% % 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 8 / Chapter 8 % % 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}{{\sf 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 ====================================== p. 1 \transparencyframe{ \setcounter{chapter}{7} \vspace*{-1.8cm} \chapter{Scaling of the hydrodynamical model} Scaling is understood in two ways: \B Ultra-relativistic HI \\ - - 1) scaling solution / scaling fluid dynamics\\ - - - basic assumption of the Bjorken model, i.e. (Ch. 6)\\ - - - invariance against a beam directed Lorentz boost\\ - - 2) Related scaling: Feynman scaling,\\ - - - universality of parton momentum distributions in Feynman $x$\\ - - - (Chapter 10. QGP). \B Scaling of measurables at diff. masses and beam energies\\ - - - Comparisions \LT definition of secondary variables\\ - - - which are invariant under the change of beam energy or mass.\\ - - - Invariances are model dependent and not necessarily exact. - - - If scaling confirmed by experiments \LT\\ - - - the approach is basically a valid approach. We will discuss this latter version of scaling. }%end tr-page \newpage % transparency ====================================== p. 2 \transparencyframe{ \vspace*{-0.8cm} \section{Similarity in classical fluid dynamics} Equations of perfect non-relativistic fluid-dynamics:\\ a) the continuity equation: \beq \frac{\partial \rho}{\partial t} + \nabla (\rho {\bf u} ) = 0 , \eeq{sc.6} which connects the mass distr., $\rho ({\bf r}, t)$, \& velocity distr. ${\bf u}({\bf r}, t)$;\\ b) the Euler equation: \beq \frac{\partial {\bf u} }{\partial t} + ({\bf u} \nabla ) {\bf u} = - \frac{1}{\rho} P , \eeq{sc.7} c) and the EOS,\\ relates, the pressure $P$ to the density, $\rho$, \& entropy density, $s=S/V$: \beq P = P(\rho ,s) \ . \eeq{sc.8} For non-viscous fluid the entropy is constant during expansion: \beq \nabla P \approx \left( \frac{\partial P}{\partial \rho} \right)_s \nabla \rho = c_s^2 \nabla \rho , \eeq{sc.9} where $c_s $ is the adiabatic sound velocity.\\ These eqs. with the initial conditions, ${\bf u}$, $\rho$, and $s$,\\ \LT the hydrodynamical evolution of the system. \B Dimensionless, scale-invariant quantities can be derived [1]\\ - - to describe the general properties of a system, and also\\ - - to compare the hydrodynamical behavior at diff. masses and energies. }%end tr-page \newpage % transparency ====================================== p. 3 \transparencyframe{ \vspace*{-1.05cm} \B Characteristic mass, $m_1 $, temperature $T_1 $, length $l_1 $, \& velocity $u_1$\\ can be introduced in a heavy ion collision as: \beq m_1 = m A , \eeq{sc.10} where $m$ is the nucleon mass and $A$ is the number of nucleons: \beq u_1 = |{\bf u}_0 | = \left( \frac{2E_0}{m} \right)^{\frac{1}{2}} , \eeq{sc.11} where $E_0 $ is the initial c.m. energy per nucleon of the projectile, and \beq l_1^3 =\frac{4}{3} \pi r_0^3 \ A , \eeq{sc.12} which represents the volume of the system.\\ \B Definitions: ${\bf r}=l_1 \tilde{r}$,\ \ $t=t_1 \tilde{t}$,\ \ $T=T_1 \tilde{T}$ (where $T_1 =2/3\ E_0$),\\ - Introduce dimensionless variables using the characteristic quantities: \beq \rho ({\bf r},t) = \frac{m_1}{l_1^3} \tilde{\rho}(\tilde{r},\tilde{t}) , \eeq{sc.13} \beq {\bf u}({\bf r},t) = u_1 \tilde{u}(\tilde{r},\tilde{t}) . \eeq{sc.14} \B These characteristic dimensionless hydrodynamical functions are\\ independent of the total mass, $A$, and energy, $E_0$. The sound velocity is of the order of the thermal nucleon velocities:\\ - - $c= u_1 \tilde{c}$, with $\tilde{c} \approx 1$, \LT\\ \B The Continuity- and Euler- equation can be cast in a dimensionless form: \beq m_1 \left[ \frac{\partial \tilde{\rho}}{\partial \tilde{t}} + S \tilde{\nabla} (\tilde{\rho} \tilde{u} ) \right] = 0 , \eeq{sc.15} \beq \frac{\partial \tilde{u} }{\partial \tilde{t}} + S ( \tilde{u} \tilde{\nabla }) \tilde{u} = - S \tilde{c}^2 \frac{\tilde{\nabla}\tilde{\rho}}{\tilde{\rho}} , \eeq{sc.16} where $S=u_1 t_1 /l_1 $ is the Strouhal number. }%end tr-page \newpage % transparency ====================================== p. 4 \transparencyframe{ \vspace*{-0.8cm} \B Small systems, e.g. nuclear collision, \LT\\ - - role of viscosity should not be neglected.\\ \B Assuming that \\ - a) bulk viscosity, $\xi$, is $\propto$ the shear viscosity, $\eta$, \\ - - - - ($\xi=q\eta$, where $q$ is a dimensionless constant) and that\\ - b) the kinematic viscosity, $\nu=\eta/\rho= $ Const. during expansion\\ \B \LT the Navier-Stokes eq. can be written in dimensionless form: \beq \frac{\partial \tilde{u} }{\partial \tilde{t}} + S ( \tilde{u} \tilde{\nabla }) \tilde{u} = - S \tilde{c}^2 \frac{\tilde{\nabla}\tilde{\rho}}{\tilde{\rho}} - \frac{S}{Re} \left[ \tilde{\nabla} \tilde{u} + (q+1/3) \tilde{\nabla} (\tilde{\nabla} \tilde{u}) \right] , \eeq{sc.17} where $Re$ is the Reynolds number: $Re=l_1 u_1 /\nu$. \\ \B With a proper choice of the time scale, $t_1 =l_1 /u_1 $:\\ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $S$ can be set equal to 1. \B \LT The solutions of FD eqs. depend on:\\ -\ \ \ \ \ \ \ \ $\tilde{r}$, $\tilde{t}$, \& Reynolds number, $Re$, only. \B Flow patterns at diff. energies and masses are similar if:\\ -\ \ \ \ \ \ the Reynolds number, $\tilde{r}$ and $\tilde{t}$ are the same. \B \LT Scale-invariant quantities can be defined!!!,\\ \B Deviation from the scale invariance indicates the onset of processes:\\ - - - which lead to a non-scale-invariant flow in FD,\\ - - - such as viscosity, a change in EOS or in reaction mechanism. }%end tr-page \newpage % transparency ====================================== p. 5 \transparencyframe{ \vspace*{-1.7cm} \subsection{Application to heavy ion collisions} \B Assume non-relativistic perfect flow, \& study the expansion stage.\\ - Initial condition: The intermediate most compressed state. Let $\rho(\vec{r},t)$ be the mass density - the normalization if $A_P + A_T =A$ is \beq \int \rho(\vec{r},t) dr = m A . \eeq{e10.1} In symmetric collisions, in the C.M. system the incoming energy is \beq E_0 = \frac{1}{2} m u_0^2 . \eeq{e10.2} The radii of the nuclei can be characterized by \beq R = R_P = R_T = r_0 A_T^{1/3} . \eeq{e10.3} Repeat the characteristic quantities, $q_1$, \& dimensionless flow variables \beqar \vec{r} & \equiv & l_1 \ \tilde{r} , \nonumber \\ t & \equiv & t_1 \ \tilde{t} , \nonumber \\ \rho(\vec{r},t) & \equiv & \frac{m_1}{l_1^3} \ \tilde{\rho}(\tilde{r},\tilde{t}) , \nonumber \\ u(\vec{r},t) & \equiv & u_1 \ \tilde{u} (\tilde{r},\tilde{t}) , \nonumber \\ T(\vec{r},t) & \equiv & T_1 \ \tilde{T} (\tilde{r},\tilde{t}) . \eeqar{e10.4} Kin. viscosity: $\nu \approx 10^{-1}$fm, \& \\ Char. time and length: $t_1 \approx 10$fm/c \& $l_1 \approx 10$fm. \LT\\ \B - - Reynolds number: $Re \approx 10 - 100$. \\ \B ! Since the dim.-less break-up time: $\tilde{t}_{BU} \approx 1$,\\ - \LT viscous effects have no time to develop, $\tilde{t}_{BU} \ll Re$.\\ - \LT viscosity is not dominating the expansion in rel. HI collisions.\\ (It may, however, cause corrections, like entropy increase. 10 \% [2].) }%end tr-page \newpage % transparency ====================================== p. 6 \transparencyframe{ \vspace*{-0.8cm} Fluid dynamical scaling is only approximate in heavy ion collisions because: \begin{itemize} \item The initial compression stage is not scale invariant\\ since the viscosity is essential in forming the shock profiles. \item Scaling works well if our EOS is similar to an IDEAL GAS.\\ Binding and compressional energy: \LT new dimensional constants\\ - which violate scaling, espec. at low energies ($<$ 100 MeV/nucl.). \item The break-up may and may not be formulated in a dimensionless way.\\ If it includes dimensional constants (e.g. $\sigma$) \\ \LT the dim.-less break-up time may not be constant: violates scaling.\\ - - However, many break-up descriptions lead to $\tilde{t}_{BU} = const.$ \item Physical processes after the break-up may violate scaling,\\ \LT preventing the FD scaling laws to be seen in the measurables. \end{itemize} \B \LT It should be determined experimentally whether scaling is present in the observables or not. }%end tr-page \newpage % transparency ====================================== p. 7 \transparencyframe{ \vspace*{-1.2cm} \section{Scaling properties of cross sections} \B How to study FD scaling laws in measured cross sections? (c.s.-s) \\ - We have to follow the way we evaluate these c.s.-s in the FD model.\\ - We stop the FD calculation at some time $t_{BU}(\vec{r})$, then:\\ - We assume that the local momentum distribution is a MB distribution\\ - \LT mom.-distribution in a collision with impact parameter $b$: \beq F_b(\vec{p},\vec{r},t) = n(\vec{r},t_{BU},b) f^{MB}[\vec{p}-\vec{p}_{flow} (\vec{r},t_{BU},b)], \eeq{e10.a1} - $n$: the baryon density,\ $ n(\vec{r},t_{BU},b)$ $ =$ $ \rho(\vec{r},t_{BU},b) / m$, and \\ - $ \vec{p}_{flow}(\vec{r},t_{BU},b)$ $ =$ $ m\ \vec{u}(\vec{r},t_{BU},b)$.\\ \B The double differential nucleon inclusive cross section is then \beq \sigma_2 \equiv \frac{d^2 \sigma}{dE d\Omega} = \frac{d\sigma}{d^3 p} \cdot \frac{d^3 p}{dE d\Omega} = \frac{d\sigma}{d^3 p} \cdot \frac{p^2 dp d\Omega}{dE d\Omega} = m \sqrt{2mE} \frac{d\sigma}{d^3 p} . \eeq{e10.3b} Thus inserting the phase space distribution yielded by the fluid dynamics \beq \sigma_2 = m \sqrt{2mE} \int_0^{b_{max}} d^2b \int d^3r\ F_b(\vec{p},\vec{r},t) , \eeq{e10.4a} where the total local distribution, $F_b$, can be factorized to\\ - the density, $n$, and a MB mom.-distribution, $f^{MB}$, (normalized to 1)\\ - - depends on the local T, and is shifted by the local flow mom., $p_{flow}$: $$ \sigma_2 = m \sqrt{2mE} \times \hspace*{12.5cm} $$ \beq \hspace*{1.3cm} \times \int_0^{b_{max}} d^2b \int d^3r \ n(\vec{r}, t_{BU}, b) \ f^{MB}(\vec{p}-\vec{p}_{flow}(\vec{r},t_{BU},b)) . \eeq{e10.4b} }%end tr-page \newpage % transparency ====================================== p. 8 \transparencyframe{ \vspace*{-0.8cm} \B Introduce the dimensionless variables \beqar b_{max} & = & l_1 \tilde{b}_{max} , \nonumber \\ \vec{r} & = & l_1 \tilde{r} , \nonumber \\ n & = & \frac{A}{l_1^3} \tilde{\rho}(\tilde{r},\tilde{t};\tilde{b}, Re) . \eeqar{e10.5} So far we did not introduce the scaling decomposition of the momentum $p$. We can use the usual procedure \beqar \vec{p} & = & p_1 \tilde{p} , \nonumber \\ f & = & \frac{1}{p_1^3} \ \tilde{f}(\tilde{p}-\tilde{p}_{flow}) . \eeqar{e10.6} The parameter $p_1$ should be fixed by the typical momentum: $$ p_1 = \sqrt{2 m E_0} . $$ \B Thus the double differential cross section can be written as: \beq \sigma_2 = m \sqrt{2mE}\ l_1^2 A \frac{1}{p_1^3} \int_0^{\tilde{b}_{max}} d^2\tilde{b} \int d^3\tilde{r} \ \tilde{\rho} (\tilde{r},\tilde{b}) \ \tilde{f}(\tilde{p}-\tilde{p}_{flow}) , \eeq{e10.7} where we have separated the dimensional and dimensionless factors.\\ Assume that $b_{max} = \alpha 2 r_0 \left({A \over 2} \right)^{1/3}$ $=$ $ l_1 \tilde{b}_{max}$,\\ - where $\alpha$ is a const. defining the cut in the impact parameters.\\ Using the definition of $l_1$ we obtain $$ \tilde{b}_{max} = \alpha (3/\pi)^{1/3} . $$ \B We can now define a dimensionless cross section as \beq \tilde{g} (\tilde{p}, Re) \equiv \frac{^3\sqrt{2}}{\tilde{b}^2_{max}} \ \int^{\tilde{b}_{max}}_0 d^2 \tilde{b} \int d^3\tilde{r} \tilde{\rho}(\tilde{r},\tilde{b}) f(|\tilde{p}-\tilde{p}_f) . \eeq{10.7a} }%end tr-page \newpage % transparency ====================================== p. 9 \transparencyframe{ \vspace*{-0.8cm} \B This expression contains only dimensionless quantities:\\ - invariant under the change of energy or mass of the system.\\ \LT The cross section can be expressed in terms of this quantity as \beq \sigma_2 = \frac{ A^{5/3} }{ E_0 } r_0^2 \alpha^2 \sqrt{X} \cdot \tilde{g}(\tilde{p},Re) \ , \eeq{10.8} where $X\equiv E/E_0$. \\ \B The double differential cross section, $\sigma_2$,\\ - - is usually measured in terms of $E$ and $\Theta$.\\ * - $\Theta$ is dim.-less and scale invariant,\\ * - $E$ can be replaced by $X$ to get a dimensionless variable.\\ \LT We can introduce another dimensionless cross section: \beq \tilde{G}(X, \Theta, Re) \equiv \alpha^2 \sqrt{X} \ \tilde{g}(\tilde{p},Re) . \eeq{10.9} - $\alpha$ characterizes the multiplicity selection.\\ * If $\alpha= 1$ this means $b_{max} = 2R$, \\ - - i.e. all collisions are taken into account, the c.s. is inclusive.\\ * If $\alpha < 1$, this choice corresponds to a selection of central events \\ - - (e.g. $\alpha= {1 \over 2} $ corresponds to $\sim$25\% of all events with the highest $M_c$).\\ \B The double differential cross section in terms of $\tilde{G}$ is \beq \sigma_2 = \frac{ A^{5/3} }{ E_0 } r_0^2 \cdot \tilde{G}(X,\Theta,Re) \ . \eeq{10.10} \B \LT From a measured c.s. at given $E$ and $A$, \\ - - - with well defined impact parameter selection, $\alpha$,\\ * we can separate the dimensional factors in the c.s. to obtain $\tilde{G}$.\\ \B This dimensionless part, $\tilde{G}$, should then be the same\\ - - for collisions at other energies and masses\\ - - if the impact parameter selection is the same, and\\ - - if the hydrodynamical scaling is not violated. }%end tr-page \newpage % transparency ====================================== p. 10 \transparencyframe{ \vspace*{-0.2cm} \B Results of numerical FD calculations show this property clearly.\\ Calculations for $250\ A \cdot MeV$ and $2.1 \ A \cdot GeV$ Ne + U [3]\\ - gave similar results in terms of $\tilde{G}$, Fig. 8.1: \vspace*{14cm} Figure 8.1 {\it Scale invariant double differential cross sections, $\tilde{G}$, taken from 3-dimensional, numerical, perfect, relativistic, fluid dynamical calculations. The break-up was assumed at one third nuclear density. From [1] } }%end tr-page \newpage % transparency ====================================== p. 11 \transparencyframe{ \vspace*{-0.8cm} \vspace*{13cm} Figure 8.2 {\it Scale invariant double differential cross sections, $\tilde{G}$, taken from two experiments at 250 and 400 A$\cdot$MeV beam energy. From [1] } \B Experimental data: similarily strong scaling behaviour, Fig. 8.2.\\ * The agreement in the transverse directions indicates that\\ - - scaling violating factors, like viscosity, do not play dominating role.\\ * The scaling violation at forward angles indicates that\\ - - pre-equilibrium, quasi-elastic scatterings,\\ - - which are important at this angle, do not follow FD scaling laws. }%end tr-page \newpage % transparency ====================================== p. 12 \transparencyframe{ \vspace*{-0.8cm} \vspace*{12cm} Figure 8.3 {\it Structure functions, $\tilde{G}$, in the C.M. system extracted from experimental proton inclusive data. Circles: 400 A MeV Ne+Ne, triangles: 800 A MeV Ne+Ne, squares: 800 A MeV Ar+Ar. Quasi elastic scattering is removed (top) and maintained (bottom) theoretically. The error bars indicate the uncertainties in the subtraction-procedure of the quasi-free component. From [1] } \B 3 sets of experimental data were compared (Fig. 8.3) also \\ - - by removing the quasi-elastic component from the c.s. [4]:\\ \LT 800 MeV c.s.-s are not affected by the quasi-free scattering, because\\ - - - at this energy the direct scattering is forward-backward peaked,\\ - - - and so it does not influence the 90$^0$ cross section. }%end tr-page \newpage % transparency ====================================== p. 13 \transparencyframe{ \vspace*{-0.8cm} \section{Scaling properties of the transverse flow} \B Collective FD effect in HI collisions.\\ - Occurs in the whole energy range dicscussed. {*} 1st: Theoretical predictions of scaling properties of measurables\\ {*} 2nd: Validity of these predictions in experimental data, \&\\ - - deviations from the scaling behavior and their causes. \subsection {Global flow tensor} Let us repeat the definition of the global flow tensor: \beq M_{\alpha\beta} = \sum_\nu \frac{1}{2m_\nu} p_\alpha(\nu)p_\beta(\nu), \eeq{11e.01} The fluid dynamical expectation value of this quantity is given by \beq \overline{M}_{\alpha\beta} = \int d^3r \ d^3p\ p_\alpha p_\beta \ n(\vec{r}, t_{BU}) f^{MB} [\vec{r},\vec{p}-\vec{p}_{flow}, T (\vec{r})] \frac{1}{2m} . \eeq{11e.02} Let us introduce the scaling parameters: $$ p_1 = \sqrt{2mE_0},\ \ \ \ \ \ T_1 = \frac{2E_0}{3},\ \ \ \ \ \ u_1 = \sqrt{\frac{2E_0}{m}}, $$ \beq l_{1}^{3} = \frac{4\pi }{3} r_{0}^{3} A,\ \ \ \ \ \ t_1 = \frac{l_1}{u_1}\ \ \ \ \ \ {\rm and}\ \ \ \ \ \ f = \frac{1}{p_{1}^{3}} \tilde{f} . \eeq{11e.03} }%end tr-page \newpage % transparency ====================================== p. 14 \transparencyframe{ \vspace*{-0.1cm} \B FD expectation value of the global flow tensor reduces to: \beq \overline{M}_{\alpha \beta } = \underbrace{ \frac{1}{2m}l_{1}^{3} p_{1}^{3} p_{1}^{2} \frac{A}{l_1^3} \frac{1}{p_1^3} }_{AE_0} \ \underbrace{ \int d^3 \tilde{r} \ \tilde{p}_\alpha \tilde{p}_\beta \ \tilde{n}(\tilde{r}) \tilde{f} (\tilde{p} - \tilde{p}_f)}_{\tilde{M}_{\alpha\beta}} . \eeq{11e.04} If:\\ - we have several fluid cells, and\\ - the thermal momentum distribution is a MB distribution\\ then: \beq M_{\alpha \beta} = \frac{1}{2} \sum _{(cell)s=1}^N A_s [m\ u_\alpha^{flow}(s)\ u_\beta^{flow}(s) + T(s)\delta_{\alpha\beta}] \ . \eeq{11e.05} }%end tr-page \newpage % transparency ====================================== p. 15 \transparencyframe{ \vspace*{-0.8cm} On the other hand the energy conservation is: \beq E_0 =\sum_ s E_s = \sum_ s \frac{A_s}{A} \left[\frac{m\ u^2(s)}{2} + \frac{3}{2}T(s)\right] . \eeq{11e.06} Introduce the partition of thermal and flow energies: \beq \tau _s = \frac{A_s}{A} \frac{3}{2} \ \frac{T(s)}{E_0} ; \ \ \lambda _s = \frac{A_s}{A} \frac{1}{2}\ \frac{ m\ u^2(s)}{E_0} . \eeq{11e.07} * \LT the energy conservation in the form: \beq \sum _{s} \tau _{s} + \lambda _{s} = 1 \ . \eeq{11e.08} Thus \beq \tilde {M}_{\alpha \beta} = \sum _{s} \lambda _s \frac{ \tilde{u}_\alpha(s) \tilde{u}_\beta(s)}{\tilde{u}^2(s)} + \tau _{s} \delta _{\alpha \beta } \frac{1}{3}\ , \eeq{11e.09} and this leads to: \beq {\rm Tr} \tilde{M} = \tilde{M}_{\alpha \alpha} = 1 . \eeq{11e.10} \B Energy conservation \LT \\ - - the trace of the dimensionless energy flow tensor is one. \B Experimentally, however, $\tilde M_{\alpha \alpha} < 1 $,\\ - - because not all particles are detected. To check the trace of the experimentally measured dimensionless energy flow tensor is a very important test of the experimental acceptance! }%end tr-page \newpage % transparency ====================================== p. 16 \transparencyframe{ \vspace*{-0.8cm} \subsection{Transverse Momentum Analysis} {*} Blast-wave model [5], \& thermal fireball model [6] \LT no Transv. Flow.\\ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (according to the Danielewicz analysis).\\ {*} `Few source' model [7] on the other hand does. \B Let us take a simple example of 3 sources:\\ - one in the c.m. (C) and \\ - two symmetrically deflected side (S) sources,\\ - - with masses $A_{c}, A_{s}, A_{s}$ \& temperatures $T_{c}, T_{s}, T_{s}$ respectively. S-sources have collective momenta per nucleon: $\vec{p}_s= (p_\perp^s, 0, p_\parallel^s)$ \& $-\vec{p}_s$,\\ - - where $[x,z]$ is the reaction plane and $z$ is the beam axis,\\ - - $p_\parallel^s$ is parallel to the beam, while\\ - - $p_\perp^s$ is orthogonal and it lies in the reaction plane. \B The transverse momentum projected to the reaction plane: \beq < \frac{p^x}{a} > = \frac{ \int d^3r\ d^3p\ n(r,t)\ f( \vec{p}, \vec{r})\ (\vec{p}\ \vec{e}_x) \left. \right|_{y =y_{0}} } {\left. A \right|_{y =y_{0}} }, \eeq{11e.11} where $\vec{e}_x$ is a transverse unit vector in the reaction plane. }%end tr-page \newpage % transparency ====================================== p. 17 \transparencyframe{ \vspace*{-0.8cm} For one central source (C) and two side or spectator sources (S)\\ the energy and mass conservation: \beqar A & = & \sum _{s} A_{s} = 2A_{s} + A_{c}, \nonumber \\ E_{0} & = & \left[ 2A_s \left( \frac{1}{2} m\ u^2(s) + \frac{3}{2} T_s \right) + A_c \frac{3}{2} T_c \right] /A , \eeqar{11e.12} where \beq \vec{u}(s) = u_\parallel + u_\perp \eeq{11e.13} * Using the scale parameters,\ \ \ $p_{1} = \sqrt {2mE_{0}}$\ \ \ \& \ \ \ $T_1 = \frac{2}{3}E_{0}$,\\ - - these take the scale invariant form: \beqar 1 &=& 2\tilde {A}_{s} + \tilde{A}_{c}, \nonumber \\ 1 &=& 2 \tilde {A}_{s} [\tilde{E}_{\perp} + \tilde{E}_{\parallel} + \tilde{T}_s] + \tilde{A}_c \tilde {T}_c = \nonumber\\ & & 2 \tilde{A}_s [\tilde{p}^{2}_{\perp } + \tilde{p}^{2}_{\parallel} + \tilde{T}_s] + \tilde{A}_{c} \tilde{T}_{c} , \eeqar{11e.14} Now let us calculate the transverse momentum, $p^x$,\\ at fixed $y = y_{0}$ or fixed $p_z = m_\perp \cosh y$: \beq p^x (y) = N \int d^2p_{\perp}\ p_\perp^s \left\{ \exp\left[-\frac{(\vec{p}- \vec{p}_{s})^2}{2mT}\right] + \exp\left[-\frac{(\vec{p}+ \vec{p}_{s})^2}{2mT}\right] \right\} \eeq{11e.15} where N is a normalization constant.\\ The center source does not contribute to $p^x$ because $ \vec{p}_{c} = 0 $. \\ Performing the integrals yields (see the solution of Assignment 7b): \beq p^x (y) = \frac{ A_{s}}{(2 \pi m T_{s})^{1/2}}\ p_{sx} \left[ e^{-\frac{(p_z -p_\parallel^s)^2}{2mT_s}} - e^{-\frac{(p_z +p_\parallel^s)^2}{2mT_s}} \right] . \eeq{11e.17} Here $p_z=p_x(y)$ is a function of the rapidity $y$.\\ Since $y \cong v_{z} = \frac{1}{m}p_z$ it follows that $p_z \cong my $. }%end tr-page \newpage % transparency ====================================== p. 18 \transparencyframe{ \vspace*{-0.8cm} The total number of nucleons $ A|_{y=y_0} $ is: $$ A|_{y=y_{0}} = \frac{ A_s}{(2 \pi m T_{s})^{3/2}} \left( e^{-\frac{ (p_z -p_\parallel^s)^2}{2mT_s}} + e^{-\frac{ (p_z +p_\parallel^s)^2}{2mT_s}} \right) \ + \hspace*{4cm} $$ \beq \hspace*{6cm} \frac{ A_c}{(2 \pi m T_c)^{3/2}} e^{-\frac{p_z^2}{2mT_c}}, \eeq{11e.19} consequently: \beq %$$ \frac{p^x (y)}{a} = %\hspace*{13.5cm} %$$ %\beq %\hspace*{2cm} \frac{ 2 \pi m T_s p_\perp^s \left( e^{-\frac{ (p_z -p_\parallel^s)^2}{2mT_s}} - e^{-\frac{ (p_z +p_\parallel^s)^2}{2mT_s}} \right) }{ \left( e^{-\frac{ (p_z -p_\parallel^s)^2}{2mT_s}} + e^{-\frac{ (p_z +p_\parallel^s)^2}{2mT_s}} \right) + \frac{A_c}{A-s} \left(\frac{T_s}{T_c}\right)^{3/2} e^{-\frac{p_z^2}{2mT_c} } . }, \eeq{11e.20} Here $p_z = p_z(y) = m y$. We will see the scaling behaviour of the transverse flow in the framework of the three source model in the next sections. }%end tr-page \newpage % transparency ====================================== p. 19 \transparencyframe{ \vspace*{-1.6cm} \subsection{Fragment flow and scaling} \B `Few source' model: composite fragments emitted at the break up,\\ - their momenta are $ \vec{p}_\kappa =\vec{p} A_\kappa $,\\ - - - ($\vec{p}$: momentum per nucl., \& $ A_\kappa$: fragment mass).\\ \B The transverse momentum per nucleon projected to the reaction plane,\\ in FD model for each type of fragment $\kappa$: \beq \left< \frac{p^x}{a} \right>_\kappa = \frac{1}{A_\kappa N_\kappa(y, \Delta y)} \int F_\kappa(\vec{p},\ \vec{r}) (\vec{p}\ \vec{e}_x)\ d^3r\ d^3p \left|_{ y_\kappa}{p_1} = \tilde{N}^{-1}_\kappa (\tilde{y}) \int \tilde{F}_\kappa (\tilde{p}, \tilde{r}) (\tilde{p} \tilde{e}_x) d^3\tilde{r} d^3\tilde{p} \left|_{\tilde{y} < \tilde{y}(\tilde {p})< \tilde{y} + \Delta \tilde{y}'} \right. , \eeq{11e.22} - the range of the momentum integral is given via the c.m. beam rapidity\\ - - $ \tilde{y} = y/y_1 = \frac {y}{y^{proj}_{CM}}$ (since we take the non-relativistic limit, $y^{proj}_{CM} = u_1$).\\ Here $ \tilde{F}_\kappa (\tilde{p}, \tilde{r}) =p_1^3 F_\kappa / A$, is normalized to $ \tilde{N}_\kappa = \frac{N_\kappa A_\kappa}{A}$, i.e. $ \sum _\kappa \tilde{N}_\kappa = 1$. \\ \B The total scale invariant transverse momentum is then: \beq \tilde{p}^x (\tilde{y} ) = \frac{ \sum_\kappa N_\kappa A_\kappa \tilde{p}^x_{\kappa}(\tilde{y}) }{ \sum_\kappa N_\kappa A_\kappa } \ . \eeq{11e.23} The scaling with beam energy is not expected to be perfect due to the same reasons that were mentioned before. }%end tr-page \newpage % transparency ====================================== p. 20 \transparencyframe{ \vspace*{-0.8cm} \subsection{Fragment flow - mass dependence (*)} Different fragments, $\kappa$, can be emitted from each source, $q=c, s_{1}, s_{2}$,\\ \LT the normalization is given by: \beq \tilde{A}_q = \sum_\kappa \tilde{A}_{\kappa q} = \sum_\kappa N_{\kappa q} A_\kappa/A \ \ \ \ \rm{and} \ \ \ \ \sum_{\kappa q }\tilde{A}_{\kappa q} = 1\ . \eeq{11e.24} - $\tilde{A}_{\kappa q}$ is the fraction of nucl. \# in source $q$, within fragments of type $\kappa$.\\ Thus the energy conservation can be expressed by: \beq \sum_{\kappa}\left[ 2\tilde{A}_{\kappa s} \left( (\tilde{p}_\perp^s)^2+(\tilde{p}_\parallel^s)^2 +\frac{\tilde{T}_s}{A_\kappa} \right) + \frac{\tilde{A}_{\kappa c} \tilde{T}_c}{A_\kappa} \right] =1 \eeq{11e.25} If the dim.-less source momentum vector of a given source is $\tilde{p}_q$\\ \LT the mom. distribution, $\tilde{F}_\kappa (\tilde{p}, \tilde{r}) d^3\tilde{r}$ , is a sum over the sources: \beq \int \ \tilde{F}_\kappa (\tilde{p}, \tilde{r}) d^3\tilde{r} = \sum_q \tilde{F}_{\kappa q}(\tilde{p}). \eeq{11e.26} Let us assume that each source has a local thermal motion,\\ - described by a MB distribution: \beq \tilde{F}_{\kappa q} (\tilde{p}) = \tilde{A}_{\kappa q} \left(\frac{ 3A_\kappa}{2\pi \tilde{T}_q}\right)^{3/2} \exp \left[ -\frac{3A_\kappa(\tilde{p} -\tilde{p}_{q})^2} {2\tilde{T}_q} \right] , \eeq{11e.27} normalized to $\tilde{A}_{\kappa q}$. }%end tr-page \newpage % transparency ====================================== p. 21 \transparencyframe{ \vspace*{-0.8cm} Now the integral (8.47) can be calculated explicitly.\\ The source, C, will not contribute to the integral due to symmetry.\\ Nevertheless, it contributes to the normalization constant $\tilde{N}_\kappa(\tilde{y}): $ $$ \tilde{p}_{t}(\tilde{y}) = \tilde{p}^x_{\kappa}(\tilde{y}) = Const. \left( \frac{\tilde{A}_{\kappa s} }{\sqrt{\tilde{T}_s} } \right) \tilde{N}_\kappa(\tilde{y})^{-1} \tilde{p}_\perp^s \hspace*{5cm} $$ \beq \hspace*{0.5cm} \left\{ \exp \left( -\frac{3A_\kappa(\tilde{y} -\tilde{p}_\parallel^s)^2} {2\tilde{T}_s} \right) - \exp \left( -\frac{3A_\kappa(\tilde{y} +\tilde{p}_\parallel^s)^2} {2\tilde{T}_s} \right) \right\} . \eeq{11e.29} where $$ \tilde{N}_\kappa(\tilde{y}) = Const. \left[ \left( \frac{\tilde{A}_{\kappa s}} {\sqrt{\tilde{T}_s}} \right) \left\{ \exp \left( -\frac{3A_\kappa(\tilde{y} -\tilde{p}_\parallel^s)^2} {2\tilde{T}_s} \right) + \right. \right. \hspace*{3cm} $$ \beq \left. \left. + \exp \left( -\frac{3A_\kappa(\tilde{y} +\tilde{p}_\parallel^s)^2}{2\tilde{T}_s} \right) \right\} + \left( \frac{\tilde{A}_{\kappa c}}{\sqrt{\tilde{T}_c}} \right) \exp \left( \frac{-3A_\kappa \tilde{y}^2}{2\tilde{T}_c} \right) \right] . \eeq{11e.30} }%end tr-page \newpage % transparency ====================================== p. 22 \transparencyframe{ \vspace*{-0.8cm} After a straightforward calculation we obtain that $$ \tilde{p}_{t}(\tilde{y}) = \tilde{p}^x_{\kappa}(\tilde{y}) = \tilde{p}_\perp^s \sinh \left( \frac{3 \tilde{y} A_\kappa \tilde{p}_\parallel^s}{\tilde{T}_s} \right) \left[ \cosh \left( \frac{3 \tilde{y} A_\kappa \tilde{p}_\parallel^s }{\tilde{T}_s} \right) \right. \hspace*{3.5cm} $$ \beq \left. + 1/2 \left( \frac{A_{\kappa c}}{A_{\kappa s}} \right) \sqrt{ \frac{\tilde{T}_s}{\tilde{T}_c} } \exp \left( 3 \tilde{A}_\kappa \frac{ \tilde{y}^2 \left( \tilde{T}_c - \tilde{T}_s \right) + (\tilde{p}_\parallel^s)^2 \tilde{T}_c } {2\tilde{T}_c \tilde{T}_s} \right) \right] . \eeq{11e.31} \B This expr. reproduces basic features of observed transverse momenta:\\ - The transverse momentum, $ \tilde{p}_{t}(\tilde{y}) = \tilde{p}^x_{\kappa}(\tilde{y}) $, as a function of the rapidity\\ - \ \ \ \ \ \ \ \ crosses the axis at the c.m. rapidity with a tangent of \beq %$$ \tilde{F}_\kappa = \frac{\partial \tilde{p}^x_{\kappa}} {\partial \tilde{y}} \left. \right|_{\tilde{y} = 0} = %\hspace*{12cm} %$$ %\beq %\hspace*{4cm} \frac{3A_\kappa \tilde{p}_\perp^s \tilde{p}_\parallel^s } { \tilde{T}_s \left( 1+ 1/2 \left( \frac{A_{\kappa c}}{A_{\kappa s}} \right) \sqrt{ \frac{\tilde{T}_s}{\tilde{T}_c} } \exp \left( \frac{3 A_\kappa (\tilde{p}^s_\parallel)^2}{2\tilde{T}_s} \right) \right) } , \eeq{11e.32} which \\ - increases with fragment mass, $A_\kappa$\\ - increases with increasing transverse momentum, $\tilde{p}_\perp^s$ and\\ - decreases with increasing dissipation (increasing $ T_{s}$ ). }%end tr-page \newpage % transparency ====================================== p. 23 \transparencyframe{ \vspace*{-0.1cm} \B Collective transverse momentum, $\tilde{p}^x_{\kappa}$, \& $\tilde{F}$:\\ - \ \ \ \ \ \ \ \ \ \ \ \ increase with the fragment mass, $ A_\kappa$.\\ - The dependence is, however, not trivial because of the denominator.\\ - The factor $\left( \frac{A_{\kappa c}}{A_{\kappa s}}\right)$ is not necessarily the same for all $\kappa$. \B If we assume: $ \tilde{T}_c > \tilde{T}_s$ \LT $ \left(\frac{A_{\kappa c}}{A_{\kappa s}}\right)$ decreases with increasing $ A_\kappa$\\ - enhancing the transverse momentum further. \B The $ \tilde{p}_\parallel^s$ dependence is not trivial either. }%end tr-page \newpage % transparency ====================================== p. 24 \transparencyframe{ \vspace*{-0.8cm} \B Use the THREE SOURCE MODEL [7] parametrization \LT\\ - for each $b$ \LT mass partition among the C- and S-sources\\ - collective kinetic energy partition \\ - thermal energy partition \beq \Theta _{cm} (b) = z_0 (1 - b/b_{max}) \pi /2 \eeq{11e.33} $$ |\vec{p}_s | = p_1 \left[ 1- \left( 1-b/b_{max}\right)^2 y_0 \right] $$ - $z_0$ is a parameter governing the transverse momentum transfer and\\ - $y_0$ is responsible for the dissipation\\ - $x_0$ governs the partition of thermal energy among the sources. \B If $ Q_h$ is the fraction of heat energy in the system\\ - $ Q_h = \sum _\kappa \frac{ \tilde{A}_{\kappa q} \tilde{T}_q}{A_\kappa}$ \LT \\ - $x = x_0 \left( 1- \frac{b}{b_{max}} \right)^2 $ is the fraction of thermal energy per nucl. in sources S$_1$ \& S$_2$: $ x = \frac{\tilde {T}_s}{A_\kappa Q_h}$. \B \LT the temp. of the sources are $\tilde{T}_s = xQ_h A_\kappa$ and $ \tilde{T}_c = \frac{Q_h - 2\tilde{T}_s \tilde{N}_s}{\tilde{N}_c}$, \\ - $\tilde{N}_q $ is the number of fragments in the source $q$. \B If composite fragments are present $\tilde{N}_q < \tilde{A}_q$, and\\ - we assume that $\tilde{N}_q=\tilde{A}_q/3$ in the followings\\ - \ (neglecting the possible differences between the C- and S- sources).\\ \ \LT $\left( \frac{A_{\kappa c}}{A_{\kappa s}}\right) = \left( \frac{A_c}{A_s}\right)$ is assumed.\\ \B Parameters: ($z_0 = 1, y_0 = 0.6, x_0 = 0.3 $). }%end tr-page \newpage % transparency ====================================== p. 25 \transparencyframe{ \vspace*{-0.8cm} \vskip 17cm Figure 8.4 {\it Scale invariant transverse momentum distribution calculated in the Few Source (FS) model for different impact parameters $ \tilde{b}=0.05-0.25$ and averaged over impact parameters in the range of $\tilde{b}=0-0.3.$ The $ \tilde{p}_{t} = \tilde{p}^x_{\kappa}(\tilde{y}) = $ distribution is different for different fragment masses $A_{\kappa}$. From [10]} }%end tr-page \newpage % transparency ====================================== p. 26 \transparencyframe{ \vspace*{-0.8cm} \B Fig. 8.4:\\ - impact parameter dependence\\ - impact parameter averaged \B Transverse momentum increases with impact parameter \\ - \ \ \ \ \ \ \ \ \ (for small impact parameters $ \tilde{b} < 0.4 $) and\\ - increases with the fragment mass $ A_\kappa $ for small rapidities.\\ - At high rapidities the heavy fragments show less transverse momentum.\\ (The Few Source model is not appropriate to describe peripherical collisions $\tilde{b} > 0.7$ similarly to the FD model.) \B Two basic features:\\ - the maximum of the transverse momentum and\\ - the tangent of $ \tilde{p}_t(\tilde{y})$ at the c.m., eq. (8.56).\\ (The latter quantity was also used in [11] to compare different colliding systems, and it was called "Flow" or $F$. ) \B The scale invariant quantity, $\tilde{F}$, is obtained by\\ - dividing $F$ by the C.M. beam momentum. \B Fig. 8.5: $\tilde{F}$ averaged over the impact parameter range $ 0 - \tilde{b}_{max}$: $\tilde{F}_{av} $.\\ - Selecting more central collisions (decreasing $ b_{max}$)\\ - \LT increase of $ \tilde{F}_{av}$ until we reach $\tilde{b}_{max}^{crit}$,\\ .\hspace*{7cm} \ \ \ \ \ \ \ where $\tilde{F}_{av}$ reaches its maximum.\\ - Further decrease of the maximum impact parameter \\ - \LT rapid decrease of the flow. \B $\tilde{b}_{max}^{crit}$ decreases with increasing fragment mass $A_\kappa$,\\ .\hspace*{12cm} \ $\tilde{b}_{max}^{crit} \approx 0.3-0.5$. }%end tr-page \newpage % transparency ====================================== p. 27 \transparencyframe{ \vspace*{-0.8cm} \vspace*{17cm} Figure 8.5 {\it Impact parameter averaged scale invariant flow $\tilde{F}$ calculated in the Few Source model (FS) for different maximum impact parameters and for different emitted fragment masses $A_{\kappa}$ From [10]} }%end tr-page \newpage % transparency ====================================== p. 28 \transparencyframe{ \vspace*{-0.8cm} \section{Scaling violations} \B Our assumptions do not always hold:\\ - i) on the scaling of the speed of sound, which is related to the EOS\\ - ii) on the smallness of the viscous effects. \B Let us start with perfect FD, and \\ - let us call $\Phi(\vec{r},t)$ any field entering these equations ($\Phi = \rho, \vec{u},$ or $T$).\\ - If $\Phi(\vec{r}, t)$ is a solution, then $\Phi(\vec{r}\alpha, t\alpha)$ is also a solution,\\ .\ \ \ \ \ \ provided that the initial and final (break-up) conditions scale. \B OK, for a restricted set of systems,\\ - e.g. at the same incident energy, and at the same $b/R$ ratio.\\ \B \LT Every dim.-less quantity is a function of the Strouhal nu., $S$, only.\\ - However, at given $b$, $\exists$ another dim.-less parameter: $b/b_{Max}$.\\ - Only those collisions yield a similar flow pattern where both\\ - the Strouhal number, $S$, \& the parameter, $b/b_{Max}$, are the same }%end tr-page \newpage % transparency ====================================== p. 29 \transparencyframe{ \vspace*{-0.8cm} This could be expressed by requiring the constancy of a dim.-less parm.: $$ \Psi = \Psi( S, b/b_{Max}). $$ Or for viscous FD, any dim.-less physical quantity can be expressed as $$ \Psi = \Psi( S,\ Re,\ b/b_{Max}). $$ \B Scaling analyses - compare e.g. 25\% of most central events each other:\\ - This selection is dimensionless: \\ - - \LT for this selected class of events scale invariance holds.\\ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (if viscous effects are negligible.) \B This follows from the fact that the Strouhal number can be fixed to 1\\ - - - \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the appropriate choice of time scale.\\ - If the break up time scales according to this time scale \LT\\ - - the experimental results scale also, i.e.\\ - - dim.-less quantities will not show $A$ or $E^{c.m.}_0$ dependence. \B Viscous effects are important \LT scale invariance cannot be expected.\\ - Constancy of Reynolds number is also required for scale invariance.\\ \LT not all events with all $A$ and $E^{c.m.}_0$ will be invariant\\ - - only those satisfying the relation $Re(A,E_0) = {\rm const.} $. }%end tr-page \newpage % transparency ====================================== p. 30 \transparencyframe{ \vspace*{-0.8cm} \subsection{Scaling violation in transverse flow} \B The transverse flow depends on the Reynolds number [12].\\ - - a theor. estimate [12] gives (for $b \lton b_{Max}/2$): \beq \tan ( <\Theta_{flow}> ) \approx \frac{1}{3} \left[ S \left( 1 - \frac{b}{b_{Max}}\right) - \frac{3}{Re}\right] \frac{\bar{\varepsilon}}{\bar{\varepsilon}_z} , \eeq{resca1} - $\bar{\varepsilon}$: average nucl. energy \\ - $\bar{\varepsilon}_z$: average nucl. energy in the $z$-direction.\\ Eq. (8.58) was compared to experimental data at 400 \amev\ [13]\\ \B \LT $Re \approx 8$ (close to the free gas value [14]) \vspace*{10cm} Figure 8.6 {\it Comparison of measured flow angle (dots) [13], with eq. (8.58) (full line), using Reynolds number, $Re=8$ for Nb+Nb. From ref. [16] } }%end tr-page \newpage % transparency ====================================== p. 31 \transparencyframe{ \vspace*{-0.8cm} \B To study the validity of scaling as assumption,\\ - 1st: express measured quantities in a scale-invariant way.\\ - \LT Introduce a scale-invariant transverse momentum per nucleon: \beq \tilde{p}^x = p^x/p^{c.m.}_{proj.} \eeq{bc1} - $p^x$ is the transverse momentum per nucl. from exp.\\ - $p^{c.m.}_{proj.}$ is the c.m. momentum of a projectile nucleon\\ Def.: Scale-invariant rapidity: \beq \tilde{y} = y^{c.m.}/y^{c.m.}_{proj.}; \eeq{bc2} Fig. 8.6: Exp. data, plotted in scale inv. form.\\ - differences arise from multiplicity selections (or $b$-s) and\\ - from different types of particles detected. \B Definition (introduced in [8,16-17]):\\ --- Slope of $\tilde{p}^x(\tilde{y})$ at mid-rapidity.\\ --- This parameter is called ``Flow'', $F$. \\ --- We denote the corresponding scale-invariant slope by $\tilde{F}$. }%end tr-page \newpage % transparency ====================================== p. 32 \transparencyframe{ \vspace*{-0.8cm} \vspace*{16cm} Figure 8.7 {\it Projected trasverse flow from three different experiment plotted in dimensionless variables. The curves lie on the top of each other due to approximate scale invariance. From [15]} }%end tr-page \newpage % transparency ====================================== p. 33 \transparencyframe{ \vspace*{-0.8cm} \B A recent, more extensive analysis [16,8]:\\ - Fig. 8.7: Resemblance between the dim.-less transverse Flow \beq \tilde{F} = \frac{d \tilde{p}^x}{d \tilde{y}}, \eeq{inv-flow} and the value of Reynolds number (viscosity from [14]).\\ \B \LT Flow properties are more sensitive to viscosity than to the EOS. \B Contour lines of constant $\tilde{F}$ and $Re$ (Fig. 8.8) \LT\\ - At medium and high energies: contour lines are similar.\\ - But, data rise somewhat more sharply than the Reynolds number.\\ ==$>$ This indicates that the viscosity increases faster with energy\\ - - - than $\sqrt{T}$ calculated in [2], \\ - - - because of other inelastic processes, pion emission, etc. \B The same processes may also lead to a softening of the EOS\\ - \LT smaller transverse flow.\\ \B At high beam energies partial transparency is expected\\ - \LT smaller transverse flow. }%end tr-page \newpage % transparency ====================================== p. 34 \transparencyframe{ \vspace*{-0.8cm} \vspace*{14cm} Figure 8.8 {\it Contour plots in the [A.E] plane. Full lines in (a) correspond to constant Reynolds number. Full lines in (b) correspond to constant $\tilde{F}$. The dotted curve indicates the expected behavior of $\tilde{F}$ at low energies. Symbols circles, squares, diamonds, dotted circles, inverted triangles, and triangles correspond to experimentally measured $\tilde{F}$-values of 0.4, 0.325, 0.275, 0.225, 0.175, 0.125, and 0.1 respectively. From [15] } }%end tr-page \newpage % transparency ====================================== p. 35 \transparencyframe{ \vspace*{-1.2cm} \subsection{Disappearance of the transverse flow} \B The most drastic difference btwn $Re=$const. \& $\tilde{F}$=const. curves\\ - - $\exists$ at low energies.\\ - - * Below $E_{nucl.}^{c.m.} \approx $ 60 - 70 MeV,\\ - - - the scale-invariant transverse flow, $\tilde{F}$, drops suddenly\\ - - - with decreasing energy, and\\ - - - below $E_{nucl.}^{c.m.} \approx $ 10 - 20 MeV it becomes NEGATIVE!! \B A given mean-field potential leads to a well defined EOS\\ - - (and vice versa) in thermal equilibrium \& for a given statistics.\\ \B Attractive mean field \LT EOS with a liquid-gas type phase transition\\ - Thus, either the EOS or the reaction mechanism should change here\\ - The EOS enters the scaling analysis via the sound speed.\\ - - - Namely, a scale-invariant flow pattern can be obtained only if\\ - - - the pressure satisfies: $\nabla p=c_s^2 \nabla \rho$\\ - - - with a sound speed $c_s$, which scales with the c.m. energy as \beq c_s=\tilde{c}_s \sqrt{2 E_{nucl.}^{c.m.}/m} = \tilde{c}_s \sqrt{2 E_0/m} . \eeq{bc3} - For ideal gas $\tilde{c}_s = $const.\\ - For an EOS with binding $\tilde{c}_s = 0.7-0.9$ for $\tilde{n} =0.3-1.0$ and $E_0=50-240$ MeV, \\ - - below $E_0=$40 MeV the sound speed diverges with decreasing $E$:\\ - - - at $\tilde{n} = 1$ \ $c_s \gg 1$,\\ - - - at $\tilde{n} = 0.6$ \ $c_s =0.7 $ and\\ - - - at $\tilde{n} = 0.3$ \ $c_s \rightarrow 0$. }%end tr-page \newpage % transparency ====================================== p. 36 \transparencyframe{ \vspace*{-0.8cm} \B This non-scaling behavior of $c_s$, at low energies \\ - is related to the liquid-gas phase transition of our EOS.\\ - Compared to the ideal gas, the phase transition decreases $P$ \& $F$.\\ - - - In eq. the pressure is small but always positive.\\ - - - In a non-eq. phase transition the pressure may become negative,\\ (if the matter expands rapidly into the supercooled liquid phase\\ not having sufficient time to establish phase equilibrium.)\\ - Attractive nuclear interactions may overcom the repulsion caused by the pressure. Such an attractive interaction is out of the scope of the fluid-dynamical scaling studies. \B The two effects,\\ - i) the softening of the EOS or the negative $P$ , and\\ -ii) the predominance of the attractive mean field \\ - - are, of course, the two sides of the same microscopic N-N interaction. \B Theoretical works in BUU or VUU approach predict ``negative'' $\Theta_{flow}$\\ - - due to the attractive nuclear mean field.\\ \B The same nuclear mean field leads to fragmentation \\ - - representing a liquid-gas phase transition.\\ \B Recently these effects were detected \& analysed by Westfall, et al.[18].\\ - - The energy where the flow vanishes is sensitive to the\\ - - in medium nucleo-nucleon cross section, [19].\\ - - i.e. on the viscosity, which determines the scaling violations also. }%end tr-page \newpage % transparency ======================================== p. 37 \transparencyframe{ \vspace*{-0.8cm} \section{Assignment 8} \begin{description} \item[8.a] Calculate the nonrelativistic scale invariant cross section in the non-relativistic Blast wave model with local Maxwell-Boltzmann distribution. Assume that at the breakup the ratio of the flow energy to the thermal energy is $\mu$. \end{description} }%end tr-page % transparency ================================================================ \end{document} % % % % % % % % % % % % % % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % % % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % % % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % % % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % % % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 % %