special relativity - Are Lorentz transformations linear transformations?

My textbook says that Lorentz transformations are linear transformations and present them as matrices. Lorentz transformations relate different coordinate systems with each other. It seems that coordinate systems are linear spaces, but coordinates are just labels for spacetime points, with no other structures attached. So what is the linearity all about?

Answer

Minkowski space is a real affine space of dimension $4$ whose space of translations is equipped with a metric of Lorentzian type.

A (real) affine space is a triple $(\mathbb A, V, \vec{})$, where $\mathbb A$ is a set whose elements are said points, $V$ is a (real) vector space and $\vec{}$ is a map $\vec{} : \mathbb A \times \mathbb A \to V$ with the following properties,

$$\forall q \in \mathbb A\:, \forall v \in V\:, \exists \mbox{ and is unique } p \in \mathbb A\quad \mbox{such that}\quad \vec{qp} = v\:,\tag{1}$$

$$\vec{pq} + \vec{qr} = \vec{pr}\quad \forall p, q,r \in \mathbb A\:.\tag{2}$$

by definition, the dimension of the affine space is that of $V$, whose elements are said translations.

From now on, if $p,q\in \mathbb A$ a $v \in V$,

$$p= q+ v$$

means

$$\vec{qp}=v\:.$$

Form (1) this notation is well posed. $q+v$ is the action of the translation $v$ on the point $q$. This action is transitive and free, its existence physically corresponds to homogeneity of both space and time in special relativity.

Assuming that $V$ is finite dimensional, if one fixes $o \in \mathbb A$ and a basis $e_1,\ldots, e_n \in V$, a Cartesian coordinate system on the affine space $\mathbb A$ with origin $o$ and axes $e_1,\ldots, e_n$ is the bijective map

$$\mathbb R^n \ni (x^1,\ldots, x^n) \mapsto o + \sum_{j=1}^n x^je_j \in \mathbb A$$ Once again, using (1) one sees that, in fact, the map above is bijectve and thus identifies $\mathbb A$ with $\mathbb R^n$.

Changing $o$ to $o'$ and the basis $e_1,\ldots, e_n$ to the basis $e'_1,\ldots, e_n'$, one obtains a different Cartesian coordinate system $x'^1,\ldots, x'^n$. It is simply proved that the rule to pass form the latter coordinate system to the former has the form

$$x'^a = c^a+ \sum_{j=1}^n {A^a}_j x^j \tag{3}$$ for $n$ constant coefficients $c^j$ and a nonsingular $n\times n$ matrix of coefficients ${A^a}_j$.

The said matrix verifies

$$e_k = \sum_{i=1}^n {A^i}_k e'_i\tag{3'}$$

whereas the coefficients $c^k$ are the components of the vector $\vec{oo'}$.

(As a matter of fact the affine structure gives rise to a natural differentiable real analytic structure on $\mathbb A$ of dimension $n$.)

A real affine space equipped with a (pseudo)scalar product in $V$ is called (pseudo)Euclidean space.

Minkowski spacetime $\mathbb M^4$ is a (real) four dimensional affine space equipped with a pseudo scalar product $g: V\times V \to \mathbb R$ of Lorentzian type.

"Of Lorentzian type" means that there exist bases, $e_0,e_1,e_2,e_3$, in $V$ such that (I adopt here the convention $-+++$)

$$g(e_0,e_0)=-1 \:,\quad g(e_i,e_i) = 1 \mbox{ if $i=1,2,3$}\:,\quad g(e_i,e_j) =0 \mbox{ if $i\neq j$.}\tag{4}$$

These bases are called Minkowskian bases. Lorentz group $O(1,3)$ is nothing but the group of matrices $\Lambda$ connecting pairs of Minkowskian bases. It is therefore defined by

$$O(1,3) := \left\{ \lambda \in M(4,\mathbb R) \:|\: \Lambda \eta \Lambda^t = \eta \right\}$$

where $\eta = diag(-1,1,1,1)$ is the matrix representing the metric $g$ in (4) in every Minkowskian basis.

A Minkowskian coordinate system on $\mathbb M^4$ is a Cartesian coordinate system whose axes are a Minkowskian basis.

Lorentz transformations are transformations of coordinates between pairs of Minkowskian coordinate systems with the same origin (so that $c^k=0$ in (3)). Thus they have the form

$$x'^a = \sum_{j=1}^n {\Lambda^a}_j x^j$$

for some $\Lambda \in O(1,3)$. If we admit different origins we obtain the so-called Poincaré transformations

$$x'^a = c^a+ \sum_{j=1}^n {\Lambda^a}_j x^j \:.$$

When viewing Lorentz transformations as transformation of coordinates, their formal linearity does not play a relevant physical role, since it only reflects the arbitrary initial choice of the same origin for both reference frames. However, these transformations are also transformations of bases (3') in the space of translations (the tangent space), in this case linearity is natural because it reflects the natural linear space structure of the translations.