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I have just begun reading Gregory's Classical Mechanics and, amazingly, he has blown my mind in the first chapter discussing nothing more than measly old vector algebra. Fascinating that Gregory was able to make basic vector algebra...fascinating.

Nevertheless, I find myself confused and want to understand, in particular, what seems to be an isomorphism between "abstract vector quantities". I will try to make this question as self-contained as possible but it will be difficult to encapsulate 10 pages (the first ten pages) of Gregory here for context. In the end, much of this question has to do with the "correspondence rules" between the physical world and the mathematical structures intended to represent them (I use the language used by Ballentine at the start of Chapter 2 in his Quantum Mechanics: A Modern Development, but I think of correspondence rule as meaning "isomorphism between the physical world and the given mathematical structure").

Gregory begins with this:

Definition 1.1 Vector quantity If a quantity $Q$ has a magnitude and a direction associated with it, then $Q$ is said to be a vector quantity. [Here, magnitude means a positive real number and direction is specified relative to some underlying reference frame that we regard as fixed.]

Now reference frame is not defined yet, though he says he will do so later. I am reminded of what Professor Steane says at the beginning of his text on relativity (Relativity Made Relatively Easy) -- that a reference frame should be thought of as a real rigid body with some origin and set of clocks.

He continues:

In order to manipulate all such quantities without regard to their physical origin, we introduce the concept of a vector as an abstract quantity. Thus, Definition 1.2 Vector A vector is an abstract quantity characterised by the two properties magnitude and direction. Thus two vectors are equal if they have the same magnitude and the same direction.

If I am here understanding correctly, if we consider the set of all possible vector quantities of this "type" (e.g. force), then for this fixed type Gregory begins referring to them as "abstract vectors". Fair enough. I will come back to this later, but one of my key questions is whether we have yet made any identification with $\mathbb{R}^3$, or not.

Next, Gregory says

It is convenient to define operations involving abstract vectors by reference to some simple, easily visualised vector quantity. The standard choice is the set of directed line segments. Each straight line joining two points ($P$ and $Q$ say, in that order) is a vector quantity, where the magnitude is the distance $PQ$ and the direction is the direction of $Q$ relative to $P$. We call this the line segment $\overline{PQ}$ and we say that it represents some abstract vector $\mathbf{a}$. Note that each vector $\mathbf{a}$ is represented by infinitely many different line segments.

I here have my first question. Call $\mathcal{D}$ the set of directed line segments and $\mathcal{A}$ the set of abstract vectors of the given type.

(1) Is Gregory defining an isomorphism between the equivalence class of all directed line segments in $\mathcal{D}$ with the same magnitude (given a choice of units for each set) and direction (and differing only in where they are in space) as some given element $\mathbf{a} \in \mathcal{A}$? If so, what is the rule for this isomorphism? Are we, as I alluded to, assigning units to each set $\mathcal{A}$ and $\mathcal{D}$ and then identifying things that have the same magnitude and direction?

Lastly, Gregory says the following:

Suppose that $O$ is a fixed point of space. Then relative to the origin O (and relative to the underlying reference frame), any point of space, such as $A$, has an associated line segment, $\overline{PQ}$, which represents some vector $\mathbf{a}$. Conversely, the vector $\mathbf{a}$ is sufficient to specify the position of the point $A$.

Here are my last questions:

(2) Here Gregory seems to be saying that when we choose an origin for our reference frame we define an isomorphism between the abstract vectors and special elements of the aforementioned equivalence classes (namely those directed segments which emanate from the chosen origin). Is this understanding correct? Gregory has specialized to "vector quantities" of the "position" type, but surely this should be general for any "vector quantity type"?

(3) Perhaps this has not been discussed yet by Gregory (but I'm not sure it will be). As of yet, when we pinned down an origin we did not specify an "orientation" of our reference frame. How are different reference frames related in the picture which has just been developed? Do all references frames with the same origin agree on the isomorphism mentioned in (2)? Now, crucially and if what I've just said is true, is the difference between different reference frames with the same origin that they disagree on the further isomorphism between the two sets in (2) and $\mathbb{R}^3$?

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I do not know Gregory's textbook. From your citations, there is an attempt to clarify the relationship between what are usually called free and bound vectors. To better understand their difference, I suggest you read an introduction to them like this Wikipedia page, although, in my view, more clarity would be possible.

Here, I can try to sketch the main conceptual points and answer your direct questions. However, a very terse introduction to the subject I know is the chapter on affine Euclidean vectors in Lichnerowicz's Elements of Tensor Calculus (a pdf can be found online).

First, although very widespread and present in the cited Wikipedia page, it is not entirely correct to define a vector as a quantity "having a magnitude and a direction." My favorite counterexample is rotations in 3D. It is possible to assign a magnitude (the value of the rotation angle) and a direction (that of the rotation axis) to each rotation. Still, rotations in 3D cannot be vectors because their composition is not commutative ( rotating a book by 90 degrees around that $x$ axis and then around the $y$ axis does not end up with the same configuration obtained by changing the order of the two rotations and said in another way, magnitude and direction are not enough. One needs to add the composition rules specifying the properties of the sum (I assume that the multiplication by a scalar is not problematic). After such an addition to the definition, verifying that the oriented line segments with a common tail are elements of a vector space is trivial.

An oriented segment can be specified by giving in a conventional order two points. For example, we can define $\vec{AB}$ as the oriented segment starting from point $A$ of an Euclidean space and ending at point $B$.

The set of oriented segments in the plane or the space is not a vector space. However, we can consider an isomorphism between the set of all the oriented segments and a set whose elements are an ordered pair made by a point and a vector. I.e., we associate to each oriented segment $\vec{AB}$ the ordered pair $(A, {\bf b})$, where ${\bf b}$ is a vector representing the displacement of point $B$ from point $A$. This, in essence, is the isomorphism Gregory is speaking about, and that can be described as the isomorphism between the equivalence class of all directed line segments in the space with the same magnitude and direction (and differing only in where they are in space). The information about "where are in the space" is the first element of the pair, the point.

In this description, the concept of reference frame results from the property that the obvious rules for oriented segment compositions require $$ \begin{align} {\vec {AB}} &= -{\vec {BA}}\\ {\vec {AB}} &= {\vec {AC}} + {\vec {CB}} \tag{1} \end{align} $$ for every triple of points $A$, $B$, and $C$. In particular, equation $(1)$ can be seen as defining the relation between the oriented segment specifying the position of $B$ with respect to the origin $C$ with the oriented segment specifying the position of the same point $B$ with respect to the origin $A$.

Different orientations of the reference frames correspond to other choices of the basis of the vector space describing the equivalence classes of the oriented segments. Such freedom can be interpreted geometrically as the orientation of the basis vectors at describing the displacement vector ${\bf b}$ at each point.

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  • $\begingroup$ The link does not work/is not displayed correctly. $\endgroup$ Commented Sep 11, 2023 at 8:19
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    $\begingroup$ @TobiasFünke Thank you for signaling it. I have corrected the link. $\endgroup$ Commented Sep 11, 2023 at 12:04
  • $\begingroup$ Your counter-example to Wikipedia's statement does not work. Composition of rotations is not achieved by adding anything. It is achieved by multiplying rotation matrices. Treating rotation as a vector is useful in explaining precession (see the section on torque-induced precession in the wikipedia article at en.wikipedia.org/wiki/Precession $\endgroup$
    – Peter
    Commented Sep 11, 2023 at 12:19
  • $\begingroup$ @Peter If you follow my example carefully, you'll realize it works as a counterexample. It is possible to assign magnitude and direction to any rotation. Is there any way to consider a vector space of rotations? If you do not like the rotation composition as a sum, no problem, although it is possible. If you have better multiplication of rotation matrices, no problem again. Still, you have to recognize that multiplication of matrices is not commutative, in general, and also this alternative "sum" fails to make the set of rotation a vector space. $\endgroup$ Commented Sep 11, 2023 at 12:51
  • $\begingroup$ Thank you for this very nice answer. If I may, I have a couple questions: (1) You write "Different orientations of the reference frames correspond to other choices of the basis of the vector space describing the equivalence classes of the oriented segments" and I want to focus on this because I think it's the crux. Are you saying that a choice of reference frame means, in particular, a choice of origin and that this choice selects a bound vector from each equivalence class of free vectors (namely, the vector emanating from this chosen origin). And this is the key: are you saying... $\endgroup$
    – EE18
    Commented Sep 11, 2023 at 14:51

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