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Spacetime diagram drawing problem

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  1. Oct 15, 2013 #1
    I was drawing a spacetime diagram to relate Doppler shift effect but i stuck at a point which i can't understand
    This is what I'm trying to draw
    An object with mirror is moving away from me with a velocity of 50% speed of light
    When each second passes in my clock i send light pulses at that object.
    I sent 4 light pulses in 4 second.
    The object starts when t=0 in my clock
    When t=1 in my clock that object will be ’half light second away’ at the same time t=1 i sent my first light pulse. So in my clock the first light pulse must hit the object when t=2 in my clock (when t=2 in my clock the object will be 1 light second away)
    When t=2 i sent my 2nd light pulse and it must hit the object when t=4 in my clock
    When t=3 i sends my 3rd light pulse & it must hit the object when t=6 in my clock
    Finally t=4 i sends my 4th light pulse and it must hit the object when t=8 in my clock
    So i draw my spacetime diagram and drew a 22.5 degree angle, it will be 50% speed of light
    In each seconds i sends a light pulse, so i drew it in 45 degree angle.
    But the intersection of light pulse and speed of the object doesn't match with time in my graph
    The first light pulse which i sent should intersect at t=2 in my graph but it intersect only below 2 second.
    So if i try to adjust the graph inorder to reach at t=2 the angle changes and it will increase the speed of the object. This is like the same for all 4 light pulses
    I know moving objects contract in length, but this isn't that
    I also know time slows in moving object, but i don't need to care, i need my spacetime graph to be correct, they may receive it in different times
    If it is my mistake help me to find it.
    The picture is below

    http://www.flickr.com/photos/sreenath666/10286909563/ [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 15, 2013 #2

    dvf

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    Try using an Arctan(1/2) = 26.6 degrees angle.
     
  4. Oct 15, 2013 #3
    That will change the speed of the object
     
  5. Oct 15, 2013 #4

    dvf

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    Oops, I meant 45 degrees - Arctan(1/2) = 18.4 degrees.
     
  6. Oct 15, 2013 #5
    anyway changing angle will increase or decrease its speed, i want it going at 50% speed of light
     
  7. Oct 15, 2013 #6

    dvf

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    Then you need Arctan(1/2) between the ct-axis and the ct'-axis, and thus 45 degrees -Arctan(1/2) between the ct'-axis and the light diagonal.
     
  8. Oct 15, 2013 #7

    robphy

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    "50% of the speed of light" is a line with slope (1/2) on an x-vs-t graph.
    Use your grid boxes to determine that slope...
    for every (say) 4 seconds, the mirror has traveled another 2 light-seconds.
    Don't use a [Euclidean] protractor on a Minkowski spacetime diagram.
     
  9. Oct 15, 2013 #8
    by the way, do someone studying for pHd draw 50% in ct x diagram like u say?
    They draw it exact at the half position, and if it is wrong in using Euclidean tools, then isn't it wrong to draw a 45 degree line in spacetime diagram? we still use protractor for that
     
  10. Oct 15, 2013 #9

    dvf

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    To get a slope 1/2 between the horizontal and a line, you do *not* bisect the 45 degrees angle. Instead you draw a line with slope 1/2, which means climbing 1 unit for every 2 units of horizontal displacement. That gives an angle of 26.6 degrees, not one of 22.5 degrees.
    See https://en.wikipedia.org/wiki/Slope
     
    Last edited: Oct 15, 2013
  11. Oct 15, 2013 #10

    robphy

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    You can use a protractor to draw a 45-degree line...
    or you can use your grid, which helps you avoid having to use any special tools.
    (It's not wrong to use a protractor...
    but it is tedious [compared to counting boxes in a grid]
    and it is unnatural since the geometry of a spactime diagram [or even a Galilean spacetime diagram [i.e, the ordinary phy 101 x-vs-t graph] ] is not Euclidean.)


    Note that the relation between slope v and Euclidean-angle is
    v=tan(theta), which implies that v and theta do not scale proportionally.
    Doubling theta does not result in doubling v.

    (Using rapidity in Minkowskian geometry, v=tanh(rapidity), which also implies that v and rapidity do not scale proportionally.)
     
  12. Oct 15, 2013 #11
    Draw me a diagram, please
     
  13. Oct 15, 2013 #12

    dvf

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  14. Oct 15, 2013 #13

    ghwellsjr

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    The second line you drew in your diagram is correct. Why do you want someone else to draw you a diagram?

    Let me offer you some suggestions to make your work easier. First, instead of using a speed of 0.5c, use 0.6c. At 0.5c, gamma is 1.1547, making it more difficult to draw in some features. At 0.6c, gamma is 1.25 which will make drawing in some features trivially easy. The same holds true for the Doppler factor which you will see when you get to that point.

    Also, it looks like you drew your diagram on a brown paper bag with a white pen and then with a black pen. It's practically illegible. Instead, here is a blank spacetime diagram that you can download and markup with your favorite drawing program such as Paint if you are using Windows. Then you can upload it directly into the forum instead of using some third party server.

    If you want, I could easily draw any diagram you want but you're doing great and I think it will be much more satisfying to you to do it on your own.

    attachment.php?attachmentid=62950&stc=1&d=1381844295.png
     

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