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Signed-off-by: Marcello Seri <[email protected]>
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\usepackage{graphicx}

\usepackage{amsmath, amssymb, amsthm}
\numberwithin{equation}{chapter}
\usepackage{physics} % documentation at https://mirror.koddos.net/CTAN/macros/latex/contrib/physics/physics.pdf
\begin{warpprint}
\usepackage{fourier}
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\end{equation}
where $m\in\mathbb{R}$ is a constant called the \emph{mass} of the point particle.
\end{theorem}
\begin{remark}
In view of our previous discussions, the statement above should be read as
\begin{quote}
the lagrangian of an isolated point particle in an inertial system of coordinates has the form~\eqref{eq:singleptlag} up to total derivatives.
\end{quote}
\end{remark}
\begin{proof}
The invariance with respect to translations implies that $L(\vb*{x} + \vb*{x}_0, \dot{\vb*{x}}, t + t_0) = L(\vb*{x}, \dot{\vb*{x}}, t)$, that is, the lagrangian must be independent from $t$ and $\vb*{x}$: the lagrangian $L = L(\dot{\vb*{x}})$ must be a function of the velocity only.

Euler-Lagrange equations~\eqref{eq:eulerlagrange} then imply that the velocity must be constant. Indeed, we have
Euler-Lagrange equations~\eqref{eq:eulerlagrange} then imply that either $L$ is the constant function altogether or the velocity must be constant. Indeed, we have
\begin{equation}
0 = \dv{t}\pdv{L}{\dot{\vb*{x}}} - \pdv{L}{\vb*{x}} = \dv{t}\pdv{L}{\dot{\vb*{x}}},
\end{equation}
that is, $\pdv{L}{\dot{\vb*{x}}} = \mathrm{const}$. Since $L$ only depends on the velocity, this means that either $L$ is the constant function, which would make Euler-Lagrange equations completely trivial, or $\dot{\vb*{x}}(t) = \mathrm{const}$.
We just derived an alternative proof of Corollary~\ref{cor:Nfl}, Newton's first law, from the Galilean principle of relativity.

The invariance from orthogonal transformations then implies that $L(\dot{\vb*{x}}) = L(G\dot{\vb*{x}})$, where $G^t G = \Id$. That is, $L$ cannot depend on the direction of the velocity, only on its magnitude:
\begin{equation}
L = L(\|\dot{\vb*{x}}\|^2), \quad \|\dot{\vb*{x}}\|^2 := \langle\dot{\vb*{x}}, \dot{\vb*{x}}\rangle.
L(\dot{\vb*{x}}) = L(\|\dot{\vb*{x}}\|), \quad \|\dot{\vb*{x}}\|^2 := \langle\dot{\vb*{x}}, \dot{\vb*{x}}\rangle.
\end{equation}
Since we expect the lagrangian to be a smooth function, we want to avoid square roots and therefore we will write
\begin{equation}
L(\dot{\vb*{x}}) = L(\|\dot{\vb*{x}}\|^2).
\end{equation}

The invariance with respect to uniform motion now implies that the lagrangian must actually be proportional to $\|\dot{\vb*{x}}\|^2$.
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\begin{equation}
\frac{m \|\dot{\vb*{x}}\|^2}2 \mapsto
\frac{m \|\dot{\vb*{x}}\|^2}2 + m\,\langle\vb*{v},\dot{\vb*{x}}\rangle + \frac{m \vb*{v}^2}2
= \frac{m \|\dot{\vb*{x}}\|^2}2 + \frac{\dd }{\dd t}\left(m\,\langle\vb*{v},\vb*{x}\rangle + \frac{m \vb*{v}^2 t}2\right),
= \frac{m \|\dot{\vb*{x}}\|^2}2 + \dv{t}\left(m\,\langle\vb*{v},\vb*{x}\rangle + \frac{m \vb*{v}^2 t}2\right),
\end{equation}
and thus the equations of motion remain invariant under the transformation.
\end{proof}

\begin{corollary}[Newton's first law]\index{Newton!first law}
\begin{corollary}[Newton's first law]\label{cor:Nfl}\index{Newton!first law}
In an inertial frame of reference, an isolated point particle either does not move or it is in uniform motion with constant velocity.
\end{corollary}
\begin{proof}
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whose solutions are all of the form $x^i(t) = x^i_0 + v^i_0 t$.
\end{proof}

Remembering the additive property of the lagrangians, we can show that for a system of $N$ particles which do not interact, the lagrangian is simply the sum
The additive property of the lagrangians now implies that for a system of $N$ particles which do not interact, the lagrangian is simply the sum
\begin{equation}\label{eq:freel}
L = \sum_{k=1}^N \frac{m_k \|\dot{\vb*{x}}_k\|^2}{2}.
\end{equation}
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\begin{remark}
A lagrangian can always be multiplied by an arbitrary constant without affecting the equations of motion;
such multiplication then amounts to a change in the unit of mass.
So, the above definition of mass becomes meaningful when we take the additive property into account:
So, the above definition of mass becomes meaningful when we take the additive property into account:
the ratios of the masses remain unchanged and it is only these ratios which are physically meaningful.
\end{remark}

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\end{equation}
This confirms the intuitive notion that the kinetic energy should increase if a force pushes the particle in the direction of the current motion and should decrease when it pulls the particle in the opposite direction.

There is more, in fact
There is more, in fact our computation above shows that
\begin{equation}
T(t_1) - T(t_0) = \int_{t_0}^{t_1} \left\langle\dot{\vb*{x}}(t), \vb*{F}(\vb*{x}(t))\right\rangle\; \dd t = \int_{\vb*{x}_0}^{\vb*{x}_1} \left\langle\vb*{F}(\vb*{x}), \dd{\vb*{x}}\right\rangle,
\end{equation}
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You can read more about this in \cite[Chapter 2.5]{book:arnold} and \cite[Theorem 6.3 and 8.1]{book:knauf}.
\end{remark}


\begin{remark}
Often in physics, the kinetic energy is instead defined as \emph{the work required to bring the mass from rest to some speed~$\dot{\vb{x}}$}.
One can then perform the computation above backwards
\begin{equation}
\int_{x_0}^{x^1} \left\langle\vb*{F}(\vb*{x}), \dd{\vb*{x}}\right\rangle
= \int_{t_0}^{t_1} \left\langle\dot{\vb*{x}}(t), \vb*{F}(\vb*{x}(t))\right\rangle\; \dd t
= \int_{t_0}^{t_1} m \langle\dot{\vb*{x}}(t), \ddot{\vb*{x}}(t)\rangle\; \dd t
= \int_{t_0}^{t_1} \dv{t} \frac{m}{2} \langle\dot{\vb*{x}}(t), \dot{\vb*{x}}(t)\rangle \; \dd t,
\end{equation},
leading to the definition of kinetic energy as $T = \frac{m}{2} \|\dot{\vb{x}}\|^2$.
\end{remark}

Forces that can be written in the form $\vb*{F} = -\frac{\partial U}{\partial \vb*{x}}$, for some function $U$, are called \emph{conservative}.\index{force!conservative}
We will understand better why in a couple of sections.

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\warpHTMLonly{<p><iframe scrolling=no title=The Lorentz force src=https://www.geogebra.org/material/iframe/id/axnqyhdx/width/1100/height/800/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/true/rc/false/ld/false/sdz/false/ctl/false width=1100px height=800px style=border:0px;> </iframe></p>
<p>In the simulation above you can see how the Lorentz force can affect a charged particle motion in a uniform magnetic field. For an easy comparison two different particles are shown, with independent parameters but immersed in the same magnetic field. The value $q = e/c$ corresponds to the charge of the particle. The standalone version of this applet is <a href=https://www.geogebra.org/m/tvbfjfct>available on geogebra</a> and was derived from an <a href=https://www.geogebra.org/m/xpRMzPgc>applet by Luca Moroni</a>.</p>}

For a beautiful geometric discussion of this problem, you can refer to \cite[Chapter 8.3]{book:amr} (this has moved to Chapter 9.3 in more recent editions).
For a beautiful geometric discussion of this problem, you can refer to \cite[Chapter 8.3 (9.3 in more recent editions)]{book:amr}.
\end{exercise}

Note that the magnetic vector potential is not unique!
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\end{example}

\section{First steps with conserved quantities}
\subsection{Back to one degree of freedom}\label{sec:bdf}
\subsection{One degree of freedom}\label{sec:bdf}

Consider the one dimensional Newton's equation for a point particle with unit mass and position $x(t)$:
\begin{equation}\label{eq:oscillator}
\ddot x = F(x), \qquad F:\mathbb{R}\to\mathbb{R}, \quad t\in \mathbb{R}.
\end{equation}
It should come as no surprise, that introducing the auxiliary variable $y(t) = \dot x(t)$, \eqref{eq:oscillator} is equivalent to the system of first order equations
Introducing the auxiliary variable $y(t) = \dot x(t)$, \eqref{eq:oscillator} is equivalent to the system of first order equations
\begin{equation}\label{eq:oscillatorfirstorder}
\left\lbrace
\begin{aligned}
Expand All @@ -909,51 +934,63 @@ \chapter*{Preface}
\right..
\end{equation}
The solutions of \eqref{eq:oscillatorfirstorder} are parametric curves $(x(t),y(t)):\mathbb{R}\to\mathbb{R}^2$ in the $(x,y)$-space.
These curves are often called \emphidx{trajectories} of the system or \emph{integral curves} of the ordinary differential equation, it depends mostly on the context and the research fields in which they appear.
These curves are often called \emphidx{trajectories} of the system or \emph{integral curves} of the ordinary differential equation.

If $y\neq0$, we can apply the chain rule, $\frac{\dd y}{\dd t} = \frac{\dd y}{\dd x} \frac{\dd x}{\dd t}$, to get
If $y\neq0$, we can apply the chain rule, $\dv{y}{t} = \dv{y}{x} \dv{x}{t}$, to get
\begin{equation}\label{eq:lef}
\frac{F(x)}y = \frac{\dot y}{\dot x} = \frac{\dd y}{\dd x}.
\frac{F(x)}y = \frac{\dot y}{\dot x} = \dv{y}{x}.
\end{equation}
Reasoning formally\footnote{For a rigorous understanding of this, you can refer to \cite[Equation (5.1) with $f=y$ and Remark 5.1.3]{lectures:aom:seri}.} for a moment, we can rephrase this as
\begin{equation}
y\,\dd y - F(x)\, \dd x = 0.
\end{equation}

Getting rid of time and considering equation \eqref{eq:lef} comes with a price, the solution is now an implicit curve $y(x)$, but also with a huge advantage: this new equation can be solved exactly!
What is important here, is that this new equation does not depend on time. Although eliminating time and considering \eqref{eq:lef} comes at a cost -- the solution is now an implicit curve $y(x)$ -- it also presents a substantial benefit: this new equation can be solved exactly!

How is that so? We can separate $x$ and $y$ and integrate to get
How is that so? After a bit of algebra and integrating both sides with respect to $x$, we can rewrite this as
\begin{align}
&\int y\,\dv{y}{x}\dd x - \int F(x)\, \dd x \notag \\
& = \int y\,\dd y - \int F(x)\, \dd x = 0 \label{eq:pp-notime-int}
\end{align}
Integrating the integral on the left, we get
\begin{equation}
\frac12 y^2 + C_y = \int F(x) dx
\frac12 y^2 + C_y = \int F(x) dx.
\end{equation}
If $U(x)$ is such that $F(x) = -\frac{\dd U}{\dd x}$, we can further simplify the equation into
\begin{equation}
%
\begin{remark}
This identity in \eqref{eq:pp-notime-int} is often expressed as
\begin{equation}
y\,\dd y - F(x)\, \dd x = 0.
\end{equation}
This may look like a formal manipulation, but it is actually a rigorous identity in terms of differential forms.
To know more, you can refer to \cite[Equation (5.1) with $f=y$ and Remark 5.1.3]{lectures:aom:seri} or any good textbook on differential forms.
For simplicity, we will limit ourselves to the formal manipulation.
\end{remark}
%

If $U(x)$ is such that $F(x) = -\dv{U}{x}$, we can further simplify the equation into
\begin{equation}\label{eq:pp-energy}
\frac12 y^2 = -U(x) + C,
\end{equation}
where $C = C_x - C_y \in\mathbb{R}$ is just a number due to the constants of integration.
We can locally invert the equation above to get
where $C = C_x - C_y \in\mathbb{R}$ is just a number due to the constants of integration. Varying $C$, \eqref{eq:pp-energy} describes a family of curves in the $(x,y)$-plane which correspond to different solutions of our original problem.
We can also locally invert the equation above to get the explicit solution
\begin{equation}
y(x) = \pm \sqrt{2(C-U(x))}.
\end{equation}

The equation above may already be familiar: $\frac12 y^2 + U(x)$ is the sum of the kinetic energy $\frac12 y^2 = \frac12 {\dot x}^2$ of the particle and its potential energy $U$. In fact, the statement above is a theorem and we can prove it without the need of the formal step with the differentials.
Equation \eqref{eq:pp-energy} may already be familiar: $\frac12 y^2 + U(x)$ is the sum of the kinetic energy $\frac12 y^2 = \frac12 {\dot x}^2$ of the particle and its potential energy $U$. In fact, the statement above is a theorem and we can prove it without the need of the formal step with the differentials.

\begin{theorem}\label{thm:ham1}
Let $H(x, y) := \frac12 y^2 + U(x)$ where $U:\mathbb{R}\to\mathbb{R}$ is such that $F(x) = -\frac{\dd U}{\dd x}$.
Let $H(x, y) := \frac12 y^2 + U(x)$ where $U:\mathbb{R}\to\mathbb{R}$ is such that $F(x) = -\dv{U}{x}$.
Then, the connected components of the level curves $H(x,y) = C$ are the integral curves of \eqref{eq:oscillatorfirstorder}.
\end{theorem}
The function $H$ is called the \emph{total energy} of the mechanical system and we are already mimicking the notation that we will use when we will describe hamiltonian systems.
The function $H$ introduced in Theorem~\ref{thm:ham1} is called the \emph{total energy}\index{energy} of the mechanical system.
\begin{proof}
The proof is surprisingly simple:
\begin{align}
\dot H & = \frac{\partial H}{\partial x}\frac{\dd x}{\dd t} + \frac{\partial H}{\partial y}\frac{\dd y}{\dd t} \\
& = \frac{\dd U}{\dd x} y + y F(x)
\dot H & = \frac{\partial H}{\partial x}\dv{x}{t} + \frac{\partial H}{\partial y}\dv{y}{t} \\
& = \dv{U}{x} y + y F(x)
= -y F(x) + y F(x) = 0.
\end{align}
\end{proof}

The fact that $H(x,y)$ remains constant on the trajectories is crucial: when this happens we say that the total energy of the system is a \emph{conserved quantity}.
A curve $(x(t), y(t))$ spanned by a solution of \eqref{eq:oscillatorfirstorder} is called a \emph{phase curve}.
The fact that $H(x,y)$ remains constant on the trajectories is crucial: when this happens we say that the total energy of the system is a \emph{conserved quantity}\index{energy!conservation (1d)}.
A curve $(x(t), y(t))$ spanned by a solution of \eqref{eq:oscillatorfirstorder} is called a \emphidx{phase curve}.

\begin{figure}[htbp]
\centering
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\end{itemize}
\end{example}

From these examples we already see that phase curves can consist of only one point. In such cases, the points are called \emph{equilibrium} points.
From these examples we already see that phase curves can consist of only one point. In such cases, the points are called \emphidx{equilibrium} points since at these points the system is at rest.

\begin{exercise}
Show that at the equilibrium points of the examples above the system is at rest, that is, $\dot x = \dot y = 0$.
\end{exercise}

\subsection{The conservation of energy}\label{sec:energy}
\subsection{Conservation of energy}\label{sec:energy}

Let's see how general is the phenomenon described in the previous section.

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and to reason about further properties of the solutions.
To grasp where this comes from, it is enough to spell out
\begin{equation}
\frac m2 \left(\frac{\dd x}{\dd t}\right)^2 = E - U(x)
\frac m2 \left(\dv{x}{t}\right)^2 = E - U(x)
\end{equation}
and perform some formal computations with the differentials.

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\begin{equation}
p_i = \frac{\partial L}{\partial \dot{q}^i}
\qquad\mbox{and}\qquad
\dot p_i = \frac{\dd }{\dd t}\frac{\partial L}{\partial \dot{q}^i} = \pdv{L}{q^i}.
\dot p_i = \dv{t}\frac{\partial L}{\partial \dot{q}^i} = \pdv{L}{q^i}.
\end{equation}
The theorem follows comparing coefficients in \eqref{eq:dHlc} and \eqref{eq:dHlc2}.
\end{proof}
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Let $\Phi_t:T^*M \to T^*M$ denote a one--parameter group of canonical transformations, i.e $\big\{\Phi_t^* f, \Phi_t^* g\big\} = \Phi_t^*\big\{f,g\big\}$ for all $t\in\mathbb{R}$.
Show that the vector field $X$ which generates the group,
\begin{equation}
X(x) := \frac{\dd }{\dd t}\Phi_t(x)\Big|_{t=0},
X(x) := \dv{t}\Phi_t(x)\Big|_{t=0},
\end{equation}
is an infinitesimal symmetry.

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\begin{tcolorbox}
Given a one--parameter group of canonical transformations $\Phi_t:T^*M\to T^*M$, we call \emph{hamiltonian generator} of the group the hamiltonian $H$ on $T^*M$ such that
\begin{equation}
\frac{\dd }{\dd t}\Phi_t(x)\Big|_{t=0} = X_H(x).
\dv{t}\Phi_t(x)\Big|_{t=0} = X_H(x).
\end{equation}
\end{tcolorbox}

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