Monday, March 12, 2018

inertial frames - Special Relativity - Regarding the Simultaneity of Events During the Train Paradox


My questions concerns that classic train paradox, wherein there is a train and a tunnel of equal length, and the train is traveling and some fraction of the speed of light towards the tunnel.


According to the Special Theory of Relativity, an observer outside the tunnel will see the train length contracted (Lorentz Contraction), whereas an observer inside will see the tunnel contracted.


Additionally, suppose that there were doors at the ends of the tunnel and that the observer outside the tunnel closed both doors instantaneously when he/she saw that the train was completely inside the tunnel.


The classic resolution of this paradox invokes the non-simultaneity of events, explaining that the observer in the train sees the far door close first, and then, once the train has begun to exit the tunnel, looks back to see the door at the beginning close. Thus, both observers agree that the train does not touch the door when they are closed for an instant.


Now my questions.


Why is it that the observer on the train sees the far door close first? It seems to me that the information coming from the far door would reach the observer on the train only after the information from the other door is reach.


Under this interpretation, the observer on the train would observe the train getting hit by the doors. What if, by some means, this event can be explained in terms of a stationary observer too? Everyone always concludes that the train remains untouched by the doors, but really the only condition that needs to be met is that both observers must agree. Why can't they both agree that the train was hit?


So, to summarize.





  1. Why is the door that is farther away from the train observed to close first?




  2. Why can't the other possible conclusion (both see a hit train) be observed?





Answer




Why is the door that is farther away from the train observed to close first?




By observe, in SR, we don't mean see, we mean essentially to record the time and place of events according to rods and (synchronized) clocks at rest. For example,



  • Assume that, at the front and back of the train, there are identical clocks that are synchronized in the frame of the train.

  • Further assume that, at the ends of the tunnel, there are identical clocks that are synchronized in the frame of the tunnel.


By the relativity of simultaneity, the clocks on the train are not synchronized in the frame of the tunnel where it is observed that the clock at the front is behind the clock at the back.


Symmetrically, the clocks in the tunnel are not synchronized in the frame of the train where it is observed that the clock at the entrance of the tunnel is behind the clock at the exit.



  • Finally, assume that the length contracted train in the frame of the tunnel just fits within the tunnel.



Thus, there is a moment, according to the tunnel clocks, that the contracted train is completely within the tunnel.


But remember, in the frame of the tunnel, the train clocks are not synchronized. In particular, since the clock at the front of the train is observed to be behind the clock at the back, it must be the case that, as recorded by the clocks on the train, the door at the exit of the tunnel closes earlier than the door at the entrance.


That is, in the frame of the train, the front of the train just reaches the exit, the door there closes for an instant without hitting the train and there is still a trailing portion of the train that is yet to enter the tunnel.


When the back of the train just clears the entrance of the tunnel, the door at the entrance closes for an instant without hitting the train and there is a leading portion of the train that has exited the tunnel.


As always, I recommend that you draw a spacetime diagram of this sequence of events to get a better 'picture' of how this works.


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