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\begin{center}
  {\bf University of Kentucky, Physics 404G \\
    Bonus: Homework \#6, Rev. B }
\end{center}

\vspace{.3 in}

{\bf 1.} Physical interpretation of the {\bf vector potential} $\bs A$.  The quantity $q\bs{A}$ has physical significance as potential momentum, just as $qV$ is potential energy.  The purpose of this problem is to explore the implications of this connection.  For more background information see \href{http://www.pa.uky.edu/~crawford/phy417_sp11/priv/semon_taylor_ajp_64_1361_1996.pdf} {Am.  J.  Phys. {\bf 64}, 1368 (1996)}.

{\bf \quad a)} Show that $\partial \rho/\partial t + \nabla\cdot\bs{J}=0$ (conservation of charge).  In 4-vector notation, this is written $\partial_\mu J^\mu = 0$, where $\partial = (\partial_t, \nabla)$, and $J=(\rho,\bs{J})$.

{\bf \quad b)}  For a continuous distribution $\rho(\bs r,t)$ of charges in a velocity field $\bs v(\bs r,t)$ so that $\bs J = \rho \bs v$, show that the {\it convective derivative} of $\rho$ (the time derivative of $\rho(\bs{r}(t),t)$ along the path $\bs{r}(t)$ with velocity $\bs{v}=d\bs{r}/dt$) is $d\rho/dt = \partial\rho/\partial t + \bs{v}\cdot\nabla\rho$, where $\partial\rho/\partial t$ is the time derivative of $\rho$ at a fixed position $\bs{r}$ and $\nabla\rho$ is the gradient of $\rho$ at a fixed time.  Use part a) to show that $d\rho/dt + \rho \nabla\cdot\bs{v}=0$.  What is the physical interpretation of $\nabla\cdot\bs{v}$ for a velocity field?

{\bf \quad c)}  Show that using the velocity-dependent potential $U=q(V-\bs{v}\cdot\bs{A})$ in Lagrange's equation produces the Lorentz force:
\begin{equation}
  \label{eq:vpot}
  \bs{F} \equiv \frac{d}{dt}\frac{\partial U}{\partial\bs{v}}
  - \frac{\partial U}{\partial\bs{r}}
  = q(\bs{E}+\bs{v}\times\bs{B}),
\end{equation}
where $\partial/\partial\bs{r} = \nabla = \bh x \partial_x + \bh y \partial_y + \bh z \partial_z$ and $\partial/\partial\bs{v} = \nabla_{\bs{v}} = \bh x \partial_{v_x} + \bh y \partial_{v_y} + \bh z \partial_{v_z}$ are gradients with respect to the independent variables $\bs{r}$ and $\bs{v}$, respectively.  Use $\bs E \equiv -\nabla V - \partial \bs A / \partial t$, which includes time dependence.  This generalizes the standard formula $\bs{F}=-\partial U / \partial {\bs r} = -\nabla U$.  Thus the electromagnetic Lagrangian is $L=T-U=\frac12 mv^2 - q(V-\bs{v}\cdot\bs{A})$.

{\bf \quad d)} Calculate the {\it canonical momentum} $\bs\pi \equiv \frac{\partial L}{\partial \bs v}$.  This generalized momentum includes contributions from both the particle and the magnetic field.

{\bf \quad e)} Calculate the {\it Hamiltonian} $H \equiv \bs\pi \cdot \bs v - L$.  Express $H$ in terms of $\bs\pi$ and $\bs r$, not $\bs{v}$.  This is the total energy in the system.  Thus the Hamiltonian and canonical momentum $(H,\bs \pi)$ are the generalized energy and momentum of a system.  Each includes a kinetic and
potential term.

{\bf \quad f)} Show that $\frac{dH}{dt} = \frac{d}{dt}(T+qV) = \frac{\partial}{\partial t}q(V - \bs v \cdot \bs A) = \frac{\partial U}{\partial t}$, where $\frac{\partial}{\partial t}$ acts on the potentials, but not $\bs r$ or $\bs v$.  Thus the total energy $H$ is conserved if the generalized potential is time-independent.
              
{\bf \quad g)} Show that $\frac{d\bs\pi}{dt} = \frac{d}{dt}(\bs p+q \bs A) = - \frac{\partial}{\partial \bs r} q(V - \bs v \cdot \bs A) = -\frac{\partial U}{\partial \bs r}$, where again $\frac{\partial}{\partial \bs r}$ acts on the potentials, but not $\bs v$.  Thus $\bs\pi$ is conserved if the generalized potential is position-independent.

{\bf \quad h)} Show that the electromagnetic fields do not change if $V\to V+\partial\chi/\partial t$ and $\bs{A}\to\bs{A}+\bs\nabla\chi$. Calculate the Noether current associated with this {\it gauge symmetry}.

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