Extrapolating from non-relativistic formulae, the pressure (P) of a potentially-relativistic ideal gas, of particle density n, in N m^-2, exerted as momentum perpendicular to some unit area, per unit time, is proportional to:

 

[math]P \propto avg \left( n \times v_{\perp} \times p_{\perp} \right)[/math]

 

[math]\longrightarrow n mc^2 avg \left( \gamma(\beta) \beta_{\perp}^2 \right)[/math]

 

[math]\longrightarrow \frac{1}{3} n mc^2 avg \left(\gamma(\beta) \beta^2 \right)[/math]

 

by isotropy, ultimately resulting from [math]<\beta^2> = <\beta_x^2 + \beta_y^2 + \beta_z^2> = 3 <\beta_{1D}^2>[/math]. Now, the "average" function requires integrating over possible momentum eigenstates, weighted by the appropriate Boltzmann factor of gas temperature:

 

[math]\longrightarrow \frac{1}{3} n mc^2 \int C 4 \pi \beta^2 d\beta \left(\gamma \beta^2 \right) e^{-\gamma \frac{mc^2}{k_B T}}[/math]

 

[math]\propto \frac{1}{3} n mc^2 \int_1^{\infty} \frac{d \gamma}{\gamma^5} \left( \gamma^2 - 1 \right)^{\frac{3}{2}} e^{-\gamma \frac{mc^2}{k_B T}}[/math]

 

employing the relativistic relation [math]\gamma = (1-\beta^2)^{1/2}[/math], and its (cumbersome) differential.

 

[math]\longrightarrow \frac{1}{3} n mc^2 \tau^4 \int_{\frac{1}{\tau}}^{\infty} \frac{dx}{x^5} \left( \left( x \tau \right)^2 - 1 \right)^{\frac{3}{2}} e^{-x}[/math]

 

seeking simplicity by substituting [math]\tau \equiv \frac{k_B T}{mc^2}[/math]. The above formula has no (obvious) closed-form solution. Is the full-fledged Relativistic equation for Pressure so complicated a function of temperature?? To solve for the integration constant requires [math]1 = C \int 4 \pi \beta^2 d\beta e^{-\frac{\gamma}{\tau}} = 4 \pi C \int \frac{dx}{x^4} \left( \left( x \tau \right)^2 - 1 \right)^{\frac{1}{2}} e^{-x}[/math]. (And all of this assumes that well-defined momenta are quantum-mechanically available to the particles.)

Did someone drop a Rainman toothpick count?

Edited October 25, 201213 yr by STeve555

This is the first time I'm seeing anything like this. From the looks of it, I suppose it might not have a closed form.