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Vector space vs affine space in Lagrangian mechanics

Physics Asked by Gideon Ilung on October 1, 2021

I want to know why is it preferred that the classical universe is seen as an affine space instead of a vector space?

From my understanding an affine space violates the zero vector in the vector space.

Because of this this is it same to assume

Let $a,bin A$ Where $A$ is an affine space

$ca in A$ is false

$a + b in A$ is false

If those are true then what makes it useful defining the classical universe as an affine space?

3 Answers

To acknowledge the fact that no point in space is a priori distinguished. In many physical models space is not even an affine space, just a manifold.

Answered by Qmechanic on October 1, 2021

First I'd like to clarify what an affine space actually is. It's a set of points $A$, together with a vector space $V$ which contains all the translations between points of $A$, and a translation map $+:Atimes Vto A,~(P,v)mapsto P+v$. This translation has some special properties, like $P+0=P$, $(P+v)+w=P+(v+w)$, etc., which codify our understanding what a translation is. That said, you can't actually do the operations you did. You can't add points, and you can't scale points. You can add vectors to points, and you can scale vectors.

Now to why affine space is a preferred description of physical space. Physical space has no special points. All points are equal! The only thing that matters is how the points relate to one another. Distances and angles between points are what matter, and you can translate or rotate (with a completely arbitrary axis of rotation, it doesn't need to go through some origin) the entire space however you want the physics don't change. A vector space can't do that for you. Its features break when you just translate everything, or rotate along an arbitrary axis. Vector spaces play nicely with linear operations, like rotations around an axis through the origin but they don't play nicely with affine transformations (translations plus something linear). But physical space does play nicely with affine transformations. And the mathematical object which plays nicely with affine transformations is an affine space, because you're not forced to preserve the origin in order to preserve its other features.

That said, you can turn any vector space into an affine space by using its vectors both as points and vectors. We routinely do so by specifying a coordinate system in physical space. We specify points and vectors both by giving their three coordinates, so they're both elements of $mathbb R^3$. But we never add points to points, and we never scale points, so it's still conceptually useful to make a distinction between points on the one hand and vectors pointing from one point to another on the other hand.

Answered by Vercassivelaunos on October 1, 2021

You have obtained your understanding from misleading treatments, which do not properly respect the mathematical definitions, and which confuse the zero vector with the origin of coordinates in a Euclidean vector space (and vectors with coordinates generally). When physicists talk of vectors having magnitude and direction, they are really talking of Euclidean vectors. When they "define" affine space by "forgetting where the origin is", they are talking of coordinates; they do not lose the zero vector. They are simply defining affine space as a Euclidean vector space in which the origin is arbitrary.

It is much easier to understand this mathematically. Mathematicicans define a vector space, $V$, of vectors, over a field, $F$, of scalars, as an Abelian group under addition, such that vectors can be multiplied by scalars and, if $x, y, z$ are vectors and $a, b$ are scalars, then three axioms hold:

  1. closure: $ax$ is a vector;

  2. compatibility: $(ab)x = a(bx)$;

  3. distributivity: $a(x + y) = ax + ay$ and $(a + b)x = ax + bx$

This is much more general than the common physics definition, and is important in mathematics generally (e.g. complex numbers and functions are understood as vectors), and in other fields like IT.

Vector space (on its own) does not include certain notions commonly attributed to vectors, like magnitude and direction. For these concepts we require an inner product. The most familiar example of an inner product is the dot product. Also, vector space is quite abstract. It does not contain any notion of position in space. Hilbert space used in quantum mechanics is a important example of an inner product space used in physics.

An affine space $mathscr A$ is a space of points, together with a vector space $V$ such that for any two points $A$ and $B$ in $mathscr A$ there is a vector $vec{AB}$ in $V$ where:

  1. for any point $A$ and any vector $v$ there is a unique point $B$ with $vec{AB}=v$

  2. for any points $A, B, C, vec{AB} + vec{BC} = vec{AC}$

(these are Weyl’s axioms. Other axiom sets can be used to define the same (isomorphically identical) structure).

We can construct an affine space from a vector space. First use vector space to define a coordinate space in one-one correspondence with vector space (this is the set of points, given by coordinates with values equal to vector components in a given basis). The origin of coordinates corresponds to the zero vector (but should not be confused with it). Now "forget where the origin is" by adding translations on coordinate space. Then any point can be chosen as an origin.

A Euclidean space is a finite-dimensional affine space over an inner product space. The affine property together with the inner product enable us to define vectors with magnitude and direction (angle) at any position in $mathscr A$.

I hope this resolves your confusion makes clear that Euclidean space (as used in classical physics) is an affine space.

Answered by Charles Francis on October 1, 2021

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