I am reading one very good notes on Maxwell-Juttner distribution, but I met some places that puzzled me a lot. Please give me a hand!
The notes:
Let $\Theta = kT/mc^2$ be the dimensionless temperature, $\vec\beta = \vec{v}/c$ is the dimensionless particle velocity, $\gamma = (1-\beta^2)^{-1/2}$ is the particle Lorentz factor (energy), and $\vec{u} = \gamma\vec\beta$ is the particle 4-velocity (momentum). The Maxwell-Juttner distribution is given by the following equivalent forms:
\begin{eqnarray} f_1(\gamma)\,{\rm d}\gamma &=& \frac{\gamma u\,{\rm d}\gamma}{\Theta\,K_2(1/\Theta)}\exp\left(-\frac{\gamma}{\Theta}\right) \\ f_2(u)\,{\rm d}u &=& \frac{u^2\,{\rm d}u}{\Theta\,K_2(1/\Theta)}\exp\left(-\frac{\sqrt{u^2+1}}{\Theta}\right) \\ f_3(\vec{u})\,{\rm d}^3u &=& \frac{{\rm d}^3u}{4\pi\Theta\,K_2(1/\Theta)}\exp\left(-\frac{\sqrt{\vec{u}^2+1}}{\Theta}\right) \end{eqnarray}
We make use of two modified Bessel functions of the second kind in the following integral form:
\begin{eqnarray} K_n(1/\Theta) &=& \frac{\sqrt\pi}{(n-1/2)!(2\Theta)^n}\int_1^\infty u^{(2n-1)}\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma\,, \\ K_1(1/\Theta) &=& \frac{1}{\Theta}\int_1^\infty u\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma\,, \\ K_2(1/\Theta) &=& \frac{1}{3\Theta^2}\int_1^\infty u^3\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma = \frac{1}{\Theta}\int_1^\infty \gamma u\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma\,. \end{eqnarray}
Normalization of the distribution is:
\begin{equation} \int_1^\infty f_1(\gamma)\,{\rm d}\gamma = \frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty \gamma u\exp\left(-\frac{\gamma}{\Theta}\right)\,{\rm d}\gamma = 1 \end{equation}
The mean particle energy is (one can also check this numerical solution):
\begin{equation} \left<\gamma\right> = \frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty\gamma^2 u\exp\left(-\frac{\gamma}{\Theta}\right)\,{\rm d}\gamma = 3\Theta + h(\Theta)\,, \end{equation}
where $h(\Theta) = K_1(1/\Theta)/K_2(1/\Theta)$ is a function plotted below.
Question 1: What I don't understand are these two places, maybe they are the same problems:
- how can we know:
\begin{equation} \frac{1}{3\Theta^2}\int_1^\infty u^3\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma = \frac{1}{\Theta}\int_1^\infty \gamma u\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma\,. \end{equation}
Answer:
\begin{eqnarray} &&\frac{1}{3\Theta^2}\int_1^\infty u^3\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma = -\frac{1}{3\Theta}\int_1^\infty u^3\;{\rm d}\exp\left(-\frac{\gamma}{\Theta}\right) \\ && = -\frac{1}{3\Theta}u^3\exp\left(-\frac{\gamma}{\Theta}\right)\Big|_0^\infty -\frac{1}{3\Theta}\int_1^\infty \exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}u^3 \\ &&= \frac{1}{\Theta}\int_1^\infty \gamma u\exp\left(-\frac{\gamma}{\Theta}\right)\;{\rm d}\gamma\,. \end{eqnarray}
The first term vanish since $u(0)=0$ and $\exp(-\infty)=0$. The second term, ${\rm d}u = {\rm d}(\gamma^2-1)^{1/2} = \gamma/u{\rm d}\gamma$.
Question 2:
- how is this derived
\begin{equation} \left<\gamma\right> = \frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty\gamma^2 u\exp\left(-\frac{\gamma}{\Theta}\right)\,{\rm d}\gamma = 3\Theta + h(\Theta)\,, \end{equation}
Answer
\begin{eqnarray} &&\frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty\gamma^2 u\exp\left(-\frac{\gamma}{\Theta}\right)\,{\rm d}\gamma \\ &&= \frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty\gamma^2u(-\Theta)\,{\rm d}\exp{-\frac{\gamma}{\Theta}} \\ && = \gamma^2u(-\Theta)\,{\rm d}\exp{-\frac{\gamma}{\Theta}}\Big|_1^\infty + \frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty\exp{-\frac{\gamma}{\Theta}}\,{\rm d}\gamma^2u \end{eqnarray}
The first term vanishes, and ${\rm d}\gamma^2u= {\rm d}(u^3-u) = 3u\gamma\,{\rm d}\gamma-{\rm d}u$.
\begin{eqnarray} &&\frac{1}{\Theta\,K_2(1/\Theta)}\int_1^\infty\exp{\left(-\frac{\gamma}{\Theta}\right)}\,{\rm d}\gamma^2u \\ &&= \frac{3}{K_2(1/\Theta)}\int_1^\infty u\gamma\exp{\left(-\frac{\gamma}{\Theta}\right)}\,{\rm d}\gamma + \frac{1}{K_2(1/\Theta)}\int_1……\infty\exp{\left(-\frac{\gamma}{\Theta}\right)}\,{\rm d}u \\ &&= 3\Theta + \frac{1}{K_2(1/\Theta)}\left[\exp{\left(-\frac{\gamma}{\Theta}\right)}u\Big|_1^\infty -\int_1^\infty u\,{\rm d}\exp{\left(-\frac{\gamma}{\Theta}\right)}\right] \\ &&= 3\Theta +h(\Theta) \end{eqnarray}
