davey2222 Posted December 24, 2010 Posted December 24, 2010 (edited) My first post! And this is a very paradoxical problem which I am unable to make head or tail. http://galileo.phys....chronizing.html From the above link... To summarize: as seen from the ground, the two clocks on the train (which is moving at v in the x-direction) are running slowly, registering only seconds for each second that passes. Equally important, the clocks—which are synchronized by an observer on the train—appear unsynchronized when viewed from the ground, the one at the back of the train reading seconds ahead of the clock at the front of the train, where L is the rest length of the train (the length as measured by an observer on the train). Note that if L = 0, that is, if the clocks are together, both the observers on the train and those on the ground will agree that they are synchronized. We need a distance between the clocks, as well as relative motion, to get a disagreement about synchronization. We agree that the person on the train observes synch-clocks and on the ground observes non-synch of the two clocks. What if the train comes to a stop and the person on the train brings out the clock to let the person on the ground to have a look. Then which version of the synchronization is correct? I mean what do the two observers see? I know it is not possible for one to see a synch-clock and the other to see non-synch-clock at the same time. What gives? Edited December 24, 2010 by davey2222
swansont Posted December 24, 2010 Posted December 24, 2010 If the train comes to rest, v=0. The rates of all the clocks will be equal then, too, though the clocks on the train will have accumulated less elapsed time on them.
michel123456 Posted December 24, 2010 Posted December 24, 2010 (edited) My first post! (...) the answer is in the statement: We need a distance between the clocks, as well as relative motion, to get a disagreement about synchronization. Edited December 24, 2010 by michel123456
davey2222 Posted December 24, 2010 Author Posted December 24, 2010 (edited) Guys, what I am trying to say is that there seems to be something impossible (perhaps I missed something?) in the theory as applied to the clock synchronization above. The two clocks are being synchronized by using light signal. According to Relativity, both observers are correct in their observations: the observer on the train (rest frame) observes both clocks to start ticking from 0-second. The observer on the ground observes the clock in the rear end of the train to start ticking from 0-second followed by the clock in the front after a time interval. As the train comes to a stop, assuming both clocks are still separated by a distance, which observer has the correct synchronization observation (or results?) if the clocks are brought together for comparison? I mean is there still a time-interval between the two clocks? Maybe the observer on the ground is wrong? Edited December 24, 2010 by davey2222
swansont Posted December 24, 2010 Posted December 24, 2010 The clocks on the train are synchronized — when v=0, the lag between them in the platform frame goes away. All of these effects are due to the finite speed of light being a constant in all inertial frames. It means that simultaneity is relative.
michel123456 Posted December 24, 2010 Posted December 24, 2010 (edited) I mean is there still a time-interval between the two clocks? No, because there will be no distance between clocks. The time interval is proportional to distance & relative speed. When there is no distance, there is no time interval. Any other paradoxal result coming from some kind of reasonning must have a mistake somewhere. Edited December 24, 2010 by michel123456
cypress Posted December 24, 2010 Posted December 24, 2010 Guys, what I am trying to say is that there seems to be something impossible (perhaps I missed something?) in the theory as applied to the clock synchronization above. The two clocks are being synchronized by using light signal. According to Relativity, both observers are correct in their observations: the observer on the train (rest frame) observes both clocks to start ticking from 0-second. The observer on the ground observes the clock in the rear end of the train to start ticking from 0-second followed by the clock in the front after a time interval. As the train comes to a stop, assuming both clocks are still separated by a distance, which observer has the correct synchronization observation (or results?) if the clocks are brought together for comparison? I mean is there still a time-interval between the two clocks? Maybe the observer on the ground is wrong? The solution to your paradox is that you specified that the clocks and the observer are brought back together. Bringing the clocks and an observer together requires relative motion as well and that action returns them all to the same reference frame.
michel123456 Posted December 24, 2010 Posted December 24, 2010 The solution to your paradox is that you specified that the clocks and the observer are brought back together. Bringing the clocks and an observer together requires relative motion as well and that action returns them all to the same reference frame. Correct. You just solved the twin paradox. (truly, not sarcastic)
IM Egdall Posted December 24, 2010 Posted December 24, 2010 (edited) Take a look at the link below. It gives a detailed explanation of the so-called Twins Paradox. Hope it helps. Go to : http://www.marksmodernphysics.com/ then click on Its Relative, archives, The Twins Paradox. Edited December 24, 2010 by I ME
davey2222 Posted December 25, 2010 Author Posted December 25, 2010 (edited) The clocks on the train are synchronized — when v=0, the lag between them in the platform frame goes away. All of these effects are due to the finite speed of light being a constant in all inertial frames. It means that simultaneity is relative. I appreciate your answer. When v=0 they appear to be able to synchronise to an observer on the platform and on the train. This I agree with anyone 100%. But the clocks on the train are indeed synchronised and on the platform they appear not-synchronised when v=/0. No, because there will be no distance between clocks. The time interval is proportional to distance & relative speed. When there is no distance, there is no time interval. Any other paradoxal result coming from some kind of reasonning must have a mistake somewhere. I 100% agree that when 'distance between clocks = 0' then there is no time-interval when the clocks are being synchronised no matter if the train is in motion or not. But if 'distance between clocks =/ 0' when 'v =/ 0' then the clocks won't be synchronised as observed by a ground observer. Take a look at the link below. It gives a detailed explanation of the so-called Twins Paradox. Hope it helps. Go to : http://www.marksmodernphysics.com/ then click on Its Relative, archives, The Twins Paradox. Thnaks. Have glanced through but is not related. The solution to your paradox is that you specified that the clocks and the observer are brought back together. Bringing the clocks and an observer together requires relative motion as well and that action returns them all to the same reference frame. Yes, bringing the clocks and the two observers together returns them all to the same reference frame. But what will the two clocks show? Perfect synchronisation or time-interval between the clocks? Edited December 25, 2010 by davey2222
davey2222 Posted December 29, 2010 Author Posted December 29, 2010 Seems like the above shows an example where the prediction of the theory of Relativity is impossible in reality.
swansont Posted December 29, 2010 Posted December 29, 2010 Seems like the above shows an example where the prediction of the theory of Relativity is impossible in reality. Your prediction is not what relativity predicts, if I am understanding you correctly. The two train clocks would be synchronized to each other. What is the problem with that?
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now