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

Minkowski-minkowski thin shell paradoxon?

  1. Aug 4, 2010 #1
    Spherically symmetric infinitesimally thin shells
    can be described via the well known junction formalism of Israel.
    Let us consider such shell in vacuum,
    this means that on both sides of the shell we have Schwarzschild spacetimes.
    One of the dynamical equations is the first component of the Einstein equation
    in the thin shell limit:

    sqrt(1-2mc/r+v^2) - sqrt(1-2(mc+mg)/r+v^2) = mr/r

    where r is the circumferential radius;
    v = dr/dtau, and tau is the proper time of the shell;
    mc is the central Schwarzschild mass parameter;
    mg is the gravitational mass of the shell, this means
    that the outer Schwarzschild mass parameter is mc+mg;
    and mr is the rest mass of the shell, mr > 0;

    Let assume the dust case,
    when all mass parameters are contant during the motion.

    In the case of Minkowski-Minkowski junction mc=0, mg=0
    and the equation reduces to

    sqrt(1+v^2) - sqrt(1+v^2) = mr/r

    therefore mr=0 is the only solution.
    This is what we expect becaue the whole space is Minkowski.

    But let us consider Minkowski-Schwarzschild junction.
    In this case mc=0 but mg is not restricted.
    The equation is

    sqrt(1+v^2) - sqrt(1-2mg/r+v^2) = mr/r

    The solution of this equation for mg is:

    mg = mr(2r*sqrt(1+v^2)-mr)/r/2

    We can see that for mr=0 we get mg=0,
    but it is possible to set mg=0 with positive mr also!
    For example with the initial condition v=0,
    and mr=2r we get mg=0.
    And since mass parameters are constant during the motion,
    we get a moving shell solution with positive rest mass
    but zero gravitational mass,
    which means that the spacetime is Minkowski both inside and outside!
    It is very strange if this is the reality.
    We have energy in the spacetime
    but the spacetime is Minkowski except on a singular hypesurface.
     
  2. jcsd
  3. Aug 5, 2010 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Suppose

    [tex]\sqrt{1 + x} - \sqrt{1 + y} = z.[/tex]

    Then, [itex]x = y[/itex] implies [itex]z = 0[/itex].

    Rearrangement and squaring, however, gives

    [tex]
    \begin{equation*}
    \begin{split}
    1 + y &= \left(\sqrt{1 + x} -z \right)^2 \\
    &= 1 +x -2z\sqrt{1 + x} + z^2 \\
    y &= x + z\left(z - 2\sqrt{1 + x}\right) .
    \end{equation*}
    \end{split}
    [/tex]

    Consequently, [itex]x = y[/itex] implies [itex]z = 0[/itex] or [itex]z = 2\sqrt{1 + x}[/itex]. What happened?
     
  4. Aug 5, 2010 #3
    Thank you, sorry about this triviality
     
  5. Aug 5, 2010 #4
    But I have another question.
    Now we assume Schwarzschild-Schwarzschild shell.
    In this case mg can be negative,
    and the positive energy theorem is not violated,
    since mc + mg > 0 and mr > 0,
    where mc is the central Schwarzschild mass parameter,
    mr is the rest mass of the shell,
    and mg is the gravitational mass of the shell.
    How can we interpret this situation?
    Above both horizon, this is a normal situation?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Minkowski-minkowski thin shell paradoxon?
  1. Minkowski metric (Replies: 8)

  2. Minkowski force (Replies: 4)

  3. Minkowski Metric (Replies: 36)

  4. Minkowski space (Replies: 28)

  5. Minkowski Force (Replies: 3)

Loading...