% % 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 6 / Chapter 6 % % 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} %\setlength{\unitlength}{1mm} %}%end tr-page %\newpage % transparency ===================================================== \transparencyframe{ \setcounter{chapter}{5} \vspace*{-1.9cm} \chapter{Simple models} For the compression phase of Rel. HI collisions\\ in the ``stopping'' region -- simple model solutions: \begin{itemize} \item Shock wave solutions, \item Detonation wave solutions. \end{itemize} Expansion is spherical or has general 3-dim. distorted shape Ultra-relativistic energy region, $E_{CM} \ge 20 -30 $A GeV:\\ reaction mechanism is different\\ (From p+p and p+A reactions:) \begin{itemize} \item The leading baryon is hardly stopped, \item In the rapidity region between the projectile and the target,\\ secondary charged particles (mesons $\pi^+, \pi^-, \pi^0, K^+, K^-$, etc.)\\ are created. \end{itemize} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.1cm} Theoretical expectations based on p+p and p+A are shown in Fig. 6.1. At ultra-relativistic energies the mid-rapidity region and the target and projectile rapidity regions are studied separately. \vspace*{6truecm} Figure 6.1 {\it Exagerated plot of observed rapidity distribution of baryons and mesons in ultra-relativistic proton-proton collisions. In reality particles in the fragmentation regions and in the central rapidity region are not separated, their rapidity distributions overlap. Heavy ion collisions are expected to yield similar rapidity distribution at very high energies.} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \section{Applicability of simple models} $\exists$ several simple models\\ (symmetry properties \LT very pleasant to handle\\ but they are not necessarily realistic.) The three most basic collective reaction models are\\ \B i) the spherical fireball, or different versions of it,\\ \B ii) the Bjorken model, and \\ \B iii) the Landau model. \setlength{\baselineskip}{9pt} {\normalsize\sf In a certain sense these are all fluid-dynamical models. All of them are applicable to central symmetric collisions, since the treatment of spectators is not incorporated in any of the basic versions of these models.} \bigskip Spherical models:\\ {\large\sf The fireball model assumes spherical symmetry.\\ Although the initial state of a HI collision is never spherically symmetric,\\ at energies in the order of 1 \agev\ the conditions are such that,\\ by the time the maximum density and compression are reached,\\ the system is thermalized and gets close to being spherically symmetric.} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Landau model:\\ As the energy increases, $\approx$ $E_{lab.} =$ 10-100 \agev\ \LT\\ Lorentz contraction of the projectile and target cannot be disregarded\\ Even when thermalization is reached rapidly:\\ the ratio of longitudinal thickness versus the diameter is $\sim$ 0.05 - 0.1,\\ {\normalsize\sf i.e. close to the aspect ratio of the projectile and target in the c.m. frame, $\gamma^{c.m.}$.}\\ \B Landau's fluid-dynamical model assumes an initial condition:\\ a static homogenous disk of such an aspect ratio.\\ \medskip Apply in this (AGS, SPS) energy range.\\ Although, for simplicity spherical fireball or spherical fluid dynamical models are some times used for characterizing these reactions, one should keep in mind that\\ spherical models are not applicable at these (BNL-AGS and CERN-SPS) energies and quantitative physical conclusions should not be drawn from such comparisons. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Bjorken model:\\ Increase the energy further to $E_{c.m.}$ = 100 \agev\ \LT the Lorentz contracted nuclei will become transparent to each other,\\ (Their valence quarks will almost maintain their original rapidities.)\\ At their interpenetration, however, these quanta may exchange color\\ which leads to the creation of a chromo-electric field,\\ similar to the electric field between two condenser plates,\\ where the condenser plates fly apart from each other. The energy density of this field is substantial.\\ (Due to the self interaction of the field it is assumed\\ to be confined in the transverse direction.) \B Bjorken's model is applicable for such a physical schenario.\\ The model is one dimensional and time dependent. \setlength{\baselineskip}{9pt} {\normalsize\sf If the transverse expansion of this flux tube is to be studied, or if the behaviour around the ends of this object is of interest, the model should be supplemented. The spherical models and Bjorken's model are 1+1 dimensional, while the Landau model is 2+1 dimensional. If one addresses 3 dimensional effects, like the directed transverse flow in the reaction plane, none of these simple models apply, due to symmetry reasons! } }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \section{The Bjorken model} \B In the mid-rapidity region: the EOS is easier - $\mu_B = 0$\\ \B $\exists$ a scaling solution for the dynamical problem\\ - the scaling hydrodynamical model of Bjorken [5]\\ - - This model has become the basis of many subsequent models The rapidity distribution of the charged secondaries is constant in the mid-rapidity region \LT \\ the energy density is also constant, Fig. 6.2 \vspace*{8truecm} Figure 6.2 {\it Space-time evolution of the mid-rapidity region in the Bjorken-model. The solution is uniform along contours of constant proper time. This ensures that the solution is invariant under Lorentz transformations in the beam direction.} If the rapidity distribution $dN_{ch}/dy =$ const. \\ ($\approx$ 3 at the CERN SPS) \ \LT \\ invariant under Lorentz transformation in the mid-rapidity region. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \B Assume that all quantities (like $n(y), \ e(y), $ etc.) \\ also have this symmetry at least at the freeze-out\\ \B Thus the density of charged particles, $n_{ch}$,\\ depends on the proper time, $\tau$, only\\ because $\tau$ is invariant under a Lorentz transformation \beq n_{ch} = n_{ch}(t,z) = n_{ch}(\tau), \eeq{nch} where \beq \tau = t/\gamma = t \sqrt{1-v^2 \ } = \sqrt{ t^2 - z^2 \ } , \eeq{tau} if all the particles originate from one point in the space-time, \LT\\ their velocity is $v=z/t$ (if the point of origin is the origo)\\ This point is then the point of impact in the space-time. Assume that this symmetry is achieved at the time of thermalization and\\ look for the solution of Relativistic Fluid Dynamics from that time on. Special boundary condition: $$ e = e_0 (\tau_0=1\ {\sf fm/c\ }) \approx 1-10\ {\sf GeV/fm}^3 . $$ In [5] a simple hydrodynamical model is developed, which is applicable for ultra-relativistic heavy ion collisions under this symmetry.\\ \B Instead of variables $x$ and $t$ $\longrightarrow$\\ rapidity $y$ and proper time, $\tau$, }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Estimate the initial energy density at the beginning of the expansion $$ {{dN_{ch}}\over{dy}}_{p+p} = 3 \ \ \ \leadsto \ \ \ {{d}\over{dy}}_{p+p} \approx 1.8\ {\sf GeV} , $$ because $dN_{tot}/dy = 3\times 2$ $\leadsto$ $ dN_{tot}/dy = d/dy$,\\ and $ \approx 0.4$GeV. Due to the Lorentz contraction, the colliding nuclei in their C.M.\\ resemble two flat disks approaching each other (of surface $A$).\\ If $N$ nucleons the average surface per nucleon is $d^2_0 \equiv A/N$,\\ \LT in a heavy ion collision $d_0 \approx 0.3 - 1$fm. If we reach thermalization at $t_0 = 1$fm/c the energy density: $$ e_0 \approx {{1\ {\sf GeV}}\over{t_0\ d_0^2}} \approx 1-10\ {\sf GeV/fm}^3 . $$ - Initial condition: variables depend on $\tau$ but not on the rapidity $$ e=e(\tau ), \ \ p=p(\tau ), \ \ T = T (\tau ), \ \ {\sf etc.} $$ The initial proper time is $\tau_0$, \LT initial condition $e(\tau_0) = e_0$. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Let us introduce a four vector $\tilde{x}_\mu = (t,0,0,-z)$ \\ (or the same in contravariant coordinates $\tilde{x}^\mu = (t,0,0,z)$).\\ Then, if all particles are at rest in their own LR \\ we can express the velocity field using eq. (\ref{tau}) as \beqar u^\mu = {1 \over {\tau}} (t,0,0,z) = {{\tilde{x}^\mu}\over{\tau}} \\ u_\mu = {1 \over {\tau}} (t,0,0,-z) = {{\tilde{x}_\mu}\over{\tau}} \eeqar{7a0} The local velocity is orthogonal to the constant proper time curve everywhere. \vspace*{8truecm} Figure 6.3 {\it The four-velocity vector of the flow in the Bjorken-model is orthogonal to the constant proper time, $\tau=const.$ hyperbolas.} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Neglect the viscosity and heat conduction and use perfect fluid dynamics: \beq T^{\mu\nu},_\nu = 0, \eeq{7.1} where $T^{\mu\nu} = (e+P) u^\mu u^\nu - P g^{\mu\nu}$. Let us observe that $$ \tau ,_\mu = {{\partial \tau}\over {\partial x^\mu}} = u_\mu \ \ \ \leadsto \ \ \ \partial \tau = u_\mu \partial x^\mu . $$ We can then rewrite eq. (\ref{7.1}) as \beq (e+P),_\mu u^\mu u^\nu - P,_\mu g^{\mu\nu} + (e+P) u^\mu ,_\mu u^\nu + (e+P) u^\mu u^\nu ,_\mu = 0 , \eeq{7.3} where the last term vanishes and the others can be rewritten as shown below. \bigskip\hrule\bigskip \begin{quotation} {\small\sf Aside: $$ u^\nu ,_\mu = \left({{\tilde{x}^\nu}\over{\tau}}\right),_\mu = {1\over{\tau}} \tilde{g}^\nu_\mu - {{\tilde{x}^\nu}\over{\tau^2}} {{\tilde{x}_\mu}\over{\tau}} = {1\over{\tau}} ( \tilde{g}^\nu_\mu - u^\nu u_\mu) , $$ where $ \tilde{g}^\nu_\mu = \delta^\nu_\mu $ if $\mu =0,3$ else $0$. I.e. $\tilde{g}^0_0 =1$ and $\tilde{g}^z_z =1$. Consequently: $$ u^\mu ,_\mu = {1\over{\tau}} (2-1) = {1\over{\tau}} . $$ } %endsmall \end{quotation} \bigskip\hrule\bigskip }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Inserting this into eq. (\ref{7.3}): $$ (e+P),_\mu u^\mu u^\nu - P,_\mu g^{\mu\nu} + (e+P) {1\over{\tau}} u^\nu + $$ $$ (e+P) \underbrace{ {1\over{\tau}} u^\mu ( \tilde{g}^\nu_\mu - u^\nu u_\mu) }_{ {1\over{\tau}} (u^\nu - u^\nu u^\mu u_\mu) }= 0 , $$ where we see that the last term vanishes and multiplying the first two terms by $u_\mu/u_\mu$ \LT $$ {{\partial (e+P)}\over {u_\mu \partial x^\mu}} u^\mu u_\mu u^\nu - {{\partial P}\over{ u_\mu \partial x^\mu}} u_\mu g^{\mu\nu} + (e+P) {1\over{\tau}} u^\nu = 0 . $$ Using the relation $u^\mu u_\mu = +1$ and the definition of $\tau$ $$ {{\partial (e+P)}\over {\partial \tau}} u^\nu - {{\partial P}\over{ \partial \tau}} u^\nu + (e+P) {1\over{\tau}} u^\nu = 0 . $$ This should be satisfied for all $u^\nu$'s, thus \beq {{\partial e}\over {\partial \tau}} = - {{e+P} \over{\tau}} . \eeq{bjorken} This is the basic differential equation of Bjorken's hydrodynamical model. $e(\tau_0) = e_0$ is given as initial condition. \vfill }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \subsection{Entropy conservation} \B In perfect fluids $S^\mu ,_\mu = 0$,\\ (where $S^\mu = s u^\mu$ and $s$ is the entropy density in the proper frame) \LT $$ {{\partial (s u^\mu)}\over{\partial x^\mu}} = {{\partial s }\over{\partial x^\mu}} u^\mu + s {{\partial u^\mu}\over{\partial x^\mu}} = {{\partial s }\over{u_\mu \partial x^\mu}} u_\mu u^\mu + s {1\over{\tau}} = 0 , $$ \LT the differential equation: \beq {{\partial s }\over{\partial \tau}} = - {s \over{\tau}} . \eeq{7entro} The solution of this equation is $$ s(\tau) = s(\tau_0) {{\tau_0}\over{\tau}}. $$ Consequently $dS/dy = $constant.\\ To solve the dynamical equations of the FD we need an EOS\\ \LT eq. (\ref{bjorken}) still allows several solutions.\\ For an ideal ultra-relativistic gas the EOS is: $e=3P$.\\ This reduces eq. (\ref{bjorken}) to $$ {{d e}\over{d\tau}} = - {4 \over 3} {e \over \tau} \ , $$ which leads to the solution $$ e(\tau) = e(\tau_0) \left( {\tau \over{\tau_0}} \right)^{-4/3} . $$ }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \subsection{Multiplicity estimate in ultra-relativistic collisions} The ``chromo-electric'' field between the projectile and target\\ considered as a flux-tube or as a string.\\ If the field is created by several constituents (partins)\\ \LT the field strength increases $\approx$ the number of strings increases. In A+A collisions a large number of hadron collisions occur.\\ This number is proportional to the surface of the projectile and target nuclei.\\ Thus the number of strings increases \LT \\ the entropy per unit rapidity increases the same way $$ \left( {{dS}\over{dy}}\right)_{A+A} = {{r_0^2 A^{2/3} \pi}\over{d_0^2}} \left( {{dS}\over{dy}}\right)_{p+p} . $$ Assuming that the pion multiplicity and the entropy are proportional\\ (this is the case for an ideal Bose gas), $N_\pi \sim S$,\\ \LT we can estimate the parameter $d_0^2$ by comparing\\ the results of p+p and $\alpha$ + $\alpha$ (from ISR 30+30 A GeV): $$ {{(dN_\pi / dy)_{A+A} } \over {(dN_\pi / dy)_{p+p} }} = \left( {{2 fm}\over{d_0}} \right)^2 A^{2/3} . $$ Inserting the experimental results into this equation we get $$ d_0 \approx 0.7 \ {\sf fm} . $$ }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} This estimate \LT\\ multiplicity increase $$ \propto A^{2/3} $$ but the cross section is bigger than one layer of nucleons would produce.\\ \B I.e., not only the 1st but the subsequent nucleons in a row\\ also contribute to the collision. Later on this problem of subsequent or better to say multiple collisions on a row of nucleons was studied extensively in a study of ``Nuclear Stopping Power''(see section 10.5). \B Bjorken estimated the multiplicity one could get in a U+U collision,\\ and his estimate was $$ {{dN_\pi}\over{dy}} \approx 800 $$ This estimate indicates the difficulty of future heavy ion experiments,\\ but as the first CERN measurements indicate,\\ the present measurement techniques can cope with this task. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \subsection{Inclusion of phase transition in the Bjorken model (*)} One of the most important questions:\\ \B\B How to estimate the initial energy density from the measurables. If the energy density is high \LT QGP is probably formed in the collision.\\ Energy density \LT the entropy density if the EOS is known,\\ and the entropy density \LT the particle multiplicities in the collision. In [19] the assumption of the initial energy density was modified\\ considering the fact that the expansion is not free.\\ The basic idea is that the expansion is very likely to be adiabatic\\ \LT the initial and final entropy is the same (per unit rapidity). On the other hand the time evolution of the energy density depends on the EOS.\\ The entropy density $s$ and the density of quanta $\tilde{n}$\\ (not necessarily the density of conserved particles)\\ are related to each other by $$ s = \xi \tilde{n} , $$ where $\xi$ is a constant, $\xi\approx 4$, \\ e.g. for quark-gluon plasma $\tilde{n} = n_q + n_{\bar{q}} + n_g $.\\ (Typical values:\\ Bose gas - $\xi$ =3.6,\\ Boltzmann gas - $\xi$ = 4.0,\\ Fermi gas - $\xi$ = 4.2, etc.) }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} It follows then that $$ {{ dS}\over{dy}} \approx 4 {{dN}\over{dy}} . $$ This was also the basic assumption in the Landau theory of multiparticle production [20]. Thus $$ {{dN}\over{dy}} = {1 \over{dy}} \int d^3x \ \tilde{n} = {1 \over{dy}} \int d^2x (\tau \ dy) \xi^{-1} s = d_0^2 \tau s / \xi = d_0^2 \tau_0 s_0 / \xi . $$ The relation between $s_0$ and $e_0$ depends on the EOS of the initial state. Using the EOS for the hadronic matter: $$ s_{h0} = {4 \over 3} e_{h0}/T_0 = {4 \over 3} \left( {3 \over {30}} \right)^{1/4} e_{h0}^{3/4} , $$ and for QCD-plasma: $$ s_{q0} = {4 \over 3}( e_{q0}-B) /T_0 = {4 \over 3} \left({{37} \over {30}} \right)^{1/4} (e_{q0}-B)^{3/4} . $$ This leads us to the conclusion that $$ e_0 \propto \left( {4 \over{\tau_0 \ d_0^2}} {{dN}\over{dy}}\right)^{4/3} , $$ somewhat larger than the original estimate of Bjorken for $e_0$. If our EOS includes a phase transition the expansion is different from the one presented earlier, at the discussion of the Bjorken model.\\ Inserting the EOS of the QGP into eq. (\ref{bjorken}) we get another solution \beq e(\tau) = B+ [e(\tau_0)-B] \left( {\tau \over{\tau_0}} \right)^{-4/3} . \eeq{plasmex} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} If we start the expansion from QGP, this is the solution we follow initially. There are several possibilities if we have a phase transition:\\ - i) we can assume that the expansion is still adiabatic,\\ \ \ \ \ i.e. no dissipation will occur due to the transition,\\ - ii) we may assume that the expansion is isoergic,\\ \ \ \ \ i.e. the internal energy remains constant, and\\ \ \ \ \ no internal energy is converted into the kinetic energy of the expansion\\ \ \ \ \ so that the matter is coasting during the transition\\ \ \ \ \ with constant speed,\\ \ \ \ \ \ \ \ or probably the most realistic assumption is that\\ - iii) the QGP supercools to some temperature $T_q < T_{cr.}$, and then \ \ \ \ it undergoes a time-like deflagration into hadronic matter [21] Let us only sketch case i) briefly:\\ The expansion starts in the plasma and\\ from eq. (\ref{7entro}) the temperature decreases as \beq T(\tau) = T(\tau_0) \left( {\tau \over{\tau_0}} \right)^{-1/3} , \eeq{plasmt} while the energy follows (\ref{plasmex}).\\ \B When during the expansion $T_c$ is reached in the QGP phase\\ the formation of hadronic matter begins.\\ For a while we will have a phase mixture,\\ during the existence of the mixture the temperature will not drop\\ but it stays at the critical $T_{cr.}$.\\ \B When the matter is converted completely into hadronic phase\\ the expansion continues according to the solution we have seen before: $$ e(\tau) = e(\tau_0) \left( {\tau \over{\tau_0}} \right)^{-4/3} . $$ }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} The time scale of the expansion is given by eq. (\ref{7entro}) and by $$ s(\tau) = \lambda(\tau) s_{q\ cr.} + [1- \lambda(\tau)] s_{h\ cr.} $$ where $\lambda$ is the volume ratio of QGP in the expanding matter\\ (it is 1 initially and 0 at the end of the transition): \beq \lambda(\tau) = {{37}\over{34}} \left[ {{T(\tau_0)}\over{T_{cr.}}}\right]^3 \left( {{\tau_0 }\over{\tau}} \right) - {{ 3}\over{34}} . \eeq{plasml} $$ {\sf The phase transition begins at\ \ } \tau_1 = [ T(\tau_0)/T_{cr.}]^3 \ \tau_0 , $$ $$ {\sf and it ends }(\lambda = 0)\ {\sf at \ \ } \tau_2 = {{37}\over 3} \tau_1 . $$ If $T_0 = T(\tau_0) = 200$MeV, $T_{cr.} = 169$MeV (i.e. $\Lambda_B = 235$MeV, or $B=0.397$GeV/fm$^3$) and $\tau_0 = 1$fm/c the characteristic times are: $$ \tau_1 = 1.66\ {\sf fm/c\ and\ \ } \tau_2 = 20.4\ {\sf fm/c.} $$ \B The relation between the initial energy density and entropy density\\ is different if we start from different phases: If we start from QGP $$ s = { 4\over 3} \left( {{ 37}\over{30}}\right)^{1/4} (e_{q0} - B)^{3/4}, $$ if we start from the mixed phase $$ s = [e_{m0} - {1 \over 3} e_h(T_{cr.})]/T_{cr.} , $$ and if we start from the hadronic phase $$ s = { 4\over 3} \left( {{ 3}\over{30}}\right)^{1/4} e_{h0}^{3/4} . $$ }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.2cm} In [19] the initial energy density was calculated for two EOS-s, $(1),\ (2)$\\ describing phase mixture between the quark and hadronic phases with bag constants B=0.74$(1)$, 0.05$(2)$ GeV/fm$^3$.\\ The relation between the initial energy density at $\tau_0$=1fm/c and the pseudo rapidity density of the emitted particles is shown in Fig. 6.4.\\ \B Curves (1) and (2) are assuming the adiabatic scenario i).\\ - The points above the black dot on the dashed lines, (1) and (2),\\ - - correspond to QGP initial states,\\ - below the dots to mixed phase initial states.\\ \B The full line corresponds to the isoergic expansion scenario ii), \\ - which involves entropy increase during the expansion,\\ - thus the initial entropy and energy densities are smaller. \vspace*{9cm} Figure 6.4 {\it Dependence of initial energy density on $dN/d\eta$ at $\tau_0$=1fm/c calculated in the framework of the Bjorken fluid dynamical picture. The transverse area is taken to be $d_0^2 = A^{2/3} r_0^2 \pi$, (with $r_0$=1.18fm). Multiplicities observed in cosmic ray events suggest that an initial energy density of 3-6 GeV/fm$^3$ can be reached in heavy ion reactions. From [19]} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.6cm} - For a complete model: one has to consider the transverse expansion,\\ - the string is not infinitely long \LT end effects Numerical solutions [22] included source terms in the FD equation\\ - in order to simulate matter - not thermalized initially.\\ \B From the many N+N collisions, the constituents (quarks and gluons)\\ - thermalize only after some proper time, $\tau_0$, has passed!\\ - The estimate for this time is $\tau_0 \approx 1$fm/c. \B Energy and momentum contribution of one nucleon nucleon collision\\ will be added to the fluid flow $\tau_0$ proper time after the collision.\\ \LT The locus of these points in the space-time lies on a hyperbola. \vspace*{11cm} Figure 6.5 {\it Space-time picture of 4-4 nucleons colliding on each other. The contribution to the subsequent flow starts at different proper time hyperbolas. From [22] } }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} In the FD equations the subsequent contributions are source terms [10,22] \beq T^{\mu\nu},_{\mu} = \Sigma^\nu , \eeq{5i1} and \beq N_B^\mu ,_\mu = \sigma_B . \eeq{5i2} The source terms from p+p collisions: $$ \Sigma_{p+p}^\nu = \Sigma_{i=N,\pi,K,...} m_i \ \rho_i(y) \ x^\nu \delta(\tau - \tau_0) , $$ where $\rho_i(y)$ is the observed rapidity dist. of particle type, $i$\\ The source term of the baryon continuity equation $$ \sigma_B = \rho_B(y) \delta(\tau - \tau_0). $$ Then for A+A collisions: \beq \Sigma^\mu = \sum_{i=coll.} \sum_{c=\pi,N} {{}\over{d_0^2}} {{dN_c(y)}\over{dy}} v^\mu_i 2 \delta(\Delta\tau_i^2-\tau_0^2) = e_c v^\mu_i 2 \delta(\Delta\tau_i-\tau_0) . \eeq{5i4} Here $v^\mu_i$ is the 4-velocity of the deposited quanta, $v^\mu_i = {{(x-x_i)^\mu}\over \tau}$, \\ which originates from the nucleon nucleon collision at ($t_i ,z_i $) ,\\ and $v_i^\mu$ is the normal unit vector of the proper time hyperbola: \beq ( t-t_i )^2 - (z-z_i )^2 = \tau^2_0 . \eeq{5i3} $\Delta\tau_i$ is the proper time difference between\\ - the space time point of a collision ($t_i ,z_i $)\\ - and another space time point ($t,z$): $$ \Delta \tau_i^2 = (t-t_i )^2 - (z-z_i )^2 . $$ }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} In order to model a heavy ion collision of a given energy in [10,22]\\ the function $dN_c(y)/dy$ has a smooth cut off, at target and projectile rapidities. In pp and pA experiments, the net baryon charge is observed:\\ - around the original target and projectile rapidities.\\ It is convenient to introduce a similar source term for the baryons $$ \sigma_B = \sum_{i=coll.} {1\over{d_0^2}} {{dN_B(y)}\over{dy}} 2 \delta(\Delta\tau_i^2-\tau_0^2) , $$ and $dN_B(y)/dy$ is chosen to be a smooth function\\ - simulating the final baryon rapidity distribution. \LT This way the baryon charge is not conserved in the calculation\\ - - initially the baryons are not present, nor is the energy. \bigskip Eqs. (\ref{5i1}) and (\ref{5i2}) were solved numerically [22-24]:\\ \LT gradual increase of the energy density was obtained\\ - until the energy of the last NN collision contributed to the flow. \B At this moment in a U + U collision with beam rapidity y=3.4, the\\ - maximum energy density reached was $e$=5.8 GeV/fm$^3$ in the center. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.1cm} In the target rapidity region, however, the energy density remained below 1$\gfm$, (less than previous estimates). \vspace*{8cm} Figure 6.6 {\it Contour lines of energy density in the rapidity y, proper time variable $\hat{t}=\ln(\tau/\tau_0)$ plane for an U + U collision. From [22]} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.6cm} \subsection{Baryon recoil in the Bjorken model (*)} To overcome the baryon conservation problem\\ \LT a model including nuclear recoil \& baryon conservation [4]. \B Recoil: from acceleration in an external field $F^{\mu\nu}$.\\ - The physical picture: chromoelectric flux tubes or strings:\\ - - projectile and target parton clouds are "charged" up \LT\\ - - covariant constant color electric field in each flux tube \LT\\ - - field energy per unit length or effective string tension $\sigma^*$.\\ - - For p+p collisions we expect $\sigma^*$ = 1 GeV/fm$^3$.\\ - - For A+A collisions (random walk in color space) \LT much larger. - - Through pair production the color fields are then neutralized,\\ - - in the final state: pions distributed uniformly in rapidity.\\ - - However, the string tension also accelerates the partons\\ - - in the fragmentation regions. \LT Mechanism of baryon recoil.\\ - It can be described by modifying eqs.(\ref{5i1}-\ref{5i4}): \beq T^{\mu\nu},_\nu = \Sigma^\mu_\pi + F^{\mu\nu} N_\nu , \eeq{5i5} \beq N^\mu ,_\mu = 0 , \eeq{5.6} where the source term $\Sigma^\mu_\pi$ is due to pions alone, and \beq F^{\mu\nu} = \left( \begin{array}{cc} 0 & -\sigma^* \\ \sigma^* & 0 \end{array} \right) . \eeq{5i7} These equations have the advantage of:\\ - incorporating longitudinal growth, as well as\\ - exact baryon flux conservation. \\ - (via the introduction of an effective external field). }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.7cm} \B In the absence of the source term, $\Sigma^\mu$,\\ - the compression is due entirely to recoil in the external field.\\ - In the "dust" limit, - - (where the internal pressure, $P$, is neglected compared to $e$)\\ - - the alculation of the recoil compression is particularly simple: \beq T^{\mu\nu} = m^* n u^\mu u^\nu . \eeq{5i8} Since $m^*,_\mu = 0$, and $(nu^\mu ),_\mu = 0$, we obtain that \beq \partial_\tau u^\mu = {1 \over{m^*}} F^{\mu\nu} u_\nu . \eeq{5i9} In the target frame, the external field is turned on when\\ - the Lorentz contracted projectile nucleus passes by: for $t>z$\\ \B The solution of (\ref{5i9}) for the fluid flow rapidity, $y=y(\tau)$: \beq y(\tau) - y(\tau_0) = (\tau-\tau_0)/\tau^* , \eeq{5i10} where $\tau^* \equiv m^* / \sigma^*$.\\ \B The fluid flow velocity is then $$ u^\mu (\tau) = (\cosh(y(\tau)), \sinh(y(\tau))) \ . $$ \LT A fluid element at $(t_j ,z_j )$ with $u^\mu (0)= (\cosh(y_0), \sinh(y_0))$ initially, moves in absence of a source, $\Sigma^\mu_\pi$, along the hyperbola: \beq (z-z_j +\tau^* u^1(0))^2 - (t-t_j +\tau^* u^0(0))^2 = (\tau^*)^2 . \eeq{5i11} \setlength{\baselineskip}{9pt} {\normalsize\sf Note that the light cone variable, $x^- =t-z$, for the trajectory of the fluid cell is bounded between $x^-(0) < x^- < x^-(0)+\tau^* e^{- y_0}$. For our problem the boundary condition is $ y=0$ on the forward light cone, i.e., $x^-(0)=0$.} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \B The acceleration ceases:\\ - when the field is neutralized by pair production.\\ \B However, the energy stored in the field must be conserved\\ - The neutralization converts the field-energy into matter fields.\\ \LT source term, $\Sigma^\mu_\pi$, in (\ref{5i5}). - We assume that each struck nucleon contributes to a string\\ - - that neutralizes along a proper time curve characterized by $\tau$. - - We thus parametrize the i-th source function by eq. (\ref{5i4}),\\ - - except the baryon contribution, because these are always present. The energy density of the matter produced by the neutralization is\\ $\propto \sigma^*$, i.e. $\propto$ the initial field energy density. Using 'stopping power' studies [2,29-30], \LT\\ that the space-time structure of the energy deposition is very important.\\ \B The interplay between the baryon recoil and the energy deposition \LT much higher energy densities in the fragmentation region\\ - - than it was expected before.\\ \B The sudden energy-momentum deposition on a hypersurface,\\ \LT discontinuity of the flow pattern. In 2-dim. space-time the surface of neutralization is giveb by (\ref{5i3}).\\ The normal four-vector to this surface is $v_i^\mu$, at point $x^\mu$.\\ This is a time-like surface because $v^\mu v_\mu =+1$ \ \LT \\ We can use the formalism of time-like discontinuities for this problem.\\ This discontinuity, however, is caused by the source $\Sigma^\mu_\pi$,\\ \LT our conservation equations across the surface of discontinuity are: \beq [ n u^\mu v_{i \mu} ] = 0, \eeq{5i12} \beq [ T^{\mu\nu}, v_{i \nu} ] = {1 \over{d_0^2}} \int_{\delta V} d^4x \Sigma^\mu_\pi = e_\pi v_i^\mu . \eeq{5i13} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.2cm} We can now introduce an invariant (time-like) baryon current [31] across the i-th surface of discontinuity: \beq j_i = n u^\mu v_{i \mu} . \eeq{5i14} According to eq. (\ref{5i12}) $j_i$ is the same on both sides.\\ \B Now making the orthogonal projection of eq. (\ref{5i13}) \\ to the surface and using the definition of $j_i$ we obtain that \beq j^2 = ( [P] + e_\pi ) / [X] , \eeq{5i15} where $X = (e+P)/n^2$ is the generalized specific volume. \\ Eq. (\ref{5i15}) differs from the similar equation for standard shock waves\\ - in the sign, because of the time-like discontinuity,\\ - and also in the additive term arising from the source term (\ref{5i13}).\\ \B The parallel projection of eq. (\ref{5i13}) leads to $j^2= [(e+P)X]/[X^2]$.\\ - Eliminating $j_i$ from the above equations, \LT\\ \ \ The usual Taub adiabat, except the source.\\ \ \ The source now contributes additively to the pressure difference \beq ([P] + e_\pi ) (X_1+X_0) = [(e+P)X]. \eeq{5i16} We derived two invariant scalar equations (\ref{5i15}-\ref{5i16}) from (\ref{5i12}-\ref{5i13}).\\ The velocity four vectors appear explicitly only in the definition of the invariant baryon current, $j_i$ . Picture: The Lorentz contracted projectile sweeps through the target.\\ Meets 1st the 1st nucleon in the row \LT chromoelectric flux-tube\\ This pulls the struck nucleon until its neutralization, $\tau_0$ later.\\ The 2nd and ... nucleons are also pulled by the chromoelectric field,\\ the projectile forms when passing them. (Fig. 6.7) }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} The field created at contact with a previous nucleon will not result in extra recoil of the nucleons sitting deeper inside the target. Thus, the nucleon recoil is uniform initially. \vspace*{14truecm} Figure 6.7 {\it Space-time picture of the recoil and color neutralization. As a result of interactions between projectile and target at depth $z_0$ and $z_1$, two incoherent strings neutralize along hyperbolas indicated. The fluid cell initially at depth $z_1$ follows the dashed world line. From [4]} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} When the first chromoelectric tube neutralizes\\ - - it distributes its energy and momentum over the whole target\\ - - on a 3-dimensional hyper-surface\\ - - (which is represented by eq. (\ref{5i3}) in our 1+1 dimensional model). Previous to this moment\\ - - the momentum of the field and \\ - - - of the recoiling target nucleons was different. At the neutralization\\ - - these two components thermalize and\\ - - the matter will obtain a uniform momentum, and its energy will change. Our task is now first to find the space-time points\\ - - where the neutralization surface reaches each target nucleon.\\ - - For the i-th neutralization surface and for the j-th nucleon\\ - - we should solve the system of eqs. (\ref{5i3}) and (\ref{5i11}).\\ - - In the absence of intermediate sources the intersection is at $\tau_c$: \beq \tau_c /\tau^* = \cosh^{-1} \left( 1+\tau^2_0 \right) /\left( 2\tau^{*2} \right) . \eeq{5i17} \B The recoil rapidity at this point is $y_r =\tau_c /\tau^*$.\\ - At this point ($t_k ,z_k$ ) we know the:\\ recoil velocity of each baryon $u^\mu$, the density $n$, and effective mass $m^*$. Using eqs.(\ref{5i12},\ref{5i15},\ref{5i17}) we can determine the values of these quantities after crossing the discontinuity. Then we propagate the target baryons according to eq. (\ref{5i11}) until they reach the next neutralization surface, and so on. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} According to the above scenario the complete fluid dynamical problem was solved on a grid specifically suited to the description of the fragmentation region [4]. \B\B The results indicate that the\\ - energy density in the target increases $\sim$ linearly with the target depth,\\ - the deposited energy increases with increasing projectile mass. For a thicker projectile the field strength is greater\\ \LT larger recoil rapidities [2,29-33]:\\ - - $\Delta y = \Delta y(\nu_p )$,\\ - - where $\nu_p$ is the average number of wounded nucleons\\ - - in the projectile per inelastic collision (eg. in p+A). Since the empirical rapidity shift $\Delta y$ is the recoil rapidity,\\ \LT eq. (\ref{5i17}) is used to set the parameter $\tau^*$\\ - - characterizing the field strength. The resulting energy density and the divergence of the flow $u^\mu ,_\mu$,\\ is plotted for a fluid element initially at 6.9 fm depth in the target in Fig. 6.8. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} The flow divergence characterizes the compression,\\ because from the continuity equation: $(nu^\mu ),_\mu =0$ \LT\\ $\dot{n}/n = - u^\mu ,_\mu$ .\\ The dashed line is without baryon recoil [22] with $\nu_p =6$.\\ The small energy density obtained in [22] is due to the lack of recoil. \vspace*{12truecm} Figure 6.8 {\it The divergence of the flow (a) and the energy density (b) as a function of proper time for a fluid element in the target at $z_0=6.9fm$. $\nu_p$ is the projectile thickness in units of mean free path. From [4] } }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.5cm} \section{Spherical expansion} Up to now linear 1-dimensional solutions were discussed. \\ Another type of simple solutions are spherical 1-dimensional solutions:\\ for central relativistic heavy ion collisions at late expansion stage. {INITIAL CONDITION}\\ Initial conditions for such an expansion are the following\\ \B The sphere is at rest in the C.M. system,\\ \B Uniform density, pressure, temperature, etc. distribution,\\ \B $v_r (t=0) =0$, i.e. there is no radial flow initially. For a given reaction at a well defined C.M. energy\\ there is one free initial parameter: \\ - either the density, $n (t=0)$, or the temperature\\ - - (or some other thermodynamical quantity). If one of these is given the others can be calculated from energy conservation if the the beam energy and the EOS are known. {FINAL BREAK-UP CONDITION}\\ There is on more (free) parameter \\ - because the validity of the fluid-dynamical approach has to break down\\ - at late stages, when particle interactions are weak to maintain loc. EQ.\\ Usually the break-up density, $n_{BU}$ is chosen as $n_{BU} \approx 0.1-0.7 \ n_0$.\\ {\bf This determines, or at least influences the observables!} We will discuss 3 simple models for spherical expansion:\\ \B - The Fireball model,\\ \B - the Blast-Wave model, and\\ \B - an approximate time dependent FD solution. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.2cm} \subsection{Fireball model} This model was the first attempt [34,35] to describe the measured cross sections of HI collisions in a collective thermal model. \B Collective flow is not included in this model. \B The matter is assumed to be {\bf globally} thermalized\\ - - by the end of the reaction, and\\ the cross section is determined from the thermal momentum distributions\\ - - of the particles present in this final heat-bath. At this stage it is {\bf assumed}:\\ - $n = n_{BU}$, and that\\ - there is no flow. \B Then cross sections are easily calculated because,\\ from the energy conservation and the EOS, $e(n,T)$, \\ the temperature of the system, $T_{BU}$, can be determined: $$ \epsilon_{inc.}^{c.m.} = \frac { e(n_{BU},T_{BU}) } { n_{BU} }, $$ where $\varepsilon_{inc.}^{c.m.}$ is the c.m. energy per particle for the incoming beam.\\ Thus $T_{BU}$ can be calculated if we assume $n_{BU}$, or vice versa. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.6cm} It was assumed that the temperature is high and the density is low enough\\ \LT the matter is close to an ideal gas at break-up.\\ \LT the momentum distribution of the nucleons, $f(\vec{p})$, is known,\\ \ \ \ - - and can be measured by the detectors.\\ \LT $T_{BU}$ was directly measured by the energy spectra (???) $$ f \propto e^{-\varepsilon/T_{BU}} \ . $$ \B It turned out, however, that this was wrong or largely oversimplified:\\ - no collective flow was assumed,\\ - ideal gas EOS was assumed. \B This showed up in form of ``experimental'' problems, such as:\\ - Pion and proton "temperatures" were different,\\ - The "entropy" did not go to 0 when $E_{beam} \rightarrow \ 0$,\\ - Transverse flow was observed in noncentral collisions. This simple model was, nevertheless, very powerful and\\ - many basic facts were interpreted correctly by the model.\\ Even nowadays the ``temperature'' extracted from the data\\ \ \ \ in this simple fashion, is frequently used. \ \ \ It is called:\\ \B slope temperature, slope parameter or effective temperature,\\ indicating the fact that the\\ - c.m. energy spectrum of particles on a logarithmic plot\\ - is frequently a linearly decreasing curve or very close to it. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.7cm} \subsection{Blast-wave model} Thermalization can be reached at high densities (before break-up),\\ then a collective and (almost) adiabatic expansion follows.\\ This lasts until the matter becomes dilute. \B \LT Break-up state thus has:\\ - a collective expansion (radial flow) $+$ superposed thermal motion.\\ - [by Siemens, Rasmussen and Kapusta, 36,37]\\ - - Spherical flow with constant radial flow velocity, $u_r$, assumed! \B Later [38]: radial expansion was calculated numerically in a\\ - 1-dimensional spherically symmetric relativistic FD.\\ - Viscosity of nuclear matter was also taken into account, (weak) \B Initial state of the expansion was a uniform sphere,\\ - Temperature, $T_0$, \& density, $n_0$, from Rankine-Hugoniot rels.(s.5.5) % Fig. 6.9-10 \vspace*{7cm} Figure 6.9 {\it Time dependence of density and temperature profiles calculated in a relativistic, viscous, spherical fluid dynamical model for the reaction $Ar+KCl$ at 800 MeV/nucleon projectile energy. The break-up is gradual in this model, it happens layer by layer. From [38]} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.9cm} At the initial state: $\exists$ form of compressional energy. \B The break-up temperature, $T_{BU}$, is smaller than the initial, $T_0$,\\ - because the expansion is close to adiabatic,\\ - the int. energy is converted into collective flow (see Fig. 6.10). \B The fireball model temperature $T_{FB}$ is higher than $T_0$, \\ - because ideal gas EOS is assumed \LT no compressional energy \vspace*{8cm} Figure 6.10 {\it Dependence of the break-up velocity, $\beta$, (dashed curves)\ \ and\ of\ the\ break - up temperature, $T$,\ \ (full curves) on the break up radius in the same spherical fluid dynamical model. The subscripts $\eta_<$ and $\eta_>$ belong to constant viscosity, $6\ MeV/(fm^2\ c)$, and to temperature dependent viscosity, $[ 6+2 (T/MeV)^{1/2} ]\ MeV/(fm^2\ c)$, respectively. $T_{BW}$ and $\beta_{BW}$ correspond to the Blast-Wave model parameters, $T_0$ is the initial temperature and $T_{FB}$ is the temperature of the Fireball model. The temperature of the Blast-Wave model, $T_{BW}$ is close to the numerically obtained break up temperatures, $T_{BU}(t)$. From [38]} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} {BASIC ASSUMPTIONS OF THE BLAST WAVE MODEL} At the break-up \\ - the radial flow velocity is constant (free parameter),\\ - the local temperature, $T$, is constant\\ \ \ \ \ (not a free parameter because of energy conservation). Now part of the energy is in collective flow.\\ \LT the observed particle velocities have two components:\\ - thermal random velocity, parametrized by $T_{BU}$ \\ - collective flow velocity, $v$, or $\beta_{flow}$ . {\normalsize\sf Assumptions: debatable, ... . E.g. possible that weakly interacting particles leave the system before the general break-up \LT these particles represent an earlier stage of expansion, (higher $T$ and smaller $v$-expansion. In the original Blast Wave model this possibility was, however, not considered.} \B Particles of different masses, e.g. $m_{\pi} \ll m_p$, \LT\\ then the flow energies are such that, $$ E^{flow}_p \gg E^{flow}_\pi \ , $$ while the thermal energies are equal, $E^{therm} = {3\over 2}T_{BU}$. \B At the same time the flow velocities are equal,\\ but the random thermal velocities of the lighter particles are larger. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} As a consequence this model explains one basic feature of the observations. \B The energy spectra of pions decrease steeper than of proton spectra \vspace*{13.5cm} Figure 6.11 {\it Inclusive cross sections, $d^3\sigma/dp^3$, at $\Theta = 90^0$ in the c.m. for the $^{20}Ne$ + $NaF$ reaction at 800 MeV/nucleon laboratory beam energy. Open circles are protons, closed circles are $\pi^-$'s. Solid lines are the results of the Blast Wave model: the free parameter the flow velocity is chosen as $\beta_{flow}=0.373$, then the resulting temperature is $T=44$ MeV. The number of charged pions is 9.4\% of the protons. From [36]} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \subsection{An approximate spherical solution} An approximate solution for adiabatic (or dissipative) expansion [39,40]\\ - an alternate to the scaling spherical expansion model (sect. 6.4.2) Assumptions:\\ \B all thermodynamical quantities are uniform during the expansion,\\ \ \ \ \ \ - $P(r,t)=P(t), \ T(r,t)=T(t), \ n(r,t)=n(t)$, etc.,\\ \B the 4-velocity is linearly increasing with the radius $$ \vec{u}_r (\vec{r},t) = \gamma_r \vec{v}_r = \frac{\dot{R}(t)}{R(t)} \vec{r} $$ Time dependence of parameters: from energy conservation + 1 constraint\\ - simplest additional constraint: adiabatic expansion.\\ - - (all internal energy is converted into collective flow, $\sigma=C$)\\ - - (or viscosity \LT entropy increase. Extr.: flow does not increase) For adiabatic expansion ($\sigma=Const.$): the energy density \beq T^{00}(r) = [ e(\sigma,n) + P(\sigma,n) ] \gamma^2(r) - P(\sigma,n) \eeq{ase1} depends on the radius $r$. The total energy of the system is then \beq V [ (e+P) < \gamma^2 > - P ] = E^{tot}_{init.} \ , \eeq{ase2} where $ < \gamma^2 >$ is the volume average, (because the other quantities are assumed to be independent of $r$.) \beq < \gamma^2 > = \frac { 4 \pi \int_0^R {1 \over {1-v^2}} r^2 dr} { 4 \pi R^3 / 3} . \eeq{ase3} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Using the assumed linear velocity profile it follows that $v^2 = x^2 / (1+x^2)$ where $x = r\ \dot{R} / R$ \LT $$ < \gamma^2 > = \left[ 4 \pi {{R^3}\over{\dot{R}^3}} \int_0^{\dot{R}} (1+x^2) x^2 dx \right]/ ( 4 \pi R^3 / 3) = $$ \beq = {3\over{\dot{R}^3}} \int_0^{\dot{R}} (1+x^2) x^2 dx = 1+{3\over 5} \dot{R}^2 . \eeq{ase4} Inserting (\ref{ase4}) into (\ref{ase2}) we get an equation for the radius \beq \dot{R} = \sqrt{ \frac{{5 \over 3} \left( {{R_{in}^3}\over{R^3}} e_{in} - e \right)} {e+P} } \ , \eeq{ase7} where $e$ and $P$ depend on the density, $n(t)={{3 A_{tot}}\over{4\pi R^3}}$, only ($\sigma=Const.$) At the break-up density we can calculate single particle cross sections, etc. (see the next chapter). \B Most dissipative, isoergic ($\varepsilon=e/n=$const.) expansion can be calculated similarly.\\ - It should be preceded by a (close to) adiabatic expansion [40],\\ so that $e_{in} R_{in}^3 \ne e(t) R^3(t)$, - - otherwise the system would not expand. Unlike in the scaling spherical expansion model (see section 6.4.2) the expansion velocity of the surface is not constant in this model. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.4cm} \section{The Landau model} % At $E_{lab.} =$ 10-100 \agev\, Lorentz contraction \LT large $\gamma^{c.m.}$.\\ \B Landau's fluid-dynamical model [20,42] - initial condition:\\ static homogenous disk of aspect ratio $\gamma^{c.m.} \ (\sim 2-8)$.\\ \LT this disk is rather flat and orthogonal to the beam direction\\ - - the pressure gradient is the largest in the beam direction. \subsection{Physical assumptions} \B Strong interactions, short relaxation times \\ \ \ \LT \ \ LOCAL STATISTICAL EQUILIBRIUM. \\ (Mean free paths are much smaller than the characteristic lengths.)\\ \LT Classical fluid governed by relativistic hydrodynamics. \B Features of the model\\ - dynamics \& equation of state\\ - initial condition and\\ - final (break-up) boundary condition \B Initial condition: initial temperature distribution\\ \ \ \ (We assume baryon free matter for simplicity)\\ \ \ \ (Determine $T_0$ from the c.m. energy \& the relevant dynamics.) }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \B Large pressure \LT the system expands and cools.\\ - 1st stage: expansion can be approximated by linear expansion\\ - 2nd stage: when the system expanded in the beam direction\\ \ \ \ \ \ \ \ to a size comparable to its transverse diameter\\ \ \ \ \ \ \ \ will the transverse expansion, 2+1 dim. expansion.\\ - This separation of the expansion to two stages is somewhat artificial,\\ \ \ it is done for the sake of developing an approximate solution[43,44]\\ (This approximate analytic solution, however, is valid only for a restricted class of EOS's with small sound speeds [45]. The linear and the spherical expansion stages can be handled similarly to the Bjorken model. \B Originally the model was developed for $p+p$ collisions, but\\ - it is well applicable to HI collisions as well.\\ - initial condition: Lorentz-contracted disk of volume, $V_0$,\\ - - $V_0 \approx V_{rest.} / \gamma^{c.m.}$ \B If the system undergoes a large expansion,\\ \LT the initial energy distribution over $V_0$ is irrelevant, and\\ - - sufficient to assume: initial energy density is constant over $V_0$ \beq e_0 = \frac{E_{c.m.}}{V_0} . \eeq{coop1} The matter is at rest initially in this highly Lorentz contracted disk. \B The expansion will be initially much stronger in the beam direction\\ - - because the pressure gradients are much larger. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.4cm} \subsection{Quasi-analytic solution} \B Introduce the rapidity coordinate in the beam ($z-$) direction, $\alpha(z,t)$ \beq \alpha \equiv \frac{1}{2} \ln\left(\frac{t+z}{t-z}\right) \eeq{coop42} and the proper time coordinate, $$ \tau \equiv \sqrt{t^2-z^2} . $$ \B Introduce a parameter, $\beta$, to characterize the proper time with respect to the initial size of the system \beq \beta \equiv \ln \left( \frac{\tau}{r_0} \right), \eeq{coop42b} where $r_0$ is characterizing the Lorentz contracted initial disk,\\ - \ \ \ \ \ \ \ \ \ \ $r_0 \approx d /\gamma^{c.m.}$,\\ and $d$ is the diameter of the colliding object.\\ For protons \ \ \ \ $d \approx m_\pi^{-1}$. \B Introduce the rapidity of the flow, $\eta$, and\\ - the rapidity of the particles emitted, $y$.\\ NOT the SAME, even at the end of the collision (thermal motion). \beq \eta \equiv {\sf arth} v_\parallel = {\sf arth} v_z \eeq{coop2} If we have a scaling solution, (e.g. Bjorken model),\\ the flow rapidity and the coordinate rapidity are equal, \\ -\ \ \ \ \ \ \ \ \ \ \ \ \ $\alpha = \eta$,\\ which means that the flow velocity satisfies the equation \beq v_z = z/t . \eeq{coop41} }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} This parametrization is also possible for a scaling {SPHERICAL} solution,\\ where we just replace the coordinate $z$, with the radial coordinate $r$.\\ In this case the scaling solution implies that $v_r = r/t$ and\\ the radial coordinate rapidity, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\alpha_r$,\\ will be equal with the radial flow rapidity $\eta_r$. In 1-dim. scaling solutions the introduction of these radial coordinates is possible. In both of these scaling cases the FD equations reduce to [45] \beq \partial_\beta \ln T + \lambda c_0^2 = 0, \ \ \ \ \partial_\alpha \ln T = 0, \eeq{coop45} or \beq {{ \partial e}\over{\partial \tau}} = - \lambda {{(e+P)}\over \tau} , \eeq{coop45b} where $$ \lambda = \left\{ \begin{array}{ll} 1, & {\sf linear\ one-dimensional\ expansion} \\ 3, & {\sf spherical\ expansion} \end{array} \right. $$ and $c_0$ is the sound speed of the EOS, $dp/de = c_0^2 $.\\ The second equation in (\ref{coop45}) expresses the assumption that\\ - the thermodynamical quantities, i.e. $T$, are independent of $\alpha$ ! For simplicity we assume that the sound speed is $dp/de = c_0^2 $= const.\\ Then eqs. (\ref{coop45}) yield the solution \beqar \frac{T}{T_0} &=& \left( \frac{\tau}{r_0}\right)^{-\lambda c_0^2} ,\nonumber \\ \frac{e}{e_0} &=& \left( \frac{\tau}{r_0}\right)^{-\lambda (1+ c_0^2)} , \eeqar{coop47} where $T_0$ and $e_0$ are the initial temperature and energy density.\\ This result is formally the generalization of the result obtained at the introduction of the Bjorken model above. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} For a 3-dim. solution eq. (\ref{coop41}) is never valid, \\ although as we will see, when the beam energy increases the solution\\ - will approach the scaling solution even in the Landau model! In Landau's approximate three dimensional solution\\ - the 1st stage of the expansion is approximated as a linear expansion\\ - when the system expanded in the beam direction to a size\\ \ \ \ \ \ \ \ comparable to the transverse diameter \LT\\ \ \ \ \ \ \ \ the transverse expansion is also considered. When the energy density reduces to the freeze-out (or break-up) density, \\ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ $e_{bu} \approx e_\pi \approx 1\ \gfm$,\\ - the number of particles becomes a well-defined quantity.\\ - the fluid "breaks up" into quasi-free final particles.\\ - (We can make this criterion more specific for a given equation of state.) \B The break-up criterion defines a space-time surface, $\sigma$,\\ - which is an isotherm, $T(\vec{r},t) = T_{bu}$.\\ - Along this surface $N_\pi$ is well defined, and\\ - the distribution of energy and particles can be calcualted. If we neglect the random thermal velocities, \LT \\ - the distribution of particle rapidities, $y$, at break up\\ - will follow the flow rapidity, $\eta$, distribution [45]: \beq \frac{dN}{d\eta} \approx \pi d^2 n_{bu} \left(\frac{e_0}{e_{bu}}\right)^{1/(1+c_0^2)} \times r_0\ \exp\left( \frac{- \eta^2}{2 |L|}\right) , \eeq{coop63} where $$ L = \frac{2 c_0^2}{1-c_0^4} \ln \left(\frac{e_0}{e_{bu}}\right) , $$ and $n_{bu}$ is the break up density of particles. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.2cm} \subsection{Numerical solution} Cooper \etal\ provided a numerical solution of the model, [45]\\ The flow pattern in the middle of the system is \ \ \ \ similar\\ - to the flow pattern of the Bjorken model,\\ - - if the initial energy density or temperature are high. \vspace*{11cm} Figure 6.12 {\it The isotherms of the flow at the break up when the temperature reaches $T_{bu}$($=T_c$ in the figure). The ratio of the initial temperature, $T_0$, to the final temperature, $T_{bu}$, is the parameter of the contours. The solution depends on the sound speed, $c_0$, and on the initial thickness of the disk, $r_0$. As the initial temperature increases the solution at break up approaches the Bjorken model solution, particularly for a soft EOS like $c_0^2 = 1/6$. From [45] } }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-1.2cm} . \vspace*{13cm} Figure 6.12 continued In the 1-dim. case an exact solution is known for the expansion [45]\\ The motion of the leading edge is a progressive wave,\\ in which $\eta$ and $T$ are related by \beq \ln(T/T_0) = - c_0 \eta. \eeq{coop3} In the 3-dim. case no such analytic solution is known. However,\\ - very short time after the beginning of the spherical expansion\\ - \ \ \ this behaviour still holds. }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} Thus the boundary conditions can be chosen somewhat arbitrarily so that no discontinuity appears in the temperature distribution after the beginning of the expansion. \\ \LT The initial conditions at $t=0$ are set as \beq T(z,0)=T_0, \ \ \ \eta=0 ,\ \ \ {\sf for} \ \ \ 0 r_0/c_0$ the whole fluid is in motion,\\ - 3-dim. expansion \LT cools very rapidly.\\ \LT early break-up ($L_{BU} \sim 3 r_0$)\\ - the critical isotherm becomes time-like after $t>r_0/c_0$, and\\ - for very high initial $T$ it looks like a hyperbola. {\normalsize\sf Similarly to the behavior of the fluid in the central region of the 1-dim. Bjorken problem. However, in the one-dimensional problem the expansion takes much longer, and the isotherm in the central region is closer to a hyperbola with asymptotes close to the light cone.} We can display the critical isotherm\\ - \ \ \ \ \ \ $T(z,t) = T_{bu}=m_\pi$\\ - for $c_0^2=1/3$ and for various $T_0$ values in Fig. 6.12a.\\ \B for $T_0 < 2m_\pi$, there is no isotherm looking like $\tau =$constant.\\ \B for $T_0 = 4 m_\pi$, we see the beginnings of such behavior.\\ \B even at $T_0 = 4 m_\pi$, the outer edge of the isotherm is\\ \ \ \ \ \ far from a hyperboloid, \\ \ \ \ \ \ and that is where most of the entropy lies. For $c_0^2 =1/6$ the cooling is slower (Fig. 6.12b)\\ \B for $T_0/T_{bu}=3.8$ and $z<3r_0$ the isotherm is almost a hyperbola.\\ - Thus for small $c_0^2$ the cooling is slow enough \LT scaling. At 15 and 60 \agev, for symmetric central HI collisions full scale numerical 3-dim. relativistic FD models reproduce the prediction of the Landau model [46]. This is with collision initial condition ! \LT\\ - fine details of the initial condition are irrelevant by the end!\\ - (if sufficient room for equilibration and thermalization) }%end-transparency \newpage % transparency ===================================================== \transparencyframe{ \vspace*{-0.8cm} \section{Assignment 6} \begin{description} \item[6.a] Calculate the temperature of the QCD plasma generated in a detonation front from normal nuclear matter (for help see ref. [3]) and determine the threshold beam energy for QCD plasma creation considering that $T>0$ is required! \item[6.b] Solve the Taub adiabat for an ultra-relativistic ideal gas with the EOS $$ P = {e \over 3} , $$ and determine the density increase, ${n \over{n_0}} $ as a function of $\gamma_{beam}^{(CM)}$ , for a given $\gamma_{beam}^{(CM)}$ of the incoming beam, if $\gamma_{beam}^{(CM)}\gg 1$. The initial state is normal nuclear matter of $n_0, e=n_0 m_N, P_0 = 0$. \item[6.c] Solve the Taub adiabat and the Rayleigh line equations for the relativistic ideal gas in the low temperature limit having the EOS: $$ e = n (m_N + {3 \over 2} T ) , $$ and show that ${n \over {n_0}} \rightarrow \infty$ if the beam energy increases. The initial state is normal nuclear matter. \item[6.d] Show that $$ {{dS}\over{dy}} = {\sf constant} , $$ for the solution of the Bjorken hydrodynamical model. \end{description} }%end tr-page % transparency ================================================================ \end{document}