New post for affine spaces

posts/mechanics_affine_spaces/index.qmd

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+---
+title: "Mechanics"
+description: "Introduction to affine spaces."
+author: "Georgios Douzas"
+date: "2024-08-22"
+categories: [Physics, Mechanics]
+image: "featured.png"
+jupyter: python3
+---
+
+![](featured.png)
+
+# Introduction
+
+It is common to define the physical space as $\mathbb{R}^3$. This is incorrect both mathematically and physically. The reason is
+that the structure of $\mathbb{R}^3$ is not invariant under displacements and under other physically important transformations
+called isometries, in contrast to the nature of Euclidean geometry’s results. The three-dimensional space and the temporal axis of
+Classical Physics (but also four-dimensional spacetime in Special Relativity) are first of all affine spaces.
+
+# Definition of an affine space
+
+A real affine space of finite dimension $n$ is a set $\mathbb{A}^n$ whose elements are called points equipped with the following
+structures:
+
+- A $n$-dimensional vector space $V$ called space of displacements or space of free vectors.
+- A map $\mathbb{A}^n \times \mathbb{A}^n \ni(P, Q) \mapsto P-Q \in V$ with the following two properties:
+    - For any pair $Q \in \mathbb{A}^n, \mathbf{v} \in V$ there is a unique $P \in \mathbb{A}^n$ such that $P-Q=\mathbf{v}$.
+    - For any triplet $P, Q, R \in \mathbb{A}^n$, the identity $P-Q+Q-R=P-R$ holds.
+
+If $Q \in \mathbb{A}^n$ and $\mathbf{v} \in V$ then the unique point $P$ in $\mathbb{A}^n$ such that $P-Q=\mathbf{v}$ is denoted
+as $Q+\mathbf{v}$.
+
+# Lines and line segments
+
+We define a line and a line segment in $\mathbb{A}^n$ as follows:
+
+- A line in $\mathbb{A}^n$ with origin $P$ and tangent vector $\mathbf{v} \in V$ is the map $\mathbb{R} \ni t \mapsto P+t
+\mathbf{u} \in \mathbb{A}^n$.
+- A line segment in $\mathbb{A}^n$ is a restriction of the above map to some interval.
+
+# Various properties
+
+The following properties hold for any $P, Q \in \mathbb{A}^n$ and $\mathbf{u}, \mathbf{v} \in V$:
+
+- $P-P=\mathbf{0}$.
+- $(Q+\mathbf{u})+\mathbf{v}=Q+(\mathbf{u}+\mathbf{v})$.
+- $P-Q=-(Q-P)$.
+- $P-Q=(P+\mathbf{u})-(Q+\mathbf{u})$.
+
+> Prove the above properties.
+
+# Coordinate systems
+
+A local coordinate system on $\mathbb{A}^n$ is a map $\psi: U \subset \mathbb{A}^n \rightarrow \mathbb{R}^n$ with the following properties:
+
+- $\psi$ is injective.
+- $\psi(U) \subset \mathbb{R}^n$ is an open set.
+
+The coordinate system is global if $U=\mathbb{A}^n$.
+
+Every affine space admits a family of global coordinate systems called Cartesian coordinate systems. A Cartesian coordinate system
+with origin $O$ and axes $\mathbf{e}_1, \ldots, \mathbf{e}_n$ is defined as follows:
+
+- Fix a point $O$ and a basis $\mathbf{e}_1, \ldots, \mathbf{e}_n$ of $V$.
+- Define a map $f: \mathbb{A}^n \rightarrow \mathbb{R}^n$ with $f(P) = \left((P-O)^1, \ldots,(P-O)^n\right)$
+
+> Prove the above map is bijective and therefore $f$ is a global coordinate system.
+
+Non-Cartesian local coordinate systems are called curvilinear coordinate systems.
+
+# Properties of cartesian coordinate systems
+
+Let $\left(\mathbb{A}^n, f\right)$ be a Cartesian coordinate system with coordinates $x^1, \cdots, x^n$
+and $\left(\mathbb{A}^n, g\right)$ another Cartesian coordinate system with coordinates $x^{\prime 1}, \cdots, x^{\prime n}$,
+origin $O^{\prime}$ and axes $\mathbf{e}^{\prime}{ }_1, \ldots, \mathbf{e}_n^{\prime}$, so that
+
+$$
+\mathbf{e}_i=\sum_j B^j{ }_i \mathbf{e}^{\prime}{ }_j
+$$
+
+Let $\left(O-O^{\prime}\right)=\sum_i b^i \mathbf{e}_i$ then the following properties hold:
+
+- The map $g \circ f^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is expressed in coordinates as $x^{\prime j}=\sum_{i=1}^n B^j{
+}_i\left(x^i+b^i\right)$.
+
+-  The map $f \circ g^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is expressed in coordinates as
+$x^i=-b^i+\sum_{j=1}^n\left(B^{-1}\right)^i{ }_j x^{\prime j}$.
+
+> Prove the above properties.
+
+# Affine transformations
+
+A map $\psi: \mathbb{A}_1^n \rightarrow \mathbb{A}_2^m$ between affine spaces $\mathbb{A}_1^n$ and $\mathbb{A}_2^n$ with spaces of
+displacements $V_1$ and $V_2$ is called an affine transformation and has the following properties:
+
+- $\psi$ is invariant under displacements i.e. $\psi(P+\mathbf{u})-\psi(Q+\mathbf{u})=\psi(P)-\psi(Q)$ for any $P, Q \in
+\mathbb{A}_1^n$ and $\mathbf{u} \in V_1$.
+- The map $d \psi: V_1 \rightarrow V_2$ with $d \psi(P - Q) = \psi(P) - \psi(Q)$ is a linear between the vector spaces $V_1$ and
+$V_2$.
+
+> Prove the map is well defined and linear.
+
+An affine transformation is called an isomorphism of affine spaces if it is bijective. Note that the inverse of an isomorphism of
+affine spaces is an affine map and hence an isomorphism of affine spaces. If $\psi: \mathbb{A}_1^n \rightarrow
+\mathbb{A}_2^n$ is an isomorphism of affine spaces then $d \psi: V_1 \rightarrow V_2$ is a vector space isomorphism.
+
+> Prove the above properties.
+
+Affine transformations map straight lines to straight lines: If $\psi: \mathbb{A}_1^n \rightarrow \mathbb{A}_2^n$ is affine and
+$P(t):=P+t \mathbf{u}$, with $t \in \mathbb{R}$, is the line in $\mathbb{A}_1^n$ with origin $P$ and tangent vector $\mathbf{v}
+\in V_1$, then $\psi(P(t))$, as $t \in \mathbb{R}$ varies, defines a line in $\mathbb{A}^m$.
+
+> Prove the above properties.
+
+# Representation of affine transformations
+
+Let $\psi: \mathbb{A}_1^n \rightarrow \mathbb{A}_2^m$ an affine transformation. Then, for any choice of Cartesian coordinate
+systems in the two spaces $\left(\mathbb{A}_1^n, f_1\right)$ and $\left(\mathbb{A}_2^m, f_2\right)$, the representation of $\psi$
+in coordinates, i.e. the map $f_2 \circ \psi \circ f_1^{-1}: \mathbb{R}^n \ni\left(x_1^1, \ldots, x_1^n\right) \mapsto\left(x_2^1,
+\ldots, x_2^m\right) \in \mathbb{R}^m$ has the form
+
+$$x_2^i=c^i+\sum_{j=1}^n L^i{ }_j x_1^j \quad(i=1,2, \ldots, m)$$ 
+
+> Prove the above property.
+
+for suitable coefficients $L^i{ }_j$ and $c^i$ depending on $\psi$ and on the coordinate systems. Above, $\left(x_2^1, \ldots,
+x_2^m\right)$ are the Cartesian coordinates of $\psi(P) \in \mathbb{A}_2^m$ and $\left(x_1^1, \ldots, x_1^n\right)$ those of $P
+\in \mathbb{A}_1^n$.
+
+Conversely, $\psi: \mathbb{A}_1^n \rightarrow \mathbb{A}_2^m$ is affine if there exist Cartesian coordinate systems on the two
+spaces in which $\psi$ has the above form in coordinates.
+
+> Prove the above property.
+
+# Group of displacements
+
+The mapping $T_{\mathbf{v}}: \mathbb{A}^n \rightarrow \mathbb{A}^n$ of $\mathbf{v} \in V$ on $\mathbb{A}^n$ is defined as
+$T_{\mathbf{v}}: P \mapsto P+\mathbf{v}$.
+
+The set $\left\{T_{\mathbf{v}}\right\}_{\mathbf{v} \in V}$ is a group under composition, called the group of
+displacements of $\mathbb{A}^n$. As $T_{\mathbf{u}} T_{\mathbf{v}}=T_{\mathbf{u}+\mathbf{v}}$ and
+$\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$, the group is Abelian.
+
+> Prove the above properties.
+
+The following properties hold:
+
+- The map $V \ni \mathbf{v} \mapsto T_{\mathbf{v}}$ is injective. It is a group isomorphism when we view $V$ as Abelian group
+under addition.
+- Only $\mathbf{v}=\mathbf{0}$ satisfies $T_{\mathbf{v}}(P)=P$ for some $P \in \mathbb{A}^n$, i.e. the action of the
+group of displacements is free. We also have the stronger condition $T_0(P)=P$ for any $P \in \mathbb{A}^n$.
+- For any pair $P, Q \in \mathbb{A}^n$ there is a displacement $T_{\mathbf{v}}$ such that $T_{\mathbf{v}}(P)=Q$ i.e.
+the group of displacements acts transitively.
+
+> Prove the above properties.
+
+# Group action
+
+Given a set $S$, and a group $G$ with neutral element $e$ and product $\circ$, an action of $G$ on $S$ is a map $A: G \times S \ni(g, s) \mapsto$ $A_g(s) \in S$, where $A_g \in \mathcal{G}_S$ and the latter is the group of bijections on $S$ under composition, such that:
+
+- $A_e=i d$,
+- $A_g A_{g^{\prime}}=A_{g \circ g^{\prime}}$ where $g, g^{\prime} \in G$.
+
+We have the following definitions:
+
+- The action is called free if $A_g(s)=s$ for some $s \in S$ implies $g=e$.
+- The action is called transitive if for any $s, s^{\prime} \in S$ there exists $g \in G$ such that $A_g(s)=s^{\prime}$.
+- The action is called faithful if $G \ni g \mapsto A_g \in \mathcal{G}_S$ is injective. 
+
+An action always determines a group homomorphism $G \rightarrow \mathcal{G}_S$, and therefore $G_S:=\left\{A_g\right\}_{g \in G}$
+is a subgroup of $\mathcal{G}_S$ and $A$ is faithful if and only if defines a group isomorphism $G \rightarrow G_S$.
+
+> Prove the above properties.