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Picturing reference frames

  1. Sep 18, 2008 #1
    I picture two reference frames for two different objects moving relative to each other as two coordinate spaces S and S'. My question is, say both spaces start of on top of each other (ie both objects are initially stationary) and then the objects start moving relative to each other, would the whole coordinate space for one of the objects move or do their origins always coincide and one just expands/shrink but remains centered on the origin?

    Is this a bad way to think of it? I've read my notes a few times but I'm still having a hard time trying to picture SR in my mind :(. Can anyone suggest any other ways to picture SR?
  2. jcsd
  3. Sep 18, 2008 #2
    The way you should think about it is that the two observers are sitting at the origins, and the reference frames are their coordinate system. Therefore if the two observers are in a relative motion to each other their coordinate system moves with them, including the origin since they are sitting at it. You can picture a frame of reference as a lattice of synchronized clocks and meter sticks. The whole point of SR is not really to “see” how the frames would look like but rather to compare “measurements” between them.
    To make the idea of the lattice more clear the following might help:
    “A way to avoid this ambiguity is to remove the observers and simply fill up space
    with a large rigid lattice of meter sticks and synchronized clocks. Different frames
    are defined by different lattices. All of the meter sticks in a given frame are at rest
    with respect to all the others, so there is no issue of length contraction within each
    frame. To measure the length of something, we simply need to determine where
    the ends are (at simultaneous times, as measured in that frame) with respect to the
    This lattice way of looking at things emphasizes that observers are not important,
    and that a frame is defined simply as a lattice of space and time coordinates.
    Anything that happens (an “event”) is automatically assigned a space and time
    coordinate in every frame, independent of any observer.” Page 523, Introduction to classical mechanics with problems and solutions, 2008, David Morin
  4. Sep 18, 2008 #3


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    What you should do is learn about spacetime diagrams, i.e. you picture one time dimension and one spatial dimension.
  5. Sep 18, 2008 #4
    Thanks guys. Another thing I was wondering is will there be some kind of proportionality between events in frames? Say three events happens at t=0, t=1 and t=2 in one frame. The first two of those events happen at t=4 and t=6 in another frame. Will the third happen at t=8 then?
    I also didn't quite get why geometry (x^2+y^2+z^2-(ct)^2) will be constant in all frames since there's length contraction? Does that mean that since the length in the direction parallel to motion has contracted, lengths in other axes must increase by the same amount? I'm assuming it would be the time x c component that increases...
    Last edited: Sep 18, 2008
  6. Sep 18, 2008 #5
    Lengths in axes orthogonal to the relative motion of the reference frames (usually taken to be the y and z axes with the relative motion of the frames parallel to the x axis) does not decrease or increase. It is the time dilation of time coordinate (ct) that adjusts to compensate for the length contraction of the x coordinate.
    Last edited: Sep 18, 2008
  7. Sep 18, 2008 #6
    You need to specify a time and a place to define an event. So let's say you have three events t=0, t=1 and t=2 in one frame that happen at the same place (x=0) from the point of view of that frame's observer. These coordinates would be denoted in (x,t) coordinates as (0,0), (0,1) and (0,2) in that frame.

    In another frame S' with a relative velocity v/c of 0.866c to original frame S, the gamma factor [tex]\gamma [/tex] would be:

    [tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2[/tex]

    The Lorentz transformations for the t and x coords in the primed frame are given by :

    [tex]t^{\prime} = \gamma \left( t - \frac{vx}{c^2} \right)[/tex]

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

    The transformation of the three (x,t) events, (0,0), (0,1) and (0,2) is:

    (x,t) --> (x' , t')

    (0,0) --> (0 , 0)
    (0,1) --> (-1.732 , 2)
    (0,2) --> (-3.464 , 4)

    The t coords of 0, 1 and 2 transform to 0, 2 and 4 in this specific example where the events all occur at x=0 in the rest frame of the original observer. It is important to note that the transformation of the time coordinate is interrelated to how the x coordinate of the events evolve in the rest frame of the original observer.

    In an every day example if you invite people to an event, you might say "You are invited to a wedding on (date of event) at (place of event).
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