Widdekind Posted October 25, 2012 Posted October 25, 2012 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.)
STeve555 Posted October 25, 2012 Posted October 25, 2012 (edited) Did someone drop a Rainman toothpick count? Edited October 25, 2012 by STeve555
elfmotat Posted November 6, 2012 Posted November 6, 2012 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.
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