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Einstein's train - When+where does side observer see lightnings to the train rider?

  1. Apr 24, 2008 #1
    Hi.

    I am learning special relativity and have decided to take Einstein's train example to practice some calculations. I have given the train speed v=1/4c and have decided on distances that make the side observer's space and time calculations straightforward.

    I have calculated everything using the invariance of the spacetime interval equation. (ST^2 = T^2 - S^2) for timelike intervals and (ST^2 = S^2 - T^2) for spacelike intervals; where S is the spatial distance, T the temporal distance and ST the spacetime interval between events.

    However, whenever I try to think about how I would approach the matter of the train rider's interpretation of the time and space coordinates of the side observer's seeing both lightning flashes, I draw a blank.

    Can you think of the easiest way to approach this using just the spacetime interval equation?


    edit: Also please note this isn't a homework question - I wanted to do this for myself. This is why I placed this thread here rather than on the homework forum.

    Thanks!
     

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    Last edited: Apr 24, 2008
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  3. Apr 24, 2008 #2

    JesseM

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    I don't understand what you mean by "using just the spacetime interval equation"--could you give an example of how you got some of the numbers in the train observer's frame using only the spacetime interval equation and the coordinates in the track observer's frame?
     
  4. Apr 24, 2008 #3
    Sure.

    Once I had all the coordinates for the side observer down, I knew the spacetime interval between every event.

    Now, I knew that the train rider would have event C (seeing front lightning) and event D (seeing rear lightning) occur on the same spatial point (0,0). Therefore I knew the spatial distance between those two events. Since I knew the spacetime interval between C and D from the side observer's coordinates, I could find out from the spacetime interval equation what the time distance between those two events is for the train rider. The answer is sq(12/5) meters.

    Since the front and rear of the train are not moving for the train rider, and now that he knows the time distance between seeing the front lightning and seeing the rear lightning is sq(12/5) meters, he knows that the rear lightning event originated sq(12/5) meters of time after the front lightning. Since the front lightning time has been deemed to be 0m, then event B (rear lightning)'s time is thus sq(12/5) meters.

    Now that we know both Event A and Event B's time coordinates for the train rider, we can use the spacetime interval between events A and B that we know from the side observer to find out the spatial distance between those events for the train rider.

    And so on...

    Like a Sudoku puzzle!
     
  5. Apr 24, 2008 #4

    JesseM

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    That's a valid method of getting answers in this particular problem, but I don't think it will really generalize very well--I don't see any simple way of using the invariant interval to get the position and time the light reaches the track-observer in the train-observer's frame, though maybe I just haven't given it enough thought. The simplest method would just be to use the Lorentz transformation, although this may not be helpful for building intuition. The best method for building intuition might be to use Lorentz contraction and time dilation...for example, if you assume both observer's clocks read 0 at the moment they are lined up on the same y-axis, then if it takes 4 seconds for light from the lightning at the back to reach the train-observer in the track-observer's frame, using the time dilation equation you know that at that moment the train observer's clock will read [tex]\sqrt{1 - 0.25^2} * 4[/tex] = 3.87298 seconds. And if the distance from the back of the train to the middle is 3 l.s. in the track-observer's frame, you know the distance must be greater by a factor of [tex]1/\sqrt{1 - 0.25^2}[/tex] in the train-observer's frame, a distance of 3.09839 light seconds. So, if the train observer-received the light at a time of 3.87298 seconds in his frame, and it took the light 3.09839 seconds to reach him in his frame, the lightning must have struck the back at a time of 3.87298 - 3.09839 = 0.77459 seconds in his frame. And so forth, just using length contraction, time dilation, and the fact that light moves at c in every frame.
     
  6. Apr 24, 2008 #5
    Thanks for that answer!

    I had considered the Lorentz transformation and length contraction/time dilation as a way of approaching this but gave it up for the very reason you proposed - building intuition. I wanted to "do the dirty work" and not take shortcuts. I believed that finding when the light reaches the track observer in the rider's frame by constraining myself to only being able to use the invariant interval would challenge me to think of logical approaches that would really build intuition.

    I'll keep thinking on it. If anybody can think of a way to find event E's coordinates for the rider frame only with the invariant interval equation, don't hesitate to reply!
     
    Last edited: Apr 24, 2008
  7. Apr 25, 2008 #6

    Ich

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    If I'm not totally mistaken, the equations
    (te-ta)²-(xe-xa)²-(4m)²=0
    (te-tb)²-(xe-xb)²-(4m)²=0
    are sufficient, at least in principle.
     
  8. Apr 27, 2008 #7
    Eienstein's train

    The train observer coordinates for event E can also be obtained from the spacetime interval graphically from a well calibrated spacetime diagram.
    The tangent of the angle between the ct and ct' axis and between the x and x' axis
    = v/c. To calibrate the x' axis a line passing through x=5 and parallel to ct' axis intercepts the x' axis at 3.098.
    The spacetime interval =5 at x = 0 from the side observer.
    So a line passing through ct= 5 and parallel to the ct' axis intercepts the x' axis at the event E coordinate and another line passing through ct =5 and parallel to the x' axis intercepts the ct' axis at the event E coordinate.
    Also a hyperbola can be generated with spacetime interval = 5 that is superimposed on the diagram- with one point on it being x=0,ct=5 and x'= -1.29 , ct' = 5.16.
     
  9. Apr 28, 2008 #8
    correct type error:
    To calibrate the x' axis a line passing through x = 3
     
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