source: trunk/documents/UserDoc/DocBookUsersGuides/PhysicsReferenceManual/latex/optical/optical.tex @ 1345

% GEANT4 Physics Reference Manual - Optical Photons
% in LaTex 2e - adopted from GEANT3 manual by P. Gumplinger

\section{Interactions of optical photons}

Optical photons are produced when a charged particle traverses:

\begin{enumerate}
\item a dielectric material with velocity above the \v{C}erenkov
threshold; 
\item a scintillating material.
\end{enumerate}

\subsection{Physics processes for optical photons}

A photon is called optical when its wavelength is much greater than the
typical atomic spacing, for instance when $\lambda \geq 10nm$ 
which corresponds to an energy $E \leq 100eV$\@. Production of an 
optical photon in a HEP detector is primarily due to:

\begin{enumerate}
\item \v{C}erenkov effect;
\item Scintillation.
\end{enumerate}

Optical photons undergo three kinds of interactions:

\begin{enumerate}
\item Elastic (Rayleigh) scattering;
\item Absorption;
\item Medium boundary interactions.
\end{enumerate}

\subsubsection{Rayleigh scattering}

For optical photons Rayleigh scattering is usually unimportant. For
$\lambda=.2\mu m$ we have $\sigma_{Rayleigh} \approx .2b$ for $N_{2}$ or
$O_{2}$ which gives a mean free path of $\approx1.7km$ in air and
$\approx1m$ in quartz. Two important exceptions are aerogel, which is
used as a \v{C}erenkov radiator for some special applications and large
water \v{C}erenkov detectors for neutrino detection. 

The differential cross section in Rayleigh scattering,
$d\sigma/d\Omega$, is proportional to $\cos^{2}\theta$, where $\theta$ 
is the polar angle of the new polarization with respect to the old 
polarization. 

\subsubsection{Absorption}

Absorption is important for optical photons because it determines the
lower $\lambda$ limit in the window of transparency of the radiator.
Absorption competes with photo-ionization in producing the signal in the
detector, so it must be treated properly in the tracking of optical
photons. 

\subsubsection {Medium boundary effects}

When a photon arrives at the boundary of a dielectric medium, its
behaviour depends on the nature of the two materials which join at that
boundary: 

\begin{itemize}
\item Case dielectric $\rightarrow$ dielectric.\\
The photon can be transmitted (refracted ray) or reflected (reflected
ray). In case where the photon can only be reflected, total internal
reflection takes place. 
\item Case dielectric $\rightarrow$ metal.\\
The photon can be absorbed by the metal or reflected back into the
dielectric. If the photon is absorbed it can be detected according to
the photoelectron efficiency of the metal. 
\item Case dielectric $\rightarrow$ black material.\\
A black material is a tracking medium for which the user has not
defined any optical property. In this case the photon is immediately
absorbed undetected.
\end{itemize}

\subsection {Photon polarization}

The photon polarization is defined as a two component vector normal to
the direction of the photon: 

\begin{displaymath}
{a_{1}e^{i\Phi_{1}} \choose a_{2}e^{i\Phi_{2}}} =
e^{\Phi_{o}}
{a_{1}e^{i\Phi_{c}} \choose a_{2}e^{-i\Phi_{c}}}
\end{displaymath}

where $\Phi_{c}= (\Phi_{1}-\Phi_{2})/2$ is called circularity and
$\Phi_{o}=(\Phi_{1}+\Phi_{2})/2$ is called overall phase. Circularity
gives the left- or right-polarization characteristic of the photon. RICH
materials usually do not distinguish between the two polarizations and
photons produced by the \v{C}erenkov effect and scintillation are
linearly polarized, that is $\Phi_{c}=0$\@. 

The overall phase is important in determining interference effects between
coherent waves. These are important only in layers of thickness comparable
with the wavelength, such as interference filters on mirrors. The effects of
such coatings can be accounted for by the empirical reflectivity factor for
the surface, and do not require a microscopic simulation. GEANT4 does not
keep track of the overall phase.

Vector polarization is described by the polarization angle $\tan \Psi =
a_{2}/a_{1}$\@. Reflection/transmission probabilities are sensitive to
the state of linear polarization, so this has to be taken into account.
One parameter is sufficient to describe vector polarization, but to
avoid too many trigonometrical transformations, a unit vector
perpendicular to the direction of the photon is used in GEANT4. The
polarization vector is a data member of \texttt{G4DynamicParticle}. 

\subsection{Tracking of the photons}

Optical photons are subject to in flight absorption, Rayleigh scattering
and boundary action. As explained above, the status of the photon is
defined by two vectors, the photon momentum ($\vec{p}=\hbar \vec{k}$) and
photon polarization ($\vec{e}$). By convention the direction of the
polarization vector is that of the electric field. Let also $\vec{u}$ be
the normal to the material boundary at the point of intersection,
pointing out of the material which the photon is leaving and toward the
one which the photon is entering. The behaviour of a photon at the
surface boundary is determined by three quantities: 

\begin{enumerate}
\item refraction or reflection angle, this represents the kinematics of
the effect; 
\item amplitude of the reflected and refracted waves, this is the
dynamics of the effect; 
\item probability of the photon to be refracted or reflected, this is
the quantum mechanical effect which we have to take into account if we
want to describe the photon as a particle and not as a wave. 
\end{enumerate}

As said above, we distinguish three kinds of boundary action, dielectric
$\rightarrow$ black material, dielectric $\rightarrow$ metal, dielectric
$\rightarrow$ dielectric. The first case is trivial, in the sense that
the photon is immediately absorbed and it goes undetected. 

To determine the behaviour of the photon at the boundary, we will at
first treat it as an homogeneous monochromatic plane wave: 

\begin{displaymath}
\vec{E} = \vec{E}_{0}e^{i\vec{k} \cdot \vec{x}-i\omega t}
\end{displaymath}
\begin{displaymath}
\vec{B} = \sqrt{\mu \epsilon} \frac{\vec{k} \times \vec{E}}{k}
\end{displaymath}

\subsubsection{Case dielectric $\rightarrow$ dielectric}

In the classical description the incoming wave splits into a reflected
wave (quantities with a double prime) and a refracted wave (quantities
with a single prime). Our problem is solved if we find the following
quantities: 

\begin{displaymath}
\vec{E}' = \vec{E}_{0}' e^{i\vec{k}'\cdot \vec{x}-i\omega t}
\end{displaymath}
\begin{displaymath}
\vec{E}'' = \vec{E}_{0}'' e^{i\vec{k}''\cdot \vec{x}-i\omega t}
\end{displaymath}

For the wave numbers the following relations hold:

\begin{displaymath}
|\vec{k}| = |\vec{k}''| = k = \frac{\omega}{c}\sqrt{\mu \epsilon}
\end{displaymath}
\begin{displaymath}
|\vec{k}'| = k' = \frac{\omega}{c}\sqrt{\mu ' \epsilon '}
\end{displaymath}

Where the speed of the wave in the medium is $v=c/\sqrt{\mu \epsilon}$
and the quantity $n=c/v=\sqrt{\mu \epsilon}$ is called refractive index
of the medium. The condition that the three waves, refracted, reflected
and incident have the same phase at the surface of the medium, gives us
the well known Fresnel law: 

\begin{displaymath}
(\vec{k} \cdot \vec{x})_{surf} = (\vec{k}' \cdot \vec{x})_{surf} =
(\vec{k}'' \cdot \vec{x})_{surf}
\end{displaymath}

\begin{displaymath}
k \sin i = k' \sin r = k'' \sin r'
\end{displaymath}

where $i, r, r'$ are, respectively, the angle of the incident, refracted
and reflected ray with the normal to the surface. From this formula the
well known condition emerges: 

\begin{displaymath}
         i  =  r'
\end{displaymath}
\begin{displaymath}
\frac{\sin i}{\sin r} = \sqrt{\frac{\mu ' \epsilon '}{\mu \epsilon}} =
\frac{n'}{n}
\end{displaymath}

The dynamic properties of the wave at the boundary are derived from
Maxwell's equations which impose the continuity of the normal components
of $\vec{D}$ and $\vec{B}$ and of the tangential components of $\vec{E}$
and $\vec{H}$ at the surface boundary. The resulting ratios between the
amplitudes of the the generated waves with respect to the incoming one
are expressed in the two following cases: 

\begin{enumerate}

\item a plane wave with the electric field (polarization vector)
perpendicular to the plane defined by the photon direction and the
normal to the boundary: 

\begin{displaymath}
\frac{E_{0}'}{E_{0}} = \frac{2n\cos i}{n \cos i = \frac{\mu}{\mu '}
n' \cos r} = \frac{2n \cos i}{n \cos i + n' \cos r}
\end{displaymath}
\begin{displaymath}
\frac{E_{0}''}{E_{0}} = \frac{n \cos i - \frac{\mu}{\mu '}
n' \cos r}{n \cos i + \frac{\mu}{\mu '}n' \cos r} =
\frac{n \cos i - n' \cos r}{n \cos i + n' \cos r}
\end{displaymath}

where we suppose, as it is legitimate for visible or near-visible
light, that $\mu/\mu ' \approx 1$;

\item a plane wave with the electric field parallel to the above
surface: 

\begin{displaymath}
\frac{E_{0}'}{E_{0}} = \frac{2n \cos i}{\frac{\mu}{\mu '}n' 
\cos i + n \cos r} = \frac{2n \cos i}{n' \cos i + n \cos r}
\end{displaymath}
\begin{displaymath}
\frac{E_{0}''}{E_{0}} = \frac{\frac{\mu}{\mu '}n' \cos i - n \cos r}
{\frac{\mu}{\mu '}n' \cos i + n \cos r} = 
\frac{n' \cos i - n \cos r}{n' \cos i + n \cos r}
\end{displaymath}

with the same approximation as above.

\end{enumerate}

We note that in case of photon perpendicular to the surface, the
following relations hold: 

\begin{displaymath}
\frac{E_{0}'}{E_{0}} = \frac{2n}{n'+n} \qquad
\frac{E_{0}''}{E_{0}} = \frac{n'-n}{n'+n}
\end{displaymath}

where the sign convention for the parallel field has been adopted. This
means that if $n'>n$ there is a phase inversion for the reflected wave. 

Any incoming wave can be separated into one piece polarized parallel to
the plane and one polarized perpendicular, and the two components
treated accordingly. 

To maintain the particle description of the photon, the probability to
have a refracted or reflected photon must be calculated. The constraint
is that the number of photons be conserved, and this can be imposed via
the conservation of the energy flux at the boundary, as the number of
photons is proportional to the energy. The energy current is given by
the expression: 

\begin{displaymath}
\vec{S} = \frac{1}{2}\frac{c}{4\pi}\sqrt{\mu \epsilon}
\vec{E} \times \vec{H} = \frac{c}{8\pi}\sqrt{\frac{\epsilon}{\mu}}
E_{0}^{2}\hat{k}
\end{displaymath}

and the energy balance on a unit area of the boundary requires that: 

\begin{displaymath}
\vec{S} \cdot \vec{u} = \vec{S}' \cdot \vec{u} - \vec{S}'' \cdot \vec{u}
\end{displaymath}
\begin{displaymath}
S \cos i = S' cos r + S'' cos i
\end{displaymath}
\begin{displaymath}
\frac{c}{8\pi}\frac{1}{\mu}nE_{0}^{2}\cos i =
\frac{c}{8\pi}\frac{1}{\mu '}n'E_{0}'^{2}\cos r +
\frac{c}{8\pi}\frac{1}{\mu}nE_{0}''^{2}\cos i
\end{displaymath}

If we set again $\mu /\mu ' \approx 1$, then the transmission
probability for the photon will be: 

\begin{displaymath}
T = (\frac{E_{0}'}{E_{0}})^{2} \frac{n' \cos r}{n \cos i}
\end{displaymath}

and the corresponding probability to be reflected will be $R=1-T$.

In case of reflection, the relation between the incoming photon
($\vec{k},\vec{e}$), the refracted one ($\vec{k}', \vec{e}'$) and the
reflected one ($\vec{k}'', \vec{e}''$) is given by the following
relations: 

\begin{displaymath}
\vec{q} = \vec{k} \times \vec{u}
\end{displaymath}
\begin{displaymath}
\vec{e}_{\perp} = (\frac{\vec{e} \cdot
\vec{q}}{|\vec{q}|}) \frac{\vec{q}}{|\vec{q}|}
\end{displaymath}
\begin{displaymath}
\vec{e}_{\parallel} = \vec{e} - \vec{e}_{\perp}
\end{displaymath}
\begin{displaymath}
e_{\parallel}' = e_{\parallel} \frac{2n \cos i}{n'\cos i + n \cos r}
\end{displaymath}
\begin{displaymath}
e_{\perp|}' = e_{\perp} \frac{2n \cos i}{n \cos i + n' \cos r} 
\end{displaymath}
\begin{displaymath} 
e_{\parallel}'' = \frac{n'}{n}e_{\parallel}' - e_{\parallel}
\end{displaymath} 
\begin{displaymath}
e_{\perp}'' = e_{\perp}' - e_{\perp}
\end{displaymath} 

After transmission or reflection of the photon, the polarization vector
is re-normalized to 1. In the case where $\sin r = n \sin i/n' > 1$ then
there cannot be a refracted wave, and in this case we have a total
internal reflection according to the following formulas: 
\begin{displaymath}
\vec{k}'' = \vec{k} - 2(\vec{k} \cdot \vec{u})\vec{u}
\end{displaymath}
\begin{displaymath}
\vec{e}'' = -\vec{e} + 2(\vec{e} \cdot \vec{u})\vec{u}
\end{displaymath}

\subsubsection{Case dielectric $\rightarrow$ metal}

In this case the photon cannot be transmitted. So the probability for
the photon to be absorbed by the metal is estimated according to the
table provided by the user. If the photon is not absorbed, it is
reflected. 

\begin{latexonly}

\begin{thebibliography}{99}
\bibitem{one} J.D.~Jackson, \emph{Classical Electrodynamics},
J.~Wiley \& Sons Inc., New York, 1975.
\end{thebibliography}

\end{latexonly}

\begin{htmlonly}

\subsection{Bibliography}

\begin{enumerate}
\item J.D.~Jackson, \emph{Classical Electrodynamics},
J.~Wiley \& Sons Inc., New York, 1975.
\end{enumerate}

\end{htmlonly}