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Rotating rigid sphere stress-energy tensor

  1. Mar 21, 2009 #1
    Hy.

    Can somebody please show me the way, how to transform stress-energy tensor for sphere in rest frame to stress-energy tensor in rotating frame using Lorentz transformations?
     
  2. jcsd
  3. Mar 21, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hy aatw! Welcome to PF! :smile:

    Whyever would you want to use a rotating frame? :confused:

    Anyway, the rotations in the group of Lorentz transformations are exactly the same as in ordinary Euclidean space. :wink:
     
  4. Mar 21, 2009 #3
    Thanks for your quick reply.
    Maybe I didn't express myself so well.
    Let's start from stress-energy tensor for sphere, which can be written as [tex]T^{ab} = (\rho + p) u^a u^b + p g^{ab}[/tex], but in rest frame where sphere doesn't rotate, it's only [tex]T^{00}[/tex] element that's nonzero (I think so!).
    But now sphere starts to rotate with angular speed [tex]\omega[/tex] around z axis and I want to use Lorentz transformation to determine [tex]T^{ab}[/tex] for rotating sphere.
     
  5. Mar 21, 2009 #4

    Mentz114

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    Have you tried plugging in this velocity 4-vector ?
    [tex]


    \left[ \begin{array}{c}
    -\sqrt{\omega^2r^2-c^2}\\\
    -\omega r\sin(\omega t) \\\
    \omega r\cos(\omega t) \\\
    0 \end{array} \right]
    [/tex]
    where [itex]r^2=x^2+y^2[/itex] into the EMT ? This seems to be in cartesian coords but easy enough to transform to polar.

    I suppose another approach would be to boost in the x and y directions with velocities as given above. I'll try that later if I have time.
     
    Last edited: Mar 21, 2009
  6. Apr 2, 2009 #5
    No, if [itex]g = \eta[/itex] then [itex]T^{ij} = P \delta^i_j[/itex] where P is the pressure between the elements (just use the formula you quoted with [itex]u^a = (1,0,0,0)[/itex] -which is the 4-velocity in the rest frame).

    ok, so just use an ordinary boost in all 3 spacial directions with [itex]v=\omega r[/itex]
     
    Last edited: Apr 2, 2009
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