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The definition of the adiabatic index of a relativistic polytrope fluid is this: $$\tag{1} \gamma = \frac{\rho + p}{p} \, \frac{dp}{d\rho}, $$ where $\rho$ is the energy density (not the mass density). The sound velocity in the fluid is restricted by the principle of causality, so (I'm using units such that $c \equiv 1$): $$\tag{2} v_s^2 \equiv \frac{dp}{d\rho} \le 1 \qquad \Rightarrow \qquad \gamma \le \frac{\rho + p}{p}. $$ For a constant $\gamma$, one solution to (1) is non-linear: $$\tag{3} \rho = \frac{p}{\gamma - 1} + (p / \kappa)^{1/\gamma}, $$ where $\kappa$ is a positive integration constant. Then, a linear EoS (equation of state) is another solution for the same value of $\gamma$: $$\tag{4} \rho = \frac{1}{\gamma - 1} \, p. $$ Clearly, this linear function has $v_s > 1$ for $\gamma > 2$.

In the special case of $\gamma = 2$, (3) can be inverted: $$\tag{5} p(\rho) = \frac{1}{4 \kappa} \bigl( \sqrt{1 + 4 \kappa \rho} - 1 \Bigr)^2. $$ This non-linear EoS gives $p(\rho) \approx \kappa \rho^2$ when $4 \kappa \rho \ll 1$, and gives $p(\rho) \approx \rho$ when $4 \kappa \rho \gg 1$. The speed of sound is always lower than 1 (except in the limiting case $\rho \rightarrow \infty$): $$\tag{6} v_s^2 = \frac{dp}{d\rho} = 1 - \frac{1}{\sqrt{1 + 4 \kappa \rho}}. $$ This could also be the case for $\gamma > 2$, when the EoS is non-linear, so from (1)-(3), how can we prove that $v_s \le 1 \; \Rightarrow \; \gamma \le 2$ even for a non-linear EoS?

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  • $\begingroup$ The stiffest causal EoS is $p=\rho$ which yields the limiting case of $\gamma=2$. $\endgroup$
    – N0va
    Commented Aug 17, 2023 at 22:52
  • $\begingroup$ @N0va, how do you derive this from the polytope expression (3)? It implies $\kappa \rightarrow \infty$ $\endgroup$
    – Cham
    Commented Aug 17, 2023 at 23:55

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I am not really sure if I understood the question correctly to be honest but I will give an answer a try: the polytropes discussed here can be expressed in terms of number density $n$ as $$ \begin{align} p(n)&=k n^\gamma,\\ \rho(n)&=m n+\frac{k n^\gamma}{\gamma-1}, \end{align} $$ which we may use to get the following closed expression for the speed of sound squared $$ v_s^2=\frac{(\gamma -1) \gamma k n^{\gamma }}{\gamma k n^{\gamma }+(\gamma -1) m n}. $$ This can be analyzed for different values of $\gamma$ and for $\gamma>1$ we may note that for large $n$ the denominator in the expression for $v_s^2$ is dominated by the term $n^\gamma$ for large $n$ and thus we find $v_s^2=\gamma-1$ for large $n$ and $\gamma>1$. Which yields $\gamma=2$ as the limit for a causal polytrope.

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  • $\begingroup$ The conclusion depends on the absence of a limit to the number density $n$. For example, we could use $\gamma = 3$ into your last equation, and find $n_{max}$ from the constraint $v_s^2 \le 1$. It gives $$n \le n_{max} = \sqrt{\frac{2 m}{3 \kappa}}.$$This allows causal polytropes with $\gamma > 2$, at the condition $n < n_{max}$. I'm not sure what to think of this. Why should $n$ be without any finite upper limit? $\endgroup$
    – Cham
    Commented Aug 18, 2023 at 18:43
  • $\begingroup$ In general, for $\gamma > 2$, we get $$v_s^2 \le 1 \qquad \Rightarrow \qquad n_{max}^{\gamma - 1} = \frac{m}{\gamma k} \Bigl( \frac{\gamma - 1}{\gamma - 2} \Bigr).$$ $\endgroup$
    – Cham
    Commented Aug 18, 2023 at 18:49
  • $\begingroup$ Getting into semantics a 'polytrope' with $\gamma>2$ would technically be a piecewise polytrope and a rather peculiar one at that: looking at Eq. (3) in the question one might notice that $\rho<p$ for small $p$ and $\gamma>2$ thus violating this "dominant energy condition". There might a gerneal connection here; a-causal EoS at high $n$ $\leftrightarrow$ violation of "dominant energy condition" at low $n$? $\endgroup$
    – N0va
    Commented Aug 18, 2023 at 20:20
  • $\begingroup$ I don't understand your comment. Violating the dominant energy condition means $p > \rho$, which implies the following inequalities for $\gamma > 2$:$$n^{\gamma - 1} > \frac{m}{k} \Bigl( \frac{\gamma - 1}{\gamma -2} \Bigr) > \frac{m}{\gamma k} \Bigl( \frac{\gamma - 1}{\gamma - 2} \Bigr) \equiv n_{max}^{\gamma - 1}.$$Thus, there's no violation of causality and no violation of the energy condition when $n < n_{max}$. But I'm still perplexed by an upper limiting value of the particles density ($n_{max}$). $\endgroup$
    – Cham
    Commented Aug 19, 2023 at 13:01
  • $\begingroup$ The upper limiting value of the particles density implies a maximal pressure (for $\gamma > 2$):$$p_{max} = k n_{max}^{\gamma} = k \Bigl[\frac{m}{\gamma k} \Bigl( \frac{\gamma - 1}{\gamma - 2} \Bigr) \Bigr]^{\frac{\gamma}{\gamma - 1}}.$$ Notice that $n_{max} \rightarrow \infty$ for $\gamma = 2$. Once a star reaches the maximal pressure at the center, it would be violating the causality condition first (before the energy condition), and would probably collapse or abruptly explode. $\endgroup$
    – Cham
    Commented Aug 19, 2023 at 13:25

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