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Relativistic density

  1. Aug 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A cube, one cm on an edge, is filled with pure water. As measured in a reference frame at rest relative to the cube, the density of the water is exactly 1000.0 kg/m3. The cube is now accelerated to a constant speed of 0.791c. What is the density of the water in the moving cube as measured in the same reference frame?


    2. Relevant equations

    D=m/v

    mrel=m/(sqrt(1-(v^2/c^2))

    3. The attempt at a solution

    I found m relative to be 1630 kg by using the second equation and if I divide it my 1m^3 the density should equal the new mass but that is incorrect. Can someone help please
     
  2. jcsd
  3. Aug 10, 2009 #2

    rock.freak667

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    I don't know much about this, but I believe the length would contract as well.
     
  4. Aug 10, 2009 #3

    diazona

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    Yes, that's the key.
     
  5. Aug 10, 2009 #4
    Agreed. The 1 cm length that's measured in the cube's frame is not the length measured in the observer's frame.
     
  6. Aug 10, 2009 #5
    So the relativistic density is 1634 kg ( at least that's what I calcuated)

    Then I converted 1cm to meters and found the relativistic length which was 6.1182e-3 m
    then i cubed it to find volume and got 2.29e-1m^3

    Next I found the density by diving the new mass by the new volume and got 7.13e3 kg/m^3

    Is that correct?
     
  7. Aug 10, 2009 #6
    The relativistic length contraction is:

    [tex]L' = \dfrac{L}{\gamma}[/tex]

    where [tex]\gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    The easy way to remember this formula is to remember that [tex]\gamma[/tex] is always greater than or equal to 1. Anyway, when calculating the new volume, remember that the cube is only contracted along the direction of motion. Therefore, in another reference frame it won't look like a cube, but rather a rectangular prism. You need to take this into account when finding the volume.

    As a sidenote, my recommendation (both as a student and a former TA) is to solve all problems symbolically before plugging in numbers. It makes it easier for you to find mistakes, and it makes it easier for the grader to give you partial credit. Not to mention the physics insight you tend to get...
     
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