Let $\begin{pmatrix}
t\\
x
\end{pmatrix}$ be a vector in $tx$-plane and $\begin{pmatrix}
t’\\
x’
\end{pmatrix}$ denote its rotation by a hyperbolic angle $\phi$. Then as we have seen here we have:
\begin{equation}
\begin{aligned}
t’&=t\cosh\phi-x\sinh\phi\\
x’&=-t\sinh\phi+x\cosh\phi
\end{aligned}\label{eq:boost1}
\end{equation}
$t’$-axis ($x’=0$) is moving at a constant speed
\begin{equation}
\label{eq:velocity1}
v=\frac{x}{t}=\frac{\sinh\phi}{\cosh\phi}=\tanh\phi\end{equation}
while $x’$-axis ($t’=0$) is moving at a constant speed
\begin{equation}
\label{eq:velocity2}
v=\frac{x}{t}=\frac{\cosh\phi}{\sinh\phi}=\coth\phi
\end{equation}
This means that the time and spacial axes scissor together and hence the speed of light remains the same regardless of the coordinate transformation as illustrated in Figure 1. The picture in Figure 1 is called the spacetime diagram.

Figure 1. Spacetime Diagram

From Euclidean perspective, $t’$ and $x’$ do not appear to be orthogonal. However, from Lorentzian perspective they are orthogonal. To see that let $e_0=\begin{pmatrix}
1\\
0 \end{pmatrix}$ and $e_1=\begin{pmatrix}
0\\
1
\end{pmatrix}$. Also let $e_0’$ and $e_1’$ be the rotations of $e_0$ and $e_1$ by the hyperbolic angle $\phi$, respectively. Then by \eqref{eq:boost1} we have
$$e_0’=\begin{pmatrix}
\cosh\phi\\
-\sinh\phi
\end{pmatrix},\ e_1’=\begin{pmatrix}
-\sinh\phi\\
\cosh\phi
\end{pmatrix}$$
Now \begin{align*}\langle {e_0}’,e_1’\rangle&=(\cosh\phi\ -\sinh\phi)\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}\begin{pmatrix}
-\sinh\phi\\
\cosh\phi
\end{pmatrix}\\&=\cosh\phi\sinh\phi-\sinh\phi\cosh\phi=0\end{align*} So $e_0’$ and $e_1’$ are orthogonal.

\eqref{eq:boost1} with \eqref{eq:velocity1} can be written as
\begin{align*}
t’&=\cosh\phi(t-vx)\\
x’&=\cosh\phi(x-vt)
\end{align*}
Using the well-known identy $\cosh^2\phi-\sinh^2\phi=1$, we find $\cosh\phi=\frac{1}{\sqrt{1-v^2}}$. Therefore, \eqref{eq:boost1} can be written in terms of $t,x,t’,x’,v$ as
\begin{equation}
\begin{aligned}
t’&=\frac{t-vx}{\sqrt{1-v^2}}\\
x’&=\frac{x-vt}{\sqrt{1-v^2}}
\end{aligned}\label{eq:boost2}
\end{equation}
Remember that we assumed the speed of light to be $c=1$ for simplicity. For the real $c$ value \eqref{eq:boost2} turns into
\begin{equation}
\begin{aligned}
t’&=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}\\
x’&=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}
\end{aligned}\label{eq:boost3}
\end{equation}
In physics textbooks, \eqref{eq:boost3} is introduced as the Lorentz transformation. If $v\ll c$ meaning $v$ is significantly slower than the speed of light (such a motion is called a non-relativistic motion), \eqref{eq:boost3} is effectively the Galilean transformation
\begin{equation}
\begin{aligned}
t’&=t\\
x’&=x-vt
\end{aligned}\label{eq:galilean}
\end{equation}
for Newtonian mechanics in Euclidean space. $t’=t$ means that time is independent of the relative motion of different observers and we already know this is the case of Newtonian mechanics. The Galilean transformation \eqref{eq:galilean} can be also obtained by taking the limit $c\to \infty$. This indicates that in Newtonian mechanics the speed of light is presumed to be infinity.

This time we consider spacetime $\mathbb{R}^{3+1}$ with the Minkowski metric $ds^2=-(dt)^2+(dx)^2+(dy)^2+(dz)^2$. (Here we set the speed of light in vacuum $c=1$). For each time slice i.e $t$=constant we get Euclidean 3-space and from what we know about Euclidean space there are rotations in the coordinate planes, the $xy$-plane, $yz$-plane, and $xz$-plane. What we have no clue about is rotations (from Euclidean perspective) in the $tx$-plane, $ty$-plane, and $tz$-plane. For that we consider $\mathbb{R}^{1+1}$ with metric $ds^2=-(dt)^2+(dx)^2$. Let $v,w$ be two tangent vectors to $\mathbb{R}^{1+1}$. Then $v,w$ are written as
\begin{align*}
v&=v_1\frac{\partial}{\partial t}+v_2\frac{\partial}{\partial x}\\
w&=w_1\frac{\partial}{\partial t}+w_2\frac{\partial}{\partial x}
\end{align*}
$ds^2$ acting on them results the inner product:
$$\langle v,w\rangle=ds^2(v,w)=-v_1w_1+v_2w_2=v^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}w$$
The last expression is obtained by considering $v$ and $w$ as column vectors (the reason we can do this is because every tangent plane to $\mathbb{R}^{1+1}$ is isomorphic to the vector space $\mathbb{R}^{1+1}$). Let $A$ be an isometry of $\mathbb{R}^{1+1}$. Let $v’=Av$ and $w’=Aw$. Then
\begin{align*}
\langle v’,w’\rangle&={v’}^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}w’\\
&=(Av)^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}(Aw)\\
&=v^tA^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}Aw
\end{align*}
In order for $A$ to be an isometry i.e. $\langle v’,w’\rangle=\langle v,w\rangle$ we require that $A^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}$. A $2\times 2$ matrix $A$ satisfying
\begin{equation}
\label{eq:lo}
A^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
\end{equation}
is called an Lorentz orthogonal matrix. More generally a $4\times 4$ Lorentz orthogonal matrix $A$ satisfies
\begin{equation}
\label{eq:lo2}
A^t\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}
\end{equation}
Let $A=\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}$ be a Lorentz orthogonal matrix. We also assume that $\det A=1$ i.e. $A$ is a special Lorentz orthogonal group. Then we get the following set of equations
\begin{equation}
\begin{aligned}
-a^2+c^2&=-1\\
-ab+cd&=0\\
-b^2+d^2&=1\\
ad-bc&=1
\end{aligned}
\label{eq:slo}
\end{equation}
Solving the equations in \eqref{eq:slo} simultaneously we obtain $A=\begin{pmatrix}
a & b\\
b & a
\end{pmatrix}$ with $a^2-b^2=1$. Hence one may choose $a=\cosh\phi$ and $b=-\sinh\phi$. Now we find a rotation matrix in the $tx$-plane
\begin{equation}
\label{eq:boost}
\begin{pmatrix}
\cosh\phi & -\sinh\phi\\
-\sinh\phi & \cosh\phi
\end{pmatrix}
\end{equation}
A friend of mine, a retired physicist named Larry has to confirm everything by actually calculating. For being a lazy mathematician I try to avoid messy calculations as much as possible, instead try to confirm things indirectly (and more elegantly). In honor of my dear friend let us play Larry. Let $\begin{pmatrix} t\\
x \end{pmatrix}$ be a vector in the $tx$-plane and $\begin{pmatrix} t’\\
x’ \end{pmatrix}$ denote its rotation by an angle $\phi$. (This $\phi$ is not really an angle and it could take any real value. It is called a hyperbolic angle.)
$$\begin{pmatrix} t’\\
x’ \end{pmatrix}=\begin{pmatrix}
\cosh\phi & -\sinh\phi\\
-\sinh\phi & \cosh\phi
\end{pmatrix}\begin{pmatrix} t\\
x \end{pmatrix}$$
i.e.
\begin{align*}
t’&=\cosh\phi t-\sinh\phi x\\
x’&=-\sinh\phi t+\cosh\phi x
\end{align*}
and
\begin{align*}
dt’&=\cosh\phi dt-\sinh\phi dx\\
dx’&=-\sinh\phi dt+\cosh\phi dx
\end{align*}
Using this we can confirm that
$$(ds’)^2=-(dt’)^2+(dx’)^2=-(dt)^2+(dx)^2=ds^2$$
i.e. \eqref{eq:boost} is in fact an isometry which is called a Lorentz boost. In spacetime $\mathbb{R}^{3+1}$ there are three boosts, one of which is
$$\begin{pmatrix}
\cosh\phi & -\sinh\phi & 0 & 0\\
-\sinh\phi & \cosh \phi& 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
This is a rotation (boost) in the $tx$-plane. In addition, there are regular rotations one of which is
$$\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & \cos\theta & -\sin\theta & 0\\
0 & \sin\theta & \cos\theta & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
This is a rotation by the angle $\theta$ in the $xy$-plane. The three boosts and three rotations form generators of the Lorentz group $\mathrm{O}(3,1)$, the set of all $4\times 4$ Lorentz orthogonal matrices, which is a Lie group under matrix multiplication. As a topological space $\mathrm{O}(3,1)$ has four connected components depending on whether Lorentz transformations are time-direction preserving or orientation preserving. The component that contains the identity transformation consists of Lorentz transformations that preserve both the time-direction and orientation. It is denoted by $\mathrm{SO}^+(3,1)$ and called the restricted Lorentz group. There are 4 translations along the coordinate axes in $\mathbb{R}^{3+1}$. The three boosts, three rotations and four translations form generators of a Lie group called the Poincaré group. Like Euclidean motion group elements of the Poincaré group are affine transformations. Such an affine transformation would be given by $v\longmapsto Av+b$ where $A$ is a Lorentz transformation and $b$ is a fixed four-vector. The Lorentz group is not a (Lie) subgroup of the Poincaré group as the elements of the Poincaré group are not matrices. Note however that the Poincaré group acts on $\mathbb{R}^{3+1}$ in an obvious way and the Lorentz group fixes the origin under the group action, hence the Lorentz group is the stabilizer subgroup (also called the isotropy group) of the Poincaré group with respect to the origin.

I will finish this lecture with an interesting comment from geometry point of view. The spacetime $\mathbb{R}^{3+1}$ can be identified with the linear space $\mathscr{H}$ of all $2\times 2$ Hermitian matrices via the correspondence
\begin{align*}
v=(t,x,y,z)\longleftrightarrow\underline{v}&=\begin{pmatrix}
t+z & x+iy\\
x-iy & t-z
\end{pmatrix}\\
&=t\sigma_0+x\sigma_1+y\sigma_2+z\sigma_3
\end{align*}
where
$$\sigma_0=\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix},\ \sigma_1=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix},\ \sigma_2=\begin{pmatrix}
0 & i\\
-i & 0
\end{pmatrix},\ \sigma_3=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
are the Pauli spin matrices. The Lie group $\mathrm{SL}(2,\mathbb{C})$ acts on $\mathbb{R}^{3+1}$ isometrically via the group action:
$$\mu: \mathrm{SL}(2,\mathbb{C})\times\mathbb{R}^{3+1}\longrightarrow\mathbb{R}^{3+1};\ \mu(g,v)=gvg^\dagger$$
where $g^\dagger={\bar g}^t$. The action induces a double covering $\mathrm{SL}(2,\mathbb{C})\stackrel{2:1}{\stackrel{\longrightarrow}{\rho}}\mathrm{SO}^+(3,1)$. Since $\ker\rho=\{\pm I\}$, $\mathrm{PSL}(2,\mathbb{C})=\mathrm{SL}(2,\mathbb{C})/\{\pm I\}=\mathrm{SO}^+(3,1)$. $\mathrm{SL}(2,\mathbb{C})$ is simply-connected, so it is the universal covering of $\mathrm{SO}^+(3,1)$.

I will discuss some physical implications of Lorentz transformations next time.

When you study theory of relativity, one of the notions you will first stumble onto is about Lorentz transformations. It is actually a pretty big deal. Without knowing Lorentz transformations, you can’t study theory of relativity. So what are Lorentz transformations? A short answer is they are isometries of spacetime. If you haven’t heard of the word isometry, it is a linear isomorphism from an inner product space to another which also preserves inner product. Let $V$, $V’$ be inner product space. A linear map $\varphi: V\longrightarrow W$ is said to preserve inner product if $\langle v,w\rangle=\langle\varphi(v),\varphi(w)\rangle$ for all $v,w\in V$. Since the length between two vectors is measured by inner product, clearly inner product preserving map also preserves length, so we have the name isometry. An isometry is clearly conformal (angle-preserving). Before we discuss isometries of spacetime, which is usually denoted by $\mathbb{R}^{3+1}$, let us visit Euclidean space that we are most familiar with and consider isometries there. For a vector in Euclidean space there are three types of basic transformations: dilations, translations, and rotations. Dilations are linear but not isometries. Translations preserve the length of a vector but not linear (they are called affine transformations). Rotations are indeed isometries. Let us check that rotations are isometries by calculation. It suffices to consider 2-dimensional Euclidean space $\mathbb{R}^2$ with metric $ds^2=dx^2+dy^2$. This is not only because of simplicity but also because what we discuss for 2-dimensional case can be easily extended for higher dimensional cases such as 3-dimensional or 4-dimensional Euclidean space. The reason is simple. No matter in what dimensional space you are in a rotation takes place only in two dimensional space (plane). A rotation of a vector $\begin{pmatrix}
x\\
y
\end{pmatrix}$ by an angle $\theta$ is given by
$$\begin{pmatrix}
x’\\
y’
\end{pmatrix}=\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}$$
that is,
\begin{align*}
x’&=\cos\theta x-\sin\theta y\\
y’&=\sin\theta x+\cos\theta y
\end{align*}
and
\begin{align*}
dx’&=\frac{\partial x’}{\partial x}dx+\frac{\partial x’}{\partial y}dy\\
&=\cos\theta dx-\sin\theta dy\\
dy’&=\frac{\partial y’}{\partial x}dx+\frac{\partial y’}{\partial y}dy\\
&=\sin\theta dx+\cos\theta dy
\end{align*}
With these you can easily check that
$$(ds’)^2=(dx’)^2+(dy’)^2=dx^2+dy^2=ds^2$$
i.e. the metric is preserved so rotations are isometries. Here I kind of cheated because I already know so well (as you would do also) about Euclidean space! Assume that the only thing we know about Euclidean space is its metric. Besides that we know nothing about its geometry whatsoever. How do we figure out things like rotations? This is important because once we can figure it out, we can apply the same method to figure out isometries of other spaces that we don’t know about. Let $v$ and $w$ be two tangent vectors to $\mathbb{R}^2$ (here $\mathbb{R}^2$ is regarded as a 2-dimensional differentiable manifold). Then they can be written as
\begin{align*}
v&=v_1\frac{\partial}{\partial x}+v_2\frac{\partial}{\partial y}\\
w&=w_1\frac{\partial}{\partial x}+w_2\frac{\partial}{\partial y}
\end{align*} We calculate
$$ds^2(v,w)=v_1w_1+v_2w_2=v^tw=v^t\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}w,$$
where the tangent vectors are identified with column vectors $v=\begin{pmatrix}
v_1\\
v_2
\end{pmatrix}$ and $w=\begin{pmatrix}
w_1\\
w_2
\end{pmatrix}$. (Recall that any
tangent plane $T_p\mathbb{R}^2$ is isomorphic to the vector space $\mathbb{R}^2$.) So, the Euclidean metric induces the usual dot product $\langle\ ,\ \rangle$. Let $A$ be a linear transformation from $\mathbb{R}^2$ to itself. Let $v’=Av$ and $w’=Aw$. Then
\begin{align*}
\langle v’,w’\rangle&={v’}^tw’\\
&=(Av)^t(Aw)\\
&=v^t(A^tA)w.
\end{align*}
Suppose that $A$ is also an isometry. In order for $A$ to be an isometry i.e. $\langle v’,w’\rangle=\langle v,w\rangle$ we require that $A^tA=I$. As you may have learned from linear algebra such a matrix is called an orthogonal matrix. Let $A=\begin{pmatrix}
a & b\\
c & d \end{pmatrix}$. In addition to assuming that $A$ is orthogonal let us also assume that $\det A=1$. Such an orthogonal matrix is called a special orthogonal matrix. Then we obtain the equations
\begin{equation}\begin{aligned}
a^2+b^2&=1\\
ab+cd&=0\\
b^2+d^2&=1\\ad-bc&=1
\end{aligned}\label{eq:so}
\end{equation}
Solving equations in \eqref{eq:so} simultaneously we get
$A=\begin{pmatrix}
a & -c\\
c & a
\end{pmatrix}$ with $a^2+c^2=1$. Hence we may choose $a$ and $c$ as $a=\cos\theta$ and $c=\sin\theta$. That is we obtained a rotational matrix only from the assumption that $A$ is an isometry without employing any familiar geometric intuition on Euclidean space. The set of all isometries of $\mathbb{R}^n$ is denoted by $\mathrm{O}(n)$. $\mathrm{O}(n)$ is the set of all $n\times n$ orthogonal matrices and it is a group with matrix multiplication. It is in fact more than an algebraic group. It is also a Lie group. Simply speaking a Lie group is a group which is also a differentiable manifold. Furthermore, the binary operation is a smooth map. The set of all isometries of $\mathbb{R}^n$ along with translations form a group with composition called the Euclidean motion group. The name obviously comes from Euclidean motions that comprise successive applications of rotations and translations. A Euclidean motion of a vector $v$ is given by $Av+b$ where $A$ is an orthogonal matrix and $b$ is a fixed vector i.e. it is an affine transformation.

Now we are ready to discuss Lorentz transformations i.e. isometries of spacetime $\mathbb{R}^{3+1}$ and we will continue this next time.