What is the real notion/feel of a tensor quantity?

Question

I have been just introduced to the term tensor while studying Rotational Dynamics, particularly about Inertia. But I just don't get a clear line separating vector from a tensor. What does someone mean when stating a particular physical quantity as a tensor quantity?

Answer 1

Since you're question is specifically in the context of Rotational Dynamics, I'll offer an answer that is specific to that context.

In that context, a vector $v$ is something whose components $v_i$ transform under rotations like $$ v_i\mapsto\sum_j R_{ij}v_j $$ where $R$ is a rotation matrix. A tensor (such as the moment-of-inertia tensor) is something whose components $I_{ij}$ transform under rotations like $$ I_{ij}\mapsto\sum_{k,\ell} R_{ik}R_{j\ell}I_{k\ell}. $$ More specifically, this is a $2$-index tensor. A vector is a $1$-index tensor. In general, an $N$-index tensor is a quantity with $N$ indices (of course) that transforms under rotations according to the pattern illistrated above, with one $R$ per index.

An easy example of a tensor with $3$ indices is $T_{ijk}=a_i b_j c_k$, where $a,b,c$ are vectors. This automatically transforms correctly under rotations because of the way $a,b,c$ transform.

Tensors often have special symmetries. For example, the moment of inertia tensor $I_{ij}$ is symmetric: $I_{ij}=I_{ji}$.

Such symmetries do not affect the general rule for how the tensor transforms under rotations, but they can lead to interesting coincidences. For example, the appropriate generalization of angular momentum to $D$-dimensional space is represented by an antisymmetric $2$-index tensor, $L_{ij}=-L_{ji}$. Because of antisymmetry, this has $D(D-1)/2$ independent components, namely those with $i<j$. (Those with $j>i$ are determined by antisymmetry, and those with $i=j$ are zero by antisymmetry.) In the physically-relevant case $D=3$, $L_{ij}$ happens to have $3$ components, like a vector. Even more interestingly (and less trivially), those three components turn out to transform like the three components of a vector under rotations, even though the general rule for transforming a two-index tensor involves two rotation matrices instead of just one! This is why most formulations of rotational dynamics represent things like angular momentum, angular velocity, and torque as though they were vectors, even though they would more properly be represented as two-index antisymmetric tensors.

Actually, even in $D=3$, there is one clear symptom that quantities like angular velocity (etc) are not really vectors: they transform like vectors when $R$ is an ordinary rotation, but not when $R$ is a reflection. Terms like "axial vector" or "pseudovector" refer to this situation. The direction of angular velocity cannot be reversed by a mirror reflection (because a two-index tensor transforms with two factors of $R$, so the minus signs cancel), but the direction of a legitimate vector (one-index tensor) can be reversed by a mirror reflection.

A more subtle symptom that angular momentum should be represented as a two-index antisymmetric tensor (instead of a vector = one-index tensor) is the way it is constructed. For example, an object with momentum $p$ (a vector) on the end of a massless rotating rod of length $r$ (a vector) is usually written as $L=r\times p$, or something like that (the sign is a matter of convention). The cross-product only makes sense in three-dimensional space. It really should be written $L_{ij}=r_i p_j-r_j p_i$ instead. In three-dimensional space, this has the same list of independent components as the cross-product, but the two-index tensor representation makes sense in any number of dimensions.

In other contexts, such as relativity (which is addressed in the possible-duplicate posts that were cited in the comments), the definition is analogous but with the group of Lorentz transformations (in special relativity) or all coordinate transformations (in general relativity) in place of the group of rotations.

Answer 2

In a way there isn't a great deal that seperates a tensor from a vector. It's essentially a higher-dimensional generalisation where instead of working in a higher dimensional space we work with a higher dimensional 'arrow'.

This was the attitude of Hermann Grassmann that introduced tensors though he did not call them that around the 1840s. In fact, although he thought about them geometrically - in essence extending an idea of his father, Johann Grassmann, he also understood that it was possible to develop them - as he put it, 'arithmetically' - that is algebraically and it was this rephrasing or conceptual translation that allowed him to think through the possibilities of higher dimensional vector spaces - up to an infinite number of dimensions. Which of course is the precursor to the notion of Hilbert space in QM.

Now, there are commonly three ways that vectors are introduced: geometrically, algebraically & transformationally. The first is common from classical mechanics when we identify a force exerted in a certain direction as a vector. The second is common in Quantum Mechanics where it's introduced as a system of axioms similar to that of the reals. For example, the associativity rule is obeyed.

The transformational approach really only appears in GR because it's the simplest case of a vector. And for the same reason, in continuum mechanics where the notion of a stress-energy tensor was taken from. It's usually simply given without much motivation. However, the physical idea behind this approach is this: although geometrically we can 'see' a vector when it comes to measure it we give it in components. However, if someone handed me a triple of numbers saying that it came from a vector that he measured in his apparatus - whatever it is. Now, how would I know that he's speaking the truth? For all I know he just picked three random numbers, and such a triple obviously does not come from a vector. We need a truth procedure to verify that these numbers came from a vector.

Now vectors don't change when we reorient our measuring apparatus but the measured numbers are different and there is a relationship between them. So we ask him to reorient his apparatus ten times and tell us what these new coordinates are. We can then check whether this coordinates obey the right transformation rule. If they do we can plausibly say that these coordinates represent a vector. In fact, we say it is a contravariant vector because the numbers change 'contravariantly' to how we change the orientation of the apparatus.

There's a subtle point here which it is the orientation of the apparatus that we change and not that of space itself. These are called passive and active coordinate changes and it's the former we require. This makes sense physically, as we can change the orientation of the apparatus but not that of space.

Now it's possible to derive the transformation law for vectors, but I won't do it here.

Now in GR tensors are often introduced with the transformational law but they need not be. Like vectors, we can introduce them geometrically and algebraically - as Grassmann did - and then derive the transformational law that they satisfy.

Algebraically, a 2d tensor in the directions of vectors $u$ and $v$ is:

$u \otimes v$

Grassmann would have introduced this as taking a point and drawing it in the direction $u$ for a certain length (in fact the length of vector $u$) hence we get a directed line in space. He then takes this whole line and draws it in the direction $v$ for the length of $v$ and so gets a parallelogram orientated in space. A tensor, then, is represented by this orientated parallelogram.

Now, vectors have an additive law and a scalar multiication law; likewise we need similar for tensors. The axioms aren't difficult and are what you might expect if you reflect on Grassmanns geometric analogy. They are:

$\alpha(u \otimes v) = (\alpha u)\otimes v = u \otimes (\alpha v)$

Also

$(u + v \otimes w) = u \otimes w + v \otimes w$

And

$(u\otimes v+w) = u \otimes v + u \otimes w$

Now if we take a basis $(e_i)$ in the space then we can write

$u = u^i.e_i$ and $v=v^j.e_j$

Hence our tensor $T$ can be written in a base as

$T:=u\otimes v$ $=u^i.e_i \otimes v^i.e_j$ $=u^i.v^j.e_i \otimes e_j$ $T^ij.e_ij$

Where I have defined $T^ij:=u^i.v^j$ and $e_ij := e_i \otimes e_j$. Now, this looks very much like how we write out a vector in components in a basis. And in fact, $(e_ij)$ is a basis for the 2d tensor space.

To finish: The transformation law is derived in a similar way as the vector case. We can define 3d, 4d tensors and so on. We can do all this dually. Then we can lift all this to vector bundles. Now the tangent bundle over a manifold is a vector bundle so all of this applies. Further, given a chart of the manifold there is a natural basis of the tangent bundle over that chart.

Then using this basis we get the usual transformation law for tensors that we see in GR.

This all looks like a lot of work but it's not much different from constructing the real number line from the integers. It merely needs to be set out properly and motivated.

Physically speaking, tensors come in useful when the appropriate unit is 2 or more dimensional. That is we represent an area by an actual area element or a volume by an actual volume element. For example charge density would be an infinitesimal volume element. Were we taking about the charge density on a conductor then this would be represented by an infinitesimal area element.

Answer 3

A $n$-tensor is a multilinear function taking $n$ vectors as input and returning a scalar as output. The key is this multilinearity, that $f(u_1 + v_1, v_2, \dots) = f(u_1, v_2, \dots) + f(v_1, v_2, \dots)$ and so forth for all of the different vector arguments that it takes on. That makes it look "simple" in the sense that we get from, say, calculus.

Some examples

The one that you have doubtless already encountered is the dot product, which is a $2$-tensor taking two vectors and returning a scalar which describes their overlap. You may have also encountered a $3$-tensor called the "orientation tensor in 3D," this is the number you get when you cross two vectors and dot that with a third.

For another example we might ask for a little area element -- both its size and orientation in space -- and then tell you how much electrical current is flowing through it. If this is a linear relationship then it is described by a 1-tensor "current density". We might however instead be operating in a crystal where the current "wants" to flow along the crystal axes $\hat e_1,\hat e_2, \hat e_3,$ but it flows along each of these with different conductivities and in some crystals those three vectors are not even orthogonal! So given an electric field $\vec E$, and the area element, I can tell you what current was induced in that direction. So in crystals conductivity is a $2$-tensor:$$dI\left(\vec E, ~d\vec A\right) = G_1 ~\left(\hat e_1\cdot \vec E\right)~\left(\hat e_1\cdot d\vec A\right) + G_2 ~\left(\hat e_2\cdot \vec E\right)~\left(\hat e_2\cdot d\vec A\right) + G_3 ~\left(\hat e_3\cdot \vec E\right)~\left(\hat e_3\cdot d\vec A\right). $$

How vectors are tensors

You may have noticed or heard that $1$-tensors are vectors; they are strictly speaking "covectors" or "one-forms" that map a vector to a scalar, but let's go through the general idea.

Given an $m$-tensor and an $n$-tensor there is an "outer product" where we produce a $m+n$-tensor. Take the $m+n$ vectors, feed the first $m$ to the first tensor and the other $n$ to the second tensor, you get two scalars, you multiply those two numbers together. As a special case, constant scalars are $0$-tensors and so you can multiply any tensor by a scalar by eventually modifying its output. On the flip side we can "partially apply" an $m$-tensor to $n<m$ vectors to create an $m-n$-tensor.

There is also a "tensor sum" where we take two $n$-tensors and create another $n$-tensor: Take $n$ vectors, feed them to both tensors, add together the two scalars.

So this general idea uses two axioms which aren't always true in strange mathematical spaces, but seem to apply in physics pretty frequently. The first is to say that some $n$-tensor is always describable as some big sum of outer products of $1$-tensors. You can see this in that conductivity tensor above where we are using a sum of outer products of two $1$-tensors like $(\hat e_1\cdot)~(\hat e_1\cdot)$ to build up the $2$-tensor. The second axiom says that any $1$-tensor can be identified with a dot product with some vector -- in other words the dot product has an inverse, so that given $\left(\vec v \cdot\right)$ and its effects on all the other vectors of your space, you can figure out uniquely what $\vec v$ is.

When both of these are true, then any $n$-tensor is really a family of $[a, b]$-tensors such that $a + b = n$. What I have just described are only the $[0,n]$-tensors, multilinear maps from $n$ vectors to a scalar, but the rest of these are multilinear maps from $a$ covectors and $b$ vectors to a scalar. You still have the "sum of outer products" weirdness, so you can't say that, say, a $[3,0]$-tensor is a triple of vectors--some are (no sum), some are not.

In that sense, any $2$-tensor can be thought of as a linear map from vectors to vectors (i.e. a matrix), the orientation tensor in 3D is a bilinear map from two vectors to a vector, and so forth.

One last idea for completeness

That axiom also gives us a notion of "contractions," a family of operations which take an $[a, b]$-tensor and give you an $[a-1, b-1]$-tensor. The idea is that you write the tensor as a sum of outer products, you have converted some of the covectors to vectors, so finally you can feed that vector as another of the arguments to the corresponding covector.

What's really helpful, notation-wise, is called "abstract index" notation. This writes any $[a,b]$-tensor as some name with "raised" and "lowered" symbols for indexes. So you choose $a+b$ distinct symbols and you put $a$ of them as upper indices and $b$ of them as lower indices. Like our conductivity above might be written as $G_{ab}$, here the symbols are $a, b$. Then, when you repeat a symbol as both a top and a bottom index, it means "expand as a sum of vectors and covectors, then contract those two indexes together, applying that covector to that vector." Since vectors are $[1,0]$-tensors the above expression is compactly written as just: $$dI = G_{ab} ~E^a~dA^b.$$We can thus see our outer products (given $u_\bullet$ and $v_\bullet$ I can make the outer product $u_\bullet ~ v_\bullet$), and we can define a $[1,1]$-tensor that just "relabels" one symbol to another, $\delta^a_b v^b = v^a$, we have our dot product as a tensor $g_{\bullet\bullet}$, and we have its inverse $g^{\bullet\bullet}$ defined so that $g^{ab}~g_{bc}~=\delta^a_c.$

In many cases you might insert an explicit choice of basis, in which case many of these choices then look very similar to "Einstein summation notation." The more proper formal way to do this involves inventing coordinate vectors $c^\alpha_{1,2,\dots D}$ for your $n$-dimensional space and their dual covectors $c^{1,2\dots D}_{\alpha}$ such that $$c^\alpha_a ~c_\alpha^b=\{1\text{ if } a=b \text{ else } 0\}.$$ And then you can define that a vector $v^\alpha$ has scalar coordinates $v^a =c^a_\alpha v^\alpha$ and recover that vector as $v^\alpha = \sum_{a=1}^n v^a~c_a^\alpha.$ In this case you can see that some of my indexes are "abstract" (the symbol $\alpha$ doesn't mean anything, it just identifies something as a vector or covector) and some are "concrete" (in all cases the variables $a,b$ are placeholders for normal integers between $1$ and the dimension $D$ of the space). And then you can get similar expressions where you just insist on "I'm not going to write these summation signs any more."