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Spacetime diagrams: ct axis and time contraction and length dilation

  1. Oct 2, 2012 #1
    Hi, this isn't really a homework problem... but I'm just wondering

    I see the time axis as c*t, now people say that its to scale the time axis so that the world line of light is 45 degrees. But if you were to multiply time by the speed of light, wouldn't you just get the units metres or.. light years, a measurement of length not time? How can you still say that the axis represents time?

    Also another question, is it possible to show both time dilation and length contraction on one spacetime diagram?
     
  2. jcsd
  3. Oct 2, 2012 #2
    On grid paper (squares grid) the t axis would look similar to what you are used to. Let's say you mark 1 second per 2 squares up. 2 seconds would be at the 4th square up and so on...

    The x-axis, you would use light-seconds. Each two squares would be 1 light-second. 4 squares 2 light-seconds and so on...

    Light travels at about 300000km/s. This means that one light-second is 300000km. So instead of using 1 light-second, you could write 300000km, 600000km and so on... but using light-second i guess is more elegant.

    If you scale your diagram as stated above, light-beams inside that diagram will necessarily have a gradient of 45° or otherwise said, for each square you move left/right on your grid paper, you move 1 square up/down.
     
  4. Oct 3, 2012 #3
    You would have to take this drawing

    and extend the two diagrams towards the bottom (negative t). Then get the light-beams going from E0/E0' to cross the blue lines(you also extend towards the bottom).

    Call the events E3/E3' and E4/E4'.
    What you want to get is a formula which let's you see by which factor MOVING clocks would run slower, seen from an observer in an inertial frame of reference, compared to clocks being at rest inside the observer's frame.

    All clocks following the blue line in system A, are moving at vrel = 0.5c in system A. In system B, they are at rest.

    getting the Δt/Δt' between E1/E3 and E1'/E3'



    Do the same for system B. This time we need a clock which moves at vrel=0,5c relative to an observer being at rest in system B. This would be the green line. After extending the diagram to the bottom along with the green line, cross some light-beams that go through E0' with the green lines.
    E5/E5' and E6/E6' where the beams cross the green lines.

    get the formulas for the Δt2/Δt2' and you should be able to solve similar to how i solved for the length contraction.


    (you only need either E3/E3'/E5/E5' OR alternatively E4/E4'/E6/E6' to arrive at the formulas)

    Just remember that because of one of the two postulates of SR:

    "The laws of physics are the same in all inertial frames of reference." implying that,

    by whatever factor (γ) Δt is modified to arrive at Δt' using Δt * γ = Δt'
    the same is true the other way around Δt2' * γ = Δt2


    I hope this is good enough to get you started. The description is sloppy but that is because it's harder to imagine than measuring distances with rulers. The clocks here are used just like rulers, to measure time-frames. But before i fry my brain, trying to explain this further, i am out of here.


    Warning: I call the two postulates, Axioms. Postulates is the better term here i believe. Also, the factor γ i use is inverse to the γ you will find on Wikipedia. This is because of the way i derived the formula, where using γ that way made perfect sense.
    If you substitute γ in my formulas with sqrt(1-(v^2/x^2)), you arrive at the same formulas as Wikipedia does, so there is not really anything wrong here other than maybe not going by the standard or being a carbon copy of Wikipedia.
    Also noteworthy. The numbering of the two Axioms/Postulates of SR as i used it, is flipped compared to how Wikipedia has it. You might want to change that in your derivation to be more in accord with the standard. Again, it seemed more reasonable to me to flip this, and goes along to how i derived it as i used, light always traveling at C in a vacuum absent of gravity seen from any arbitrary inertial reference frame, for the first part of my derivation.
     
    Last edited by a moderator: Apr 18, 2017
  5. Oct 7, 2012 #4
    Last edited by a moderator: Sep 25, 2014
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