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On length contraction (Special Relativity)

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the length contraction deforms a sphere in motion to an oblate rotational ellipsoid whose volume decreases by a factor of γ (gamma)


    2. Relevant equations
    x=x'/γ v=4∏r^3/3 (volume for sphere) v=4∏(a^2)b/3 (volume for prolate and oblate spheroid)


    3. The attempt at a solution
    The question is taken from Special relativity for beginners by Jurgen Freund. Firstly, I am confused as to why the question claims that the sphere would deform into an oblate spheroid rather than a prolate spheroid since length contraction does not contract transversely. Anyway by length contraction, the radius (r) of the sphere would contract by a factor of γ, Therefore:
    r=a/γ where a is the equatorial radius and b is the polar radius. By subbing r=a/γ into the formula of sphere, through some algebraic manipulation, I get a spheroid whose radius decreased by a factor of γ^3 instead. How should I go about doing it and where are my errors ?

    Please help.

    Thank you for viewing.
     
  2. jcsd
  3. Jun 4, 2012 #2
    So what happens to the sphere with the length contraction? The direction parallel to motion is contracted, and the two perpendicular to it are not, right? So which of a and b are transformed?
     
  4. Jun 4, 2012 #3
    The equatorial radius (a) of the prolate spheroid would be contracted by a factor of γ. (i.e a=r/γ). Therefore the volume of the sphere is V=4∏(r^3)/3 and the volume of the prolate spheroid would be V'=4∏(r^2)b/3(γ^2). Finally, taking the volume of the sphere and divide it by the volume of the prolate spheroid gives (V/V')=r(γ^2)/b ??

    Sorry if I am slow at catching ideas. Please bear with me.

    Thank you very much
     
  5. Jun 4, 2012 #4

    nrqed

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    Watch out. It is b which is contracted whereas "a" remains equal to the initial radius.
     
  6. Jun 5, 2012 #5
    Why?
     
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