% vim: tw=50 % 20/01/2022 - 10AM \section{Introduction} Books: \begin{itemize} \item Classical Mechanics - Douglas Gregory (more examples) \item Classical Mechanics - Tom Kibble \& Frank Berkshire (more ``chatty'') \end{itemize} Lecture Notes - David Tong PDF online. \\ Lecture on Saturday 5th March will be pre-recorded \& online only. \subsubsection*{Newtonian Mechanics - Definitions} \begin{definition*}[Particle] A \emph{particle} is an object of insignificant size. For now, its only attribute is its position. \end{definition*} \noindent For large objects, we take the centre of mass to define the position and treat them like a particle. \\ To describe the position, we pick a \emph{reference frame}: \begin{definition*}[Reference frame] A \emph{reference frame} is a choice of origin and 3 coordinate axes. \end{definition*} \noindent With respect to this frame, a particle sweeps out a trajectory $\bf{x}(t)$. (sometimes, we may write $\bf{r}(t)$). \begin{center} \begin{tsqx} (0,0)->>(1,0) (0,0)->>(0,1) (0,0)->>(-0.4,-0.6) label $x$ @ (1.1,0) label $z$ @ (0,1.1) label $y$ @ 1.1*(-0.4,-0.6) (0,-0.8)..(0.3,-0.3)..(0.35,0.3)..(0.8,0.8) (0,0)->>(0.35,0.3) label $\mathbf{x}(t)$ @ (0.35,0.3)+(0.1,0) ! dot((0, 0)); \end{tsqx} \end{center} \noindent The \emph{velocity} of the particle is $\bf{v} = \dot{\bf{x}} := \dfrac{\bf{x}}{t}$. The \emph{acceleration} of the particle is $\bf{a} = \ddot{\bf{x}} := \dfrac[2]{\bf{x}}{t}$. \begin{note*} The derivative of a vector is defined using components: \[ \dfrac{\bf{x}}{t} = \begin{pmatrix} \dfrac{x}{t} \\ \dfrac{y}{t} \\ \dfrac{z}{t} \end{pmatrix} .\] \end{note*} \noindent Given two vector functions $\bf{f}(t)$ and $\bf{g}(t)$, \[ \dfrac{}{t}(\bf{f}, \bf{g}) = \dfrac{\bf{f}}{t} \cdot \bf{g} + \bf{f} \cdot \dfrac{\bf{g}}{t} \] and \[ \dfrac{}{t} (\bf{f} \times \bf{g}) = \dfrac{\bf{f}}{t} \times \bf{g} + \bf{f} \times \dfrac{\bf{g}}{t} .\] (proof by components) \subsubsection*{Newton's Laws of Motions} The framework of Newtonian mechanics rests on three axioms, known as Newton's Laws: \begin{enumerate}[N1] \item Left alone, a particle moves with constant velocity. \item The rate of change of momentum is proportional to the force. \item Every action has an equal and opposite reaction. \end{enumerate} \subsubsection*{Inertial Frames and The First Law} For many reference frames, N1 isn't true! It only holds for frames that are not themselves accelerating. Such frames are called \emph{intertial frames}. In an intertial frame, $\ddot{\bf{x}} = \bf{0}$ when left alone ($\bf{F} = \bf{0}$). \bigskip \noindent A better framing of the first law is: \begin{enumerate}[N1'] \item Inertial frames exist. \end{enumerate} Inertial frames provide the setting for much of what follows in this course. For most purposes, this room approximates an inertial frame. \subsubsection*{Galilean Relativity} Inertial frames are not unique. Given an inertial frame $S$, in which a particle has coordinate $\bf{x}$, we can construct another inertial frame $S'$ in which the coordinates of the particle are given by $\bf{x}'$. \begin{itemize} \item Translations: $\bf{x}' = \bf{x} + \bf{a}$, where $\bf{a}'$ is a constant. \item Rotations and reflections: $\bf{x}' = R\bf{x}$ where $R$ is a $3 \times 3$ matrix with $R^\top R = I$ (orthogonal matrix). \item Boosts: $\bf{x}' = \bf{x} + \bf{v}t$ where $\bf{v}$ is a constant. \end{itemize} For each of these, if there is no force on a particle, we have that $\dfrac[2]{\bf{x}}{t} = 0$ since $S$ is an inertial frame, which implies that $\dfrac[2]{\bf{x}}{t} = 0$ so $S'$ is an inertial frame. The \emph{Galilean principle of relativity} tells us that the laws of physics stay the same \begin{itemize} \item At every point in space. \item No matter which direction you face. \item No matter what constant velocity you move at. \item At all moments in time. \end{itemize} (these are experimentally tested facts). \bigskip \noindent There is no such thing as ``absolutely stationary'', but notice that acceleration is absolute - you don't have to accelerate relative to something.