## You are here

# Scalar done wrong

## Primary tabs

When I was updating my very brief notes on tensors, it occurred to me to post on iMechanica a request for recommendation of textbooks on linear algebra. I was delighted to see Arash respond. I then asked for his opinion about the definition of tensor. He responded again, and we seemed to agree. Then Amit joined the discussion, and then others. That thread has become very interesting and very long.

But I have another issue with the way we use linear algebra. I wish to get your opinion. The issue is about scalars.

**Vector space**. First some background. A vector space involves two sets and four operations. One set is the number field F, along with two operations: addition of two elements in F gives another element in F, and multiplication of two elements in F gives another element in F. The two operations follow the usual arithmetic rules. See a formal definition of number field. For our purpose, F can be the field of real numbers. There is nothing more to it. The other set V is a set of vectors, along with two more operations: addition of two elements in V gives another element in V, and multiplication of an element in F and an element in V gives an element in V. The two operations follow the usual rules for vectors. See a formal definition of vector space.

A space V is said to be n-dimensional if there exist n linearly independent elements in V, but every n + 1 elements of the space are linearly dependent.

In particular, for a one-dimensional vector space S on a number field F, any two elements in S are linearly dependent. Let u be an arbitrary, but none-zero, element in S. All other elements in S takes the form au where a is an element in F.

**Is scalar a synonym of number?** In many textbooks on linear algebra, the word *scalar* is a synonym to the word *number*, an element in the field F. The word scalar is also commonly used in physics to indicate quantities like mass, energy and entropy. The two usages of the word scalar are incompatible in several ways. First, a physical property like mass is more than just a number; it has a unit. Second, the multiplication defined on a field makes no sense to mass: the multiplication of two elements in F gives yet another element in F, but the multiplication of two masses does not give another mass. Third, if we regard both mass and entropy as elements in the field F, then we need to assign a meaning to the addition of mass and entropy. What does that even mean?

Thus, we will call an element in the field F a number, and will reserve the word scalar for physical quantities.

**The set of pieces of a substance of all sizes.** Given a substance (e.g., gold), pieces of all different amounts of the substance form a set. We can define the addition of the pieces, but we do not have a sensible definition for the multiplication of the pieces. Thus, this set is not a number field. This set, however, is a one-dimensional vector space. We stipulate the two operations in a natural way. The addition of two pieces of the substance is another piece of the substance, and multiplication of a real number and a piece of the substance is another piece of the substance.

**Extensive property of a substance.** A piece of a substance has many physical properties, such as volume, shape, color, temperature, mass, energy, entropy. A physical property is extensive if it is proportional to the amount of the substance. Volume, mass, energy, and entropy are extensive properties. Shape, color, temperature are not extensive properties.

**Extensive scalar**. We can use a one-dimensional vector space S to model a physical property such as mass. In this model, F is the field of real numbers. We call this one-dimensional vector space a scalar set. We use a particular element in S as the unit for this quantity. For example, for the scalar set S of all masses, the unit mass, kg, is a block of metal located in Sevres, France. All other masses equal this unit times a real number.

By contrast, temperature cannot be represented as a one-dimensional vector space. The addition of two temperatures does not give another temperature.

**Linear form**. In textbooks of linear algebra, a linear form is commonly defined as follows. Let V be a vector space on a number field F. A linear form is a linear map that maps an element in V to an element in F.

I believe that this definition is inconsistent with how mechanicians use linear form.

Here is my modified definition. Let V be a vector space and S be a scalar set, both on the same number field F. A linear form is a linear map that maps an element in V to an element in S.

**Dual space**. The set of all linear forms from V to S is also a vector space. We denote this space by V', and call it the dual space of V with respect to the scalar set S. The vector space V and its dual space V' have the same dimensions.

**Work, displacement, and force. ** Here is an example how linear form arises in mechanics. Work cannot form a number field: the multiplication of two amounts of work does not give another work. Work, however, is a scalar. If the displacement is small, we know the work is linear in the displacement. That is, there exists a linear map that maps the displacement to work. The displacement is an element of a three-dimensional vector space U, and the work is an element of a scalar set. The linear map fits the definition of the linear form, and we call the linear map the force. The force is an element of the dual space U' with respect to the scalar set of work.

**Tensor**. We have had a long thread on the definition of tensor. Here is a definition that I like. Let V1, V2..., Vp and V be vector spaces on the same field F. These vector spaces may represent objects of different kinds, and may even have different dimensions. A tensor is a multilinear map that maps an element in V1, an element in V2,..., and an element in Vp to an element in V.

**Remark 1**. Tensor is a generalization of linear form.

**Remark 2**. The set all multilinear maps from V1, V2..., Vp to V is a vector space. This vector space generalizes dual space.

**Remark 3**. We can use this new vector space to generate other tensors. To be consistent, a vector is a special case of tensor, so is a scalar. They live in different vector spaces.

This is the main theme of linear algebra: vector spaces, and linear maps between them. Linear maps themseves form new vector spaces. We map the maps. The never-ending strory of linear algebra.

- Zhigang Suo's blog
- Log in or register to post comments
- 5642 reads

## Comments

## Re: Scalar done wrong

Dear Zhigang,

I also have thought about this issue, and make a mention of it in my class. Since the courses which I teach are not much mathematically oriented, I usually tell the students to be conscious about the physical meaning of the operations they are performing: if they are multiplying vectors or tensors with "pure" numbers, there is no need for much concern. However, if the quantity is a scalar with a dimension, the students are told to be conscious about physical meaning of the operations they are performing.

What if the scalar quantity does not have a unit (say, first principal strain)? Can we consider it to be a pure number?

I feel the issue is not very specific to scalars. For example, an inner product may be defined in a vector field, but trying to do it in a field of force vectors is probably be meaningless...

Trying to include all these in mathematics itself might unnecessarily complicate the subject, which is already tough enough for many of us, with all those abstractions. In my opinion, it is better for physicists and engineers to take care of the physical meaning and validity of the operations they are performing than putting the burden on mathematicians.

Regards,

Jayadeep

## Mathematicians need to learn to do no harm

Dear Jayadeep: Thank you very much for your comment. You raise an excellent point. In teaching, it is wise to divide labor, and keep mathematical structure clean, decoupled from applications.

However, the mathematical structure itself should be general enough for significant applications, even if the applications are unmentioned in the mathematics course itself.

In this particular case, I believe that many textbooks on linear algebra have done wrong in two places.

Here is an example of the last statement. Let V be the three-dimensional space of position vectors, and F be the field of real numbers. We have already used work as a one-dimensional vector space in the original post. The dual space of V with respect to work is the vector space of force. Now we wish to use electric potential as the one-dimensional vector space. The dual space of V with respect to electric potential is the vector space of electric field.

Thus, I am asking teachers of linear algebra to do less, not more. Do not do extra work to do harm.

## Re: Re: Scalar done wrong

Dear Jayadeep:

Regarding your question: "What if the scalar quantity does not have a unit (say, first principal strain)? Can we consider it to be a pure number?" I would say no. Can you multiply two principal strains? Zhigang?

Regards,

Arash

## Re: Can we multiply principal strains?

I think we can, and it gets done when multiplying two symmetric tensors (think in spectral form). The result is independent of bases.

In certain instances this can even be given a sound physical meaning - homogeneous deformation by two successive pure stretch deformations, or a pure stretch followed by an invertible homogeneous deformation....

## Re: Re: Can we multiply principal strains?

Hi Amit,

Yes, but isn't this too special? For arbitrary successive deformations principal stretches are not multiplicative.

In some failure criteria we would see products of principal strains (or stresses). But those are just some phenomenological criteria. What do you think?

Regards,

Arash

## Re: Re: Re: Can we multiply principal strains?

Arash: It seems to me that there is always a physical question that you can attach to the multiplication of principal values of two tensors (let's keep it to symmetric tensors for simplicity, but not necessarily co-axial). It is the product of the stretch ratio of the corresponding principal directions of the two tensors. Why one would need such a product is another question, but it is a well-posed physical question and, of course, independent of bases.

## Re: The physicality of the product of principal strains

"Why one would need such a product is another question"

The one physical quantity I can imagine here is, in some indirect way, the strain energy density (implicitly assuming that a number for elastic modulus by way of unit conversion has been assumed/given). Is there some other physical quantity more directly attributable to such a product?

--Ajit

- - - - -

[E&OE]

## Re: Scalar done wrong

Dear Zhigang:

Thank you for this nice post. You're posing some excellent questions. I think we should emphasize that under any change of coordinates the value of a scalar remains unchanged.

Regards,

Arash

## When do you use the linear map V-->F?

Dear Arash: Happy new year! Thank you so much for your comment. I have not thought through many things, but will add to this post as comments as aspects become clear.

One thing is quite clear, the two definitions of a linear form have different algebraic structures

The multiplication of two elements in F gives another element in F, but the multiplication of two elements in S is undefined. S has a simple structure of a one-dimensional vector space: Given one element u in S, all elements in S take the form au, where a is any element in F.

I have given several examples in which we scientists and engineers use defintion II. I cannot think of any physical situation in which we use definition I. Can you? It will help if we can identify any use of definition I at all.

## Multiplication of particle masses (electric charges)

Dear Zhigang:

Happy new year to you too! By the way, to calculate electrostatic force (or gravitational force) we multiply electric charges (or masses). Not sure if this would imply anything?

Regards,

Arash

## Re: Multiplication of particle masses (electric charges)

Dear Arash: Excellent question. In the case of Coulomb's law, we can derive the law from the Maxwell equations. Of course, many textbooks will start with Coulomb's law as an empirical observation, and then formulate the Maxwell equations to be consistent with it. The approach follows historical development of electrostatics. But we have the Maxwell equations now, and we can choose to start with them, and derive Coulomb's law. This is the apporach I will take here.

The Mawell equations are linear. This fact itself is interesting, but here I will take it as given. Once you solve the boundary value problem and find that the electric field is linear in charge, you need to calculate force. Most people will use the Lorentz force, where force equals charge times electric field. Thus, force involves charge times charge.

But Lorentz force looks mysterious. You don't need his equation. You can find this quadratic dependence in charge from calculating energy. In the following review article, I describe a simple approach in Section 3.

Zhigang Suo. Theory of dielectric elastomers. Acta Mechanica Solida Sinica 23, 549-578 (2010).

To sum up, Maxwell equations are linear. The force is qudratic in charge because energy is involved.

Perhaps the case of gravity can be rationalized in a similar way, but I am not sure at this point.

## Re: When do you use the linear map V-->F?

Zhigang: Your definition I gets used when one defines components of a vector w.r.t. a basis (and similarly for tensor components, whichever way you define it). The act of obtaining components of a vector w.r.t. a basis in V (not necessarily orthonormal) is a linear functional on V to F. In fact this set of n linear mappings from V_n to F, given a basis in V_n, is key in proving the existence of a dual basis in V_n corresponding to the chosen 'primal' basis, and then on to covariant and contravariant components, raising and lowering indices etc..

Of course, since these linear maps depend on a basis, they do not have intrinsic physical significance, in the sense of being invariant w.r.t. choice of bases. On the other hand components are useful and most computations with vectors and tensors utilize them..

## Re:Re: When do you use the linear map V-->F?

Thank you Amit, and happy new year! I fully agree with you on the two points you made:

This example perhaps illustrates a general point: Definition I is less desirable because the map takes an element from a vector space to a number field, which is different from a vector space. To study base-inddiferent behavior, one should focus on maps between vector spaces.

## Re:Re:Re: When do you use the linear map V-->F?

Hi Zhigang: I agree with you that students of mechanics must understand the difference between physical scalars and 'just scalars' (your numbers), the elements of the number field, from the gut; importantly that your sets S and F need not always be the same, as you have tried to spell out.

I would add though that I think you were a little harsh on mathematicians when you say they do harm etc. - my preference is to be a little more tolerant. It does not take much to see the adjustment required to mathematical definitions to fit mechanics needs and vice versa (it is also a good exercise in real understanding for a student), and I don't think any of us really think anyone else intentionally tries to mislead, so why not just appreciate the subtleties that come from both specificity and generality and get on with life?....

On that philosophical note, with my best wishes for the New Year,

- Amit

## Do no harm

Dear Amit: Thank you for the note. I shoud have made a link to the Hippocratic Oath for physicians. The oath meant, I quote from this wikipedia page, "given an existing problem, it may be better not to do something, or even to do nothing, than to risk causing more harm than good."

The Oath was not intended to be harsh to physicians. Likewise, my reference to the phrase was not intended to be harsh to mathematicians.

## How to think of force linear functional as force vector?

Zhigang: Suppose as you say above, we completely sever the connection between elements of the sets S and F (which I would like to go along with......). Then in your example with work, displacement, and force, as yet I cannot see how to construct a vector (something with direction and magnitude) analog of force from what you call the force linear functional.

Below, f is the linear map of force and {e_i, i = 1 to 3} is an orthonormal basis for the vector space of displacements. We assume that the space of displacements has an inner product so that an orhonormal basis can always be constructed and used, wlog.

I hit a stumbling block when, in constructing this vector, I need to interpret f(e_i) for each i = 1 to 3 as pure numbers or, in other words, as elements of F, so that looking at

f(e_i)e_i (sum on repeated indices implied) would yield the force

corresponding to the force linear functional f.

Now physically, I think we would probably agree that we should be able to interpret force as something also with a direction and magnitude (mesurable attributes) and not only some abstract thing that produces work on displacements?

Of course, I have tried only one mode of attack and it is conceivable that there is some other way to do this successfully. It would be good to see an explicit construction of what we might think of the force vector in this case, or is it that if we sever that connection between F and S that we lose the Newtonian conception of force? Or is there some clever way of using the DND mappings here?...... I think we need some youngsters on imechanica to do some thinking here......

## Re: How to think of force linear functional as force vector?

Dear Amit: Thank you for the comment. As you noted, we define force by the linear map f: V-->E, where V is the vector space of displacements, and E is the one-dimensional vector space (i.e., the scalar set) of energy. (We have to careful here, E should really be change in energy, or work.)

In terms familiar to us mechanicians, this map is the same as saying

force times displacement = work

We know work is a scalar, and displacement is a vector. Thus, force is a vector.

We can write the above in terms of linear algebra.

The collection of all linear maps from V to S is itself a vector field over the number field F. We denote this vector space by V', and call it the dual space of V with respect to S. V' and V have the same dimension.

In mechanics, we say that

In algebra, we say that

The words "conjugate" and "dual" have the same meaning.

Here is the Wikipedia page on dual space. The page is OK except that they it still uses V-->F. The definition has the shortcomings as discussed before.

We would like to use V-->S, so that S can be energy, charge, or electric potential. Depending on the choice of S, we can have different dual spaces for fixed V and F. The significance of this flexibility is described in another comment titled the linear algebra of the world.

## Re:^2 How to think of force linear functional as force vector?

Zhigang: Thanks for trying to explain and the link to the Wiki page. There is a subtlety here.

I will call your number field R instead of F (I'll need F for Newtonian force below).

The dual space of V w.r.t. S that you are defining is simply the space of all linear transformations, say L(V,S), between V and S. I agree that it can be proved (and it does require a proof) that L(V,S) is again a vector space, but just from its definition it is also clear that that vector space does not contain elements of V.

Just for the moment, think that R = S. The whole job of the representation theorem for linear functionals in linear algebra (and the Riesz representation theorem in Functional Analysis) is to show that corresponding to each element f of L(V,S) (in functional analysis, L(V,S) would have to be the set of bounded linear functionals) there exists a unique element A(f) in V such that the action of f belonging to L(V,S) on *any* u in V is given by the inner-product (in V) of A(f) and u. Thus A(f) is the representation in V of f in L(V,S). In mechanics, this (A) would be the force vector corresponding to the force linear functional (f). And it is this force vector that I was trying to construct.

Why is such a construction important? Let's say I want to solve a problem in classical particle mechanics F = ma. Let's say it is also mentioned that F is conservative and so generated from the negative spatial gradient of a potential. The potential field is specified as a given function, say psi, of position in ambient Euclidean 3-d space. What is the meaning of the gradient? Clearly the inertial force and the acceleration gets its direction from what we specify for F; F better be a force vector, a member of V or a close cousin that looks and smells like V with the right physical dimensions of force. What CANNOT suffice in this context is to say that F is a member of L(V,S), even though L(V,S) has the right dimension (both vector space meaning and physical units meaning), and setting it equal to ma which is a member of V (or the close 'force' cousin of V). Alternatively, you would have to raise ma to be a member of L(V,S) somehow.

Now given psi as a function of position, do its derivative. What is the derivative of psi at any given point of space? It is a member of L(V,S), by definition. Is the gradient the derivative? No. The gradient is the representation in V of the derivative in L(V,S), when S = R. If you wrote psi as a function say of rectangular Cartesian coordinates, and just did what we would think of doing for a 'derivative', what do you get? One gets the components of the gradient vector of psi on the orthonormal basis corresponding to the Rectangular Cartesian coordinates chosen (all this can be done in arbitrary coordinates). Now I ask, give me a similar 'formula' for the derivative. One will be hard-pressed to come up with something that is not contrived, even though most would accept that the derivative exists. It is in this sense I find the algebraic dual space of a vector space an 'abstract' object.

So, my point is in applications, the representation of linear functionals in the vector space matters a great deal and in what has been spelled out until now, how this can work out is not clear. This is what I was trying to do in what I described. When S = R (your F) all this works out fine, but when S is not R it seems problematic.

My feeling is that some notion of multiplication between elements of S and elements in V is required at the very least (to set up the 'close cousin' vector space of forces), but a satisfactory resolution seems to require some careful thinking. Making S separate from R is a good idea, but coming up with a satisfactory alternate framework requires more than the definitions we have seen on imechanica until now (keep in mind we have poked with just one example as yet). It would be good if we could get the help of some people good at both algebra and mechanics.

## Re:^3 How to think of force linear functional as force vector?

Dear Amit: I have not been thinking about analysis. Just algebra at the moment.

By definition, a scalar set S is a one-dimensional vector space over the field of real number. Thus, once we choose any one non-zero element u in S, all elements in S take the form au, where a is a real number.

In linear algebra, u is a basis of S

In physics, u is a unit for the scalar in S.

Thus, in terms of analysis, S has the same structure as the field of real number. Also the position space V is still an Euclidean space.

Does this solve your issue?

## Re:^4 How to think of force linear functional as force vector?

Hi Zhigang: Everything I am saying has to do with algebra (the special example from particle mechanics involved a little calculus and I made a few side remarks about functional analysis, a subject that may be viewed as infinite dimensional linear algebra). As to the question you ask, being able to allow an algebraic rule for 'multiplying' elements of S and V would help I think in the problem I was considering, but I am not sure that by itself will solve the problem and it needs more work than I did for a clean answer. Moreover, similarly, it is not clear to me that there are not other important questions where the fact that your number field is actually an algebra (roughly, vector space that allows 'multiplication' of elements) becomes important too. I simply don't know. I think it needs a real expert with deep knowledge of the ramifications of the interaction of the number field and the vector space......

## Re:^5 How to think of force linear functional as force vector?

Dear Amit: A question you posted in an earlier thread gives a focus to this discussion. Here is your question again:

"...why conventional mechanics gets away with treating position vectors, displacements, velocities, area vectors, tractions etc. as all belonging to the same vector space (when physically we cannot really add displacements and velocities, e..g.)..."

The answer, it seems to me, is that all these objects are generated from a single 3D vector space: the space of position vectors.

As I noted in a previous comment , the linear algebra of our world has the following basic building blocks:

We can then use these basic parts to generate objects like velocity, force, area vector, traction, etc, by using linear maps among V and various scalar sets. We can then map the maps to generate even more objects; for example, stress is a linear map that maps area vector to force. And then the set of all stresses is a vector space, ready to be mapped.

In the above list of the basic building blocks, only a single 3D vector space V appears. We use linear maps to generate other objects. Thus, a change of basis in V will change the components of all the objects so generated, according to the rules of tensor.

## Answer to question posed + sundry remarks

1) Since I do not want to have the dubious distinction of "raiser of questions without trying to answer", my answer to the question I posed appeared some ways down in that long thread, so here is a link to it.

2) Somewhere in these threads I read about the importance of linearity. I just want to point out, may be only as a side remark, that while an object like a derivative Df(x) of a scalar potential f: V --> R is by definition a linear map on the underlying vector space

i.e. it acts like Df(x)[au + bv] = aDf(x)[u] + bDf(x)[v], for all a,b in R and u,v in V and (to keep things simple) say x in V

Df's dependence on x can be completely nonlinear. Therefore the gradient of f at x can also be completely nonlinear in x.

3) My feeling is linear PDEs of physics should not play a big role in our discussions. As my friend Luc Tartar likes to think, may be at the smallest scales our physical world is at best a semilinear system of hyperbolic PDE, but the smallest scale does not physics make...

In particular, just based on rotational invariance, we know our beloved Elasticity cannot be a linear theory.

4) Discussion on temperature - much like a point space and its translation space, the space of temperatures may be considered something close to a vector space by introducing an 'origin' - so differences of temperatures make sense (otherwise we would have to say bye-bye to the heat equation, e.g.). However, it is only 'close to a vector space' because even temperature differences would not be closed under addition (there is a physical meaning to absolute zero)?

## Re: Answer to question posed + sundry remarks

Dear Amit: Thank you. I just came back from shoveling snow.

What is the "DND mapping" that you mentioned in your previous comment?

Indeed, temperature is a troublesome variable. This fact has troubled me for some years. The temperature from 50K to 70K is different from 100K to 120K. The physics in the two intervals are different.

We'd love to have a additive quantity, but temperature is not. I agree with you: it is still troublesome even if we choose an origin. Then we need to say how far away this origin is from the absolute zero. Thus, temperature does not have the "translational symmetry", unlike time and spatial point.

Entropy may also have no translational symmetry.

Why does energy have translational symmetry? Perhaps energy does not have this "translational symmetry", at some fundamental level. We are just operating at a point so far away from its "absolute zero", whatever that may be.

Random questions for which I have no answers now.

## Re:^2 Answer to question posed + sundry remarks

Hi Zhigang: The DND map from say the traction vector space to the non-dimensional 3d vector space would be generated as follows:

One would have already (arbitrarily) chosen some fundamental system of units for say mass, time, and length, and this system would be common to all the 'DND maps' from the various dimensional 3d vector spaces to the non-dimensional one. Now simply nondimensionalize each of the traction vectors in the dimensional vector space. Then I think each dimensional traction vector corresponds to an element of the canonical nondimensional vector space chosen. Now one can do the same thing for say the vector space of position vectors and velocities etc. It is amusing to note that in effect a set of these DND maps (parametrized by a choices of the fundamental unit set) allows one to associate the velocity vector representing say 30deg NE of magnitude 2 m/s with some traction vector (not to be construed as anything physical of course).

Then the stress tensor, which we understand in the dimensional context as a mapping of a certain area vector to a certain traction vector can be translated to a linear mapping from V_3 to V_3 where I now think of V_3 being the canonical nondimensional space just as we want in continuum mechanics...... and so on and so forth for other tensors....

Of course, all this requires careful checking about preservation of linearity, and whether these are really 1-1 maps etc.

And yes, I agree with you, with these things the deeper you dig the murkier it gets. There's probably a philosophical reason behind mathematics (and physics) not being decidable if they are logically consistent.....

In this context, I sometimes really like David Mermin's comment - "shut up and calculate" about quantum mechanics.

There's also been talk of Strang here - a great saying of his is "adept at abstraction, inept at computation" I am very happy to be an engineer, so that we come up with some answers, however imperfect....

And while I am at it with random comments, here is a plug for one of Strang's books that I think is absolutely terrific - it is "Introduction to Applied Mathematics"

## Re^2: Answer to question posed + sundry remarks

How does this sound? Strain is a troublesome variable. The strain in a steel specimen is different before the yield point and after it. The physics in the two intervals are different.

The absolute zero itself would be the sought "origin," even if temperature can't be modeled as a vector. But, yes, it should be possible to think of addition of temperatures. I addressed this issue in an earlier comment somewhere in this thread. And, that way, you never add electric potentials, anyway; you only displace charges from one point to another point. Yet, we always regard potentials as quantities that can be added to each other. If so, why not temperature?

The third law only prohibits

goingfrom a non-zero temperature to the absolute zero temperature; it does not prohibit some thing/some region somewhere in the universe possiblyalreadybeingatthe absolute zero---the law leaves that possibility open. The issue is similar to c. Just because massive objects cannot be accelerated from v < c to v = c does not mean that objects cannot move at c---photons do. Of course, to my knowledge, we haven't yet actually run into something which already was at absolute zero. But no known law prohibits our possibly running into it. If a prohibitive principle is found, it will be a new knowledge; a new law, unknown as of today.There were certain reports of negative (absolute) temperatures being reached in experiments. These were gimmicks (like certain other reports of light going faster than c). Barring these, an object at absolute zero simply means that it cannot give up its energy; that it can only absorb some. Known laws and empirical data do not rule out such a possibility.

Of course, to define mathematical object, it's only necessary to know that there is no physical constraint to at all have 0K on the scale. So, the scale can indeed be defined, and used for additions/subtractions of temperatures.

See you sometime tomorrow (IST).

--Ajit

- - - - -

[E&OE]

## Re:^7 How to think of force linear functional as force vector?

Hi Zhigang: I don't see why my previous question is relevant in some crucial way to this discussion.

## Re:^6 How to think of force linear functional as force vector?

Dear Amit: I have just looked again at the book Mechanics by Landau and Lifshitz. Chapter 1 describes the principle of least action, basis-free. The book defines force by the gradient of energy.

At some point the isotropy of the space is used. Also the magnitude of velocity enters. These would rely on an inner product of the position space.

## Re: Is scalar a synonym of number?

Dear Zhigang,

Thanks for this post; you bring to the fore some really tricky issues.

That way, I hardly have had any background in abstract algebra, but still, guess (true to my form) I could (still) brainstorm a bit about the points you raise in the paragraph on scalars.

You note:

"First, a physical property like mass is more than just a number; it has a unit. Second, the multiplication defined on a field makes no sense to mass: the multiplication of two elements in F gives yet another element in F, but the multiplication of two masses does not give another mass.

Third, if we regard both mass and entropy as elements in the field F, then we need to assign a meaning to the addition of mass and entropy. What does that even mean?"

Excellent notings!

In response, what I think might be happenning here is the following (and I also refer to Arash's reply above, regarding products of charges or masses):

Let's view your second point, in the light of your first point, and this is what I think we get: The multiplication of two elements in

oneF gives a unique element inanotherF.Just the way you can define a number field for mass, so it should also be possible for you to define

anothernumber field for their product (mass X mass). The algebraic properties of the second F arguably still are those of a number field; it's just that it's not the same instance of a number field---the physical units to be attached to each are different. The operation of addition might conceivably close into the same instance of the same kind of an algebraic field (here, a number field). For instance, if you stick two handfuls of the crazy putty together, you do get a ball of a crazy putty that has for the quantity of its mass the sum of the masses of the two balls. But the operation of multiplication does not close into the same instance of the same kind of a field. The result of the product operation might have a meaning, but only in relation to some other physical considerations (e.g., as a factor in calculation of gravitational force).Inversely, for a similar reason, mass and energy each does obey the laws of a number field, but they therefore are not elements of the same instance of number field---they belong to two different number fields. Only the abstract algebraic structure is in common; but the actual mathematical sets---their instances---are different.

Thus, a scalar is an element of a concrete instance of a number field. Each instance may be specified in reference to the physical unit.

...

Oh well, I am not sure how the idea of the multiple instances of the same algebraic structure, goes by the mathematicians.

...Hello mathematicians---or those who by any chance are reading this thread, anyway---happy new year to you, too (!), and, am I talking any sense? (If it helps: I just got these ideas not because I know abstract maths but simply because I am familiar with certain concepts we programmers use in object-oriented programming---things like abstract data types (ADTs), classes and instances, genericity, and all that.)

And, of course, before closing, let me wish a happy and a prosperous new year to all the iMechanicians, too!

--Ajit

- - - - -

[E&OE]

## Re:Re: Is scalar a synonym of number?

Dear Ajit: In the same spirit as you have just outlined, I suggest that we represent each scalar quantity by a one-dimensional vector space, which I call a scalar set.

Thus, we have the scalar set of mass, a scalar set of energy, etc.

The nice thing about the vector space is that it is not closed under multiplication. The one-dimensional vector space has the structure that we are looking for.

## Re:Re:Re: Is scalar a synonym of number?

Dear Zhigang,

1. Looks like I used a wrong term. I meant to keep a scalar quantity only to a simple real number field [^], whereas a vector quantity would properly belong to a vector space [^].

So, please substitute "real number field" wherever I have used the term "number field" in my above reply.

2. The main point I then made is that the real number field for mass would not be the same as the real number field for the quantity: mass X mass.

Bothwould be real number fields, but since their physical units differ, they would belong to two differentinstancesof real number fields. Neither would be a vector space.3. The structure of a vector space, even if it be only a one-dimensional vector space, brings in the additional baggage of a basis set.

A basis set is nothing but, what else, just a minimal set of

vectorsthat can be used to span that entire space. For 1D space, the basis would consist of just one vector, but it would still be avector! Yet, when we talk of a scalar like mass, we don't think in terms of some real number multiplied by a "unit vector of a mass," where the latter notion supposedly might mean going from an origin point of a zero mass to a point of a unit mass---the very ideas of a "from" and "to" don't apply for a scalar quantity like mass.This extra conceptual baggage that the basis set introduces is neither necessary for scalar quantities nor does it make any sense for them.

4. Yes, it is true that a vector space is not closed under multiplication. But the converse is not true (I am not sure if "converse" is the word I want here). The idea of a vector space is not necessarily an answer for every operation that is not closed under multiplication.

As I pointed out, the answer is:

multipleinstances, identified by physical units, of a certain "scalar-like" structure. In my first reply above, I called it the number field. Looks like I should have called it the "real number field." The key idea I introduced is, however, themultipleinstances of thesamestructure, not an introduction ofanotherstructure.I would like to know what other mechanicians think of these points, too.

(PS: Signing off for now, will be back tomorrow morning IST, i.e. after about 15 hours.)

--Ajit

- - - - -

[E&OE]

## Re: Scalars. I was being wrong (or, may be, half right!)

Dear Zhigang, and others,

Guess I got the problem

now(!)---or so I think (!!)So, I only now realize that, no, when it comes to scalars, even the ordinary "real number field" wouldn't be suitable either. The real number field by definition

isclosed under multiplication and we need a structure that isn't so. (I appreciate this part better only now!!)But then, for the reasons mentioned above, it won't be a vector space either---whether 1D or otherwise.

So, at least for scalars like mass and energy, it would be neither a vector space nor a real number field, but a new structure that is like the real number field except for a further constraint that it isn't closed under multiplication. Suppose we call this new structure S. Now, inasmuch as the physical units of mass, mass^2 and mass^4 do differ (not to mention mass^3), the idea of there being different instances of this new structure each for the

operandsand theresultsof the multiplication operation, might still hold. (I would be half-right only in this sense.)And, I don't know what to think of adding two temperatures. Shouldn't addition be a valid operation if the

absolutetemperature scale is used? I tend to think so, but now am no longer sure of anything (LOL!)(Ok, now, see you tomorrow, really!)

--Ajit

- - - - -

[E&OE]

## Dear Ajid and others, I'm

Dear Ajid and others,

I'm not familiar with abstract algebra (still on my to-do list!) but I totally see the Your point. I agree that if field of physical quantities is not closed under multiplication then other structure should be formulated. One could think of physical quantities as a special structure (S) created in following way:

S consists of two elements (so it's similar to vectors) but with one element being formulated over real number field (F) and the other over physical units (let's call it PU). Now we can postulate that addition on S is defined only if Si(2)=Sj(2) -> second terms must match, then Si+Sj = [Si(1)+Sj(1), Si(2)]. while under multiplication terms are multiplied respectively. I don't know if that makes sense but that's the way I see it.

PS1. Wikipedia differentiate between scalars in mathematics and physics.

PS2. This is more philosophy discussion than engineering one (again, my opinion)

PS3. Addition of temperatures is meaningless as temperature only indicates direction of energy flow. Temperature has no meaning if there aren't at least 2 bodies. So what would be adding temperature mean? If you join two bodies they won't have added temperature altogether, but rather their difference will determine heat flux (I guess). On the other hand difference is subtraction (so we got addition here) and this quantity is used for example in Newton's Law of Cooling.

## Re: Alecsander's suggestion

Hi Alec (hope this way of calling you is OK by you), and others:

Re. Your suggestion of S = F + PU: Yes, I think that's a very neat suggestion. (And you are right in noting that a constraint also needs to be put on more operations like addition.) There is no reason why a mathematical object or a structure cannot be a composite. And, yes, the physical units is an essential part of the consideration here.

But, no, this is not a

philosophicaldiscussion. It's about the logical part ofmathematics, and aboutfoundationalissues ofmechanics. You could call it philosophy of science/mechanics, if you wish. But inasmuch as science, including engineering science, requires precision in not only data but alsoconcepts, there is a definite place for such discussions, right in the realm of science itself---not of philosophy.Yes, we engineers do usually rely on mathematicians to supply us with the most precise definitions, but if their offering falls short, it sure is our responsibility to point out our requirements to them. At least I see it that way. So, it's very much a discussion in science. (Think what a Kant, a Hume, or, for that matter, a Budhist will say by way of a discussion or an answer to this same question, and you can immediately see that this is not philosophy per say.)

Re. temperature: Epistemologically, it sure is true that concepts are formed in reference to two or more instances/objects. (A notable exception is the concept of existence, or in physics, of the physical universe taken as a whole: what it refers to is a single object, in a way.) But once a concept like temperature is formed, I think that it also is epistemologically proper to regard it as a property or attribute of just single bodies taken one at a time. A

changeof temperature would sure take two bodies, but right there we first assume that each body does have that attribute of temperature in the first place.It is heat that means energy in transit, and thus requires at least two bodies, not temperature.

Inasmuch as temperature means the level of the energy content of a (single) body, I think that we can think of adding (or subtracting) temperatures. Can you add two heights? Yes. Two potentials? Yes. If so, then on similar lines, adding (or subtracting) two temperatures also

shouldmake sense. To understand temperature as a level, it is useful to make reference to the kinetic theory (as a measure of the kinetic energy/phononic energy), but I also think that this procedure is not entirely necessary---you can define temperature "purely" in continuum terms, too, and even if defined this way, it means a level of a kind.Coming back to your suggestion, let me share some further thoughts:

Mathematicians have never cared for physical units; indeed, progressing further in their abstractions, they often don't care even for constants, ratios, or sometimes even limits (in fractal theory) or measures like lengths/areas (in topology). And, their pursuit of ideas like these certainly have been useful, too; I won't deny that. But that does not mean that they should

nevercare for physical units. It indeed is their task to ponder over thetheoreticalnecessity to have physical units, and then suggest proper logico-mathematical ways to handle it.As an aside: In earlier, better times, e.g. around 19th century, mathematics was not divorced from physical reality, and innovations like dimensional analysis were possible. The law of dimensional homogeneity (first put forth, I think, by Fourier) is usually taken as a physical law, but inasmuch as it applies to

equations---anyphysically valid equation---there also is an essentialmathematicalaspect to it; it also is, I think, a mathematical law of a kind. (But, of course, the nature of the relation between physics and mathematics is way out of scope here.) In a similar vein, mathematicians should not shirk from tackling the kind of issues that have been raised in this thread in general, and of physical units or dimensions in particular---suggesting proper mathematical structures/forms/objects to encapsulate these concerns.--Ajit

- - - - -

[E&OE]

## A scalar set is defined as a 1D vector space over real numbers

Dear Aleksander and Ajid: In the initial post, I defined a scalar set as follows

Definition. A scalar set is a one-dimensional vector space, S, over the field of real numbers, R.I followed the definition of number field and the defintion of vector space. This definition involves two sets S and R, and four binary maps:

These binary maps follow the familar rules, as listed in the definitions of vector space and number field. Each element in S is called a scalar. Each element in R is called a number.

Let u be a nonzero element in S. Each element s in S scales with u by a real number r, namely, s = ru. We call u a unit of S, and r the magnitude of s relative to the unit u. These terms parallel those in a higher dimensional vector space: scalar~vector, unit~basis, magnitude~component.

The definition is abstract, but is intended to model objects like energy, mass, charge, entropy. For example, for the scalar set of masses, 1.7kg means a mass, which is 1,7 times the mass of a particular block of metal. The mass of the particular block of metal is an element in the scalar set of masses and is chosen as a unit. 1.7kg is another element in the set.

The purpose of this definition is to understand the linear algebra of our physical world, as noted in another comment.

Is this definition the same as yours?

## Re: A scalar set is defined as a 1D vector space over real numbe

Dear Zhigang,

0. Paraphrasing Ian Fleming [^] (and changing the order of the (extended) operands of the operator phrase in question), the name is Ajit---Ajit R. Jadhav.

1. No, the definitions are different. The object that both Alecsander and I agree on would be a composite object that is neither a vector space nor derived from one.

2. Your fourth operation is an

externalbinary map [^]. This is analogous to the reason why we would like to make it acompositeobject.3. The terms do

parallelto those used in the vector spaces, but that still does not mean that it should be called avectorspace.Let me suggest a new name:

physical unitspace, or, if mathematicians feel repelled by it, the PU space. (The moniker "unit space" would make it too general---even the real number field could arguably be called that.)4. Making it a composite object

seemsto also automatically take care of the idea that there should be multiple instances of F. I have to think more on this aspect.5. Indeed, as I was writing my above replies in this thread, though I didn't mention it, I had also thought that also the vector and tensor spaces would have multiple instances---same abstract mathematical structure but different objects, instantiated according to their physical units.

Re. your recent reply to Amit on this thread here [^]. I did not know that Amit had mentioned this point (concerning different instances of the same structure) before me, because I had not read the previous thread. Thus, I got this point later than him, but, sure, also independently of him. (I came from the OO programming and ADT perspective.) I also got the point regarding the law of the dimensional homogeneity independently of him---I now realize that he does mention dimensional analysis, too. So, let me first go through the previous thread.

--Ajit

- - - - -

[E&OE]

## vector space vs. scalar set

Dear Ajit: Happy new year! I apologize for misspelling your name.

I am trying to construct a linear map between two vector spaces. f: V-->S. Here V is a 3D vector space, S is a 1D vector space, both over the same number field R. A linear combination of elements in V is an element in V. All elements in S scale with each other. f is a linear map. Being linear is crucial to address an issue raised by Amit.

I also dislike the term one-dimensional vector space, although the object S does have the same structure as the one-dimensional vector space. The phrase is too long. Calling a scalar a 1D vector is confusing.

I have proposed to call S a scalar set over the field of real numbers. Thus, we have the parallel between a vector space and scalar set:

The last pair of parallel terms are worth commenting. Because the 3D vector space V is an inner-product space, the chosen basis is usually orthonormal. A change of basis invove a charge in the orinetation of the basis, not the length of the each base vector. In changing the unit of a scalar set, however, we change the unit form one scalar to a scalar of a different magnitude.

## Re: vector space vs. scalar set

Dear Zhigang,

Happy new year to you, too! (Now, continuing in the jocular vein, let me say: apologies not accepted . More seriously: An apology here was neither necessary nor expected. )

With your noting that calling S a vector space is confusing, guess we essentially are in agreement.

Yes, the change of basis for vectors (and tensors) involves rotation whereas speaking of rotation is not meaningful for scalars because there is no direction associated with them in the first place.

--Ajit

- - - - -

[E&OE]

## Application of Vector space

Dear Prof. Suo,

I agree with your revised definitions of tensors, scalars. But as a student, I'm much more concerned about the applications of these definitions. I hope to see, for example, with the introduction of new definition of linear funtionals, i.e. linear map from a general vector space to one-dimension scalar space, the classical concepts in mechanics will be refined which will lead to a renewed derivations of the subsequent concepts.

However, although the concept of vector space is fundamental to mechanics, I found most books on Continuum Mechanics, even for the well-known book written by German mathematicians Truesdell&Noll, they didn't introduce tensors with linear maps.

Could you introduce some applications of the renewed definitions? I think, tha would be more straightforward to the audience. Thanks very much.

## The linear algebra of the world

Thank you, Jing, for an excellent question. My son Michael taught me how to use twitter yesterday. My twitter account is @zhigangsuo. He said twitter is becoming main stream. On matters of the Internet, he speaks, I listen. He was the one who gave me the initial idea that lead to the launch of iMechanica.

As a trial tweet, I listed the algebraic structure of the physical world:

Then the space of each tweet is small. I tweet again by saying We then construct linear maps.

Here I can add more details

These maps correspond to commonly used physical quantities.

Linear algebra provides a structure to generate linear maps from the following buiding blocks:

{one number field, one vector space, many scalar sets}.

A choice of basis for each scalar set gives a unit to the scalar. We fix the bases of all the scalar sets.

A choice of a basis of the vector space of positions also gives the bases for all other vectors and tensors, including force, electric field, stress...

I'll add a new tweet link to this comment.

## The trend for the development of rational mechanics

Dear Prof. Suo,

If this is the case, the physical world could have a rigorous and beautiful mathematical structure. We could use the revised definitions to analyze and redefine classical mechanical concepts. Hence, in this aspect, it seems to deserve a lot of work??

It's pity that at present I don't have direct access to you comments on twitter since I'm in Zhejiang University, China. Just as I suggested in the e-mail that I sent to you, I think imechanica could consider to incoporate latex in order to enable its users to edit math formulas directly. Technically,it seems not to hard for a website to be compatible with Latex.

With the help of compatibility with latex , maybe we can describe the applications of the linear maps in detail.

## Re: The trend for the development of rational mechanics

Dear Jing: The adoption of scalar sets cleans up our practice, but it does not introduce any new mechancs. We are looking into the use of Latex on iMechanica.

## Not quite clear where the problem is....

For "mass", for example, is not a "scalar"; mass, volume, etc, are measures defined on a measurable space (the real line, a body in 3D, etc).

Happy New Year !

## Scalars and Numbers

Dear Zhigang,

If we are going to use numbers for scalars then what can we call

individual elements of tensors they are not scalars because they change under

coordinate trasformation we cannot call them numbers as well because we are

going to substitute them for scalars.

Regards

Mohsen