In relativity, "stiff matter" is described by the relation $p = \rho$, where $p$ is the fluid's pressure and $\rho$ is its total energy density. The sound velocity in stiff matter equals the velocity of light in vacuum (I'm using natural units, so $c \equiv 1$): \begin{equation}\tag{1} c_s = \sqrt{\frac{dp}{d\rho}} = 1. \end{equation} For a general polytrop fluid (pressure $p = \kappa \, \rho_{\text{mass}}^{\gamma}$), we can prove the following expression: \begin{equation}\tag{2} c_s = \sqrt{\frac{\gamma \, p}{\rho + p}}. \end{equation} I need a confirmation that the adiabatic index $\gamma$ of stiff matter is $\gamma = 2$ ($p = \rho$ in expression (2) gives $\gamma = 2$, when $c_s = 1$). I find this value puzzling, since the adiabatic index of a polytrop fluid is usually the exponent of the following state equation ($\rho_{\text{mass}}$ is the proper mass density, not the total energy density): \begin{equation}\tag{3} p = \kappa \, \rho_{\text{mass}}^{\gamma}. \end{equation} So for $\gamma = 2$, we get $p = \kappa \, \rho_{\text{mass}}^2$ for stiff matter? Is that right? I feel there's an inconsistency somewhere.
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I now believe that stiff matter cannot be described as a polytrop fluid, so (2) and (3) doesn't apply to it. It's not even a "perfect fluid", since there are strong interactions between its microscopic constituents. – Cham Jan 29 '20 at 18:06
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Perfect fluid is not about interactions of microscopic constituents but about absence of shear stresses: stress–energy tensor in appropriate frame has only $\rho$ and three equal $p$ components. – A.V.S. Jan 29 '20 at 18:49
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@A.V.S., I don't think this is enough to define a perfect fluid. – Cham Jan 29 '20 at 18:56
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If you add some consistency requirement (e.g. fluid must be barotropic) then it is enough. In particular there is no need to even postulate some sort of microscopic structure. – A.V.S. Jan 29 '20 at 19:51
2 Answers
The equation $$ p = \kappa \rho_{\rm mass}^{\gamma},$$ with $\gamma = 2$, does represent a perfect fluid with $p = \rho$ in the limit that $\rho$ becomes very large.
This is shown for example by Chavanis (2014, see section II).
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I find it very strange that stiff matter is a perfect fluid, if $\rho = \rho_{\text{mass}} + \rho_{\text{int}} \approx \rho_{\text{int}}$ when $\rho_{\text{int}} \gg \rho_{\text{mass}}$ (internal energy much larger than proper mass energy) while not an ultra-relativistic fluid $p = \frac{1}{3} , \rho \approx \frac{1}{3},\rho_{\text{int}}$. Internal interactions must be "hard" and long range to get a stiff material, so it's not properly a "perfect fluid"! Physically, a "perfect fluid" is one without significant internal interactions (except at point collisions) or long rang internal forces. – Cham Jan 29 '20 at 23:17
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1@Cham: Physically, a "perfect fluid" is one without significant internal interactions here is the root of your misconceptions/confusion. “Perfect fluid” in relativity is a model of stress-energy tensor structure only, it does not requires the existence of some “free” dynamics. – A.V.S. Jan 30 '20 at 08:24
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While I agree with you, a perfect fluid is defined in statistical mechanics as the one without microscopic interactions (free particles), so we get $p V = N k T$ (valid even in special relativity). From this, we can find the polytropic relation (assuming adiabatic reversibility) : $p = \kappa,\rho_{\text{mass}}^{\gamma}$ and $p = (\gamma-1),\rho_{\text{int}}$. The special relativistic stress-energy tensor assumes implicitely the same since there's no shear and viscosity. Maybe the SR way has more room for other fluids which aren't "perfect", like the Van der Waals gas and variants... – Cham Jan 30 '20 at 11:25
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The point is that the relation $p = \rho$ implies some microscopic interactions, so the polytrop state relation $p = \kappa , \rho_{\text{mass}}^{\gamma}$ isn't strictly valid for stiff matter. It may apply only as an approximation or some limiting case, when $\rho_{\text{int}} \gg \rho_{\text{mass}}$ (internal energy much larger than proper mass energy). – Cham Jan 30 '20 at 11:51
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2@Cham: we are not in statistical mechanics, it is normal for the concepts shared between different areas of physics to have different meanings that agree in some specific cases and disagree in others. The equation of state $p=\kappa \rho_m^γ$ is something that makes sense in relativity, but for example interpreting $γ$ as heat capacity ratio is wrong because the temperature isn't really defined here. – A.V.S. Jan 30 '20 at 16:24
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@A.V.S., I started a new question about terminology here: https://physics.stackexchange.com/questions/528144/what-is-a-perfect-fluid-really. I think this is the essence of my confusion. What really is a perfect fluid and an ideal fluid? – Cham Jan 30 '20 at 16:25
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@A.V.S., oh I agree with the comment on the heat capacity ratio being wrong! I always hated this definition in thermodynamics! – Cham Jan 30 '20 at 16:27
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1@Cham: …polytrop state relation $p=κρ^γ_\text{mass}$ isn't strictly valid for stiff matter. One way to reconcile our stances would be to abandon $ρ\text{mass}$ as “proper mass density” for strongly interacting systems. In GR we only have the requirement of conservation $\nabla\mu j^\mu=0$ for the current transporting the density $\rho$, it does not necessarily mean that it must correspond to a “proper mass” of some noninteracting system for specific EOS that do not admit a weakly coupling limit (like stiff matter). – A.V.S. Jan 30 '20 at 16:44
My answer follows Hawking & Ellis (section 3.3, Example 4: Isentropic perfect fluid) except for notation which was taken from the answer.
Isentropic perfect fluid is described by proper mass density $\rho_\text{mass}$ and an elastic potential $e$ (which is a function of $\rho_\text{mass}$). Then the energy density is $$ \rho = \rho_\text{mass} (1+e),$$ while the pressure is $$ p = \rho^2_\text{mass} \frac{\mathrm{d}\, e}{\mathrm{d} \rho_\text{mass}}. $$
Assuming $e = \kappa \,\rho_\text{mass} - 1$ we find that $$ \rho =\kappa \, \rho^2_\text{mass},\qquad p=\kappa \, \rho^2_\text{mass} = \rho. $$
Thus, there is no inconsistency: the equation of state corresponds to stiff matter for all values of pressure/energy.
The difference with Chavanis' paper and Rob Jeffries' answer comes from the choice of constant $A$ in Chavanis' equation $(4)$. By setting $A$ to $1$ Chavanis ensures that $\rho\approx \rho_\text{mass}$ when $\rho_\text{mass}\to 0$. But if we set $A=0$ we would ensure that $p=\rho$ for all values of pressure.
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Why? Do you have the book of H&E? The equations are the same as in Chavanis only instead of starting with $P(\rho_m)$ we start with Lagrangian with $e(\rho_m)$ (which is $\int P(\rho_m)/\rho^2_m d\rho_m$ in Chavanis). The difference is the only the choice of integration constant which in Chavinis is based on unmotivated requirement of Newtonian limit. – A.V.S. Jan 30 '20 at 08:00