%%
%%
%%
\chapter{Lagrangians and Noether's Theorem}
\label{ch:Lagrangians&NoethersTheorem}

If the form of a system of dynamical equations does not change under spatial translations then the momentum is a conserved quantity.  When the form of the equations is similarly invariant under time translations then the total energy is a conserved quantity (a constant of the equations of motion).  Time and space translations are examples of 1-parameter groups of transformations. \emph{Invariance under a group of transformations} is precisely what we mean by a symmetry in group theory.  So symmetries of a dynamical system give conserved quantities or conservation laws.  The rigorous statement of all this is the content of \emph{Noether's theorem}.

%%
%%
%%
\section{Time Translation}

To handle time translations we need to replace our paths $q:[t_0,t_1]\rightarrow Q$ by paths $q:\mathbb{R}\rightarrow Q$, and then define a new space of paths,
\[
 \Gamma = \{q:\mathbb{R}\rightarrow Q\}.
\]
The bad news is that the action
\[
 S(q) = \int_{-\infty}^{\infty}L\Bigl(q(t),\dot{q}(t)\Bigr)\,dt
\]
typically will not converge, so $S$ is then no longer a function of the space of paths.  Nevertheless, if $\delta q=0$ outside of some finite interval, then the functional variation,
\[
 \delta S \defeq \int_{-\infty}^{\infty} \left.\frac{d}{ds}
L\Bigl(q_s(t),\dot{q}_s(t)\Bigr)\right|_{s=0}\, dt
\]
\emph{will} converge, since the integral is smooth and vanishes outside this
interval.  Moreover, demanding that this $\delta S$ vanishes for all such
variations $\delta q$ is enough to imply the Euler-Lagrange equations:
\begin{align*}
\delta S &=
\int_{-\infty}^{\infty}\left.\frac{d}{ds}
L\Bigl( q_s(t),\dot{q}_s(t)\Bigr)\right|_{s=0}\, dt \\
&= \int_{-\infty}^{\infty}\left(\frac{\pa L}{\pa q_i}\delta q_i 
 + \frac{\pa L}{\pa\dot{q}_i}\delta\dot{q}_i\right)\,dt \\
&= \int_{-\infty}^{\infty}\left(\frac{\pa L}{\pa q_i} 
 - \frac{d}{dt}\frac{\pa L}{\pa\dot{q}_i}\right)\delta q_i\,dt
\end{align*}
where again the boundary terms have vanished since $\delta q=0$ near
$t=\pm\infty$.  To be explicit, the first term in
\[
 \frac{\pa L}{\pa\qdot^i}\delta\qdot^i = \frac{d}{dt}\left(\frac{\pa L}{\pa\qdot^i}\delta q^i\right) - \left(\frac{d}{dt}\frac{\pa L}{\pa\qdot^i}\right)\delta q
\]
vanishes when we integrate.  Then the whole thing vanishes for all compactly supported smooth $\delta q$ iff
\[
  \frac{d}{dt}\frac{\pa L}{\pa\dot{q}_i} = \frac{\pa L}{\pa q_i}.
\]
Recall that,
\begin{align*}
\frac{\pa L}{\pa\dot{q}_i} = p_i,\quad
  &\text{is the generalized momentum, by defn.}\\
\frac{\pa L}{\pa q_i}=\dot{p}_i,\quad
  &\text{is the force, by the E-L eqns.}
\end{align*}

Note the similarity to Hamilton's equations---if you change $L$ to $H$ you need
to stick in a minus sign, and change variables from $\dot{q}$ to $p_i$ and
eliminate $\dot{p}_i$.

%%
%%
\subsection{Canonical and Generalized Coordinates}

In light of this noted similarity with the Hamilton equations of motion, let's spend a few moments clearing up some terminology (I hate using jargon, but sometimes it's unavoidable, and sometimes it can be efficient---provided everyone is clued in).

%%
%%
\subsubsection{Generalized Coordinates}

For Lagrangian mechanics we have been using \emph{generalized coordinates}, these are the $\{q_i,\dot{q}_i\}$.  The $q_i$ are generalized positions, and the $\dot{q}_i$ are generalized velocities.  The full set of independent generalized coordinates represent the degrees of freedom of a particle, or system of particles.  So if we have $N$ particles then we'd typically have $6N$ generalized coordinates (the ``6'' is for 3 space dimensions, and at each point a position and a momentum).  These can be in any reference frame or system of axes, so for example, in a Cartesian frame, with two particles, in 3D space we'd have the $2\times 3=6$ position coordinates, and $2\times 3=6$ velocities,
\[
  \{x_1,y_1,z_1,x_2,y_2,z_2\}, \{u_1,v_1,w_1,u_2,v_2,w_2\}
\]
where say $u=v_x$, $v=v_y$, $w=v_z$ are the Cartesian velocity components.  This makes $12=6\times2=6N$ coordinates, matching the total degrees of freedom as claimed.   If we constrain the particles to move in a plane (say place them on a table in a gravitational field) then we get $2N$ fewer degrees of freedom, and so $4N$ d.o.f.\ overall.  By judicious choice of coordinate frame we can eliminate one velocity component and one position component for each particle.

It is also handy to respect other symmetries of a system, maybe the particles move on a sphere for example, one can then define new positions and momenta with a consequent reduction in the number of these generalized coordinates needed to describe the system.

%%
%%
\subsubsection{Canonical Coordinates}

In Hamiltonian mechanics (which we have not yet fully introduced) we will find it more useful to transform from generalized coordinates to canonical coordinates.  The canonical coordinates are a special set of coordinates on the cotangent bundle of the configuration space manifold $Q$.  They are usually written as a set of $(q^i,p_j)$ or $(x^i,p_j)$ with the $x$'s or $q$'s denoting the coordinates on the underlying manifold and the $p$'s denoting the conjugate momentum, which are 1-forms in the cotangent bundle at the point $q$ in the manifold.

It turns out that the $q^i$ together with the $p_j$, form a coordinate system on the cotangent bundle $T^{\ast}Q$ of the configuration space $Q$, hence these coordinates are called the \emph{canonical coordinates}.

We will not discuss this here, but if you care to know, later on we'll see that the relation between the generalized coordinates and the canonical coordinates is given by the Hamilton-Jacobi equations for a system.

%%
%%
%%
\section{Symmetry and Noether's Theorem}

First, let's give a useful definition that will make it easy to refer to a type
of dynamical system symmetry.  We want to refer to symmetry transformations
(of the Lagrangian) governed by a single parameter.

\begin{definition}[one-parameter family of symmetries]
\label{def:1-parameter-family}
A 1-parameter family of symmetries of a Lagrangian system
$L:TQ\rightarrow\mathbb{R}$ is a smooth map,
\begin{align*}
 F:\mathbb{R}\times\Gamma &\longrightarrow\Gamma  \\
  (s,q) & \longmapsto q_s,\qquad\text{with $q_0=q$} 
\end{align*}
such that there exists a function $\ell(q,\dot{q})$ for which
\[
 \delta L = \frac{d \ell}{dt}
\]
for some $\ell:TQ\rightarrow\mathbb{R}$, that is,
\[
 \left.\frac{d}{ds}L\Bigl(q_s(t),\dot{q}_s(t)\Bigr)\right|_{s=0}
  = \frac{d}{dt}\ell\Bigl(q_s(t),\dot{q}_s(t)\Bigr)
\]
for all paths $q$.
\end{definition}

%%
\paragraph{Remark:}
The simplest case is $\delta L=0$, in which case we really have a way of moving
paths around ($q\mapsto q_s$) that doesn't change the Lagrangian---i.e., a
symmetry of $L$ in the most obvious way.  But $\delta L = \frac{d}{dt}\ell$ is a
sneaky generalization whose usefulness will become clear.

%%
%%
\subsection{Noether's Theorem}
\label{sec:NoethersTheorem}

Here's a statement of the theorem.  Note that $\ell$ in this theorem is the function associated with $F$ in definition~\ref{def:1-parameter-family}.

\begin{theorem}[Noether's Theorem]
Suppose $F$ is a one-parameter family of symmetries of the Lagrangian system,
$L:TQ\rightarrow\mathbb{R}$.  Then,
\[
 p^i \delta q_i - \ell
\]
is conserved, that is, its time derivative is zero for any path $q\in\Gamma$
satisfying the Euler-Lagrange equations.  In other words, in boring detail,
\[
\frac{d}{dt}\left[\frac{\pa L}{\pa y^i}\bigl(q(s)\dot{q}(s)\bigr)
\left.\frac{d}{ds}q_s^i(t)\right|_{s=0} 
- \ell\bigl(q(t),\dot{q}(t)\bigr)\right] = 0
\]
\end{theorem}
\begin{proof}
\begin{align*}
   \frac{d}{dt}\Bigl(p_i\delta q^i - \ell\Bigr) &= \dot{p}_i\delta q^i 
   + p_i\delta\dot{q}^i - \frac{d}{dt}\ell \\
   &= \frac{\pa L}{\pa q^i}\delta q^i 
   +\frac{\pa L}{\pa\dot{q}^i}\delta\dot{q}^i - \delta L  \\
   &= \delta L - \delta L = 0.
\end{align*}
\end{proof}


``Okay, big deal'' you might say.  
Before this can be of any use we'd need to find a
symmetry $F$.  Then we'd need to find out what this $p^i\delta q_i-\ell$
business is that is conserved.  So let's look at some examples.

%%
%%
\subsubsection{Example}

\begin{enumerate}
\item \textbf{Conservation of Energy.}  (The most important example!)

All of our Lagrangian systems will have time translation invariance (because
the laws of physics do not change with time, at least not to any extent that we
can tell).  So we have a one-parameter family of symmetries
\[
 q_s(t) = q(t+s)
\]
This indeed gives,
\begin{align*}
   \delta L &= \dot{L} \\
   \intertext{for}
   \left.\frac{d}{ds}L(q_s)\right|_{s=0} &= \frac{d}{dt}L = \dot{L}
\end{align*}
so here we take $\ell = L$ simply!  We then get the conserved quantity
\[
 p_i\delta q^i - \ell = p_i\dot{q}^i - L
\]
which we normally call the \emph{energy}.  For example, if $Q=\mathbb{R}^n$,
and if
\[
 L = \frac{1}{2}m\dot{q}^2 - V(q)
\]
then this quantity is
\[
 m\dot{q}\vdot\dot{q} - \Bigl(\frac{1}{2}m\dot{q}\vdot\dot{q}-V\Bigr) 
 = \frac{1}{2}m\dot{q}^2 + V(q)
\]
The term in parentheses is $K-V$, and the left-hand side is $K+V$.
\end{enumerate}

Let's repeat this example, this time with a specific Lagrangian.  It doesn't
matter what the Lagrangian is, if it has 1-parameter families of symmetries
then it'll have conserved quantities, guaranteed.  The trick in physics is to
write down a correct Lagrangian in the first place! (Something that will accurately describe the system of interest.)

%%
%%
%%
\section{Conserved Quantities from Symmetries}

We've seen that any 1-parameter family
\begin{align*}
 F_s:\Gamma &\longrightarrow\Gamma \\
 q&\longmapsto q_s
\end{align*}
which satisfies
\[
 \delta L = \dot{\ell}
\]
for some function $\ell=\ell(q,\dot{q})$ gives a conserved quantity
\[
 p_i\delta q^i - \ell
\]
As usual we've defined
\[
 \delta L \defeq \left.\frac{d}{ds}L\Bigl(q_s(t),\dot{q}_s(t)\Bigr)\right|_{s=0}
\]
Let's see how we arrive at a conserved quantity from a symmetry.

%%
%%
\subsection{Time Translation Symmetry}

For any Lagrangian system, $L:TQ\rightarrow\mathbb{R}$, we have a 1-parameter
family of symmetries
\[
 q_s(t) = q(t+s)
\]
because
\[
 \delta L = \dot{L}
\]
so we get a conserved quantity called the \emph{total energy} or
\emph{Hamiltonian},
\begin{equation}
  H = p_i\dot{q}^i - L
\end{equation}
(You might prefer ``Hamiltonian'' to ``total energy'' because in general we
are not in the same configuration space as Newtonian mechanics, if you are doing
Newtonian mechanics then ``total energy'' is appropriate.) 

For example: a particle on $\mathbb{R}^n$ in a potential V has
$Q=\mathbb{R}^n$, $L(q,\dot{q})=\frac{1}{2}m\dot{q}^2-V(q)$.  This system has
\[
p_i\dot{q}^i = \frac{\pa L}{\pa\dot{q}^i}\dot{q}^i = m\dot{q}^2 = 2K
\]
so
\[
  H = p_i\dot{q}^i - L = 2K - (K - V) = K+V
\]
as you'd have hoped.

%%
%%
\subsection{Space Translation Symmetry}

For a free particle in $\mathbb{R}^n$, we have $Q=\mathbb{R}^n$ and
$L=K=\frac{1}{2}m\dot{q}^2$.  This has \emph{spatial translation} symmetries,
so that for any $v\in\mathbb{R}^n$ we have the symmetry
\[
  q_s(t) = q(t) + s\, v
\]
with
\[
  \delta L = 0
\]
because $\delta\dot{q}=0$ and $L$ depends only on $\dot{q}$ not on $q$ in this
particular case.  (Since $L$ does not depend upon $q^i$ we'll call $q^i$ an
\emph{ignorable coordinate}; as above, these ignorables always give symmetries,
hence conserved quantities.  It is often useful therefore, to change
coordinates so as to make some of them ignorable if possible!)

In this example we get a conserved quantity called \emph{momentum in the $v$
direction}:
\[
  p_i\delta q^i = m\dot{q}_i v^i = m\dot{q}\vdot v
\]

%%
\paragraph{Aside:}  Note the subtle difference between two uses of the term
``momentum''; here it is a conserved quantity derived from space translation
invariance, but earlier it was a different thing, namely the momentum $\pa
L/\pa\dot{q}^i = p_i$ conjugate to $q^i$.  These two different ``momentum's''
happen to be the same in this example!

Since this is conserved for all $v$ we say that $m\dot{q}\in\mathbb{R}^n$ is
conserved.  (In fact that whole Lie group $G=\mathbb{R}^n$ is acting as
a translation symmetry group, and we're getting a
$\mathfrak{q}(=\mathbb{R}^n)$-valued conserved quantity!)


%%
%%
\subsection{Rotational Symmetry}

The free particle in $\mathbb{R}^n$ also has \emph{rotation} symmetry. 
Consider any $X\in\mathfrak{so}(n)$ (that is a skew-symmetric $n\times n$
matrix), then for all $s\in\mathbb{R}$ the matrix $e^{sX}$ is in $SO(n)$,
that is, it describes a rotation.  This gives a 1-parameter family of
symmetries
\[
  q_s(t) = e^{sX} q(t)
\]
which has
\[
\delta L = \frac{\pa L}{\pa q^i}\delta q^i
 + \frac{\pa L}{\pa\dot{q}^i}\delta\dot{q}^i = m\dot{q}_i\delta\dot{q}^i
\]
now $q_i$ is ignorable and so $\pa L/\pa q^i = 0$, and $\pa L/\pa \dot{q}^i =
p_i$, and
\begin{align*}
  \delta\dot{q}^i &= \left.\frac{d}{ds}\dot{q}_s^i\right|_{s=0} \\
  &= \left.\frac{d}{ds}\frac{d}{dt}\Bigl(e^{sX}q\Bigr)\right|_{s=0} \\
  &= \frac{d}{dt}X\, q \\
  &= X\,\dot{q}
\end{align*}

So,
\begin{align*}
  \delta L &= m\dot{q}_i X^i_j \dot{q}^j \\
  &= m\dot{q}\vdot(X\,\dot{q}) \\
  &= 0
\end{align*}
since $X$ is skew symmetric as stated previously ($X\in\mathfrak{so}(n)$).  So we get a conserved quantity, the \emph{angular momentum} in the $X$ direction. 

(Note: this whole bunch of math above for $\delta L$ just says that the
kinetic energy doesn't change when the velocity is \emph{rotated}, without
changing its magnitude.)

We write,
\[
  p_i\delta q^i = m\dot{q}_i\vdot(X\,q)^i
\]
($\delta q^i = Xq$ just as $\delta\dot{q}^i=X\dot{q}$ in our previous calculation), or if $X$ has zero entries except in $ij$ and $ji$ positions, where it's $\pm 1$, then we get
\[
 m(\dot{q}_i q^j - \dot{q}_j q^i)
\]
the ``$ij$ component of angular momentum''.  If $n=3$ we write these as,
\[
m\dot{\vect{q}}\cross \vect{q}
\]

Note that above we have assumed one can construct a basis for $\mathfrak{so}(n)$ using matrices of the form assumed for $X$, i.e., skew symmetric with $\pm 1$ in the respectively $ij$ and $ji$ elements, otherwise zero.

I mentioned earlier that we can do mechanics with any Lagrangian, but if we want to be useful we'd better pick a Lagrangian that actually describes a real system.  But how do we do that?  All this theory is fine but is useless unless you know how to apply it.  The above examples were for a particularly simple system, a free particle, for which the Lagrangian is just the kinetic energy, since there is no potential energy variation for a free particle.  We'd like to know how to solve more complicated dynamics. 

The general idea is to guess the kinetic energy and potential energy of the particle (as functions of your generalized positions and velocities) and then let,
\[
 L = K - V
\]
So we are not using Lagrangians directly to tell us what the fundamental physical laws should be, instead we plug in some assumed physics and use the Lagrangian approch to solve the system of equations.  If we like, we can then compare our answers with experiments, which indirectly tells us something about the physical laws---but only provided the Lagrangian formulation of mechanics is itself a valid procedure in the first place.


%%
%%
%%
\section{Example Problems}

% \noindent (Week 3, Apr. 11, 13, 15.)\vspace{2ex}

To see how the formalisms in this chapter function in practise, let's do some problems.  It's vastly superior to the simplistic $F=ma$ formulation of mechanics.  The Lagrangian formulation allows the configuration space to be any manifold, and allows us to easily use any coordinates we wish.

%%
%%
\subsection{The Atwood Machine}

Consider a frictionless pulley with two masses, $m_1$ and $m_2$, hanging from it:
\begin{figure}[ht]
	\centering
	\psfrag{x}{$x$} \psfrag{l-x}{$\ell-x$} \psfrag{m1}{$m_1$} \psfrag{m2}{$m_2$}
	\includegraphics[width=.15\textwidth]{images/atwood-pulley}
	%\caption{The Atwood machine.}
	%\label{fig:atwood-pulley}
\end{figure}

\noindent We have
\begin{align*}
 K &= \frac{1}{2}(m_1+m_2)(\frac{d}{dt}(\ell-x))^2  = 
\frac{1}{2}(m_1+m_2)\dot{x}^2 \\
 V &= -m_1 g x - m_2 g(\ell-x) \\
 \intertext{so}
 L &= K-V = \frac{1}{2}(m_1+m_2)\dot{x}^2 + m_1gx+m_2g(\ell-x)
\end{align*}
The configuration space is $Q=(0,\ell)$, and $x\in(0,\ell)$ (we could use the ``owns'' symbol $\ni$ here and write $Q=(0,\ell) \ni x$ ).  Moreover $TQ=(0,\ell)\times\mathbb{R}\ni(x,\dot{x})$.
As usual $L:TQ\rightarrow\mathbb{R}$.  Note that solutions of the Euler-Lagrange equations will only be defined for \emph{some} time $t\in\mathbb{R}$, as eventually the solutions reaches the ``edge'' of $Q$.

The momentum is:
\[
 p = \frac{\pa L}{\pa\dot{x}} = (m_1+m_2)\dot{x}
\]
and the force is:
\[
 F = \frac{\pa L}{\pa x} = (m_1-m_2)g
\]
The Euler-Lagrange equations say
\begin{align*}
 \dot{p} &= F \\
 (m_1+m_2)\ddot{x} &= (m_1-m_2)g \\
 \ddot{x} &= \frac{m_1-m_2}{m_1+m_2}g
\end{align*}
So this is like a falling object in a downwards gravitational acceleration $a=\left(\frac{m_1-m_2}{m_1+m_2}\right)g$.


It is trivial to integrate the expression for $\ddot{x}$ twice (feeding in some appropriate initial conditions) to obtain the complete solution to the motion $x(t)$ and $\dot{x}(t)$.  Note that $\ddot{x}=0$ when $m_1=m_2$, and $\ddot{x}=g$ if $m_2=0$.

%%
%%
\subsection{Disk Pulled by Falling Mass}

Consider next a disk pulled across a table by a falling mass.  The disk is free to move on a frictionless surface, and it can thus whirl around the hole to which it is tethered to the mass below.
\begin{figure}[ht]
	\centering
	\psfrag{r}{$r$} \psfrag{l-r}{$\ell-r$} \psfrag{m1}{$m_1$} \psfrag{m2}{$m_2$}
	\psfrag{no swinging}{no swinging allowed!}
	\includegraphics[width=.33\textwidth]{images/disk-string-mass}
	%\caption{Disk dragged by gravitationally falling mass on a string.}
	%\label{fig:disk-string-mass}
\end{figure}
\begin{align*}
\text{Here }Q &=\text{ open disk of radius $\ell$, minus its center} \\
 &= (0,\ell)\times S^1\,\ni (r,\theta)
\end{align*}
\[
 TQ = (0,\ell)\times S^1\times\mathbb{R}\times\mathbb{R}
  \,\ni (r,\theta,\dot{r},\dot{\theta})
\]
\begin{align*}
 K &= \frac{1}{2}m_1(\dot{r}^2+r^2\dot{\theta}^2) 
 + \frac{1}{2}m_2(\frac{d}{dt}(\ell-r))^2 \\
 V &= gm_2(r-\ell) \\
 L &= \frac{1}{2}m_1(\dot{r}^2+r^2\dot{\theta}^2) +\frac{1}{2}m_2\dot{r}^2
 +gm_2(\ell-r)
\end{align*}
having noted that $\ell$ is constant so $d/dt(\ell-r)=-\dot{r}$.  For the momenta we get,
\begin{align*}
 p_r &= \frac{\pa L}{\pa\dot{r}} = (m_1+m_2)\dot{r} \\
 p_\theta &= \frac{\pa L}{\pa\dot{\theta}} = m_1r^2\dot{\theta}.
\end{align*}
Note that $\theta$ is an ``ignorable coordinate''---it doesn't appear in $L$---so there's a symmetry, rotational symmetry, and $p_\theta$, the conjugate momentum, is conserved.

The forces are,
\begin{align*}
 F_r &= \frac{\pa L}{\pa r} = m_1 r\dot{\theta}^2 - gm_2 \\
 F_\theta &= \frac{\pa L}{\pa\theta} = 0, \quad\text{($\theta$ is ignorable)}
\end{align*}

Note: in $F_r$ the term $m_1 r\dot{\theta}^2$ is recognizable as a centrifugal force, pushing $m_1$ radially \emph{out}, while the term $-gm_2$ is gravity pulling $m_2$ down and thus pulling $m_1$ radially \emph{in}.

So, the Euler-Lagrange equations give,
\begin{align*}
 \dot{p}_r &= F_r, \quad (m_1+m_2)\ddot{r} = m_1r\dot{\theta}^2-m_2g \\
 \dot{p}_\theta &=0, \quad p_\theta=m_1r^2\dot{\theta} = J =\text{ a constant}.
\end{align*}

Let's use our conservation law here to eliminate $\dot{\theta}$ from the first equation:
\begin{align*}
 \dot{\theta} &= \frac{J}{m_1 r^2} \\
\intertext{so}
 (m_1+m_2)\ddot{r} &= \frac{J^2}{m_1r^3}-m_2g
\end{align*}
Thus effectively we have a particle on $(0,\ell)$ of mass $m=m_1+m_2$ feeling a force
\[
 F_r = \frac{J^2}{m_1 r^3} - m_2 g
\]
which could come from an ``effective potential'' $V(r)$ such that $dV/dr=-F_r$. So integrate $-F_r$ to find $V(r)$:
\[
 V(r) = \frac{J^2}{2m_1r^2}+m_2 g r
\]
this is a sum of two terms that look like Fig.~\ref{fig:disk-string-mass_potential}
\begin{figure}[ht]
	\centering
	\psfrag{r}{$r$} \psfrag{r0}{$r_0$} \psfrag{V(r)}{$V(r)$}
	\psfrag{attractive grav. potnl}{\ssz{attractive gravitational potential}}
	\psfrag{repulsive centrifugal}{\ssz{repulsive centrifugal potential}}
	\psfrag{no swinging}{no swinging allowed!}
	\includegraphics[width=.5\textwidth]{images/disk-string-mass_potential}
	\caption{Potential function for disk pulled by gravitating mass.}
	\label{fig:disk-string-mass_potential}
\end{figure}

If $\dot{\theta}(t=0)=0$ then there is no centrifugal force and the disk will be pulled into the hole until it gets stuck.  At that time the disk reaches the hole, which is topologically the center of the disk that has been removed from $Q$, so then we've hit the boundary of $Q$ and our solution is broken.

At $r=r_0$, the minimum of $V(r)$, our disc mass $m_1$ will be in a stable circular orbit of radius $r_0$ (which depends upon $J$).  Otherwise we get orbits like Fig.~\ref{fig:disk-string-mass_orbits}.
\begin{figure}[ht]
	\centering
	\includegraphics[width=.33\textwidth]{images/disk-string-mass_orbits}
	\caption{Orbits for the disc and gravitating mass system.}
	\label{fig:disk-string-mass_orbits}
\end{figure}



%%
%%
\subsection{Free Particle in Special Relativity}
\label{ssec:RelativisticFreeParticle}

In relativistic dynamics the parameter coordinate that parametrizes the particle's path in Minkowski spacetime need not be the ``time coordinate'', indeed in special relativity there are many allowed time coordinates.

Minkowski spacetime is,
\[
 \mathbb{R}^{n+1} \ni (x^0,x^1,\ldots,x^n)
\]
if space is $n$-dimensional.  We normally take $x^0$ as ``time'', and $(x^1,\ldots,x^n)$ as ``space'', but of course this is all relative to one's reference frame.  Someone else traveling at some high velocity relative to us will have to make a Lorentz transformation to translate from our coordinates to theirs.

This has a Lorentzian metric
\begin{align*}
 g(v,w) &= v^0w^0-v^1w^1 -\ldots - v^nw^n \\
 &=\eta_{\mu\nu}v^\mu w^\nu
\end{align*}
where
\[
 \eta_{\mu\nu} = \left(
                  \begin{array}{ccccc}
                   1      & 0       & 0      & \ldots & 0 \\
                   0      & -1      & 0      & \ldots & 0 \\
                   0      & 0       & -1     &        & 0 \\
                   \vdots & \vdots  &        & \ddots & \vdots \\
                   0      & 0       & 0      & \ldots & -1 
                  \end{array}\right)
\]

In special relativity we take \emph{spacetime} to be the configuration space of a single point particle, so we let $Q$ be Minkowski spacetime, i.e., $\mathbb{R}^{n+1}\ni(x^0,\ldots,x^n)$ with the metric $\eta_{\mu\nu}$ defined above.  Then the path of the particle is,
\[
 q:\mathbb{R}(\ni t) \longrightarrow Q
\]
where $t$ is a completely arbitrary parameter for the path, not necessarily $x^0$, and not necessarily \emph{proper time} either.  We want some Lagrangian $L:TQ\rightarrow\mathbb{R}$, i.e., $L(q^i,\dot{q}^i)$ such that the Euler-Lagrange equations will dictate how our free particle moves at a constant velocity.  Many Lagrangians do this, but the ``best'' ones(s) give an action that is \emph{independent} of the parameterization of the path---since the parameterization is ``unphysical'' (it can't be measured).  So the action
\[
 S(q) = \int_{t_0}^{t_1} L\Bigl(q^i(t),\dot{q}^i(t)\Bigr)\,dt
\]
for $q:[t_0,t_1]\rightarrow Q$, should be independent of $t$.  The obvious candidate for $S$ is mass times arclength,
\[
 S = m\int_{t_0}^{t_1} \sqrt{\eta_{ij}\dot{q}^i(t)\dot{q}^j(t)}\,dt
\]
or rather the Minkowski analogue of arclength, called \emph{proper time}, at least when $\dot{q}$ is a timelike vector, i.e., $\eta_{ij}\dot{q}^i\dot{q}^j>0$, which says $\dot{q}$ points into the future (or past) lightcone and makes $S$ \emph{real}, in fact it's then the time ticked off by a clock moving along the path $q:[t_0,t_1]\rightarrow Q$.
\begin{figure}[ht]
	\centering
	\includegraphics[width=.25\textwidth]{images/lightcone}
	%\caption{Lightcone at a point in spacetime, showing timelike, spacelike and null (lightlike) vectors.}
	%\label{fig:lightcone}
\end{figure}
By ``obvious candidate'' we are appealing somewhat to physical intuition and generalization.  In Euclidean space, free particles follow straight paths, so the arclength or pathlength variation is an extremum, and we expect the same behavior in Minkowski spacetime.  Also, the arclength does not depend upon the parameterization, and lastly, the mass $m$ merely provides the correct units for `action'.

So let's take
\begin{equation}
\label{eq:relativistic-Lagrangian}
 L = m \sqrt{\eta_{ij}\dot{q}^i\dot{q}^j}
\end{equation}
and work out the Euler-Lagrange equations.  We have
\begin{align*}
 p_i = \frac{\pa L}{\pa\dot{q}^i} 
 &= m\frac{\pa}{\pa\dot{q}^i}\sqrt{\eta_{ij}\dot{q}^i\dot{q}^j} \\
 &= m\frac{2\eta_{ij}\dot{q}^j}{2\sqrt{\eta_{ij}\dot{q}^i\dot{q}^j}} \\
 &= m\frac{\eta_{ij}\dot{q}^j}{\sqrt{\eta_{ij}\dot{q}^i\dot{q}^j}} 
  \, = \frac{m\dot{q}_i}{\norm{\dot{q}}}
\end{align*}
(Note the numerator is ``mass times 4-velocity'', at least when $n=3$ for a real single particle system, but we're actually in a more general $n+1$-dim spacetime, so it's more like the ``mass times $n+1$-velocity'').  
Now note that this $p_i$ doesn't change when we change the parameter to accomplish $\dot{q}\mapsto \alpha\dot{q}$.  The Euler-Lagrange equations say,
\[
 \dot{p}_i = F_i = \frac{\pa L}{\pa q^i} = 0
\]

The meaning of this becomes clearer if we use ``proper time'' as our parameter (like parameterizing a curve by its arclength) so that
\[
 \int_{t_0}^{t_1}\norm{\dot{q}}dt = t_1-t_0, \quad \forall\,t_0,t_1
\]
which fixes the parametrization up to an additive constant.  This implies $\norm{\dot{q}}=1$, so that
\[
 p_i = m\frac{\dot{q}_i}{\norm{\dot{q}}} = m\dot{q}_i
\]
and the Euler-Lagrange equations say
\[
 \dot{p}_i =0 \Rightarrow m\ddot{q}_i = 0
\]
so our (free) particle moves unaccelerated along a straight line, which is as we desired (expected).

%%
%%
\subsubsection{Comments}

This Lagrangian from Eq.\eqref{eq:relativistic-Lagrangian} has lots of symmetries coming from \emph{reparameterizing the path}, so Noether's theorem yields lots of conserved quantities for the relativistic free particle.  This is in fact called ``the problem of time'' in general relativity.  Here we see it starting to show up in special relativity.

These reparameterization symmetries work as follows.  Consider any (smooth) 1-parameter family of reparameterizations, i.e., diffeomorphisms
\[
 f_s:\mathbb{R}\longrightarrow\mathbb{R}
\]
with $f_0=\mathbb{1}_\mathbb{R}$.  These act on the space of paths $\Gamma = \{q:\mathbb{R}\rightarrow Q\}$ as follows: given any $q\in\Gamma$ we get
\[
 q_s(t) = q\bigl(f_s(t)\bigr)
\]
where we should note that $q_s$ is \emph{physically indistinguishable} from $q$.  Let's show that
\[
 \delta L = \dot{\ell},\quad\text{(when E-L eqns.\ hold)}
\]
so that Noether's theorem gives a conserved quantity
\[
 p_i\delta q^i - \ell
\]
Here we go then:
\begin{align*}
 \delta L &= \frac{\pa L}{\pa q^i}\delta q^i 
 + \frac{\pa L}{\pa\dot{q}^i}\delta\dot{q}^i \\
 &= p_i\delta\dot{q}^i \\
 &= \frac{m\dot{q}_i}{\norm{\dot{q}}}
 \left.\frac{d}{ds}\dot{q}^i\Bigl(f_s(t)\Bigr)\right|_{s=0} \\
 &= \frac{m\dot{q}_i}{\norm{\dot{q}}}
 \left.\frac{d}{dt}\frac{d}{ds} q^i\Bigl(f_s(t)\Bigr)\right|_{s=0} \\
 &= \frac{m\dot{q}_i}{\norm{\dot{q}}}
 \left.\frac{d}{dt}\dot{q}^i\Bigl(f_s(t)\Bigr)\frac{f_s(t)}{ds}\right|_{s=0} \\
 &= \frac{m\dot{q}_i}{\norm{\dot{q}}}
 \frac{d}{dt}\left(\dot{q}^i\delta f_s\right) \\
 &= \frac{d}{dt}\left(p_i\dot{q}^i\delta f\right)
\end{align*}
where in the last step we used the E-L eqns., i.e.\ $\frac{d}{dt}p_i=0$, so $\delta L=\dot{\ell}$ with $\ell=p_i\dot{q}^i\delta f$.


So to recap a little: we saw the free relativistic particle has
\[
 L = m\norm{\dot{q}} = m\sqrt{\eta_{ij}\dot{q}^i\dot{q}^j}
\]
and we've considered reparameterization symmetries
\[
 q_s(t) = q\Bigl(f_s(t)\Bigr), \qquad f_s:\mathbb{R}\rightarrow\mathbb{R}
\]
we've used the fact that
\[
 \delta q^i \defeq \left.\frac{d}{ds}q^i\Bigl(f_s(t)\Bigr)\right|_{s=0}
 = \dot{q}^i\delta f
\]
so (repeating a bit of the above)
\begin{align*}
 \delta L &= \frac{\pa L}{\pa q^i}\delta q^i 
 + \frac{\pa L}{\pa\dot{q}^i}\delta\dot{q}^i \\
 &= p_i\delta\dot{q}^i,\quad\text{(since $\pa L/\pa q^i =0$, 
 and $\pa L/\pa\dot{q}^i = p$)} \\
 &= p_i\delta\dot{q}^i \\
 &= p_i\frac{d}{dt}\delta q^i \\
 &= p_i\frac{d}{dt}\dot{q}^i\delta f \\
 &= \frac{d}{dt} p_i\dot{q}^i\delta f,
    \quad\text{and set $p_i\dot{q}^i\delta f=\ell$ }
\end{align*}
so Noether's theorem gives a conserved quantity
\begin{align*}
 p_i\delta q^i - \ell &= p_i\dot{q}^i\delta f - p_i\dot{q}^i\delta f \\
 &= 0
\end{align*}


So these conserved quantities \emph{vanish}!  In short, we're seeing an example of what physicists call \emph{gauge symmetries}.  This is a good topic for starting a new section.

%%
%%
%%
\section{Electrodynamics and Relativistic Lagrangians}
\label{sec:RelativisticLagrangians}

We will continue the story of symmetry and Noether's theorem from the last section with a  few more examples. We use principles of least action to conjure up Lagrangians for our systems, realizing that a given system may not have a unique Lagrangian but will often have an obvious natural Lagrangian.  Given a Lagrangian we derive equations of motion from the Euler-Lagrange equations.   Symmetries of $L$ guide us in finding conserved quantities, in particular Hamiltonians from time translation invariance, via Noether's theorem.

This section also introduces gauge symmetry, and this is where we begin.

%%
%%
%%
\subsection{Gauge Symmetry and Relativistic Hamiltonian}

What are gauge symmetries?
\begin{enumerate}
 \item These are symmetries that permute different mathematical descriptions of the same physical situation---in this case reparameterizations of a path.\label{i:same-phys}
 \item These symmetries make it impossible to compute $q(t)$ given $q(0)$ and $\dot{q}(0)$: since if $q(t)$ is a solution so is $q(f(t))$ for any reparameterization $f:\mathbb{R}\rightarrow\mathbb{R}$.  We have a high degree of non-uniqueness of solutions to the Euler-Lagrange equations. \label{i:non-unique_E-L_solns}
 \item These symmetries give conserved quantities that work out to equal zero!\label{i:conserved-quant-vanish}
\end{enumerate}

Note that (\ref{i:same-phys}) is a subjective criterion, (\ref{i:non-unique_E-L_solns}) and (\ref{i:conserved-quant-vanish}) are objective, and (\ref{i:conserved-quant-vanish}) is easy to test, so we often use (\ref{i:conserved-quant-vanish}) to distinguish \emph{gauge} symmetries from \emph{physical} symmetries.

%%
%%
\subsection{Relativistic Hamiltonian}

What then is the Hamiltonian for special relativity theory?  We're continuing here with the example problem of \S\ref{ssec:RelativisticFreeParticle}. Well, the Hamiltonian comes from Noether's theorem from \emph{time translation} symmetry,
\[
 q_s(t) = q(t+s)
\]
and this is an example of a reparametrization (with $\delta f=1$), so we see from the previous results that the Hamiltonian is \emph{zero}!
\[
 H = 0.
\]
Explicitly, $H=p_i\delta\qdot^i-\ell$ where under $q(t)\rightarrow q(t+s)$ we have $\delta\qdot^i=\qdot^i\delta f$, and so $\delta L=d\ell/dt$, which implies $\ell=p_i\delta q^i$.  The result $H=0$ follows.


Now you know why people talk about ``the problem of time'' in general relativity theory, its glimmerings are seen in the flat Minkowski spacetime of special relativity.  You may think it's nice and simple to have $H=0$, but in fact it means that there is no temporal evolution possible!  So we can't establish a dynamical theory on this footing!  That's bad news.  (Because it means you might have to solve the static equations for the 4D universe as a whole, and that's impossible!)

But there is another conserved quantity deserving the title of ``energy'' which is \emph{not} zero, and it comes from the symmetry,
\[
 q_s(t) = q(t) + s \,w
\]
where $w\in\mathbb{R}^{n+1}$ and $w$ points in some timelike direction.
\begin{figure}[ht]
	\centering
	\psfrag{w}{$w$}\psfrag{q}{$q$}\psfrag{qs}{$q_s$}
	\includegraphics[width=.2\textwidth]{images/timelike-translation}
	%\caption{A timelike translation.}
	%\label{fig:timelike-translation}
\end{figure}

In fact \emph{any} vector $w$ gives a conserved quantity,
\begin{align*}
 \delta L &= \frac{\pa L}{\pa q^i}\delta q^i
 + \frac{\pa L}{\pa\dot{q}^i}\delta\dot{q}^i \\
 &= p_i\delta\dot{q}^i,
 \quad\text{(since $\pa L/\pa q^i=0$ and $\pa L/\pa\dot{q}^i=p_i$)} \\
 &= p_i 0 \, = 0
\end{align*}
since $\delta q^i = w^i$, $\delta\dot{q}^i=\dot{w}^i=0$.  This is our $\dot{\ell}$ from Noether's theorem with $\ell=0$, so Noether's theorem says that we get a conserved quantity
\[
 p_i\delta q^i - \ell = p_ w^i
\]
namely, the \emph{momentum in the $w$ direction}.  We know $\dot{p}=0$ from the Euler-Lagrange equations, for our free particle, but here we see it coming from spacetime translation symmetry;
\begin{align*}
 p = &(p_0,p_1,\ldots,p_n) \\
     & p_0\text{ is energy},\quad (p_1,\ldots,p_n)\text{ is spatial momentum}.
\end{align*}

We've just about exhausted all the basic stuff that we can learn from the free particle.  So next we'll add some external force via an electromagnetic field.


%%
%%
%%
\section{Relativistic Particle in an Electromagnetic Field}
\label{sec:RelativisticParticle_EM-field}

The electromagnetic field is described by a 1-form $A$ on spacetime, $A$ is the \emph{vector potential}, such that
\begin{equation}
  dA = \fvect{F}
\end{equation}
is a 2-form containing the electric and magnetic fields,
\begin{equation}
 \fvect{F}_{ij} = \frac{\pa A_j}{\pa x^i} - \frac{\pa A_i}{\pa x^j} 
\end{equation}
We'd write (for $Q$ having local charts to $\mathbb{R}^{n+1}$),
\begin{align*}
 A &= A_0dx^0 + A_1dx^1 +\ldots A_ndx^n \\
 \intertext{and then because $d^2 =0$}
 dA &= dA_0dx^0 + dA_1dx^1 +\ldots dA_ndx^n
\end{align*}
and since the ``$A_j$'' are just functions,
\[
 dA_j = \pa_i A_jdx^i
\]
using the summation convention and $\pa_i := \pa/\pa x^i$.  The student can easily check that the components for $F =F_{01}dx^0\wedge dx^1 + F_{02}dx^0\wedge dx^2 + \ldots$, agrees with the matrix expression below (at least for 4D).

So, for example, in 4D spacetime
\[
 F = \left(\begin{array}{cccc}
      0    &  E_1 &  E_2 &  E_3 \\
      -E_1 &   0  &  B_3 & -B_2 \\
      -E_2 & -B_3 &   0  &  B_1 \\
      -E_3 &  B_2 & -B_1 &   0 \\
     \end{array}\right)
\]
where $E$ is the electric field and $B$ is the magnetic field.  The action for a particle of charge $e$ is
\[
  S = m\int_{t_0}^{t_1}\norm{\dot{q}}\,dt + e \int_{q} A
\]
here
\begin{align*}
 &\int_{t_0}^{t_1}\norm{\dot{q}}\,dt = \text{ proper time}, \\
 &\int_{q} A = \text{ integral of $A$ along the path $q$}.
\end{align*}
Note that since $A$ is a 1-form it can be integrated (it is a linear combination of some basis 1-forms like the $\{dx^i\}$).

\vspace{2ex}\noindent (Week 4, April 18, 20, 22.)\vspace{2ex}

Note that since $A$ is a 1-form we can integrate it over an oriented manifold, 
but one can also write the path integral using time $t$ as a parameter, with 
$A_i\qdot^i\,dt$ the differential, after $dq^i=\qdot^idt$.

The Lagrangian in the above action, for a charge $e$ with mass $m$ in an electromagnetic potential $A$ is
\begin{equation}
\label{eq:Lagrangian-relativistic-EM}
 L(q,\dot{q}) = m\norm{\dot{q}} + eA_i\dot{q}^i 
\end{equation}
so we can work out the Euler-Lagrange equations:
\begin{align*}
 p_i = \frac{\pa L}{\pa\dot{q}^i} &= m\frac{\dot{q}_i}{\norm{\dot{q}}} + eA_i\\
 &= m v_i + e \,A_i
\end{align*}
where $v\in\mathbb{R}^{n+1}$ is the velocity, normalized so that $\norm{v}=1$.  Note that \emph{now momentum is no longer mass times velocity}!  That's because we're in $n+1$-d spacetime, the momentum is an $n+1$-vector.  Continuing the analysis, we find the force
\begin{align*}
 F_i = \frac{\pa L}{\pa q^i} &= \frac{\pa}{\pa q^i}\Bigl(e\,A_j\dot{q}^j\Bigr)\\
 &= e\frac{\pa A_j}{\pa q^i} \dot{q}^j
\end{align*}
So the Euler-Lagrange equations say (noting that $A_i=A_j\Bigl(q(t)\Bigr)$:
\begin{align*}
 \dot{p} &= F \\
 \frac{d}{dt}\Bigl(mv_i+eA_i\Bigr) &= e\frac{\pa A_j}{\pa q^i}\dot{q}^j\\
 m\frac{d v_i}{dt} &= e\frac{\pa A_j}{\pa q^i}\dot{q}^j - e\frac{d A_i}{dt}\\
 m\frac{d v_i}{dt} &= e\frac{\pa A_j}{\pa q^i}\dot{q}^j 
 - e\frac{\pa A_i}{\pa q^j}\dot{q}^j\\
 &= e\left(\frac{\pa A_j}{\pa q^i} - \frac{\pa A_i}{\pa q^j}\right)\dot{q}^j
\end{align*}
the term in parentheses is $\fvect{F}_{ij}=$~the electromagnetic field, $F=dA$.  So we get the following equations of motion
% \begin{equation}
% \boxed{
%  \raisebox{15pt}[20pt][0pt]{} \gapleft
%  m\frac{d v_i}{dt} = e\fvect{F}_{ij}\dot{q}^j, \quad\text{(Lorentz force law)}
%  \gapright
%  \raisebox{-15pt}[0pt][15pt]{}
% }
% \end{equation}
\begin{equation}
\boxed{
 \eqngapabove \gapleft
 m\frac{d v_i}{dt} = e\fvect{F}_{ij}\dot{q}^j, \quad\text{(Lorentz force law)}
 \gapright \eqngapbelow
}
\end{equation}

\noindent (Usually called the ``Lorenz'' force law.)

%%
%%
%%
\section{Alternative Lagrangians}

We'll soon discuss a charged particle Lagrangian that is free of the reparameterization symmetry.  First a paragraph on objects other than point particles!

%%
%%
\subsection{Lagrangian for a String}

So we've looked at a point particle and tried
\[
 S = m\cdot(\text{arclength}) + \int A
\]
or with `proper time' instead of `arclength', where the 1-from $A$ can be integrated over a 1-dimensional path. A generalization (or specialization, depending on how you look at it) would be to consider a Lagrangian for an extended object.

In string theory we boost the dimension by +1 and consider a string tracing out a 2D surface as time passes (Fig.~\ref{fig:string-worldtube}).
\begin{figure}[ht]
	\centering
	%\psfrag{becomes}{becomes}
	\includegraphics[width=.25\textwidth]{images/string-worldtube}
	\caption{Worldtube of a closed string.}
	\label{fig:string-worldtube}
\end{figure}


Can you infer an appropriate action for this system?  Remember, the physical or physico-philosophical principle we've been using is that the path followed by physical objects minimizes the ``activity'' or ``aliveness'' of the system.  Given that we presumably cannot tamper with the length of the closed string, then the worldtube quantity analogous to arclength or proper time would be the area of the worldtube (or worldsheet for an open string).  If the string is also assumed to be a source of electromagnetic field then we need a 2-form to integrate over the 2D worldtube analogous to the 1-form integrated over the pathline of the point particle.  In string theory this is usually the ``Kalb-Ramond field'', call it $B$.  To recover electrodynamic interactions it should be antisymmetric like $A$, but its tensor components will have two indices since it's a 2-form.  The string action can then be written
\begin{equation}
\label{eq:string-action}
 S = \alpha\cdot(area) + e \int B
\end{equation}

We've also replaced the point particle mass by the string tension $\alpha$ [mass$\cdot$length$^{-1}$] to obtain the correct units for the action (since replacing arclength by area meant we had to compensate for the extra length dimension in the first term of the above string action).

This may still seem like we've pulled a rabbit out of a hat.  But we haven't checked that this action yields sensible dynamics yet!  But supposing it does, then would it justify our guesswork and intuition in arriving at Eq.\eqref{eq:string-action}?  Well by now you've probably realized that one can have more than one form of action or Lagrangian that yields the same dynamics.  So provided we supply reasonabe physically realistic heuristics then whatever Lagrangian or action that we come up with will stand a good chance of describing some system with a healthy measure of physical verisimilitude.


That's enough about string for now.  The point was to illustrate the type of reasoning that one can use in conjuring up a Lagrangian.  It's particularly useful when Newtonian theory cannot give us a head start, i.e., in relativistic dynamics and in the physics of extended particles.

%%
%%
\subsection{Alternate Lagrangian for Relativistic Electrodynamics}

In \S~\ref{sec:RelativisticParticle_EM-field}, Eq.\eqref{eq:Lagrangian-relativistic-EM} we saw an example of a Lagrangian for relativistic electrodynamics that had awkward reparametrization symmetries, meaning that $H=0$ and there were non-unique solutions to the Euler-Lagrange equations arising from applying gauge transformations.  This freedom to change the gauge can be avoided.

Recall Eq.\eqref{eq:Lagrangian-relativistic-EM}, which was a Lagrangian for a charged particle with reparametrization symmetry
\[
 L = m\norm{\dot{q}} + eA_i\dot{q}^i
\]
just as for an uncharged relativistic particle.  But there's another Lagrangian we can use that doesn't have this gauge symmetry:
\begin{equation}
 L = \frac{1}{2}m\dot{q}\vdot\dot{q} + eA_i\dot{q}^i
\end{equation}
This one even has some nice features.
\begin{itemize}
 \item It looks formally like ``$\frac{1}{2}mv^2$'', familiar from nonrelativistic mechanics.
 \item There's no ugly square root, so it's everywhere differentiable, and there's no trouble with paths being timelike or spacelike in direction, they are handled the same.
\end{itemize}

What Euler-Lagrange equations does this Lagrangian yield?
\begin{align*}
 p_i &= \frac{\pa L}{\pa\dot{q}^i} = m\dot{q}_i + eA_i \\
 F_i &= \frac{\pa L}{\pa q^i} = e\frac{\pa A_j}{\pa q^i}\dot{q}^j
\end{align*}
Very similar to before!  The E-L eqns.\ then say
\begin{align*}
 \frac{d}{dt}\Bigl(m\dot{q}_i+eA_i\Bigr) &= e\frac{\pa A_j}{\pa q^i}\dot{q}^j\\
 m\ddot{q}_i &= e\fvect{F}_{ij}\dot{q}^j
\end{align*}
almost as before.  (I've taken to using $\fvect{F}$ here for the electromagnetic field tensor to avoid clashing with $F$ for the generalized force.)  The only difference is that we have $m\ddot{q}_i$ instead of $m\dot{v}_i$ where $v_i=\dot{q}_i/\norm{\dot{q}}$.  So the old Euler-Lagrange equations of motion reduce to the new ones if we pick a parametrization with $\norm{\dot{q}}=1$, which would be a parametrization by proper time for example.

Let's work out the Hamiltonian for this
\[
 L = \frac{1}{2}m\dot{q}\vdot\dot{q} + eA_i\dot{q}^i
\]
for the relativistic charged particle in an electromagnetic field.  Recall that for our reparametrization-invariant Lagrangian
\[
 L = m\sqrt{\dot{q}_i\dot{q}^i} + eA_iq^i
\]
we got $H=0$, time translation was a gauge symmetry.  With the new Lagrangian it's not!  Indeed
\[
 H = p_i\dot{q}^i -L
\]
and now
\begin{align*}
 p_i &= \frac{\pa L}{\pa\dot{q}^i} = m\qdot^i + e A_i \\
 \intertext{so}
 H &= (m\dot{q}_i + eA_i)\dot{q}^i - (\tfrac{1}{2}m\dot{q}_i\dot{q}^i + eA_i\dot{q}^i) \\
 &= \tfrac{1}{2}m\dot{q}_i\dot{q}^i
\end{align*}

%%
\paragraph{Comments.}  This is vaguely like how a nonrelativistic particle in a potential $V$ has
\[
 H=p_i\dot{q}^i-L=2K-(K-V)=K+V,
\]
but now the ``potential' $V=eA_i\dot{q}^i$ is \emph{linear in velocity}, so now
\[
 H=p_i\dot{q}^i-L=(2K-V)-(K-V)=K.
\]

As claimed $H$ is not zero, and the fact that it's conserved says $\norm{\dot{q}(t)}$ is constant as a function of $t$, so the particle's path is parameterized by proper time up to rescaling of $t$.  That is, we're getting ``conservation of speed'' rather than some more familiar ``conservation of energy''.  The reason is that this Hamiltonian comes from the symmetry
\[
 q_s(t) = q(t+s)
\]
instead of spacetime translation symmetry
\[
 q_s(t) = q(t) + s\,w,\quad w\in\mathbb{R}^{n+1}
\]
the difference is illustrated schematically in Fig.~\ref{fig:propertime-rescale-vs-translation}.

\begin{figure}[ht]
	\centering
	\psfrag{qs=q(t+s)}{$q_s(t)=q(t+s)$}
	\psfrag{qs=q+sw}{$q_s(t)=q(t)+sw$}
	\psfrag{w}{$w$}
	\includegraphics[width=.45\textwidth]{images/propertime-rescale-vs-translation}
	\caption{Proper time rescaling vs spacetime translation.}
	\label{fig:propertime-rescale-vs-translation}
\end{figure}

Our Lagrangian
\[
 L(q,\dot{q}) = \tfrac{1}{2}m\norm{\dot{q}}^2 + A_i(q)\dot{q}^i
\]
has time translation symmetry iff $A$ is translation invariant (but it's highly unlikely a given system of interest will have $A(q)=A(q+sw)$).  In general then there's no conserved ``energy'' for our particle corresponding to translations in time.

%%
%%
%%
\section{The General Relativistic Particle}
\label{sec:GR-particle}

In GR spacetime, $Q$, is an $(n+1)$-dimensional Lorentzian manifold, namely a smooth $(n+1)$-dimensional manifold with a \emph{Lorentzian metric} $g$.  We define the metric as follows.
\begin{enumerate}
 \item For each $x\in Q$, we have a bilinear map
 \begin{align*}
  g(x):T_x Q\times T_x Q &\longrightarrow\mathbb{R} \\
   (v,w) &\longmapsto g(x)(v,w)
 \end{align*}
 or we could write $g(v,w)$ for short.
 \item With respect to some basis of $T_x Q$ we have
 \begin{align*}
  g(v,w) &= g_{ij}v^iw^j \\
  g_{ij} &=\left(\begin{array}{cccc}
                  1      &  0  & \ldots &  0   \\
                  0      & -1  &        &  0   \\
                  \vdots &     & \ddots & \vdots \\
                  0      &  0  & \ldots & -1
                 \end{array}\right)
 \end{align*}
 Of course we can write $g(v,w)=g_{ij}v^iw^j$ in any basis, but for different bases $g_{ij}$ will have a different form.
 \item $g(x)$ varies smoothly with $x$.
\end{enumerate}

%%
%%
\subsection{Free Particle Lagrangian in GR}

The Lagrangian for a free point particle in the spacetime $Q$ is 
\begin{align*}
 L(q,\dot{q}) &= m\sqrt{g(q)(\dot{q},\dot{q})} \\ 
 &= m\sqrt{g_{ij}\dot{q}^i\dot{q}^j}
\end{align*}
just like in special relativity but with $\eta_{ij}$ replaced by $g_{ij}$.  Alternatively we could just as well use
\begin{align*}
 L(q,\dot{q}) &= \tfrac{1}{2}m g(q)(\dot{q},\dot{q}) \\ 
 &= \tfrac{1}{2}m g_{ij}\dot{q}^i\dot{q}^j
\end{align*}

The big difference between these two Lagrangians is that now spacetime translation symmetry (and rotation, and boost symmetry) is gone!  So there is no conserved energy-momentum (nor angular momentum, nor velocity of center of energy) anymore!

Let's find the equations of motion.  Suppose then $Q$ is a Lorentzian manifold with metric $g$ and $L:TQ\rightarrow\mathbb{R}$ is the Lagrangian of a free particle,
\[
 L(q,\dot{q}) = \tfrac{1}{2}mg_{ij}\dot{q}^i\dot{q}^j
\]
We find equations of motion from the Euler-Lagrange equations, which in this case start from
\[
 p_i = \frac{\pa L}{\pa\dot{q}^i} = mg_{ij}\dot{q}^j
\]
The velocity $\dot{q}$ here is a tangent vector, the momentum $p$ is a cotangent vector, and we need the metric to relate them, via
\begin{align*}
 g:T_q\mathcal{M}\times T_q\mathcal{M}&\longrightarrow\mathbb{R}\\
 (v,w)&\longmapsto g(v,w)
\end{align*}
which gives
\begin{align*}
 T_q\mathcal{M} &\longrightarrow T^\ast_q\mathcal{M} \\
 v &\longmapsto g(v,-).
\end{align*}
In coordinates this would say that the tangent vector $v^i$ gets mapped to the cotangent vector $g_{ij}v^j$.  This is lurking behind the passage from $\dot{q}^i$ to the momentum $mg_{ij}v^j$.

Getting back to the E-L equations,
\begin{align*}
 p_i = \frac{\pa L}{\pa\dot{q}^i} &= mg_{ij}\dot{q}^j \\
 F_i = \frac{\pa L}{\pa q^i} 
 &= \frac{\pa}{\pa q^i}\Bigl(\tfrac{1}{2}mg_{jk}(q)\dot{q}^j\dot{q}^k\Bigr)\\
 &=\tfrac{1}{2}m\pa_{i}g_{jk}\dot{q}^j\dot{q}^k,
 \quad(\text{where } \pa_i=\frac{\pa}{\pa q^i}).
\end{align*}
So the Euler-Lagrange equations say
\[
 \frac{d}{dt}mg_{ij}\dot{q}^j = \tfrac{1}{2}m\pa_{i}g_{jk}\dot{q}^j\dot{q}^k.
\]
The mass factors away, so the motion is independent of the mass!  Essentially we have a geodesic equation.

We can rewrite this \emph{geodesic equation} as follows
\begin{align*}
 \frac{d}{dt}g_{ij}\dot{q}^j &= \tfrac{1}{2}\pa_{i}g_{jk}\dot{q}^j\dot{q}^k \\
 \hspace{-3ex}\therefore\quad
 \pa_{k}g_{ij}\dot{q}^k\dot{q}^j + g_{ij}\ddot{q}^j 
 &= \tfrac{1}{2}\pa_{i}g_{jk}\dot{q}^j\dot{q}^k \\
 \hspace{-3ex}\therefore\quad
 g_{ij}\ddot{q}^j &= \bigl(\tfrac{1}{2}\pa_{i}g_{jk}-\pa_{k}g_{ij}\bigr)\dot{q}^j\dot{q}^k \\
 &= \tfrac{1}{2}\bigl(\pa_{i}g_{jk}-\pa_{k}g_{ij}-\pa_{j}g_{ki}\bigr)\dot{q}^j\dot{q}^k
\end{align*}
where the last line follows by symmetry of the metric, $g_{ik}=g_{ki}$.
Now let,
\[
 \Gamma_{ijk} = -\bigl(\pa_{i}g_{jk}-\pa_{k}g_{ij}-\pa_{j}g_{ki}\bigr)
\]
the minus sign being just a convention (so that we agree with everyone else).  This defines what we call the Christoffel symbols $\Gamma^i_{jk}$.  Then
\begin{align*}
 \ddot{q}_i = g_{ij}\ddot{q}^j &= -\Gamma_{ijk}\dot{q}^j\dot{q}^k \\
 \hspace{-3ex}\therefore\quad
 \ddot{q}^i &= -\Gamma^i_{jk}\dot{q}^j\dot{q}^k.
\end{align*}
So we see that $\ddot{q}$ can be computed in terms of $\dot{q}$ and the Christoffel symbols $\Gamma^i_{jk}$, which is really a particular type of \emph{connection} that a Lorentzian manifold has (the Levi-Civita connection), a \emph{connection} is just the rule for \emph{parallel transporting} tangent vectors around the manifold.

Parallel transport is just the simplest way to compare vectors at different points in the manifold.  This allows us to define, among other things, a \emph{covariant derivative}.


%%
%%
\subsection{Charged particle in EM Field in GR}

We can now apply what we've learned in consideration of a charged particle, of charge $e$, in an electromagnetic field with potential $A$, in our Lorentzian manifold.  The Lagrangian would be
\[
 L = \tfrac{1}{2}m g_{jk}\dot{q}^j\dot{q}^k + e A_i \dot{q}^i
\]
which again was conjured up be replacing the flat space metric $\eta_{ij}$ by the metric for GR $g_{ij}$. Not surprisingly, the Euler-Lagrange equations then yield the following equations of motion,
\[
 m\ddot{q}_i = -m\Gamma_{ijk}\dot{q}^j\dot{q}^k + e \fvect{F}_{ij}\dot{q}^j.
\]

If you want to know more about Lagrangians for general relativity we recommend the paper by Peldan~\cite{kn:Peldan94}, and also the ``black book'' of Misner, Thorne \& Wheeler~\cite{kn:MisnerThorne&Wheeler}.



%%
%%
%%
\section{The Principle of Least Action and Geodesics}
\label{sec:PrincipleOfLeastAction}

\noindent (Week 4, April 18, 20, 22.) \vspace{2ex}

%%
%%
\subsection{Jacobi and Least Time vs Least Action}

We've mentioned that Fermat's principle of least time in optics is analogous to the principle of least action in particle mechanics.  This analogy is strange, since in the principle of least action we fix the time interval $q:[0,1]\rightarrow Q$. Also, if one imagines a force on a particle resulting from a potential gradient at an interface as analogous to light refraction then you also get a screw-up in the analogy (Fig.~\ref{fig:leasttime-action-analogy}).
\begin{figure}[ht]
	\centering
	\psfrag{light}{light} 	\psfrag{particle}{particle}
	\psfrag{faster}{faster}	\psfrag{slower}{slower}
	\psfrag{V high}{$V$ high}	\psfrag{V low}{$V$ low}
	\psfrag{n high}{$n$ high}	\psfrag{n low}{$n$ low}
	\includegraphics[width=.75\textwidth]{images/leasttime-action-analogy}
	\caption{Least time versus least action.}
	\label{fig:leasttime-action-analogy}
\end{figure}

Nevertheless, Jacobi was able to reinterpret the mechanics of a particle as an optics problem and hence ``unify'' the two minimization principles.  First, let's consider light in a medium with a varying index of refraction $n$ (recall $1/n\propto$ speed of light).  Suppose it's in $\mathbb{R}^n$ with its usual Euclidean metric.  If the light s trying to minimize the \emph{time}, its trying to minimize the arclength of its path in the metric
\[
 g_{ij} = n^2\delta_{ij}
\]
that is, the index of refraction $n:\mathbb{R}^n\rightarrow (0,\infty)$, times the usual Euclidean metric
\[
 \delta_{ij} = \left( \begin{array}{ccc}
               1 &        & \text{\raisebox{-2ex}{\huge{0}}} \\
                 & \ddots &   \\
 \text{\huge{0}} &        & 1 \\
 \end{array} \right)
\]
% \[
% \begin{matrix}
%  1 &        & \text{\huge{0}} \\
%                  & \ddots &   \\
%  \text{\huge{0}} &        & 1
% \end{matrix}
% \]

This is just like the free particle in general relativity (minimizing its proper time) except that now $g_{ij}$ is a Riemannian metric
\begin{align*}
 g(v,w) &= g_{ij}v^i w^j \\
 \text{where}\quad g(v,v) &\ge 0
\end{align*}
So we'll use the same Lagrangian:
\[
 L(q,\dot{q}) = \sqrt{g_{ij}(q)\dot{q}^i\dot{q}^j}
\]
and get the same Euler-Lagrange equations:
\be
\label{eq:E-L_free-GR}
 \frac{d^2 q^i}{d t^2} + \Gamma^i_{jk}\dot{q}^j\dot{q}^k = 0
\ee
if $q$ is parameterized by arclength or more generally
\[
\lVert\dot{q}\rVert=\sqrt{g_{ij}(q)\dot{q}^i\dot{q}^j} = \,\text{constant}.
\]
As before the Christoffel symbols $\Gamma$ are built from the derivatives of 
the metric $g$.

Now, what Jacobi did is show how the motion of a particle in a potential could be viewed as a special case of this.  Consider a particle of mass $m$ in Euclidean $\mathbb{R}^n$ with potential $V:\mathbb{R}^n\rightarrow\mathbb{R}$.  It satisfies $F=ma$, i.e.,
\be\label{eq:Newton-2}
 m\frac{d^2 q^i}{d t^2} = -\pa_i V
\ee
How did Jacobi see \eqref{eq:Newton-2} as a special case of \eqref{eq:E-L_free-GR}?  He considered a particle of energy $E$ and he chose the index of refraction to be
\[
 n(q) = \sqrt{\frac{2}{m}\bigl(E-V(q)\bigr)}
\]
which is just the \emph{speed} of a particle of energy $E$ when the potential energy is $V(q)$, since
\[
 \sqrt{\frac{2}{m}(E-V)} = \sqrt{\frac{2}{m}\frac{1}{2}m\lVert\dot{q}\rVert^2} = \lVert\dot{q}\rVert. 
\]
Note: this is precisely backwards compared to optics, where $n(q)$ is proportional to the \emph{reciprocal} of the speed of light!!  But let's see that it works.
\begin{align*}
 L &= \sqrt{g_{ij}(q)\qdot^i\qdot^j} \\
   &= \sqrt{n^2(q)\qdot^i\qdot^j} \\
   &= \sqrt{2/m (E-V(q))\qdot^2}
\end{align*}
where $\qdot^2 =\qdot \vdot\qdot$ is just the usual Euclidean dot product, $v\vdot w =\delta_{ij}v^iw^j$.  We get the Euler-Lagrange equations,
\begin{align*}
 p_i = \frac{\pa L}{\pa\qdot^i} &= \sqrt{\frac{2}{m} (E-V)}\,\cdot\frac{\qdot}{\norm{\qdot}} \\
 F_i = \frac{\pa L}{\pa q^i} &= \pa_i\sqrt{\frac{2}{m} (E-V(q))}\,\cdot\norm{\qdot} \\
 &=\frac{1}{2} \frac{-2/m \pa_i V}{\sqrt{2/m (E-Vq)}}\,\cdot\norm{\qdot}
\end{align*}
Then $\pdot=F$ says,
\[
 \frac{d}{dt}\sqrt{2/m (E-V(q))}\,\cdot\frac{\qdot_i}{\norm{\qdot}} = -\frac{1}{m}\pa_i V\frac{\norm{\qdot}}{\sqrt{2/m (E-V)}}
\]
Jacobi noticed that this is just $F=ma$, or $m\ddot q_i = -\pa_i V$, that is, provided we reparameterize $q$ so that,
\[
 \norm{\qdot} = \sqrt{2/m (E-V(q))}.
\]
Recall that our Lagrangian gives reparameterization invariant Euler-Lagrange equations!  This is the unification between least time (from optics) and least action (from mechanics) that we sought.


%%
%%
\subsection{The Ubiquity of Geodesic Motion}

We've seen that many classical systems trace out paths that are geodesics, i.e., paths $q:[t_0,t_1]\rightarrow Q$ that are critical points of
\[
 S(q) = \int_{t_0}^{t_1}\sqrt{g_{ij}\qdot^i\qdot^j}\,dt
\]
which is proper time when $(Q,g)$ is a \emph{Lorentzian manifold}, or arclength when $(Q,g)$ is a \emph{Riemannian manifold}. We have
\begin{enumerate}
 \item The metric at $q\in Q$ is,
 \begin{align*}
   g(q):T_qQ\times T_qQ &\rightarrow\mathbb{R}  \\
   (v,w) &\mapsto g(v,w)
 \end{align*}
 and it is bilinear.
 \item  w.r.t a basis of $T_qQ$
 \[
  g(v,w) = \delta_{ij}v^iw^j
 \]
 \item $g(q)$ varies smoothly with $q\in Q$.
\end{enumerate}
An important distinction to keep in mind is that Lorentzian manifolds represent \emph{spacetimes}, whereas Riemannian manifolds represent that we'd normally consider as just \emph{space}.

We've seen at least three important things.
\begin{itemize}
 \item[(1)] In the geometric optics approximation, light in $Q=\mathbb{R}^n$ acts like particles tracing out geodesics in the metric
 \[
  g_{ij} = n(q)^2\delta_{ij}
 \]
 where $n:Q\rightarrow(0,\infty)$ is the index of refraction function.
 \item[(2)] Jacobi saw that a particle in $Q=\mathbb{R}^n$ in some potential $V:Q\rightarrow\mathbb{R}$ traces out geodesics in the metric
 \[
  g_{ij} = \frac{2}{m}(E-V)\delta_{ij}
 \]
 \emph{if} the particle has energy $E$ (where\footnote{The case $V>E$, if they exist, would be classically forbidden regions.} $V<E$).
\item[(3)] A free particle in general relativity traces out a geodesic on a Lorentzian manifold $(Q,q)$.
\end{itemize}

\noindent In fact all three of these results can be generalized to cover every problem that we've discussed!


\begin{itemize}
 \item[(1$^\prime$)] Light on \emph{any} Riemannian manifold $(Q,q)$ with index of refraction $n:Q\rightarrow(0,\infty)$ traces out geodesics in the metric $h=n^2g$.
 \item[(2$^\prime$)] A particle on a Riemannian manifold $(Q,q)$ with potential $V:Q\rightarrow\mathbb{R}$ traces out geodesics w.r.t the metric
 \[
  h = \frac{2}{m}(E-V) g
 \]
 \emph{if} it has energy $E$.  Lots of physical systems can be described this way, e.g., the Atwood machine, a rigid rotating body ($Q=SO(3)$), spinning tops, and others.  All of these systems have a Lagrangian which is a quadratic function of position, so they all fit into this framework.
\item[(3$^\prime$)] \textbf{Kaluza-Klein Theory.}  A particle with charge $e$ on a Lorentzian manifold $(Q,q)$ in an electromagnetic vector potential follows a path with
\[
 \ddot{q}_i = -\Gamma_{ijk}\qdot^j\qdot^k + \frac{e}{m}\fvect{F}_{ij}\qdot^j
\]
where
\[
 \fvect{F}_{ij} = \pa_iA_j - \pa_jA_i
\]
but this is actually geodesic motion on the manifold $Q\times U(1)$ where $U(1)=\{e^{i\theta}:\theta\in\mathbb{R}\}$ is a circle.
\end{itemize}

\noindent Let's examine this last result a bit further.  To get the desired equations for motion on $Q\times U(1)$ we need to given $Q\times U(1)$ a cleverly designed metric built from $g$ and $A$ where the amount of ``spiralling''---the velocity in the $U(1)$ direction is $e/m$.
\begin{figure}[ht]
	\centering
	\psfrag{QxU(1)}[c]{$Q\times U(1)$}
	\psfrag{a geodesic}{a geodesic}
	\psfrag{Q}{$Q$}
	\psfrag{the apparent path}{the apparent path}
	\includegraphics[width=.6\textwidth]{images/QxU(1)_geodesic}
	%\caption{Geodesic for a Kaluza-Klein metric.}
	\label{fig:QxU1_geodesic}
\end{figure}
The metric $h$ on $Q\times U(1)$ is built from $g$ and $A$ in a very simple way.  Let's pick coordinates $x^i$ on $Q$ where $i\in \{0,\ldots,n\}$ since we're in $n+1$-dimensional spacetime, and $\theta$ is our local coordinate on $S^1$.  The components of $h$ are
\begin{align*}
 h_{ij} &= g_{ij} +A_{i}A_{j} \\
 h_{\theta i} &= h_{\theta i} = -A_i \\
 h_{\theta\theta} &= 1
\end{align*}
Working out the equations for a geodesic in this metric we get
\begin{align*}
 \ddot q_i &= -\Gamma_{ijk}\qdot^j\qdot^k + \frac{e}{m}\fvect{F}_{ij}\qdot^j \\
 \ddot q_{\theta} &= 0, \\
 &\text{if $\qdot_\theta=e/m$}
\end{align*}
since $\fvect{F}_{ij}$ is part of the Christoffel symbols for $h$.

To summarize this section on least time versus least action we can say that every problem that we've discussed in classical mechanics can be regarded as geodesic motion!


