Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Density Increasing

  1. Apr 22, 2009 #1


    User Avatar
    Gold Member

    When an object moves by you, I know that its density is supposed to increase (at least that's what I was told by my physics teacher); however, I don't understand why. I understand why the density is greater in a moving reference frame relative to an observer at rest, but why does it continue to increase as it moves by you?

  2. jcsd
  3. Apr 22, 2009 #2
    Your physics teacher is probably thinking that if a box of mass M and dimension x by y by z travelling in the z direction has its z dimension reduced by the Lorentz contraction. So the volume of the box has apparently shrunk. But mass is conserved.
  4. Apr 22, 2009 #3


    User Avatar
    Gold Member

    I don't quite follow. I understand that the observer at rest would see the length of the box contract in its direction of motion, and that, assuming work was done on the box, the mass of the box appears to increase as well. However, I do not understand how the density continues to increase as the box approaches the observer.
  5. Apr 22, 2009 #4


    Staff: Mentor

    If it approaches inertially then the density will be constant (but higher than the density in the rest frame).
  6. Apr 22, 2009 #5


    User Avatar
    Gold Member

    That's what I thought. Perhaps that's what my teacher meant by density increasing as the block passes by you. Thanks.
  7. Apr 23, 2009 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    As others have said, the mass-energy density rho' in the observer's frame is constant (if velocity is constant) and greater than the density rho as measured in a frame in which the stuff is at rest.

    For completeness, I'll make things a bit quantitative, even though jgens does not need an answer at this level.

    The two densities differ by two factors of gamma = 1/sqrt(1 - V^2 / c^2), roughly, one for Lorentz contraction and one for the transformation of energy between the two frames. Consequently,

    rho' = rho/(1 - v^2 / c^2).

    A more sophisticated way to look at it is that density is the zero-zero component of the stress-energy tensor, and there is a factor of gamma for each of the two indices.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook