in your scenario they can only be in the same stationary frame if they aren't moving relative to each other. But you say one of them is moving toward the origin, so it can't be at the origin and it can't be in a stationary frame. Pete02-04-11, 10:04 PMIn step 3 #117, you claimed (x',t') = (-k, k/v) when t'=t=0 with k > 0. Assume a clock is located at -k in the primed system. Now, let's shoot a light beam at that clock when the origins are the same. Hence, while light moves c from the origin, x' moves toward the light at v cutting the distance light must travel. ct = vt + k/γ The clock is at -k, not at k, and the light is moving in the negative direction (is going at -c). So: -ct = vt - k/\gamma t = \frac{k}{\gamma(c+v)} x = -k/\gamma LT... t' = (t - \frac{vx}{c^2})\gamma t' = (\frac{k}{\gamma(c+v)} + \frac{vk}{\gamma c^2})\gamma t' = \frac{k(v+1)}{c(v+1)} t' = k/c Which is, of course, consistent with light moving at -c from x'=0 at t'=0 to x'=-k at t'=k/c. chinglu02-04-11, 11:50 PMThe clock is at -k, not at k, and the light is moving in the negative direction (is going at -c). So: -ct = vt - k/\gamma t = \frac{k}{\gamma(c+v)} x = -k/\gamma LT... t' = (t - \frac{vx}{c^2})\gamma t' = (\frac{k}{\gamma(c+v)} + \frac{vk}{\gamma c^2})\gamma t' = \frac{k(v+1)}{c(v+1)} t' = k/c Which is, of course, consistent with light moving at -c from x'=0 at t'=0 to x'=-k at t'=k/c. Looks Ok, should have checked my work. that is because that part is not relevent. The clock at x, corresponding to x'=-k had t=-γkv/c², when the origins were the same. Either way, it is a contradiction since you would have applied the wrong t in LT. So, you backed a losing position and find a +- error in my calculations. Thanks for your attention to the detail. Pete02-05-11, 03:06 AMRight, I was hasty and didn't check things through. There's some really dodgy arithmetic in there! This step is embarrassingly wrong: t' = (\frac{k}{\gamma(c+v)} + \frac{vk}{\gamma c^2})\gamma t' = \frac{k(v+1)}{c(v+1)} So, let's try again, carefully: Assume a clock is located at -k in the primed system. Now, let's shoot a light beam at that clock when the origins are the same. Hence, while light moves c from the origin, x' moves toward the light at v cutting the distance light must travel. We've fixed this bit: ct = vt + k/γ t = k/(γ(c-v)) -ct = vt - k/\gamma t = \frac{k}{\gamma(c+v)} But there's another mistake here: x = -k/γ The clock at x'=-k was at x=-k/γ when the t clock there had t=0 (i.e. in the unprimed frame, this is when the light beam started.) But, we just figured out that the light doesn't reach that clock until t = k/γ(c+v) By that time, the clock at x'=-k has moved from x=-k/γ to: x = -k/\gamma + vt x = -k/\gamma + \frac{vk}{\gamma(c+v)} According to LT, t' = ( t - vx/c² )γ Plug in the numbers. Plugging in the numbers suddenly got a lot harder! chinglu02-05-11, 06:12 PMRight, I was hasty and didn't check things through. There's some really dodgy arithmetic in there! This step is embarrassingly wrong: So, let's try again, carefully: We've fixed this bit: -ct = vt - k/\gamma t = \frac{k}{\gamma(c+v)} But there's another mistake here: The clock at x'=-k was at x=-k/γ when the t clock there had t=0 (i.e. in the unprimed frame, this is when the light beam started.) But, we just figured out that the light doesn't reach that clock until t = k/γ(c+v) By that time, the clock at x'=-k has moved from x=-k/γ to: x = -k/\gamma + vt x = -k/\gamma + \frac{vk}{\gamma(c+v)} Plugging in the numbers suddenly got a lot harder! Interesting post. If you will look at JamesR post #117 #2 in the stationary frame/unprimed frame, you will find an initial condition at x=-k/γ with a spooky at a distance desynchronization on the stationary frame clock which is not valid under SR. I would be curious of your interpretation of #2 before I proceed. James R02-05-11, 07:13 PMNothing new here, chinglu, except a bunch of new errors from you. I won't bother replying to your new errors. I'll wait for you to admit your old errors. chinglu02-05-11, 07:18 PMNothing new here, chinglu, except a bunch of new errors from you. I won't bother replying to your new errors. I'll wait for you to admit your old errors. Well, we have figured out in step 2 of the stationary frame of your post #117, you have added a spooky at a distance time to the clock there even though clocks in the stationary frame are assumed to be Einstein synched at t'=t=0. Can you explain that? James R02-05-11, 08:39 PMWell, we have figured out in step 2 of the stationary frame of your post #117, you have added a spooky at a distance time to the clock there even though clocks in the stationary frame are assumed to be Einstein synched at t'=t=0. Can you explain that? You first. Pete02-05-11, 10:21 PMInteresting post. If you will look at JamesR post #117 #2 in the stationary frame/unprimed frame, you will find an initial condition at x=-k/γ with a spooky at a distance desynchronization on the stationary frame clock which is not valid under SR. I would be curious of your interpretation of #2 before I proceed. I don't understand where you think the spooky at a distance stuff comes in. Are you objecting to this event transformation? 2. (x',t') = (-k,0) ... 2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) It looks fine to me. It just means that: if all the primed clocks are einstein synchronized with each other, and the primed clock at x'=0 and the unprimed clock at x=0 read t'=0 and t=0 respectively as they pass each other, then when the primed clock at x'=-k passes the unprimed clock at x=-\gamma k, the primed clock shows t'=0 and the unprimed clock shows t=-\frac{\gamma k v}{c^2}. Why do you think that is that valid under SR? What do you think that SR says those clocks read as they pass each other, given the specified synchronization? chinglu02-06-11, 05:41 PM2. (x,t)=(-\gamma k, -\frac{\gamma k v}{c^2}) To both posts above. We have taken the unprimed frame as stationary. The time component in (2) above implies the clocks are not Einstein synched in the unprimed frame. Pete02-06-11, 07:28 PMHow do you figure that? James R02-06-11, 11:19 PMThe time component in (2) above implies the clocks are not Einstein synched in the unprimed frame. We've already been through all this. Two spatially-separated events that are simultaneous in one frame CANNOT be simultaneous in any other frame. Do you agree? Yes or no? Therefore, if two events occur simultaneously in the primed frame (as your problem specified), then they cannot occur simultaneously in the unprimed frame. Do you agree? Yes or no? Pete02-07-11, 01:22 AMWe've already been through all this. Two events that are simultaneous in one frame CANNOT be simultaneous in any other frame. More specifically (I think): If two clocks, T1 and T2, are: Einstein synchronized with each other, and T1 is present at an event S1, and T1 is present at an event S2, and the time T1 records for S1 is the same time that T2 records for S2, and if two more clocks, T1' and T2', are: Einstein synchronized with each other, and T1' is present at S1, and T2' is present at S2, Then the time T1' records for S1 must be a different time that T2' records for S2. This is clearly much more wordy, but it avoids misunderstandings about what "simultaneous" means. chinglu02-07-11, 07:31 PMI can't understand why this is so complex. We are taking the view of the unprimed frame for this exercise. When the origins are the same, t'=t=0 at the origins. t=0 in all parts of the unprimed frame. You two are confused. Pete02-07-11, 09:34 PMI can't understand why this is so complex. We are taking the view of the unprimed frame for this exercise. When the origins are the same, t'=t=0 at the origins. t=0 in all parts of the unprimed frame. You two are confused. well, (x', t')=(-k,0) doesn't happen at t=0. chinglu02-09-11, 05:54 PMwell, (x', t')=(-k,0) doesn't happen at t=0. It doesn't make any difference. If you like, put 1,000,000,00 BC on the clock for the initial condition. LT is about elapsed times, ie dt. Pete02-09-11, 07:44 PMSorry chinglu, I don't know what you're thinking. Perhaps you could explain your difficulty with James's post #117 more explicitly? chinglu02-09-11, 07:45 PMLet's get factual. Any moving clock coming toward the rest origin is time expanded, not time dilated. Any moving clock moving away from the rest origin is time dilated chinglu02-09-11, 07:46 PMSorry chinglu, I don't know what you're thinking. Perhaps you could explain your difficulty with James's post #117 more explicitly? I have explained this. Let's get your misunderstanding. Is LT based on dt? Pete02-09-11, 08:27 PMNo, you haven't explained it. James's post is clear, your objection is not. To spell out the step you are objecting to: if all the primed clocks are einstein synchronized with each other, and the primed clock at x'=0 and the unprimed clock at x=0 read t'=0 and t=0 respectively as they pass each other, then when the primed clock at x'=-k passes the unprimed clock at x=-\gamma k, the primed clock shows t'=0 and the unprimed clock shows t=-\frac{\gamma k v}{c^2}. Where is the 'spooky action at a distance'? James R02-10-11, 04:12 AMLet's get factual. Any moving clock coming toward the rest origin is time expanded, not time dilated. No. All observers see moving clocks running slower than their own clocks. chinglu02-10-11, 07:45 PMNo, you haven't explained it. James's post is clear, your objection is not. To spell out the step you are objecting to: if all the primed clocks are einstein synchronized with each other, and the primed clock at x'=0 and the unprimed clock at x=0 read t'=0 and t=0 respectively as they pass each other, then when the primed clock at x'=-k passes the unprimed clock at x=-\gamma k, the primed clock shows t'=0 and the unprimed clock shows t=-\frac{\gamma k v}{c^2}. Where is the 'spooky action at a distance'? James R added the t=-\frac{\gamma k v}{c^2} to the elapsed time at x. This is invalid. chinglu02-10-11, 07:47 PMNo. All observers see moving clocks running slower than their own clocks. No, it is possible to conclude t'=t and in SR t' and t mean dt' and dt. So, the only way to make that happen is time expansion + time dilation. AlexG02-10-11, 09:16 PMBut nothing can be measured and reconciled until both frames are brought together in the same inertial framework. So it is not possible to conclude anything without taking into consideration the accelerated motion that at least one, or both frames must experience, to synchronize their relative velocities. James R02-10-11, 09:21 PMJames R added the t=-\frac{\gamma k v}{c^2} to the elapsed time at x. This is invalid. Nothing was "added". That result is a simple application of the Lorentz transformation to an event that YOU specified. No, it is possible to conclude t'=t and in SR t' and t mean dt' and dt. So, the only way to make that happen is time expansion + time dilation. It's hard to make any sense of this bizarre statement. If you have the same object at two different events, then the proper time is always the shortest time interval that can be measured between those two events. Every frame other than the rest frame of the object will measure a longer time for the same two events. Do you agree? Pete02-11-11, 12:02 AMJames R added the t=-\frac{\gamma k v}{c^2} to the elapsed time at x. This is invalid. Why do you think it is invalid? It comes straight out of the Lorentz transform: t' = 0, x' = -k t = \gamma(t' + \frac{vx'}{c^2}) t=-\frac{\gamma k v}{c^2} This just tells that the primed clock at x'=-k doesn't show t'=0 until it is passing an unprimed clock reading t=-\frac{\gamma k v}{c^2} What do you not understand about this? Pete02-11-11, 12:22 AMNo, it is possible to conclude t'=t and in SR t' and t mean dt' and dt. So, the only way to make that happen is time expansion + time dilation. http://knowyourmeme.com/i/7917/original/u-r-wrong.jpg?1249909893 From KnowYourMeme.com (http://knowyourmeme.com/memes/youre-doing-it-wrong) Also... http://scienceblogs.com/insolence/Science.jpg From Respectful Insolence (http://scienceblogs.com/insolence/2008/05/how_do_you_know_when_youre_doing_science.php) chinglu02-11-11, 05:42 PMNothing was "added". That result is a simple application of the Lorentz transformation to an event that YOU specified. It's hard to make any sense of this bizarre statement. If you have the same object at two different events, then the proper time is always the shortest time interval that can be measured between those two events. Every frame other than the rest frame of the object will measure a longer time for the same two events. Do you agree? No, this is not the case that this was the problem I set up. We were in the context of the unprimed frame and you added a time to the clock in the unprimed frame. Next, I will go ahead and ignore that and assume you wanted to add some time on the clock in the primed frame as I see Pete has done in the post following yours. First, it does not matter the start time of any clock. LT is about elapsed times. So, any start time you claim is not part of the answer. For example, assume Pete's initial condition start time on the clock at k in the negative x direction. Now, when the frames are the same, assume a light pulse is emiitted. If it is you contention that this "start time" is part of the problem, then try it out with the time of LT with the light sphere + "your start time" and you will find the light like space time interval is not invariant. That proves your method is wrong. Pete02-11-11, 09:51 PMchinglu, you said: Assume clock is located at x' < 0 in the primed coordinates. Assume standard configuration and time interval begins for both frame when origins same. Assume time interval ends when x' and unprimed origin same. What is this time interval in each frame. Note that you asked for the time interval in each frame, which is exactly what James did in post #117. In the primed frame, the time interval begins when t'=0, right? And at x'=-k and t'=0, the unprimed clock passing by reads t=-\frac{\gamma k v}{c^2} What do you not understand? James R02-11-11, 10:52 PMNo, this is not the case that this was the problem I set up. We were in the context of the unprimed frame and you added a time to the clock in the unprimed frame. Obviously you're simply mistaken about this. Please review posts #117, #118, #165 and #177 of the current thread. If you find any errors in those, please point them out by quoting the relevant parts of those posts and explaining exactly where the mistake was made. Right now you're just making false claims about what I wrote. My entire explanation of where you went wrong is in those four posts. If you want to discuss this further, you'll have to step through those posts line by line and show an error. I note that I have invited you to do this explicitly about 10 times or more by now, and you have NEVER even started to try to do so. I regard you as a troll with nothing interesting to say. First, it does not matter the start time of any clock. LT is about elapsed times. So, any start time you claim is not part of the answer. Your entire opening post and the entire point of this thread is about the time intervals between various events. To specify a time interval you need a start time and an end time - two events in spacetime. The Lorentz transformation first and foremost transforms spacetime events, not intervals. What do you not understand about this? For example, assume Pete's initial condition start time on the clock at k in the negative x direction. There's no assumption. YOU, chinglu, specified which events would be used, not me. Now, when the frames are the same, assume a light pulse is emiitted. What do you mean by "when the frames are the same"? The frames are never the same. If it is you contention that this "start time" is part of the problem, then try it out with the time of LT with the light sphere + "your start time" and you will find the light like space time interval is not invariant. That proves your method is wrong. Why don't YOU "try it out"? Show me your maths. Show me your calculations. I'm sick of doing all your thinking for you. I'm sick of pointing out the same mistakes of yours over and over again. Get up off your backside, switch on your brain, and start thinking and actually doing something, rather than continuing to make idiotic claims. chinglu02-13-11, 11:41 AMShow me your maths. Show me your calculations. I have made this simple. There is a clock at k in the context of the primed frame with k < 0. The question is from the time the origins are the same to the time k and the unprimed origin is the same, what will be the time on the clock at k and what will be the time on the unprimed origin clock. You then introduce some clock at k/ γ in the unprimed frame when that clock is not even in the problem. I assume you understand when the origins are the same, t'=t=0. Next, from the view of the primed frame, the unprimed origin is a distance k from the clock at the unprimed origin when the two origins are the same. Hopefully, you can understand, the clock at k will elapse -k/v for the unprimed origin to reach k. Note all measurements are in the context of the primed frame so this should be simple with not disagreements. Next, the final question is what is the elapsed time on the unprimed origin clock? 1) From the view of the primed frame, this is standard time dilation so, the primed frame concludes the unprimed origin clock elapses -k/(γv) 2) From the view of the unprimed frame, since the clock at k is measured in primed frame coordinates, then when the origins are the same, the unprimed frame concludes primed clock at k is a distance k/γ from the unprimed origin. Since the distance is k/γ to the unprimed origin and the speed is v, then the elapsed time from the view of the unprimed origin on the unprimed origin clock is -k/(γv). Note how both frames agree on the time of the unprimed origin clock. There is no disagreement. So far, we have both frame agree the unprimed origin clock will elapse -k/(γv) from the time the origins are the same to the time the clock at k meets the unprimed origin. We have the primed frame claiming the clock at k will elapse -k/v. Last, we need the conclusion of the unprimed frame for the clock at k. This is standard LT. k us measured in light seconds and v is measured in terms of c. t = -k/(γv) x = 0 t' = ( t - vx/c² )γ =( -k/(γv) - 0)γ = -k/v. Let's makes sure we have the correct x'. x' = ( x - vt )γ = ( 0 - v(-k/(γv))γ = k. Yes, we have the correct x'. Therefore, both frames agree the clock at k will elapse -k/v. Both frames agree the clock at the unprimed origin will elapse -k/(γv). Finally, we take the view of the unprimed origin and the moving clock at k coming toward the origin beats time expanded as agreed on both frames. And frame agreement is necessary because of the invertibility of LT. To summarize your error, you consider a clock in the unprimed frame at the same location as k when the origins were the same and placed some time on it. That clock at that location is not in any way involved in this problem and has no affect on the conclusions. We are only concerned with two clocks, the clock at the unprimed origin and the clock at k in the primed frame. Post ReplyCreate New Thread