## Are Scalar Quantities in Physics really 1-D Affine Spaces in disguise?

What do you mean when you say that geometric vectors require an additional binary operation? Are you referring to a metric? If I had to define what I mean by vector, I would say the elements of the tangent space of a manifold (or some technical extension of that to algebraic varieties/non-smooth manifolds).

Re: functional analysis. Scheme is a strong word for my internet ramblings in between semesters . My only scheme is to try to refer to things in the most specific way possible for the problem at hand.
 The vector space axioms include the binary operation of vector addition of two elements of the space as necessary. The binary operation of multiplication of two elements (vectors) is not part of the axiom set {either interior (dot) or exterior (cross) products}. Some vector spaces have one or both, some have neither. You require both for the vector calculus used by physicists in abundance.
 Ah I see what you mean. I guess this just serves to illustrate the futility of isolating exactly what is a vector. Perhaps a mathematician's "vector" is not a physicists. Here is a physical example of a vector in which neither type of product plays a role. Consider a physical system consisting of two particles in space subject to a force that depends only on the distance between the particles. According to Newton's second law we get a system of 6 2nd order ODE. One for each component of the position of each particle. By introducing velocity or momentum, we can transform it into a system of 12 first order ODE. So we have a 12 dimensional phase space which is the set of all possible positions and velocities. And on that space, we have the 1st order differential equation. X'=G(X), The thing on the right side is a vector field, precisely because it tells us the infinitesimal displacements of X(t).
 But in order to define a vector derivative (or integral) you need a multiplication (actually division) operation with vectors do you not?
 No. Nothing there requires vector multiplication. Only scalar multiplication.

Thank you Studiot and Vargo for stepping in with your insights.

 Quote by Studiot It is unfortunate that different scientific disciplines sometimes have different definitions for terms. Vector is one such term where there are at least five different definitions in use.
I ran across that hurdle long ago. As I learned more, I began to see a larger tapestry of Vector Spaces, not just a term that is used in different ways. To me, there is a minimal or generic Vector (aka, Linear) Space which defines a basic architecture or blueprint as a set of abstract member vectors which obey certain linearity axioms. But it is not really implemented in any useful way in this form and the abstract vectors and their linear behavior are simply standins for what may be installed in their place. The generic Vector Space is just a starting point.

It's sort of like the idea of a generic car (4 tires, engine, seats, etc.). The car can be implemented as a Ford fiesta model, Volkswagen bug model, etc, but lots of details must be worked out, such as which Goodyear tires will go on it, etc. Different cars implemented for different needs. Vector Spaces are extended for implementation in different situations with different needs as well. We may specify a finite, countably infinite, or uncountably infinite Set of Vectors for the Space. We can assign the Vectors to be Physical units of measure such as delta_meters (displacement vectors) or delta_meters/sec (velocity vectors). We could assign the vectors to be linear functionals, tensors, polynomial or fourier (sin, cos, e) functions, rows or columns of a matrix or some other discrete functions, normally distributed random variables, gradient and other linear operators; anything that obeys the linearity axioms. We may even assign the vectors to be the Real numbers. Name other examples if you can think of them.

We may require 2, 3, or uncountably infinite dimensions, so we throw that in. Some folks need an inner product or metric of some kind. I'm sure there are other ways to specialize our generic Vector Space. If you think of them, name them. We could find unique names for every variation, but I fear it would be too long of a catalogue. Car companies as a collective only produce a finite number of fully implemented car models and they each get unique names. In Vector Spaces, we should probably have some systematic naming scheme as Organic Chemistry does with its IUPAC names. That might clear up some of the confusion.

 Quote by Studiot 4)The definition used in biological and medical science as a carrier.
Yeah. They should be shot for this one. This one confused the hell out of me too. Apparently the earliest uses of the term Vector actually means carrier or transporter. Maybe we are the ones that need to be shot.

 Quote by Studiot 5)The definition used in computer science as an row or column element of a table.
The computer guys have their own little world, don't they? But they are very, very good at explicitly and unambiguously defining and organizing data structures, no matter how complicated. We should be so good at organizing mathematical structures.

 Quote by Vargo Many of the quantities you mentioned do not have this property at all. Kinetic energy is not affine because the range of acceptable values form a ray, not a line. KE=0 is physically different from KE=1. There is a kind of scaling symmetry to the universe that renders KE=1 physically the same as KE=2, but in order to realize that symmetry you have to change your scale of mass or length or time. But there is never any symmetry of the universe that will make KE=0 equivalent to KE=1. And we could say the same thing for mass or temperature. Going back to the geometric definition of affine, you cannot arbitrarily translate a point on a ray and still end up with a point on the ray.
Hmmm. My understanding of Affine must be in error. Please identify any flaws you see in my thinking. I think of an Affine Set as a collection of points with labels specifically and systematically inherited from the Vector Space defined at the origin. The physical meaning that has been assigned to each point stays with the point under any relabeling via linear transformation of the origin's Vector Space or translation of the origin itself. The origin may be arbitrarilly assigned, but the physical meaning of the underlying point stays with the point, even if the origin is assigned elsewhere. So the origin gets a new meaning under translation, because it takes on the physical meaning of the new point. In the example of a 2D Vector Space with Vectors assigned displacement in delta_meters, the Affine Space becomes a 2D Euclidean plane in R^2 because of its inheritance of meaning from the Displacement Vector Space connected to the origin. At every other point in the plane, there is a similar (actually, identical twin) Displacement Vector Space, but they are dormant as far as labeling is concerned, until the origin is changed, at which time the new origin's Displacement Vector Space supplies new coordinate labels.

The Axioms define a vector's existence in the Affine Space and tell us that any sum of the Vectors from affine points a to b, b to c, and c back to a is a zero vector (it forms a loop). They also tell us that if we know the vector and its starting (ending) point in the Affine Space, we know the ending (starting) point. I can't see where the Affine Space ceases to be an Affine Space if the plane is trimmed to become finite or if the origin is moved to the edge or corner of the finite plane. After all, any straight line segment in the plane is still a straight line segment under Affine transformation (the coordinate labels change, but the points don't and are all still lined up in a row in the same order). That usually confirms the characterization as an Affine Space. I do recognize that a large number of the Displacement Vectors at the edge or corner points won't be useful, but I am unsure if defuncting a subset of an Affine Set disqualifies the remaining part from behaving as an Affine Space. We can just say the domain of the measurement is a subset of the Affine Space.

Oh wait, it does have Displacement Vectors adding to points that lead nowhere, which violates one of the stated axioms. However, if we were to say the measurements are Reals, there is a violation of closure there as well because the domain of measurements exclude a large portion of the Reals, even in theory. Maybe we need some mixture of Affine attributes combined with the idea of rays to meet the need. Any thoughts? I still would like to know, in any case, if the zero point of Scalar Quantities can be assigned arbitrarily. Again, any thoughts? It would be nice to unify the structures that characterize all Physical Quantities found in Physics.

 Quote by Vargo I say call it as it is. Mass/Kinetic energy/temperature/speed are numbers, not vectors (unless they are in fact vectors in modern physics ).
Yes. They are numbers. The scale units for assigning those numbers are arbitrary, but agreed upon by all to use. The choice of zero point and of unit size may be changed. In measurements, we are assigning numbers to things that have no numbers until we do so. My question is, is there a deeper mathematical structure then numbers that characterizes the quantification scheme of the fundamental Physical Quantities we find in the study of Physics? I propose Affine structure - knowing that a piece of it will not be used. And what do we make of changes in those quantities: delta_meters, delta_time, delta_mass, delta_temp. Do they qualify as "Displacement" Vectors in a Vector Space?

 Quote by Vargo "vectors" are fundamentally related to spatial displacements.
I'm willing to go along with this naming scheme, though I'm uncertain if it is a universally standard presumption of the meaning of the term Vector, even within the Physics community.

 Quote by Vargo Perhaps we could agree to the following convention. From now on, what are now called Vector spaces, we shall henceforth call Linear spaces. And what are now called vectors, we shall call.... um....... Linear combinationables (combinables?) :). The term vector shall exclusively mean exactly what we thought it meant before we learned about linear algebra.
I vote for Linear Elements as the term for elements of a Linear Space. And I would like to call measurements of Scalar Quantities (scalar values) Affinitors or something similar if Numbers turns out not to be the best model for such measurements.
 I'm going to complicate things. In Physics, Scalars are often assumed to be a part of a Scalar Field, i.e., a Scalar function over a coordinate system (the coordinate space is mapped to a subset of the Reals - i.e., a set of Numbers). Scalars are defined as Physical Quantities that remain unchanged when the underlying coordinate system is changed (i.e., the labels on the coordinate points are relabeled) by certain Affine transformations: rotation or translation. A Rank 0 tensor is also called a Scalar which I believe implies a Scalar is a 1-D Vector Space. Is a Scalar also a 1-D Vector Space (which the Reals is)? This makes some sense since the Basis Vector would be the Physical Units, e.g., Kelvins or Kilograms, and the Real coefficient would specify the amount of the Basis Unit. Temperature and Mass are also introduced as Scalars, but not specifically in connection with any coordinate system - just as measurable Physical Quantities with their own 1-D space with Real Numbers on the scale. So are Temperature and Mass both 1-D Vector Vpaces? What about all other Scalar Quantities? If so, what are all the delta quantities (delta_Kelvins, delta_kilograms, etc.)? Are all these definitions identical? Is there a problem characterizing Temperature and Mass as 1-D vector spaces since they exclude half the Vector Space and violate the requirement of closure for the addition and scaling of Vectors? Since there are Temperature scales that do have differing zero points, doesn't that disqualify Temperature as a 1-D Vector Space and give it more of an Affine character?
 Seems like somebody at Wiki had the same thoughts: http://en.wikipedia.org/wiki/Dimensi...s_displacement

 Quote by Rising Eagle A Rank 0 tensor is also called a Scalar which I believe implies a Scalar is a 1-D Vector Space. Is a Scalar also a 1-D Vector Space (which the Reals is)?
I do not believe that physics belongs in a math discussion, so I am going to ignore everything else but this comment. In the context of a vector space over a field F, the scalar field consists solely of the elements of F. So if we regard F as a vector space over itself, then the scalar field is just a 1-dimensional vector space.

 Quote by jgens I do not believe that physics belongs in a math discussion, so I am going to ignore everything else
No. This topic does indeed belong in the math section. I believe you are missing the point of this thread. This topic is not about the laws of physics or physical interpretation of some phenomenon. It is about a mathematical definition of an Affine Space and whether or not it is properly applied. The context does not nullify this thread's mathematical content. In fact, math applied to physics is kind of the whole purpose of a mathematics section at a website called physicsforums.com

Physicists couldn't care one wit about the axioms of an Affine Space, only how to understand and use its features. This is very much a mathematics topic. Please be accepting of these kinds of questions. We need all the help we can get to clarify some of these details.

Yes, the discussion has gone on some tangents, but they are necessary and relevant to the topic at hand. I would ask that you weigh in with your insight on the original question. We can benefit from your experience.
 Of course, "Field", is another word where physicists and mathematicians have different ideas. One view (a physicist's) of a field is a region of space where we can assign the value(s) of some property of interest at every point. These values may be a scalar, a vector, a table ( physics tensor). A mathematicians field must conform to the field axioms. My physicist's field also conforms to these axioms, but excludes some mathematical fields which are not distributed in space. The interesting thing about physicist's fields is that the dimensions of scalars and vectors will 'fit' into the dimensions of the region of space, but tensors will not.

 Quote by Studiot Of course, "Field", is another word where physicists and mathematicians have different ideas.
I am not sure this is a relevant distinction. In mathematics you have the algebraic structure that we call a field, but we also have things like vector fields and tensor fields, and these objects seem to encompass most of the fields that physicists work with. A physicist might work with a bit more intuitive notion of these concepts, but I am not convinced that there is a disconnect between the usage in mathematics and physics.
 With the exclusion of the zero element a mathematical field is commutative under both binary operations. Now tell me that the moment product of two vectors in physics is commutative.

 Quote by Studiot With the exclusion of the zero element a mathematical field is commutative under both binary operations.
Both operations are commutative with the zero element included as well.

 Now tell me that the moment product of two vectors in physics is commutative.
Perhaps I am being dense, but what is the relevance of the moment product here?

Edit: Let me elaborate on why I am asking about the relevance. In mathematics there are all sorts of non-commutative products on vector fields and tensor fields and we can define all sorts of non-commutative 'products' on algebraic fields as well. So that is hardly a distinguished feature in a physicists use of the word field. It is also worth noting, that many of the fields that physicists work with are special cases of mathematical fields. For example, the electric fields and magnetic fields can be realized as vector/tensor fields (I forget which) on a manifold.

 Both operations are commutative with the zero element included as well.
Let x, 1, 0 belong to field F such that 0*x = 1 = x*0

This is why zero is excluded.

 Quote by Studiot Let x, 1, 0 belong to field F such that 0*x = 1 = x*0
You are confused. In a field you have the forced convention $0 \neq 1$ and also that for all $x \in F$ the equality $0x = 0 = x0$ holds. So both operations are commutative (if you do not trust me, then consult any book on abstract algebra).

Edit: It is worth noting that if you have a ring with $1 = 0$, then it is the trivial ring $\{0\}$ and the addition and multiplication on this ring is clearly commutative over the whole ring.

I am not at all confused.

Collins Reference Dictionary of Mathematics

 Field n A set of entities subject to two binary operations, usually referred to as addition and multiplication, such that the set is a commutative group under the addition and the set excluding the zero element is also a commutative group under the multiplication.....
It says the same thing in my textbook on algebra (Archbold, Oxford University) in more esoteric terms.

My last post was a light hearted way to say this is to avoid division by zero.

go well