# Tensors in the context of engineering mechanics: can they be explained in an intuitive way?

I've spent a few weeks scouring the internet for a an explanation of tensors in the context of engineering mechanics. You know, the ones every engineering student know and love (stress, strain, etc.). But I cannot find any explanations of tensors without running into abstract formalisms like "homomorphisms" and "inner product spaces". I'm not looking for an explanation of tensors using abstract algebra or infinite, generalized vector spaces. I just want some clarification on what they actually mean and are doing in the nice 3D, Euclidean space, especially in the context of mechanics. There are a few questions that have been bugging me that I'm hoping all you smart people here can answer:

1. What's the difference between a linear transformation and a tensor? Somehow they can both be represented by a $3\times 3$ matrix, but they do different things when acting on a vector? Like the columns of a $3 \times 3$ matrix of a linear transformation tell you where the basis vectors end up, but the same columns of a tensor don't represent basis vectors at all?

2. Furthermore, a linear transformation transforms all of space but a tensor is defined at every point in space? Does a tensor act on vectors the same way as linear transformations do?

3. What is the difference between a tensor product, dyadic product, and outer product and why are engineering tensors like the Cauchy stress built from the tensor product of two vectors (i.e. traction vector and normal vector)?

4. Is it true that scalars and vectors are just $0^\mathrm{th}$ order and $1^\mathrm{st}$ order tensors, respectively? How are all these things related to each other?

5. What topics and/or subtopics of linear algebra are essential to grasp the essence of tensors in the context of physics and engineering? Are they really just objects that act on vectors to produce other vectors (or numbers) or are they something more?

I have plenty more questions, but I figure the answers to these could already be enough to fill a whole textbook. Just to note, I have already searched Math.StackExchange for tensors but haven't found any explanations that make sense to me yet.

Thanks!

• If you are looking at expressions which only involve vectors from finite dimensional spaces, you can substitute the word tensor for multilinear map without running into any problems. Multilinear maps aren't too bad. I have a feeling that this is what is usually done in 'applied' contexts. – Bernard W Jun 10 '17 at 1:49
• Tensors are how you multiply in linear algebra. Every other 'product' you've encountered can be expressed as a two-step calculation: take the tensor product, then apply a linear transformation. – user14972 Jun 10 '17 at 2:00
• It is also important to distinguish vectors and covectors (differentials). The former describe the small change in position (i.e. displacement), and the latter describe the small change of quantity along that displacement. Then linear maps describe how one vector transforms to other (thus comprising (1,1)-tensors), and bilinear tensors describe how a quantity changes along two independent displacements (thus comprising (0,2)-tensors). – Sangchul Lee Jun 10 '17 at 5:29
• By differentials do you mean the "del operator"? Like in the continuity equation for incompressible flows, the del operator is the "covector" and the velocity would be the vector? Also, I somewhat understand the tensor notation (n,m) as being the number of covariant and contravariant components correct? Why then does a (2,2) 4th order elasticity tensor operate on a second order (0,2) strain tensor? Is there some rule for what order tensor a higher order tensor can operate on? – RobbieFresh Jun 10 '17 at 6:19
• For what it's worth, I've been trying to understand tensors on and off for over 25 years now. I have a BS in mathematics but that hasn't helped at all. My dad has a BS in EE and he seems to understand but I don't understand his explanations. I've read about tensors in countless books and web sites, but it always seems like the explanations skip a step or five. I hope someone provides an answer here that helps, but at this point I'm not optimistic. – Todd Wilcox Jun 10 '17 at 6:42

## 9 Answers

Although its a bit lengthy subject but i'll try to give you the exact mathematical information in as short an introduction as possible. Let's start directly with ....

DEFINITION OF TENSOR --- A ($p$,$q$)-tensor $T$ (with p and q integers) is a multilinear transformation $$T:\underbrace{V^*\times V^*\times\dots V^*}_{p\text{ times}}\times\underbrace{V\times V\times\dots V}_{q\text{ times}}\to\mathbb R$$ where $V$ is a vector space, $V^*$ is its dual vector space and $\mathbb R$ is the set of real numbers. The integer $p+q$ is the rank of the tensor.

Example, A (1,1) tensor is a multilinear transformation, $T:V^*\times V\to \mathbb R$. Using the same information we can construct an object $T:V\to V$ as shown later. We recognise this as a simple linear trasnformation of vectors, represented by a matrix. Hence a matrix is a (1,1) tensor.

What does that mean? It means that a Tensor takes p covectors and q vectors and converts them multilinearly to a real number. The main thing to understand here is the difference between a vector (member of a vector space) and a covector (member of the dual vector space). If you already know about this, you can skip this section. A vector is defined as a member of a vector space which itself is defined as a set with a addition and scalar multiplication following certain axioms.* A covector is defined as follows:

Definition (Dual space) The set of all linear transformations $\boldsymbol\omega:V\to\mathbb R$ is called the dual vector space and denoted by $V^*$. The members of the dual vector space are called covectors.

Theorem (without proof) The dual of a dual space of a finite dimesnional vector space $V$ is $V$. i.e., $$(V^*)^*=V$$

We usually denote vectors by $\boldsymbol v$ and covectors by $\boldsymbol \omega$. Also by convention, vectors have indices up and covectors have indices down. (The indices representing coordinates)

$$\boldsymbol{v}=v^i\boldsymbol e_i, \quad\boldsymbol\omega=\omega_i \boldsymbol \epsilon^i$$ Here $e^i$ are the basis vectors. Whenever you see an index up and the same index down you have to sum over that index, like in the above equations, ($\boldsymbol v = \sum_i v^i \boldsymbol e_i$).

Notice that a covector is a (0,1) tensor and a real number is a (0,0) tensor. This can bee seen readily from the definition. We can show that a vector is a (1,0) tensor, using the above mentioned theorem, although it is not very obvious.

How to represent tensors in a basis? Let's say we want to represent a (1,2)-tensor in a given basis. We apply it to an arbitrary input: $$T(\boldsymbol \omega, \boldsymbol v, \boldsymbol w)=T(\omega_a \boldsymbol\epsilon^a,v^b\boldsymbol e_b,w^c\boldsymbol e_c)=\omega_a v^b w^c T(\boldsymbol\epsilon^a,\boldsymbol e_b,\boldsymbol e_c)$$

Here the objects $T(\boldsymbol\epsilon^a,\boldsymbol e_b,\boldsymbol e_c)$ are simply real numbers and they can be labelled as $T^a_{bc}$. Hence a tensor can be represented by a set of $(\dim V)^{p+q}$ numbers. A tensor T of type (p,q) and rank (p+q) has p indices up and q indices down.

Theorem In the definition mentioned above we can transfer $V$ or $V^*$ to the other side by removing or adding a $*$ to V.

Consider a (1,1) tensor. It is an object $T^a_b$ which takes a vector and a covector and converts it to a real number, like so: $$T:V^*\times V\to\mathbb R$$ $$T^a_b \omega_a v^b = r, \,\, r\in\mathbb R.$$ However the same object can be used like so: $$T^a_bv^b = w^a,$$ here it has converted a vector to another, $$T:V\to V.$$

A matrix can do the same things, just think row vector = covector, column vector = vector, and matrix (NxN) = Tensor. Then: covector * Matrix * vector = real number, while Matrix * vector = vector. The entries of the matrix are precisely the numbers $T^a_b$.

Hence, a matrix is simply a (1,1) tensor. However the notation of matrices requires us to use it in some particular ways, like you can do covector * Matrix * vector but not Matrix * vector * covector.

Footnotes:

*the axioms are CANI ADDU - For addition: Commutativity, Associativity, Neutral element (0 vector) exists, Inverse elements exist. For scalar multiplication and addition: Associativity, two Distributivities, Unit element (1*v=v)

Please note this is only a mathematical introduction to this subject. I have not answered all your questions but this is an attempt to make things precise so you dont learn something wrong and probably you will now be able to answer thos questions yourself.

• @mehrdad sorry editing mistake . . Right now i'm on a mobile, i'll correct it tommorow. Right now you can just ignore that paragraph as its explained in the end. – Kartik Jun 10 '17 at 15:47

I will try to answer only part of your questions.

A rank $k$ covariant tensor on $\Bbb R^3$, or simply $k$-tensor, is a multilinear function $T:(\Bbb R^3)^k\to\Bbb R$, that is $T(v_1,...,cu_i+v_i,...,v_k)=c(T(v_1,...,u_i,...,v_k))+T(v_1,...,v_i,...,v_k)$ for each $i=1,...,k$. The set of all $k$-tensors form a vector space with addition and scalar multiplication defined as $(T_1+T_2)(x):=T_1(x)+T_2(x)$ and $(cT_1)(x):=c(T_1(x))$. This vector space is known to have dimension $3^k$.

When $k=2$, the dimension of this vector space equals the dimension of space of all $3\times 3$ matrices, so that we can represent a $2$-tensor by a $3\times 3$ matrix. While it is represented as a matrix, it does not mean that it can act on a vector. Not all tensors have to be able to act on a vector. It highly depends on the type of tensor. I will explain "type of tensor" later.

A rank $k$ contravariant tensor is a multilinear function $T:((\Bbb R^3)^*)^k\to\Bbb R$. If you know that double dual of a finite-dimensional vector space is canonically isomorphic to the original space, then you should know that a rank $1$ contravariant tensor can be represented by a usual vector in $\Bbb R^3$.

A type $(p,q)$ tensor (or $(q,p)$, this depends on author's choice, but we stick to $(p,q)$ here) is a multilinear function $T:((\Bbb R^3)^*)^p\times(\Bbb R^3)^q\to\Bbb R$.

Each linear transformation from $\Bbb R^3$ to itself can be regarded as a type $(1,1)$ tensor. The reason is that for a $3\times 3$ matrix M representing a linear transformation, we define a bilinear function $T_M:(\Bbb R^3)^*\times\Bbb R^3\to\Bbb R$ by $T_M(f,v):=f(Mv)$. This draws a connection between linear transformations and type $(1,1)$ tensors. This type of tensor may act on a vector, but others may not necessarily so.

A type $(1,0)$ tensor is simply a vector, while a type $(0,0)$ tensor, according to Wikipedia, is simply a scalar, and a type $(0,1)$ tensor is simply a linear functional. I am not sure why the second one is the case, most probably a convention. As you see, type $(1,0)$ tensor cannot really act on a vector, but type $(0,1)$ can. In fact, every type $(p,1)$ tensor can act on a vector to give a $(p,0)$ tensor, i.e. a rank $p$ contravariant tensor.

When you see that a tensor is defined everywhere in space, it is actually not a tensor, but is a tensor field, which is a function that assigns to each point a tensor.

As on Wikipedia, there is no difference between tensor product, dyadic product and outer product.

These are good questions. I'll try to keep the maths to a minimum, below.

Zero-th question: why do we need tensors at all?

A lot of things in engineering and physics can be represented by a function (which is nothing more than a number, or a magnitude) which takes a different value at different points in space – think of the pressure at different points in a room. A lot of other things can be represented by a magnitude and a direction – think of the electric field at different points in a room. A few things are best represented by a quantity which involves a magnitude and two directions. The first lot of things are represented by a scalar field, the second by a vector field, and the third lot by (rank 2) tensors. I can't think of any physical quantities which are modelled by rank-3 tensors.

The air pressure in a room is a scalar, $p$ (we're thinking of this as a field, so there's an implicit dependence on position, $p(\mathbf r)$). If you want to know the pressure at a particular point, your answer is: ‘it's $p$’. Simple.

Now think of the electrostatic force $\mathbf F$ on a charged particle, which is moved through a displacement $\mathbf s$ (remember $\mathbf F$ is a field, so it takes a different value at different points). How much work is done during this displacement? To answer this, think of $\mathbf F$ as a function which takes the displacement as an argument: the work done is $\mathbf F(\mathbf s)$. That's a long way of giving the answer you were doubtless about to give, namely the inner product of the two vectors $\mathbf F\cdot\mathbf s$.

Now think of the stress tensor $\sigma$. Given a surface within the body with normal $\mathbf n$ and a unit reference direction $\mathbf e$, the magnitude of the stress in the direction $\mathbf e$, when you're considering a surface $\mathbf n$, is $\sigma(\mathbf n,\mathbf e)$ – a number. If we postpone thinking about the reference direction $\mathbf e$, and only apply one of the two arguments, we get $\sigma(\mathbf n, \cdot)$. That's a thing which is waiting for a single further vector argument: you may be more used to thinking of that as the stress vector, $\mathbf T^{(n)}$: the number $\sigma(\mathbf n,\mathbf e_x)$ is just $\mathbf T^{(n)}(\mathbf e_x) = \mathbf T^{(n)}\cdot\mathbf e_x = T^{(n)}_x$ (I may have slightly garbled the definition of $\sigma$ above; that's not important).

The short version:

• A rank-0 tensor is a scalar: it's a field with zero vector arguments.
• A rank-1 tensor is a vector: it's a field with one vector argument, and the way that that field acts on its argument is what we are more used to calling the inner product.
• A rank-2 tensor is what is commonly called just ‘a tensor’; it has two vector arguments. If you give it only one argument, then you're left with a thing which has a single remaining vector argument, such as $\mathbf T^{(n)} = \sigma(\mathbf n,\cdot)$.

The above is true for flat euclidean space – that is, the space of our ordinary experience. If you want to handle these mathematical ideas ‘properly’, or in odd coordinates, or in non-flat spaces (eg, general relativity), then you have to worry about vectors versus co-vectors, metrics, inner products, contractions, and observe a couple more distinctions which I've glossed over here, but the core intuitions are as above.

So, returning to your questions...

1. What's the difference between a linear transformation and a tensor?

They're very different, but unfortunately they look very similar, because they're both represented by a matrix of numbers.

If you ask for the components of a vector, the answer is a set of numbers $\mathbf n=(n_x,n_y,n_z)$ (three of them, in 3D) with respect to a particular set of coordinate axes. If you change your mind about the axes, then the same vector $\mathbf n$ will have different components $\mathbf n=(n_x',n_y',n_z')$, which are systematically related to $(n_x,n_y,n_z)$ by the change-of-basis linear transformation matrix.

Exactly analogously, if you ask for the components of a (rank 2) tensor, then the answer is a $3\times3$ matrix of numbers, again with respect to a particular set of axes. If you change your mind about the axes, the components are a different matrix of numbers, again systematically related to the original set via (two applications of) the transformation matrix.

1. Does a tensor act on vectors the same way as linear transformations do?

See 1: no, a tensor is a very different thing from a linear transformation matrix.

But also see point zero: you can use a (rank 2) tensor to turn one vector into another one. For example, $\mathbf T^{(n)}=\sigma(\mathbf n,\cdot)$ effectively relates the vector $\mathbf n$ and the vector $\mathbf T^{(n)}$.

1. What is the difference between a tensor product, dyadic product, and outer product and why are engineering tensors like the Cauchy stress built from the tensor product of two vectors (i.e. traction vector and normal vector)?

The tensor/outer/dyadic product (I hadn't heard the last name before!) are different names for the same thing (I expect a mathematician would quibble about this in full generality, but let's not worry about them).

The outer product is one way of making a rank-2 tensor from a couple of handy rank-1 tensors (ie, vectors). Any tensor can be written as a sum of outer products.

1. Is it true that scalars and vectors are just 0th order and 1st order tensors, respectively?

Yes.

1. What topics and/or subtopics of linear algebra are essential to grasp the essence of tensors in the context of physics and engineering?

If you're going to study linear algebra in any mathematical context, then tensors are going to be at the beginning of that. In a more applied context, if you have a clear idea of the relationship between a vector, a set of basis vectors, and the vector's components with respect to that basis, then you're off to a good start.

In my experience with students, the idea that ‘a (rank 2) tensor models a physical quantity which depends on two directions’ is a bit of an aha! moment.

Vectors need one subscript. $V_x, V_y, V_z$ are all components of the vector $V$. Tensors may have more than one. $V_{xy}$ is an example of a component of a rank $2$ tensor. For example, the stress on a face may have a tensor representation, $V_{xy}$ can represent the shear stress on the $x$ face in the $y$ direction.

Each vector component has a unit vector basis, each tensor component has multiple unit vector bases. From this you may see why a vector is actually a rank $1$ tensor, a matrix is a rank $2$ tensor, and so on.

• So what's the difference between a linear transformation and a rank 2 tensor? Nothing? And what about tensor components that use superscripts for some indices (covariant versus contravarisnt, right?) – Todd Wilcox Jun 10 '17 at 6:32
• Pretty sure "rank-1 tensor" is not synonymous with "vector", and rank-2 with matrix, etc. – user541686 Jun 10 '17 at 8:00

Preface: As an old student of Physics and someone trying to re-grasp Applie Math, I will admit my own notion of tensors is shaky at best as I haven't been exposed to a text yet that has given a decent definition of "tensor". But in my exposure so far to tensors, I do have a slight bit of an intuitive notion of what they are (further buttressed by Wikipedia). I can't answer all your questions, but maybe I can help with my shoddy explanation to give a better sense and I invite anyone else with any constructive criticisms to comment as I would like to be on firmer ground with this myself....

1) I believe that a linear transformation can be considered a "subset" of a tensor in a way. From what I've found, tensors can represent linear transformations, but they aren't restricted to being linear transformations for just vectors in physical scenarios. From what I gather, linear transformations are a specific case (mapping one vector space to another) of mapping just plain mathematical objects in one space to another (which tensors do in the general case). To put this in more physical terms, something like the rotation tensor maps a point in a coordinate system to a new point in a different coordinate system that may have resulted via multiple transformations.

2) Don't have a great enough handle to take a stab at this one.

3) Same as 2

4) From the literature I've read, this is indeed the case. The relation comes from tensors needing multiple degrees to designate one single component of the whole tensor. Ever notice that vectors usually are designated by one index. Let $\beta = \{v_1 , v_2, ... , v_n\}$ be a basis for some vector space V (it may help to think of it as the basis for the n-th dimensional Cartesian Space ${\Re}^n$. Since any vector in V is a linear combination of the vectors in the basis, only one index is needed to specify the constants that correspond to the elements in the basis for v (i.e. $v = {x_i}{v_i}$ where Einstein Summation notation is used here with the index i ranging from 1 to n). Note that with tensors however, in order to specify a component of a tensor, you need more than one index to get to that component. Here we see the "generalization" aspect of it again.

5) To be honest, I'm not sure. I have come across tensors very seldom in my study of Physics (please note I have yet to get to an advanced EM, or intro Quantum Mech class yet).

I hope that helps a bit. If there's something mirky about one of my explanations, let me know, and I'll do my best to clarify.

I also found this highly confounding when first starting the continuum mechanics (CM). I think it is easiest to understand tensors using the "classical" approach from old Riemannian geometry, which is closer to how they are used in CM. Let me just clear a few things up first.

The CM tensors are treated sort of like a special case of the tensors from classical physics and Riemannian geometry.

Note: in other fields, especially computer science, a tensor is defined as a multidimensional array of numbers; but in physics and mathematics, these are usually arrays of functions that change over space (or spacetime). These are more accurately called tensor fields (a generalization of vector fields), but let's just call them tensors.

First fact: although you can always write a tensor as an array of numbers, the actual numbers present depend on the coordinate system you choose. This is quite similar to matrices, of course. However, there are certain invariants of tensor fields (e.g. the trace of stress tensor), that do not change as the coordinate system changes.

Second fact: you can index into the arrays in two different ways, covariant (lower indices) or contravariant (upper indices). E.g. $R_{ijk}^\ell$ is type-(1,3), meaning 1 contravariant index and 3 covariant indices. The stress tensor is type-(0,2), for instance, although it should probably really be considered type-(1,1). You can raise or lower indices with the metric tensor and its inverse, but this is not so important in CM (see the link just above).

Definition: how do we know a given array of functions defines a tensor? The classic definition is that, given a tensor $T_{\ell_1,\ldots,\ell_n}^{i_1,\ldots,i_m}$, and two coordinate systems $x^i$ and $y^i$, with Jacobian $J_{j}^i=\partial y^i/\partial x^j$. $$\hat{T}_{\ell_1',\ldots,\ell_n'}^{i_1',\ldots,i_m'} = J_{i_1}^{i_1'}\ldots J_{i_m}^{i_m'} {T}_{\ell_1,\ldots,\ell_n}^{i_1,\ldots,i_m} (J^{-1})_{\ell_1'}^{\ell_1}\ldots (J^{-1})_{\ell_m'}^{\ell_m}$$ using the Einstein summation convention (where seeing the same index twice, one lower and one upper, means summing over all possible values of those indices) to transform from $x$ coordinates to $y$ coordinates.

So, here the defining property of a tensor is how it transforms when the coordinates change. Note that the Jacobian $J$ is not a tensor by this definition, even though we write it like one!

Intuition: a tensor represents an entity that captures information that is a geometric invariant, i.e. does not change as the coordinate system changes. For instance, the Jacobian obviously depends on the coordinate system, so it is not a tensor. But the stress or strain tensors can be used to measure or compute physical quantities that will remain the same across coordinate systems (e.g. the strain tensor can measure changes in length, which is a scalar invariant).

Let's go back to CM now. Usually, in CM, we tend to ignore the type of the tensor. Why? Because normally one changes the index types of a tensor using the metric tensor $g_{ab}$ and its inverse $g^{\alpha\beta}$, e.g. $\sigma_j^i=\sigma_{ik}g^{kj}$. But often in CM it's assumed that $g=I$ (i.e. $g_{ij}=\delta_{ij}$). So even though the stress tensor is written $\sigma_{ij}$, it is actually often used as $\sigma_i^j$, e.g. to take its trace. This is a source of great confusion.

In general, one can just assume that most of the CM tensors are type-(1,1), and in $\mathbb{R}^3$, so they can be written as $3\times 3$ arrays. In these cases, they can be treated as linear maps OR as bilinear forms (under the assumptions about index movement above)! One exception is the fourth-order elasticity tensor.

(And, yes, this is indeed horrifying and confusing from the notational point of view)

Ok, now for your actual questions.

What's the difference between a linear transformation and a tensor? Somehow they can both be represented by a 3×3 matrix, but they do different things when acting on a vector? Like the columns of a 3×3 matrix of a linear transformation tell you where the basis vectors end up, but the same columns of a tensor don't represent basis vectors at all?

Some tensors are linear transformations i.e. can be written as matrices in the classical way. As noted above, this is often the case for most CM tensors. For instance, the Cauchy stress tensor $\sigma(x)$ takes in a direction $v(x)$ at a specific point $x$, and outputs a new vector $T(x)=\sigma(x)v(x)$, which is the traction (using matrix notation rather than tensor indices). So, here, indeed, it is a linear transformation.

Furthermore, a linear transformation transforms all of space but a tensor is defined at every point in space? Does a tensor act on vectors the same way as linear transformations do?

As mentioned above, CM usually refers to tensor fields, which you can think of as matrix fields (given that you specify the coordinate system). The second question is harder, but for CM specifically, generally, yes.

What is the difference between a tensor product, dyadic product, and outer product and why are engineering tensors like the Cauchy stress built from the tensor product of two vectors (i.e. traction vector and normal vector)?

They are the same. See here and here.

As for the second part, I'm not sure what you mean. One often writes $T^{(n)}_j := \sigma_{ij} n_i$ (in the unfortunate CM index notation) to define the $\sigma$. Also, I could be wrong, but if the stress tensor was always defined by a tensor product of two vectors, then it would always be rank 1 and have $\det(\sigma)=0$, which is not true (indeed the third invariant is $I_3(\sigma)=\det(\sigma)$)!

Is it true that scalars and vectors are just 0th order and 1st order tensors, respectively? How are all these things related to each other?

Yes, scalars are order $0$ and vectors are type-$(1,0)$. Intuitively, note that a scalar is automatically a geometric invariant (e.g. surface area), and hence a tensor.

(Aside that is not necessary for CM: note that this index notation makes understanding covectors easier. If $v^i$ is a vector field, then $s=v^ia_i$ is a scalar [remember the Einstein convention!]. So any type-$(0,1)$ tensor [i.e. covector] can be thought of as a way to map a vector $v$ to a scalar $s$ using an array of numbers $a_j$)

What topics and/or subtopics of linear algebra are essential to grasp the essence of tensors in the context of physics and engineering? Are they really just objects that act on vectors to produce other vectors (or numbers) or are they something more?

Tensors are a subfield of an area called multilinear algebra, which might be worth studying. However, for CM, this is unnecessary. My suggestion is looking at "classical" (i.e. using index notation) tensor theory and Riemannian geometry, e.g. looking at Lovelock and Rund's Tensors, Differential Forms, and Variational Principles. On the other hand, CM has frankly developed its own confusing notation that doesn't match any other field (that I know of) and uses them in their own way, ignoring all the nuances of other areas of physics, such as contravariant vs covariant or covariant derivatives, (where curvilinear spaces are more common) or the more powerful abstractions now used in mathematics. So, just read the continuum mechanics literature instead.

Whew, well, normally I don't answer when there are plenty of other answers present, but here I felt you might like an answer focusing a bit more on CM and its idiosyncrasies :)

This is an inescapable question for almost every student of continuum mechanics with engineering background. After teaching these subjects at undergrad and grad level for several years I came to the following conclusion:

• for practically minded students it is better to say, according to Gurtin and many others, that tensors "are" linear transformations;

• for more conceptually minded students, it is better to avoid shortcuts and adopt, without discounts, the only definition that include all other ones as a special cases, namely: "second order tensors over a vector space $V$ are elements of a tensor product of vector spaces $V \otimes V$". Of course this requires some effort to learn the definition of tensor product of vector spaces (see e.g. Birkhoff-Mac Lane "Algebra" p. 319), which is hard for engineering students, but this effort is the worth price to pay to hope to really understand the problem.

The crucial aspect is then to emphasize that we are dealing with "a tensor product" instead of "the tensor product". In my opinion, most of the confusion about tensors for engineering students, comes from the failure to realize that spaces of tensors are defined by means of a universal property hence up to an isomorphism. Namely, there is no a single entity that deserves to be called $V \otimes V$ there are infinitely many, but all of them are linearly isomorphic.

The second crucial aspect to emphasize is that a tensor product is not defined by a vector space alone but by a vector space $Z$ plus a bilinear map $\otimes : V \times V \rightarrow Z$ which defines the meaning of simple tensors. It is the pair $(Z, \otimes)$ that deserves to be denoted $V \otimes V$. Therefore the same vector space can determine different tensor products with different definition of simple tensors.

This is the only way to understand that questions like "what is the difference between a linear transformation and a tensor ?" have little meaning.

In fact, ignoring variance issues that play little role at this level (for example by identifying $V^*=V$) the vector space $Lin(V,V)$ of linear maps $V \rightarrow V$ is a tensor product $V \otimes V$ hence linear transformations are tensors wihtout quotation marks.

At the same time (again to within variance issues):

• the vector space $R^{n \times n}$ of square matrices is a tensor product $V \otimes V$ too
• the vector space of bilinear maps $BiLin(V,V;R)$ is a tensor product $V \otimes V$ too

and so on for any other definition which can be found in various textbooks. Clearly this applies immediately to tensors of any order and variance.

In conclusion, once one grasps that spaces of tensors are defined up to isomorphism, it should be comfortable with the fact that it is equally true to say that a second order tensor is a linear transformation, a matrix or a bilinear tranformation. These are all elements of a tensor product $V \otimes V$ and each tensor product is linearly isomorphic to each other, therefore it is only a matter of taste to chose one definition or another.

A tensor is a geometric object independent of any coordinate system. A vector is such an object.

Take the gradient of a vector field. Not the divergence. For example, the velocity field of a fluid. Consider the velocity at a point very close to the point where you differentiated. Why does multiplying the 3x3 matrix of partial derivatives by the displacement provide a good approximation of the difference of the velocity at the tail and head of the displacement?

Don't confuse a vector space with physical space.

The operation of a tensor on a vector is linear. So multiplying a vector by the matrix representation of a rank 2 tensor is a linear transformation.

In linear algebra, all vectors more-or-less radiate from the origin. In physics and differential geometry, vectors have "points of application" in the "real" world.

Find a discussion of the stress tensor that is heavy on drawings, and light on mathematical jargon.

• You're right, but "real" space is a special kind of vector space correct? And in this space, vectors are typically described by a tuple of 3 real numbers (e.g. position, velocity, etc.). The tensors that we study (e.g. stress, strain, deformation gradient, etc.) "operate" on these vectors in some mathematical way, independent of their physical meaning. I guess what I'm really looking for is what these tensors are actually doing and represent from this mathematical standpoint and why we describe them in the way we typically learn in physics classes. – RobbieFresh Jun 10 '17 at 6:12
• "Real" space is probably better thought of as a manifold than as a vector space. At every point in the manifold there is a "tangent" vector space. – Steven Thomas Hatton Jun 10 '17 at 7:46
• Work the example of the gradient of a velocity field. – Steven Thomas Hatton Jun 10 '17 at 10:58

Tensors can be used to represent things. For example shapes in geometry or orientation in coordinate systems. Stress, strain, conductivity and many other things are properties which are useful to be able to encode orientation.

For example:

1. a rank 1 tensor can encode the plane of insulation or the rod of electrification. Typical example : a thin insulated metallic wire.

2. a rank 2 tensor in 3 Dimensions can encode the orientation of the local surface of a 3D object.

Then these representations can sometimes make it easier to manage and define functions or operators for what we want to calculate.

• -1 I don't think this really answers the question. Tensors can be used to represent things, but so can matrices. Or sets. Or strings & alphabets. What they can represent is hardly distinguishing or useful information. – user541686 Jun 10 '17 at 7:58
• The asker seemed to want a more practically oriented answer towards explanation for usage in engineering and that is what I tried to give. – mathreadler Jun 10 '17 at 8:04
• Didn't the asker start by saying that tensors represent stress, strain, etc.? What gave you the impression he didn't already know that tensors can be used to represent things? – user541686 Jun 10 '17 at 8:05
• What? I'm evaluating your answer, that's how this network of sites works. The community evaluates answers, voting them up or down accordingly. – user541686 Jun 10 '17 at 8:08
• I don't understand what you're saying... – user541686 Jun 10 '17 at 8:17