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Lorentz - dont get it

  1. Apr 22, 2008 #1
    I am new here

    didnt find the reply to this before

    I hope I get lorentz transformations right

    he said, that not only we have to shift placement axis, but also shift time axis

    I know that the slope for velocity and time dalation must be same, but what formula to use to prove that the slopes are the same. (time delation formula doesnt work)
     
  2. jcsd
  3. Apr 22, 2008 #2
    I am talking about the slopes for the axes for moving body in coordinate system impied by lorentz transformations

    en.wikipedia.org /wiki/Lorentz_transformation

    I get not equal slopes using time delation formula
     
  4. Apr 22, 2008 #3

    JesseM

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    Science Advisor

    You mean, the slopes for the time and space axes of the second frame, as seen in the first frame? The slopes should not be equal, but inverse, like 2 and 1/2, when expressed in units where c=1 (like x=light-seconds and t=seconds). And what do you mean by "time dilation formula"? How would you calculate slope from the time dilation formula?
     
  5. Apr 23, 2008 #4
    I understand how to draw velocity axis as seen in first frame, but about time axis I jut know it must be inverse... OK but why is it inverse
     
  6. Apr 23, 2008 #5

    JesseM

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    Science Advisor

    What do you mean by "velocity axis" and "time axis"? Is the velocity axis the x'-axis of the second frame as seen in the first frame, while the time axis is the t'-axis of the second frame? The t'-axis is very simple, it behaves just like an object at rest in that frame (since the t'-axis is a line of constant x' coordinate, and an object at rest in that frame will have a constant x' coordinate)...if the second frame is moving at 0.6c relative to the first, then you'd just draw a line that moves sideways 0.6 light-seconds for every 1 second you go upward, so the slope is just 0.6. The x'-axis is the inverse because it goes sideways 1 light-second for every 0.6 seconds you go upward. You can see this by picking two points on the x'-axis, like (x'=0, t'=0) and (x'=10, t'=0) and then using the Lorentz transform to see where they'd be when drawn in the first frame's (x, t) coordinate system:

    x = gamma*(x' + vt')
    t = gamma*(t' + vx'/c^2)

    (x'=0, t'=0) would be mapped to (x=0, t=0), i.e. the origin, while (x'=10, t'=0) would be mapped to:

    x = gamma*(10)
    t = gamma*(0.6*10) = gamma*(6)

    Since gamma is the same in both cases (it's equal to 1/sqrt(1 - v^2/c^2)), we can see the point will be less far from the origin on the t-axis than it is on the x-axis, by a factor of 0.6. So, the x'-axis drawn in the first coordinate system must be a straight line through the origin which goes up 0.6 seconds for every 1 light-second you move sideways.
     
  7. Apr 23, 2008 #6

    jtbell

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    Staff: Mentor

    To put it a more mathematical way:

    For simplicity, use units in which c = 1 so the Lorentz transformation between inertial reference frames S and S' is

    [tex]x^{\prime} = \gamma (x - vt)[/tex]

    [tex]t^{\prime} = \gamma (t - vx)[/tex]

    In S, the x-axis is the set of points that have t = 0. To find the corresponding set of points in S':

    [tex]x^{\prime} = \gamma (x - v \cdot 0)[/tex]

    [tex]t^{\prime} = \gamma (0 - vx)[/tex]

    Combining these two equations gives

    [tex]t^{\prime} = -vx^{\prime}[/tex]

    which is a straight line through the origin in S', with slope -v.

    Returning to S, the t-axis is the set of points that have x = 0. Proceeding similarly to the above, the corresponding set of points in S' is

    [tex]t^{\prime} = -x^{\prime} / v[/tex]

    which is a straight line through the origin in S', with slope -1/v. The two slopes are reciprocals.
     
    Last edited: Apr 23, 2008
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