I would like to compute the first relativistic correction to the heat capacity of free relativistic gas. At first I would chose the following Hamiltonian:

$$H(p,q)=\sum_{a=0}^N \sqrt{m^2c^4+c^2\vec{p}_a^2}$$

My problem is now the approach. My first idea was to take the statistical sum $Z_N$ and insert the Hamiltonian. Through $Z_N$ I would be able to derive the heat capacity from the Helmholz free energy $F(T,V,N)=-k_BT\ln Z_N$. My thoughts behind this idea are that the Hamiltonian of free relativistic gas is the correction itself. But I am not sure about that, because it is not really a correction. Unfortunately, I did not find any literature on the first relativistic correction.

I just know the result should be $\propto \frac{k_B T}{mc^2}$.

I am grateful for any helpful approaches or comments.


The trick is first to realize that the Hamiltonian you have there is the sum of the so-called rest energy plus kinetic energy, so one needs to first subtract the rest energy in order to remain only with KE = Internal Energy:

$$H= \sum_{i=1}^{N} mc^2 \left\{\left[1+\left(\frac{\vec{p}_i}{mc}\right)^2\right]^{\frac{1}{2}}-1\right\} \tag{1}$$

Then indeed the canonical partitition function is calculated in the no-self-interaction approximation, i.e.

$$Z(T,V,N) = \frac{1}{N!}\left[Z(T,V,1)\right]^N \tag{2}$$

Now (after integrating the phase space coordinates and converting to spherical coordinates in momentum space)

$$Z(T,V,1) = \frac{4\pi V}{h^3}\exp\left(\frac{mc^2}{k_B T}\right) \int_0^\infty p^2 ~ dp \exp\left[-\frac{mc^2}{k_B T}\left(1+\frac{p^2}{m^2 c^2}\right)^{\frac{1}{2}}\right] \tag{3}$$

The integration is easily done with Kelvin functions and you eventually obtain $F(T,V,N)$ as

$$ F(T,V,N) = -N\beta^{-1} \left[\ln\left(\frac{4\pi V}{N}\left(\frac{mc}{h}\right)^3 \frac{K_2 (\beta mc^2)}{\beta mc^2}\right)-1\right]- Nmc^2 \tag{4},$$ where $\beta=(k_B T)^{-1}$. From $(4)$ you can compute any thermodynamical quantities. Here, as you can see, the calculation is exact, in the end results for $C_V, C_p, C_T$ you can consider various approximations (series expansions).

  • $\begingroup$ I have three questions. The first is why you have in equation (1) $((p_i^2/(mc))^2$ or is it a mistake and it should be only $p_i$ in the brackets? In equation (4) I would have $(1+p^2/c^2m^2)$ instead of $(1+p^2/2m)$. Is there a mistake too or I am wrong? And to solve the integral (3) did you use the series expansion for $e^x$? $\endgroup$ – StefanBoltzmann Jun 13 at 20:47
  • $\begingroup$ Typos corrected, now. The integral is a substitution to bring it to a known value in terms of modified Bessel functions: just plug $\frac{p}{mc} = \sinh x$. $\endgroup$ – DanielC Jun 13 at 22:20

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