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Minkowski Diagram

  1. Mar 31, 2009 #1
    In the twin paradox, if B moves away from A at constant high speed what would the minkwoski diagram look like in the frame of A and in the frame of B.

    well in the frame of A. if B starts next to A i.e. at O and is moving at close to c to begin with the world line of B will just be a straight line making an angle of just less than 45 degrees with the positive ct axis, yes?

    but how do i draw the return journey?


    and in the frame of B the diagram should end up as just a reflection of the above in the ct axis (and i also have to relabel my axes to x' and ct'), yes?

    cheers.
     
  2. jcsd
  3. Mar 31, 2009 #2

    tiny-tim

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    Hi latentcorpse! :smile:

    You're not asked to draw the return journey …

    but if you did, wouldn't it be the same slope, but down instead of up ? :wink:

    And how could you draw the return journey "in B's frame", when he has two frames?
    Yes … if they're using their common x-axis, then B uses -x where A uses x. :smile:
     
  4. Mar 31, 2009 #3
    so in the frame of A if x is horizontal axis and ct is vertical axis it would look like

    \
    \
    \
    /
    /
    /
    hopefully that makes sense i.e. he returns back to his x=0 position with teh same velocity

    i dont understand what you mean by "B has two frames"
    also i just made this question up so its not very rigorously worded haha
     
  5. Mar 31, 2009 #4

    tiny-tim

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    ah :redface: … i should have guessed! :biggrin:
    (use the CODE tag:
    Code (Text):
     \
      \
       \
       /
      /
     /
    :wink:)


    Yes (though I'd prefer t along the bottom axis :wink:).
    I mean you can't do the same thing for B, because he has a different velocity, and therefore a different frame, on the way out, as he does on the way back. :smile:
     
  6. Mar 31, 2009 #5
    interesting. hadn't even thought of that. so on his way out, A has a world line making just less than -45 degrees with the positive ct axis.

    than i was thinking about jumping over to the positive quadrant (i.e. a discontinuity at the point where he turns round)???? not sure though
     
  7. Mar 31, 2009 #6

    tiny-tim

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    Noooo … in Special Relativity, you need to keep the same frame. :smile:
     
  8. Mar 31, 2009 #7
    but you said B has a different fram on the way back - doesn;t this contradict special relativity itself?

    shall i draw this on two seperate minkowski space time diagrams then?
     
  9. Mar 31, 2009 #8
    oh and why do we need the same frame?
     
  10. Mar 31, 2009 #9

    tiny-tim

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    No … why should it? :confused:
    If it makes you happy …

    but I don't see what good it will do you.
    How can you combine results for, say, x' t' and x'' t'' if you use different frames?

    You'd never dream of using two different frames for a single calculation in any other branch of maths or physics :wink:
     
  11. Apr 1, 2009 #10
    ok. im getting confused!
    2 main problems:

    (i)"I mean you can't do the same thing for B, because he has a different velocity, and therefore a different frame, on the way out, as he does on the way back."

    "Noooo … in Special Relativity, you need to keep the same frame"

    so in the first quote your saying i should use different frames and in the second one i shouldn't.

    (ii) we'd relate them by a lorentz transformation - do you mean to transform the world line for B's return journey in his new frame back into the original frame using a lorentz transformation so i can plot them both on the same diagram?
     
  12. Apr 1, 2009 #11

    tiny-tim

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    ah … no, it's more that I'm saying that you would need two frames, but of course you can't.
    uhh? :confused: when did I say that?

    you can't plot them both on the same diagram

    as I said …
     
  13. Apr 1, 2009 #12
    so the only way to do this would be to use two seperate diagrams - one for the outgoing journey and one for the retuen and as you said, this would not be very useful.
     
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