\documentclass{book} \usepackage[dvips]{epsfig,color} \usepackage[dvips]{graphicx, color} \renewcommand\baselinestretch{1.5} \setlength{\textwidth}{6.5in} \setlength{\textheight}{9.00in} \setlength{\oddsidemargin}{0.01in} \setlength{\evensidemargin}{0.01in} \begin{document} \tableofcontents \date{29 October 2000} \chapter {Pion Nucleon scattering} \label{chappn}%======================== Let us begin with a quick summary of partial waves. \section{Review of partial waves}%======================== \subsection{Spin 0 $\times $ Spin 0}%======================== That is for spinless target and projectile. An excellent theoretical treatment of the main topics in this chapter can be found in \cite{QMII} The radial part of \index{Schr\"odinger equation} Schr\"odinger equation: \begin{equation} \Bigl ( {d^2\over dr^2}+k^2\Bigr )\,u_l(kr)=\Bigl [ U(r)+ {l(l+1)\over r^2}\Bigr ]\,u_l(kr), \end{equation} The solution for a V=0 potential are given in terms of spherical Bessel and Neumann functions: \begin{equation} F_l(kr)=k\,r\,j_l(kr), \ \ \ \ \ \ \ \ G_l(kr) = -k\,r\,n_l(kr). \end{equation} Asymtotically these functions go as: \begin{equation} \sin\,(kr-l\pi/2), \ \ \ \ \ {\rm and} \ \ \ \ \ \ \,\cos(kr-l\pi/2). \end{equation} respectively. $F_l$ is called the regular solution and $G_l$ is the irregular one. \index {Regular solution} \index{Irregular solution} \index{Phase shifts} \index{Free particle} For finite potentials the asymtotic wave function has the form: \begin{equation} u_l(kr) \sim e^{i\delta_l}\,\sin(kr-l\pi/2+\delta_l). \label{phasesh} \end{equation} Comparing the two forms of the asymtotic wave function with and without potential, equation (\ref{phasesh}) shows that the wave fucntions has been shifted by and angle $\delta_l$, that is why the $\delta_l$'s are called phase shifts. \index {Radial equation} The $\delta$'s can be complex so that can be expresed as: $\delta_l=\delta_{l,R}+i\,\delta_{l,I}$ and we define the absorption parameter $\eta$ as: \begin{equation} \eta_l=e^{-2\,\delta_{l,I}}. \end{equation} There are groups in some countries in the world that analyze thousand of available data of scattering experiments and find the phase shifts for different particle-particle or particle-nucleus interactions. Once the phase shifts are available, the scattering amplitude $f_E(\theta)$ can \index{Scattering amplitude} for particles with no spin scattered by spinless targets, be obtained: \begin{equation} f_E(\theta)=\sum_{l=0}^{\infty} (2\,l+1)\,P_l(\cos\,\theta){(\eta_le^{2i \delta_l}-1)\over 2ik}, \end{equation} and the differetnial cross section: \begin{equation} {d\sigma \over d\Omega}=\vert f_E(\theta)\vert^2. \end{equation} The total cross section $\sigma_t$ can also be found using: \begin{equation} \sigma_t={4\pi \over k^2}\,\sum_{l=0}^{\infty}(2l+1)[1-\,\eta_l\,\cos 2\,\delta_l]. \end{equation} \subsection{Spin 0 $\times $ Spin 1/2}%======================================== There are are now two pieces for the scattering amplitude, the spin-nonflip amplitude $f(\theta)$, and spin-flip amplitude $g(\theta)$, they are: \begin{equation} f(\theta)= {1\over k}\sum_{l=0}^{\infty} \Bigl [(l+1)\,e^{\delta_l^{l+1/2}} \sin\,\delta_l + l\,e^{{i\delta_l}^{l-1/2}}\,\sin\, \delta_l\Bigr ]P_l(\cos \,\theta), \end{equation} and: \begin{equation} g(\theta)={1\over k}\,\sum^{\infty}_{l=1}\Bigl [ e^{i\delta_l^{l+1/2}} \,\sin\,\delta_l^{l+1/2}-e^{i\delta_l^{l-1/2}}\,\sin\,\delta_l^{l-1/2} \Bigr ] {dP_l(\cos\,\theta)\over d(\cos\,\theta)}. \end{equation} \section{Pion-nucleon phase shifts}%======================================== \subsection{Argand diagrams}%======================================== Figure (\ref{argand}) illustrates an Argand diagram; it is a diagram in the complex plane, the vertical axis is the Imaginary part of the $f_l$ amplitude, \index{Argand diagram} and the horizontal axis is the Real its real part. The phase shift for the l-th partial wave, $\delta_l$ multiplied by two is the central angle as shown in figure (\ref{argand}). The absortion parameter $\eta$/2 is the line shown in the same figure. When the absorption parameter has its maximum value 1, $f_l$ is at the circumference. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{argand.eps}} \end{center} \caption{Parameters involved in an Argand diagram, and the geometrical interpretation. Central angle is $2\delta_l$. The absorption parameter $\eta$ and the amplitude $f_l$ are shown in this graph. This figure is {\tt argand.eps} in LatexM} \label{argand} \end{figure} \subsection{Almehed and Lovelace phase shifts}%===================== We use file {\tt delal} which contains the phaseshifts and absorption parameters $\eta$ for $\pi$-nucleon scattering for different center of mass energies; \index{Pion-nucleon phaseshifts} the file was written with the parameters given by \cite{almehed}. This file contains 71 sets of $\eta$'s and $\delta$'s (in degrees) for the same number of energies. The order in which the wave parameters appear is (from the Lpott code): \begin{verbatim} c nwave code c 1-----s31 2-----p31 3-----p33 4-----s11 c 5-----p11 6-----p13 7-----d33 8-----d35 c 9-----f35 10----f37 11----d13 12----d15 c 13----f15 14----f17 c \end{verbatim} To understand some of the physics involved, we wrote a program {\tt chap1.f} that reads file {\tt delal}, converts $\delta$'s in radians and then calls routine {\tt argand}: \begin{verbatim} Subroutine argand(eta,d,retf,aimtf,nes) c ecm, eta, d, nes are inputs to the routine c Returns args. retf and aimtf and also c the output is a file "chap1ar.dat" with c the 2 columns real and imag. parts of partial c amplitudes f's. With this file you can c plot Argand diagrams. The f's for each wave are c f=-0.5i(eta*exp(2id -1)) c separating the real and imaginary parts c Compare with S.Almehed and C Lovelace p 163 c i=0 :for Argand diagram c select below the nwave you like (from 1 to 14) Implicit Real*8 (a-h,o-z) Real*8 mpi Dimension eta(14,100),d(14,100),retf(14,100) Dimension aimtf(14,100) pi=3.141592 mpi=139.578 Open(6,file='chap1ar.dat', status='unknown') Do nwave=1,14 Do ne=1,nes twodel=2.0*d(nwave,ne) retf(nwave,ne)=0.5*eta(nwave,ne)*sin(twodel) aimtf(nwave,ne)=0.5*(1.0-eta(nwave,ne)*cos(twodel)) c retf(nwave,ne)=0.5*sin(twodel) ! eta=1 c aimtf(nwave,ne)=0.5*(1.0-cos(twodel)) !eta=1 End Do End Do nwave=2 Do ne=1,nes-1 Write(6,100)retf(nwave,ne),aimtf(nwave,ne)!argand dia End Do 100 Format(1h ,e15.5,e15.4) Close(6) Return End \end{verbatim} which computes separately the real and imaginary parts of the partial amplitudes for diferent energies: (see for example Judah M Eisenberg and E S. Koltun, (1980), ``Theory of meson interactions with nuclei'', Wiley-Interscience, New York.) \begin{equation} f= {1 \over 2i}\,(\eta \exp \,2i\delta -1), \label{argand} \end{equation} the routine writes the output in a disk file {\tt chap1ar.dat}. The output contains two columns, the real and imaginary parts of eq. (\ref{argand}), with this file you can plot $Im(f)$ versus $Re(f)$ and the figure is called an {\em Argand diagram}. In this routine you can select the partial wave you like in line \begin{verbatim} nwave=2 \end{verbatim} changing the 2 by any number 1 to 14 (see the code for partial waves above). It is convenient to plot in the same Argand diagram, a circle with radius 0.5 and centered at point (0.5,0.5) so that scales the $x$ and $y$ axis properly and shows the diagrams as the look in papes (otherwise it plots the part of the diagrams between the limit extreme points, not necessarily (0.0,0.0), (-0.5,0.5), (0.0,1.0),(0.5,0.5). {\bf Exercise 1} Each time you get an output file, observe the first and the last pairs of x and y points, to have an idea of where the graphs starts and where it ends. In figures (\ref{ars31}), (\ref{arp31}) and (\ref{arp33}) we have the Argand diagrams for waves S31, P31 and P33. From the data files for each output you can see that S31 starts its drawing at (-0.219,0.50) and ends at (-0.085,0.56), that means the graph is plot in the clockwise sense. For P31, the diagram starts at (-0.075,0.0057) and ends at (-0.172,0.5539), looking at the diagram we conclude that is also plot in the clockwise sense, but diagram P33 is special, it starts at point (0.33,0.87) and ends at (-0.0799,0.22) that means that the diagram is plot in the counterclockwise sense, something characteristic of a resonance, also, the graph passes very quickly through the point (0.0,1.0). {\bf Exercise 2} Find other waves which have resonances. This figure shows the original Argand diagrams from S. Almehed and C. Lovelace to be compared with your graphs. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{S31.eps}} \end{center} \caption{Argand diagram for S31 wave M this figure is S31.eps in latex M} \label{ars31} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{P31.eps}} \end{center} \caption{Argand diagram for P31 wave this figure is P31.eps in Latex } \label{arp31} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{P33.eps}} \end{center} \caption{Argand diagram for P33 wave M This figure is P33.eps in Latex M } \label{arp33} \end{figure} The inelasticity parameter $\eta$, it is in the range: $0\leq\eta \leq 1$ determines the fraction of flux that is lost from the elastic channel (in which flux is conserved and $\eta=1$.). Then it can be fun to replace in the code {\tt eta(nwave,ne)} by 1 and plot again to see how an elastic channel appears (of course it has no physical meaning to use these data from real experiments but just to illustrate an elastic channel), f \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{S31el.ps}} \end{center} \caption{Argand diagram for ``elastic'' scattering S31 wave} \label{ars31el} \end{figure} you can see from fig. (\ref{ars31el}) that the plot is mainly on the the circunference. In figure (\ref{rfp33}) we have the real part of the amplitud for the partial wave P33, as a function of the center of mass energy. And in figure (\ref{ifp33}) we see the imaginary part of the same wave as function of the com energy. Comparing the two graphs we see at the energy in which the real part of the partial wave is 0, the imaginary part is at its maximum, and this is the point where resonance occurs as also shown in figure (\ref{arp33}). \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{rfP33.eps}} \end{center} \caption{Real part of the amplitude for wave P33 MThis figure is {\tt rfP33.eps} in Latex} \label{rfp33} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{ifP33.eps}} \end{center} \caption{Imaginary part of the amplitude for wave P33 This figure is {\tt ifP33.eps} in Latex } \label{ifp33} \end{figure} Since in textbooks the figures for the total cross section for $\pi$-nucleon scattering are shown as functions of the laboratory pion momentum $p_{\pi}$, (or its kinetic energy) it is necessary to find this momentum in terms of the center of mass energies $E_{cm}$ which are read by our program. The expression for the total cross section can be computed from relations (28) and (107a) in Lpott: \begin{equation} \sigma^{tot}={4\pi \over \kappa_0^2} \, \sum_{L=0}^{\infty} [(L+1)\,Im\,T_{I=3/2 \, J=L+1/2} +L\,Im\,T_{I=3/2\, J=L-1/2}], \end{equation} with \begin{equation} {\rm T}_{I\,L\pm}= {\eta_{I\,L\pm}\exp\,2i\delta_{R\, IL\pm}-1 \over 2i} \end{equation} with I=3/2 for $\pi-p$ and $\kappa$ is the center of mass pion momentum, which also needs to be found in term of the given center of mass energy $E_{cm}$, and the masses $m_N$ for the proton and $m_{\pi}$ for the pion, with the relations: \begin{equation} s=E_{cm}^2 \end{equation} \begin{equation} \kappa=\sqrt{(s-(m_N+m_{\pi})(m_N+m_{\pi}))(s-(m_N-m_{\pi})(m_N-m_{\pi}))}, \end{equation} We now want find the center pion laboratory momentum $p{\pi}$ in terms of the center of mass energy $E_{cm}$, and the pion and nucleon masses. Using the invariance of the four momentum squared in laboratory and center of mass reference frames: \begin{equation} ((p_{\pi}+p_N, E_{\pi}+E_N)^L) ^2=(({\bf p}_{\pi}+{\bf p}_N=0,E_{cm})^{cm})^2, \end{equation} or: with $E_N^L=m_N$ and $E_{\pi}^L=\sqrt{p_{\pi}^2+m_{\pi}^2}$, we have: \begin{equation} E_{\pi}^L= {E_{cm}^2-m_{\pi}^2-m_N^2\over 2m_N}, \end{equation} The pion laboratory kinetic energy is then $E_{\pi}^L-m_{\pi}^2$. and finally: \begin{equation} p_{\pi}=\sqrt{E_{\pi}-m_{\pi}^2}. \end{equation} The routine that computes the total cross section or any partial wave contribution to total cross sections is below: \begin{verbatim} Subroutine crossect(ecm,retf,aimtf,nener) c input: ecm aimtf and c nener: number of com energies and aimmtf's needed c the output is a file "chap1st.dat" c that contains either the total cross sect. versus pi-lab momemt. c or the contribution of any partial wave to the cross section c or tpilab and reft for a wave c or tplilab and aimtf for a wase c depending of your choice o i value c write(6,100) statement Implicit Real*8 (a-h,o-z) Real*8 kappa, mpi Dimension ecm(100), retf(14,100),aimtf(14,100) pi=3.141592 hbarc=197.3289 mpi=139.578 ! pion mass c fact : to give total cross section in mb Open(6,file='chap1st.dat', status='unknown') c call kinemat gives com enery (come) and c gets kappa: com pion momentum c pp: lab. pion momentum c tpilab: kinetic energy of the pion in lab system c i=0 tpilab and total cross section c i=1 tpilab and cntrbtn of any partial wave to crossection c i=2 tpilab and retf for a wave c i=3 tpilab and imtf for a wave i=3 Do ne=1,nener come=ecm(ne) Call kinemat(come,kappa,pp) tpilab=sqrt(pp*pp+mpi*mpi)-mpi fact=10.0*4.0*pi*(hbarc/kappa)**2 swve=aimtf(1,ne) pwve=2.0*aimtf(3,ne)+aimtf(2,ne) dwve=3.0*aimtf(8,ne)+2.0*aimtf(7,ne) fwve=4.0*aimtf(10,ne)+3.0*aimtf(9,ne) sigtot=fact*(swve+pwve+dwve+fwve) c first write for sigma total c second write for contribution of a wave to sigma total c uncomment one and comment the other according to your c choice if(i.eq.0)Write(6,100) tpilab,sigtot if(i.eq.1)Write(6,100)tpilab,fact*2.0*aimtf(3,ne) if(i.eq.2)Write(6,100)tpilab,retf(3,ne) if(i.eq.3)Write(6,100)tpilab,aimtf(3,ne) End Do Close(6) 100 Format(1h ,e15.5,e15.4) Return End \end{verbatim} \begin{itemize} \item Statement: {\tt write(6,100)} selects the output according to your choice it plots either the total cross section as a function of the laboratory pion momentum or the contribution of any partial wave to the total cross sections also as a function of the pion momentum in the lab system . \item This routine call routine {\tt kinemat} that computes the com pion momentum ({\tt kappa}) in terms of the center of mass energy, pion and nucleon masses, and also gives the pion momentum ({\tt pp}) in the lab system. \end{itemize} In figure (\ref{sigtot}) we have the total cross section for $\pi$-proton scattering as a function of the pion laboratory momentum in (Mev/c). This figure shows the features: \begin{itemize} \item The total cross section for low energies (momenta) are not present, the phase shifts are not given for low center of mass energies. \item There is a big resonance in the vicinity of a laboratory pion momentum of 300 (MeV/c). \item The cross section depends greatly on the laboratory momentum. \item the resonance occurs at the same laboratory pion momentum in which the zero occurs in the real part of the partial wave amplitude P33 and the maximum of the imaginary part of the partial wave amplitude P33. See figures (\ref{rfp33}) and (\ref{ifp33}). \item Then is interesting to see how is the contribution of the partial wave P33 to the total cross section. \item There are other two resonances in this graph. \end{itemize} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{sigtot.eps}} \end{center} \caption{Total crosssection for $\pi$-proton scattering M This figure is {\tt sigtot.eps} in Latex M} \label{sigtot} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{sigp33.eps}} \end{center} \caption{Contribution of wave P33 to total crosssection for $\pi$-proton scattering M This graph is {\tt sigp33.eps} in Latex M } \label{sigp33} \end{figure} From figure (\ref{sigp33}) you can see that the resonance is due mainly to the P33 wave. \subsection{Rowe, Salomon and Landau phase shifts}%====================== These three scientists supplied the phase shits for $\pi$-nucleon scattering at low energies. See Fig. (6) in Lpott, and G Rowe, M. Salomon and R. H. Landau Phys. Rev. C18 (1978) 445. The following is a routine that calls the original routine {\tt tpirsl} and outputs the new partial wave parameters: $\eta$'s and $\delta$ 's and the inclusion of new energies in the first 25 values for {\tt ecm} now called {\tt cecm}. \begin{verbatim} Subroutine rosalan(come,ceta,cd,cecm) c Rowe, Salomon, Landau partial amplitudes c Phys.Rev C 18, (1978), 584-589 c input: come=ecm(1) c output: ceta(14,100), cd(14,100) and cecm(100) c eta, delta and ecm with its first 25 data including c the RSL phase shifts and corresponding com energies c parts taken from routine tpicm from lpott c the output is a file "chap1rsl.dat" with c first column: pp: lab pion momentum c second column: cd: phase shift in degrees for a c freely selected wave (1 to 14), 3 in this case Implicit Real*8 (a-h,o-z) Real*8 kappa,kaph,dumkap Dimension ceta(14,100),cd(14,100),cecm(100) Open(6,file='chap1prsl.dat', status='unknown') pi=3.141592 Call kinemat(come,kappa,pp) If (kappa.le.5.) kappa = 6. kaph = (kappa-5.)/20. kappa = 0. c Do nkap=1,25 ! as in tpicm c comment last line and uncomment next to plot phashifts Do nkap=1,51 !for com pi momentum, up to 51 for P33graph kappa = kappa+kaph If (nkap.le.6) kappa = (nkap*1.5) Do ni=1,14 !for partial waves Call tpirsl (kappa,ni,ddum,w,tr,ti) c w is given by last call is com energy ceta(ni,nkap) = 1. cd(ni,nkap) = ddum*180./pi c tangent jumps at 90 prom + to -, correct to sum phases c for ni=3 if((ni.eq.3).and.(cd(3,nkap).lt.0.0)) cd(3,nkap)= 1 180.0+cd(3,nkap) End Do cecm(nkap) = w c input w in kinemat and use pp: lab pion kin. ener. Call kinemat(w,dumkap,pp) c To plot S31 P31,P33, P11 phaseshits in degrees Write(6,*)pp,cd(3,nkap) !here End Do Close(6) Return End \end{verbatim} \begin{itemize} \item The routine {\tt rosalan} in program {\tt chap1.f} outputs the phaseshifts at low energies. \item It calls routine {\tt tpirsl} (taken from {\tt lpott}) which produces the phase shifts for waves S11, S31, P11, P13, P31, P33, D13 and D15, but does not change other waves. They are in the same numerical order we are using. \item \item The routine corrects the jumps that are originated when the angles goes trough 90$^o$ originated by the fact that tangent changes sign abruptly when crosses 90$^o$, so that the angles beyod 90 degrees are adden, with out showing the sing change. \item The original routine {\tt tpirsl} was left without changes. This routine takes into account the order of the partial waves, given at the beginning of this chapter and fits the experimental data with a formula. Also computes the pion center of mass energy in terms of the momenta of the pion and the nucleon in the center of mass system. \end{itemize} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{p33rsl.eps}} \end{center} \caption{$\delta_{33}$ phaseshift in degrees as a function of lab $\pi$ momentum this figure is {\tt p33rsl.eps} in Latex } \label{p33rsl} \end{figure} As an {\bf exercise} Plot the phaseshifts corresponding to waves S31 and P31 (in the same graph). Figure (\ref{p33rsl}) shows the P33 wave phase shitfs in degrees versus the the center of mass pion momentum. This graph can be compared with another in G Rowe, M. Salomon and R. H. Landau, Phys. Rev, C18 (1978) 584. To see the effects of the RSL phase shifts, we draw again in Figure (\ref{arp33rsl}) the Argand diagram using these phase shifts. \begin{itemize} \item Remember that this graph goes in the counterclockwise sense, so that the low energy data begin close to (0,0) and we see that since $\eta$ was taken as 1 (no absorption) the graph draws a circunference, but as the energy increases, the absorption increases and the graph leaves the circunference. \item Notice the difference with Figure (\ref{arp33}) in which the low energy phase shifts were not included. \item The effects of the low energy phase shifts can also be seen in Figure (\ref{sigtorsl}). The total cross section exhibits its shape at low pion momenta. \end{itemize} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfysize=2.6in \epsfbox{arp33rsl.eps}} \end{center} \caption{Argand diagram for P33 wave with RSL phase shifts M figure arp33rsl.eps in Latex} \label{arp33rsl} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{sigtorsl.eps}} \end{center} \caption{Total cross section for $\pi$-nucleon, as a function of pion laboratory momentum; computed with RSL phase shifts M this figure is {\tt sigtorsl.eps} in Latex M } \label{sigp33rsl} \end{figure} {\bf Exercise} Once the total cross section is found is easy to use formulas (lpott 29 and 30) to find the elastic and inelastic cross sections: \begin{equation} \sigma^{el}={4\pi\over k_0^2}\,\sum_{L=0}^{\infty}[(L+1)\vert T_{L+}\vert^2 +L\vert T_{L-}\vert^2], \end{equation} \begin{equation} \sigma_{{\rm inel}}=\sigma^{{\rm tot}}-\sigma^{{\rm el}}. \end{equation} \chapter {Analyzing input form factors} \label {formfac}%================= \section{Proton and neutron charge form factor}%======================= LPOTT needs also as input, the phase shifts and electromagnetic form factors for the nucleus for proton and neutron, and for the three nucleon system. \subsection{Cross section measurements}%============================= The theoreticians arrived at the electron-nucleon elastic scattering differential cross section, known as the Rosenbluth formula: \begin{equation} {d\sigma\over d\Omega}=A\,\Bigl [ {G_E^2+\tau G_M^2\over 1+\tau} \, {\rm ctg}^2\,(\theta/2)\,+ 2\tau G_M^2\Bigr ], \end{equation} where \begin{equation} A=\Bigl ({e^2\over 2E_0}\Bigr )^2 {1\over \sin^2(\theta/2)(1+ (2E_0/M)\sin^2(\theta/2))} =\Bigl ( {d\sigma\over d\Omega}\Bigr )_{NS}{\rm tg} {\theta\over 2}, \end{equation} and \begin{equation} \tau= {-q^2\over 4 M^2}. \end{equation} In the last formula $\theta$ is the scattering angle for the electron, $E_0$ its incident energy, $M$ is the nucleon mass and $G_E$, $E_M$ are the electric and magnetic form factors, respectively. For the electric charge form factor for protons and neutrons, LPOTT uses the fits by: B. Bartoli, F. Felicetti and V. Silvestrini (Rivista del Nuovo Cimento,{\bf 2} (1972) 214). The routine that computes these fits follows: \begin{verbatim} Subroutine ffpn (q2,fcp,fcn) c *** see RH Landau, Program lpott, 1981 c proton charge ff (also p mag and n mag If nrom(0)=1) c neutron charge form factor c q in f-1 Implicit Real*8 (a-h,m,o-z) c data x,un,mn/18.2,-1.91315,4.7548/ fcp = 1./((1.+q2/x)**2) fcn = -(un*q2*0.25/mn/mn)*fcp Return End \end{verbatim} \begin{itemize} \item Uses as input $q^2$, with $q$ the momentum transfer. \item The experimental data came from electron proton scattering. \item The output are the charge form factors for proton and neutron, {\tt fp} and {\tt fn} respectively. \item A program: {\tt ch2haj.f} was written to run this routine (and other more), the output is a disk file {\tt rhop.dat} which contains different sets ot data depending on what is writen in it. If {\tt ffpn} is called for a series of values of $q^2$ a graph like Figure (\ref{pnff}) is obtained. \end{itemize} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{pnff.eps}} \end{center} \caption{Proton and neutron charge form factors this is graph {\tt pnff.eps} in latex } \label{pnff} \end{figure} \section{Electric and magnetic form factors for three nucleon systems} %==================================================================== One of the options of program LPOTT is the use of several fits to the electromagnetic form factor. One of the possibilities is the use of the C. Hadjimichael, B. Goulard and R. Bornais, (Phys. Rev. C27, (1983),831) fits to the electromagnetic form factor for the three nucleons system. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{hehff.eps}} \end{center} \caption{$^3$He and $^3$H charge form factors M this figure is {\tt hehff.eps} in Latex } \label{hehff} \end{figure} In Figure (\ref{hehff}) we have the electric form factors for $^3$He and $^3$N and in Figure (\ref{hehmff}) their magnetic form factors, in both cases we are using the Hadjimichael fits in the impulse approximation. Compare with graphs (3) and (4) from Hadjimichael et. al. paper. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{hehmff.eps}} \end{center} \caption{$^3$He and $^3$H magnetic form factors m thisfigure is {\tt hehmff} in Latex } \label{hehmff} \end{figure} The matter and spin form factors are evaluated in terms of the charge and magnetic form factor for $^3$He, $^2$H, protons and neutrons using Lpott formulas (80)-(84) (expressed in a different way in routine {\tt ffhadj.f}). Figure (\ref{h3chmff} shows the spin and magnetic form factors for $^3$He. The absolute values of $\rho_{{\rm matter}}$ and $\rho_{{\rm spin}}$ form factor are plotted as functions of the moment transfer $q^2 ({\rm fm}^{-2}$, and can be compared to the ones in Rubin H. Landau (Annals of Physics, {\bf 92}, 205-231, Figure 1). \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{h3chmff.eps}} \end{center} \caption{$^3$He matter and spin form factors this figure is {\tt h3chmff.eps} in Latex } \label{h3chmff} \end{figure} The following program present in file {\tt ch2hadj.f} computes proton and neutron electric form factors. $^3He$ and $^3$H charge and magnetic form factors depending of the choice of the option {\tt nopt}. See at the beginning of the program the options. \begin{footnotesize} \begin{verbatim} program hadji c main program to run routine ffhadj c Hadjimichael c The output if file "rhop.dat" c The contents depends of nopt below c nopt=1 :proton and neutron electric ff c 2 :He3 and H3 charge ff c 3 :He3 and H3 magnetic ff c 4 :Charge He3 form factor Implicit Real*8 (a-h,o-z) q2mev=0.0 Open(6,FILE='rhop.dat', Status='Unknown') nif=1 !it is nifty(19) no MEC nopt=4 q=0.05 Do i=1,200 q2mev=(q*197.329)**2 !to give (Mev/c)**2 units qfer2=q*q If(nopt.eq.1) Then Call ffpn(qfer2,fp,fn) Write(6,*)q*q,fp,fn q=q+0.05 End If If((nopt.eq.2).or.(nopt.eq.3))Then Call ffmec(qfer2,fche3,fch3,fmhe3,fmh3,nif) if(nopt.eq.2)Write(6,*) q,abs(fche3),abs(fch3) if(nopt.eq.3)Write(6,*) q,abs(fmhe3),abs(fmh3) q=q+0.05 End If if(nopt.eq.4) Then Call ffhadj(q2mev,rp,rn,rpsp,rnsp,nif) Write(6,100) qfer2,abs(rp),abs(rn),abs(rpsp),abs(rnsp) q=q+0.024 End If End Do 100 Format(1h , 5(e13.5)) Close(6) Stop 'finished' End Subroutine ffhadj(q2mev,rp,rn,rpsp,rnsp,nif) c************************************************************************** c * c calculates the matter and spin form factors of the protons and neutron * c in helium-3 from the charge and magnetic form factors of he-3 and h-3 * c as calculate by hadjimichael et. al. in the impulse approximation. * c * c thus no meson currents are in the charge and magnetic form factors. * c * c************************************************************************** Implicit Real*8 (a-h,o-z) data up,un/2.79278,-1.91315/ c************************************************************************** c * c q2mev must be converted to inverse Fermis squared * c * c************************************************************************** qfer2 = q2mev/(197.329**2) c c c************************************************************************** c * c now calculate the form factor of the proton * c * c************************************************************************** c c Call ffpn(qfer2,fp,fn) c c c************************************************************************** c * c now calculate the charge and magnetic form factors of he-3 and h-3 * c * c************************************************************************** c c Call ffmec(qfer2,fche3,fch3,fmhe3,fmh3,nif) c c c************************************************************************** c * c now calculate matter and spin form factors using formulas of mach that * c relate these to the charge and magnetic form factors of he-3 and h-3. * c LPOTT eqs. 80- 83 * c************************************************************************** c c rp = fche3/fp rn = fch3/fp denom = fp*((up**2)-(un**2)) rpsp = ((up*un)*(fmhe3-fmh3))/(2*denom) up2 = up**2 un2 = un**2 rnsp = (up2*fmh3-un2*fmhe3)/denom Return End Subroutine ffmec(q2f2,ff1,ff2,ff3,ff4,n) c************************************************************************** c * c this Subroutine calculates the charge and magnetic form factors of * c he-3 and h-3 using the impulse approximation of hadjimichael et. al. * c (phys rev c vol 27, no.2, p831, feb 1983). * c * c************************************************************************** c****************************************************************************** c * c up dated on sept 9, 1984 to enable the calculation of the form factors * c from hadjimicheal et al. using the total charge, and magnetic form factors * c * c****************************************************************************** Implicit Real*8 (a-h,o-z) Real*8 lpt Dimension formfs(4,20),y(4),q2hadj(20) Dimension form1(20),form2(20),form3(20),form4(20) Dimension ftot1(20),ftot2(20),ftot3(20),ftot4(20) Dimension fhadj(4),a2(4) Dimension a2tot(4) c c data q2hadj/ 0.0, 0.25, 1.0, 2.25, 4.0, 6.25, 9.0, 12.25, 16.0, 1 20.25, 25.0, 30.25, 36.0, 42.25, 49.0, 56.25, 2 64.0, 72.25, 81.0, 90.25/ c data a2/ 39.474, 3.012, 16.667, 18.204/ c data a2tot/28.667, 21.928, 26.864, 83.324/ c data form1/1.0,0.865,0.571,0.308,0.141,0.552d-01,0.174d-01, 10.332d-02,-0.919d-03,-0.150d-02,-0.112d-02,-0.673d-03,-0.349d-03, 2-0.162d-03,-0.659d-04,-0.223d-4,-0.445d-05,0.133d-05,0.251d-05, 30.203d-5/ c data form2/1.0,0.868,0.604,0.347,0.170,0.719d-1,0.252d-01, a 0.612d-02, 1-0.294d-03,-0.171d-02,-0.152d-02,-0.102d-02,-0.595d-03, 2-0.316d-03,-0.154d-03,-0.696d-04,-0.285d-04,-0.101d-04,-0.261d-05, 3-0.691d-06/ c data form3/1.0,0.843,0.522,0.246,0.873d-01,0.163d-01,-0.719d-02, 1-0.107d-01,-0.818d-02,-0.494d-02,-0.257d-02,-0.116d-02,-0.419d-03, 2-0.826d-04,0.480d-04,0.798d-04,0.735d-04,0.554d-04,0.379d-04, 30.241d-04/ c data form4/1.0,0.861,0.566,0.298,0.130,0.462d-01,0.113d-01, a-0.313d-03, 1-0.274d-02,-0.232d-02,-0.143d-02,-0.740d-03,-0.321d-03,-0.107d-03, 2-0.104d-04,0.233d-04,0.296d-04,0.253d-04,0.188d-04,0.125d-04/ c data ftot1/1.0,0.849,0.556,0.294,0.129,0.445d-01,0.890d-02, a-0.276d-02,-0.480d-02,-0.376d-02,-0.232d-02,-0.123d-02, b-0.549d-03,-0.185d-03,-0.124d-04,0.532d-04,0.679d-04, c0.608d-04,0.475d-04,0.344d-04/ c data ftot2/1.0,0.868,0.604,0.347,0.169,0.705d-01,0.236d-01, a0.462d-02,-0.162d-02,-0.279d-02,-0.235d-02,-0.163d-02, b-0.102d-02,-0.611d-03,-0.355d-03,-0.206d-03,-0.121d-03, c-0.735d-04,-0.462d-04,-0.303d-04/ c data ftot3/1.0,0.863,0.576,0.317,0.151,0.625d-01, a0.221d-01,0.587d-02,0.358d-03,-0.990d-03,-0.975d-03, b-0.665d-03,-0.376d-03,-0.183d-03,-0.711d-04,-0.153d-04, c0.840d-05,0.148d-04,0.138d-04,0.978d-05/ c data ftot4/1.0,0.874,0.599,0.345,0.172,0.770d-01, a0.307d-01,0.108d-01,0.301d-02,0.349d-03,-0.343d-03, b-0.401d-03,-0.287d-03,-0.171d-03,-0.882d-04,-0.390d-04, c-0.132d-04,-0.160d-05,0.247d-05,0.276d-05/ c c c************************************************************************** c * c If q2f2 outside of table then extrapolate with * c * c f(q2) = f(q02)exp -(q2-q02)/a2 * c * c where f(q02) = last point in hadjimichaels table (ie. q02=90.25 or * c q = 9.5) * c * c************************************************************************** c c If (n .eq. 2) GoTo 10 Do 70 j=1,20 formfs(1,j) = form1(j) formfs(2,j) = form2(j) formfs(3,j) = form3(j) formfs(4,j) = form4(j) 70 Continue GoTo 15 10 Continue Do 75 j=1,20 formfs(1,j) = ftot1(j) formfs(2,j) = ftot2(j) formfs(3,j) = ftot3(j) formfs(4,j) = ftot4(j) 75 Continue 15 Continue If (q2f2 .lt. 90.25) GoTo 40 lpt = 90.25 If (n .eq.2) GoTo 25 Do 20 i=1,4 q2a2 = (q2f2-lpt)/a2(i) If (q2a2 .gt. 150.) q2a2 = 150. fhadj(i) = formfs(i,20)*exp(-q2a2) 20 Continue GoTo 30 25 Continue Do 26 i=1,4 q2a2 =(q2f2-lpt)/a2tot(i) If (q2a2 .gt. 150.) q2a2 = 150. fhadj(i) = formfs(i,20)*exp(-q2a2) 26 Continue 30 Continue ff1 = fhadj(1) ff2 = fhadj(2) ff3 = fhadj(3) ff4 = fhadj(4) Return 40 Continue If ((q2f2 .gt. 81.0) .or. (q2f2 .lt. 0.25)) GoTo 50 c c c************************************************************************* c * c for q2f2 less than 81.0 use four point lagrangian interpolation. * c * c************************************************************************* c c Call lagrng(q2f2,q2hadj,y,formfs,20,4,4,20,4) ff1 = y(1) ff2 = y(2) ff3 = y(3) ff4 = y(4) Return c c 50 Continue c c c************************************************************************** c * c for q2f2 less than 90.25 but greater than 81.0 must use 2-point * c lagrangian interpolation since in between the last two data points * c of table. * c * c************************************************************************** c c Call lagrng(q2f2,q2hadj,y,formfs,20,4,2,20,4) ff1 = y(1) ff2 = y(2) ff3 = y(3) ff4 = y(4) Return End Subroutine ffpn (q2,fcp,fcn) c *** see RH Landau, Program lpott, 1981 c proton charge ff (also p mag and n mag If nrom(0)=1) c neutron charge form factor c q in f-1 Implicit Real*8 (a-h,m,o-z) c data x,un,mn/18.2,-1.91315,4.7548/ fcp = 1./((1.+q2/x)**2) fcn = -(un*q2*0.25/mn/mn)*fcp Return End Subroutine lagrng (x,arg,y,val,ndim,nfs,nptx,maxarg,maxfs) c *** see RH Landau, Program lpott, 1981 c lagrange interpolation,unequally spaced points c npts=2,3,4,5,6, nfs functions(y*s) simultaneously interpltd c x= value of argument, arg is the tabulated x*s c y= a vector of interpolad functions, from tabulted val c ndim= Dimension of table c nfs= = of functions simult interpolated c maxarg and maxfs are maximum values of subscripts,ie Dimensions Implicit Real*8 (a-h,o-z) Dimension arg(maxarg), val(maxfs,maxarg), y(maxfs) c -----find x0 the Closest point to x npts = nptx c 2=st ni = 1 nf = ndim 10 If ((x.le.arg(ni)).or.(x.ge.arg(nf))) GoTo 30 If ((nf-ni+1).eq.2) GoTo 70 nmid = (nf+ni)/2 If (x.gt.arg(nmid)) GoTo 20 nf = nmid GoTo 10 20 ni = nmid GoTo 10 c ------ x is one of the tabltd values 30 If (x.le.arg(ni)) GoTo 60 nn = nf 40 nused = 0 Do 50 n=1,nfs y(n) = val(n,nn) 50 Continue Return 60 nn = ni GoTo 40 c ------- 2 pts left, choose smaller one 70 n0 = ni nn = npts-1 GoTo (140,110,100,90,80), nn 80 Continue If (((n0+3).gt.ndim).or.((n0-2).lt.1)) GoTo 90 nused = 6 GoTo 150 90 Continue If ((n0+2).gt.ndim) GoTo 110 If ((n0-2).lt.1) GoTo 100 nused = 5 GoTo 150 100 Continue If (((n0+2).gt.ndim).or.((n0-1).lt.1)) GoTo 110 nused = 4 GoTo 150 110 If ((n0+1).lt.ndim) GoTo 130 c n0=ndim,special case 120 nn = ndim GoTo 40 130 nused = 3 If ((n0-1).lt.1) nused = 2 GoTo 150 140 If ((n0+1).gt.ndim) GoTo 120 nused = 2 150 Continue c at least 2 pts left y0 = x-arg(n0) y1 = x-arg(n0+1) y01 = y1-y0 c0 = y1/y01 c1 = -y0/y01 If (nused.eq.2) GoTo 160 c at least 3 pts ym1 = x-arg(n0-1) y0m1 = ym1-y0 ym11 = y1-ym1 cm1 = -y0*y1/y0m1/ym11 c0 = c0*ym1/y0m1 c1 = -c1*ym1/ym11 If (nused.eq.3) GoTo 180 c ------at least 4 pts y2 = x-arg(n0+2) ym12 = y2-ym1 y02 = y2-y0 y12 = y2-y1 cm1 = cm1*y2/ym12 c0 = c0*y2/y02 c1 = c1*y2/y12 c2 = -ym1*y0*y1/ym12/y02/y12 If (nused.eq.4) GoTo 200 c at least 5 pts ym2 = x-arg(n0-2) ym2m1 = ym1-ym2 ym20 = y0-ym2 ym21 = y1-ym2 ym22 = y2-ym2 cm2 = ym1*y0*y1*y2/ym2m1/ym20/ym21/ym22 cm1 = -cm1*ym2/ym2m1 c0 = -c0*ym2/ym20 c1 = -c1*ym2/ym21 c2 = -c2*ym2/ym22 If (nused.eq.5) GoTo 220 c at least 6 pts y3 = x-arg(n0+3) ym23 = y3-ym2 ym13 = y3-ym1 y03 = y3-y0 y13 = y3-y1 y23 = y3-y2 cm2 = cm2*y3/ym23 cm1 = cm1*y3/ym13 c0 = c0*y3/y03 c1 = c1*y3/y13 c2 = c2*y3/y23 c3 = ym2*ym1*y0*y1*y2/ym23/ym13/y03/y13/y23 GoTo 240 160 Continue Do 170 n=1,nfs y(n) = c0*val(n,n0)+c1*val(n,n0+1) 170 Continue GoTo 260 180 Continue Do 190 n=1,nfs y(n) = cm1*val(n,n0-1)+c0*val(n,n0)+c1*val(n,n0+1) 190 Continue GoTo 260 200 Continue Do 210 n=1,nfs y(n) = cm1*val(n,n0-1)+c0*val(n,n0)+c1*val(n,n0+1)+c2*val(n,n0+ 1 2) 210 Continue GoTo 260 220 Continue Do 230 n=1,nfs y(n) = cm2*val(n,n0-2)+cm1*val(n,n0-1)+c0*val(n,n0)+c1*val(n,n0 1 +1)+c2*val(n,n0+2) 230 Continue GoTo 260 240 Continue Do 250 n=1,nfs y(n) = cm2*val(n,n0-2)+cm1*val(n,n0-1)+c0*val(n,n0)+c1*val(n,n0 1 +1)+c2*val(n,n0+2)+c3*val(n,n0+3) 250 Continue 260 Return End \end{verbatim} \end{footnotesize} \chapter{Running the program for $\pi^{\pm} -^4He$} The first action that LPOTT takes is to read information to run setup variables, like the laboratory kinetic energy of the incident particle, the target nucleus, and a variety of possibilities of interactions and outputs that the program offers. \section{Input}%======================================== A typical input (for $\pi^+$$^4$He) looks like: \begin{verbatim} 1 10 110.0 16 22 20000. -3 -2. -3. 1.362 0.316 0.00 1.362 0.316 0.00 3.0 7. 2 4 0 1 0 1 7 0 5 2 0 3 0 0 0 0 0 3 0 0 0 0 0 0 0 0 59 14 0. 0. 0. 0. \end{verbatim} \begin{itemize} \item The explanations for the possibilities of inputs are written in programs Lpott. \item {\bf Note}:since the program was written in Fortran, the reading format is very strict and any misplacement can be fatal to run the code. Every space is important and care has to be taken when this file is modified to write the data in the right places. \end{itemize} Since this we are going to run the program for positive and negative pions scatteed by Helium isotopes, in this part we concentrate our explantions to these cases. \begin{enumerate} \item The first line reads the equation to solve, {\bf nr}=1 means Schr\"odinger equation with relativistic kinematics, and {\bf lmax}=10 the number of $\pi-$A partial waves. \item The second is for the kinetic energy of the incident particle in the laboratory system of reference. \item The third line of input is to consider parameter for integrals that need to be computed and for some plotting routines in text style. \item The fourth line is for some proton and neutron distribution parameters characteristic of each nucleus and for Coulomb interaction parameters that depends on the nucleus. \item The fifth line contains the atomic number Z and the number of protons plus neutrons that the nucleus has (Z=2 for Helium, and we are considering the Helium isotope that has 2 neutrons (and two protons of course),the A=4. It contains 20 integers that are related to many combination of possibilities that the program offers (they are called Nifty()in the code). \begin{itemize} \item The selection of Nifty(1)=1 if for the scattering of $\pi^+$ and \item Nifty(1)=2, if for ${\pi^-}$ scattering. \end{itemize} \item The sixth and last line reads the number ({\bf nes}) of energies that the program reads from a table, the number of two body eigenchannels,(14) for pions, other parameters are related to pion interaction. \end{enumerate} As far as this part is concerned, the only parameters that you are going to change are: The laboratory kinetic energy of the pion and Nifty(1), 0 for ${\pi^+}$ scattering and 1 for ${\pi^-}$ scattering. \section{Running the code}%======================================== In OSU the input file is present in a directory called {\tt in} and the output will be in a file in a directory called {\tt out}, will have the same name as the input. Both are subdirectories of directory {\tt lpott} and there is a script file called {\tt run} that takes care of using the input and output in the proper way, will erase some temporary disk files. To run the program you give the commands like (in the directory where {\tt run} is located): \begin{verbatim} run lpott he3.30 \end{verbatim} Where {\tt lpott} is the executable file, and {\tt he4.30} is an arbitrary name given by someone to the input file. The output will contain a great amount of information, it writes again the parameters of the input, explaining what the options are, like: \begin{verbatim} nifty( 1)= 0 pi+ nifty( 2)= 1 *5=be shift nifty( 3)= 0 no w.f. nifty( 4)= 1 nlsp-g(p) nifty( 5)= 7 e3b, aay nifty( 6)= 0 no spin ms nifty( 7)= 5 abs-old,sp nifty( 8)= 2 del,no fld nifty( 9)= 0 no Pauli nifty(10)= 3 exact coul nifty(11)= 0 pi0,pi-,k+ shift nifty(12)= 0 pi+,ko shift nifty(13)= 0 pi-channel shift nifty(14)= 0 pi+channel shift nifty(15)= 0 not used shift nifty(16)= 3 he/d nifty(17)= 0 kmt nifty(18)= 0 full amps nifty(19)= 0 use nif-16 nifty(20)= 0 no t10 plt nes,nwaves,b0r,b0i,c0r,c0i= 59 14 .00000 .00000 .00000 .00000 nes,nwaves,b0r,b0i,c0r,c0i= 59 14 -.04000 .04000 .00000 .08000 ********** amass= 4.0000********* \end{verbatim} \section{Differential cross sections for $\pi^{\pm}$ }%========= Since the pions and the $^4He$ nucleus do not have spin, no spin flip spin orbit interactions have to be considered. The differential cross section depends of both the nuclear and Coulomb interactions and is given by: \begin{equation} {d\sigma \over d\Omega }=\vert f(\theta) \vert^2, \end{equation} with \begin{equation} f(\theta)=f_c^{{\rm pt}}(\theta)+f_{NC}(\theta), \end{equation} That is a Coulomb point charge and a nuclear term in the presence of the Coulomb interaction. \begin{equation} f_c^{{\rm pt}}=-{\gamma\over 2k_0\,\sin^2\theta/2}\exp(2i\sigma_0-\gamma \ln \,\sin\theta/2), \end{equation} \begin{equation} f_{\theta}={1\over ik_ 0}\sum_{L=0}^\infty(2L+1)e^{2i\sigma_l}[e^{2i\delta^{NC}_L-1]P_L(\cos \theta}). \end{equation} \subsection{Low energy Coulomb nuclear interference}%=============== You will repeat the calculations by R.H. Landau and A.W. Thomas (Nucl. Phys. A {\bf A302} (1978) 461-492), to see how Coulomb and nuclear forces between the $\pi$ and the nucleus interfere. This effect can be observed at low pion kinetic energies. \begin{itemize} \item Run the program for $\pi^+-^4$He for pion laboratory kinetic energies 110, 75, 50, 40, 30 and 24 MeV. \item Look at each output, before the graphs and you will see a title like: \begin{verbatim} th-c.m. cos(cm) dsig/dw-cm ds/dw(b.a.) t th-lab ...... \end{verbatim} \item You are interested in the first and third columns; the first is the angle ${\theta}$, the angle that the momentum of the scattered pion makes with its incident momentum, in the center of mass system, and the third column is the differential cross section ${d\sigma\over d\Omega }$ in mb. \item Extract these two columns from the output and draw a semilog plot of the differential cross section versus ${\theta}$, if possible for the complete set of energies, in the same graph. \item Repeat the same procedure for the same energies but for $\pi^--^4He$. \item Observe the differences between the two graphs at low angles. \end{itemize} Tables of the differential cross sections for the lab energies specified. (since there are few points greater than 1000 at the beginning, is better to not use them). \begin{footnotesize} \begin{verbatim} angle 110 75 50 40 30 24 .0 .127E+14 .245E+14 .501E+14 .750E+14 .127E+15 .192E+15 3.0 .118E+04 .263E+04 .582E+04 .876E+04 .125E+05 .218E+05 6.0 .401E+02 .104E+03 .301E+03 .462E+03 .382E+03 .113E+04 9.0 .207E+02 .912E+01 .418E+02 .669E+02 .638E+02 .220E+03 12.0 .269E+02 .297E+01 .738E+01 .127E+02 .914E+02 .115E+03 15.0 .311E+02 .461E+01 .125E+01 .223E+01 .126E+03 .103E+03 18.0 .327E+02 .643E+01 .430E+00 .311E+00 .149E+03 .106E+03 21.0 .327E+02 .762E+01 .716E+00 .330E+00 .163E+03 .110E+03 24.0 .316E+02 .823E+01 .115E+01 .730E+00 .170E+03 .114E+03 27.0 .297E+02 .840E+01 .150E+01 .111E+01 .173E+03 .117E+03 30.0 .274E+02 .825E+01 .173E+01 .137E+01 .173E+03 .118E+03 33.0 .247E+02 .786E+01 .185E+01 .151E+01 .171E+03 .119E+03 36.0 .219E+02 .731E+01 .187E+01 .155E+01 .167E+03 .120E+03 39.0 .190E+02 .664E+01 .181E+01 .151E+01 .162E+03 .120E+03 42.0 .162E+02 .591E+01 .170E+01 .142E+01 .157E+03 .120E+03 45.0 .135E+02 .514E+01 .155E+01 .129E+01 .151E+03 .119E+03 48.0 .109E+02 .437E+01 .137E+01 .113E+01 .144E+03 .118E+03 51.0 .866E+01 .362E+01 .117E+01 .959E+00 .137E+03 .117E+03 54.0 .666E+01 .292E+01 .976E+00 .789E+00 .130E+03 .116E+03 57.0 .495E+01 .228E+01 .783E+00 .627E+00 .123E+03 .115E+03 60.0 .355E+01 .172E+01 .604E+00 .477E+00 .116E+03 .114E+03 63.0 .244E+01 .124E+01 .444E+00 .346E+00 .109E+03 .112E+03 66.0 .161E+01 .854E+00 .309E+00 .237E+00 .102E+03 .111E+03 69.0 .104E+01 .565E+00 .204E+00 .154E+00 .953E+02 .110E+03 72.0 .693E+00 .373E+00 .132E+00 .968E-01 .888E+02 .109E+03 75.0 .542E+00 .273E+00 .951E-01 .685E-01 .825E+02 .107E+03 78.0 .553E+00 .262E+00 .948E-01 .695E-01 .764E+02 .106E+03 81.0 .693E+00 .332E+00 .131E+00 .100E+00 .707E+02 .105E+03 84.0 .934E+00 .476E+00 .204E+00 .160E+00 .652E+02 .104E+03 87.0 .125E+01 .683E+00 .311E+00 .250E+00 .601E+02 .103E+03 90.0 .161E+01 .946E+00 .451E+00 .368E+00 .553E+02 .102E+03 93.0 .199E+01 .125E+01 .622E+00 .513E+00 .509E+02 .101E+03 96.0 .239E+01 .160E+01 .820E+00 .685E+00 .468E+02 .997E+02 99.0 .278E+01 .197E+01 .104E+01 .881E+00 .431E+02 .989E+02 102.0 .315E+01 .236E+01 .129E+01 .110E+01 .398E+02 .982E+02 105.0 .350E+01 .276E+01 .155E+01 .134E+01 .368E+02 .976E+02 108.0 .382E+01 .316E+01 .182E+01 .160E+01 .342E+02 .970E+02 111.0 .410E+01 .357E+01 .211E+01 .187E+01 .319E+02 .965E+02 114.0 .435E+01 .396E+01 .240E+01 .216E+01 .300E+02 .960E+02 117.0 .456E+01 .435E+01 .270E+01 .246E+01 .284E+02 .957E+02 120.0 .473E+01 .472E+01 .300E+01 .276E+01 .271E+02 .954E+02 123.0 .487E+01 .508E+01 .330E+01 .307E+01 .261E+02 .951E+02 126.0 .498E+01 .542E+01 .360E+01 .338E+01 .254E+02 .949E+02 129.0 .505E+01 .573E+01 .389E+01 .369E+01 .248E+02 .948E+02 132.0 .510E+01 .603E+01 .417E+01 .400E+01 .245E+02 .947E+02 135.0 .513E+01 .630E+01 .444E+01 .429E+01 .243E+02 .947E+02 138.0 .514E+01 .656E+01 .470E+01 .458E+01 .243E+02 .946E+02 141.0 .513E+01 .679E+01 .495E+01 .485E+01 .244E+02 .946E+02 144.0 .511E+01 .699E+01 .518E+01 .511E+01 .246E+02 .947E+02 147.0 .508E+01 .718E+01 .540E+01 .536E+01 .248E+02 .947E+02 150.0 .504E+01 .735E+01 .560E+01 .558E+01 .251E+02 .948E+02 153.0 .500E+01 .750E+01 .579E+01 .579E+01 .255E+02 .949E+02 156.0 .496E+01 .763E+01 .595E+01 .597E+01 .258E+02 .949E+02 159.0 .492E+01 .774E+01 .610E+01 .614E+01 .261E+02 .950E+02 162.0 .488E+01 .784E+01 .623E+01 .628E+01 .264E+02 .951E+02 165.0 .485E+01 .792E+01 .634E+01 .640E+01 .267E+02 .952E+02 168.0 .482E+01 .798E+01 .643E+01 .650E+01 .269E+02 .953E+02 171.0 .479E+01 .803E+01 .650E+01 .658E+01 .271E+02 .953E+02 174.0 .477E+01 .806E+01 .655E+01 .663E+01 .273E+02 .953E+02 177.0 .476E+01 .808E+01 .658E+01 .667E+01 .274E+02 .953E+02 180.0 .476E+01 .809E+01 .660E+01 .668E+01 .274E+02 .954E+02 \end{verbatim} \end{footnotesize} In figure (\ref{pi+4he}) we have the $\pi^+-^4He$ differential cross section for different pion kinetic energies in the laboratory reference system. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{pi+4he.eps}} \end{center} \caption{Differential cross section for $\pi^+-^4$He At several pion kinetic laboratory energies this is figure {\tt pi+4he.eps}in Latex M} \label{pi+4he} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{pi-4he.eps}} \end{center} \caption{Differential cross section for $\pi^--^4$He M this is graph {\tt pi-4he.eps} in Latex m} \label{pi-4he} \end{figure} Graph (\ref{pi-4he}) used the following set of data: (The columns are in same order as for $\pi^+ - ^4$He \begin{footnotesize} \begin{verbatim} 6.000 2.78e+02 3.36e+02 5.24e+02 7.60e+02 2.10e+03 2.00e+03 9.000 1.32e+02 1.15e+02 1.43e+02 2.02e+02 8.63e+02 6.10e+02 12.000 9.23e+01 6.49e+01 6.58e+01 8.99e+01 5.68e+02 3.37e+02 15.000 7.45e+01 4.55e+01 3.96e+01 5.26e+01 4.52e+02 2.47e+02 18.000 6.34e+01 3.56e+01 2.76e+01 3.59e+01 3.92e+02 2.07e+02 21.000 5.52e+01 2.93e+01 2.10e+01 2.69e+01 3.54e+02 1.86e+02 24.000 4.82e+01 2.48e+01 1.69e+01 2.12e+01 3.26e+02 1.72e+02 27.000 4.21e+01 2.13e+01 1.39e+01 1.73e+01 3.05e+02 1.63e+02 30.000 3.65e+01 1.83e+01 1.17e+01 1.44e+01 2.86e+02 1.56e+02 33.000 3.14e+01 1.57e+01 9.86e+00 1.21e+01 2.69e+02 1.51e+02 36.000 2.66e+01 1.34e+01 8.34e+00 1.02e+01 2.53e+02 1.46e+02 39.000 2.23e+01 1.13e+01 7.03e+00 8.58e+00 2.38e+02 1.42e+02 42.000 1.85e+01 9.41e+00 5.89e+00 7.18e+00 2.24e+02 1.38e+02 45.000 1.50e+01 7.75e+00 4.88e+00 5.96e+00 2.10e+02 1.35e+02 48.000 1.20e+01 6.28e+00 3.99e+00 4.89e+00 1.97e+02 1.32e+02 51.000 9.36e+00 4.99e+00 3.21e+00 3.96e+00 1.84e+02 1.29e+02 54.000 7.13e+00 3.88e+00 2.53e+00 3.15e+00 1.71e+02 1.26e+02 57.000 5.27e+00 2.94e+00 1.94e+00 2.46e+00 1.59e+02 1.23e+02 60.000 3.77e+00 2.15e+00 1.44e+00 1.86e+00 1.48e+02 1.20e+02 63.000 2.59e+00 1.52e+00 1.03e+00 1.36e+00 1.36e+02 1.18e+02 66.000 1.70e+00 1.02e+00 6.97e-01 9.55e-01 1.26e+02 1.16e+02 69.000 1.08e+00 6.59e-01 4.40e-01 6.31e-01 1.16e+02 1.13e+02 72.000 6.96e-01 4.15e-01 2.56e-01 3.84e-01 1.06e+02 1.11e+02 75.000 5.10e-01 2.80e-01 1.38e-01 2.09e-01 9.71e+01 1.09e+02 78.000 4.93e-01 2.43e-01 8.21e-02 1.00e-01 8.87e+01 1.08e+02 81.000 6.12e-01 2.94e-01 8.27e-02 5.31e-02 8.09e+01 1.06e+02 84.000 8.40e-01 4.21e-01 1.35e-01 6.19e-02 7.37e+01 1.04e+02 87.000 1.15e+00 6.15e-01 2.33e-01 1.22e-01 6.71e+01 1.03e+02 90.000 1.51e+00 8.66e-01 3.73e-01 2.29e-01 6.11e+01 1.02e+02 93.000 1.90e+00 1.16e+00 5.50e-01 3.79e-01 5.56e+01 1.01e+02 96.000 2.31e+00 1.50e+00 7.58e-01 5.67e-01 5.08e+01 9.95e+01 99.000 2.71e+00 1.86e+00 9.93e-01 7.89e-01 4.65e+01 9.86e+01 102.000 3.10e+00 2.25e+00 1.25e+00 1.04e+00 4.27e+01 9.78e+01 105.000 3.46e+00 2.65e+00 1.53e+00 1.32e+00 3.95e+01 9.72e+01 108.000 3.79e+00 3.05e+00 1.82e+00 1.62e+00 3.68e+01 9.66e+01 111.000 4.08e+00 3.46e+00 2.12e+00 1.94e+00 3.46e+01 9.61e+01 114.000 4.34e+00 3.87e+00 2.43e+00 2.28e+00 3.28e+01 9.58e+01 117.000 4.55e+00 4.26e+00 2.74e+00 2.63e+00 3.15e+01 9.55e+01 120.000 4.72e+00 4.64e+00 3.05e+00 2.98e+00 3.05e+01 9.53e+01 123.000 4.85e+00 5.01e+00 3.36e+00 3.34e+00 2.98e+01 9.51e+01 126.000 4.95e+00 5.36e+00 3.67e+00 3.70e+00 2.94e+01 9.51e+01 129.000 5.02e+00 5.69e+00 3.97e+00 4.06e+00 2.93e+01 9.51e+01 132.000 5.05e+00 6.00e+00 4.27e+00 4.41e+00 2.95e+01 9.51e+01 135.000 5.07e+00 6.29e+00 4.55e+00 4.75e+00 2.97e+01 9.52e+01 138.000 5.07e+00 6.55e+00 4.82e+00 5.08e+00 3.02e+01 9.53e+01 141.000 5.05e+00 6.79e+00 5.08e+00 5.39e+00 3.07e+01 9.55e+01 144.000 5.02e+00 7.01e+00 5.32e+00 5.68e+00 3.13e+01 9.56e+01 147.000 4.98e+00 7.21e+00 5.55e+00 5.96e+00 3.20e+01 9.58e+01 150.000 4.93e+00 7.38e+00 5.76e+00 6.21e+00 3.27e+01 9.60e+01 153.000 4.88e+00 7.54e+00 5.96e+00 6.45e+00 3.33e+01 9.62e+01 156.000 4.83e+00 7.68e+00 6.13e+00 6.66e+00 3.40e+01 9.64e+01 159.000 4.79e+00 7.80e+00 6.29e+00 6.84e+00 3.46e+01 9.66e+01 162.000 4.74e+00 7.90e+00 6.42e+00 7.00e+00 3.51e+01 9.67e+01 165.000 4.70e+00 7.98e+00 6.54e+00 7.14e+00 3.56e+01 9.69e+01 168.000 4.67e+00 8.05e+00 6.64e+00 7.25e+00 3.60e+01 9.70e+01 171.000 4.64e+00 8.10e+00 6.71e+00 7.34e+00 3.63e+01 9.71e+01 174.000 4.62e+00 8.13e+00 6.76e+00 7.40e+00 3.66e+01 9.72e+01 177.000 4.61e+00 8.16e+00 6.80e+00 7.44e+00 3.67e+01 9.72e+01 180.000 4.60e+00 8.16e+00 6.81e+00 7.45e+00 3.68e+01 9.72e+01 180.000 4.60e+00 8.16e+00 6.81e+00 7.45e+00 3.68e+01 9.72e+01 \end{verbatim} \end{footnotesize} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{dsigcoul.eps}} \end{center} \caption{Differential cross section for $\pi^--^4$He pure Coulomb this graph is {\tt dsigcoul.eps} in Latex M} \label{dsigcoul} \end{figure} Figure (\ref{dsigcoul.eps}) shows the differential cross section for pure Coulomb scattering of $\pi^+$ by $^4He$; no nuclear interactions included; was drawn for $T_\pi^{Lab}$=50 MeV, and the angles are in the center of mass system. Witn a little bit of abstraction you can see it at low angles in Figure (\ref{pi+4he}), for the graph corresponding to 50MeV. The order of the columns is the same as in the previous table. In figure (\ref{pi+4he}) You can see a minimum at $\simeq$ $25^o$ followed by a maximum at about $40^o$ for all energies above 30 MeV. This is due to the interference of Coulomb (repulsive) and nuclear (attractive for $\pi^+$) amplitudes. As the energy decreases from 110MeV the minimum moves out in angle and at $T_{\pi}$=30MeV, the Coulomb minimum merges with the $\pi$N-P wave at $\simeq$ 70$^o$. At 24MeV the destructive interference structure completely vanishes for $\pi^+$, and the cross section rises because the nuclear amplitude changes sign around 60$^o$ and produce constructive interference. In contrast, the $\pi^{-} -^4He$ cross section simply show a monotonic decrease with energy. For similar results in Carbon, see R.R. Johnston, T. G. Masterson, K.L Erdman A. W. Thomas and R.H. Landau,Nucl. Phys. {\bf A296} (1978) 44-460 \subsection{$\pi^+$-nucleus total cross sections}%================ Now comes the question: the $\pi^+$-nucleon total cross section exhibits a resonance as in figure (\ref{sigtot}), what consequences has this resonance on $\pi^{+}- ^{4}$He total cross sections? To see the effect extract the total cross sections from the $\pi^+-^4$He output that you already have , looking next to the tables of the differential cross sections in the output tables and you will find the total cross sections; make a table of the pion ${\rm T_{LAB}}$ kinetic energies versus the total cross sections for the same energies you already have, and plot a graph of the total cross section versus the pion kinetic energies. The table of pion lab momentum, $T_{\pi}^{Lab}$ and the total cross sections follows: \begin{footnotesize} \begin{verbatim} pi Lab T lab sigma momentum Mev mb 112.9878 40.0 41.353 128.2880 50.0 48.900 162.9776 75.0 59.641 206.8989 110.0 194.631 230.6302 130.0 268.368 253.7191 150.0 346.136 276.3268 170.0 399.546 287.4858 180.0 405.169 298.5626 190.0 405.002 195.0 395.390 309.5661 200.0 377.606 210.0 362.317 331.3824 220.0 328.170 352.9836 240.0 255.907 385.0612 270.0 187.242 416.8295 300.0 154.611 469.2596 350.0 133.821 521.2124 400.0 101.691 572.8178 450.0 86.865 \end{verbatim} \end{footnotesize} With these data we plot the total cross sections versus the laboratory pion kinetic energy. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{pi+he4tcs.eps}} \end{center} \caption{Total cross section for $\pi^--^4$He M this is {\tt pi+he4tcs.eps}in latex M} \label{pi+he4tcs} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{k2sig.eps}} \end{center} \caption{$k^2$ times the total cross section for $\pi^--^4$He. With $k$ the center of mass pion momentum. The graph was smoothed with a 30 point spline this is {\tt k2sig.eps} in Latex } \label{k2sig} \end{figure} \subsection{Effects of $\pi^- -$N on $\pi^- - ^{12}$C}%==================== \begin{footnotesize} \begin{verbatim} tpi stot epilab ecm(Sqrts) kcm plab 90.0000 445.226 229.5780 11487.2037 178.6547 182.2746 100.0000 521.634 239.5780 11497.0010 190.6896 194.7193 110.0000 598.415 249.5780 11506.7898 202.4448 206.8989 120.0000 668.452 259.5780 11516.5704 213.9643 218.8578 130.0000 734.778 269.5780 11526.3427 225.2824 230.6302 140.0000 791.244 279.5780 11536.1067 236.4260 242.2434 150.0000 839.352 289.5780 11545.8624 247.4169 253.7191 160.0000 873.209 299.5780 11555.6099 258.2731 265.0754 170.0000 893.834 309.5780 11565.3492 269.0091 276.3268 180.0000 896.225 319.5780 11575.0802 279.6372 287.4858 190.0000 881.547 329.5780 11584.8032 290.1679 298.5626 200.0000 833.553 339.5780 11594.5179 300.6100 309.5661 210.0000 783.646 349.5780 11604.2245 310.9709 320.5039 220.0000 761.058 359.5780 11613.9230 321.2574 331.3824 230.0000 665.579 369.5780 11623.6135 331.4750 342.2074 240.0000 611.394 379.5780 11633.2958 341.6287 352.9836 250.0000 572.515 389.5780 11642.9701 351.7229 363.7156 260.0000 545.361 399.5780 11652.6364 361.7615 374.4069 \end{verbatim} \end{footnotesize} With the first two columns of this table we plot (smoothed with splines) figure (\ref{pi-c12tcs}). See R. H. Landau, S. C. Phatak, and F. Tabakin, Ann. Phys., {\bf 78} (1973), 299, and for similar results with Oxygen, see S.C. Phatak and F. Tabakin Phys. Rev. C7, (1973) 1803 \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{pi-c12tcs.eps}} \end{center} \caption{Total cross section for elastic scattering $\pi^--^{12}$C. This is {\tt pi-c12tcs.eps} in Latex M } \label{pi-c12tcs} \end{figure} And again we now plot $k^2 \sigma$, with $k$ the center of mass momentum \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{k2sigC12.eps}} \end{center} \caption{$k^2$ times the total cross section for $\pi^--^{12}$C. With $k$ the center of mass pion momentum. this is graph {\tt k2sigC12.eps} in latex } \label{k2sigC12} \end{figure} Let us take a look at the Argand diagrams for some of the partial waves that contribute to the scattering, Figures (\ref{arg0} to \ref{arg4}). \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{arg0.eps}} \end{center} \caption{Argand diagram for wave L=0 in $\pi^- -^{12}$C scattering this is {\tt arg0.eps} in Latex Mmm} \label{arg0} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{arg1.eps}} \end{center} \caption{$k^2$ times the total cross section for $\pi^--^{12}$C. With $k$ the center of mass pion momentum. this is {tt arg1.eps} in Latex } \label{arg1} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{arg2.eps}} \end{center} \caption{Argand diagram for wave L=2 in $\pi^- -^{12}$C scattering M this is {\tt arg2.eps} in Latex } \label{arg2} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{arg3.eps}} \end{center} \caption{Argand diagram for wave L=3 in $\pi^- -^{12}$C scattering this is {\tt arg3.eps} in Latex M} \label{arg3} \end{figure} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{arg4.eps}} \end{center} \caption{Argand diagram for wave L=4 in $\pi^- -^{12}$C scattering M this is {\tt arg4.eps} in latex M} \label{arg4} \end{figure} The graphs were made with the data of this table: \begin{footnotesize} \begin{verbatim} tpilab l=0 l=1 l=2 l=3 l=4 MeV re T im T re im re im re im re im 050 .137E+00 .148E+00 .417E+00 .290E+00 .238E+00 .668E-01 .659E-01 .448E-02 .107E-01 .862E-05 060 .132E+00 .141E+00 .373E+00 .248E+00 .248E+00 .846E-01 .836E-01 .881E-02 .215E-01 .557E-03 070 .215E+00 .236E+00 .380E+00 .304E+00 .290E+00 .134E+00 .115E+00 .189E-01 .365E-01 .187E-02 080 .270E+00 .367E+00 .377E+00 .359E+00 .321E+00 .197E+00 .147E+00 .349E-01 .522E-01 .403E-02 090 .272E+00 .511E+00 .365E+00 .418E+00 .339E+00 .271E+00 .122E+00 .414E-01 .714E-01 .868E-02 100 .234E+00 .626E+00 .343E+00 .478E+00 .340E+00 .349E+00 .210E+00 .990E-01 .902E-01 .159E-01 110 .178E+00 .704E+00 .306E+00 .538E+00 .323E+00 .424E+00 .237E+00 .153E+00 .110E+00 .284E-01 120 .128E+00 .749E+00 .257E+00 .593E+00 .292E+00 .489E+00 .251E+00 .222E+00 .129E+00 .482E-01 130 .812E-01 .773E+00 .194E+00 .639E+00 .248E+00 .544E+00 .247E+00 .299E+00 .147E+00 .780E-01 140 .408E-01 .780E+00 .128E+00 .668E+00 .196E+00 .586E+00 .222E+00 .374E+00 .161E+00 .120E+00 150 .272E-02 .776E+00 .667E-01 .682E+00 .137E+00 .617E+00 .179E+00 .438E+00 .162E+00 .174E+00 160 -.348E-01 .758E+00 .117E-01 .681E+00 .729E-01 .634E+00 .122E+00 .483E+00 .146E+00 .233E+00 170 -.622E-01 .733E+00 -.329E-01 .671E+00 .116E-01 .639E+00 .600E-01 .513E+00 .114E+00 .287E+00 180 -.783E-01 .706E+00 -.668E-01 .655E+00 -.416E-01 .634E+00 -.196E-02 .527E+00 .706E-01 .328E+00 190 -.104E+00 .677E+00 -.102E+00 .634E+00 -.919E-01 .616E+00 -.647E-01 .521E+00 .144E-01 .348E+00 200 -.104E+00 .618E+00 -.120E+00 .587E+00 -.128E+00 .577E+00 -.120E+00 .496E+00 -.426E-01 .348E+00 210 -.623E-01 .617E+00 -.953E-01 .587E+00 -.122E+00 .561E+00 -.136E+00 .495E+00 -.686E-01 .335E+00 220 -.909E-01 .611E+00 -.131E+00 .589E+00 -.163E+00 .558E+00 -.182E+00 .471E+00 -.125E+00 .333E+00 230 -.697E-01 .568E+00 -.118E+00 .546E+00 -.159E+00 .505E+00 -.184E+00 .419E+00 -.137E+00 .296E+00 250 -.525E-01 .569E+00 -.101E+00 .546E+00 -.149E+00 .490E+00 -.178E+00 .392E+00 -.150E+00 .270E+00 270 -.780E-01 .579E+00 -.120E+00 .554E+00 -.166E+00 .493E+00 -.188E+00 .387E+00 -.162E+00 .264E+00 290 -.987E-01 .569E+00 -.136E+00 .541E+00 -.175E+00 .479E+00 -.193E+00 .375E+00 -.164E+00 .253E+00 310 -.128E+00 .563E+00 -.157E+00 .537E+00 -.185E+00 .476E+00 -.197E+00 .379E+00 -.167E+00 .262E+00 330 -.163E+00 .525E+00 -.184E+00 .503E+00 -.205E+00 .448E+00 -.211E+00 .362E+00 -.183E+00 .256E+00 \end{verbatim} \end{footnotesize} \subsection{Resonance and differential cross section}%======================================== Now that we have seen the influence of the $\pi^+-N$ pion nucleon resonance on the $\pi^+ -^4$He total cross section we may ask ?`and what is the effect on the differential cross sections for $T_{\pi}^{Lab}$ near the values for resonance? Nothing better than a graph to visualize its effects. Figure (\ref{dif+res}) shows differential cross sections for $\pi^+-^4$He at $T_{\pi}^{Lab}$ energies near the occurrence of resonance. As the pion kinetic energy in the lab system approaches 240 MeV, the dip is more pronounced and is very low, this is because the p-wave is dominant in the scattering. A p-wave in the differential cross section is multiplied by Legendre polynomial $P_1(\cos \theta)=\cos (\theta)$ and when theta is 90$^o$ the value of the polynomial is zero, and that causes a dip in the differential cross section. The input that we use to produce the differential cross sections were like: \begin{verbatim} 1 10 240.0 16 22 20000. -3 -2. -3. 1.362 0.316 0.00 1.362 0.316 0.00 3.0 7. 2 4 0 1 0 1 7 0 5 2 0 3 0 0 0 0 0 3 0 0 0 0 0 0 0 0 59 14 0. 0. 0. 0. \end{verbatim} \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{dif+res.eps}} \end{center} \caption{Differential cross sections at various $T_{\pi}^{Lab}$-He energies this is {\tt dif+res.eps} in Latex } \label{dif+res} \end{figure} \subsection{Effect of spin flip in $\pi^--^3$He Scattering}%==================== For unpolarized nuclei (the spins are not pointing in a prefered direction ``up'' or ``down''), the $\pi-A$ differential cross sections can be computed with: (see Lpott eqs (21)-(26) \begin{equation} f(\theta)={1\over k_0}\,\sum_{L=0}^\infty [(L+1)T_{L+}+LT_{L-}]P_L(cos\,\theta), \end{equation} \begin{equation} g(\theta)={1\over k_0}\sum_{L=1}^{\infty}[T_{L+}-T_L]P'_L(cos\,\theta), \end{equation} the differential cross sections is then: \begin{equation} {d\sigma\over d\Omega}(\theta)\vert_{{\rm unpol}}=\vert f(\theta)\vert^2 +\sin^2 \,\theta \,\vert f(\theta)\vert^2. \end{equation} where \begin{equation} T_{L\pm}=\{[\exp(2i\delta_{L\pm}(k_0))-1]/2i\}/(-\rho_E), \end{equation} and \begin{equation} \rho_E=2k_0\mu_{\pi A}=2k_0\,{E_{\pi}(k_0)\,E_A(k_0)\over [E_{\pi}(k_0)+E_A(k_0)]}. \end{equation} $f(\theta)$ and $g(\theta)$ are the conventional spin non flip and spin-flip scattering amplitudes. We like to see the effect of each of spin flip and non flip in the differential cross for $\pi^--^3He$ scattering and 120 MeV, and the results can be compared with figure (6) in Rubin H. Landau, Phys. Rev. C15, 2127,(1977). The input parameters for the file are: \begin{verbatim} 1 9 120. 16 22 20000. -3 -2. -4. 0.6540 0.4560 0.8210 0.6750 0.366 0.8360 2.16 6. 2 3 1 1 0 1 7 3 3 2 0 1 0 0 0 0 0 3 0 0 1 0 59 14 0.0 0.0 0.0 0.0 \end{verbatim} \begin{itemize} \item The output contains the data for the differential cross section, the spin flip and spin non flip differential cross sections. \item Extract the table with the differential cross section from the output, \item make a data file with it \item Use the following program in C to extract the information you need, i.e. the angle, the differential cross section, the spin flip and non flip differential cross sections \begin{verbatim} #include main() { // Read from several files diff cross sections //and writes in a file. float ang, dumb1, diff,d2,d3,d4,d5,d6,d7,d8,d9,d10; float dum,nfl,nflb,pol,flip,flib; int i; FILE *f1; f1=fopen("he3-sf.dat","r"); for(i=0;i<62;i++){ fscanf(f1,"%f %f %e %e %e %f %f %e %e %f %f %e",&ang,&dumb1,&diff,&d2,&d3, &d4,&d5,&d6,&d7,&d8,&d9,&d10); fscanf(f1,"%e %e %e %e %e",&nfl,&nflb,&pol,&flip,&flib); printf("%5.3lf %10.2e %10.2e %10.2e\n",ang,diff, nfl,flip); } fclose(f1); } \end{verbatim} \item the output of the program can be piped to a file using something like: {\tt a.out >spinflip.dat} \end{itemize} The data contain the angle, the differential cross section, the non flip and spin flip differential cross sections respectively (The data for 0 degress was eliminated, spoils the graph) \begin{verbatim} 3.000 1.67e+03 3.49e+01 1.43e-02 6.000 1.87e+02 3.43e+01 5.67e-02 9.000 7.98e+01 3.33e+01 1.25e-01 12.000 5.37e+01 3.19e+01 2.17e-01 15.000 4.28e+01 3.03e+01 3.29e-01 18.000 3.65e+01 2.83e+01 4.55e-01 21.000 3.20e+01 2.62e+01 5.92e-01 24.000 2.83e+01 2.40e+01 7.34e-01 27.000 2.51e+01 2.16e+01 8.78e-01 30.000 2.22e+01 1.93e+01 1.02e+00 33.000 1.96e+01 1.70e+01 1.15e+00 36.000 1.71e+01 1.48e+01 1.27e+00 39.000 1.48e+01 1.27e+01 1.38e+00 42.000 1.28e+01 1.07e+01 1.47e+00 45.000 1.09e+01 8.90e+00 1.54e+00 48.000 9.21e+00 7.29e+00 1.59e+00 51.000 7.73e+00 5.86e+00 1.63e+00 54.000 6.43e+00 4.62e+00 1.64e+00 57.000 5.32e+00 3.56e+00 1.64e+00 60.000 4.37e+00 2.67e+00 1.62e+00 63.000 3.58e+00 1.95e+00 1.58e+00 66.000 2.94e+00 1.38e+00 1.53e+00 69.000 2.42e+00 9.38e-01 1.47e+00 72.000 2.03e+00 6.20e-01 1.41e+00 75.000 1.73e+00 4.07e-01 1.33e+00 78.000 1.52e+00 2.82e-01 1.25e+00 81.000 1.39e+00 2.32e-01 1.17e+00 84.000 1.31e+00 2.42e-01 1.09e+00 87.000 1.29e+00 3.00e-01 1.01e+00 90.000 1.30e+00 3.93e-01 9.30e-01 93.000 1.34e+00 5.12e-01 8.51e-01 96.000 1.39e+00 6.47e-01 7.74e-01 99.000 1.47e+00 7.91e-01 6.99e-01 102.000 1.54e+00 9.38e-01 6.29e-01 105.000 1.62e+00 1.08e+00 5.61e-01 108.000 1.70e+00 1.22e+00 4.99e-01 111.000 1.78e+00 1.36e+00 4.40e-01 114.000 1.85e+00 1.48e+00 3.87e-01 117.000 1.92e+00 1.60e+00 3.38e-01 120.000 1.98e+00 1.70e+00 2.94e-01 123.000 2.04e+00 1.80e+00 2.55e-01 126.000 2.09e+00 1.88e+00 2.20e-01 129.000 2.14e+00 1.96e+00 1.89e-01 132.000 2.18e+00 2.03e+00 1.61e-01 135.000 2.22e+00 2.09e+00 1.37e-01 138.000 2.25e+00 2.15e+00 1.16e-01 141.000 2.28e+00 2.20e+00 9.69e-02 144.000 2.31e+00 2.24e+00 8.02e-02 147.000 2.33e+00 2.27e+00 6.56e-02 150.000 2.34e+00 2.30e+00 5.28e-02 153.000 2.36e+00 2.32e+00 4.16e-02 156.000 2.37e+00 2.34e+00 3.20e-02 159.000 2.37e+00 2.36e+00 2.39e-02 162.000 2.38e+00 2.37e+00 1.71e-02 165.000 2.38e+00 2.38e+00 1.16e-02 168.000 2.38e+00 2.38e+00 7.29e-03 171.000 2.38e+00 2.39e+00 4.03e-03 174.000 2.38e+00 2.39e+00 1.77e-03 177.000 2.38e+00 2.39e+00 4.39e-04 180.000 2.38e+00 2.39e+00 1.69e-14 180.000 2.38e+00 2.39e+00 1.69e-14 \end{verbatim} With these data you can plot the graph (\ref{spinflip}). \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{spinflip.eps}} \end{center} \caption{$\pi^{-}-{ }^3$He Effect of Spin Flip for the elastic scattering pf $\pi^-$ from an unpolarized $^3$He. this is {\tt spinflip.eps} in Latex } \label{spinflip} \end{figure} \subsection{Polarization}%======================================== Next we have figure (\ref{p-hepol.eps}) with the polarization of the recoiling nucleus in $\pi^--^3He$ elastic scattering at $T_{\pi}^{Lab}=98$ MeV, see Rubin H. Landau, {\bf C15} (1977) 2127. \begin{figure} \begin{center} \leavevmode \hbox{% \epsfxsize=3.5in \epsfbox{p-hepol.eps}} \end{center} \caption{He Polarization of the recoiling nucleus is elastic scattering $\pi^{-}-^3$He. is {\tt p-hepol.eps} in Latex M} \label{p-hepol} \end{figure} \chapter{$\pi^- -^{12}$C Wavefunction}%======================================== \section{Introduction}%======================================== Program LPOTT has the possibility of obtaining the wavefunction for the particle-nucleus scattering. Its equation (150) gives the wave function: \begin{equation} \psi_{\bf k}^{(+)}({\bf r})=\sum_{L=0}^{\infty}U_{L}(r)i^L(2L+1) P_L(cos \, \theta), \label{wavpsi} \end{equation} With \begin{equation} U_{L<{\rm lstore}}(r)={e^{i\delta_L}\over kr}\{\cos\,\delta_L\, F_L (kr) + \sin\,\delta_L\, G_L (kr)\}, r>R_{cut}, \end{equation} \begin{equation} =\sum_{m=1}^{N1} j_L(k_mr)\Omega_T(k_{N+1},k_m)Z_L, \ \ \ r