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\chapter{Equations of Motion}
\label{ch:Equations_of_Motion}

%\noindent (Week 2, April 4, 6, 8.)\vspace{2ex}

In this chapter we'll start to look at the Lagrangian equations of motion in more depth.  We'll look at some specific examples of problem solving using the Euler-Lagrange equations.  First we'll show how the equations are derived.


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\section{The Euler-Lagrange Equations}

We are going to start thinking of a general classical system as a set of points in an abstract \emph{configuration space} or phase space\footnote{The tangent bundle $TQ$ will be referred to as \emph{configuration space}, later on when we get to the chapter on Hamiltonian mechanics we'll find a use for the cotangent bundle $T^\ast Q$, and normally we call this the \emph{phase space}.}.  So consider an arbitrary classical system as living in a space of points in some manifold $Q$.  For example, the space for a spherical double pendulum would look like Fig.~\ref{fig:double-pendulum-space}, where $Q=S^2\times S^2$.
\begin{figure}[ht]
	\centering
	 \psfrag{Q=S2xS2}{$Q=S^2\times S^2$}
	\includegraphics[width=.33\textwidth]{images/double-pendulum-space}
	\caption{Double pendulum configuration space.}
	\label{fig:double-pendulum-space}
\end{figure}
So our system is ``a particle in $Q$'', which means you have to disabuse yourself of the notion that we're dealing with real particles, we are not, we are dealing with a \emph{single quasi-particle} in an abstract higher dimensional space.  The single quasi-particle represents two real particles if we are talking about the classical system in Fig.~\ref{fig:double-pendulum-space}.  Sometimes to make this clear we'll talk about ``the system taking a path'', instead of ``the particle taking a path''.  It is then clear that when we say, ``the system follows a path $q(t)$'' that we're referring to the point $q$ in configuration space $Q$ that represents all of the particles in the real system.

So as time passes ``the system'' traces out a path
\[
 q:[t_0,t_1]\longrightarrow Q
\]
and we define its \emph{velocity},
\[
 \dot{q}(t)\in T_{q(t)}Q
\]
to be the tangent vector at $q(t)$ given by the equivalence class $[\sigma]$ of curves through $q(t)$ with derivatives $\dot{\sigma}(t)=dq(s)/ds|_{s=t}$.  We'll just write is as $\dot{q}(t)$.

Let $\Gamma$ be the space of smooth paths from $a\in Q$ to $b\in Q$,
\[
 \Gamma = \{q:[t_0,t_1]\rightarrow Q| q(t_0)=a, q(t_1)=b\}
\]
($\Gamma$ is an infinite dimensional manifold, but we won't go into that for now.)  Let the \emph{Lagrangian}=$L$ for the system be \emph{any} smooth function of position and velocity (not explicitly of time, for simplicity),
\[
 L: TQ\longrightarrow \mathbb{R}
\]
and define the \emph{action}, $S$:
\[
 S:\Gamma \longrightarrow \mathbb{R}
\]
by
\begin{equation}
 S(q) \defeq \int_{t_0}^{t_1} L(q,\dot{q})\,dt
\end{equation}
The path that the quasi particle will actually take is a critical point of $S$, in accord with D'Alembert's principle of least action.  In other words, a path $q\in\Gamma$ such that for any smooth 1-parameter family of paths $q_s\in\Gamma$ with $q_0=q$, we have
\begin{equation}
\label{eq:leastaction}
 \frac{d}{ds}S(q_s)\Bigl.\Bigr|_{s=0} = 0
\end{equation}
We abbreviate
\[
 \frac{d}{ds}\Bigl.\Bigr|_{s=0} 
\]
as $\delta$, the \textbf{variational derivative}, so 
that Eq.\eqref{eq:leastaction} can be rewritten
\begin{equation}
 \delta S = 0
\end{equation}

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\subsection{Comments}

What is a ``1-parameter family of paths''?  Well, a path is a curve, or a 1D manifold.  So the 1-parameter family is nothing more nor less than a set of well-defined paths $\{q_s\}$, each one labeled by a parameter $s$.  So a smooth 1-parameter family of paths will have $q(s)$ everywhere infinitesimally close to $q(s+\epsilon)$ for an infinitesimal hyperreal $\epsilon$.
\begin{figure}[ht]
	\centering
	\psfrag{q0}{$q_0$} \psfrag{q_s}{$q_s$}
	\includegraphics[width=.15\textwidth]{images/1-param-family}
	\caption{Schematic of a 1-parameter family of curves.}
	\label{fig:1-param-family}
\end{figure}
So in Fig.~\ref{fig:1-param-family} we can go from $q_0$ to $q_s$ by smoothly varying $s$ from $s=0$ to $s=s$

What does the condition $\delta S=0$ imply?  Patience, we are just getting to that.  We will now start to explore what $\delta S=0$ means for our Lagrangian.

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\subsection{Lagrangian Dynamics}

We were given that $Q$ is a manifold, so it admits a covering of coordinate charts.  For now, let's pick coordinates in a neighborhood $U$ of some point $q(t)\in Q$.  Next, consider only variations $q_s$ such that $q_s=q$ outside $U$.  A cartoon of this looks like Fig.~\ref{fig:local-path-variation}
\begin{figure}[ht]
	\centering
	\psfrag{Q}{$Q$} \psfrag{U}{$U$} \psfrag{a}{$a$} \psfrag{b}{$b$}
	\includegraphics[width=.5\textwidth]{images/local-path-variation}
	\caption{Local path variation.}
	\label{fig:local-path-variation}
\end{figure}
Then we restrict attention to a subinterval $[t_0',t_1']\subseteq[t_0,t_1]$ such that $q_s(t)\in U$ for $t_0'\leq t\leq t_1'$.

Let's just go ahead and rename $t_0'$ and $t_1'$ as `` $t_0$ and $t_1$'' to drop the primes.  We can use the coordinate charts on $U$,
\begin{align*}
 \varphi : U &\longrightarrow \mathbb{R}^n \\
 x & \longmapsto \varphi(x)=(x^1,x^2,\ldots,x^n)
\end{align*}
and we also have coordinates for the 1-forms,
\begin{align*}
 d\varphi : TU &\longrightarrow T\mathbb{R}^n 
 \cong \mathbb{R}^n\times\mathbb{R}^n \\
 (x,y) & \longmapsto d\varphi(x,y)=(x^1,\ldots,x^n,y^1,\ldots,y^n)
\end{align*}
where $y\in T_xQ$.  We restrict $L:T\mathcal{M}\rightarrow\mathbb{R}$ to $TU\subseteq T\mathcal{M}$, in our case the manifold is $\mathcal{M}=Q$, and then we can describe it, $L$, using the coordinates $x^i,y^i$ on $TU$.  The $^i$ are generalized \emph{position} coordinates, the $y^i$ are the associated generalized \emph{velocity} coordinates.  (Velocity and position are in the abstract configuration space $Q$).  Using these coordinates we get
\begin{align*}
 \delta S &= \delta \int_{t_0}^{t_1} L(q(t),\dot{q}(t))\,dt \\
 &= \int_{t_0}^{t_1} \delta L(q,\dot{q})\,dt \\
 &= \int_{t_0}^{t_1} \Bigl(\frac{\pa L}{\pa x^i}\delta q^i
 + \frac{\pa L}{\pa y^i}\delta\dot{q}^i \Bigr) \,dt \\
\end{align*}
where we've used the given smoothness of $L$ and the Einstein summation convention for repeated indices $i$.  Note that we can write $\delta L$ as above using a local coordinate patch because the path variations $\delta q$ are entirely trivial outside the patch for $U$.  Continuing, using the Leibniz rule
\[
 \frac{d}{dt}\Bigl(\frac{\pa L}{\pa y}\delta q\Bigr) 
 = \frac{d}{dt}\frac{\pa L}{\pa y}\delta q 
 + \frac{\pa L}{\pa y}\delta \dot{q}
\]
we have,
\begin{align*}
 \delta S &= \int_{t_0}^{t_1} \Bigl(\frac{\pa L}{\pa x^i}
 - \frac{d}{dt}\frac{\pa L}{\pa y^i}\Bigr)\delta q^i(t)\,dt \\
 &= 0.
\end{align*}
If this integral is to vanish as demanded by $\delta S=0$, then it must vanish for all path variations $\delta q$, further, the boundary terms vanish because we deliberately chose $\delta q$ that vanish at the endpoints $t_0$ and $t_1$ inside $U$.  That means the term in brackets must be identically zero, or
\begin{equation}
 \frac{d}{dt}\frac{\pa L}{\pa y^i} - \frac{\pa L}{\pa x^i} = 0
\end{equation}
This is necessary to get $\delta S=0$, for all $\delta q$, but in fact it's also sufficient.  Physicists always give the coordinates $x^i$, $y^i$ on $TU$ the symbols ``$q^i$'' and ``$\dot{q}^i$'', despite the fact that these also have another meaning, namely the $x^i$ and $y^i$ coordinates of the quantity,
\[
 \bigl(q(t),\dot{q}(t)\bigr) \in TU.
\]

So in any case, physicists write,
\[
\boxed{
	\gapleft \gapabove
	\frac{d}{dt}\frac{\pa L}{\pa \dot{q}^i} = \frac{\pa L}{\pa q^i}
	\gapright \gapbelow
}
\]
and they call these the Euler-Lagrange equations.


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\section{Interpretation of Terms}

The derivation of the Euler-Lagrange equations above was fairly abstract, the terms ``position'' and ``velocity'' were used but were not assumed to be the usual kinematic notions that we are used to in physics, indeed the only reason we used those terms was for their analogical appeal.  Now we'll try to illuminate the E-L equations a bit by casting them into the usual position and velocity terms.

So consider,
\begin{align*}
 Q &= \mathbb{R} \\
 L(q,\dot{q}) &= \frac{1}{2}m\dot{q}\vdot\dot{q} - V(q) \\
 &= \frac{1}{2}m\dot{q}^i\dot{q}_i - V(q) 
\end{align*}

\begin{tabular}{ccc}
& & \\
TERMS & MEANING           & MEANING \\
      & (in this example) &  (in general)\\
\gapabove $\dfrac{\pa L}{\pa\dot{q}^i}$ & $m\dot{q}$ & the momentum $p_i$\\
\gapabove $\dfrac{\pa L}{\pa q^i}$ & $-(\grad V)_i$ & the force $F_i$\\
& & \\
\end{tabular}

When we write $\pa V/\pa q^i = \grad V$ we're assuming the $q^i$ are Cartesian coordinates on $Q$.

So translating our example into general terms, if we conjure up some abstract Lagrangian then we can think of the independent variables as generalized positions and velocities, and then the Euler-Lagrange equations can be interpreted as equations relating generalized concepts of \emph{momentum} and \emph{force}, and they say that
\begin{equation}
 \dot{p} = F
\end{equation}

So there's no surprise that in the mundane case of a single particle moving in $\mathbb{R}^3$ under time $t$ this just recovers Newton~II.  Of course we can do all of our classical mechanics with Newton's laws, it's just a pain in the neck to deal with the redundancies in $F=ma$ when we could use symmetry principles to vastly simplify many examples.  It turns out that the Euler-Lagrange equations are one of the reformulations of Newtonian physics that make it highly convenient for introducing symmetries and consequent simplifications.  Simplifications generally mean quicker, shorter solutions and more transparent analysis or at least more chance at insight into the characteristics of the system.  The main thing is that when we use symmetry to simplify the equations we are reducing the number of independent variables, so it gets closer to the fundamental degrees of freedom of the system and so we cut out a lot of the wheat and chaff (so to speak) with the full redundant Newton equations.

One can of course introduce simplifications when solving Newton's equations, it's just that it's easier to do this when working with the Euler-Lagrange equations.  Another good reason to learn Lagrangian (or Hamiltonian) mechanics is that it translates better into quantum mechanics.








