Lets say you are given a distribution function $f(p)$ and you want to define a temperature, $T_f$, for this distribution. (I assume $\mu = 0$.)

It is then natural to define a temperature the following way: \begin{equation} T_f \equiv \frac{ \int d^3p \ G(p) f(p)}{\int d^3p \ f(p)}, \end{equation} where $G(p)$ is defined by the following equation \begin{equation} T = \frac{ \int d^3p \ G(p) f_{eq}(p,T)}{\int d^3p \ f_{eq}(p,T)}, \end{equation} where $f_{eq}(p,T)$ is the equilibrium thermal distribution function.

I know that if $f_{eq}$ is given by the Maxwell-Boltzmann distribution, then $$G_{MB}(p) = \frac{p^2}{3E},$$ where $E = \sqrt{p^2 + m^2}$.

What I need is to find an expression for $G(p)$ if $f_{eq}$ is the Bose-Einstein or Fermi-Dirac distribution $$ f_{eq} = \frac{1}{e^{E(p)/T} \pm 1}.$$

I do not need an analytic expression for $G(p)$, an integral that I can solve numerically is sufficient. It seems to me that this should be possible to do, but I just can't think of how.

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    $\begingroup$ why does it make sense to ask for a temperature if the system is not in a equilibrium? $\endgroup$ – tonydo May 9 '16 at 10:57
  • $\begingroup$ Well, it would not exactly be the temperature, per se, but it would be a property of the gas that is similar to temperature. Often, we have close to but not exactly an equilibrium distribution, and then it is helpful to fint the "temperature". $\endgroup$ – Ihle May 9 '16 at 11:19
  • $\begingroup$ Is $E(p)$ in the definition of $f_{eq}$ relativistic, or simply $p^2/2m$? $\endgroup$ – tonydo May 9 '16 at 15:04
  • $\begingroup$ $E(p) = \sqrt{p^2 +m^2}$. $\endgroup$ – Ihle May 9 '16 at 15:41

I think such a function may only exist in the Maxwell-Boltzmann limit. Here's why:

For simplicity let us parametrize everything in terms of $\beta = 1/T$ and denote $Z(\beta) = \int{d^3p\; f_{eq}(p, \beta)}$. Rewrite the latter as $$ Z(\beta) = 4\pi \int_0^\infty{dp\;\frac{p^2}{e^{\beta E_p}\pm 1}} = 4\pi \int_m^\infty{dE\;\frac{E\sqrt{E^2-m^2}}{e^{\beta E}\pm 1}} = \\ =\frac{4\pi}{3} \int_m^\infty{dE\;\left[\frac{d}{dE}(E^2-m^2)^{3/2}\right]\frac{1}{e^{\beta E}\pm 1}} $$ and upon integrating by parts, $$ Z(\beta) = \beta \frac{4\pi}{3} \int_m^\infty{dE\; (E^2-m^2)^{3/2} \frac{e^{\beta E}}{\left(e^{\beta E}\pm 1\right)^2}} $$ Now, for given $E$ let $e^{\beta E}/\left(e^{\beta E}\pm 1\right)^2$ be the Laplace transform of $\Lambda_\pm(\epsilon; E)$, such that $$ \frac{e^{\beta E}}{\left(e^{\beta E}\pm 1\right)^2} = \int_0^\infty{d\epsilon \; \Lambda_\pm(\epsilon; E)e^{-\beta \epsilon}} $$ (I know the phrasing is awkward, but I'm trying to avoid complex plane integration issues with the inverse Laplace transform.) If $\Lambda_\pm(\epsilon; E)$ exist, substitute in $Z(\beta)$, rearrange, and obtain $$ \frac{Z(\beta)}{4\pi\beta} = \frac{1}{3}\int_0^\infty{d\epsilon \; \left[ \int_m^\infty{dE\; (E^2-m^2)^{3/2}\Lambda_\pm(\epsilon; E) }\right]e^{-\beta \epsilon}} $$ Basically, this gives us an expression for the Laplace transform of $Z(\beta)/(4\pi\beta)$. Keeping it in mind, let us look for a function $G(p )$ such that $$ \frac{1}{\beta} = \frac{1}{Z(\beta)} \int{d^3p\;G(p )f_{eq}(p,\beta)} $$ or $$ \frac{Z(\beta)}{\beta} = \int{d^3p\;G(p )f_{eq}(p,\beta)} = 4\pi\int_0^\infty{dp\; \frac{p^2G(p )}{e^{\beta E_p}\pm 1} } = 4\pi\int_m^\infty{dE\; \frac{{\bar G}(E )E \sqrt{E^2 - m^2}}{e^{\beta E}\pm 1} } $$ where ${\bar G}(E) = G(p )$. As before, let $1/\left(e^{\beta E}\pm 1\right)$ be the Laplace transform of $\Lambda^0_\pm(\epsilon; E)$, such that $$ \frac{1}{e^{\beta E}\pm 1} = \int_0^\infty{d\epsilon \; \Lambda^0_\pm(\epsilon; E)e^{-\beta \epsilon}} $$ and obtain $$ \frac{Z(\beta)}{4\pi\beta} = \int_0^\infty{d\epsilon \; \left[ \int_m^\infty{dE\; E(E^2-m^2)^{1/2}{\bar G}(E)\Lambda^0_\pm(\epsilon; E) }\right]e^{-\beta \epsilon}} $$ This is yet another expression for the Laplace transform of $Z(\beta)/(4\pi\beta)$. Identifying with the one obtained previously gives $$ \int_m^\infty{dE\; E(E^2-m^2)^{1/2}{\bar G}(E)\Lambda^0_\pm(\epsilon; E) } = \frac{1}{3} \int_m^\infty{dE\; (E^2-m^2)^{3/2}\Lambda_\pm(\epsilon; E) } $$ or $$ \int_m^\infty{dE\; (E^2-m^2)^{1/2} \left[ E \;{\bar G}(E)\Lambda^0_\pm(\epsilon; E) - \frac{1}{3} (E^2-m^2) \Lambda_\pm(\epsilon; E) \right]} = 0 $$ But ${\bar G}(E)$ has to satisfy this identity for all $\epsilon \ge 0$. This effectively implies that $\Lambda_\pm(\epsilon; E) = \chi(E) \Lambda^0_\pm(\epsilon; E)$ for some suitable $\chi(E)$ and in turn means $$ \frac{e^{\beta E}}{\left(e^{\beta E}\pm 1\right)^2} = \chi(E) \frac{1}{e^{\beta E}\pm 1}\;\;\; \Rightarrow \;\;\; \chi(E) = \frac{e^{\beta E}}{e^{\beta E}\pm 1} $$ Since the lhs above is always temperature-independent while the rhs is only in the low-temperature limit, it seems that a proper function ${\bar G}(E)$ can exist only in the same limit, when $\Lambda_\pm(\epsilon; E) \rightarrow \Lambda^0_\pm(\epsilon; E)$ and $$ {\bar G}(E) = \frac{E^2 - m^2}{3E} = \frac{p^2}{3E} $$ as you already know.

  • $\begingroup$ Thanks alot for the great answer! Do you mean that $\Lambda = \Lambda^0$ in the low temperature limit? Since this is when the MB solution is valid. (You seem to say in the high-temperature limit.) $\endgroup$ – Ihle May 12 '16 at 9:08
  • $\begingroup$ Wait, $\Lambda$ is not a function of temperature, right? So why is not $\bar G(E) = \frac{p^2}{3E} \frac{\Lambda}{\Lambda^0}.$ I guess we would need the inverse Laplace transform in order to find the $\Lambda$'s, but as long as this is an integral I can solve numerically, then I am satisfied. $\endgroup$ – Ihle May 12 '16 at 9:22
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    $\begingroup$ Welcome. You are right about low temp, it shows I don't do relativistic kinetics very often. About $\Lambda/\Lambda^0$: the reasoning is that ${\bar G}(E)$ cannot depend on the expansion parameter $\epsilon$, so $\Lambda/\Lambda^0$ shouldn't either, hence $\Lambda/\Lambda^0 = \chi(E)$. Now if the ratio happens to have regions of shallow dependence on $\epsilon$ other than $\beta \rightarrow \infty$, you might be able to find nontrivial solutions for ${\bar G}(E)$ around there. $\endgroup$ – udrv May 12 '16 at 11:42
  • $\begingroup$ The expression in the parenthesis should vanish for any value of $\epsilon$, right? If we just fix $\epsilon$ to a value, can't we then get an expression for $\bar G(E)$, that holds for all $E$? $\endgroup$ – Ihle May 12 '16 at 11:49
  • $\begingroup$ Well, you can always try it. Or use some sort of mean value. $\endgroup$ – udrv May 12 '16 at 16:09

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