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\begin{document}
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\vspace*{2cm}
\begin{center}
  \textbf{\LARGE{\uwave{Reviews \& Exercises}}}
  \vspace{1cm}

  \textbf{\texttt{http://olympus.het.brown.edu/summer04/}}
\end{center}
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\vspace{2cm}
\tableofcontents
\newpage
%
\section{Vector Spaces}
%
\subsection{Review}
%
A \emph{vector space} $\mathbb{E}$ is a set whose elements are called \emph{vectors} and
where two operations are defined: \emph{addition} (to each pair of vectors $u, v\, \in
\mathbb{E}$ makes a new vector $(u+v)\, \in \mathbb{E}$ called the \emph{sum} of $u$ and
$v$), and \emph{multiplication by a real number (aka scalar)} (to each number
$\alpha\in\mathbb{R}$ and for each vector $v\in\mathbb{E}$ corresponds a new vector
$\alpha\cdot v = \alpha v$ called the \emph{product} of $\alpha$ by $v$).

These operations \textbf{must} satisfy --- for every $\alpha, \beta\in \mathbb{R}$ and
$u,v\in\mathbb{E}$ --- the following conditions, called the \emph{axioms} of vector space
($\alpha, \beta\in\mathbb{R}$ and $u, v, w\in\mathbb{E}$):

\begin{description}
\item[comutativity] $u + v = v + u$ ;
\item[associativity] $(u + v) + w = u + (v + w)$ and $(\alpha\beta)\, v = \alpha(\beta\,
  v)$ ;
\item[null vector] $\exists\; \mathbf{0}\in \mathbb{E}$, called \textbf{null vector} (or
    \emph{zero vector}), such that $v + \mathbf{0} = \mathbf{0} + v = v$ for all
    $v\in\mathbb{E}$ ;
\item[additive inverse] for each $v\in\mathbb{E}$ there exists a vector
  $(-v)\in\mathbb{E}$, called the \emph{additive inverse} (or \emph{symmetric}), such that
  $(-v) + v = -v + v = v + (-v) = v - v = \mathbf{0}$ ;
\item[distributivity] $(\alpha + \beta)\, v = \alpha\, v + \beta\, v$ and $\alpha\,(v + u)
  = \alpha\, u + \alpha\, v$ ; \&
\item[multiplication by 1] $1\cdot v = 1 \, v = v$.
\end{description}
%
\subsection{Exercises}
%
\begin{enumerate}
\item Verify the above for $\mathbb{R}^n$, called the $n$-dimensional Euclidean vector
  space, whose elements are given by ordered lists of real numbers: $v =
  (v_1,v_2,\dotsc,v_n)$.
\item Verify the above for the space $\mathbb{M}^{m\times n}$, i.e., the space of the $m$
  by $n$ \textsl{matrices} denoted by $\mathbf{A} = [a_{i\, j}]$, where $1\leqslant i
  \leqslant m$ and $1\leqslant j \leqslant n$,
  \begin{equation*}
    \mathbf{A} = 
    \begin{bmatrix}
      a_{11} & a_{12} & \dots & a_{1n} \\
      a_{21} & a_{22} & \dots & a_{2n} \\
      \vdots & \vdots & \vdots & \vdots \\
      a_{m1} & a_{m2} & \dots & a_{mn}
    \end{bmatrix} \; .
  \end{equation*}
  (When $m = n$ the matrices are called \emph{square matrices}.) Addition is defined via:
  $\mathbf{A} = [a_{i\, j}]$ and $\mathbf{B} = [b_{i\, j}]$ such that $\mathbf{A} +
  \mathbf{B} = [a_{i\, j} + b_{i\, j}]$. And multiplication by a scalar is defined via:
  $\alpha\, \mathbf{A} = [\alpha\, a_{i\, j}]$. The \textbf{null} matrix is one formed by
  all entries being equal to zero and the additive inverse is given by: $-\mathbf{A} =
  [-a_{i\, j}]$. Graphically, one has:
  \begin{equation*}
    \mathbf{A} + \mathbf{B} = 
    \begin{bmatrix}
      a_{11} + b_{11} & a_{12} + b_{12} & \dots & a_{1n} + b_{1n} \\
      a_{21} + b_{21} & a_{22} + b_{22} & \dots & a_{2n} + b_{2n} \\
      \vdots & \vdots & \vdots & \vdots \\
      a_{m1} + b_{m1} & a_{m2} + b_{m2} & \dots & a_{mn} + b_{mn}
    \end{bmatrix} \; ,\; 
    \alpha\, \mathbf{A} = 
    \begin{bmatrix}
      \alpha\,a_{11} & \alpha\,a_{12} & \dots & \alpha\,a_{1n} \\
      \alpha\,a_{21} & \alpha\,a_{22} & \dots & \alpha\,a_{2n} \\
      \vdots & \vdots & \vdots & \vdots \\
      \alpha\,a_{m1} & \alpha\,a_{m2} & \dots & \alpha\,a_{mn}
    \end{bmatrix} \; .
  \end{equation*}(Note that the definition of matrix multiplication --- which is the
      multiplication between two vectors in this matrix space --- is \textbf{not} needed!)
\item Find $t$ such that the matrix below is equal to the null matrix:
  \begin{equation*}
    \mathbf{M} = 
    \begin{bmatrix}
      t^2 - 1 & t^2 - t \\
      t^3 - 1 & t^2 - 3\, t + 2
    \end{bmatrix} \; .
  \end{equation*}
\item Determine the vectors $u, v\in \mathbb{R}^4$ knowing that the coordinates of $u$ are
  all equal, that the last coordinate of $v$ is $3$ and that $u + v = (1,2,3,4)$.
\item Define the \textsf{average} $u\ast v$, between two vectors $u,v\in\mathbb{E}$, by
  $u\ast v = \tfrac{1}{2}\, u + \tfrac{1}{2}\, v$. Prove that $(u\ast v)\ast w = u\ast
  (v\ast w)$ if, and only if, $u = w$.
\end{enumerate}
%
\section{Linear Transformations}
%
\subsection{Review}
%
Let $\mathbb{E}$ and $\mathbb{F}$ be vector spaces. A \textbf{linear transformation}
$\mathcal{A}: \mathbb{E} \rightarrow \mathbb{F}$ is a correspondence that associates to
each vector $v\in\mathbb{E}$ another vector $\mathcal{A}(v) = \mathcal{A} \cdot v =
\mathcal{A}\, v \in \mathbb{F}$ such that, for all $u,v\in\mathbb{E}$ and
$\alpha\in\mathbb{R}$, the following relations are valid:

\begin{align*}
  \mathcal{A}(u + v) &= \mathcal{A}\,u + \mathcal{A}\, v \; ;\\
  \mathcal{A}(\alpha\, v) &= \alpha \cdot \mathcal{A}\, v \; .
\end{align*}

In fact, the following are consequencies of the above: $\mathcal{A}(0) = 0$,
$\mathcal{A}(\alpha u + \beta\, v) = \alpha\, \mathcal{A}(u) + \beta\, \mathcal{A}(v)$,
$\mathcal{A}(-v) = -\mathcal{A}\, v$ and $\mathcal{A}(u - v) = \mathcal{A}\, u -
\mathcal{A}\, v$. Furthermore, the \emph{sum} of two linear transformations and the
\emph{product} by a scalar are linear transformations in their own right.
%
\subsection{Exercises}
%
\begin{enumerate}
\item Show that if $\mathcal{A}, \mathcal{B}: \mathbb{E} \rightarrow \mathbb{F}$ are
  linear transformations and if $\alpha\in\mathbb{R}$, then $\mathcal{A} + \mathcal{B}$
  and $\alpha\, \mathcal{A}$ are linear transformations. (Note that
  $(\mathcal{A}+\mathcal{B}): \mathbb{E} \rightarrow \mathbb{F}$ is given by
  $(\mathcal{A}+\mathcal{B})(v) = \mathcal{A}\, v + \mathcal{B}\, v$ and that $\alpha\,
  \mathcal{A}: \mathbb{E} \rightarrow \mathbb{F}$ is given by $(\alpha\, \mathcal{A})(v) =
  \alpha\cdot \mathcal{A}\, v$.)
\item Let $\mathcal{R}, \mathcal{S}: \mathbb{R}² \rightarrow \mathbb{R}²$ be,
  respectively, the rotation of $30^{\circ}$ around the origin and the reflection with
  respect to the $x$-axis. Given the vector $v=(2,5)$, determine the values of
  $\mathcal{R}\, v$ and of $\mathcal{S}\, v$.
\item Given $u_1 = (2,-1), \, u_2 = (1,1), \, u_3 = (-1,-4), \, v_1 = (1,3), \, v_2 =
  (2,3), \, v_3 = (-5,-6)$, decide whether there exists a linear operator such that
  $\mathcal{A}: \mathbb{R}² \rightarrow \mathbb{R}²$ such that $\mathcal{A}\, u_1 = v_1,
  \, \mathcal{A}\, u_2 = v_2,\, \mathcal{A}\, u_3 = v_3$. Repeat the above for $v_3 =
  (5,-6)$ and for $v_3 = (5,6)$.
\item The general expression of a linear operator $\mathcal{A}: \mathbb{R}² \rightarrow
  \mathbb{R}²$ is $\mathcal{A}(x,y) = (a\, x + b\, y, c\, x + d\, y)$. Determine the value
  of the constants $a, b, c, d$ such that $\mathcal{A}$ transforms the vectors $u = (1,2)$
  and $v = (3,4)$ in the vectors $\mathcal{A}\, u = (1,1)$ and $\mathcal{A}\, v = (2,2)$.
\item Let $\mathcal{A}: \mathbb{R}² \rightarrow \mathbb{R}²$ be defined by
  $\mathcal{A}(x,y) = (5\, x + 4\, y, -3\, x -2\, y)$. Find the non-vanishing vectors $u =
  (u_1, u_2)$ and $v = (v_1, v_2)$ such that $\mathcal{A}\, u = u$ and $\mathcal{A}\, v =
  2\, v$. Are those solutions unique? Is it possible to find $w \neq 0, w\in\mathbb{R}²$
  such that $\mathcal{A}\, w = \alpha\, w$, where $\alpha\neq 1, \alpha\neq 2$?
\end{enumerate}
%
\section{Product of Linear Transformations}
%
\subsection{Review}
%
The product of linear transformations is a concrete example of an algebraic structure that
has interesting phenomena, not found in the operations among numbers or vectors.

Given the linear transformations $\mathcal{A}: \mathbb{E} \rightarrow \mathbb{F}$ and
$\mathcal{B}: \mathbb{F} \rightarrow \mathbb{G}$, where the domain of $\mathcal{B}$
coincides with the image of $\mathcal{A}$, the \emph{product} of them is defined as
$\mathcal{BA}: \mathbb{E} \rightarrow \mathbb{G}$, where $(\mathcal{BA})\, v =
\mathcal{B}(\mathcal{A}\, v),\; \forall\, v\in\mathbb{E}$. That is, the diagram below
commutes:

\begin{equation*}
  \xymatrix{
    \mathbb{E} \ar[r]^{\mathcal{A}} \ar@{>}[dr]_{\mathcal{BA}= \mathcal{B}\circ
    \mathcal{A}} & \mathbb{F} \ar[d]^{\mathcal{B}} \\
      & \mathbb{G}
  }
\end{equation*}

It is not hard to show that $\mathcal{BA}$ is linear. Note that $\mathcal{BA}$ is nothing
more than the composition of the operators $\mathcal{B}$ and $\mathcal{A}$: $\mathcal{BA}
= \mathcal{B}\circ \mathcal{A}$. Therefore, one can establish the following properties:

\begin{description}
\item[associativity] $\mathcal{C}\, (\mathcal{BA}) = (\mathcal{CB})\, \mathcal{A} =
  \mathcal{CBA}$ ;
\item[left distributive] $(\mathcal{B} + \mathcal{C})\, \mathcal{A} = \mathcal{BA} +
  \mathcal{CA}$ ;
\item[right distributive] $\mathcal{C}\, (\mathcal{A} + \mathcal{B}) = \mathcal{CA} +
  \mathcal{CB}$ ; \&
\item[homogeneity] $\mathcal{B}(\alpha\, \mathcal{A}) = \alpha\, (\mathcal{BA})$.
\end{description}
%
\subsection{Exercises}
%
\begin{enumerate}
\item Verify \textsl{explicitly} that the product $\mathcal{BA}$ of linear transforms is a
  linear transform on its own right.
\item Consider the linear operators $\mathcal{R}_{\alpha},\, \mathcal{R}_{\beta}:
  \mathbb{R}² \rightarrow \mathbb{R}²$ the rotations around the origin of the angles
  $\alpha$ and $\beta$, respectively. Show that $\mathcal{R}_{\alpha}\,
  \mathcal{R}_{\beta} = \mathcal{R}_{\alpha + \beta}$.
\item If $\mathcal{S}: \mathbb{R}² \rightarrow \mathbb{R}²$ is the reflection on the
  $x$-axis, show that $\mathcal{S}\, \mathcal{S} = \mathds{1}$ ($\mathds{1}$ is the
  identity operator). (Can this be generalized to reflections about \textbf{any} line?
  Would the result above still be valid?)
\item For $\mathcal{R}$ and $\mathcal{S}$ as defined above, show that $\mathcal{RSR} =
  \mathcal{S}$ and that $\mathcal{RS} \neq \mathcal{SR}$.
\item Let $\mathcal{R}_{\alpha},\, \mathcal{R}_{\beta}$ be the rotations around the origin
  of the angles $\alpha$ and $\beta$. Starting from the fact that the product
  $\mathcal{R}_{\alpha}\, \mathcal{R}_{\beta} = \mathcal{R}_{\alpha + \beta}$ is the
  rotation by an angle $\alpha + \beta$, use the construction done in class in order to
  obtain the classical formul\ae: $\cos(\alpha+\beta) = \cos(\alpha)\, \cos(\beta) -
  \sin(\alpha)\, \sin(\beta)$, and $\sin(\alpha+\beta) = \sin(\alpha)\, \cos(\beta) +
  \sin(\beta)\, \cos(\alpha)$. (Note that to solve this exercise you \textbf{will} need to
  know how to \textbf{multiply} $2\times 2$-matrices and you will have to know the
  general form of a rotation matrix, deduced in class.)
\end{enumerate}
%
\section{Inner Product}
%
\subsection{Review}
%
The \emph{inner product} is a concept that completes and enriches the structure of a
vector space, allowing the use of a geometrical language highly suggestive.

The axioms of vector space are not sufficient to define certain geometrical notions such
as angle, perpendicularism, length, distance, etc\ldots\/ This is only possible with the
introduction of an inner product.

An \emph{inner product} on a vector space $\mathbb{E}$ is a bi-linear symmetric and
positive functional on $\mathbb{E}$. More precisely, it is a function $\mathbb{E}\times
\mathbb{E} \rightarrow \mathbb{R}$ that associates to each pair of vectors $u, \, v\in
\mathbb{E}$ a \textbf{real} number $\ip{u}{v} = u\cdot v$, called the inner product of $u$
by $v$, such that the following properties are valid ($\alpha,\beta\in\mathbb{R}$ and
$u,u',\, v, v'\, \in\mathbb{E}$):

\begin{description}
\item[bilinearity] $\ip{\alpha\, u + \beta\, u'}{v + v'} = \alpha\, \ip{u}{v} + \alpha\,
  \ip{u}{v'} + \beta\, \ip{u'}{v} + \beta\, \ip{u'}{v'}$ ;
\item[symmetry (commutativity)] $\ip{u}{v} = \ip{v}{u}$ ; \&
\item[positivity] $\ip{u}{u} > 0$ and $\ip{u}{u} = 0$ if, and only if, $u = 0$.
\end{description}

The positive number $|u| = \sqrt{\ip{u}{u}}$ is called the \emph{norm} or the
\emph{length} of the vector $u$.

Furthermore, it is also true that: $|\ip{u}{v}| \leqslant |u|\, |v|$ (Schwarz's
inequality) and $|u + v| \leqslant |u| + |v|$ (triangle inequality).

In addition to this, considering a \textsf{generic} basis
$\{\mathbf{e}_1,\dotsc,\mathbf{e}_n\}$ and making $\ip{\mathbf{e}_i}{\mathbf{e}_j} =
g_{i\, j}$, the inner product of two vectors $u, v\in\mathbb{E}$ is given by:

\begin{align*}
  u = (u_1,\dotsc,u_n) &= u_1\, \mathbf{e}_1 + \dotsb + u_n\, \mathbf{e}_n = \sum_{i=1}^{n}
    u_i\, \mathbf{e}_i \; ;\\
  v = (v_1,\dotsc,v_n) &= v_1\, \mathbf{e}_1 + \dotsb + v_n\, \mathbf{e}_n = \sum_{j=1}^{n}
    v_j\, \mathbf{e}_j \; ;\\
  \ip{u}{v} &= \sum_{i,j=1}^{n} g_{i\, j}\, u_i\, v_j \; .
  \intertext{In matrix notation:}
  \ip{u}{v} &= u\, \mathbf{g}\, v \; .
\end{align*}

The matrix $\mathbf{g} = [g_{i\, j}] \in \mathbb{M}^{m\times n}$ is \textbf{symmetric},
ie, $g_{ij} = g_{ji}$, once $\ip{\mathbf{e}_i}{\mathbf{e}_j} =
\ip{\mathbf{e}_j}{\mathbf{e}_i}$. Also, the matrix $\mathbf{g}$ is \textbf{positive}, ie,
$\ip{x}{x} = \sum_{i=1}^{n}g_{ij}\, x_i\, x_j \geqslant 0$ (the matrix $\mathbf{g}$ is
called the \emph{metric}).
%
\subsection{Exercises}
%
\begin{enumerate}
\item (Pythagorean theorem) Given that $|u + v|² = \ip{u+v}{u+v} = |u|² + |v|² + 2\,
  \ip{u}{v}$, what is the condition on the angle between $u$ and $v$ so that the
  Pythagorean theorem is valid?
\item In a vector space with an inner product, the angle between two [non-vanishing]
  vectors, $u,v$, is defined to be $\theta = \varangle (u,v)$, such that
  $0\leqslant\theta\leqslant 180^{\circ}$ and $\cos(\theta) = \ip{u}{v}/|u|\, |v|$. Having
  said that, given the vectors $u = (3,4)$, $v = (1,-1)$ and $w = (-1,1)$, sort
  ascendingly the angles $\varangle(u,v), \varangle(u,w), \varangle(v,w)$.
\item Let $u=(a,b,c)\in\mathbb{R}³$ be a unit vector with $a\, b\, c \neq 0$. (Remember
  that if $u$ is a \emph{unit} vector then $|u| = 1 = \sqrt{\ip{u}{u}}$.) Find $t$ such
  that, with $v = (-b\, t, a\, t, 0)$ and $w = (a\, c\, t, b\, c\, t, -1/t)$, the vectors
  $u, v, w$ are unitary and pairwise orthogonal.
\item For each pair of vectors $u = (u_1, u_2)$ and $v = (v_1, v_2)$ in $\mathbb{R}²$,
  define $[u,v] = 2\, u_1\, v_1 - u_1\, v_2 - v_1\, u_2 + 2\, u_2\, v_2$. Prove that this
  defines an inner product in $\mathbb{R}²$.
\item Given the [standard, already defined] inner product, $\ip{u}{v}$, in $\mathbb{E}$,
  prove that $|u + v|² + |u - v|² = 2\, (|u|² + |v|²)$ for all
  $u,v\in\mathbb{E}$. Interpret this equality geometrically.
\item Given that $\ip{u}{v} = \sum_{i,j=1}^{n} g_{ij}\, u_i\, v_j$ is an inner product in
  $\mathbb{R}^n$, show that $g_{11} > 0, \dotsc, g_{nn} > 0$. (\emph{Hint:} Use the
  canonical basis vectors.)
\item For any natural number $n\in\mathbb{N}$, prove that the norm of the vector $v = (n,
  n+1, n(n+1))\in\mathbb{R}³$ is a natural number, ie, $|v| = m\in\mathbb{N}$.
\item Show that for all $u,v\in\mathbb{E}$: $||u| - |v||\leqslant|u - v|$.
\item The \emph{cosine} of the angle between two [non-vanishing] vectors $u, v$ is defined
  as being $\cos(u,v) = \ip{u}{v}/|u|\, |v|$. Prove that if $u$ and $v$ are orthogonal and
  non-vanishing, $\cos²(u,u-v) + \cos²(v,u-v) = 1$. (The sum of the squares of the cosines
  of the acute angles of a rectangular triangle is equal to 1.)
\end{enumerate}
%
\section{Physics}
%
\subsection{Review}
%
\subsubsection{Electromagnetic Radiation (Maxwell's equations)}
%
\begin{itemize}
\item mid-nineteenth ($\sim1850$s) century: well established connection between electricity
  and magnetism (generators and motors had already been invented)
\item theory lagged behind practice/experimentation (Faraday)
\item ($\sim1860$s) Maxwell did the mathematical analysis of electromagnetism! In 1864 came
  the Maxwell's equations!
\item Maxwell's eqs \textbf{imply} wave equation/phenomena with the speed of light, $c$,
  which turns out to be a \textbf{constant}. (The \textbf{constancy} of the speed of light
  is \textbf{the} key for special relativity!)
\item light \textbf{is} eletromagnetic radiation!
\item Einstein: trust Maxwell's eqs and modify Newton's laws.
\item Even though Maxwell believed in the Ether, his eqs survived the death of this
  hypothesis.
\end{itemize}
%
\subsubsection{The Ether --- Absolute Motion}
%
\begin{itemize}
\item light $\equiv$ wave $\Rightarrow$ ``something'' \textbf{must} be moving/waving!
  (This is the mechanistic viewpoint: something must ``shake'' in order for something to
  propagate; material motion.)
\item Therefore: ``whatever filled the vacuum'' $\equiv$ \textbf{ether}!
  (\emph{Luminiferous ether}: extremely tenuous gas that is more rigid than steel.)
\item \textbf{Michelson-Morley} experiment (interferometer): 1880s, method of determining
  absolute motion. Ether's motion would imply carrying light with it, which in turn
  implies that the speed of light \textbf{must} vary! (Remember Maxwell\ldots\/)
  \begin{center}
    \input{mm.pstex_t}
  \end{center}
\item MM´s interferometer: $t_1/t_2 = 1/\sqrt{1 - (v/c)²}$. Only reported
  \textbf{negative} results! (Realize that if you solve the first exercise below you will
  be proving this. Furthermore, once this is proved, Special Relativity is out in the
  open, ie, once you have found the \emph{magical} factor of $\sqrt{1 - (v/c)²}$, Special
  Relativity becomes straightforward! \smiley)
\item Note that the above means/implies: $T = T_0\, \sqrt{1 - (v/c)²}$, ie, time
  contraction.
\item \textbf{FitzGerald contraction}: $L = L_0\, \sqrt{1 - (v/c)²}$, ie, space dilation.
\item \textbf{Lorentz} ($\sim 1900$): $m = m_0/\sqrt{1 - (v/c)²}$ in order to account for
  phenomena happening in cyclotrons. Moreover, this gain of mass is independent of
  direction! Thus, \textbf{all} experiments designed to measure absolute motion have
  \textbf{\textcolor{Red}{failed}}!
\item Note that: $m = m_0/\sqrt{1 - (v/c)²} = m_0\, \bigl(1 - (v/c)²\bigr)^{-1/2} \approx
  m_0\, \bigl(1 + \tfrac{1}{2}\, (v/c)²\bigr)$, where we have used the fact (known as
  \emph{binomial expansion}) that $(1 + x)^n \approx 1 + n\, x ,\; \forall\,
  n\in\mathbb{Z}$. Thus, reminding ourselves that $\tfrac{1}{2}\, m\, v² = E$, we have
  that: $E = \Delta m\, c²$, with $\Delta m = (m - m_0)$, ie, the difference between the
  ``relativistic mass'', $(m)$, and the ``rest mass'', $(m_0)$.
\item \textbf{\textcolor{Red}{Red Flag}}: $v > c \Rightarrow L,T,m \in\mathbb{C}$, ie, for
  velocities greater than that of light, length and time would be complex numbers! (While
  we are talking about \texttt{ether} and \texttt{absolute motion}, there is
  \textbf{nothing} that prevents us from exceding $c$.)
\end{itemize}
%
\subsubsection{The Special Theory of Relativity (1905)}
%
\paragraph{\S\/ Postulates}
%
\begin{description}
\item[(a)] The \textbf{measured} speed of light, $c$, is \textbf{constant} and the maximum
  possible velocity in the Universe; \&
\item[(b)] \textbf{All} motion is \textbf{relative}, ie, there is \textbf{no} special
  reference frame, there is \textbf{no} system of coordinates which is ``more at rest''
  than any other system of coordinates!
\end{description}
%
\paragraph{\S\/ Deductions from the Theory}
%
The above postulates/axioms/assumptions imply the following. (Upshot: Dropping the
\emph{ether hypothesis} leads us to explain all the phenomena that is seen in the experiments.)
\begin{itemize}
\item FitzGerald contraction (length contraction: $L = L_0\, \sqrt{1 - (v/c)²}$).
\item Time dilation: $T = T_0\, \sqrt{1 - (v/c)²}$.
\item Lorentz mass gain: $m = m_0/\sqrt{1 - (v/c)²}$.
\item Mass-Energy \textbf{equivalence}: $E = m\, c²$ (follows from the above).
\item Lorentz transformations: $y = y', z = z'$ and $x' = \tfrac{x - v\, t}{\sqrt{1 -
  (v/c)²}}, t' = \tfrac{t - v\, x/c²}{\sqrt{1 - (v/c)²}}$.
\item \textsl{Relativity of Simultaneity}: Two events, $(t_0,x_1)$ and $(t_0,x_2)$. This
  implies that $t'_2 - t'_1 = \tfrac{v\, (x_1 - x_2)/c²}{\sqrt{1 - (v/c)²}}$:
  \emph{failure} of simultaneity!
\item \textbf{Four-vectors \& Minkowski Space}: $s² = x² + y² + z² - (c\, t)²$. This is
  called an \textsl{interval} on the frame of reference $S$. This \textbf{generalizes} our
  Euclidean concept of \emph{distance}! Thus, our four-vectors are given by: $u = (c\,t,
  x, y, z)$. In the same fashion, our \textbf{inner product} is, $\ip{u}{v} = u_x\, v_x +
  u_y\, v_y + u_z\, v_z - c\,(u_t\, v_t)$.
\item \emph{Relativity and the philosophers; Poincaré}: ``According to the principle of
  relativity, the laws of physical phenomena must be the same for a fixed observer as for
  an observer who has a uniform motion of translation relative to him, so that we have
  not, nor can we possibly have, any means of discerning whether or not we are carried
  along in such a motion.''
\item Velocities addition (has to change in order to accomodate the fact that $c$ is the
  maximum speed in the Universe). Begin by writing the Lorentz transformations in the
  following form: $y = y', z = z'$ and $x = \tfrac{x' + v\, t'}{\sqrt{1 - (v/c)²}}, t =
  \tfrac{t' + v\, x'/c²}{\sqrt{1 - (v/c)²}}$, where $v$ is the velocity between the two
  reference frames. Then, note that $v_{x'} = x'/t'$, ie, the velocity in the primed
  coordinate system is equal to the ratio between the primed space and primed time
  variables. Thus, $x' = v_{x'}\, t'$, which, in turn, can be substituted in our previous
  formul\ae\/ to yield:
  \begin{align*}
    x &= \frac{v_{x'}\, t' + v\, t'}{\sqrt{1 - (v/c)²}} \; ; \\
    t &= \frac{t' + v\, v_{x'}\, t'/c²}{\sqrt{1 - (v/c)²}} \; . \\
    \intertext{Using the above and the fact that $v_x = x/t$ (ie, the velocity in the
    unprimed reference frame), we find that:}
    v_x & = \frac{x}{t} = \frac{v + v_{x'}}{1 + v\, v_{x'}/c²} \; . \\
    \intertext{Furthermore, using the binomial expansion, it is easy to see that:}
      v_x &= (v + v_{x'})\, \biggl(1 + \frac{v\, v_{x'}}{c²}\biggr)^{-1} \approx (v +
        v_{x'})\, \biggl(1 - \frac{v\, v_{x'}}{c²}\biggr) + \dotsb \; \\
    \therefore\, \lim_{c \rightarrow\infty} v_x &= v + v_{x'} \; .
    \intertext{Even though we can always arrange our coordinate system so that the motion
      is 1-dimensional, here is the formula in its full generality:}
    \vec{v}_x &= \frac{\vec{v} + \vec{v}_{x'}}{1 + \frac{\ip{\vec{v}}{\vec{v}_{x'}}}{c²}} \; .
  \end{align*}
\item Relativistic Energy: $E² - p²\, c² = m²\, c^4$.
\end{itemize}
%
\subsection{Exercises}
%
\begin{enumerate}
\item Given the Michelson-Morley experimental setup (see above), show that $t_1/t_2 =
  1/\sqrt{1 - (v/c)²}\,$, where $v$ is the velocity of the light source with respect to
  the ether.
\item Show that $L = L_0\, \sqrt{1 - (v/c)²}$ and $T = T_0\, \sqrt{1 - (v/c)²}$ imply the
  more mathematical: $y = y', z = z'$ and $x' = \tfrac{x - v\, t}{\sqrt{1 - (v/c)²}}, t' =
  \tfrac{t - v\, x/c²}{\sqrt{1 - (v/c)²}}$.
\item Show that the Lorentz transformation above can be written in the following form: $y
  = y', z = z'$ and $x = \tfrac{x' + v\, t'}{\sqrt{1 - (v/c)²}}, t = \tfrac{t' + v\,
  x'/c²}{\sqrt{1 - (v/c)²}}$.
\item Verify that the Lorentz transformations can be written in terms of the hyperbolic
  trigonometric functions. (\emph{Hint:} See the appendix below.)
\item $\boldsymbol\star$ \textbf{Extra}: Show the the momentum 4-vector (aka 4-momentum)
  is orthogonal to the force 4-vector (aka 4-force). (Note: you will have to use a formula
  derived in class for $\tfrac{dm}{dt}$, where $m = m_0/\sqrt{1 - (v/c)²}$; or you can use
  the fact that $\langle\vec{F},v\rangle = c²\, \tfrac{dm}{dt}$. See the appendix below
  for the definitions of the 4-vectors above.) The fact that the 4-force is orthogonal to
  the 4-momentum shows that the 4-force is the derivative of the 4-momentum, just like
  what happens in 3-dimensional space! \smiley
\end{enumerate}
%
\appendix
%
\section{Physics Extras}
%
\begin{itemize}
\item Newtonian physics $\rightarrow$ interaction of particles: the potential energy
  (which is the ``object'' that determines the interaction among particles) is a function
  of the \textbf{[space] coordinates}! This implies/assumes \textbf{instantaneous}
  propagation!
\item Experiments: there is \textbf{no} \emph{instantaneous} interactions in Nature!
  \begin{equation*}
    \begin{pmatrix}
      \text{velocity of} \\
      \text{propagation of} \\
      \text{interaction}
    \end{pmatrix}
     = \frac{\text{distance between the 2 bodies}}{\text{time interval for change}}
  \end{equation*}
  Therefore, there can be \textbf{no} velocity bigger than this!
\item \textbf{Principle of [Einstein's] Relativity:} the velocity of propagation of the
  interactions is the \textbf{same} in \textbf{all} inertial systems of references. That
  is, this velocity is a \emph{universal constant}, called $c$. (It turns out that this is
  the speed of light! \smiley)
\item \textbf{Galileo's Relativity:} \emph{infinite} velocity of propagation [of
  interactions]. \emph{Absolute} time; its properties are independent of the frame of
  reference. This is why its ``group of symmetries'' is the group of the 3 spatial
  rotations.
\item \textbf{Einstein's Relativity:} \emph{finite} velocity of propagation [of
  interactions]. Its group of symmetries is the group of 6 spacetime rotations.
\item \textbf{Proper Time (generalized clock):} $s² = (c\, t)² - x² - y² - z²$. Taking its
  \emph{differential form}, one gets: $ds² = (c\, dt)² - dx² - dy² - dz²$. Now, say that
  you have another frame of reference at rest, you get: $ds² = c²\, dt'²$. Therefore,
  equalling the above two expressions, you have: $ds² = (c\, dt)² - dx² - dy² - dz² = c²\,
  dt'²$. At this point, let's find the differential form of the time dilation formula, ie:
  $t = t'\, \sqrt{1 - (v/c)²} \, \Rightarrow\, dt' = dt\, \sqrt{1 - (v/c)²}$. Putting it
  all together: $dt' = \tfrac{ds}{c} = dt\, \sqrt{1 - (v/c)²}$. This is called the
  \emph{proper time}. The reason it is so, is because while the clock at rest (primed
  reference frame) measures $(t_2 - t_1)$, the moving clock measures $(t'_2 - t'_1) =
  \int_{t_1}^{t_2} \sqrt{1 - (v/c)²}\, dt$. It is not difficult to see that this integral
  is maximum for a clock at \textbf{rest}.
\item \textbf{Hyperbolic Rotations:} A rotation on the $tx$-plane must leave $(c\, t)² -
  x² = s²$ unchanged. Remember that: $\cosh²(\theta) - \sinh²(\theta) = 1$. Therefore, we
  can use this to simplify and clarify the meaning of our rotation on the $tx$-plane.
  \begin{equation*}
    \begin{pmatrix}
      x \\
      c\, t
    \end{pmatrix}
    =
    \begin{pmatrix}
      \cosh(\psi) & \sinh(\psi) \\
      \sinh(\psi) & \cosh(\psi)
    \end{pmatrix}
    \,
    \begin{pmatrix}
      x' \\
      c\, t'
    \end{pmatrix}
    =
    \begin{pmatrix}
      x'\, \cosh(\psi) & c\, t'\, \sinh(\psi) \\
      c\, t'\, \sinh(\psi) & x'\, \cosh(\psi)
    \end{pmatrix}
    \; .
  \end{equation*}
  As a special case [of the above], you can consider $x = c\, t'\, \sinh(\psi)$ and $c\, t
  = \cosh(\psi)$. This implies that $\tanh(\psi) = \tfrac{x}{c\, t} = v/c$. Thus, it is
  not hard to see, from this, that: $\cosh(\psi) = 1/\sqrt{1 - (v/c)²}$ and $\sinh(\psi) =
  (v/c)/\sqrt{1 - (v/c)²}$.
\item \textbf{Four-momentum:} The 4-momentum vector is given by $p = (m\, c, m\, v_x, m\, v_y,
  m\, v_z) = m\, (c, \vec{v})$, where $m = m_0/\sqrt{1 - (v/c)²}$ and $\vec v$ is the
  3-dimensinal velocity. Therefore, it is not hard to see that: $\ip{p}{p} = p² = m²\, (c²
  - \vec{v}²) = (m\, c)² - (m\, \vec{v}²) = E²/c² - \vec{p}²$, where $\vec p$ is the
  3-dimensional momentum. Another way to write the 4-momentum is: $p = (E/c,
  \vec{p})$.
\item \textbf{Four-force:} The 4-force vector is: $f = \gamma\, (c\, \tfrac{dm}{dt},
  \vec{F})$, where $\gamma = \bigl(1 - (v/c)²\bigr)^{-1/2}$ and $\vec F$ is the
  3-dimensional force.
\item \textbf{Maxwell's Equations (1864):} Mathematical analysis of the electromagnetic
  field. Let $\vec{E}$ and $\vec{B}$ denote the electric and magnetic vector fields,
  respectively. Also, let $\rho$ denote the density of charge and $\vec{j}$ denote the
  dislocation current. Then, Maxwell's eqs are given by (in the \textsf{CGS} system of
  units):
  \begin{align*}
    \vec\nabla\cdot\vec{E} = \dv(\vec E) = \frac{\partial E_x}{\partial x} +
      \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = 4\, \pi\, \rho &\; ; \\
    \vec\nabla\cdot\vec{B} = \dv(\vec B) = \frac{\partial B_x}{\partial x} +
      \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0 &\; ; \\
    \vec\nabla\times\vec{E} = \rt(\vec E) = \biggl(\frac{\partial E_z}{\partial y} -
      \frac{\partial E_y}{\partial z}\biggr)\vec{e}_x + \biggl(\frac{\partial
      E_x}{\partial z} - \frac{\partial E_z}{\partial x}\biggr)\vec{e}_y +
      \biggl(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial
      y}\biggr)\vec{e}_z &=  -\frac{1}{c}\, \frac{\partial\vec B}{\partial t} \; ; \\
    \vec\nabla\times\vec{B} = \rt(\vec B) = \biggl(\frac{\partial B_z}{\partial y} -
      \frac{\partial B_y}{\partial z}\biggr)\vec{e}_x + \biggl(\frac{\partial
      B_x}{\partial z} - \frac{\partial B_z}{\partial x}\biggr)\vec{e}_y +
      \biggl(\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial
      y}\biggr)\vec{e}_z &= \frac{1}{c}\, \frac{\partial\vec E}{\partial
      t} + \frac{4\, \pi}{c}\, \vec{j} \; . \\
  \end{align*}
  The above eqs imply that the fields $\vec{E}$ and $\vec{B}$ obey the \textbf{wave}
  equation! Their wave eq is given by:
  \begin{align*}
    \biggl(\frac{\partial²\vec{E}}{\partial x²} + \frac{\partial²\vec{E}}{\partial y²} +
      \frac{\partial²\vec{E}}{\partial z²}\biggr) -
      \frac{1}{c²}\frac{\partial²\vec{E}}{\partial t²} &= 0 \; ; \\
    \biggl(\frac{\partial²\vec{B}}{\partial x²} + \frac{\partial²\vec{B}}{\partial y²} +
      \frac{\partial²\vec{B}}{\partial z²}\biggr) -
      \frac{1}{c²}\frac{\partial²\vec{B}}{\partial t²} &= 0 \; . \\
  \end{align*}
  The Lorentz transformations are a symmetry of the wave equation, ie, if the coordinates
  are changed as in a Lorentz transformation in the wave equation, after some algebra you
  can show that the final result is the same as the original one. So, the wave equation
  has the same \textbf{form} in \textbf{any} reference frame! This is called in the
  jargon: ``The wave equation is \emph{covariant} with respect to the Lorentz
  transformations.'' That is, even though Maxwell was a firm believer in the ether, his
  eqs show undoubtedly that it does not exist! \smiley

  Furthermore, Maxwell's eqs imply that $\vec{E}$ and $\vec{B}$ are orthogonal to each
  other.
\end{itemize}
%
\section{Complex Numbers}
%
A \emph{complex number} is formed out of \textbf{two} real numbers, just like a vector in
$\mathbb{R}²$. It is usually denoted $z = x + i\, y$, where $i$ is called the
\emph{imaginary unit} and $i² = -1$. The relation between the complex numbers and the
2-dimensional plane should be clear, however, the complex plane is called the
\emph{Argand} plane. Furthermore, the axis in the Argand plane are called \textbf{real}
and \textbf{imaginary}.

There is one major difference between the complex numbers and vectors in $\mathbb{R}²$:
there is no such thing like multiplication nor division of vectors [by other vectors]. On
the other hand, one can perform those operations on complex numbers! \smiley\/ Therefore,
the fancy jargon name for it is \textbf{Complex Algebra}, because an algebra is a
mathematical setting, just like a vector space, where the operations of multiplication and
division are defined. (Remember that we can only divide and multiply vectors by scalars.)

Sure enough, one could go ahead and \textbf{define} the multiplication of two vectors and
the division of two vectors according to their analogues in the complex plane. Doing so
means that one is defining an \emph{algebraic structure} on the 2-dimensional vector
space.

But, enough of this abstract theory/talk! Let's move on to more complex numbers. \smiley\/
Here are the definitions of the operations with complex numbers. In what follows, consider
$z, w \in\mathbb{C}$ and $a, b, c, d\in\mathbb{R}$.

\begin{description}
\item[Real \& Imaginary parts] $z = a + i\, b \Rightarrow \Re(z) = a\, ,\; \Im(z) = b$;
\item[addition \& subtraction] $z \pm w = (a + i\, b) \pm (c + i\, d) = (a + c) \pm i\, (b
  + d)$. (Note that this is equivalent to the parallelogram rule.);
\item[complex conjugation] $z = a + i\, b \Rightarrow \bar{z} = z^{*} = a \boldsymbol{-}
  i\, b$ (This is a reflection about the real axis.);
\item[absolute value] $|z|² = |x|² + |y|²$, by the Pythagorean theorem. Still making use
  of the analogy with 2-dimensional vectors, it is not hard to see and prove the
  \emph{triangle's inequality} for complex numbers: $|z + w| \leqslant |z| + |w|$. (There
  are two clear ways to go about proving this: (1) geometrically, using the analogy with
  2-dimensional vectors, and (2) algebraically, using the definition of the absolute value
  and the triangle inequality as applied to real numbers --- you will have to use the
  triangle inequality for real numbers twice, once for the real part and once for the
  imaginary part.);
\item[multiplication] $z\, w = (a + i\, b)\, (c + i\, d) = (a\,c - b\, d) + i\, (a\, d +
  b\, c)$. Thus, it is easy to see that $|z|² = z\, \bar{z}$. Furthermore, this operation
  is \textbf{commutative} and $|z\, w| = |z|\, |w|$;
\item[division] $\tfrac{z}{w} = \frac{z\, \bar{w}}{|w|²}$. Also, given that $z\, \bar{z} =
  |z|² \Rightarrow 1/z = \bar{z}/|z|²$, from which our formula for the division is quite
  clear.
\item[polar coordinates] In analogy with polar coordinates in the 2-dimensional plane, the
  same can be done for the Argand plane. Let $\rho = |z| = \sqrt{z\, \bar{z}\,{}} =
  \sqrt{a² + b²\,{}}$, for $z = a + i\, b$. (As usual, $\rho$ is called the \emph{radius}
  of the complex number $z$.) Now, as for the angle $\theta$, called the \emph{argument}
  of $z$ and denoted $\arg(z) = \theta$, it will be measured with respect to the real axis
  (just like it is measured with respect to the $x$-axis in $\mathbb{R}²$) and its value
  will be given by $\tan(\theta) = b/a$. Thus, our canonical picture of projections on a
  plane are kept untouched: $a = \rho\, \cos(\theta)$ and $b = \rho\, \sin(\theta)$. So,
  it is easy to see that: $z = a + i\, b = \rho\, \bigl(\cos(\theta) + i\,
  \sin(\theta)\bigr)$. This, in turn, implies that $|z|² = |\rho|² \, \bigl|\cos(\theta) +
  i\, \sin(\theta)\bigr|² = |\rho|² = \rho²$, once $\bigl|\cos(\theta) + i\,
  \sin(\theta)\bigr|² = \cos²(\theta) + \sin²(\theta)\equiv 1$.
\item[Euler formula] $e^{i\, \theta} = \cos(\theta) + i\, \sin(\theta)$. In the particular
  case where $\theta = \pi$ we get the beautiful identity $e^{i\, \pi} + 1 = 0$! (This
  identity involves the 5 most important numbers! \smiley) Therefore, our complex numbers
  can be written as $z = \rho\, e^{i\, \theta}$. Moreover, it is easy to deduce the fact
  that $| e^{i\, \theta} | \equiv 1$!
\item[Trigonometric \& Hyperbolic functions] From the above formula, it is not hard to see
  that the following holds for trigonometric functions: $\sin(x) = (e^{i\, x} - e^{-i\,
  x})/2\, i$, $\cos(x) = (e^{i\, x} + e^{-i\, x})/2$. Their hyperbolic cousins are defined
  via the following equations: $\cosh(x) = (e^{x} + e^{-x})/2$ and $\sinh(x) = (e^{x} -
  e^{-x})/2$. That is, $\sinh(x) = -\sinh(-x)$, $\cosh(x) = \cosh(-x)$, $\sinh(i\, x) =
  i\, \sin(x)$, $\cosh(i\, x) = \cos(x)$, $\cosh²(x) - \sinh²(x) = 1$, $e^x = \cosh(x) +
  \sinh(x)$ (this last formula being the hyperbolic analog of Euler's formula above).
\item[Hyperbolic rotations] Also known as the a Lorentz transformation or Procrustian
  stretch, a hyperbolic transformation leaves each branch of the hyperbola $x'\, y' = x\,
  y$ invariant and transforms circles into ellipses with the same area: $x' = x/\mu$ and
  $y' = \mu\, x$.
\end{description}
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\end{document}
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