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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

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]

and

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

??

[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]

and

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

??

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