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

jgens
Gold Member
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.

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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.

jgens
Gold Member
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.

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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.

jgens
Gold Member
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.

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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.

jgens
Gold Member
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.

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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

jgens
Gold Member
I am not at all confused.
Notice that it says that $F \setminus \{0\}$ is an abelian group. It does not say that the multiplication operation is only commutative if we exclude $0$. Addition and multiplication on a field are provably commutative over the whole field: Clearly multiplication is commutative over $F \setminus \{0\}$ and $0x = 0 = x0$ for all $x \in F$. So I really do not understand what your objection is to saying that multiplication is commutative over the entire field.

My last post was a light hearted way to say this is to avoid division by zero.
Requiring that $F \setminus \{0\}$ be a group under multiplication is to avoid division by zero. But we can multiplication can be commutative (and is commutative) over the entire field and we still avoid division by zero.

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.
I believe we can add to the list Linear Time Invariant and Linear Shift Invariant systems as studied in engineering, random processes, matrices, many different types of functionals, and, interestingly homogeneous linear and differential equations too. I never thought of it before, but equations themselves can be added and scaled and so qualify. Maybe they don't even have to be linear or homogeneous either; just any general equation. Not sure about inequalities or greater than/less than relationships, though.

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.
Another possibility is to call the elements of a Linear Space a Linear Form as a take off on 1-form or differential form. Any linear element in general would be a Linear Form.

If you don't exclude zero you allow the following paradox.

It is axiomatic that for any a, c in F there exists a 'b' in f such that

a*b = b*a = c

1 is in F
0 is in F

set a = 0, c =1

therefore there exists a 'b' such that

0*b = b*0 = 1

leading to b is the inverse or reciprocal of zero.

jgens
Gold Member
If you don't exclude zero you allow the following paradox.
Do you understand the difference between saying that the multiplication on $F$ is commutative and saying that the multiplication on $F$ makes $F$ into an abelian group? The first statement is true and the second statement is false; they are not analogous.

There is no paradox in saying that the multiplication on $F$ is commutative. If you think there is, then find the flaw in this argument: Clearly multiplication is commutative on $F \setminus \{0\}$ by hypothesis and for all $x \in F$ the equality $0x = 0 = x0$.

This area of pure maths (in fact any area) is not my speciality and we are off to the seaside for the day so perhaps Frederik or Hurkyl will explain.

I think the reasoning is as follows:

Yes composition by zero is commutative.

However you need to specifiy this separately not inherit it from group properties.

I will post a more complete explanation if no one else has done so.

jgens
Gold Member
This area of pure maths (in fact any area) is not my speciality and we are off to the seaside for the day so perhaps Frederik or Hurkyl will explain.
Luckily this happens to be the area of math that I am best at.

However you need to specifiy this separately not inherit it from group properties.
Actually no. If $(F,+,\cdot)$ is a field, then the multiplication on $F$ is a map $\cdot:F \times F \rightarrow F$. This internal law of composition is clearly commutative and is defined on all of $F$ and not just on $F \setminus \{0\}$. So multiplication on $F$ is commutative even if we include $0$.

If we want to talk about group structures, then this is a different story. The pair $(F,\cdot)$ is not a group since $0$ has no inverse while the pair $(F \setminus \{0\},\cdot)$ is a group by the definition of a field. So you would be correct if you were claiming that we need to exclude zero in order to get a group under multiplication; however, we were not talking about getting a group under multiplication, and instead were talking about whether or not the multiplication operation is commutative over all of $F$.

If you want a neat math term for $(F,\cdot)$, then you could always say that it's a commutative monoid. :tongue2:

But jgens is right: saying that $(F,\cdot)$ is commutative is perfectly alright. However, this fact does not immediately follow from $(F\setminus \{0\},\cdot)$ being an abelian group!!

and instead were talking about whether or not the multiplication operation is commutative over all of F.
You might have been, I never have been.

I stated my position clearly in post#30.

You added a further case in which multiplication is commutative in post #31

I offered an explanation as to why this case was normally excluded in post#32

From post#33 onwards you seem to want to argue that I am saying multiplication including the zero element is not commutative.

I never have said this. In fact in post#39 I said the opposite.

All I have been doing is seeking an explanation as to why respected mathematical dictionaries and most algebra, group theory and galois theory textbooks that I have access to exclude multiplicative zero in the definition of a field.

I don't know if my explanation is correct or if there is another one.

Can you provide an explanation please?

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