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Homework Help: Length Contraction Question from Rindler

  1. Jul 13, 2009 #1
    Hi, this isn't exactly homework, as I'm teaching myself, but I thought this forum would be more appropriate for it.

    1. The problem statement, all variables and given/known data
    (Page 36, Introduction to Special Relativity by Wolfgang Rindler)

    An 18-foot pole, while remaining parallel to the x-axis, moves with velocity [itex](v,-w,0)[/itex] relative to frame S, with [itex]\gamma(v) = 3[/itex], and u, w positive. The centre of the pole passes the centre of a 9-foot hole in a plate that coincides with the plane y = 0. Explain, from the point of view of the usual second frame S' moving with velocity v relative to S, how the pole gets through the (now 3-foot) hole.

    2. Relevant equations

    [tex]\vec{u} = (v, -w, 0)[/tex]

    [tex]\vec{u'} = (u_{1}', u_{2}', u_{3}')[/tex]

    Velocity transformation equations:

    [tex]u_{1}' = \frac{u_1 - v}{1-\frac{u_1 v}{c^2}} = 0[/tex]

    [tex]u_{2}' = \frac{u_2}{\gamma(1-\frac{u_1 v}{c^2})} = -\gamma w[/tex]

    [tex]u_{3}' = \frac{u_3}{\gamma(1-\frac{u_1 v}{c^2})} = 0[/tex]

    So, [tex]\vec{u'} = (0, -\gamma w, 0)[/tex]

    3. The attempt at a solution

    The proper length of the rod is 18ft. Relative to an observer in S', the rod has a length of 9 ft and is moving along the y'-axis with a velocity of [itex]-3w \hat{y}[/itex] ft/s. Also, relative to this observer, the hole is now a 3 ft hole moving along the negative x'-axis with a velocity [itex]-v \hat{x}[/itex].

    In the S' frame then, the 9 ft rod is moving down along the y' axis, trying to enter a hole that is moving along the x'-axis in the negative x' direction. This is as far as I've gotten.

    How does one explain this observation?

    Attached Files:

    Last edited: Jul 13, 2009
  2. jcsd
  3. Jul 13, 2009 #2


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    Is there any "rotation"? What about simultaneity? Is the pole entering the hole a single event?
  4. Jul 14, 2009 #3
    I don't see how it is simultaneous. But maybe I don't understand your question..
  5. Jul 14, 2009 #4


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    OK, I was just trying to stimulate your thought process. Try this: draw the worldlines of the edges of the rod and the edges of the hole.
  6. Jul 14, 2009 #5
    i am not sure i get it. it says that the x-component of the velocity is v for the rod. i.e. the same as dat of S'. Hence, there should technically be no contraction observed from S' reference frame. As the rod is going in the negative y-direction, if at all contraction is observed from S' it should be in the y direction (the thickness). Correct me if I am wrong.
  7. Jul 14, 2009 #6


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    I believe that you are correct.
  8. Jul 14, 2009 #7
    In that case there is a small paradox created. The length of the pole as observed from the original frame will be 18/3 = 6 feet along x axis. The hole is in S frame and has a proper length of 9 feet. Hence for an observer in S frame, the rod will appear to pass through the hole. But for an observer in S', the rod is 18 feet. The size of the hole is 9/3 = 3 feet. So the rod can't pass.

    :zzz: :bugeye:

  9. Jul 14, 2009 #8


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    You're making an (incorrect) assumption, and the point of this excericise (and the many other similar excercises such as the bug-rivot paradox, pole-barn paradox etc.) is to identify the assumption, and then to avoid it by using the full power of the special relativity. I don't want to give it away ...

    Here's a hint: What do you mean by "proper length"? And, more importantly, is your definition Lorentz invariant?
  10. Jul 15, 2009 #9
    Thanks for replying, turin and anant.

    This is precisely what I am weak at...visualizing problems on time-space diagrams.

    I was trying to make something out from your earlier post. Proper length is the length measured in the rest frame of the rod. The S' frame is not the rest frame of the rod, so the length measured in S' must be measured simultaneously, i.e. the coordinates [itex]x_{L}'[/itex] and [itex]x_{R}'[/itex] of the two ends of the rod must be measured at the same time instant. The difference [itex]|x_{R}'-x_{L}'|[/itex] would be the length observed in S'. Is this what you were trying to point out?

    (PS -- Sorry for the late reply, I forgot about this thread.)
  11. Jul 15, 2009 #10


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    OK, but you probably should try. Otherwise, you're missing perhaps the most important point of modern relativity.

    That's part of it. That being your thought process to arive at |xL-xR'|. Now, what about the hole ...

    Hint: In order to simplify things even further, consider only the right end of the rod and only the right edge of the hole. Then, consider the left.
  12. Jul 18, 2009 #11
    Maybe you can suggest some good exercises to sharpen the visualization...I've always thought of relativity algebraically, and I haven't "solved" any problem using the space time diagram approach. Would Wheeler's book be a good idea?
  13. Jul 18, 2009 #12

    Doc Al

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    Hint: In the original frame, the rod is horizontal. Is that true in the second frame? :wink:
  14. Jul 18, 2009 #13


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    You can handle this algebraically, too, but you need to use 4-vectors (well, actually, 3-vectors in this problem, where 1 dimension is time and the other two are space). You can describe a worldline algebraically. There is a concept of proper velocity, which is defined as:


    So, you can determine the spatial position of a massive point as a function of time, in terms of the proper velocity of the point:

    xμ - xμ0 = ∫ uμ √{1-β2} dt

    Since all of the uμ's and β's are constants in this problem (in a particular frame of reference), you find that:

    xμ - xμ0 = uμ √{1-β2} t

    To get you started, in the frame S, the hole moves with a proper velocity uμ = (c,0,0), and the rod moves with a proper veolocity uμ = (c,v,-w)/√{1-(v/c)2-(w/c)2}. The first component is the time component, the second is the x component, and the third is the y component. I would choose the origin of spacetime as the time when the rod just enters the hole, and the spatial point in the center of the hole.

    If you understand that, you should be capable of the algebraic approach. However, it is a bit tedious, and it is probably better to approach this problem geometrically. Unfortunately, I don't know of a good book about this.
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