% % Lecture presentation aid for the textbook: % % Laszlo P. Csernai: " Introduction to Relativistic Heavy Ion Collisions" % (John Wiley and Sons Ltd, Chicester, New York, Brisbane, Toronto, % Singapore, 1994; ISBN - 0-471-93420-8) % % Transparencies for Lecture 3 / Chapter 3 % % The following file is containing transparencies in plain LaTeX format. % Figures are usually not included, they can be pasted from the printed % textbook. Empty space is left for this purpose. Some LaTeX figures are % included in the files. References and Solutions of the assignments are % also not included! % \documentstyle[12pt]{book} % Transparencies in frame keretben \textheight=21.5cm \textwidth=19cm \oddsidemargin=10mm \evensidemargin=10mm %\voffset=-0.50cm \hoffset=-0.15cm % \newcommand{\beq}{\begin{equation}} \newcommand{\beqar}{\begin{eqnarray}} \newcommand{\eeq}[1]{\label{#1} \end{equation}} \newcommand{\eeqar}[1]{\label{#1} \end{eqnarray}} %\newcommand{\beq}{$$} %\newcommand{\eeq}{$$} \newcommand{\B}{$\bullet$ } \newcommand{\LT}{$\leadsto$ } %\renewcommand{\vec}[1]{{\bf #1}} \newcommand{\D}{\bigtriangledown} \newcommand{\etal}{{\it et al.,\/}} \newcommand{\lton}{$\raisebox{-.54ex}{$\stackrel{<}{\sim}$}$} \newcommand{\gton}{$\raisebox{-.54ex}{$\stackrel{>}{\sim}$}$} \newcommand{\gfm}{{\rm GeV/fm}^3} \newcommand{\ipint}{\int \frac{d^3 p}{p^0}} \newcommand{\dpi}{ \frac{d^3 p_i}{p_i^0}} \newcommand{\dpj}{ \frac{d^3 p_j}{p_j^0}} \newcommand{\dpk}{ \frac{d^3 p_k}{p_k^0}} \newcommand{\dpl}{ \frac{d^3 p_l}{p_l^0}} \newcommand{\dpkp}{ \frac{d^3 p'_k}{p^{'0}_k}} \newcommand{\dplp}{ \frac{d^3 p'_l}{p^{'0}_l}} \newcommand{\ip}{\int^{\infty}_{-\infty}dp} \newcommand{\pint}{\int d^3 p} \newcommand{\juttner}{ {1\over{(2 \pi \hbar)^3}} \exp \left( {{\mu - p^\mu u_\mu}\over{T}} \right) } \newcommand{\jut}{ e^{ \left( - {{ p^\mu u_\mu}\over{T}} \right) }} \newcommand{\pmum}{p^\mu u_\mu} \newcommand{\amev}{A$\cdot$MeV} \newcommand{\agev}{A$\cdot$GeV} \newcommand{\transparencyframe}[1]{ \phantom{a} \setlength{\unitlength}{1mm} \begin{picture}(10,10)(30,200) \thicklines \put(95,240){\oval(188,18)[t]} \put(95,20){\oval(188,18)[b]} \put(1,20){\line(0,1){220}} \put(189,20){\line(0,1){220}} \thinlines \put(95,240){\oval(186,16)[t]} \put(95,20){\oval(186,16)[b]} \put(2,20){\line(0,1){220}} \put(188,20){\line(0,1){220}} \put(95,8){\scriptsize\sf L.P. Csernai: Inroduction to Relativistic Heavy Ion Collisions (Wiley, 1994)} \put(10,235){\parbox[t]{170mm}{\Large\sf \parskip=0.3cm #1}}\end{picture} }%end transparencyframe % \begin{document} \pagestyle{plain} % transparency =============================================================== %\setlength{\unitlength}{1mm} %}%end tr-page %\newpage % transparency ===================================================== \transparencyframe{ \setcounter{chapter}{2} \vspace*{-1.8cm}\chapter{Relativistic Boltzmann Transport Equation} The relativistic Boltzmann Transport Equation (BTE) describes the time evolution of the single particle distribution function $f(x,p)$. Based on the following assumptions: \begin{itemize} \item{1} Only two-particle collisions are considered, the so called binary collisions. \item{2} ``Sto\ss zahlansatz'' or assumption of ``Molecular Chaos'': Number of binary collisions at $x$ is proportional to $f(x,p_1) \times f(x,p_2)$. \item{3} $f(x,p)$ is a smoothly varying function compared to the mean free path (m.f.p.) \end{itemize} \section{Particle conservation} Number of particles in a 3-dim. volume element at $\vec{x}$ \& time $t$, $\Delta N(x)$,\\ is identical to the number of world-lines crossing a 3-dim. (time-like) hyper surface: \beq \Delta N(x) = \int_{\Delta^3 \sigma} d^3\sigma_\mu \ N^\mu (x') = \int_{\Delta^3 \sigma} \int_{\Delta^3 p} d^3\sigma_\mu \ {{d^3p}\over{p^0}} \ p^\mu \ f(x',p). \eeq{3.1} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace{-1.0cm} Here $ d^3\sigma_\mu $ is a time-like 4-vector, orthogonal to the surface $\Delta^3 \sigma$.\\ In the (LR) $ d^3\sigma_\mu = (d^3x',0,0,0)$. In the (LR) $\Delta N(x) = \int_{\Delta^3 x} \int_{\Delta^3 p} d^3x' \ d^3p \ f(x',p)$ is the number of world-lines crossing $\Delta^3 \sigma$, with momenta in the range $\Delta^3 p$ around $p$. %\begin{figure}[hbtp] \vbox{ Total surface of the\\ 4-volume element $\Delta^4 x$ \setlength{\unitlength}{0.3mm} \begin{picture}(400,150)(-90,0) \put(150,50){\vector(1,0){150}} \put(155,45){\vector(0,1){100}} \put(160,55){\vector(-1,-1){60}} \put(303,50){$z$} \put(158,145){$t$} \put(85,0){$x$} \end{picture} \setlength{\unitlength}{1mm} \hskip 9cm Edges are negligibly small } %\caption[F:CS91] Figure 3.1 {\it World-lines penetrating through a 4-volume element} %\label{f2.1} %\end{figure} If particles are conserved, the number of world-lines is constant, i.e. they cross both $\Delta^3 \sigma$ and later $\Delta^3 \sigma '$. (see Figure 3.1). Therefore \beq \int_{\Delta^3 \sigma} \int_{\Delta^3 p} d^3\sigma_\mu \ {{d^3p}\over{p^0}} \ p^\mu \ f(x',p) - \int_{\Delta^3 \sigma '} \int_{\Delta^3 p} d^3\sigma_\mu \ {{d^3p}\over{p^0}} \ p^\mu \ f(x',p) = 0. \eeq{3.2x} Since the edges of the 4-volume element are negligibly small \LT the integral over the total surface of the volume element, $S$, should vanish: \beq \int_{S(\Delta^4 x)} \int_{\Delta^3 p} d^3\sigma_\mu \ {{d^3p}\over{p^0}} \ p^\mu \ f(x',p) = 0. \eeq{3.3} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1cm} Using the Gauss theorem, \\ ($\frac{\partial}{\partial x^\mu} (p^\mu f)$ $=$ $(p^\mu f),_\mu = p^\mu f,_\mu$\ , because $p^\mu$ does not depend on $x$): \beq \int_{\Delta^4 x} \int_{\Delta^3 p} d^4 x \ {{d^3p}\over{p^0}} \ p^\mu \ f,_\mu(x,p) = 0. \eeq{3.3x} Notation $\partial_\mu \equiv ,_\mu \equiv {\partial \over {\partial x^\mu}} \equiv (\partial_t, \nabla ) $. \\ Since $x,p$ and $\Delta^4 x $ , $\Delta^3 p $ are arbitrary: \beq p^\mu f,_\mu = 0 , \ \ \ or \ \ \ p^\mu \partial_\mu f(x,p) = 0. \eeq{3.4} \B This is the relativistic transport equation for the collisionless case in non-relativistic notation. Dividing eq. (3.5) by $p^0$,: \beq (\partial_t + \vec{v} \ \nabla_x ) f(x,p) = 0, \eeq{3.6} since $\vec{v} = \vec{p} / p^0$. This is the known form of the continuity equation. (No external forces.) }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace{-1.5cm}\section{Collisions} Collisions among particles lead to a change of $f(x,p)$.\\ The number of particles in $\Delta^4 x $ , $\Delta^3 p $ changes by $\Delta^4 x {{\Delta^3 p}\over{p^0}} C(x,p) $, \\ where $C(x,p)$ is the collision integral.\\ Momenta in a general binary collision: \setlength{\unitlength}{0.25mm} \begin{picture}(200,100)(-400,0) \put(20, 10){\vector(1,1){40}} \put(100,10){\vector(-1,1){40}} \put(58,50){$\bullet$} \put(60,50){\vector(1,1){40}} \put(60,50){\vector(-1,1){40}} \put(27,10){$p^\mu$} \put(105,10){$p^\mu_1$} \put(25,90){$p'^\mu$} \put(105,90){$p'^\mu_1$} \end{picture} \setlength{\unitlength}{1mm} The number of such collisions is proportional to: \begin{itemize} \item[(i)] The number of nucleons (particles) around $\vec{p}$: $\Delta^3p\ f(x,p)$. \item[(ii)] The number of nucleons (particles) around $\vec{p}_1$: $\Delta^3p_1\ f(x,p_1)$. \item[(iii)] The final state and configuration volume intervals $\Delta^3p',\ \Delta^3p'_1,\ $ and $\Delta^4x\ $. The proportionality factor is: \beq {{W(p,p_1 | p',p'_1)}\over{ p^0\ p_1^0\ p^{'0}\ p^{'0}_1}} \eeq{3.7} The quantity $W(p,p_1 | p',p'_1)$ is the {\em transition rate}. \\ (Invariant scalar; $x$ dependence is assumed to be weak). \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} Number of particles scattering out of $\Delta^3p \ \Delta^4x\ $ is \beq {1\over 2} \Delta^4x\ {{\Delta^3p}\over{p^0}} \int {{d^3p_1}\over{p_1^0}} {{d^3p'}\over{p^{'0}}} {{d^3p'_1}\over{p^{'0}_1}} \ f(x,p) \ f(x,p_1) W(p,p_1 | p',p'_1). \eeq{3.8} The factor ${1\over 2} $: symmetry under the exchange $p,p_1 \leftrightarrow p'_1,p'$ , and we correct for double counting.\\ Similarly the change due to the gain term is: \beq {1\over 2} \Delta^4x\ {{\Delta^3p}\over{p^0}} \int {{d^3p_1}\over{p_1^0}} {{d^3p'}\over{p^{'0}}} {{d^3p'_1}\over{p^{'0}_1}} \ f(x,p') \ f(x,p'_1) W(p',p'_1 | p,p_1). \eeq{3.10} \B Thus the total transport equation is: \beq p^\mu \partial_\mu f(x,p) = C(x,p), \eeq{3.11} where $$ C(x,p) = \hspace*{13cm} $$ \beq \hspace*{1cm} {1\over 2} \int {{d^3p_1}\over{p_1^0}} {{d^3p'}\over{p^{'0}}} {{d^3p'_1}\over{p^{'0}_1}} \left[ f' f'_1 W(p',p'_1 | p,p_1) - f f_1 W(p,p_1 | p',p'_1) \right], \eeq{3.12} and \beq f' \equiv f(x,p'), \ \ f'_1 \equiv f(x,p'_1), \ \ f \equiv f(x,p), \ \ f_1 \equiv f(x,p_1) . \eeq{3.78} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\section{Non-relativistic limit} Rewrite the equation in another form $$ (\partial_t + \vec{v} \ \nabla_x ) f(x,p) = \hspace*{12cm} $$ \beq \hspace*{1cm} {1\over 2} \int d^3p_1 \ d^3p' \ d^3p'_1 \left[ f' f'_1 w(p',p'_1 | p,p_1) - f f_1 w(p,p_1 | p',p'_1) \right], \eeq{3.13} where \beq w = {{W(p,p_1 | p',p'_1)}\over{ p^0\ p_1^0\ p^{'0}\ p^{'0}_1}} \eeq{dummy} Since $W$ is a Lorentz scalar depending on $p^\mu$ 's of a collision,\\ it may depend on the invariants in a collision ( $p^\mu +p^\mu_1 = p'^\mu +p'^\mu_1 $ ).\\ \B $\exists$ only 2 independent invariants: \beqar s & \equiv & (p+p_1)^2 , \\ t & \equiv & (p-p')^2 . \eeqar{3.15} In the c.m. frame of the collision if $P^\mu \equiv p^\mu + p^\mu_1 = p'^\mu + p'^\mu_1$, then $P^\mu = (\sqrt{s},0,0,0)$; i.e. $\sqrt{s}$ is the total c.m. energy. $t$ is related to the scattering angle $\Theta$ \beq \cos \Theta \equiv \left[ {{(\vec{p} \ \vec{p} ')}\over {|\vec{p}| \ |\vec{p}'|}} \right]_{cm} = {{(p^\mu - p_1^\mu) (p'_{\mu} - p'_{\mu_1})} \over {(p-p_1)^2}} = {{2t}\over{s-4m^2}}+1. \eeq{3.15x} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} Thus the transition rate can be expressed as \beq W(p,p_1 | p',p'_1) = s \ \sigma(s,\Theta) \ \delta^{(4)}(p+p_1-p'-p'_1) \eeq{3.16} where the delta function enforces energy conservation and $s$ is introduced to have the proper non-relativistic limit as we will see it later;\\ (By dimensional analysis: [W] = [fm /GeV ]; from eq. (3.10-11), and $c=1$.)\\ Here $\sigma(s,\Theta)$ is some function having the dimension of fm$^2$ . We will see that $\sigma(s,\Theta)$ is in fact the differential cross-section in the non-relativistic limit. Insert eq. (3.18) into eq. (3.11), and e.g. for the gain term of the collision integral: \beq I = \int {{d^3p_1}\over{p_1^0}} {{d^3p'}\over{p^{'0}}} {{d^3p'_1}\over{p^{'0}_1}} \ f(x,p') \ f(x,p'_1) \ s \ \sigma(s,\Theta) \ \delta^{(4)}(p+p_1-p'-p'_1) . \eeq{16x} In the c.m. system $P^\mu = p^\mu + p^\mu_1 = p'^\mu + p'^\mu_1$ $ = (\sqrt{s},0,0,0)$, so the integration over ${{d^3p'_1}\over{p^{'0}_1}}$ can be carried out. There is only the $^0$th component left then: \beq \int {{d^3p'}\over{p^{'0}}} {{d^3p'_1}\over{p^{'0}_1}} \ \delta^{(4)}(p+p_1-p'-p'_1) = {1 \over 2} \int {{d^3p'}\over{p^{'0}}} \ {1\over{p^{'0}}} \ \delta({1\over 2} \sqrt{s}-p^{'0}) , \eeq{3.16x} i.e. $p'^0={1\over 2} \sqrt{s} $ in the c.m. frame. Then \beq I = 2 \int d^3p' {{d^3p_1}\over{p_1^0}} \ \delta({1\over 2} \sqrt{s}-p^{'0}) \sigma(s,\Theta) \ \ f(x,p') \ f(x,p'_1) \ , \eeq{3.16y} where $d^3p' = |\vec{p}\, '|^2 \ dp' \ d\Omega$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} Now using the relations for $\delta(\phi(x))$ (see Lecture 2) \beq \delta({1\over 2}\sqrt{s} - \sqrt{(\vec{p}\, ')^2 + m^2}) = \sqrt{ {s \over{s-4m^2}} } \ \ \delta(|\vec{p}\, '| - {1\over 2}\sqrt{s-4m^2} ). \eeq{3.16z} Thus the integral is: \beq I = 2 \int dp' {{d^3p_1}\over{p_1^0}} d\Omega \ |\vec{p}\, '|^2 \sqrt{ {s \over{s-4m^2}} } \delta(|\vec{p}\, '| - {{\sqrt{s-4m^2}} \over 2} ) \sigma(s,\Theta) f' f'_1 \eeq{3.16f} \beq = \int {{d^3p_1}\over{p_1^0}} d\Omega \ \ \underbrace{ {1\over 2} \sqrt{s(s-4m^2)} }_{\equiv F} \sigma(s,\Theta) f' f'_1 , \eeq{3.16e} where the part indicated by $F$ is the so called {\em Invariant Flux}.\\ The flux was $|\vec{v}_2 - \vec{v}_1|$ in the non-relativistic transport theory.\\ F corresponds to this quantity $$ F = {1\over 2} \sqrt{ s (s-4m^2) } = \sqrt{(p_1^\mu p_\mu)^2 - m^4}_{|_{in\ c.m.}} = \left| { \vec{p}\over {p^0}} - {\vec{p_1}\over {p_1^0}} \right| p^0 p^0_1 = \hspace*{1.0cm} $$ \beq \hspace*{4cm} \left| \vec{v} - \vec{v_1} \right|_{c.m.} p^0 p^0_1. \eeq{3.16p} The two $0^{\rm th}$ components of the momenta are present in the relativistic expression\\ to balance the invariant scalar volume elements of the integration, $ {{d^3p}\over{p^0}}$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} Using this quantity \beq I = \int {{d^3p_1}\over{p_1^0}} \ d\Omega \left| \vec{v} - \vec{v_1} \right|_{c.m.} p^0 p^0_1 \sigma(s,\Theta) \ f(x,p') \ f(x,p'_1) . \eeq{3.16m} Finally, using the symmetry relations of $W$ or $\sigma$ the complete collision integral and then the full BTE: $$ p^\mu f,_\mu = \hspace*{12cm} $$ \beq {1\over 2} \int d^3p_1 \ d\Omega \left| \vec{v} - \vec{v_1} \right|_{c.m.} p^0 \sigma(s,\Theta) \ [f(x,p') f(x,p'_1) - f(x,p) f(x,p_1) ] \eeq{3.16d} By expressing the conmoving derivative in the usual non-relativistic way $$ (\partial_t + \vec{v} \ \nabla) f(x,p) = \hspace*{10cm} $$ \beq {1\over 2} \int d^3p_1 \ d\Omega \left| \vec{v} - \vec{v_1} \right|_{c.m.} \sigma(s,\Theta) \ [f(x,p') f(x,p'_1) - f(x,p) f(x,p_1) ] \eeq{3.16s} \B This is the form of the BTE we learned in the statistical physics course.\\ The definition of the cross section via the transition rate is then \beq \sigma(s,\Theta) d\Omega = {1\over F} W(p,p_1 | p',p'_1) {{d^3p'}\over{p^{'0}}} {{d^3p'_1}\over{p^{'0}_1}} \eeq{3.16w} Important difference is the factor ${1\over 2}$! \\ {\small\sf In the non-relativistic theory there is no such factor. This means that in the classical theory $\sigma_{cl} = {1\over 2} \sigma$, \ and the factor one half takes into account the identical particles. }\\ The total cross section should be calculated here as\\[0.2ex] $\sigma_{tot} = {1\over 2} \int d\Omega \sigma(s,\Theta)$, while in the classical theory $\sigma_{tot} = \int d\Omega \sigma_{cl}(s,\Theta)$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm}\section{An example for the solution} Taken from the work of J. Randrup [2]. - A spatially uniform distribution is assumed.\\ - The example may represent the overlap region of a HI collision.\\ - The initial condition chosen accordingly, Fig. 3.2.\\ \B Question: how fast a thermalized momentum distribution is reached. %\begin{figure}[hbtp] \setlength{\unitlength}{0.5mm} \begin{picture}(200,80)(50,0) \put(84,30){\circle{40}} \put(96,40){\circle{40}} \put(150,45){Time dependence of $f(p)$ in the overlap region} \end{picture} \setlength{\unitlength}{1mm} %\caption[New] Figure 3.2 {\it The overlap region of two nuclei in a heavy ion collision } %\label{f2.2} %\end{figure} - The Pauli principle: At a given location $x$ the $p_1,p \rightarrow p'_1, p'$ scattering is forbidden to populate a momentum state which is already occupied.\\ Thus the rate is not proportional to $ff_1$, but to $$ \underbrace{f(x,p)}_{\equiv f} \underbrace{f(x,p_1)}_{\equiv f_1}\ \underbrace{[1-f(x,p')]}_{\equiv \bar{f}'} \underbrace{[1-f(x,p'_1)]}_{\equiv \bar{f}'_1}. $$ - Introduced by Nordheim [3] and by Uehling and Uhlenbeck [4].\\ Randrup then solved numerically the equation: \beq p^0 f,_0 = Const. \int d^3p_1 \ d\Omega \left| \vec{v} - \vec{v_1} \right| p^0 \sigma(s,\Theta) \ [f' f'_1 \bar{f} \bar{f}_1 - f f_1 \bar{f}' \bar{f}'_1 ]. \eeq{3.16k} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Initial state: two-peaked momentum distribution $\sim$ heavy ion collision before the thermalization. Fig. 3.3. %\begin{figure}[hbtp] \setlength{\unitlength}{0.4mm} \begin{picture}(380,100)(-30,0) \put(10,20){\vector(1,0){340}} \put(360,5){$p_\|$} \put(200,10){\vector(0,1){80}} \put(205,90){$f(p)$} \end{picture} %\caption[New] Figure 3.3 {\it Momentum distribution in the longitudinal direction in the overlap region } \setlength{\unitlength}{1mm} %\label{f2.3} %\end{figure} \B Time development of momentum distribution: %\begin{figure}[hbtp] \vspace*{10.5cm} %\caption[F:Ra79-f.2a] Figure 3.4 {\it The time development of longitudinal momentum distribution $f(p_{\|})$ which is obtained from the phase occupancy $f(p)$ by projecting onto the beam axis. The normalization is arbitrary but common. $p_\|$ runs from $0$ to the c.m. beam momentum, $p_0$. } %\label{f2.4} %\end{figure} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \B Entropy: calculated by using the definition the entropy 4-current: \beq S^\mu \equiv - \int p^\mu {{d^3p}\over{p^0}} \left[ f(x,p) \ln f(x,p) - f(x,p) \right]\ , \eeq{3.19} where the term $-f(x,p)$ ensures the appropriate entropy constant for joining smoothly low temperature quantum statistical results (See also assignment 3.e).\\ - The argument of the function $\ln(z)$ should be dimensionless: This is ensured by inserting the elementary phase space volume next to $f$ in the argument, $(2\pi \hbar)^3$. In eqs. (3.31-32) we drop this factor.\\ To take into account the Pauli principle in high temperature limit \beq S^\mu \equiv - \int p^\mu {{d^3p}\over{p^0}} \left[ f \ln f + \bar{f} \ln \bar{f} \right]. \eeq{3.29} \B We will see later that these definitions coincide with the thermodynamic definitions for equilibrium systems! \\ - Entropy density (Lorentz scalar): \beq s = S^\mu u_\mu, \eeq{3.49} - Specific entropy \beq \sigma = {s \over n}. \eeq{3.59} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Calculated time dependence of entropy production (the speed of thermalization): %\begin{figure}[hbtp] \vskip 13truecm %\caption[F:Ra79-f.4] Figure 3.5 {\it Specific entropy production as a function of time. The role of Pauli principle in the process of equilibration is shown in the figure: dashed lines are without, full lines with Pauli principle. From [2]} %\label{f2.5} %\end{figure} The role of Pauli principle is not too large in entropy production! }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Population of the momentum space: qualitatively different in elastic and in inelastic collisions.\\ \B Elastic collisions populate a sphere in the c.m. momentum space, because of energy conservation: %\begin{figure}[hbtp] \vskip 10truecm %\caption[F:Ra79-f.8] Figure 3.6 {\it Contour plots of the distribution of nucleons in rapidity space after their first collision. The upper portion is for elastic collisions only. In the lower portion inelasticity is included via delta formation. The nucleons resulting from the isotropic decay of the deltas have been added to the plot. From [2]} %\label{f2.6} %\end{figure} \B Inelastic collisions populate mid rapidities \LT fast thermalization!! Collective mean field: in BUU, VUU or Landau-Vlasov models. Effective up to a few 100 MeV/nucleon colliding energy. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm} \section{Relativistic Boltzmann equation for mixtures} $N$ components, their distribution is described by \beq f_k (x,p_k), \ \ \ k=1,2, ... ,N. \eeq{4.1} \B The BTE is then: \beq p_k^\mu f_{k,\mu} = \sum^N_{l=1} C_{kl} (x,p_k) \eeq{4.1a} where $$ C_{kl} (x,p_k) = (1-{1 \over 2} \delta_{kl}) \int \dpl \dpkp \dplp \left[ f'_k f'_l W_{kl}(p'_k,p'_l|p_k,p_l) - \right. $$ \beq \hspace*{9cm} \left. f_k f_l W_{kl}(p_k,p_l|p'_k,p'_l) \right] . \eeq{4.2} BTE for elastic collisions only. \\ - The $\delta_{kl}$ in front of the collision integral: to distinguish mixtures composed of identical particles, $k=l$, and of non-identical particles, $k \neq l$.\\ - For inelastic collisions the final states may belong to different components: $k+l \ \rightarrow \ i+j$. (Here we neglect the possibility of $ k+l \ \rightarrow \ i+j+s+... $.) The transition rate is $W_{kl \rightarrow ij}$ or $W_{kl|ij}$, and the collision integral: $$ C_{kl} (x,p_k) = {1 \over 2} \sum_{i,j=1}^N \int \dpl \dpi \dpj \left[ f_i f_j W_{ij|kl}(p_i,p_j|p_k,p_l) - \right. $$ \beq \hspace*{7cm}\left. f_k f_l W_{kl|ij}(p_k,p_l|p_i,p_j) \right] . \eeq{4.3} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} We omit the arguments of $W$, keeping the indices only. (Like for $f$.)\\ \B $W$s have some simple symmetries:\\ - the sequence of the final states is irrelevant: \beq W_{kl|ij} = W_{kl|ji}, \eeq{4.3a} - the time reversal symmetry of the microscopic processes $$ W_{ij|kl} = W_{kl|ij} . $$ \section{Conservation laws} - In macroscopic (phenomenological) theories the quantities, $n, e, P, ...$ change according to given dynamical equations (e.g. in hydrodynamics) which are {\bf postulated}. \\ - In transport theory {\bf conservation laws}: connect the {\em microscopic} properties of the system to the equations governing the development of the {\em macroscopic} quantities. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} \subsection*{LEMMA} 1. $\Psi_k$ is a microscopic quantity determined by the particle type ($k$), position ($x$) and momentum ($p_k$) as: $$ \Psi_k (x,p_k) = a_k(x) + b_\mu(x) p^\mu_k. $$ \noindent 2. This quantity is conserved in a binary collision $kl \rightarrow ij$, such that $$ \Psi_k + \Psi_l = \Psi_i + \Psi_j . $$ {\small\sf Example for such a conserved quantity is the mass in an elastic collision, $a_k = m_k$, and then the energy and momentum conservation leads to $b_\mu = 1$, because $p^\mu$ is a conserved quantity also.} {\bf Statement}: If assumptions 1 and 2 are valid then the quantity, $F$, is \beq F = \sum_{k,l=1}^N \int \dpk \Psi_k C_{kl}(x,p_k) = 0, \eeq{4.6} where $$ C_{kl} (x,p_k) = {1 \over 2} \sum_{i,j=1}^N \int \dpl \dpi \dpj \left[ f_i f_j W_{ij|kl}(p_i,p_j|p_k,p_l) - \right. $$ \beq \hspace*{7cm} \left. f_k f_l W_{kl|ij}(p_k,p_l|p_i,p_j) \right] . \eeq{4.6a} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Let us calculate $F$ \beq F = {1 \over 2} \sum_{i,j,k,l=1}^N \int \dpi \dpj \dpk \dpl \Psi_k \left[ f_i f_j W_{ij|kl} - f_k f_l W_{kl|ij} \right] . \eeq{4.7} Let us exchange the summation and integration variables in the loss term as $k,l \leftrightarrow i,j$. This yields \beq F = {1 \over 2} \sum_{i,j,k,l=1}^N \int \dpi \dpj \dpk \dpl (\Psi_k - \Psi_i) f_i f_j W_{ij|kl} . \eeq{4.8} Using the symmetry $W_{ij|kl} = W_{ij|lk}= W_{ji|lk}$, \ $F$ can be written as: \beq F = {1 \over 2} \sum_{i,j,k,l=1}^N \int \dpi \dpj \dpk \dpl (\Psi_l - \Psi_j) f_j f_i W_{ij|kl} . \eeq{4.9} Summing up the last two expressions and multiplying by ${1 \over 2}$ \beq F = {1 \over 4} \sum_{i,j,k,l=1}^N \int \dpi \dpj \dpk \dpl (\Psi_k + \Psi_l - \Psi_i - \Psi_j) f_i f_j W_{ij|kl} . \eeq{4.10} Since $\Psi$ is conserved in a collision the expression in parentheses vanishes, and so $F=0.$ q.e.d. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} \subsection{Conservation of particle number} In this case $\Psi_k=1$. Let us take the BTE \beq p_k^\mu f_{k,\mu} = \sum^N_{l=1} C_{kl} (x,p_k) , \eeq{4.11} multiply it with $\Psi_k$, sum it over $k$ and integrate it over $\int \dpk$: \beq \sum_{k=1}^N \int \dpk p_k^\mu f_{k,\mu} = \sum^N_{k,l=1} \int \dpk C_{kl} (x,p_k) . \eeq{4.11a} Lemma \LT right hand side vanishes. Since the particle four current is $N^\mu = \int \dpk p_k^\mu f_k$ equation (\ref{4.11a}) means that the 4 divergence of the particle current vanishes: \beq N^\mu,_\mu = \sum_{k=1}^N N_k^\mu,_\mu = 0. \eeq{4.13} \B This is the Continuity Equation.\\ It expresses the fact that the particle number is conserved if it is conserved in a microscopic collision. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ %\vspace*{-1.0cm} \subsection{Conservation of charge} We can choose any conserved charge, like baryon charge, strangeness, or electric charge. In this case $\Psi_k=q_k$ can be taken as the charge. The corresponding 4-current is then \beq Q_k^\mu = \int \dpk \, q_k \, p_k^\mu \ f_k. \eeq{4.14} \B Thus the Charge Conservation can be expressed similarly to the previous case as \beq Q^\mu,_\mu = \sum_{k=1}^N Q_k^\mu,_\mu = 0. \eeq{4.15} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} \subsection{Conservation of energy and momentum} Choose $\Psi_k=p_k^\nu$. Let us take the BTE again \beq p_k^\mu f_{k,\mu} = \sum^N_{l=1} C_{kl} (x,p_k) , \eeq{4.16} multiply it with $\Psi_k$, sum it over $k$ and integrate it over $\int\dpk$: \beq \sum_{k=1}^N \int \dpk p_k^\mu p_k^\nu f_{k,\mu} = \sum^N_{k,l=1} \int \dpk p_k^\nu C_{kl} (x,p_k) . \eeq{4.16a} Lemma \LT right hand side vanishes. Since the energy-momentum tensor is $T^{\mu\nu} = \int \dpk p_k^\mu p_k^\nu f_k$, equation (\ref{4.16a}) means that the 4 divergence of the energy-momentum tensor vanishes: \beq T^{\mu\nu},_\mu = \sum_{k=1}^N T_k^{\mu\nu},_\mu = 0. \eeq{4.17} \B This is the energy and momentum conservation. Equations (3.48-53) are also the equations of the {\em relativistic fluid dynamics}.\\ - In fluid dynamics these equations are postulated and not derived. As a matter of fact equations (3.48-53) are {\em not a closed set of equations}, because the energy-momentum tensor and the particle 4-current should be defined too.\\ In the transport theory this is done through the distribution function, which is known only if the solution of the BTE is known. Thus within the transport theory these equations do not provide us the solution of a dynamical problem. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \B In {\bf Eulerian (perfect)} fluid dynamics we postulate that the form of the energy-momentum tensor is given by $$ T^{\mu\nu\ (0)} $$ (see Section 2.3.4) and the {\bf Equation of State (EOS)} gives the relation $P=P(e,n)$. This provides a closed set of solvable partial differential equations. \B In the case of {\bf Navier-Stokes (viscous)} fluid dynamics we assume that the energy-momentum tensor contains $$ T^{\mu\nu \ (1)} $$ also, and not only the EOS but the {\bf transport coefficients} that occur in the dissipative part of the energy momentum tensor also have to be given. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} \section{Boltzmann H-theorem} The definition of the entropy 4-current in transport theory is \beq S^\mu = - \sum_k \int \dpk p_k^\mu f_k [ \log f_k - 1]. \eeq{4.18} ($c=\hbar=k=1$) The entropy should be a non-decreasing function of time, i.e. $S^\mu,_\mu \geq 0$. We want to see if this is a consequence of the BTE or not.\\ Calculate the 4 divergence of the entropy current according to the definition \beq S^\mu,_\mu = - \sum_k \int \dpk p_k^\mu [ \log f_k ] f_k,_\mu . \eeq{4.19} From the Boltzmann Transport Equation $p_k^\mu f_k,_\mu = \sum_l C_{kl} (x,p_k)$. Inserting this into the equation above \beq S^\mu,_\mu = - \sum_k \int \dpk [ \log f_k ] C_{kl} (x,p_k) . \eeq{4.20} Repeat the steps of the Lemma in the previous section, with $\Psi_k = \log f_k$. $\log f_k$ is not a collision invariant: the integral will not necessarily vanish, but we still can get it into a symmetrized form by using the same steps: \beq S^\mu,_\mu = - \sum_{ijkl} {1 \over 4} \int \dpi \dpj \dpk \dpl \left[ \log {{f_k f_l}\over{f_i f_j}} \right] f_i f_j W_{ij|kl} . \eeq{4.20q} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ Aside: Transition rate has also the following symmetry: \beq \sum_{ij} \int \dpi \dpj W_{kl|ij} = \sum_{ij} \int \dpi \dpj W_{ij|kl} . \eeq{4.21} This is a consequence of the unitarity of the scattering matrix.\\ Multiply the above equation by $f_k f_l$, sum it over $k$ and $l$, and integrate it over $\dpk \, \dpl$: \beq \sum_{ijkl} \int \dpi \dpj \dpk \dpl ( f_k f_l W_{kl|ij} - f_k f_l W_{ij|kl} ) = 0 . \eeq{4.22} In the first term make an index change: $i,j \leftrightarrow k,l$:. \beq {1 \over 4} \sum_{ijkl} \int \dpi \dpj \dpk \dpl ( f_k f_l - f_i f_j ) W_{ij|kl} = 0 . \eeq{4.22a} \bigskip\hrule }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} Summing up equations (3.57) and (3.60) we obtain: \beq S^\mu,_\mu = {1 \over 4} \sum_{ijkl} \int \dpi \dpj \dpk \dpl [ {{f_k f_l}\over{f_i f_j}} - \log {{f_k f_l}\over{f_i f_j}} -1 ] f_i f_j W_{ij|kl} . \eeq{4.23} The expression in parentheses is a function, $g(x) = x - \log(x) -1$, which is depicted in Figure 3.7. %\begin{figure}[hbtp] \begin{center} \setlength{\unitlength}{0.5mm} \begin{picture}(300,100)(-20,0) \put(5,10){\vector(1,0){250}} \put(10,5){\vector(0,1){90}} \put(255,15){$x$} \put(15,95){$g(x)$} \put(100,10){\line(0,1){6}} \put(100,2){1} \put(60,55){$-\log x$} \put(170,55){$x-1$} \end{picture} \setlength{\unitlength}{1mm} \end{center} %\caption[New] Figure 3.7 {\it The $g(x) = x - \ln (x) -1$ function } %\label{f2.7} %\end{figure} Since the distribution function is never negative, the argument of function $g(x)$ is also non-negative, thus it follows that $g(x) \geq 0$. \B The above equation then yields \beq S^\mu,_\mu \geq 0 , \eeq{4.24} and it vanishes only if $x=1$, i.e. if $ f_k f_l = f_i f_j$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ CONSEQUENCE:\\ - If irreversible processes are present the entropy increases.\\ - In equilibrium the distribution, $f=f^{eq.}$, is constant, so $S^\mu,_\mu =0$, i.e. the entropy is constant, but it reached its maximum when the equilibrium was reached. - Special case for one component only:\\ Since $S^\mu,_\mu = 0$ it follows \beq f^{eq.} (x,p) f^{eq.} (x,p_1) = f^{eq.} (x,p') f^{eq.} (x,p'_1), \eeq{4.26} so, $\log f$ is a collision invariant: \beq \log f^{eq.} (x,p) + \log f^{eq.} (x,p_1) = \log f^{eq.} (x,p') +\log f^{eq.} (x,p'_1). \eeq{4.27} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm} \section{Equilibrium distribution function} In a collision $p,p_1 \rightarrow p',p'_1$, where $p^\mu + p^\mu_1 = p'^\mu + p'^\mu_1$ the most general collision invariant is $\Psi = a(x) + b_\mu(x) p^\mu$. \LT \\ If $\log f^{eq.}(x,p)$ is a collision invariant it should be expressible via $\Psi$, $$ \log f^{eq.}(x,p) = a(x) + b_\mu(x) p^\mu, $$ so that \beq f^{eq.}(x,p) = \exp( a(x) + b_\mu(x) p^\mu). \eeq{4.28} If there is no external force the distribution should be homogeneous: \beq f^{eq.}(x,p) = f^{eq.} (p). \eeq{4.29} Since there are no gradients it follows that there are no transports.\\ - This way Landau's and Eckart's definitions are equivalent.\\ Determine constants $a$ and $b^\mu$ by calculating macroscopic expectation values from this distribution $f^{eq.}(p)$. \\ - Start with the flow velocity, $u^\mu = const. \times N^\mu$, where $$ N^\mu = const. \times \ipint p^\mu \exp(a+b_\mu p^\mu) = $$ \beq const. \times \ipint p^\mu \exp(b_\mu p^\mu). \eeq{4.30} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} This integral converges only if $b^\mu$ is a time-like 4-vector (since $p^\mu$ is time-like, then $b_\mu p^\mu < 0$).\\ Then $N^\mu$ will be parallel to $b^\mu$. \LT \beq b^\mu = const. \times u^\mu = - {1 \over T} u^\mu, \eeq{4.31} where $T$ is just a constant. This yields \beq f^{eq.}(p) = const. \times e^{-u_\mu p^\mu / T}, \eeq{4.32} where the normalization constant can be obtained from the requirement $$n = u_\mu N^\mu$$ and it yields $$ const. =n / [ 4 \pi m^2 T K_2(m/T)]$$ \B This is just the J\"uttner distribution that was introduced in Section 2.4. \B Thus we have seen that the J\"uttner distribution is the stationary solution of the relativistic Boltzmann equation. If $f$ is a solution of the BTE, it $\longrightarrow$ $f^{Juttner}$. See example in Section 3.4. \B This is the relativistic extension of the Boltzmann distribution. The approach to equilibrium is fast: 3-8 fm/c! Most of the time the local distribution $f(x,p)$ is close to $f^{Juttner}$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm} \section{Zeroth order approximation} \begin{center} {\bf Perfect fluid dynamics} \end{center} ASSUMPTIONS:\\ - (1) Our system is not homogeneous, but the gradients are small, so local distributions can be written as \beq f(x,p) = {1 \over{(2 \pi \hbar)^3}} \exp \left( {{\mu_{ch}(x) - p^\mu u_\mu (x) }\over {T(x)}} \right), \eeq{4.34} - (2) Local $n(x), P(x), e(x), s(x)$ are also known (from $f(x,p)$ by using the definitions)\\ - (3) We assume (!) that in the (LR), $ T^{\mu\nu}$ is diagonal. ( This is also a consequence of assumption (1), since we have neglected the gradients of the flow velocity and of the thermodynamical variables: \beq T^{\mu\nu}_{LR} = T^{\mu\nu\ (0)}_{LR} = (e+P) u^\mu_{LR} u^\nu_{LR} - P g^{\mu\nu} = \left( \begin{array}{cccc} e & 0 & 0 & 0 \\ 0 & P & 0 & 0 \\ 0 & 0 & P & 0 \\ 0 & 0 & 0 & P \end{array} \right)_{LR} \eeq{4.35} The equations of {\bf Perfect Fluid Dynamics} are the conservation laws under the assumption that $f(x,p) = f ^{Juttner}(x,p)$ \beq N^\mu ,_\mu = 0 \ \ \ \ {\rm or} \ \ \ \partial_\mu (n u^\mu) = 0, \eeq{4.36} and \beq T^{\mu\nu}_{,\mu} = 0 \ \ \ \ {\rm or} \ \ \ \partial_\mu (T^{\mu\nu}) = 0. \eeq{4.37} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Using $u^\mu = (\gamma, \gamma \vec{v}), \ $\\ $ T^{ik} = w \gamma^2 v_i v_k + P \delta_{ik}, \ $\\ $ T^{0i} = -T_{0i} = w \gamma^2 v_i, \ $ $ T^{00} = T_{00} = (e + P v^2) \gamma^2 $ , $(i,k= 1, 2, 3)$,\\ the equation of continuity takes the form \beq \partial_t (n \gamma) + {\rm div}(n \gamma \vec{v}) = 0, \eeq{4.37a} or \beq (\partial_t +\vec{v}\, {\rm grad}) (n\gamma) + n \gamma {\rm div} \vec{v} = 0. \eeq{4.37b} Now introducing the apparent density \beq {\cal N} \equiv n \gamma = {\sf n} , \eeq{4.37c} the continuity equation takes the familiar form \beq (\partial_t + \vec{v} \, {\rm grad} ) {\cal N} = - {\cal N} {\rm div}\vec{v}. \eeq{4.38} Similarly introducing \beq \vec{\cal M} \equiv T^{0i} = w \gamma^2 \vec{v}, \eeq{4.38a} \beq {\cal E} \equiv T^{00} = (e+P \vec{v}\,^2) \gamma^2 , \eeq{4.38b} the energy and momentum conservation will take the form \beqar (\partial_t + \vec{v} \, {\rm grad} ) \vec{\cal M} & = & - \vec{\cal M} ({\rm div}\vec{v}) - {\rm grad} P , \\ (\partial_t + \vec{v} \, {\rm grad} ) {\cal E} & = & - {\cal E} {\rm div}\vec{v} - {\rm div} (P \vec{v}). \eeqar{4.39} \B Last two equations are the {\bf Euler equation} of fluid dynamics and the {\bf energy conservation}.\\ - Equations (3.77, 80, 81) have the familiar form of the equations of non-relativistic perfect fluid dynamics. - Important: (!) ${\cal N}$, ${\cal E}$, $\vec{\cal M}$ are not related directly to the EOS, \\ but one has to solve a set of algebraic equations, (\ref{4.37c},\ref{4.38a},\ref{4.38b}) to obtain the thermodynamical quantities. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \section{Assignment 3} \begin{description} \item[3.a] Show that, if $u^\mu $ is the flow velocity: $u^\mu u_\mu , _\nu = 0$. (The notation is such that: \ $,_\mu \equiv {{\partial}\over{\partial x^\mu}} \equiv \partial_\mu \equiv (\partial_t, \nabla_{\vec{r}})$. \item[3.b] Prove that $\Delta^{\mu\nu} \Delta_{\nu\sigma} = \Delta^\mu_\sigma$, and $\Delta^\mu_\mu = 3$. \item[3.c] Determine the chemical potential $\mu$ from the normalization $n = N^\mu u_\mu$. (Will be done in class too!) \item[3.d] Determine the energy-momentum tensor for the equilibrium distribution function $f^{Juttner}(x,p)$, (i.e.: determine $e$ and $p$). Show that $T$ tends to the temperature in the non-relativistic limit (i.e.: $e \approx n (m + {3\over 2} T + ....))$. \item[3.e] Determine the entropy density and show that the basic thermodynamical relation $$ T s = e + P - \mu n $$ holds for the J\"uttner distribution. \end{description} }%end tr-page % transparency ================================================================ \end{document}