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Length contraction problem

  1. May 15, 2012 #1
    Consider two observers in relative motion.At x'=0, t'=0, x=0, t=0. There's a box at distance x metres in the unprimed frame. Since the primed frame moves with a relative velocity v, he observes the length x to be contracted by a factor gamma. Does this imply that when the obsever in the unprimed frame sees the observer in the primed frame to be at x, the one in the primed frame has travelled farther from the box as seen from his own frame?
     
  2. jcsd
  3. May 15, 2012 #2

    HallsofIvy

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    No. Since the primed frame is moving relative to the unprimed frame, all distances in the primed frame, including the distance the second observer moved, are shortened.
     
  4. May 15, 2012 #3
    "When" as determined by whom?

    This is a very useful applet:

    http://www.reagenix.com/personal/sci/space_time/test.html

    You are in the unprimed frame so draw a vertical line offset from the origin by x metres for the edge of the box and another through the origin at a slant for the primed frame observer. Then move the slider to view the worldlines in the primed frame. Right-click to add dots at times to mark events.
     
  5. May 15, 2012 #4
    Good enough.
     
  6. May 15, 2012 #5
    The unprimed observer.
    Why is the graph of a uniformly accelerated object a hyperbola? Can that be derived?
     
  7. May 15, 2012 #6

    DrGreg

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  8. May 16, 2012 #7
    I have another stupid question(this should probably be in the FAQ section)
    Two observers initially at the same position. Then one accelerates constantly, till it attains some velocity. As he accelerates, he observes the lengths in the other observer's frame to contract progressively according to the varying factor gamma. So now this means that an object of length x in the frame that didn't accelerate contracts to x/γ in time Δt as measured by the accelerating ob. If the length x is sufficiently large (or if γ is) the farther end of the object could be seen to contract at a speed greater than that of light.
     
  9. May 17, 2012 #8
    DrGreg, how did you derive eq.7 and 8? In eq.5, the v0 should be omitted and c^2 should be c
     
    Last edited: May 17, 2012
  10. May 17, 2012 #9

    DrGreg

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    You are quite right about that. I did attempt to correct this in post #28 of the same thread, although my correction wasn't quite right either.

    It looks like I quoted the wrong equation number. Putting (2) and (3) into (5) to get (7)...
    [tex]\begin{align}
    dt' &= dt \, \cosh \phi_0 - \frac{dx}{c} \sinh \phi_0 \\
    &= (\cosh \phi \,\,\, dt) \, \cosh \phi_0 - (\sinh \phi \,\,\, dt) \, \sinh \phi_0 \\
    &= \cosh (\phi - \phi_0) \, dt
    \end{align}
    [/tex]and similarly (2) and (3) into (6) gives (8).
     
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