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Perfect fluid, stress energy tensor in the rest frame,

  1. Jun 18, 2007 #1
    If I had a perfect fluid I could write the stress energy tensor in the rest frame of the fluid as a nice diagonal tensor with [rho,p,p,p] and use this to solve the field equations.

    Pressure is just random internal motions, so what happens if all the particles in my perfect fluid decide to move to the left? Clearly, now I could transofrm into a rest frame with T=[rho,0,0,0] and this would suggest that the resulting spacetime structure we get would be different?

    Is this correct?
     
  2. jcsd
  3. Jun 18, 2007 #2

    Mentz114

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    Gold Member

    If all the particles move left, they must be under the influence of a strong field which would certainly have changed the space-time.

    But if you moved along the same geodesic as the particles then you would observe T=[rho,0,0,0].

    It's unlikely this could happen in reality - not outside star trek.
     
  4. Jun 18, 2007 #3

    pervect

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    Staff Emeritus
    Science Advisor

    I'm not positive I understand the question. I think what you may be trying to ask is this:

    Suppose we have a swarm of particles

    -> -> ->
    <- <- <-

    in the rest frame of the particles, half of them move to the left with velocity v and half move to the right with velocity v. What is the stress energy tensor for this swarm of particles in the continuum limit?

    The answer will be that it is of the form diag(rho, P, 0, 0). P here represents an anisotropic pressure, i.e.

    [tex]
    \left[
    \begin{array}{cccc}

    \rho&0&0&0\\
    0&P&0&0\\
    0&0&0&0\\
    0&0&0&0\\

    \end{array}
    \right]
    [/tex]


    You may be asking instead:

    Suppose we have a swarm of particles, all of which move with a uniform velocity v to the right. What is the stress energy tensor of this swarm of particles in the continuum limit?

    The answer to this is

    [tex]
    \left[
    \begin{array}{cccc}

    \rho&p&0&0\\
    p&0&0&0\\
    0&0&0&0\\
    0&0&0&0\\

    \end{array}
    \right]
    [/tex]

    here p (not capitalized) is a momentum density rather than a pressure

    Note that GR is presented as a classical theory, so you can't have the particles "decide" to do something improbable, if that was the point of your question.
     
    Last edited: Jun 18, 2007
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