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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?

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  • $\begingroup$ Reduction in degrees of freedom? Wouldn't it be an increase? $n=3\to n=6$? $\endgroup$ – Kyle Kanos May 10 '18 at 16:35
  • $\begingroup$ I agree, hence why the explanation appeared wrong. $\endgroup$ – Mathews24 May 10 '18 at 18:00
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    $\begingroup$ These common "degrees of freedoms" rules are a dangerous business! It only works in certain limits! For example at low temperatures excitations with an energy gap (e.g. oscillation) freeze out, when approaching zero temperature even the translational degrees of freedom no longer have a constant heat capacity. And also in the case that the temperature gets too high (and the gas gets relativistic) the degrees of freedom approximation gets wrong or must be adapted. $\endgroup$ – Sebastian Riese May 10 '18 at 18:14
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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.

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  • $\begingroup$ Thank you for the quantitive derivation. To perhaps make the qualitative picture clear, could you describe physically why the adiabatic index increases for ultra-relativistic atoms? $\endgroup$ – Mathews24 May 10 '18 at 18:06
  • $\begingroup$ I can not give simple qualitative explanation of why the adiabatic index increases. Albeit I think the formulas are simple enough. $\endgroup$ – Gec May 10 '18 at 18:18

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