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Ultra-Relativistic sound speed problem

  1. Jun 15, 2012 #1
    I'm trying to understand how to use the special relativistic sound formula for a perfect fluid :

    [itex]c_s = c \, \sqrt{\frac{dp}{d\rho}},[/itex]

    where [itex]p[/itex] is the isotropic pressure and [itex]\rho[/itex] is the total energy density (not the internal energy density [itex]\rho_{int}[/itex] or mass density [itex]\rho_{mass}[/itex]).

    In the case of an ultra-relativistic fluid, we have [itex]p(\rho) = \frac{1}{3} \rho[/itex], so we get

    [itex]c_s = \frac{c}{\sqrt{3}} \approx 58\% \, c.[/itex]

    This is given in Weinberg's book on general relativity and looks clear to me.

    Now the problem is this : what about the following equation of state ?

    [itex]p(\rho) = \kappa \, \rho^{\gamma}.[/itex]

    For an ultra-relativistic perfect fluid, we have the adiabatic index [itex]\gamma = \frac{4}{3}[/itex], so the previous formula doesn't give the same speed as Weinberg's :

    [itex]c_s = c \, \sqrt{\frac{dp}{d\rho}} = c \, \sqrt{\frac{\gamma \, p}{\rho}} \ne \frac{c}{\sqrt{3}}[/itex]

    What am I doing wrong here ?

    I suspected it's because the variable isn't the same here (symbol confusion ?).
    I know that the total energy density, mass density and internal density are related by this relation :

    [itex]\rho = \rho_{mass} + \rho_{int},[/itex]

    but then, what density should I use in the equation of state [itex]p(\rho) = \kappa \, \rho^{\gamma}[/itex] ?
    [itex]\rho_{tot} \equiv \rho[/itex] ? [itex]\rho_{mass}[/itex] ? or [itex]\rho_{int} = \rho - \rho_{mass}[/itex] ? And if it's [itex]\rho_{mass}[/itex], how can I define [itex]\rho_{int}[/itex] as a function of [itex]\rho_{mass}[/itex] ?

    And I don't understand why we have two equations of states for the perfect ultra-relativistic fluid :

    [itex]p(\rho) = \frac{1}{3} \rho[/itex]


    [itex]p(\rho) = \kappa \, \rho^{4/3}.[/itex]

    Last edited: Jun 15, 2012
  2. jcsd
  3. Jun 16, 2012 #2
    Never mind, I've found the solution to my problem.

    In the equation of state

    [itex]p = \kappa \, \rho^{\gamma},[/itex]

    [itex]\rho[/itex] is actually the proper mass energy density ; [itex]\rho_{mass}[/itex], not the total energy density.

    We can also write this equation of state

    [itex]p = (\gamma - 1) \, \rho_{int},[/itex]

    with the relation

    [itex]\rho_{tot} = \rho_{mass} + \rho_{int}.[/itex]

    Here, [itex]\rho_{int}[/itex] is the internal energy density. Equating both equations of state gives a simple relation between [itex]\rho_{tot}[/itex] and [itex]\rho_{mass}[/itex], and allows the calculation of the sound velocity in full relativistic form :

    [itex]c_s = c \, \sqrt{\frac{dp}{d\rho}} = c \, \sqrt{\frac{\gamma \, p}{\rho \, + \, p}},[/itex]

    where [itex]\rho \equiv \rho_{tot}[/itex] is the total energy density. That's what I was looking for.

    In the ultra-relativistic regime, we have by definition [itex]p = \frac{1}{3} \rho_{tot}[/itex]. Putting this into the previous velocity formula, I get [itex]c_s = \frac{c}{\sqrt{3}}[/itex] only if [itex]\gamma = \frac{4}{3}[/itex], which is fine !

    The velocity formula given above is great since it's valid for all regimes, from non-relativistic up to the ultra-relativistic regime !
    I never saw that formula elsewhere in any book or document on the web. I'm now wondering if it's already known somewhere...
    Last edited: Jun 16, 2012
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