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Relativistic Mass Density

  1. Nov 10, 2005 #1
    I've been thinking of something for quite a while I'd would really appricate some opinions of the matter. I was wondering if the mass density of an object changes when it approaches the speed of light, just like charge density changes and causes the lorentz force. Could this change in mass density effect an objects gravitational field in anyway? I tried to formulate an equation that would equate the gravitational field emitted by an object (like a star) with the rate of its spin, but saddly I don't posess the mathematical skills necisssary to accomplish such a task. If anyone has any relavent thoughts I'd appricate your input.
  2. jcsd
  3. Nov 10, 2005 #2
    Yes. The mass density is a function of speed and varies as

    [tex]\rho = \gamma^2 \rho_0[/tex]

    And yes, the mass density can effect the gravitational field too. E.g. consider a rod lying at rest in frame S on the x-axis. Now boost to a frame S' which is moving in the +x direction with respect to frame S. In frame S' the gravitational field will have a greater value given the same distance to the axis the rod is lying on.

  4. Nov 11, 2005 #3
    Interesting. I tried to apply that to a linear rod when I first though about this (athough I used the wrong equation to get mass density), because it's a hell of a lot simplier than a spinning sphere. Now if that equation can be applied by a spinning sphere... Damn, I know I've derived the equation that calcuates the magnetic field of a rotating charge in past, I just can't remember how I did it. This should be to terribly different.
  5. Nov 11, 2005 #4


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    Computing the mass of a relativistically rotating sphere (or disk) is going to be a lot more difficult than it appears.

    The sci.physics.faq, for instance, does not compute the mass or a relativistically rotating disk, and asks anyone who has such a calculation to contribute it.

    Because the space-time metric around a rotating sphere or disk should be flat, the Komar mass should apply.

    This means that one has to integrate not the energy density [itex]\rho[/itex], but rahter [itex]\rho + 3P[/itex], where P is the pressure, assuming a nearly Minkowskian metric, and geometric units.

    Because the rotating disk/sphere will be in tension, the contribution of the pressure terms to the mass intergal will be negative, making it weigh less than a naieve analysis would suggest.
  6. Nov 12, 2005 #5
    Or you can just forget this and examine the spacetime outside the sphere and note that it is identical to the spacetime of a rotating black hole whose metric is well known.

  7. Nov 12, 2005 #6


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    Yes, but you still have to do the interior solution, presumably a constant density interior solution. I've seen those for the Schwarzschild metric, but not the rotating Kerr metric. It's still probably the easiest route to solving the problem for a rotating sphere.

    Actually, as I think about it, I'm not quite sure what shape you are going to get, it may not be spherical. You are going to have to pick whatever shape gives you the correct match-up to the Kerr metric (for some specified angular momentum and mass) over some surface outside the event horizon of the black hole represented by the Kerr metric. I'm not sure what the shape of that surface is going to have to be - if it's not spherical, it may be difficult to compute the shape, making this approach difficult as well. In the Schwarzschild case it is obvious that the correct shape is a sphere, it's not obvious that this is true for the rotating case.
    Last edited: Nov 12, 2005
  8. Nov 12, 2005 #7
    Yeah, that's way out of my league. I don't have any real experience with metrics or tensors.
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