Ultra-Relativistic sound speed problem

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SUMMARY

The discussion centers on the application of the special relativistic sound speed formula for a perfect fluid, specifically in the context of ultra-relativistic fluids. The correct formula for sound speed is derived as c_s = c \, \sqrt{\frac{dp}{d\rho}}, where p is the isotropic pressure and \rho is the proper mass energy density, not the total energy density. The confusion arises from the different equations of state for ultra-relativistic fluids, specifically p(\rho) = \frac{1}{3} \rho and p(\rho) = \kappa \, \rho^{4/3}. The final resolution confirms that the sound speed formula is valid across all regimes, including non-relativistic and ultra-relativistic conditions.

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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}.

??
 
Last edited:
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Never mind, I've found the solution to my problem.

In the equation of state

p = \kappa \, \rho^{\gamma},

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

We can also write this equation of state

p = (\gamma - 1) \, \rho_{int},

with the relation

\rho_{tot} = \rho_{mass} + \rho_{int}.

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

c_s = c \, \sqrt{\frac{dp}{d\rho}} = c \, \sqrt{\frac{\gamma \, p}{\rho \, + \, p}},

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

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