Why is the relativistic adiabatic index 4/3?

Question

I was told that in the relativistic limit the adiabatic index approaches 4/3 for a monoatomic gas instead of 5/3 in the non-relativistic case. I was told this occurs due to a reduction in degree of freedom but this may be incomplete and does not quite explain the new expression since $ \gamma = (n + 2)/n$ where $n$ is the # of degrees of freedom. Thus I am wondering both quantitatively and qualitatively, why does the adiabatic index decrease, and to 4/3 specifically, in the relativistic regime for a monoatomic gas?

Answer 1

I don't know what a reduction in degrees of freedom is in this case. But statistical physics has well known methods to compute properties of ideal gases. Ultra-relativistic particles dispersion relation has the form $$ \varepsilon(\vec{p}) = c|\vec{p}| $$ Correspondent partition function of a monatomic classical gas is $$ Z = \frac{V^N}{N!(2\pi\hbar)^{3N}} \left(\int e^{-c|\vec{p}|/\theta} d\vec{p}\right)^N = \frac{V^N}{N!(2\pi\hbar)^{3N}} \left(\frac{8\pi\theta^3}{c^3}\right)^N $$ Using standard thermodynamic relations we obtain equations of state (all specific values are computed per one particle) $$ c_v = 3, \quad p = \frac{\theta}{v} $$ Further $$ c_p = c_v - \theta \left(\frac{\partial p}{\partial \theta}\right)^2_v\Bigg/\left( \frac{\partial p}{\partial v}\right)_\theta $$ gives $$ c_p = c_v + 1 = 4 $$ Due to ideal gas's thermal equation of state $p = \theta/v$ and $c_v = const$ the adiabatic process equation is $$ pv^\gamma = const, $$ where $\gamma = c_p/c_v = 4/3$.

The talk about degrees of freedom is relevant when the energy of a molecule is equal to a sum of quadratic terms. For an ultra-relativistic atoms this is not a case.