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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

{\bf Set 7:} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

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\item Exercise: Electric and magnetic fields for an infinitely long, 
      straight wire.  \smallskip

\noindent (a) We consider a cylinder of length $L$ and radius $\rho$, 
which is centered around the $\hat{z}$ axis. The charge in the cylinder 
is $Q=L\,\lambda$ and the surface area of its mantel is 
$A=2\pi\,\rho\,L$. By symmetry reasons Gauss's law gives then
$$ E(\rho)\,2\,\pi\,\rho\,L = 4\pi\,Q = 4\pi\,L\,\lambda\ \Rightarrow\
   E(\rho) = 2\,\frac{\lambda}{\rho} $$
and as vector
$$ \vec{E}=E(\rho)\,\hat{\rho}=2\,\frac{\lambda}{\rho}\,\hat{\rho}\ .$$
\noindent (b) Lorentz transformation gives
$$\vec{E}^{\,'}=\gamma\,\vec{E}~~{\rm and}~~
  \vec{B}^{\,'}=\gamma\,E(\rho)\,\frac{v}{c}\,\hat{z}\times\hat{\rho}
               =\gamma\,E(\rho)\,\frac{v}{c}\,\hat{\phi}
  =\gamma\,\frac{v}{c}\,\frac{2\,\lambda}{\rho}\,\hat{\phi}\ . $$
Introducing the current $I=v\,\lambda$ and taking the non-relativistic 
limit $\gamma\to 1$, we obtain the familiar result of magnetostatics
(a special case of the \index{Biot-Savart law}Biot-Savart law)
$$ \vec{B}^{\,'}=\frac{2}{c}\,\frac{I}{\rho}\,\hat{\phi}\ .$$
\bigskip


\item Exercise: Charge density from a potential (hydrogen atom).
\smallskip

\noindent 
Laplace's equation for the problem is
$$ \nabla^2\Phi(r) = -4\pi\,\rho(\vec{r})~~~{\rm with}~~~
   \Phi(r) = \frac{q}{r}\,e^{-\alpha r}\,
             \left(1+\frac{\alpha r}{2}\right)\,.  $$
For $r\ne 0$ we use the Laplace operator in spherical coordinates 
and have for the continuous part
$$ \nabla^2\Phi(r) = \frac{1}{r^2}\frac{d\ }{dr}\,
   \left(r^2\frac{d\Phi}{dr}\right)\,. $$
It is straightforward, but a bit lengthy, to calculate the derivatives 
of the potential (algebraic programs like Maple and Mathematica are 
well-suited for such tasks). The results are
$$ r^2\frac{d\Phi}{dr} = -q\,e^{-\alpha r}\,\left(1 + \alpha r +
   \alpha^2 r^2/2 \right)\,, $$
$$ \frac{1}{r^2}\frac{d\ }{dr}\,\left(r^2\frac{d\Phi}{dr}\right) 
   = +\frac{\alpha^3 q}{2}\,e^{-\alpha r}\,.$$
The singular part is:
$$ q\,e^{-\alpha r}\,\left(1+\frac{\alpha r}{2}\right)\,\nabla^2\,
   \frac{1}{r} = q\,e^{-\alpha r}\,\left(1+\frac{\alpha r}{2}\right)\,
   \left[-4\pi\,\delta\left(\vec{r}\,\right)\right] =
   -4\pi\,q\,\delta(\vec{r})\,.$$
Combining the results gives
$$ \rho(\vec{r}) = q\,\delta\left(\vec{r}\right) - 
   \frac{\alpha^3q}{8\pi}\,e^{-\alpha r}\,. $$
The nucleus has the point charge $+q$, which is surrounded by a
continuous charge distribution of charge 
$$ -q = -\int d^3x\,\frac{\alpha^3q}{8\pi}\,e^{-\alpha r}\,. $$
The charge density is the one of the H$_2$ wave function, 
$\rho \sim |\psi|^2$. This separation into a singular and a
smooth part is used for Ewald sums in molecular dynamics
simulations. See, for instance, chapter~12 of Frenkel and
Smit.
\bigskip

\item Exercise: Symmetry of Green functions which fulfill Dirichlet BCs.  
\smallskip

\noindent Choosing $d^3y$ as integration variable, Green's theorem 
% \ref({green2})
reads
$$  \int_V (\phi\, \nabla^2_y \psi - \psi\, \nabla^2_y \phi)\, d^3y 
  = \int_S (\phi\, {\partial\psi\over\partial n} -
        \psi\, {\partial\phi\over\partial n})\, da\ . $$
Let $\phi=G_D(\vec{y},\vec{x}\,)$ and $\psi=G_D(\vec{y},\vec{x}^{\,'})$,
so that
$$ \nabla^2_y \phi = \delta(\vec{y}-\vec{x})\,,\qquad
   \nabla^2_y \psi = \delta(\vec{y}-\vec{x}^{\,'}) $$
holds. We find
$$ \int_V \left[ G_D(\vec{y},\vec{x}\,)\,\delta(\vec{y}-\vec{x}^{\,'})
- G_D(\vec{y},\vec{x}^{\,'})\,\delta(\vec{y}-\vec{x}) \right]\,d^3y = 0 $$
because for $\vec{y}\in S$ (i.e., $\vec{y}$ on the surface)
$G_D(\vec{y},\vec{x}\,)= G_D(\vec{y},\vec{x}^{\,'})=0$. Performing
the integration over $d^3y$ gives the desired result
$$ G_D(\vec{x}^{\,'},\vec{x}\,) = G_D(\vec{x},\vec{x}^{\,'})\ .$$
\bigskip

\end{enumerate}
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