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what is the relation between mass density $\rho$ and pressure $P$ for a perfect fluid ?

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up vote 4 down vote accepted

There is no restriction. The simplest choice is $P = \kappa \rho$, where $\kappa$ is a real constant, and that is often used to get simple results. $\kappa=0$ is dust, $\kappa=-1$ is cosmological constant, and $\kappa=\frac{1}{3}$ is a gas of photons. A proper analysis also needs an equation describing the flow of internal energy and its relationship to temperature.

As a concrete example, note that $PV=NkT$ can be rewritten as $P = \frac{N}{V}kT = \frac{\rho}{m_{p}}kT$, where $m_{p}$ is the mass of the particle making up the ideal gas. Thus, if the fluid is held at a constant temperature, we get $\kappa = \frac{kT}{m_{p}}$. If the fluid is not at a constant temperature, we need to use the First Law of Thermodynamics and knowledge of the process (isoentropic, adiabatic, etc) the system is undergoing to get a relationship between the density and pressure.

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@Jerry Schirmer: In euler equations, the third one corresponding to law of conservation of energy seems to relate pressure with $\rho$ and momentum distribution $p$. What is its usefulness – Rajesh Dachiraju Nov 30 '10 at 19:20
Yes, there's a relation there, but you have more unknowns than equations in the Navier-Stokes equation--you have two scalar equations and one vector equation, but you don't know the three components of the fluid velocity, the density, the pressure, and the energy density, which is five unknowns. – Jerry Schirmer Nov 30 '10 at 19:32
@Jerry Schirmer: What is the name of $\kappa$. It seems to be an intrinsic property of the matter itself. To avoid particle nature of matter it seems that it is safe to assume $\kappa = 0$. Is it useful to make such an assumption ? – Rajesh Dachiraju Nov 30 '10 at 19:35
@Jerry Schirmer: in the third one corresponding to law of conservation of energy, if we assume that the internal energy per unit volume $e$ to be zero, should we also assume $P = 0$ ? – Rajesh Dachiraju Nov 30 '10 at 19:40
@Jerry: you've got a typo in saying $k=-1$ for cosmological constant (should be $\kappa$). – David Z Nov 30 '10 at 20:16

I see you are at the moment interested in fluid dynamics and its motion under an external force. For the given question, you are searching the Euler equations which should be introduced in any standard fluid dynamics textbook and most of the books on continuum mechanics. I suggest you to pick your favorite one and try to understand from there. The community will be glad to help you in any related questions. Sincerely,


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Thank you.I am working on it. – Rajesh Dachiraju Nov 30 '10 at 12:51

This page gives the energy-momentum tensor of a perfect fluid:

$$T^{\mu\nu} = (\rho + p) \, U^\mu U^\nu + p \, \eta^{\mu\nu}$$

Knowing the energy and momentum densities, it is then trivial to relate them to mass density in either the classical or relativistic cases.

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Yet-to-be-postulated. Perfect fluid is a set of assumptions that does not touch compressibility (ok, it brings some limitation to some very esoteric cases, but this does not change the meritum).

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