# How to prove that $\gamma \le 2$ for *causal* relativistic polytrope fluids?

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?

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

## 1 Answer

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.

• 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?
– Cham
Commented Aug 18, 2023 at 18:43
• In general, for $\gamma > 2$, we get $$v_s^2 \le 1 \qquad \Rightarrow \qquad n_{max}^{\gamma - 1} = \frac{m}{\gamma k} \Bigl( \frac{\gamma - 1}{\gamma - 2} \Bigr).$$
– Cham
Commented Aug 18, 2023 at 18:49
• 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 $n$?
– N0va
Commented Aug 18, 2023 at 20:20
• 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 ($n_{max}$).
– Cham
Commented Aug 19, 2023 at 13:01
• 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.
– Cham
Commented Aug 19, 2023 at 13:25