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{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.

Advertisements

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.