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.