### Classical mechanics

# 4. Stresses in fluids

Stress is an important parameter characterising a fluid. It is vectorial but is not as straightforward to describe as, say, the velocity of an elemental volume of a fluid. In the latter case all we need are a magnitude and a direction; in the former case we first need to define our system in terms of the area around the point on which a stress acts, a procedure which often introduces several variables into our problem.

# Defining stress

Stress $\sigma$ is defined as the ratio of a force $\mathbf{F}$ to the area $A$ over which that force acts. Say we have an area element, in our fluid, along the $\hat{n}$ direction and located at $\mathbf{r}$, as shown in fig. 1, on which some force $\mathbf{F}$ acts.

We could define the stress right away as $F/A$ but recall that this is really pressure^{1} which acts over an area; stress, by contrast, acts on a point on this area and not over the entire area. Stress is, therefore, a quantity best defined for small areas $\delta A$ and, consequently, in relation to small forces $\delta \mathbf{F}$; and it is most accurately defined when the area reduces to a point. $$ \begin{equation} \boldsymbol{\sigma} = \lim_{\delta A \rightarrow 0} {\delta \mathbf{F} \over \delta A} \label{eq:sigmaFA} \end{equation} $$

The thing exerting the force, generally speaking, is a neighbouring volume element in the fluid. As $\delta A \rightarrow 0$ it is clear that $\boldsymbol{\sigma}$ starts to depend only on the direction of the area element, $\hat{n}$, and becomes independent of the magnitude $A$. The relation between $\boldsymbol{\sigma}$ and $\hat{n}$ is readily seen from fig. 1 as being in terms of the angle between the two vectors, i.e. $\boldsymbol{\sigma}$ depends on the *orientation* of the force with respect to $\hat{n}$. Effectively $\boldsymbol{\sigma} \equiv \boldsymbol{\sigma} ( \mathbf{r}, \hat{n}, t ) $

# Pascal’s principle

The dependence of stress on multiple variables^{2} can actually be reduced if we bring in certain constraints. If we define our elemental fluid surface area—this is completely in our control—to be perpendicular to the stress^{3}, i.e. if $\mathbf{F} \parallel \hat{n}$, the stress simply becomes the pressure or, more accurately, the normal stress (which we shall call $\mathbf{p}$) giving us the simple form $$ \begin{equation} \boldsymbol{\sigma} = -p \hat{n} \end{equation} $$

where the negative sign tells us that $\hat{p}$ is antiparallel to $\hat{n}$. Observe that by defining our system in this manner we have effectively eliminated any stresses acting obliquely to $\hat{n}$. Such a stress is known as a Shear stress and no shear stress exists in our system anymore as we restricted it exclusively to Normal stress. This gives rise to **Pascal’s principle** which states—

In the absence of shearing forces the stress at any point in a fluid is normal to the surface on which it acts, and its magnitude is independent of the orientation of said surface. The direct implication of this principle is that $\boldsymbol{\sigma}$ can be expressed purely in terms of a scalar quantity in $p(\mathbf{r},t)$.

# The stress tensor

Viscous liquids without shear forces do not exist, which means we are not lucky enough when such liquids are involved to write down our stress in terms of $p$ alone. What we can write it as is a tensor whose components arise from $\mathbf{r}$, $\hat{n}$ and $t$.

## Equal and opposite stresses

The first observation we make about our system in fig. 1 is that if our element was a new orientation $-\hat{n}$ it would experience a stress that would be related to the original orientation as $$ \begin{equation} \boldsymbol{\sigma} (\mathbf{r},\hat{n},t) = - \boldsymbol{\sigma} (\mathbf{r},-\hat{n},t) \label{eq:opposingStresses} \end{equation} $$

Consider our system in fig. 1 with two antiparallel forces acting on its two faces. Say the cylindrical disc in that figure has an infinitesimal height of $\delta h$. We know from (chapterlink: Newton's second law) that the force on our elemental volume can be written as $\mathbf{F} = m\mathbf{a} = \rho V \mathbf{a}$. The volume of our cylinder, if its height is some $\delta h$ is $\delta h \delta A$ which gives us the force $$\mathbf{F} = \rho \delta h \delta A \textrm{d}{\mathbf v} / \textrm{d} t$$

where $\mathbf{v}$ is the velocity of the elemental volume of our fluid.

All of that constitutes the **inertial force** on the one hand. On the other, from eq. \eqref{eq:sigmaFA}, we have a net force in terms of the stresses: $$\boldsymbol{\sigma}(\mathbf{r},\hat{n},t) \delta A + \boldsymbol{\sigma}(\mathbf{r},-\hat{n},t) \delta A + \rho \delta h \delta A \mathbf{G} / m$$

The last term here could seem curious; what is the point of writing $\rho \delta h \delta A \mathbf{G} / m$ which essentially reduces to $\mathbf{G}$ anyway? This is called the Body force on an object: the force^{4} per unit mass $\mathbf{G}/m$ can be rewritten as the force per unit volume as $\rho\mathbf{G}/m$ so the force, or *body force*, on some volume $\delta V$ is $\rho\mathbf{G}\delta V/m$ which in our system is nothing but $\rho \delta h \delta A \mathbf{G} / m$. We write the body force, of course, because we recognise that there could be a force acting on our volume element besides the stress forces $\mathbf{F}$.

We now equate these two forces, the inertial force and the body force, as $$\begin{equation} \rho \delta h \delta A \left( {\textrm{d}{\mathbf v} \over \textrm{d} t} \right) = \boldsymbol{\sigma}(\mathbf{r},\hat{n},t) \delta A + \boldsymbol{\sigma}(\mathbf{r},-\hat{n},t) \delta A + \rho \delta h \delta A \left( {\mathbf{F} \over m} \right) \label{eq:eq-motion-mould} \end{equation} $$

and realise that if $\delta h \rightarrow 0$ we get eq. \eqref{eq:opposingStresses} as expected.

## Building a specific stress tensor

Follow the next description carefully to physically understand the stress tensor. Consider three mutually perpendicular surfaces elements forming a cubical volume element $\delta V$ as shown in fig. 2. Align their faces along our $\hat{i}$, $\hat{j}$, $\hat{k}$ basis—or define a basis to aline with the faces. Say stresses $\boldsymbol{\sigma}_x$, $\boldsymbol{\sigma}_y$ and $\boldsymbol{\sigma}_z$ act on these faces (the direction of $\boldsymbol{\sigma}$ does not matter: as we showed earlier $\boldsymbol{\sigma}$ is antiparallel in nature so the only difference between it going into a surface and emerging from a surface is a positive or negative sign. We can now expand each $\boldsymbol{\sigma}_\mu$ into its own components as follows: $$ \begin{align*} \boldsymbol{\sigma}_x &= \tau_{xx} \hat{i} + \tau_{yx} \hat{j} + \tau_{zx} \hat{k} \\[.5em] \boldsymbol{\sigma}_y &= \tau_{xy} \hat{i} + \tau_{yy} \hat{j} + \tau_{zy} \hat{k} \\[.5em] \boldsymbol{\sigma}_z &= \tau_{xz} \hat{i} + \tau_{yz} \hat{j} + \tau_{zz} \hat{k} \end{align*} $$

To explain this set-up in one sentence, consider the following: the stress $\boldsymbol{\sigma}_i$ acting in the $i$ direction can be broken up into three components in terms of $\sum_j\tau_{ji}$ which represent the $j$ component of the $i$ directional stress.

With this we already have a tensorial representation for stress but it is for the specific case of fig. 2. so we now proceed to generalise the problem to address all scenarios: any stress $\boldsymbol{\sigma}$ may be represented in terms of the three stresses $\sigma_i$, $\sigma_j$ and $\sigma_k$ and, in turn, in terms of their components $\tau_{ij}$ going across $i = x, y, z$ and $j=x,y,z$ each.

## Generalising the stress tensor

Say we have some surface of area $\delta A_i$ defined by some $\hat{n}_i$ and forming an angle $\theta_i$ with the $\hat{i}$ axis. That is to say our surface is at an arbitrary orientation. All the forces $\boldsymbol{\sigma}_i \delta A_i$ acting on a volume element located at $\mathbf{r}$ must always balance each other if we must reduce this volume to a point. If they do not then the reduced point will no longer be located at $\mathbf{r}$ but will, instead, be translated along the direction of the net force by a displacement proportional to the magnitude of the net force. We can, as we normally do, represent the fact that the net force is zero using $$ \sigma_x \delta A_x + \sigma_y \delta A_y + \sigma_z \delta A_z = 0$$

We can always reduce a surface of area $A$ into three components $A_i$ along three spatial directions. We can, in other words, write any area $A$ as $\sum_i \hat{i} A$ where each $\hat{i} A$ refers to an $A_i$ component of $A$ along the $\hat{i}$ direction. We can proportionally replace all $A_i$ in the above equation with just $\hat{i}$ without altering the equation itself in any manner. Therefore, $$ \sigma_x \hat{n}_x + \sigma_y \hat{n}_y + \sigma_z \hat{n}_z = \boldsymbol{\sigma} \hat{n}$$

We already know what $\boldsymbol{\sigma}_x$, $\boldsymbol{\sigma}_y$ and $\boldsymbol{\sigma}_z$ are so we can write $\boldsymbol{\sigma} \hat{n}$ it is clear that we can in fact write any stress in terms of the nine tensor components discussed earlier: $$ \begin{align*} \sigma_x \hat{n} &= \tau_{xx} n_x + \tau_{yx} n_y + \tau_{zx} n_z \\[.5em] \sigma_y \hat{n} &= \tau_{xy} n_x + \tau_{yy} n_y + \tau_{zy} n_z \\[.5em] \sigma_z \hat{n} &= \tau_{xz} n_x + \tau_{yz} n_y + \tau_{zz} n_z \end{align*} $$

Plucking the $\tau_{ij}$ components from these equations we get our stress tensor as $$ \boldsymbol{\tau} = \left[ \begin{array}{ccc} \tau_{xx} & \tau_{yx} & \tau_{zx} \\ \tau_{xy} & \tau_{yy} & \tau_{zy} \\ \tau_{xz} & \tau_{yz} & \tau_{zz} \end{array} \right] $$

Of course this can—and often is—shortened using Einstein notation to $$\begin{equation} \sigma_i (\mathbf{r},\hat{n},t) = \tau_{ij} (\mathbf{r},t) n_j \label{eq:stress-tensor} \end{equation}$$

This is an extremely important measure since, unlike in solids where stresses exist at a perfectly discrete point, stresses in fluids exist *everywhere* in the fluid and in various possible orientations, all of which must be accounted for. We can use infinite equations to do this or just one tensor—the choice is pretty simple.

The stress tensor is a function of space and time and, for every point we choose in our fluid, it gives us the three components of each of the three components into which any stress can be resolved: a total of nine as we saw above. This might appear to all be in a Cartesian coördinate system alone but the coördinate system itself does not matter: when you transform from one coördinate system to another the three sums $\tau_{ij} n_j$ will also transform as components of a vector $\sigma_i \hat{n}$. We say such a tensor as $\boldsymbol{\tau}$ is a tensor of *rank two*.

- Pressure does imply only the perpendicular component of the force acting on an area but we shall keep things simple for now. ↩
- Keep in mind that $\boldsymbol{\sigma} (\mathbf{r}, \hat{n}, t)$ is dependent on variables like $\mathbf{r}$ and $\hat{n}$ which are themselves described by multiple components. ↩
- Or, alternatively, if the force is somehow in our control we could apply it antiparallel to $\hat{n}$. ↩
- We use $\mathbf{G}$ here to distinguish it from $\mathbf{F}$ purely as a notation. ↩