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%TCIDATA{Created=Mon Nov 04 19:58:42 1996}
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\markright{{\sf EM2 test, 7 Nov 95}}
\begin{document}


\begin{center}
\textbf{Phys 743 - EM II}

\textbf{In-class test 2 - 4 Nov. 1996}
\end{center}

\smallskip

\noindent \textbf{1.}

A rectangular opening\textbf{\ }defined by $x=\pm (a/2),\;y=\pm (b/2)$
exists in a flat screen filling the $xy$ plane. When light of intensity $%
I_{0}$ and wavelength $\lambda <<a$ is incident normally on the screen, the
transmitted amplitude is approximately proportional to the diffraction
integral
\[
\int \frac{e^{ikR}}{R}\,da^{\prime }
\]
where $R$ is the distance between the observation point, $\mathbf{x}$,%
\textbf{\ }and a point in the opening,\textbf{\ }$\mathbf{x}^{\prime }$. In
this approximation, evaluate explicitly the transmitted power as a function
of suitably defined angles $\theta ,\,\phi $ when$\ R>>b\geq a.$

\smallskip \ \ \ 

\noindent \textbf{2.}

Light is incident on a small dielectric sphere.\textbf{\ }What are the
angular distribution and the polarization of the scattered radiation if the
incident light is circularly polarized?

\smallskip \ \ \ 

\noindent \textbf{3.}

A relativistic starship emitting light of frequency $\omega _{0}$ appears on
the horizon and flies overhead at height $b$ with constant velocity $\mathbf{%
v}$. What frequency do we receive, as a function of our time $t$? 

\smallskip\ \ \ 

\noindent \textbf{4.}

A $\gamma $ ray of energy $\hbar \omega $ is incident on a proton at rest. 

\textbf{a. }What is the COM\ velocity?

\textbf{b. }What is the COM\ energy? 

\textbf{c. }What is the threshold energy of the $\gamma $ ray for
photoproduction of a $\pi ^{0}$ meson ($\gamma p\rightarrow \pi ^{0}p$)? 

\noindent (You may answer parts (b) and (c) independently of part (a).
Recall that $m_{\pi ^{0}}=135.0\,$MeV, $m_{p}=938.5$\thinspace MeV).

\begin{center}
\newpage\ \textbf{Solutions}
\end{center}

\smallskip

\noindent \textbf{1.}

The diffraction integral is 
\[
F(\theta ,\phi )=\frac 1{S^{\prime }}\int_{S^{\prime }}\exp (-i\mathbf{%
k\cdot r}^{\prime })\,da^{\prime } 
\]
where $S^{\prime }$ is the area of the opening (I have thrown in the factor $%
1/S^{\prime }$ to make $F$ dimensionless). In our case 
\[
\mathbf{k\cdot r}^{\prime }=k\sin \theta (x^{\prime }\cos \phi +y^{\prime
}\sin \phi ) 
\]
and we find 
\[
F(\theta ,\phi )=\frac{\sin (\frac 12ka\sin \theta \cos \phi )}{\frac
12ka\sin \theta \cos \phi }\,\frac{\sin (\frac 12kb\sin \theta \sin \phi )}{%
\frac 12kb\sin \theta \sin \phi } 
\]
The transmitted power is proportional to $F^2.$ See \textbf{Comment 1} for
all the other factors (not required).

\smallskip\ \ \ 

\noindent \textbf{2.}

From the Lorentz transformation 
\begin{eqnarray*}
dx &=&\gamma (dx^{\prime }+Vdt^{\prime }) \\
dt &=&\gamma (dt^{\prime }+V\,dx^{\prime }/c^2)
\end{eqnarray*}
and also $dy=dy^{\prime },\;dz=dz^{\prime },$ get 
\[
\frac{dx}{dt}=\frac{\dfrac{dx^{\prime }}{dt^{\prime }}+V}{1+\dfrac V{c^2}\,%
\dfrac{dx^{\prime }}{dt^{\prime }}\,}\quad \quad \quad \quad \quad \quad
\quad \frac{dy}{dt}=\frac{\dfrac 1\gamma \dfrac{dy^{\prime }}{dt^{\prime }}}{%
1+\dfrac V{c^2}\,\dfrac{dx^{\prime }}{dt^{\prime }}\,} 
\]
and similarly for $dz/dt$. In Fizeau's experiment, $dx^{\prime }/dt^{\prime
} $ is the speed of light in a medium, seen by a comoving observer. The
observer in the lab sees light moving with speed 
\[
\frac{dx}{dt}=\frac{\dfrac cn+V}{1+\dfrac V{nc}\,}\simeq \left( \dfrac
cn+V\right) \left( 1-\dfrac V{nc}\right) \simeq \dfrac cn+V\left( 1-\dfrac
1{n^2}\right) 
\]

\smallskip\ \ \ 

\noindent \textbf{3.}

\textbf{(a)} $\sin ^2\Theta ,$ where $\Theta $ is the angle from the
incident polarization vector $\mathbf{\epsilon }_{0.}$ If $\mathbf{\epsilon }%
_0=(\cos \phi _0,\sin \phi _0,0)$ 
\[
\sin ^2\Theta =\sin ^2(\phi -\phi _0)+\cos ^2\theta \cos ^2(\phi -\phi _0) 
\]

(\textbf{b) }$\frac 12(1+\cos ^2\theta )$, where $\theta $ is the angle from
the incident direction (as in \textbf{(a)}).

\smallskip\ \ \ 

\noindent \textbf{4.}

\textbf{(a) }$\mathbf{E}^2-\mathbf{B}^2$ is a Lorentz scalar, equal to $%
\frac 12\,F^{\alpha \beta }F_{\beta \alpha }$

\textbf{(b) }$\mathbf{E}^2+\mathbf{B}^2$ is not a Lorentz scalar. If it
were, $\mathbf{E}^2$ and $\mathbf{B}^2$ would each be Lorentz scalars, due
to (\textbf{a). }One can note that $(1/8\pi )(\mathbf{E}^2+\mathbf{B}^2)$ is
the energy density, which is the $T_{00}$ component of the 4-d stress tensor.

\textbf{(c) }$\mathbf{E}\cdot \mathbf{B}$ is a Lorentz scalar, equal to $%
-\frac 14\,F^{\alpha \beta }\mathcal{F}_{\alpha \beta }=-\frac 12\,\epsilon
_{\alpha \beta \gamma \delta }F^{\alpha \beta }F^{\gamma \delta }$\quad
\quad \quad

\textbf{(d) }$\mathbf{E}\times \mathbf{B}$ is not a Lorentz scalar, because
it is a 3-d vector. One can note that $(c/4\pi )\mathbf{E}\times \mathbf{B}$
is the Poynting vector.

\smallskip\ \ \ 

\noindent \textbf{5.}

\textbf{(a) }$E_\mu $ is maximum when $\mu $ and $\pi ^0$ go off in the
opposite directions, and $E_\nu =0$ (see \textbf{Comment} \textbf{2}).
Squaring the four-vector relation $\,p_K-p_\mu =p_\pi \,$ we obtain 
\[
E_\mu =\frac{m_K^2+m_\mu ^2-m_\pi ^2}{2m_K}=240\,\mathrm{MeV} 
\]

(\textbf{b) }$E_\mu $ is minimum when $\pi ^0$ and $\nu $ go off in opposite
directions and $\mu $ is created at rest. Then \textbf{\ }$E_\mu =m_\mu .$ 
\textbf{\ }

\begin{center}
\newpage\ 
\end{center}

\textbf{Comment 1. }\textsl{Diffraction with all the factors }(Jackson's
problem 9.19, with $\beta =0.$ See Estevez and Suen, page 255)

Kirchhoff-Smythe is given by Jackson's Eq. (9.156): 
\[
\mathbf{E}(\mathbf{r)=-}\frac{ike^{ikr}}{2\pi r}\,\mathbf{n}\times (\mathbf{%
n\times E}_0)\int \exp [-ik\sin \theta (x^{\prime }\cos \phi +y^{\prime
}\sin \phi )]da^{\prime } 
\]
with 
\[
\mathbf{-n}\times (\mathbf{n\times E}_0)=\mathbf{E}_0-\mathbf{n}(\mathbf{%
n\cdot E}_0)=E_0\left[ 
\begin{array}{c}
-\sin ^2\theta \sin \phi \cos \phi \\ 
1-\sin ^2\theta \sin ^2\phi \\ 
-\sin \theta \cos \theta \sin \phi
\end{array}
\right] 
\]
Plain Kirchhoff is given by Eqs. (9.125) and (9.7): 
\[
\psi (\mathbf{r)=-}\frac{ike^{ikr}}{4\pi r}(1+\cos \theta )\psi _0\int \exp
[-ik\sin \theta (x^{\prime }\cos \phi +y^{\prime }\sin \phi )]da^{\prime } 
\]
In either case we need to compute the diffraction factor 
\[
F(\theta ,\phi )=\int_{-a/2}^{a/2}\exp (-ikx^{\prime }\sin \theta \cos \phi )%
\frac{dx^{\prime }}a\int_{-b/2}^{b/2}\exp (-iky^{\prime }\sin \theta \sin
\phi )\frac{dy^{\prime }}b 
\]
We find 
\[
F(\theta ,\phi )=\frac{\sin (\frac 12ka\sin \theta \cos \phi )}{\frac
12ka\sin \theta \cos \phi }\;\frac{\sin (\frac 12kb\sin \theta \sin \phi )}{%
\frac 12kb\sin \theta \sin \phi } 
\]
For Kirchhoff-Smythe we compute the transmitted power per unit solid angle $%
dP/d\Omega =(c/8\pi )r^2|E|^2$ and divide by the incident power $P_0=(c/8\pi
)abE_0^{\,2}.$ Using 
\[
\left[ \mathbf{E}_0-\mathbf{n}(\mathbf{n\cdot E}_0)\right] ^2=E_0^{\,2}-(%
\mathbf{n\cdot E}_0)^2=E_0^{\,2}[1-\sin ^2\theta \sin ^2\phi ] 
\]
we find 
\[
\frac 1{P_0}\frac{dP}{d\Omega }=\frac{k^2ab}{4\pi ^2}\,[1-\sin ^2\theta \sin
^2\phi ]F^2 
\]
Estevez and Suen write the ``obliquity factor'' in square brackets as $\cos
^2\theta +\sin ^2\theta \cos ^2\phi .$

For Kirchhoff we have similarly 
\[
\frac 1{P_0}\frac{dP}{d\Omega }=\frac{k^2ab}{4\pi ^2}\left[ \tfrac 12(1+\cos
\theta )\right] ^2F^2 
\]

\begin{center}
\hrulefill
\end{center}

\textbf{Comment 2.}\textsl{\ End points of decay spectra}

In the answer to question \textbf{5a}, the condition $E_\nu =0$ may need
some comment. In general, when a particle of mass $M$ decays into several
particles of masses $m_1,\,m_2,\,m_3,\dots $ the maximum of $E_1$ occurs
when all the other particles come out with the same velocity $v$ in the
direction opposite to particle 1. If $m_2$ is finite and $m_3=0$ (for
instance) there is an apparent contradiction because the speed of particle 3
must be $c,$ but the speed of particle 2 must be less than $c.$ To figure
out what happens we take the limit $m_3\rightarrow 0.$ Because $E_3=m_3c^2/%
\sqrt{1-v^2/c^2}$, we see that $E_3\rightarrow 0.$

In principle the maximum $E_\mu $ in $K\rightarrow \mu +\pi ^0+\nu \,$ can
be used to decide if the neutrino has a finite mass. In practice it is hard
to attain the necessary accuracy (even in other, more favorable cases). By
the way, the neutrino involved here is the $\mu $ neutrino or antineutrino,
according to 
\[
K^{\,+}=\mu ^{+}+\pi ^0+\nu _\mu \quad \quad \quad \quad K^{\,-}=\mu
^{-}+\pi ^0+\overline{\nu }_\mu 
\]

\end{document}
