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Length contraction of a square

  1. Aug 7, 2006 #1
    If you have a drawing of a square (labeled ABCD starting at lower left and going clockwise) and it is first at rest in an inertial frame. When this square moves, length contraction changes its shape. In each of the following three situations the speed is the same, and the velocity is in the plane of the square. when is the area of the object the least if the velocity is directed paralled to:
    a) side AB
    b) diagonal AC or
    c) side AD.
    In which of these situations is the area of the object the least?

    I know that length contraction occurs only along the direction of the motion. Dimensions perpendicular to the motion are not contracted. That is why choice (a) has the larges area and is not the answer.
    The answer is (b), but I don't understand why. Please expain.
  2. jcsd
  3. Aug 7, 2006 #2
    Well, if it's directed along diagonal AC, length would contract along that axis... which kind of scales down the square... It'll look more like a parallelogram or rhombus or something. (Can't remember much of my geometry)
    But since, in the case of motion along diagonal AC, both sides are contracted, since you have something like this:
    Which, apparently (according to your answer) has less of an area than a square that's contracted to a rectangle..
    Last edited by a moderator: Apr 22, 2017
  4. Aug 8, 2006 #3
    How is it that "the case of motion along diagonal AC, both sides are contracted" ?
    Isn't it one side being contracted, the one parallel to motion?
    I'm having trouble picturing this.
  5. Aug 8, 2006 #4


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    Staff: Mentor

    In this situation, neither side is parallel to the direction of motion, or perpendicular to the direction of motion. One diagonal is parallel to the direction of motion, and that one gets contracted. The other diagonal is perpendicular to the direction of motion, and is not contracted. Draw the new diagonals, and connect the corners to get the new shape of the diagonally-contracted square.
  6. Aug 8, 2006 #5
    thanks, I get it now.
  7. Aug 8, 2006 #6


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    Homework Helper

    The areas are the same whether the direction of travel is parallel to a diagonal or parallel to a side. Consider that the Lorentz contraction will give lengths scaled by a factor 0<k<1, now suppose we have begin with a unit square, traveling parallel to a side gives one side shrunk to length k, and the other untouched, so the area is k*1=k. Now if the direction of travel is parallel to a diagonal originally of length [tex]\sqrt{2}[/tex], then the contracted diagonal is of length [tex]k\sqrt{2}[/tex] while the other diagonal is still of length [tex]\sqrt{2}[/tex]. Consider the triangles on either side of the unchanged diagonal: they each have a base of [tex]\sqrt{2}[/tex] and a height of [tex]\frac{k\sqrt{2}}{2}[/tex] and hence they each have an area of (1/2*base*height) = [tex]\frac{1}{2}*\sqrt{2}*\frac{k\sqrt{2}}{2} = \frac{k}{2}[/tex] and since there are two triangles the total area is [tex]2*\frac{k}{2}=k[/tex]. So the areas are the same either way.
    Last edited: Aug 8, 2006
  8. Aug 8, 2006 #7

    Doc Al

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    Staff: Mentor

    I agree with benorin: The direction that the square moves doesn't matter. Whatever the direction, one linear dimension will shrink by the same fraction, thus the change in area will be the same.

    The correct answer is not listed.
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