Barnak
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I'm trying to understand how to use the special relativistic sound formula for a perfect fluid :
c_s = c \, \sqrt{\frac{dp}{d\rho}},
where p is the isotropic pressure and \rho is the total energy density (not the internal energy density \rho_{int} or mass density \rho_{mass}).
In the case of an ultra-relativistic fluid, we have p(\rho) = \frac{1}{3} \rho, so we get
c_s = \frac{c}{\sqrt{3}} \approx 58\% \, c.
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 ?
p(\rho) = \kappa \, \rho^{\gamma}.
For an ultra-relativistic perfect fluid, we have the adiabatic index \gamma = \frac{4}{3}, so the previous formula doesn't give the same speed as Weinberg's :
c_s = c \, \sqrt{\frac{dp}{d\rho}} = c \, \sqrt{\frac{\gamma \, p}{\rho}} \ne \frac{c}{\sqrt{3}}
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 :
\rho = \rho_{mass} + \rho_{int},
but then, what density should I use in the equation of state p(\rho) = \kappa \, \rho^{\gamma} ?
\rho_{tot} \equiv \rho ? \rho_{mass} ? or \rho_{int} = \rho - \rho_{mass} ? And if it's \rho_{mass}, how can I define \rho_{int} as a function of \rho_{mass} ?
And I don't understand why we have two equations of states for the perfect ultra-relativistic fluid :
p(\rho) = \frac{1}{3} \rho
and
p(\rho) = \kappa \, \rho^{4/3}.
??
c_s = c \, \sqrt{\frac{dp}{d\rho}},
where p is the isotropic pressure and \rho is the total energy density (not the internal energy density \rho_{int} or mass density \rho_{mass}).
In the case of an ultra-relativistic fluid, we have p(\rho) = \frac{1}{3} \rho, so we get
c_s = \frac{c}{\sqrt{3}} \approx 58\% \, c.
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 ?
p(\rho) = \kappa \, \rho^{\gamma}.
For an ultra-relativistic perfect fluid, we have the adiabatic index \gamma = \frac{4}{3}, so the previous formula doesn't give the same speed as Weinberg's :
c_s = c \, \sqrt{\frac{dp}{d\rho}} = c \, \sqrt{\frac{\gamma \, p}{\rho}} \ne \frac{c}{\sqrt{3}}
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 :
\rho = \rho_{mass} + \rho_{int},
but then, what density should I use in the equation of state p(\rho) = \kappa \, \rho^{\gamma} ?
\rho_{tot} \equiv \rho ? \rho_{mass} ? or \rho_{int} = \rho - \rho_{mass} ? And if it's \rho_{mass}, how can I define \rho_{int} as a function of \rho_{mass} ?
And I don't understand why we have two equations of states for the perfect ultra-relativistic fluid :
p(\rho) = \frac{1}{3} \rho
and
p(\rho) = \kappa \, \rho^{4/3}.
??
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