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Motion in 2D

  1. Nov 9, 2013 #1

    PeroK

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    Hi,

    I've just started looking at SR and have got stuck with the first problem that considers 2D motion. I'm trying to work out the angle of tilt of a bar which is horizontal in one frame but moving horizontally and vertically wrt a second frame. I can't seem to pin down where the bar is at a point in time in the second frame.

    Any advice on the best way to look at this problem?

    Thanks

    Pero
     
  2. jcsd
  3. Nov 9, 2013 #2

    Dale

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    I would approach it by writing an equation representing the worldline of each part of the bar. Suppose that in the bar's rest frame it makes an angle, ##\theta## with the x axis and is of length L then the worldline of any point on the bar is:
    ##(t, r\;cos(\theta), r\;sin(\theta),0)##
    where ##0<r<L##

    Then just Lorentz transform that and you have the world line in any other frame. Express the resulting equations in terms of the transformed time and then you can get the angle and length in the other frame.
     
  4. Nov 10, 2013 #3
    Just to add to Dalespam's post, this is how you should set up the problem to make life simpler for yourself. Let 's say you have a bar of proper length r, at rest in S that is parallel to the x axis in the x y plane and want to transform to frame S' where the velocity of S' relative to S is at some angle ##\theta## to the x axis. Lets also say the coordinates of the ends of the bar in S are (t0,x0,y0) = (0,0,0) and (t1,x1,y1) = (0,r,0) respectively. While it possible to do this directly with the completely general version of the Lorentz transform, it is much simpler to rotate the bar by ##-\theta## and obtain new coordinates for the bar as described by Dalespam then boost by -v parallel to the x axis. The new initial coordinates in S would then be (t0,x0,y0) = (0,0,0) and [itex](t1,x1,y1) = (0, r \cos(-\theta), r \sin(-\theta))[/itex]. After the transformation the coordinates of the ends of the bar will be (t0',x0',y0') = (0,0,0) and (t1',x1',y1') = (t1',x1',y1) but will not be simultaneous in S'. To remedy this you can note that the velocity of the rod in S' is v in the x direction, so you can retard the position of the front of the rod back in time to where it was simultaneous with t0'. The simultaneous coordinates of the ends of the rod in S' would then be (0,0,0) and (0,x1'-vt1', y1). The new angle in S' would then be ##\theta' = \tan^{-1}(y1/(x1'-vt1'))##. You can get away with this trick because the velocity is constant.
     
    Last edited: Nov 10, 2013
  5. Nov 10, 2013 #4

    PeroK

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    Thanks. I think the change of co-ordinates does the trick. But, I don't understand the term x1'-vt'. This seems to imply that the rod is doubly fore-shortened: I guess x1' is x1 dilated by the motion, but then why take away vt' and how do I calculate t'?

    Can't I just say that in the new reference frame, the bar is fore-shortened in the x direction, but not in the y direction, so the bar appears more tilted in S'?

    I got:

    [itex]\theta ' = tan^{-1}(\frac{tan \theta}{\gamma})[/itex]
     
    Last edited: Nov 10, 2013
  6. Nov 10, 2013 #5
    The rod is not doubly foreshortened. This is because x1' is actually greater than x1 because that end of the rod is further ahead in time in S'. Taking away vt1' fixes that. You calculate t1' and x1' from the Lorentz transformations:

    [itex]t' = \gamma(t - vx)[/itex],

    [itex]x' = \gamma(x - vt)[/itex]

    [itex]y'=y[/itex],

    [itex]z'=z[/itex]

    using units of c=1. See https://www.physicsforums.com/library.php?do=view_item&itemid=19. I assumed you were attempting to use the Lorentz transformations and that was where you discovered the issue with the coordinates not being simultaneous in the new reference frame. I was trying to show what actually happens with the transforms and how to fix the issue.

    Yes, that approach works too, if you are only interested in how the angle transforms, but there is a small error in your result. It should be:

    [itex]\theta ' = tan^{-1}(\gamma *tan \theta)[/itex]

    This is because [itex]\theta ' = tan^{-1}\left(\frac{y1}{x1/\gamma}\right)[/itex]

    The other advantage of using the transforms is that they take care of simultaneity, time dilation and length contraction and sometimes it easy to miss things when calculating free-form. However, in this case your approach is perfectly adequate and straight forward, because the length contraction formula already takes care of the simultaneity issue.
     
    Last edited: Nov 10, 2013
  7. Nov 10, 2013 #6

    PeroK

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    Thanks. I haven't covered Lorentz transforms yet (next chapter!). So, bear with me.
     
  8. Nov 10, 2013 #7
    Your welcome ;)
     
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