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Relativistic hydrostatic

  1. Jul 6, 2010 #1
    Hi everybody,

    When I have a perfect fluid within a constant gravitational field g, the pressure at depth z simply reads P=\rho g z, where \rho is the fluid density (incompresible).

    If now my g is still constant, but my fluid is relativistic. Not because it's moving (everything is still) but because it is hot with k_B T \sim mc^2. What happens to P = \rho g z ?

    Thanks!
     
  2. jcsd
  3. Jul 7, 2010 #2

    Mentz114

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    Perfect fluids do not contain or conduct heat, so you have some other kind of fluid, perhaps ?
     
  4. Jul 7, 2010 #3

    Ich

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    They do.
    P= k_B T\rho_0/m + \rho g z with \rho=\gamma \rho_0.

    You can use [ tex ] and [ /tex ] (click on the formula):
    [tex]P=k_B T\rho_0/m + \rho g z[/tex]
     
  5. Jul 7, 2010 #4

    Mentz114

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    If you're right, I have been misinformed.
     
  6. Jul 7, 2010 #5

    Ich

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    Ok, my comment was too general.
    What I meant is that it makes sense to ask about a perfect fluid with a certain temperature in the way antoinebret did. From his data you can calculate energy density and pressure and answer the question with the perfect fluid behaviour, the extreme case being a photon gas.
    I agree that there's no heat conduction, though.
     
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