# What are Lorentz Transformations? 2

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

\label{eq:lo}
A^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}

is called an Lorentz orthogonal matrix. More generally a $4\times 4$ Lorentz orthogonal matrix $A$ satisfies

\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}

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{aligned}
-a^2+c^2&=-1\\
-ab+cd&=0\\
-b^2+d^2&=1\\
\end{aligned}
\label{eq:slo}

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

\label{eq:boost}
\begin{pmatrix}
\cosh\phi & -\sinh\phi\\
-\sinh\phi & \cosh\phi
\end{pmatrix}

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
$$(ds’)^2=-(dt’)^2+(dx’)^2=-(dt)^2+(dx)^2$$
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.

# What are Lorentz Transformations? 1

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{aligned}
a^2+b^2&=1\\
ab+cd&=0\\
\end{aligned}\label{eq:so}

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.

# E-mail issue

I just noticed that I am not getting any e-mail notifications and I am sure neither are you. I ran diagnostics on my mail server and it is working fine without any problem. So I concluded that certain ports (like port 25 and port 587) that are required by mail server are blocked. I opened those ports on my router and the issue still persists. I believe that my ISP is actually blocking those ports in which case it does not matter whether I open them on my router. I am currently working on a workaround such as using an alternative SMTP (like Gmail SMTP) instead of my mail server. I will update you if it works (and I surely hope it does). I am sorry for the inconvenience. In the meantime, if you need an assistance that requires an e-mail notification such as changing your password, just e-mail me your request.

Update: The workaround was successful and the e-mail issue has been resolved.

# Determinants as Area and Volume

The Area of a Parallelogram

Let $v=(v_1,v_2)$ and $w=(w_1,w_2)$ be two linearly independent vectors in $\mathbb{R}^2$. Then they span a parallelogram as shown in Figure 1.

Figure 1. Parallelogram spanned by two vector v and w.

The area $A$ of the parallelogram is
\begin{align*}
A&=||v||||w||\sin\theta\\
&=||v\times w||\\
&=\left|\begin{array}{cc}
v_1 & v_2\\
w_1 & w_2
\end{array}\right|.
\end{align*}
In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the area.

Exercise. Given two vectors $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ in $\mathbb{R}^3$, show that $||v||||w||\sin\theta=||v\times w||$ where $\theta$ is the angle between $v$ and $w$.

Hint. Note that $||v||^2||w||^2\sin^2\theta=||v||^2||w||^2-(v\cdot w)^2$.

The Volume of a Parallelepiped

Let $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$, $w=(w_1,w_2,w_3)$ be three linearly independent vectors in $\mathbb{R}^3$. Then they span a parallelepiped as shown in Figure 2.

Figure 2. Parallelepiped spanned by vectors u, v and w.

The volume $V$ of the parallelepiped is
\begin{align*}
V&=||u||\cos\theta ||v\times w||\\
&=u\cdot (v\times w)\\
&=\left|\begin{array}{ccc}
u_1 & u_2 & u_3\\
v_1 & v_2 & v_3\\
w_1 & w_2 & w_3
\end{array}\right|.
\end{align*}
In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the volume.

# Inverse of a Matrix

Let $A$ be an $n\times n$ matrix with $\det A\ne 0$. (A square matrix whose determinant is not equal to $0$ is called non-singular.) Let $X=(x_{ij})$ be an unknown $n\times n$ matrix such that
$AX=I$. Then
$$x_{1j}A^1+\cdots+x_{nj}A^n=E^j.$$
This is a system of linear equations and as we studied here, it can be solved by Cramer’s Rule as
\begin{align*}
x_{ij}&=\frac{\det(A^1,\cdots,E^j,\cdots A^n)}{\det A}\\
&=\frac{1}{\det A}\left|\begin{array}{ccccc}
a_{11} & \cdots & 0 & \cdots & a_{1n}\\
\vdots & & \vdots & & \vdots\\
a_{j1} & \cdots & 1 & \cdots & a_{jn}\\
\vdots & & \vdots & & \vdots\\
a_{n1} & \cdots & 0 & \cdots & a_{nn}
\end{array}\right|\\
&=\frac{1}{\det A}(-1)^{i+j}\det(A_{ji})
\end{align*}
for $i=1,\cdots,n$. If we show that $XA=I$ as well, then $X$ would be the inverse $A^{-1}$ and
$$A^{-1}=\frac{1}{\det A}{}^t((-1)^{i+j}\det(A_{ij})).$$
$\det({}^tA)=\det A\ne 0$, so we can find an $n\times n$ matrix $Y$ such that ${}^tAY=I$. Taking transposes, we obtain ${}^tYA=I$. Now,
$$I={}^tYA={}^tYIA={}^tY(AX)A={}^tYA(XA)=XA.$$

Example. Let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ with $\det A=ad-bc\ne 0$. Then $A^{-1}$ is given by
$$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}.$$

Example. Find the inverse of the matrix
$$A=\begin{pmatrix} 3 & 1 & -2\\ -1 & 1 & 2\\ 1 & -2 & 1 \end{pmatrix}.$$

Solution. $\det A=16$. We find
\begin{align*}
A_{11}=\begin{pmatrix}
1 & 2\\
-2 & 1
\end{pmatrix},\ A_{12}=\begin{pmatrix}
-1 & 2\\
1 & 1
\end{pmatrix},\ A_{13}=\begin{pmatrix}
-1 & 1\\
1 & -2
\end{pmatrix},\\
A_{21}=\begin{pmatrix}
1 & -2\\
-2 & 1
\end{pmatrix},\ A_{22}=\begin{pmatrix}
3 & -2\\
1 & 1
\end{pmatrix},\ A_{23}=\begin{pmatrix}
3 & 1\\
1 & -2
\end{pmatrix},\\
A_{31}=\begin{pmatrix}
1 & -2\\
1 & 2
\end{pmatrix},\ A_{32}=\begin{pmatrix}
3 & -2\\
-1 & 2
\end{pmatrix},\ A_{33}=\begin{pmatrix}
3 & 1\\
-1 & 1
\end{pmatrix}.
\end{align*}
Hence by the formula we obtain
\begin{align*}
A^{-1}&=\frac{1}{16}{}^t\begin{pmatrix}
\det(A_{11}) & -\det(A_{12}) & \det(A_{13})\\
-\det(A_{21}) & \det(A_{22}) & -\det(A_{23})\\
\det(A_{31}) &-\det(A_{32}) & \det(A_{33})
\end{pmatrix}\\
&=\frac{1}{16}{}^t\begin{pmatrix}
5 & 3 & 1\\
3 & 5 & 7\\
4 & -4 & 4
\end{pmatrix}\\
&=\frac{1}{16}\begin{pmatrix}
5 & 3 & 4\\
3 & 5 &-4\\
1 & 7 & 4
\end{pmatrix}.
\end{align*}

We introduce the following theorem without proof:

Theorem. For any two $n\times n$ matrices $A$, $B$,
$$\det(AB)=\det A\det B.$$

As a special case, we obtain:

Corollary. For an invertible matrix $A$,
$$\det(A^{-1})=\frac{1}{\det A}.$$

Proof. $AA^{-1}=I$, so by the theorem
$$1=\det(AA^{-1})=\det A\det (A^{-1}).$$
Thus proving the formula for the inverse.