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Length contraction and the direction of motion

  1. May 30, 2010 #1
    Can length contraction be treated as a vector?

    E.g.

    Suppose I am traveling 10 degrees north of east. Will my length contraction with respect to the east direction be cos 10 times the appropriate Lorentz transformation (with respect to the direction of motion)?


    Thanks.
     
  2. jcsd
  3. May 30, 2010 #2
    I just uploaded a file into my blog that answers your exact question.
     
  4. May 30, 2010 #3

    JesseM

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    This page mentions that a moving sphere is contracted into an ellipsoid (with the cross section being an ellipse), so the semiminor axis would be the length-contracted radius of the original sphere, and the semimajor axis would be the uncontracted radius. And this page gives the "radius" of an ellipse as a function of angle (relative to the semimajor axis) in polar coordinates:

    [tex]r = \frac{ab}{\sqrt{(b cos(\theta))^2 + (a sin(\theta))^2}}[/tex]

    So, if you imagine a wheel with spokes of equal radius L in its rest frame, then in a frame that sees the wheel moving, I think a spoke at angle [tex]\theta[/tex] relative to the uncontracted spoke that's perpendicular to the direction of motion (with the angle measured in the frame that sees the wheel in motion) should have a length given by this formula, with a = L and b = L / gamma. For example, a spoke oriented 10 degrees away from the direction of motion will be 80 degrees from the uncontracted spoke perpendicular to the direction of motion, and this trigonometry calculator gives sin(80) = 0.98481 and cos(80) = 0.17365, so its length should be (L^2/gamma)/sqrt[(0.17365*L/gamma)^2 + (0.98481*L)^2]. For example, at v=0.6c and 1/gamma=0.8, this would give a length of (0.8*L^2)/sqrt[0.0193*L^2 + 0.9699*L^2] = 0.804*L, just slightly longer than the length of the spoke exactly parallel to the direction of motion which is contracted to 0.8*L.

    Alternatively, if you want to define [tex]\phi[/tex] as the angle relative to the direction of motion, the formula should be:

    [tex]r = \frac{L^2 / \gamma}{\sqrt{((L / \gamma)*sin(\phi))^2 + (L*cos(\phi))^2}}[/tex]
     
    Last edited: May 30, 2010
  5. May 30, 2010 #4
    What does [itex]\sum \Delta x'^2[/itex] mean? Is it the squared length of the rod in S', this being the sum of the squares of its components?

    Why can’t we simply multiply matrices representing boosts in each of the component directions? Maybe I made a mistake, but when I tried this, the result didn’t look anything like your matrix for a general boost. It wasn’t even symmetric.

    [tex]\Lambda_x\Lambda_y\Lambda_z=\begin{bmatrix}\gamma_x \gamma_y \gamma_z & -\beta_x \gamma_x & -\beta_y \gamma_x \gamma_y & 0\\ -\beta_x \gamma_y \gamma_z & \gamma_x & \beta_x \beta_y \gamma_x \gamma_y & 0\\ -\beta_y \gamma_y \gamma_z & 0 & \gamma_y & 0\\ -\beta_z \gamma_z & 0 & 0 & \gamma_z \end{bmatrix}[/tex]
     
  6. May 30, 2010 #5
    Here is an excellent animation illustrating your above post.
     
  7. May 30, 2010 #6
    Yes


    Here is how the transform is derived.
     
  8. May 30, 2010 #7

    DrGreg

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    If you multiply two non-parallel boosts, surprisingly (perhaps) you don't get a boost, you get a boost multiplied by a spatial rotation, i.e. your x,y,z axes have rotated relative to how they started. See Thomas rotation.
     
  9. May 31, 2010 #8
    Thanks for the tips! I remember now I looked at the general boost formula last year, and even copied it down, but couldn't really make sense of it. But now that I've read about dyadics, I can finally see what they're doing in the Wikipedidia derivation: that part where they put v vT into the space-space part of the matrix. I think I must have misread it as vT v before. Right, now to Thomas rotation... Robert Littlejohn's introduction, linked to from the Wiki page, looks promisingly detailed.
     
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