# Intermezzo: Tensors

Before we continue on with our study of fluid mechanics, it’s best to make a quick pit stop and make sense of that strange concept I mentioned previously; that of a tensor, which we saw in the form of that strange multiplication $\rho \vec{v} \otimes \vec{v}$. The “proper” way of defining a tensor is hotly debated by mathematicians and physicists of different fields, so my idea here is not necessarily to rigorously define them, but to give you a brief sense of what a tensor is and how we use them in fluid mechanics. In my experience, that is best illustrated by revisiting the concept of a vector, like the velocity $\vec{v}$.

Knowing the velocity of an object tells us two things; the speed of the object, which is just some number like 20 miles per hour, and the direction in which it is traveling. But as anyone that’s said “No, I meant my left” can attest to, direction is relative—whenever we talk about velocity in its most general sense, we usually describe it in terms of three independent components representing the speeds in specific, perpendicular directions that we’ve arbitrarily defined in the 3-dimensional space we live in (usually denoted with $x, y, z$). Changing those arbitrarily defined reference directions must then necessarily change the expressions for the components, commonly referred to as projections, even though it doesn’t physically change the velocity of the object itself.

That being said, velocity doesn’t intrinsically possess three directions—as we stated above, it possesses just one, the direction in which the object is instantly traveling in. With that in mind, a tensor is just a mathematical object that possesses any number of directions. People usually refer to the amount of directions a tensor possesses as its order, and so we can straightforwardly spot that the velocity is a first-order tensor because of its single “natural” direction.

If the velocity is a first-order tensor, then it might seem obvious to state that the mathematical object $\rho \vec{v} \otimes \vec{v}$, which we said represented the momentum density flux, possesses two directions, and is as such a second-order tensor. In this case, those directions are straightforward to spot; the direction of the momentum density $\rho \vec{v}$, and the direction of the flux, which is just the velocity $\vec{v}$. As a matter of principle, you can always spot the directions of any tensor defined through a sequence of outer products of vectors by looking at the direction of each vector. In fact, both of those directions are the same in this specific case, since $\rho \vec{v} \otimes \vec{v}$ and $\vec{v}$ point in the same direction!

So now that we know what that tensor means, we can think a bit more concretely about how to represent it mathematically. Well, just like the velocity can be represented using three components with reference to some arbitrary coordinate system, the momentum density flux is the product of every component of the 3-D vector $\rho \vec{v}$ with every component of the 3-D vector $\vec{v}$, so that the tensor $\rho \vec{v} \otimes \vec{v}$ is represented by $3\times3 = 9$ independent components. To make sure we distinguish which direction is which, mathematicians usually visually represent second-order tensors using a visual “square” of numbers called a matrix.

This isn’t the end of the story on tensors, though; remember that the Cauchy momentum equation we saw before includes the term $\nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right)$, not just the momentum density flux. Hence, we need to be able to understand the effects of trying to take derivatives, like the divergence and gradient, on tensors. To do that, let’s repeat our formula on testing it out on the velocity—which we know is a first-order tensor—and see how to extrapolate from there.

The divergence of the velocity at a point, from a purely mathematical standpoint, tells you what the rate of change of the velocity is at that point as you move in each of those predefined spatial coordinates, and then adds all of those rates of changes up into a single direction-less number that tells you if the flow is “accumulating” or “diminishing” at that point. Make sure to notice that the divergence of the velocity is truly directionless; if you looked at a flow upside down, the direction of the flow would certainly change, but the amount by which the flow is “accumulating” at a point doesn’t. This is easily extrapolated to a general, tensorial case; the divergence of a tensor takes the spatial rate of change of the tensor in the direction of one of the tensor’s directional elements, usually defined to be the “last” element, and adds them up. This causes a tensor to “lose” a direction when acted upon by the divergence—meaning that the second-order tensor $\rho \vec{v} \otimes \vec{v}$, when acted on by the divergence, gives you a first-order tensor/vector, just as the momentum transport equation required.

Lastly, let’s take a look at the gradient of the velocity $\nabla \vec{v}$. Here we are still obtaining the spatial rates of change of the velocity, but this time we’re not adding anything up; in fact, we’re obtaining the spatial rate of change of every component of the velocity in each and every one of those arbitrarily defined directions of space. 3 different possible rates of change for each of the three components indicates that the velocity gradient possesses 9 distinct pieces of information, revealing that the gradient of the velocity is a second-order tensor. As such, we can extrapolate to the general case and state that the gradient of a tensor is another tensor of a single higher order.

Explicitly defining the direction that gets added as a result of taking the gradient is trickier to spot than in the outer product case, where the directions of the resulting tensor are just the directions of each component of the tensor; but the direction exists, and the vector representing it can be found by solving the following equation for the vector $\vec{m}$:

$\nabla \vec{v} \cdot \vec{m} = \vec{v}$

(The dot product here is just representing standard matrix-column vector multiplication.) This vector doesn’t really have a name, nor a more intuitive description other than the above as far as I can tell; but I call it the gradial vector for book-keeping, and would love to find a nice application for it.

These explanations are by no means rigorous or comprehensive; there are many incredible and beautiful aspects to the study of tensors that I sadly have to exclude for the sake of coherence, but I invite you to read up on them if you’re interested in the subject. My current favorite book on the topic is Henry Block’s Introduction to Tensor Analysis, which is tragically out-of-print but has been preserved electronically by the thoughtfulness of faculty members in the Theoretical & Applied Mechanics field at Cornell. In any case, this is just about everything we need to know about tensors to get a handle on all the mathematical machinery abound in introductory fluid mechanics; and although the principal language of fluid mechanics is vector calculus, it’s good to get a handle on tensors to deal with an often-ignored part of the world of fluid mechanics—the process by which molecules exert forces on each other when a fluid moves, known as the study of rheology.

2. If a fluid is incompressible, how does the expression for $\nabla \cdot \left(\rho \vec{v} \otimes \vec{v}\right)$ simplify? Remember the incompressibility equation, $\nabla \cdot \vec{v} = 0$.