% % 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 2 / Chapter 2 % % 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 ===================================================== \transparencyframe{ \setcounter{chapter}{1} \vspace*{-1.5cm}\chapter{Introduction to Relativistic Kinetic Theory} Most reaction models are based on kinetic theory. Familiarity with basic concepts of special relativity, elementary statistical physics, and some initial knowledge of basic classical nuclear and/or particle physics are assumed. Our discussion is based on the book of S.R. de Groot, W.A. van Leeuwen and Ch.G. van Weert [1] \begin{itemize} \item Microscopic variables : particle mass, momentum, energy, position \item Macroscopic variables : flow, velocity, density, temperature, etc. \end{itemize} Kinetic theory establishes a relationship between macroscopic and microscopic properties, by using a one-particle distribution function $f(x,p)$. \section{Basic definitions of microscopic quantities} Most of the time we use the convention $c=k=1$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm} \subsection{Phase-space variables} \begin{itemize} \item SPACE TIME COORDINATES.\\ (3 spatial coordinates and time)\ \LT 4 dimensional space:\\ the space-time. Coordinates are time $t$ and position 3-vector $\vec{r}$. See Fig.~2.1. \noindent %\begin{figure}[hbtp] \setlength{\unitlength}{0.7mm} \begin{picture}(400,150)(100,0) \thicklines \put(200,10){\vector(0,1){125}} \put(100,60){\vector(1,0){200}} \put(200,145){t} \put(310,60){x} \thinlines \put(150,10){\vector(1,1){125}} \put(250,10){\vector(-1,1){125}} \put(290,140){lightcone} \thicklines \put(200,60){\vector(3,1){20}} \put(223,68){$x^\mu$ \ \ {\em space-like} 4-vector} \end{picture} %\caption[F:CS91] Figure 2.1 {\it Coordinates of a point (or event) in space-time are denoted by a 4-vector: $ x^\mu = (t,\vec{r})$. From [2]} %\label{f1.1} %\end{figure} \end{itemize} \setlength{\unitlength}{1mm} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1cm} \item FOUR-MOMENTUM. \\ The four momentum of a particle in space-time is: $ p^\mu = (p^0, \vec{p})$, \\ where $p^0 = \sqrt{ (\vec{p})^2 + m^2 } $, \ \ ($p^0 = E$). The 4-momentum is a time-like vector with the normalization: $(p^0)^2 - (\vec{p})^2 = m^2$. This can be written in the form: \beq p^\mu p_\mu \equiv \sum_\mu p^\mu p_\mu = m^2, \eeq{pnorm} where $p^\mu$'s are called the {\em contravariant},\\ and \ \ \ $p_\mu$ -s are the {\em covariant} components of a 4-vector. \beq p_\mu = g_{\mu\nu} p^\nu, \eeq{pcov} where $g_{\mu\nu} = g^{\mu\nu} = $ \ diag $(1,-1,-1,-1)$, \ is the {\em metric tensor}. \LT \noindent $p^\mu = (p^0,\vec{p})$, and $ p_\mu = (p^0, -\vec{p})$. \noindent The square of the 3-momentum, $(\vec{p})^2$, $p^2 \ ( =(\vec{p})^2 )$, but $p^\mu p_\mu = m^2$. \noindent Four vectors may be space-like, $q_\mu q^\mu < 0$ or time-like $q_\mu q^\mu > 0$. \item FOUR-VELOCITY.\\ (The 3-velocity is $\vec{v} \equiv \vec{p}/p^0$.)\\ The 4-velocity is a unit vector. It points in the direction of the motion. It is a {\em time-like} unit vector: \beq u^\mu u_\mu = +1, \eeq{unorm} \beq u^\mu = ( \gamma, \gamma \vec{v} ), \ \ {\rm and} \ \ u_\mu = ( \gamma, -\gamma \vec{v} ) ; \ \ \ \ \gamma = 1/ \sqrt{1-\vec{v}\, ^2} \eeq{ucov} This velocity 4-vector satisfies the normalization, i.e.:\\ $ u^\mu u_\mu = \gamma^2 (1-\vec{v}\, ^2) = +1$, \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1cm} \item WORLD-LINE OF A PARTICLE IN SPACE-TIME The world-line of a particle in space-time is illustrated in Fig.~2.2 %\begin{figure}[hbtp] \setlength{\unitlength}{0.7mm} \begin{picture}(400,150)(100,0) \thicklines \put(200,10){\vector(0,1){125}} \put(100,60){\vector(1,0){200}} \put(200,145){t} \put(310,60){x} \put(280,90){$u^\mu$} \put(275,90){\vector(1,3){6}} \thinlines \put(150,10){\vector(1,1){125}} \put(250,10){\vector(-1,1){125}} %\put(310,140){lightcone} \thicklines \end{picture} %\caption[F:CS91] Figure 2.2 {\it $u^\mu$ is tangent to the world-line. This means that $u^\mu = {dx^\mu}/ {d\tau}$. From [2]} %\label{f1.2} %\end{figure} \setlength{\unitlength}{1mm} For a {\em space-like} unit four vector $\Lambda^\mu$: $\Lambda^\mu \Lambda_\mu = -1$,\\ while for a {\em time-like} unit vector $\Omega^\mu$: $\Omega^\mu \Omega_\mu = +1$. \\ Space-like and time-like vectors cannot be Lorentz transformed into each other! \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1cm} \item COORDINATE SYSTEM. \\ In particle and nuclear physics: special coordinate system.\\ Spatial $z$-axis is parallel to the beam of the accelerator. \\ In general, not exactly central (not head on) collisions, the 3-vector connecting the centers of a beam particle and a target particle points out an other direction. The component of this vector orthogonal to the beam is the impact vector $\vec{b}$, which is a two dimensional vector.\\ The direction of this vector is denoted usually as the $x$ direction.\\ These two axes, $x$ and $z$, span the so called reaction plane of a given collision $[x,z]$. See Fig. 2.3. %\begin{figure}[hbtp] \setlength{\unitlength}{0.5mm} \begin{picture}(400,150)(70,0) \thicklines \put(200,5){\vector(0,1){125}} \put(100,55){\vector(1,0){250}} \put(50,80){\vector(1,0){15}} \put(50,85){Beam} \put(50,72){$v_\|$} \put(200,140){$x$} \put(360,55){$z_{axis}$} \put(270,55){\vector(0,1){25}} \put(276,65){$\vec{b}$ --- impact parameter} \thinlines %\put(200,55){\circle{40}} \put(110,80){\circle{25}} \put(204,60){T} \put(110,81){P} \put(80,80){\line(1,0){250}} \end{picture} %\caption[F:CS91] Figure 2.3 {\it The $z$ component of a vector is also denoted as its parallel, ($\|$), component. Thus, any vector (3-vector) can be decomposed as: $\vec{v} = ( v_\parallel , \vec{v}_\perp)$.\ Here $\vec{v}_\perp$ is a 2- vector which can rotate in the {\em azimuthal}, ($\phi$) direction. From [2]} %\label{f1.3} %\end{figure} \setlength{\unitlength}{1mm} \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1cm} \item RAPIDITY. \\ The rapidity is a generalization of the velocity. \beq y \equiv {\rm arcth}(v_\parallel) \ \ = {\rm arcth} (\frac{p_\parallel}{p^0}) \ = \ \ \frac{1}{2} \ln ( \frac{p^0 + p_\parallel}{p^0 - p_\parallel}) \eeq{rapidity} This definition uses the components of vectors $\vec{v}$ and $\vec{p}$ \end{itemize} \subsection{Properties of the rapidity} For small velocities: $y \approx v_\parallel $.\\ Particle after the collision moves with $\vec{v}$:\\ It is customary to decompose $\vec{v}$ to coordinates $( y, \vec{p}_\perp / m)$.\\ $\vec{p}$ can also be decomposed as $ p^\mu \ = (p^0 , p_\parallel , \vec{p}_\perp )$.\\ The limit of rapidity coordinates for non-relativistic velocities:\\ $( y, \vec{p}_\perp / m) \ \longrightarrow \ (v_\parallel, \vec{v}_\perp )$. \\ While the velocity is limited to 1 $(c)$,\\ the rapidity $y$ may vary between $(-\infty, \infty)$. See Fig.~2.4. %\begin{figure}[hbtp] \setlength{\unitlength}{0.5mm} \begin{picture}(400,120)(60,0) \thicklines \put(200,5){\vector(0,1){100}} \put(100,55){\vector(1,0){250}} \put(200,110){$v_\parallel$} \put(360,55){$y$} \thinlines \put(198,15){\line(1,0){4}} \put(198,95){\line(1,0){4}} \put(189,13){-1} \put(189,93){1} \put(197.2,52.2){\begin{picture}(300,100) \put( 0, 0){$\bullet$} \put( 5, 5.0){$\bullet$} \put( 10, 9.8){$\bullet$} \put( 15,14.3){$\bullet$} \put( 20,18.5){$\bullet$} \put( 25,22.2){$\bullet$} \put( 30,25.4){$\bullet$} \put( 35,28.2){$\bullet$} \put( 40,30.5){$\bullet$} \put( 45,32.4){$\bullet$} \put( 50,33.9){$\bullet$} \put( 55,35.2){$\bullet$} \put( 60,36.2){$\bullet$} \put( 65,37.0){$\bullet$} \put( 70,37.7){$\bullet$} \put( 75,38.2){$\bullet$} \put( 80,38.6){$\bullet$} \put( 85,38.9){$\bullet$} \put( 90,39.1){$\bullet$} \put( 95,39.3){$\bullet$} \put( - 5, -5.0){$\bullet$} \put( -10, -9.8){$\bullet$} \put( -15,-14.3){$\bullet$} \put( -20,-18.5){$\bullet$} \put( -25,-22.2){$\bullet$} \put( -30,-25.4){$\bullet$} \put( -35,-28.2){$\bullet$} \put( -40,-30.5){$\bullet$} \put( -45,-32.4){$\bullet$} \put( -50,-33.9){$\bullet$} \put( -55,-35.2){$\bullet$} \put( -60,-36.2){$\bullet$} \put( -65,-37.0){$\bullet$} \put( -70,-37.7){$\bullet$} \put( -75,-38.2){$\bullet$} \put( -80,-38.6){$\bullet$} \put( -85,-38.9){$\bullet$} \put( -90,-39.1){$\bullet$} \put( -95,-39.3){$\bullet$} \end{picture} } \end{picture} %\caption[F:CS91] Figure 2.4 {\it The beam-parallel component of the velocity as function of the rapidity $y$. From [2]} %\label{f1.4} \setlength{\unitlength}{1mm} %\end{figure} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \B Lorentz transformation properties of rapidity: \\ If $ y_1$ is the rapidity of a particle in frame $ K_1$, \\ and $ y_2$ is the rapidity of frame $ K_1$ in frame $ K_2$,\\ {\em then}: $ y=y_1 +y_2$ is the rapidity of the particle in frame $K_2$. \\ Assignment No. 2.a is to prove this. Solution to assignment No. 2 also includes the definitions of further important quantities like the transverse mass, the pseudo-rapidity and the light-cone variables! }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\subsection{Some typical rapidities} Here we give some examples for nucleons $(m=0.939 $GeV). The beam energies for some accelerators are given in Table 2.1. In case of colliders the energies of both beams are given separately as 30+30 $A $GeV. \vspace*{-0.4cm} %\begin{table} \begin{center} \begin{tabular}{rrrrrc} \hline $E_{Lab}$ & $p^0$ &$ \mid \vec{p} \mid$ & $\Delta y$ & $v$ & Place \\ $[A$GeV$]$ & $[$GeV$]$& $[$GeV/c$]$ & - & $[$c$]$ & - \\ \hline 0.1 & 1.03 & 0.445 & 0.458 & 0.428 & MSU-K800 \\ 0.5 & 1.44 & 1.09 & 0.99 & 0.758 & LBL-BEVALAC\\ 1.0 & 1.94 & 1.69 & 1.35 & 0.875 & LBL-BEVALAC\\ 2.0 & 2.94 & 2.78 & 1.81 & 0.947 & LBL-BEVALAC\\ 4.1 & 5.04 & 4.95 & 2.36 & 0.982 & DUBNA \\ 10 & - & - & 3.06 & - & BNL-AGS \\ 14 & - & - & 3.4 & - & BNL-AGS \\ 60 & - & - & 4.9 & - & CERN-SPS \\ 200 & - & - & 6.0 & - & CERN-SPS \\ 1800 (30+30) & - & - & 8.2 & - & BNL-RHIC$^*$ \\ (100+100) & - & - & 10.7 & - & BNL-RHIC$^*$ \\ (900+900) & - & - & 15.1 & - & ($p \bar{p}$) FNAL-Tevatron$^*$ \\ (3500+3500) & - & - & 17.8 & - & CERN-LHC$^*$ \\ (8TeV+8TeV) & - & - & 19.5 & - & ($pp$) CERN-LHC$^*$ \\ (20TeV+20TeV)& - & - & 21.2 & - & ($pp$) SSC$^*$ \\ \hline \end{tabular} \end{center} \noindent %\caption[New] Table 2.1 {\it Typical rapidities and beam energies of some proton or heavy ion accelerators. The planned accelerators are colliders. At higher energies, $E_{Lab} \gg m_p$, the velocity tends to the velocity of light, and $p^0$ and $|\vec{p}|$ approach each other. $^*$planned } %\label{t1.1} %\end{table} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\section{Basic definitions of macroscopic quantities} Elementary macroscopic quantities and their definitions. \begin{itemize} \item[(i)] LOCAL DENSITY ${\sf n}={\sf n}(\vec{r},t)={\sf n}(x)$, function of the space-time coordinate $x$ ($\equiv x^\mu$); \ \ \ \ $(N={\sf n} \ \Delta^3x)$. \\ \B The density, {\sf n}, is not an invariant scalar because the 3-volume element, $\Delta^3 x$, is not invariant under Lorentz transformation.\\ \B The total number of particles in a fixed volume element, $N$, is obviously independent of the reference frame, so it is an invariant scalar. {\sc EXAMPLE: DENSITY} (Fig. 2.5). Density profile of a nucleus at rest\\ ${\sf n}(r), \ \ {\sf n}_0=0.145/fm^3$\\ $(1\ fm\ =\ 1\ F\ =\ 1\ fermi\ =\ 10^{-13}\ cm)$\\ $(10\ mbarn\ = 1\ fermi^2\ )$\\ $R\approx\ 7\ fm$, for Pb. %\begin{figure}[hbtp] \setlength{\unitlength}{0.5mm} \begin{picture}(400,110)(60,0) \thinlines \put(100,5){\vector(0,1){80}} \put(90,15){\vector(1,0){260}} \put(105,85){\sf n($r$)} \put(360,25){$r$} \put(100,35){\vector(1,0){80}} \put(140,37){R} \thicklines \put(85,53){${\sf n}_0$} \put(100,35){\oval(160,40)[tr]} \put(200,35){\oval(40,40)[bl]} \end{picture} %\caption[New] Figure 2.5 {\it Schematic density profile of a nucleus } %\label{f1.5} %\end{figure} \setlength{\unitlength}{1mm} The bulk of the matter is at density ${\sf n}_0$ except the relatively narrow surface region. \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1cm}\item[(ii)] LOCAL PARTICLE FLOW $\vec{j}=\vec{j}(\vec{r},t)=\vec{j}(x)$, i.e.: particle current across unit area in unit time. {\sc EXAMPLE: FLOW} (Fig. 2.6). The current $\vec{j}$, depends on $(\vec{r},t)$. %\begin{figure}[hbtp] \setlength{\unitlength}{0.5mm} \begin{flushleft} $\vec{j} = (0,0,-j_z)$\\ $\vec{j} = (0,0,0)$\\ $\vec{j} = (0,0,j_z)$\\ \end{flushleft} \begin{picture}(400,100)(50,0) \thinlines \put(100,5){\vector(0,1){70}} \put(90,15){\vector(1,0){260}} \put(103,75){$x$} \put(360,25){$z$} \thicklines %\put(230,60){\circle{50}} %\put(200,45){\circle{50}} \put(245,70){\vector(-1,0){10}} \put(185,35){\vector(1,0){10}} \end{picture} %\caption[New] Figure 2.6 {\it Baryon currents in different spatial regions of a heavy ion collision in the center of mass (c.m.) system.} %\label{f1.6} %\end{figure} \setlength{\unitlength}{1mm} The currents are directed along the beam (z-axis) and they are opposite to each other in the projectile and target regions.\\ In the overlap region (hatched area) the two currents cancel each other and the resulting current vanishes. \item[(iii)] PARTICLE FOUR FLOW. From the above two quantities create a 4-vector.\\ At this stage we do not know its transformation properties yet. \beq N^\mu (x) = ( {\sf n}(x), \vec{j}(x)) \eeq{ptc4flow} \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1cm}\item[(iv)] PARTICLE DISTRIBUTION, $f$, in $\mu$-space (6-dim. $(x,p)$-space)\\ It gives the number of particles, $N$, in a phase space volume element: \beq f(x,p) \ : \ \ \ \ N = f(x,p) \ \Delta^3x \ \Delta^3p. \eeq{fxp} \end{itemize} {\sc Example $f(x,p)$} (Fig. 2.7). %\begin{figure}[hbtp] \setlength{\unitlength}{0.45mm} \begin{picture}(400,120)(5,0) \thinlines \put(100,5){\vector(0,1){90}} \put(115,85){$f$} \put(90,15){\vector(1,0){260}} \put(360,25){$p_\|$} \put(247.2,15) {\unitlength 0.5pt \begin{picture}(300,50) \put(-105.0, 0.0){$\bullet$} \put(-97.5, 0.1){$\bullet$} \put(-90.0, 0.2){$\bullet$} \put(-82.5, 0.4){$\bullet$} \put(-75.0, 0.8){$\bullet$} \put(-67.5, 1.6){$\bullet$} \put(-60.0, 3.0){$\bullet$} \put(-52.5, 5.5){$\bullet$} \put(-45.0, 9.5){$\bullet$} \put(-37.5,15.4){$\bullet$} \put(-30.0,23.1){$\bullet$} \put(-22.5,32.2){$\bullet$} \put(-15.0,41.0){$\bullet$} \put(-7.50,47.6){$\bullet$} \put( 0.0,50.0){$\bullet$} \put( 7.5,47.6){$\bullet$} \put( 15.0,41.0){$\bullet$} \put( 22.5,32.2){$\bullet$} \put( 30.0,23.1){$\bullet$} \put( 37.5,15.4){$\bullet$} \put( 45.0, 9.5){$\bullet$} \put( 52.5, 5.5){$\bullet$} \put( 60.0, 3.0){$\bullet$} \put( 67.5, 1.6){$\bullet$} \put( 75.0, 0.8){$\bullet$} \put( 82.5, 0.4){$\bullet$} \put( 90.0, 0.2){$\bullet$} \put( 97.5, 0.1){$\bullet$} \put(105.0, 0.0){$\bullet$} \end{picture} } \put(90,45){\vector(1,0){260}} \put(360,55){$p_\|$} \put(217.2,45) {\unitlength 0.5pt \begin{picture}(300,50) \put(-10,44.0){$\bullet$} \put(-20,41.0){$\bullet$} \put(-30,32.2){$\bullet$} \put(-40,23.1){$\bullet$} \put(-50,15.4){$\bullet$} \put(-60, 9.5){$\bullet$} \put(-70, 5.5){$\bullet$} \put(-80, 3.0){$\bullet$} \put(-90, 1.6){$\bullet$} \put(-100,0.8){$\bullet$} \put(-110,0.4){$\bullet$} \put(-120,0.2){$\bullet$} \put(-130,0.1){$\bullet$} \put(-140,0.0){$\bullet$} \put(0.0,46.0){$\bullet$} \put( 10,44.0){$\bullet$} \put( 20,41.0){$\bullet$} \put( 30,32.2){$\bullet$} \put( 40,23.1){$\bullet$} \put( 50,15.4){$\bullet$} \put( 60, 9.5){$\bullet$} \put( 70, 5.5){$\bullet$} \put( 80, 3.0){$\bullet$} \put( 90, 1.6){$\bullet$} \put(100, 0.8){$\bullet$} \put(110, 0.4){$\bullet$} \put(120, 0.2){$\bullet$} \put(130, 0.1){$\bullet$} \put(140, 0.0){$\bullet$} \end{picture} } \put(90,75){\vector(1,0){260}} \put(360,85){$p_\|$} \put(187.2,75) {\unitlength 0.5pt \begin{picture}(300,50) \put(-105.0, 0.0){$\bullet$} \put(-97.5, 0.1){$\bullet$} \put(-90.0, 0.2){$\bullet$} \put(-82.5, 0.4){$\bullet$} \put(-75.0, 0.8){$\bullet$} \put(-67.5, 1.6){$\bullet$} \put(-60.0, 3.0){$\bullet$} \put(-52.5, 5.5){$\bullet$} \put(-45.0, 9.5){$\bullet$} \put(-37.5,15.4){$\bullet$} \put(-30.0,23.1){$\bullet$} \put(-22.5,32.2){$\bullet$} \put(-15.0,41.0){$\bullet$} \put(-7.50,47.6){$\bullet$} \put( 0.0,50.0){$\bullet$} \put( 7.5,47.6){$\bullet$} \put( 15.0,41.0){$\bullet$} \put( 22.5,32.2){$\bullet$} \put( 30.0,23.1){$\bullet$} \put( 37.5,15.4){$\bullet$} \put( 45.0, 9.5){$\bullet$} \put( 52.5, 5.5){$\bullet$} \put( 60.0, 3.0){$\bullet$} \put( 67.5, 1.6){$\bullet$} \put( 75.0, 0.8){$\bullet$} \put( 82.5, 0.4){$\bullet$} \put( 90.0, 0.2){$\bullet$} \put( 97.5, 0.1){$\bullet$} \put(105.0, 0.0){$\bullet$} \end{picture} } \thicklines %\put(50,80){\circle{50}} %\put(40,45){\circle{50}} \put(65,90){\vector(-1,0){10}} \put(25,35){\vector(1,0){10}} \end{picture}\vskip 0.5truecm %\caption[F:CS91] Figure 2.7 {\it Longitudinal component of the momentum distribution in different spatial regions of a heavy ion collision} %\label{f1.7} %\end{figure} \setlength{\unitlength}{1mm} Density and current can be expressed in terms of the distribution: \beq {\sf n}(x) \ = \ \ \pint \ f(x,p) , \eeq{nfxp} \beq \vec{j}(x) \ = \ \ \pint \ \vec{v} \ f(x,p) . \eeq{jfxp} Transformation properties: We know $\vec{v} = \vec{p} / p^0$, \LT (2.8) \& (2.9) can be united as $N^\mu(x) = ({\sf n}(x), \vec{j}(x))$ \beq N^\mu(x) = \ipint \ p^\mu \ f(x,p). \eeq{Nfxp} If we want $N^\mu$ to be a 4-vector also, then $f(x,p) \times \ipint$ should be an invariant scalar. As we will see later this is satisfied. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ ASIDE: \\ Show that $d^3p / p^0$ is invariant scalar:\\ In 4-dim. $p^\mu$ space $(d^3p)^\mu$ is the normal four vector of a surface element satisfying: $ p^\mu p_\mu = m^2 $.\\ This hypersurface element is then represented by its normal vector (4-vector). It is pointing into the direction of $p^\mu$. %\vskip 1.truecm \noindent 2-dim. hypersurface in a 3-dim space $\longrightarrow$\\ Sphere with constant radius: $ p^\mu p_\mu =m^2$.\\ \setlength{\unitlength}{0.6mm} \begin{picture}(350,60)(10,0) \thinlines \put(220,50){\vector(1,0){50}} \put(260,92){$(d^3 p)^\mu$} \put(220,50){\vector(0,1){50}} \put(220,50){\vector(-1,-1){30}} %\put(220,50){\circle{40}} \end{picture} \setlength{\unitlength}{1mm} We can visualize the invariance of the phase space volume element as follows:\\ $d^3 p$ is the $ 0$th component of the normal vector, $(d^3 p)^\mu$.\\ Divide it by the $0$th component of another 4-vector $p^\mu$ parallel to it,\\ then the ratio is an invariant scalar: $d^3p/p^0$. \medskip Strictly we can see this in the following way: $$ \int \ 2 \ \delta(p^\mu p_\mu - m^2) d^4 p \ = \ \ \ \ \hspace*{9cm} $$ \beq \ \ \ \ \ \ \ \ \ \ \ip^3 \ \ip^2 \ \ip^1 \ \ip^0\ 2 \ \delta(p^\mu p_\mu - m^2) \eeq{asd1} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ Using \beq \delta [ \phi(x)] = \sum_i \frac{1}{\mid \phi ' (a_i) \mid} \delta (x-a_i), \eeq{asd2} where $ a_i $ is the root of $\phi(x)$, (i.e. $\phi(a_i)= 0$), \\ we can cast the $\delta$-function in the form $ \delta(p^\mu p_\mu - m^2) \ = \ \delta((p^0)^2 - (\vec{p})^2 - m^2) \ = \ \delta((p^0)^2 - \varepsilon^2) \ = \ \frac{1}{2 \varepsilon} \ [\delta(p^0 - \varepsilon) \ + \ \delta(p^0 + \varepsilon) ] $\\ where $ \varepsilon= \sqrt{(\vec{p})^2 + m^2}$. Thus: \beq \int^\infty_{-\infty} \ 2 \ \frac{1}{2 \varepsilon} \ [\delta(p^0 - \varepsilon) \ + \ \delta(p^0 + \varepsilon) ] \Theta(p^0)\ d\varepsilon = 1/p^0 , \eeq{asd4} where $$ \Theta(p^0) = \left\{ \begin{array}{rl} 1: &\ {\rm if}\ \ p^0 \geq 0\\ 0: &\ {\rm if}\ \ p^0 < 0 \end{array} \right. $$ \hrule \medskip Therefore, \beq N^\mu(x) = \int d^4 p \ 2 \ \delta(p^2 - m^2) p^\mu f(x,p) \ \Theta(p^0) = \ \ipint \ p^\mu \ f(x,p) , \eeq{asd5} where $\Theta(p^0)$ is the step function.\\ Conclusion:\\ $ f(x,p)$ is an invariant scalar, and consequently $N^\mu$ is a contravariant 4-vector. Furthermore ${\sf n}(x)$ and $\vec{j}(x)$ are transformed as components of a 4-vector. Other macroscopic quantities are more involved: Introduced via the distribution function $f(x,p)$ in sections (2.3-4). }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\section{Energy-momentum tensor} The energy-momentum tensor: macroscopic quantity\\ \B $T^{00}(x)$ is the energy density. Since the energy of a particle is $p^0$ in the kinetic theory \beq T^{00}(x) = \pint \ p^0 \ f(x,p) \eeq{2.1} \B The energy flow is: $ c T^{0i}$ and the momentum density: ${1\over c} T^{i0},$ \\ $(i=1,2,3).$ These expressed via $f(x,p)$: \beq \begin{array}{lclcl} T^{0i} & = & \pint \ p^0 \ v^i \ f(x,p) & : &{\sf energy\ flow}, \\ T^{i0} & = & \pint \ p^i \ f(x,p) & : & {\sf momentum\ density} . \end{array} \eeq{2.2-3} where $v^i$ is the $i^{th}$-component of the flow 3-velocity. The momentum flow tensor or pressure tensor can be written as: \beq T^{ik} = \pint \ \ p^i \ v^k \ f(x,p) \eeq{2.4} \B $\Longrightarrow$ These can be combined into the form (by using $\vec{v} = {{\vec{p}}/ {p^0}} $ ): \beq \begin{array}{lclll} T^{\mu\nu} (x) & = & \ipint & p^\mu p^\nu & f(x,p) , \\ tensor & & scalar & tensor & scalar \\ \end{array} \eeq{2.5} {\it i.e. } the energy-momentum tensor is the second moment of the distribution function $f(x,p)$. It is symmetric: $ T^{\mu\nu}=T^{\nu\mu}.$ (No fields and potential energy. Only the rest mass and the kinetic energy.) }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\subsection{Hydrodynamic flow} The flow velocity of a medium, $u^\mu$, is a time-like unit vector.\\ Parallel to the world-line of the particles\\ (if there are particles in the matter,\\ otherwise is parallel to the energy flow as we will see this later !!) %\begin{figure}[hbtp] \setlength{\unitlength}{0.6mm} \begin{picture}(400,150)(40,0) \thicklines \put(200,10){\vector(0,1){125}} \put(100,60){\vector(1,0){200}} \put(200,145){t} \put(310,60){x} \thinlines \put(150,10){\vector(1,1){125}} \put(250,10){\vector(-1,1){125}} \put(280,140){lightcone} \thicklines \put(150,80){\vector(1,3){6}} \put(153,85){$u^\mu$} \put(53,85){$u^\mu u_\mu = +1$} \end{picture} %\caption[New] Figure 2.8 {\it Flow velocity} %\label{f1.8} %\end{figure} \setlength{\unitlength}{1mm} \bigskip Now define a projector, $\Delta^{\mu\nu}$,\\ which projects a 4-vector into the plane (3-dimensional hypersurface) orthogonal to $u^\mu$. \ }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ Aside: \\ Construct $\Delta^{\mu\nu}$ orthogonal projection to any time-like 4-vector $u^\mu$.\\ The unit vector parallel to $u^\mu$ \ is: ${{u^\mu}\over {(u^\mu u_\mu)^{1/2}}}$. \\ The projected length of any vector $A^\mu$ in this direction is ${{A_\mu u^\mu}\over {(u^\mu u_\mu)^{1/2}}}$,\\ and so the vector projection into the direction of $u^\mu$ is \beq {{A_\mu u^\mu}\over {(u^\mu u_\mu)^{1/2}}} \times {{u^\tau}\over {(u^\tau u_\tau)^{1/2}}}. \eeq{qwe} We have to subtract this from the total $A^\mu$ \\ in order to get the orthogonal projection. Thus the orthogonal projector is: \beq \Delta^{\mu\nu} \equiv g^{\mu\nu} - {{u^\mu u^\nu}\over {(u^\nu u_\nu)}}. \eeq{2.6} The projector acting on the flow velocity thus yields: $\Delta^{\mu\nu} u_\mu = 0.$ \bigskip\hrule \noindent\bigskip \subsection{Local Rest frame: (LR)} This is the reference frame where $u^\mu = u^\mu_{(LR)} = (1,0,0,0)$. Since $u^\mu$ is time-like there is always a Lorentz transformation which leads to the Local Rest frame (LR). In this frame: \beq \Delta^{\mu\nu}_{(LR)} = \Delta_{\mu\nu\ (LR)} = diag \ (0,-1,-1,-1), \eeq{2.7} and \beq \Delta^\mu_{\nu\ (LR)} = diag \ (0,1,1,1). \eeq{2.8} How do we find the Local Rest frame? There are two usual ways! }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} ECKART's DEFINITION Local Rest frame is tied to conserved particles, or conserved charges\\ like baryon charge (if there are any!)\\ Particle 4-flow: \beq N^\mu = \ipint \ p^\mu \ f(x,p) , \eeq{2.9} Unit vector in this direction \beq u^{\mu} = {{N^\mu}\over {(N^\nu N_\nu)^{1/2}}}. \ \ \ ({\sf Eckart}) \eeq{2.10} Decompose the four components of the flow velocity: $u^\mu = (\gamma, \gamma \vec{v})$:\\ See that the 3-vector of the flow velocity, $\vec{v}$, is \\ parallel to the particle current, $\vec{j}$, according to Eckart's definition. $N^\mu$ is always a time-like 4-vector.\\ Consequently, (for Eckart's definition) there is no particle flow in LR in the spatial directions \beq N^i_{(LR)}= 0; \ \ i=1,2,3. \ \ \ ({\sf Eckart}) \eeq{2.11} Or $\Delta_{\mu\nu} N^\mu = 0$.\\ Not OK for ultra-relativistic heavy ion reactions or early universe:\\ radiation energy density is high, and the baryon density is low or zero. \medskip (If $n \longrightarrow 0$ flow velocity, eq. (2.24), becomes ill defined.) (Coherent flow of particles and antiparticles \LT vanishing flow.) }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} LANDAU's DEFINITION According to this definition the LR is tied to energy flow:\\ in the LR frame the spatial component of the energy flow should vanish: \beq T^{0i}_{(LR)} = T^{i0}_{(LR)} = 0; \ \ i=1,2,3. \ \ \ ({\sf Landau}) . \eeq{2.12} Since the energy-flow 4-vector is $T^{\mu\nu} u_\nu$ and this should be parallel to $ u^\mu$ according to eq.(\ref{2.6}), consequently \beq \Delta_{\sigma\mu} T^{\mu\nu} u_\nu = 0 \ \ \ ({\sf Landau}) \eeq{2.13} This definition is hardly usable (because $ u^\mu$ is implicit), but it shows that $ u^\mu$ is the normalized eigenvector of $T^{\mu\nu} $, since $T^{\mu\nu} u_\nu$ is parallel to $ u^\mu$ , consequently: \beq u^\mu = Const. \times T^{\mu\nu} u_\nu \eeq{2.14} where from the normalization of $u^\mu$ \ \ the $ Const. = (u_\rho T^{\rho\nu} u_\nu)^{-1/2}$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\subsection{Other macroscopic quantities} \begin{itemize} \item[(i)] Invariant scalar density $n$: \beq n \equiv N^\mu u_\mu , \eeq{2.15} is the density in the local rest frame. \beq n = N^0_{(LR)} . \eeq{2.16} {\sf n} was not an invariant scalar, equals to $n$ only in LR. \item[(ii)] Invariant scalar energy density $e$: \beq e \equiv u_\mu T^{\mu\nu} u_\nu , \eeq{2.17} so that in the local rest frame \beq e = T^{00}_{(LR)}. \eeq{2.18} Notation $\epsilon$ for the energy density is frequently used ($\epsilon = e$).\\ Note that the specific energy is $\varepsilon = e/n = \epsilon / n$. \item[(iii)] Pressure tensor $P^{\mu\nu}$: \beq P^{\mu\nu} \equiv \Delta^\mu_\sigma T^{\sigma\tau} \Delta^\nu_\tau \eeq{2.19} so that $ P^{00}_{(LR)} = P^{i0}_{(LR)} = P^{0i}_{(LR)} = 0,$ and $ P^{ij}_{(LR)} = T^{ij}_{(LR)} $ for $ i,j=1,2,3 .$ We will see later that $ P^{\mu\nu}$ can be split up into two parts: \beq \begin{array}{lclcl} P^{\mu\nu} & = & -P \Delta^{\mu\nu} & + & \Pi^{\mu\nu} , \\ & & hydrostatic & & viscous \\ & & pressure & & stress \\ & & tensor & & tensor \end{array} \eeq{2.20} where $P$ is the hydrostatic pressure. \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace*{-1.0cm}\item[(iv)] Heat flow: \beq \begin{array}{rlclll} I^\mu_q \equiv & [ u_\nu T^{\nu\sigma}&-&({{e+P}\over{n}}) & N^\sigma] & \times \ \ \Delta^\mu_\sigma \\ & energy & & enthalpy & particle & \\ & 4-current & & per & current & \\ & vector & & particle & vector & \end{array} \eeq{2.21} The enthalpy per particle or specific enthalpy is denoted sometimes by $h/n$, sometimes by $w/n$ in the literature, $h=w=e+P$. In the local rest frame the heat flow has spatial components only \beq I^{0}_{q\ (LR)} = 0; \ \ \ I^{i}_{q\ (LR)} = T^{0i}_{(LR)} - \left( {w \over n} \right) N^i_{(LR)}, \eeq{2.22} the covariant expression is: \beq I^\mu_q u_\mu = 0 , \eeq{2.23} {\it i.e.}\/ the heat flow, $I^\mu_q$, and the four velocity, $u^\mu$, are orthogonal! \paragraph{SPECIAL CASES} \subparagraph{Eckart's definition} $(\Delta^\mu_\sigma N^\mu = 0)$: \beq I^{\mu}_q = u_\nu T^{\nu\sigma} \Delta^\mu_\sigma , \eeq{2.24} \subparagraph{Landau's definition} $(u_\mu T^{\mu\nu} \Delta_\nu^\sigma = 0)$: \beq I^{\mu}_q = - ({{e+P}\over{n}}) N^\sigma \Delta^\mu_\sigma , \eeq{2.25} \end{itemize} {\em Transport coefficients:} shear and bulk viscosity, $\eta$, and $\zeta$, and the heat conductivity, $\kappa$ \ \ \ \ \ \ \LT \ \ \ \ $\Pi^{\mu\nu}$, and $I_q^\mu$. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\subsection{Decomposition of the energy-momentum tensor} - Previously we defined $T^{\mu\nu}$ and $N^\mu$ based on microscopic quantities. - If $T^{\mu\nu}$ and $N^\mu$ are known the energy and baryon density can be obtained: Let us divide the energy momentum tensor into a reversible and an irreversible (dissipative) part. \beq \begin{array}{rlcl} T^{\mu\nu} = & T^{\mu\nu \ (0)} & + & T^{\mu\nu \ (1)} \\ & reversible, & & irreversible \\ & part & & (dissipative) \ part \end{array} \eeq{2.26} where \beq T^{\mu\nu \ (0)} = e u^\mu u^\nu - P \Delta^{\mu\nu} = (e+P) u^\mu u^\nu - P g^{\mu\nu} \eeq{2.27} In the Local Rest frame yields \beq T^{\mu\nu \ (0)}_{(LR)} = \left( \begin{array}{cccc} e & 0 & 0 & 0 \\ 0 & P & 0 & 0 \\ 0 & 0 & P & 0 \\ 0 & 0 & 0 & P \end{array} \right). \eeq{2.2x} The dissipative part is then \ \ \ \ \ \ \ \ \ \ $ T^{\mu\nu \ (1)} = $ \beq \left[ \underbrace{\left( I^\mu_q + ({{e+P}\over{n}}) N^\sigma \Delta^\mu_\sigma \right) }_{ \equiv Q^\mu} u^\nu + \underbrace{\left( I^\nu_q + ({{e+P}\over{n}}) N^\sigma \Delta^\nu_\sigma \right) }_{ \equiv Q^\nu} u^\mu \right] + \Pi^{\mu\nu} \eeq{2.28} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} Similarly the particle four flow can also be divided into two parts \beq N^\mu \ \ \ = \ \ \ N^{\mu\ (0)} + N^{\mu\ (1)} \ \ \ = \ \ \ n\ u^\mu - {n \over w} I_q^\mu \ . \eeq{2.28c} \B If $ T^{\mu\nu}$ and $u^\mu$ are given this decomposition can be constructed: \beqar e & \equiv & u_\mu T^{\mu\nu} u_\nu \\ -P \Delta^{\mu\nu} + \Pi^{\mu\nu} & \equiv & \Delta^\mu_\sigma T^{\sigma\tau} \Delta^\nu_\tau \\ Q^\mu & \equiv & u_\nu T^{\nu\sigma} \Delta^\mu_\sigma \eeqar{2.29} SPECIAL CASES The (1) component of the energy-momentum tensor: {\bf Landau}: \beq T^{\mu\nu \ (1)} = \Pi^{\mu\nu} \ \ \ \ N^{\mu\ (1)} = - {n \over w} I_q^\mu \ . \eeq{2.30} {\bf Eckart}: \beq T^{\mu\nu \ (1)} = I^\mu_q u^\nu + I^\nu_q u^\mu + \Pi^{\mu\nu} \ \ \ \ N^{\mu\ (1)} = 0\ . \eeq{2.31} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\section{J{\"u}ttner distribution} {\small\sf A question frequently asked is how can $f(x,p)$ be an invariant scalar if the volume element, $\Delta^3 x$, is Lorentz contracted? Yes, $\Delta^3 x$, is Lorentz contracted indeed, but the momentum space, $\Delta^3 p$, is Lorentz elongated! See the following:} Introduce the {\bf ``Relativistic - Boltzmann''}, or {\bf ``J\"uttner''} distribution function: \beq f^{Juttner}(p) = \juttner , \eeq{2.32} ($\mu$ is the chemical potential) Note: $ (\pmum )_{(LR)} = p^0 = E_{(LR)} $, and since $\pmum$ is an invariant scalar it is the same in all reference frames. \subsection{Normalization} If we know the invariant scalar density \beq n = N^\mu u_\mu = u_\mu \ipint p^\mu f(x,p) = \ipint \pmum \ \jut {{e^{\mu/T}}\over{(2 \pi \hbar)^3}} , \eeq{2.33} we may determine the chemical potential $\mu$. Since $ n $ is an invariant scalar it can be evaluated in any frame ($n$ = const.). In the (LR), $ u^\mu = (1,0,0,0)$, so \beq n = {{e^{\mu/T}}\over{(2 \pi \hbar)^3}} \ipint \ p^0 \ e^{-p^0/T} = {{e^{\mu/T}}\over{(2 \pi \hbar)^3}} 4 \pi \int^\infty_0 dp \ (\vec{p})^2 \ e^{-{{\sqrt{m^2+\vec{p}^2}}\over{T}} } . \eeq{2.33x} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.5cm} Let us introduce: \\ $\tau = {1\over T} \sqrt{m^2+\vec{p}^2}$,\\ $d\tau = {1\over T} {{|\vec{p}|}\over{\sqrt{m^2+\vec{p}^2}}} dp$ and \\ $p^2 = T^2 \tau^2 - m^2$,\\ then $$ n = {{4\pi \ e^{\mu/T}}\over{(2 \pi \hbar)^3}} T^3 \int^\infty_{m/T} d\tau \ \tau \sqrt{\tau^2 -\left({m\over T}\right)^2} e^{-\tau} = \hspace*{4cm} $$ \beq \hspace*{4cm} {{e^{\mu/T}}\over{(2 \pi \hbar)^3}} 4 \pi m^2 T \ K_2({m\over T}), \eeq{2.34} where $K_2$ is Modified Bessel function of the second kind. }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm}\subsection{Transformation properties of f(x,p).} IN THE CONFIGURATION SPACE Lorentz contraction: Disregard the momentum dependence of $f(x,p)$:\\ - Either assume that $f(x,p)$ is constant as a function of $p$, \\ - or integrate it over the momentum space ${\sf n}(x) = \pint \ f(x,p)$. \LT \\ Distribution function in the configuration space, ${\sf n}={\sf n}(x)$: %\begin{figure}[hbtp] Longitudinal direction at $x_\perp = 0$: \setlength{\unitlength}{0.3mm} \begin{picture}(400,100) \put(0,0){\begin{picture}(200,100) \thinlines \put( 20,20){\vector(1,0){160}} \put( 30,10){\vector(0,1){80}} \put( 35,95){\sf n} \put(185,25){$x_{\|}$} \put(100,77){2R} \put(100,75){\vector(1,0){30}} \put(100,75){\vector(-1,0){30}} \thicklines \put(100,35){\oval(60,30)[t]} \put( 55,35){\oval(30,30)[br]} \put(145,35){\oval(30,30)[bl]} \end{picture} } % Right hand side of the picture \put(350,80){Boosted} \put(200,0){\begin{picture}(200,100) \thinlines \put( 20,20){\vector(1,0){160}} \put( 30,10){\vector(0,1){80}} \put( 35,95){\sf n} \put(185,25){$x_{\|}$} \put(86,54){2R/$\gamma$} \put(100,50){\vector(1,0){15}} \put(100,50){\vector(-1,0){15}} \thicklines \put(100,50){\oval(30,60)[t]} \put( 70,50){\oval(30,60)[br]} \put(130,50){\oval(30,60)[bl]} \end{picture} } \end{picture} Transverse direction at $x_\| = < x_\| >$: \begin{picture}(400,100)(10,0) \put(0,0){\begin{picture}(200,100) \thinlines \put( 20,20){\vector(1,0){160}} \put( 30,10){\vector(0,1){80}} \put( 35,95){\sf n} \put(185,25){$x_{\perp}$} \put(100,77){2R} \put(100,75){\vector(1,0){30}} \put(100,75){\vector(-1,0){30}} \thicklines \put(100,35){\oval(60,30)[t]} \put( 55,35){\oval(30,30)[br]} \put(145,35){\oval(30,30)[bl]} \end{picture} } % Right hand side of the picture \put(350,80){Boosted} \put(200,0){\begin{picture}(200,100) \thinlines \put( 20,20){\vector(1,0){160}} \put( 30,10){\vector(0,1){80}} \put( 35,95){\sf n} \put(185,25){$x_{\perp}$} \put(100,77){2R} \put(100,75){\vector(1,0){30}} \put(100,75){\vector(-1,0){30}} \thicklines \put(100,35){\oval(60,30)[t]} \put( 55,35){\oval(30,30)[br]} \put(145,35){\oval(30,30)[bl]} \end{picture} } \thinlines \end{picture} %\caption[F:CS91] Figure 2.9 {\it Lorentz transformation in the configuration ($x$) space. From [2]} %\label{f1.9} %\end{figure} \setlength{\unitlength}{1mm} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace{-1cm} IN THE MOMENTUM SPACE: Assume that $f$ is uniform in $x$ or integrate the distribution over $x$.\\ For the latter case: $ f(x,p) \propto \jut $, where $u^\mu =(\gamma,\gamma v_\|,0,0)$ \LT \beq \pmum = p^0 \gamma - p_\| \gamma v_\| = (\sqrt{m^2 + (\vec{p}_\perp)^2 +(p_\|)^2} - p_\| v_\|) / \sqrt{1-(v_\|)^2} . \eeq{2.35} The momentum distribution, $F(p)= \int d^3 x f(x,p)$ $ \propto \jut$,\\ plotted for $m=1$ GeV, $T=0.1$ GeV at $p_\perp = 0$, for different velocities: %\begin{figure}[hbtp] \setlength{\unitlength}{0.3mm} \begin{picture}(400,100)(-50,0) \thinlines \put( 0,20){\vector(1,0){380}} \put( 50,10){\vector(0,1){80}} \put( 55,95){F} \put(385,25){$p_\|$ [GeV/c]} \put(320,60){$v_\| = 0.$} \put(125,20){\line(0,1){3}} \put(200,20){\line(0,1){6}} \put(275,20){\line(0,1){3}} \put(350,20){\line(0,1){6}} \put(125,5){1.0} \put(200,5){2.0} \put(275,5){3.0} \put(350,5){4.0} \put(47.17,20) {\begin{picture}(300,50) \put(-52.5, 5.5){$\bullet$} \put(-45.0, 9.5){$\bullet$} \put(-37.5,15.4){$\bullet$} \put(-30.0,23.1){$\bullet$} \put(-22.5,32.2){$\bullet$} \put(-15.0,41.0){$\bullet$} \put(-7.50,47.6){$\bullet$} \put( 0.0,50.0){$\bullet$} \put( 7.5,47.6){$\bullet$} \put( 15.0,41.0){$\bullet$} \put( 22.5,32.2){$\bullet$} \put( 30.0,23.1){$\bullet$} \put( 37.5,15.4){$\bullet$} \put( 45.0, 9.5){$\bullet$} \put( 52.5, 5.5){$\bullet$} \put( 60.0, 3.0){$\bullet$} \put( 67.5, 1.6){$\bullet$} \put( 75.0, 0.8){$\bullet$} \put( 82.5, 0.4){$\bullet$} \put( 90.0, 0.2){$\bullet$} \put( 97.5, 0.1){$\bullet$} \put(105.0, 0.0){$\bullet$} \end{picture} } \end{picture} \begin{picture}(400,100)(-50,0) \thinlines \put( 20,20){\vector(1,0){360}} \put( 50,10){\vector(0,1){80}} \put( 55,95){F} \put(385,25){$p_\|$ [GeV/c]} \put(320,60){$v_\| = 0.5$} \put(125,20){\line(0,1){3}} \put(200,20){\line(0,1){6}} \put(275,20){\line(0,1){3}} \put(350,20){\line(0,1){6}} \put(125,5){1.0} \put(200,5){2.0} \put(275,5){3.0} \put(350,5){4.0} \put(47.17,20) {\begin{picture}(300,50) \put(-30.0, 0.4){$\bullet$} \put(-22.5, 1.1){$\bullet$} \put(-15.0, 2.7){$\bullet$} \put(-7.50, 5.6){$\bullet$} \put( 0.0,10.6){$\bullet$} \put( 7.5,17.9){$\bullet$} \put( 15.0,26.9){$\bullet$} \put( 22.5,36.2){$\bullet$} \put( 30.0,44.0){$\bullet$} \put( 37.5,48.9){$\bullet$} \put( 45.0,49.9){$\bullet$} \put( 52.5,47.4){$\bullet$} \put( 60.0,42.2){$\bullet$} \put( 67.5,35.6){$\bullet$} \put( 75.0,28.7){$\bullet$} \put( 82.5,22.1){$\bullet$} \put( 90.0,16.5){$\bullet$} \put( 97.5,11.9){$\bullet$} \put(105.0, 8.4){$\bullet$} \put(112.5, 5.8){$\bullet$} \put(120.0, 3.9){$\bullet$} \put(127.5, 2.6){$\bullet$} \put(135.0, 1.7){$\bullet$} \put(142.5, 1.1){$\bullet$} \put(150.0, 0.7){$\bullet$} \put(157.5, 0.4){$\bullet$} \put(165.0, 0.3){$\bullet$} \put(172.5, 0.2){$\bullet$} \put(180.0, 0.1){$\bullet$} \put(187.5, 0.1){$\bullet$} \put(195.0, 0.0){$\bullet$} \end{picture} } \end{picture} \begin{picture}(400,100)(-50,0) \thinlines \put( 20,20){\vector(1,0){360}} \put( 50,10){\vector(0,1){80}} \put( 55,95){F} \put(385,25){$p_\|$ [GeV/c]} \put(320,60){$v_\| = 0.9$} \put(125,20){\line(0,1){3}} \put(200,20){\line(0,1){6}} \put(275,20){\line(0,1){3}} \put(350,20){\line(0,1){6}} \put(125,5){1.0} \put(200,5){2.0} \put(275,5){3.0} \put(350,5){4.0} \put(47.17,20) {\begin{picture}(300,50) \put( 22.5, 0.0){$\bullet$} \put( 30.0, 0.1){$\bullet$} \put( 37.5, 0.2){$\bullet$} \put( 45.0, 0.6){$\bullet$} \put( 52.5, 1.4){$\bullet$} \put( 60.0, 2.9){$\bullet$} \put( 67.5, 5.1){$\bullet$} \put( 75.0, 8.3){$\bullet$} \put( 82.5,12.4){$\bullet$} \put( 90.0,17.3){$\bullet$} \put( 97.5,22.8){$\bullet$} \put(105.0,28.5){$\bullet$} \put(112.5,33.9){$\bullet$} \put(120.0,38.9){$\bullet$} \put(127.5,43.2){$\bullet$} \put(135.0,46.4){$\bullet$} \put(142.5,48.6){$\bullet$} \put(150.0,49.8){$\bullet$} \put(157.5,49.9){$\bullet$} \put(165.0,49.2){$\bullet$} \put(172.5,47.7){$\bullet$} \put(180.0,45.5){$\bullet$} \put(187.5,42.9){$\bullet$} \put(195.0,39.9){$\bullet$} \put(202.5,36.8){$\bullet$} \put(210.0,33.6){$\bullet$} \put(217.5,30.4){$\bullet$} \put(225.0,27.3){$\bullet$} \put(232.5,24.3){$\bullet$} \put(240.0,21.5){$\bullet$} \put(247.5,18.9){$\bullet$} \put(255.0,16.6){$\bullet$} \put(262.5,14.4){$\bullet$} \put(270.0,12.5){$\bullet$} \put(277.5,10.8){$\bullet$} \put(285.0, 9.3){$\bullet$} \put(292.5, 7.9){$\bullet$} \put(300.0, 6.8){$\bullet$} \put(307.5, 5.7){$\bullet$} \put(315.0, 4.9){$\bullet$} \put(322.5, 4.1){$\bullet$} \end{picture} } \end{picture} %\caption[F:CS91] Figure 2.10 {\it Longitudinal momentum distributions Lorentz-boosted with velocity, $v_\|$. The boosted distribution is not symmetric! This is due to the fact that the width of the original distribution is not negligible compared to the boost.} %\label{f1.10} %\end{figure} \setlength{\unitlength}{1mm} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace{-1.5cm}\section{Mixtures} In non-equilibrium systems flow velocity may be problematic: it does not reflect the essential physical features of our system.E.g.: \begin{itemize} \item[(i)] HI reactions: {\bf target} and {\bf projectile} may penetrate before the collisions slow down the matter. If the random velocities are smaller than the relative velocity of the projectile \& target: \setlength{\unitlength}{0.3mm} \begin{picture}(400,100)(-20,0) \thinlines \put( 20,20){\vector(1,0){360}} \put( 50,10){\vector(0,1){80}} \put( 55,90){F} \put(385,25){$p_\|$ } \put(125,20){\line(0,1){3}} \put(200,20){\line(0,1){6}} \put(275,20){\line(0,1){3}} \put(350,20){\line(0,1){6}} \put(130,20) {\unitlength 0.9pt \begin{picture}(300,50) \put(-105.0, 0.0){$\bullet$} \put(-97.5, 0.1){$\bullet$} \put(-90.0, 0.2){$\bullet$} \put(-82.5, 0.4){$\bullet$} \put(-75.0, 0.8){$\bullet$} \put(-67.5, 1.6){$\bullet$} \put(-60.0, 3.0){$\bullet$} \put(-52.5, 5.5){$\bullet$} \put(-45.0, 9.5){$\bullet$} \put(-37.5,15.4){$\bullet$} \put(-30.0,23.1){$\bullet$} \put(-22.5,32.2){$\bullet$} \put(-15.0,41.0){$\bullet$} \put(-7.50,47.6){$\bullet$} \put( 0.0,50.0){$\bullet$} \put( 7.5,47.6){$\bullet$} \put( 15.0,41.0){$\bullet$} \put( 22.5,32.2){$\bullet$} \put( 30.0,23.1){$\bullet$} \put( 37.5,15.4){$\bullet$} \put( 45.0, 9.5){$\bullet$} \put( 52.5, 5.5){$\bullet$} \put( 60.0, 3.0){$\bullet$} \put( 67.5, 1.6){$\bullet$} \put( 75.0, 0.8){$\bullet$} \put( 82.5, 0.4){$\bullet$} \put( 90.0, 0.2){$\bullet$} \put( 97.5, 0.1){$\bullet$} \put(105.0, 0.0){$\bullet$} \end{picture} } \put(190,20) {\unitlength 0.9pt \begin{picture}(300,50) \put(-105.0, 0.0){$\bullet$} \put(-97.5, 0.1){$\bullet$} \put(-90.0, 0.2){$\bullet$} \put(-82.5, 0.4){$\bullet$} \put(-75.0, 0.8){$\bullet$} \put(-67.5, 1.6){$\bullet$} \put(-60.0, 3.0){$\bullet$} \put(-52.5, 5.5){$\bullet$} \put(-45.0, 9.5){$\bullet$} \put(-37.5,15.4){$\bullet$} \put(-30.0,23.1){$\bullet$} \put(-22.5,32.2){$\bullet$} \put(-15.0,41.0){$\bullet$} \put(-7.50,47.6){$\bullet$} \put( 0.0,50.0){$\bullet$} \put( 7.5,47.6){$\bullet$} \put( 15.0,41.0){$\bullet$} \put( 22.5,32.2){$\bullet$} \put( 30.0,23.1){$\bullet$} \put( 37.5,15.4){$\bullet$} \put( 45.0, 9.5){$\bullet$} \put( 52.5, 5.5){$\bullet$} \put( 60.0, 3.0){$\bullet$} \put( 67.5, 1.6){$\bullet$} \put( 75.0, 0.8){$\bullet$} \put( 82.5, 0.4){$\bullet$} \put( 90.0, 0.2){$\bullet$} \put( 97.5, 0.1){$\bullet$} \put(105.0, 0.0){$\bullet$} \end{picture} } \put(160,20) {\unitlength 0.6pt \begin{picture}(300,50) \put(-105.0, 0.0){$\circ$} \put(-97.5, 0.1){$\circ$} \put(-90.0, 0.2){$\circ$} \put(-82.5, 0.4){$\circ$} \put(-75.0, 0.8){$\circ$} \put(-67.5, 1.6){$\circ$} \put(-60.0, 3.0){$\circ$} \put(-52.5, 5.5){$\circ$} \put(-45.0, 9.5){$\circ$} \put(-37.5,15.4){$\circ$} \put(-30.0,23.1){$\circ$} \put(-22.5,32.2){$\circ$} \put(-15.0,41.0){$\circ$} \put(-7.50,47.6){$\circ$} \put( 0.0,50.0){$\circ$} \put( 7.5,47.6){$\circ$} \put( 15.0,41.0){$\circ$} \put( 22.5,32.2){$\circ$} \put( 30.0,23.1){$\circ$} \put( 37.5,15.4){$\circ$} \put( 45.0, 9.5){$\circ$} \put( 52.5, 5.5){$\circ$} \put( 60.0, 3.0){$\circ$} \put( 67.5, 1.6){$\circ$} \put( 75.0, 0.8){$\circ$} \put( 82.5, 0.4){$\circ$} \put( 90.0, 0.2){$\circ$} \put( 97.5, 0.1){$\circ$} \put(105.0, 0.0){$\circ$} \end{picture} } \put(290,65){\vector(1,0){10}} \put(310,65){\circle{40}} \put(355,50){\vector(-1,0){10}} \put(335,50){\circle{40}} \end{picture} \setlength{\unitlength}{1mm} \vspace*{-0.5cm} If \ $x$\ is in the overlap region\ the distribution is \\ $ f(x,p)$\ $=$\ $f_{Projectile}(x,p)$\ $+$\ $f_{Target}(x,p)$ $+$ $f_{Therm.}(x,p)$.\\ \LT two peaks in momentum space. One single flow velocity is possible, but (!???) Later thermalized momentum distribution develops. Solution: introducing more fluid components \item[(ii)] In {\bf Electron-Ion plasmas} in TOKAMAK's: Similar situation.\\ - The two plasmas show different temperatures and move with different velocities due to external electric and magnetic fields.\\ - Here the two components never get thermalized with each other. \\ - The two fluid dynamics (or magneto hydrodynamics) is a well known theoretical method in this field. \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \begin{itemize} \vspace{-1.5cm} \item[(iii)] The introduction of more fluid components is not new in nuclear physics either. Classical collective excitation of nuclei: protons and neutrons oscillate against each other. \\ This is the well known {\bf Giant Dipole Resonance}. In the {\em Jensen-Steinwedel} model this resonance is described by a two component distribution function, i.e. one for the protons $ f_p (x,p)$, and one for the neutrons $f_n (x,p)$. \setlength{\unitlength}{0.5mm} \begin{picture}(400,80)(30,0) \thicklines \put(160,40){\vector(1,0){10}} \put(160,40){\vector(-1,0){10}} \put(220,40){\vector(1,0){10}} \put(220,40){\vector(-1,0){10}} \put(185,40){\circle{40}} \put(195,40){\circle{40}} \put(100,40){protons} \put(250,40){neutrons} \end{picture} \setlength{\unitlength}{1mm} \vspace*{-1.0cm} \item[(iv)] {\bf Nucleons and light fragments} at the final stage of a Relativistic Heavy Ion Collision (RHIC) may be two components.\\ - Assumed: they are all in thermal and chemical equilibrium. Temperature measurements rely on this assumption! \\ - This assumption is, however, not true, it should only be considered as an approximation. Different fragments, and the single nucleons may have very {\em different temperatures}. \item[(v)] {\bf Coexisting phases in a gas or a fluid}. \\ The kinetic motion is thermalized due to the enhanced cross sections at the phase transition.\\ - Thus, common flow velocity and temperature. - The chemical potential may, however, be different. \end{itemize} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} In kinetic theory a mixture of several components may be described also. The distribution function of more (N) components is denoted as: \beq f_k (x,p_k) \ \ \ \ \ \ \ \ \ \ \ k = 1,2,3,...,N \eeq{2.w3} The rest masses of the components may be different: $p^\mu_k p_{\mu k} = m_k^2 $. The four flow of component $k$ is \beq N^\mu_k = \int {{d^3 p_k}\over{p^0_k}} p^\mu_k f_k (x,p_k ), \eeq{2.36} and similarly to the one component case the energy momentum tensor can be evaluated for the components separately \beq T^{\mu\nu}_k = \int {{d^3 p_k}\over{p^0_k}} p^\mu_k p^\nu_k \ f_k (x,p_k ). \eeq{2.37} \B Important: there is always one common flow velocity $u^\mu$, (although the component distributions may be centered around different velocities $u^\mu_k$.) \\ - This common flow velocity can be connected to the {\em total particle current} or to the {\em total energy momentum tensor}: \beqar N^\mu & = & \sum^N_{k=1} N^\mu_k, \\ T^{\mu\nu} & = & \sum^N_{k=1} T^{\mu\nu}_k . \eeqar{2.38} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} If particles of a component are not conserved,\\ but there is one (or more) conserved charge, $q_k$, (like baryon charge)\\ we can introduce a total current for this charge: \beqar Q^\mu_k = \int {{d^3 p_k}\over{p^0_k}} q_k \ p^\mu_k \ f_k(x,p_k), \ \ \ \ \ \ \ Q^\mu & = & \sum^N_{k=1} Q^\mu_k . \eeqar{2.39} In mixtures one can define the scalar density as $n = \sum n_k$, and the concentrations as $x_k = n_k /n$, which satisfy $\sum x_k = 1$.\\ Note: these are valid only if the flow velocities of the components are the same or if $n_k$ is defined as $n_k \equiv N^\mu_k u_\mu$ and not via the component velocities $u^\mu_k$. If the components have different flow velocities we can introduce the diffusion current: \beq D^\mu_k \equiv N^\mu_k - x_k N^\mu \eeq{2.40} so that \beq \sum_{k=1}^N D^\mu_k = 0. \eeq{2.41} }%end tr-page \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.0cm} \section{Assignment 2} \subsection*{Properties of the Rapidity} \begin{description} \item[2.a] Prove that the rapidity is additive under Lorentz transformation \item[2.b] Show that the energy of a particle of rapidity, $ y$, and transverse momentum, $ p_\perp$, is $E (= p^0 ) = m_\perp \ \cosh(y) $, where $ m_\perp $ is the transverse mass : $m_\perp = \sqrt{m^2 + p_\perp^2}$. \item[2.c] Show that if $ E \rightarrow \infty $ then $ y \rightarrow \eta \ = \ \ln ( \cot \frac{\Theta}{2})$ (pseudorapidity), where $\Theta $ is the polar angle of the emitted particle. \item[2.d] Plot the contour lines belonging to $ E $ $=$ $ 2, 4, 8$\ GeV/nucleon energy in the $[y,p_\perp/m]$ plane. ($m = 1\ GeV$). Plot also the lines in the same plane, corresponding to constant polar angles $ \Theta = 0^o , 30^o , 60^o , 90 ^o$. \end{description} }%end tr-page % transparency ================================================================ \end{document}