%         Created on  11-JANUARY-1996 by Phillip Gutierrez
%-------------------------------------------------------------------
\tolerance 1000
\documentclass [11pt]{article}
\usepackage{amsmath}
\usepackage{fancyheadings}
\usepackage{epsfig}
\textwidth 6.5in
\textheight 8.5in
\topmargin 0in
\oddsidemargin 0in
\evensidemargin 0in
%\parskip 1.5ex
\baselineskip 3.0ex
%\parindent 0in
%\thispagestyle{myheadings}
%\markright{\today}
%
%
\pagestyle{fancy}
\lhead{\today}
\chead{}
\rhead{P. Gutierrez}
\lfoot{}
\cfoot{Lecture 10-\thepage}
\rfoot{}
%
\def\reff{r_{\text{eff}}}
\input{my_macs.tex}
\input{d0_style.tex}
\begin{document}

\begin{center}
\LARGE\bf  Physics 4213/5213\\ Lecture 10\\
\end{center}
%
\section{Introduction}
That the neutrino associated with the electron and that associated with the
muon are different, required direct experimental verification. The first 
evidence
came from the non-observation of the muon decay through the channel 
$\mu^- \rightarrow e^-+\gamma$---branching ratio (probability) 
$< 4.9\times 10^{-11}$. 
This channel satisfies, energy-momentum conservation,
charge conservation, plus lepton number conservation, assuming that only the
total number of leptons needs to be conserved. This method is indirect, in that
there maybe some other unknown reason (other than specific flavor lepton number
being conserved) for the reaction not occurring.

In 1961 a direct measurement of the muon type neutrino was attempt. The goal
of the experiment was to show that muon-type neutrinos, in charged current
reactions, only produce muons and not electrons---note in 1961, only two 
leptons
were known. The experiment required a beam of $\nu_\mu$, that was intense 
enough
to allow for the possibility of some number interacting and being 
detected---note
neutrinos have a very low probability of interacting with matter. 
The $\nu_\mu$
are produced from pion decay $\pi^+ \rightarrow \mu^++\nu_\mu$---branching
ratio $(99.98782\pm0.00014)\%$ while $BR(\pi^+\rightarrow e^++\nu_e) =
(1.218\pm0.014)\times 10^{-4}$. Finally, the pions are produced through proton
nucleon interactions.

This lecture will describe the experiment, along with an introduction to 
cross sections
and the probability of interaction plus decay probabilities.
\section{Pion Production}
As stated in the introduction, the pions are produced through proton nucleon
interactions $p+N\rightarrow p+N + n\pi$ where n is the number of charged
pions produced
and must be a multiple of three---note, for every two 
charged pions
produced a neutral pion is also produced; in these
reactions to conserve charge, $\pi^+\pi^-$ pairs must be produced. 
To simplify the calculations, it
will be assumed that both the neutron and proton have the same mass and 
that the
mass is \nunit{1}{GeV}. The experiment performed used \nunit{15}{GeV} protons
incident on a beryllium target (collection of protons and neutrons).

To determine how many pions are produced, calculate the center of mass energy
\begin{equation}
  s= (P_1+P_2) = 2 m_p^2 + 2 E_{p}^i m_p \qquad \Rightarrow \sqrt{s} = 
5.7\,\text{GeV}
\end{equation}
Assuming that all particles are produced at rest in the center of mass frame
(equal momenta in the lab frame), this energy is required to produce the 
two protons
plus the $n$ pions. Since the pion mass is \nunit{140}{MeV} (note the 
neutral pion
has a mass of \nunit{135}{MeV}) and pions are produced in triples, there 
can be a
maximum of 9 pions produced or three of each charge species. Since this 
configuration
is not likely (all particles produced at rest in the center of mass frame), 
the number
of pions is more likely to be 3 or 6 meaning that there will be 1 to 2 $\pi^+$
per collision.

The next question that requires an answer is, given a beam of protons 
what is the
probability or how many will interaction in the beryllium. Consider one 
incident
proton from a beam of cross sectional area $A$, which is less than the 
cross sectional
area of the target. If there are $N$ nucleons scattered over the surface of an
infinitesimally thin target over the cross sectional area $A$, then the 
probability
of the proton interacting is $P=N\sigma /A$. This is rewritten in terms of a 
volume density, since the target will have a finite thickness;
\begin{equation}
  P= \frac{N\sigma}{A \ell}\ell
\end{equation}
where $\ell$ is the thickness of the target. Next convert the number 
density into
a mass density since this is what is measurable and given in tables:
\begin{equation}
  P= N_A \rho \sigma\ell \label{eq:prob}
\end{equation}
where $N_A$ is the number of particles per moles---note that 
one mole is equal 
to a
mass equal to the atomic number of the element in grams;
in this case the atomic number 
is divided
into the density to get the number of beryllium atoms and then multiplied 
through to
get the number of nucleons. Finally the number of protons that interact 
is given by
multiplying equation \ref{eq:prob} by the number of protons incident on 
the target
\begin{equation}
  N_{\text{int}}= N_p N_A \rho \sigma\ell
\end{equation}

For the case of \nunit{15}{GeV} protons incident on beryllium, 
$\sigma = 50\,\text{mb}$---$1\,\text{mb} = 1\times 10^{-27}\,\text{cm}^2$. The
probability for a single proton to interact in \nunit{7.6}{cm} thick beryllium
is $P=0.41$---the density of Be is \nunit{1.848}{g/cm$^3$}.
\section{Production of $\nu_\mu$}
As stated in the introduction, the production of $\nu_\mu$ comes from the decay
of the pions; $\pi^+\rightarrow \mu+\nu_\mu$. The pion lifetime is 
\nunit{2.6\times 10^{-8}}{sec}, which corresponds to a half-life, that is 
not every
pion decays after this time. The probability as a function of time for 
any pion
to decay is derived by assuming that the probability is independent of 
how long it
has lived. In this case (assume $N$ pions) the number of pions left after any
interval $dt$ is independent of any previous interval $-dN = \lambda N 
dt$---the
number that are lost (decay) is proportional to the number in the time 
interval.
Solving for $N(t)$ gives;
\begin{equation}
  \frac{dN}{N} = -\lambda dt \qquad \Rightarrow \qquad N(t)=N_0 e^{-t/\tau}
\end{equation}

In the experiment, a gap of \nunit{21}{m} was left for the pions to decay in.
To determine what fraction of pions decay in this space, an assumption of their
momentum has to be made. Assume that 6 pions are produced and that all 
particles
in the final state share equal momentum. Since all the particles have to carry
\nunit{15}{GeV} of momentum, each particle must carry \nunit{1.9}{GeV} 
under our
assumption. Transforming the life-time to the lab frame from the rest frame
gives
\begin{equation}
  \tau_{\text{lab}} = \tau_{\text{rest}} \gamma =
 \tau_{\text{rest}} \frac{E}{m_\pi} = 3.5\times10^{-7}\,\text{sec}
\end{equation}
Given that for this momentum, the pion is traveling at the speed of light, 
the
pion travels \nunit{105}{m} and the probability of a single pion decaying
in \nunit{21}{m} is
\begin{equation}
  P=1- e^{-x/x_0}=1-e^{-21/105}=.18
\end{equation}
{\em Taking pions of \nunit{1}{GeV} increases the probability to $0.31$.}
%
\section{Neutrino Interactions}
Earlier in the semester, it was shown that the cross section of a 
neutrino interacting with
an electron (same for a quark) is proportional to the energy of the neutrino
($\sigma_{\nu\text{-}q} \propto E_\nu$). The exact form, with the 
proportionality
constant is $\sigma_{\nu p}=0.67\times 10^{-38}E\, \text{cm}^2$ where the energy
is in GeV---note that this is 12
orders of magnitude smaller than the proton-proton cross section for an 
energy of
\nunit{15}{GeV}.

To make an estimate of the $\nu$ energy, first determine the energy in the
pion rest frame. Start by using energy conservation and realizing that
the energy and momentum are equal for the neutrino;
\begin{equation}
  m_\pi = E_\mu + E_\nu \qquad \Rightarrow \qquad
  m_\pi = \sqrt{E_\nu^2 + m_\mu^2} + E_\nu \qquad \Rightarrow \qquad
  E_\nu = \frac{m_\pi^2 - m_\mu^2}{2 m_\pi}
\end{equation}
were momentum conservation is used in the second expression. Inserting the
masses into the equation above gives an energy for the $\nu_\mu$ of
\nunit{30}{MeV}. Next assume that the neutrino is emitted in the same
direction as the pion was traveling, also assume that the pion momentum
is \nunit{2.5}{GeV}. The neutrino momentum is given by;
\begin{equation}
  \begin{pmatrix}
    E_{\text{lab}} \\ p_{\text{lab}}
  \end{pmatrix} = 
  \begin{pmatrix}
    \gamma & \gamma \beta \\ \gamma \beta & \gamma
  \end{pmatrix}
  \begin{pmatrix}
     E_{\text{rest}} \\ p_{\text{rest}}  
  \end{pmatrix}
\qquad \gamma \approx \gamma \beta = \frac{E_\pi}{m_\pi} = 13.57
\qquad \Rightarrow \qquad E_\nu = \nunit{.814}{GeV}
\end{equation}
Based on this neutrino energy $\sigma_{\nu}= 
\nunit{5\times10^{-39}}{cm}^2$, which is 13 orders of magnitude smaller
than the proton cross section. 

For the experiment, a \nunit{13.5}{m}
thick iron sheild ($\rho = \nunit{7.87}{g/cm$^3$}$)
was placed after the decay path. The probability for
a neutrino to interact in this sheild was $4.4\times 10^{-11}$ while
for a hadron the probability is 1. Beyond this sheild, \nunit{90}{inches}
of aluminum ($\rho=\nunit{2.7}{g/cm$^3$}$)
was placed to provide a means for the neutrinos to interact
and create muons; probability of neutrino interacting is $1.8\times 10^{-12}$.
(The proton beam intesity was $\approx \nunit{10^{16}}{pulse}^{-1}$; 
pulse every \nunit{1.2}
{sec}.)
The muons were detected by ionization detectors placed 
between the aluminum plates.
%\begin{thebibliography}{9}
%  \bibitem{r1} R
%\end{thebibliography}
\end{document}