\documentclass[12pt]{article}
\usepackage{epsfig}
%\topmargin=+0.0in
%\oddsidemargin=0.25in
\textwidth=6.0in
%\textheight=8in
%\footskip=6ex
%\footheight=2ex
\begin{document}
\begin{center}
{\bf Lecture 6}
\end{center}
\bigskip
\noindent {\bf The Born Approximation}
\bigskip
The Born approximation can be derived from time dependent perturbation theory.
The probability for a transition to occur from an initial state
$\psi_i \propto e^{ik_i \cdot r}$ to a final state $\psi_f \propto e^{ik_f \cdot r}$
(where $k = p/\hbar$) is proportional to the matrix element
\begin{equation}
<\psi_f |V|\psi_i> \; \propto \int_0^\infty V(r) e^{i(\vec{k}_f-\vec{k}_i)\cdot \vec{r}} d^3 \vec{r}
\end{equation}
\noindent This integral, which is the Fourier transform of V(r) with respect to the
momentum transfer, produces a scattering diffraction pattern. The kinematics
enters the problem only with the momentum transfer, which is most conveniently defined
as $\vec{q} = \vec{k}_f - \vec{k}_i = (\vec{p}_f - \vec{p}_i)/\hbar$.
\begin{equation}
<\psi_f |V|\psi_i> \; \propto \int_0^\infty V(r) e^{i \vec{q}\cdot \vec{r}} d^3 \vec{r}
\end{equation}
At first glance, the integral looks difficult. However, if we choose the coordinate
system for the integral such that the z-axis is directed along $\vec{q}$, then
$\vec{q} \cdot \vec{r} = |q|\; |r|\; cos(\alpha)=qr\;cos(\alpha)$, where
$\alpha$ is the angle between $\vec{q}$ and $\vec{r}$. With this
choice of coordinate system, the integral becomes:
\begin{equation}
\int_0^\infty V(r) e^{i \vec{q}\cdot \vec{r}} d^3 \vec{r} =
\int_0^\infty V(r)\int_0^\pi e^{iqr cos(\alpha )} 2 \pi sin(\alpha ) \; d\alpha r^2 dr
\end{equation}
\newpage
\noindent The angular part of the integral can be evaluated since $sin(\alpha)$ is the derivative of
$cos(\alpha)$:
\begin{eqnarray*}
\int_0^\infty V(r) e^{iqr cos(\alpha )} 2 \pi sin(\alpha )r^2 \;d\alpha dr & = &
\int_0^\infty V(r)2\pi ({{-e^{iqrcos(\alpha)}} \over {iqr}})|_0^\pi r^2 dr \\
& = & \int V(r)2\pi \; {{e^{iqr}-e^{-iqr}} \over {iqr}} \; r^2 dr \\
& = & {{4 \pi} \over q} \int_0^\infty V(r) r \; sin(qr) dr
\end{eqnarray*}
\noindent which is the integral you need to solve numerically.\\
Note that the angle enters through $q$, since $q=(2p/\hbar)sin(\theta /2)$. When integrating
over $r$, sin(qr) can change sign and the integral can be zero. That is, there can be
angles where this integral will be zero. The cancelation in the integral is in essence
destructive interference from the particle scattering off different parts of the
target. One sees the same effect in double slit interference. As with the
double slit interference pattern, the interference occurs at smaller angles
for larger targets. This effect is seen in the comparison of the $K^+$
scattering data off $^{12}C$ versus $^{40}Ca$.
The diffraction interference occurs at around $28^\circ$ for $^{12}C$ and
around $17^\circ$ for $^{40}Ca$ for kaons with the same momentum.\\
If $V(r)$ is proportional to the density of the target, this integral is often refered to
as the {\bf form factor} of the interaction. We note that the Born approximation
is an approximation of the complete calculation. An "accurate-numerical" non-relativistic
calculation of the cross section can be done by solving the Schroedinger equation with
scattering boundary conditions. We will talk about this solution in our next
assignment. This assignment introduces you to parameters used in scattering experiments,
numerical integration, and relativistic kinematics, which we discuss next.
\newpage
\noindent {\bf Relativistic Kinematics}\\
\bigskip
The experimentalists measure the kaon momentum in the laboratory frame.
However, it is better to do the analysis (theory) in the center of mass
frame. The results we obtain from the Born approximation holds in
the center-of-mass reference frame. So for a proper comparison, we
need to use the kaon's momentum in the center of mass frame of the
kaon and $^{12}C$. $800 \; MeV/c$ is the kaon's momentum in the lab.
I obtain $740 \; MeV/c$ for the kaon's momentum in the c.m. frame,
and this value should be used in our equation. Let' see how we
calculate the c.m. momentum from the lab value.\\
It might have been a while since you covered relativistic kinematics in
your classes, and since high energy interactions is in the relativistic energy
and momentum regime, this is a good time to refresh our memory on the
topic.
The total relativistic energy as well as the total relativistic momentum
for a system of particles are conserved quantities. The relationship
between a particle's (say particle "1") energy, $E_1$, and it's
momentum, $\vec{p}_1$ is:
\begin{equation}
E_1^2 = m_1^2c^4 + p_1^2c^2
\end{equation}
\noindent where $p_1^2 = \vec{p}_1 \cdot \vec{p}_1$. For a system that
starts with two particles, $a$ and $b$, and ends with two particles,
$c$ and $d$, the conservation of total energy means:
\begin{equation}
E_a + E_b= E_c + E_d
\end{equation}
\noindent The conservation law holds if $a$ and $c$, and/or $b$ and $d$ are
different particles. Similarly, momentum conservation for two particle
interactions means:
\begin{equation}
\vec{p}_a + \vec{p}_b = \vec{p}_c + \vec{p}_d
\end{equation}
\newpage
As an example, one can determine the momentum and energy of the pions
in the rho meson decay. The rho meson, $\rho$, can decay into
two pions:
\begin{equation}
\rho \rightarrow \pi^+ + \pi^-
\end{equation}
\noindent In the rest frame of the $\rho$, the pions will travel in opposite
direction with equal momentum. This is true, since momentum is conserved.
The conservation of energy requires that:
\begin{eqnarray*}
E_\rho & = & E_{\pi^-} + E_{\pi^+} \\
m_\rho c^2 & = & \sqrt{m_\pi^2c^4 + p^2c^2} + \sqrt{m_\pi^2c^4 + p^2c^2}
\end{eqnarray*}
\noindent This equation can be solved for $p$ to give
\begin{equation}
p = c \sqrt{(m_\rho /2)^2 - m_\pi^2}
\end{equation}
\noindent Using $m_\rho = 775 \; MeV/c^2$, and $m_\pi = 139 \; MeV/c^2$, one
obtains $p \approx 362 \; MeV/c$.\\
In addition to conserved quantities, invarient quantities, or invarients, are also
important. Invarients are expressions that are the same for all observers.
For example, for a particle $E^2 - p^2c^2$ will equal $m^2c^4$ for any
observer. The energy $E$ and the momentum $p$ will depend on one's reference
frame, but the combination $E^2-p^2c^2$ will give the same value in all
inertial reference frames.\\
A useful invarient quantity in particle physics is the invarient total energy, $\sqrt{s}$,
of the two particles, since it is the same in all reference frames. For two particles,
the invarient total energy squared, $s$, is given by:
\begin{equation}
s = (E_a + E_b)^2 - (\vec{p}_a + \vec{p}_b)^2c^2
\end{equation}
\noindent where the squaring of the vector sum means the scalar product with itself.
\newpage
To see the usefulness of the quantity $s$, or $\sqrt{s}$, consider the following example.
Suppose you wanted know what the threshold energy is to produce a particle of
mass (rest mass) $M$ by scattering a particle of mass $m_1$ at a target of
mass $m_2$, where $m_2$ is initially at rest. Note: $M>m_1+m_2$:
\begin{equation}
m_1 + m_2 \rightarrow M
\end{equation}
\noindent The experiment is carried out in the lab reference frame. In this frame
the total invariant energy squared is
\begin{eqnarray*}
s & = & (E_1+E_2)^2 - (\vec{p}_1 + \vec{p}_2)^2c^2 \\
& = & (E_1+m_2c^2)^2 - p_1^2c^2 \\
& = & E_1^2 + 2 E_1 m_2c^2 + m_2^2c^4 - p_1^2c^2 \\
s & = & 2E_1m_2c^2 + m_1^2c^4 + m_2^2c^4
\end{eqnarray*}
\noindent The total invarient energy is both conserved and invarient.
Since it is conserved, its value is the same before the interaction as after.
Since it is invarient, we can equate our expression to $s$ in the center of
mass frame. Thus, {\bf due to its invarience and conservation, $s$ in the lab
frame before the interaction will equal $s$ in the center of mass frame
after the interaction}. After the interaction, there is only one particle
of mass $M$ and it is at rest in the center of mass frame. The particle
$M$ will be at rest, since we are interested in the threshold
energy $E_1$ to produce the particle. Hence, $s = M^2c^4$ after the collision:
\begin{eqnarray*}
2E_1m_2c^2 + m_1^2c^4 + m_2^2c^4 & = & M^2c^4 \\
& or & \\
E_1 & = & {{M^2-m_1^2-m_2^2} \over {2m_2}} c^2
\end{eqnarray*}
\newpage
Suppose we want to produce a delta particle, mass $M=m_\Delta = 1232 \; MeV/c^2$,
by scattering a pion, mass $m_\pi = 139 \; MeV/c^2$, off a proton,
mass $m_p = 940 \; MeV/c^2$. Then, to produce the $\Delta (1232)$ particle
the pion needs to have a total energy of
\begin{equation}
E_\pi = {{1232^2-139^2-940^2} \over {2(940)}} \approx 327 MeV
\end{equation}
\noindent or a kinetic energy in the lab frame of $K.E. \approx 327-139 = 188 MeV$.
If the pion has an energy less than $327 \; MeV$, then the $\Delta$ particle cannot
be produced. In this case, most likely the pion would scatter elastically off the
proton. If the pion has an energy a little greater than $327 \; MeV$, then the
$\Delta$ could be formed, which would decay back into a pion plus proton. If
the pion had enough energy, then a $\Delta$ plus another particle (a pion) could
be the final particles produced. \\
Now let's consider the kinematics involved in our assignment. We know the laboratory
momentum of the kaon, $p_{lab}=800 \; MeV/c$. We need to know the momentum of the kaon
in the center of mass reference frame of the kaon and the nucleus ($^{12}C$ in our case).
The energy of the kaon in the lab frame is
\begin{equation}
E_K = \sqrt{m_K^2c^4 + p_{lab}^2c^2}
\end{equation}
\noindent It will be convenient to work with the total invarient energy.
From the previous calculation we know
\begin{equation}
s = m_K^2c^4 + m_N^2c^4 + 2E_Km_Nc^2
\end{equation}
\noindent where $m_N$ is the mass of the nucleus.
Since $s$ is an invarient, it's value is the same in all inertial
reference frames. Thus, we can equate $s$ in the lab frame
to $s$ in the center of mass frame. This equality will allow us
to calculate the kaon's momentum in the center of mass frame.
\newpage
For two particle systems, the two particles will have equal and opposite
momenta in the center-of-mass reference frame, $\vec{p}_1 = -\vec{p}_2$.
Letting $|\vec{p}_1| \equiv p$, we have
\begin{equation}
s = (\sqrt{m_1^2c^4 + p^2c^2} + \sqrt{m_2^2c^4 + p^2c^2})^2 - 0^2
\end{equation}
\noindent The last term is zero, since $\vec{p}_1+\vec{p}_2 = 0$. After
a bit of algebra, one can solve for $p$ in the center of mass frame
\begin{equation}
p^2 = {{(s-(m_1c^2+m_2c^2)^2)(s-(m_1c^2-m_2c^2)^2)} \over {4sc^2}}
\end{equation}
\noindent in terms of the total invarient energy squared, $s$.\\
With these equations, we can calculate the kaon's momentum in the center
of mass frame in terms of its lab frame momenta, $p_{lab}$. First we
determine $E_K$ in the lab frame:
\begin{equation}
E_K = \sqrt{m_K^2c^4 + p_{lab}^2c^2}
\end{equation}
\noindent Then we calculate $s$ in the lab frame.
\begin{equation}
s = m_K^2c^4 + m_N^2c^4 + 2E_Km_Nc^2
\end{equation}
\noindent Finally, we solve for $p$ in the center-of-mass frame using
\begin{equation}
p^2 = {{(s-(m_Kc^2+m_Nc^2)^2)(s-(m_Kc^2-m_Nc^2)^2)} \over {4sc^2}}
\end{equation}
\noindent For $m_K=493 \; MeV/c^2$, $m_N=12(940) \; MeV/c^2$, and $p_{lab}=800 \; MeV/c$,
I obtain $p = 740 \; MeV/c$. I didn't use my calculator, I just had the
computer do the work with the following code in my homework 3 code:\\
{\bf
\noindent mkaon=493.0;\\
mnuc=12.0*940.0;\\
plab=800.0;\\
ekaon=sqrt(mkaon*mkaon+plab*plab);\\
s=mkaon*mkaon+mnuc*mnuc+2.0*ekaon*mnuc;\\
pc2=(s-(mkaon+mnuc)*(mkaon+mnuc))*(s-(mkaon-mnuc)*(mkaon-mnuc))/4.0/s;\\
pc=sqrt(pc2);\\
}
\newpage
\centerline{\bf Graphing in Root}
\bigskip
Below is a code from the CERN ROOT tutorials. It is a simple
code which can be used to graph one data set:\\
\bigskip
\noindent void Ggraph() \\
{ \\
//Draw a graph with error bars\\
\noindent TCanvas *c1 = new TCanvas("c1","A Simple Graph with error bars",200,10,700,500);\\
c1$\rightarrow$SetFillColor(0);\\
// c1$\rightarrow$SetGrid();\\
c1$\rightarrow$GetFrame()$\rightarrow$SetFillColor(0);\\
c1$\rightarrow$GetFrame()$\rightarrow$SetBorderSize(0);\\
\noindent const Int\_t n = 10;\\
Float\_t x[n] = {-0.22, 0.05, 0.25, 0.35, 0.5, 0.61,0.7,0.85,0.89,0.95};\\
Float\_t y[n] = {1,2.9,5.6,7.4,9,9.6,8.7,6.3,4.5,1};\\
Float\_t ex[n] = {.05,.1,.07,.07,.04,.05,.06,.07,.08,.05};\\
Float\_t ey[n] = {.8,.7,.6,.5,.4,.4,.5,.6,.7,.8};\\
\noindent TGraphErrors *gr = new TGraphErrors(n,x,y,ex,ey);\\
gr$\rightarrow$SetTitle("TGraphErrors Example");\\
gr$\rightarrow$SetMarkerColor(4);\\
gr$\rightarrow$SetMarkerStyle(21);\\
gr$\rightarrow$Draw("ALP");\\
\noindent c1$\rightarrow$Update();\\
}\\
\newpage
\begin{figure}
\includegraphics[width=14cm]{tgraph1.png}
\end{figure}
\begin{figure}
\includegraphics[width=14cm]{tgraph2.png}
\end{figure}
\end{document}