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