Classical mechanics

5. Equation of fluid motion


Continuing from our discussions on the fluid stress tensor in chapter 4 we now examine some advanced ideas of this tensor. Our objective now is to take a closer look at the components of the stress tensor, which we simply represented as $\tau_{ij}$ in eq. (4.5). We will end up proving that there exist only six, and not nine, components in the stress tensor $\boldsymbol{\tau}$.

Components of the stress tensor

Consider an infinitesimal cubic volume of fluid $\delta x \delta y \delta z$. The reason we require our this volume to be infinitesimal is because this ensures uniformity across all parameters: density, velocity etc. For all intents and purposes then we can treat this volume of fluid like a rigid body if we only ever consider its movements over infinitesimal time intervals.

Say this volume of fluid rotates. We can relate its angular velocity $\dot{\theta}$ and the moment of inertia of the cube $I$ with some simple mathematics. The linear velocity corresponding to $\dot{\theta}$ is $v_\theta = \dot{\theta} r$ where $r = \sqrt{x^2 + y^2}$ in two dimensions.

Fig. 1: Stress tensor components of a rotating infinitesimal volume of fluid.

As shown in fig. 1 the area in the $\pm\hat{i}$ direction, i.e. the face pointing in the $\pm\hat{n}_i$ direction, has a torque $\pm\tau_{ii}$ acting along $\pm\hat{i}$ and $\pm\tau{ij}$ acting along the $\pm\hat{j}$ direction perpendicular to $\hat{i}$. Each face is naturally only along two mutually perpendicular axes forming a plane, and fig. 1 shows only one of six planes of our cube in direct sight.

The moment of inertia of our cube, which is normally given by $mr^2$ for an object of mass $m$ and a distance $r$ from the centre of mass at which a torque acts, is $\left(\rho \d x \d y \d z \right) r^2$ with the mass written in terms of the volume density. This is for an infinitesimal volume, so for the entire fluid, or at least for a larger volume of fluid, we have the simple integral1 $$ \begin{align*} I_z &= \int\limits_{\delta x} \int\limits_{\delta y} \int\limits_{\delta z} \rho \left(x^2 + y^2 \right) \d x \d y \d z \end{align*}$$

$$ \begin{align*} I_z &= \rho \int\limits_{\delta y} \int\limits_{\delta z} \left( {x^3 \over 3} + xy^2 \right)_{-\delta x/2}^{\delta x/2} \d y \d z \\[.5em] &= \rho \int\limits_{\delta y} \int\limits_{\delta z} \left( { (\delta x)^3 \over 12} + \delta x y^2 \right) \d y \d z \\[.5em] &= \rho \int \limits_{\delta z} \left( {(\delta x)^3 \delta y \over 12} + {\delta x (\delta y)^3 \over 12} \right) \d z \end{align*} $$

$$\begin{align*} \implies I_z &= \rho \left( {(\delta x)^2 + (\delta y)^2 \over 12}\right) \delta x \delta y \delta z\end{align*}$$

which gives us the moment of inertia in the $z$ direction. Subsequently the torque $\mathbf{T}$, or specifically its $z$-component $T_z$, arising from the stresses acting on the cube2 is written as $$ \begin{align} T_z &= I_z \ddot{\theta}_z \nonumber\\[.5em] &= \rho \ddot{\theta}_z \left( {(\delta x)^2 + (\delta y)^2 \over 12}\right) \delta x \delta y \delta z \label{eq:torque-one} \end{align}$$arising from the standard definition of torque as the product of angular acceleration and moment of inertia. We have then substituted our value for $I_z$ from that.

Generally speaking the torques per unit volume $\tau_{yx}$ and $-\tau_{xy}$ act along the two opposite faces of our cube3. These will naturally be opposite in direction if we must have rotation (which is our requirement)—hence the opposite signs. We therefore have $\tau_{yx}\delta x \delta y \delta z$ on the $\hat{x}$ face and $-\tau_{xy} \delta x \delta y \delta z$ on the $\hat{y}$ face. Therefore the net torque turns out to be $$\begin{equation} T_z =  \left( \tau_{yx} - \tau_{xy} \right) \delta x \delta y \delta z \label{eq:torque-two} \end{equation}$$

Equating eq. $\eqref{eq:torque-one}$ and $\eqref{eq:torque-two}$ and letting $\delta x \rightarrow 0$ and $\delta y \rightarrow 0$ we have $\tau_{yx} = \tau_{xy}$ or, generally, for the even permutation $i \rightarrow j \rightarrow k$ we can demand that the off-diagonal elements of our stress tensor $\boldsymbol{\tau}$ must satisfy $$\begin{equation} \tau_{ji} = \tau_{ij} \label{eq:off-diag-condition} \end{equation}$$

In other words the off-diagonal elements of our stress tensor are equal to each other, i.e. the fluid stress tensor is symmetric. This cuts our three of our nine elements in the matrix—either the upper triangle or the lower—leaving us with three off-diagonal and three diagonal terms, or a total of six independent elements out of nine.

An equation of motion

While we tended to a point by restricting $\delta i \rightarrow 0$ above, the stresses in a fluid really act on the entire infinitesimal cubic volume $\delta x \delta y \delta z$, where each $\delta i$ corresponds to the side length along the $\hat{i}$ direction as shown in fig. 1. The stress at the surface of the cube is then naturally different from that on a face. It varies by $\pm \left( \partial \tau_{ij}/\partial j \right) \delta j / 2$ depending on the face, with opposite faces taking opposite signs as we discussed above.

The physical reasoning behind this claim is simple: the change in the stress component $\tau_{ij}$ along the $\hat{j}$ direction is its gradient, or spatial rate of change (which is one dimensional and along $\hat{j}$ here), times the total length over which that change occurs—from a face to the centre this is half the total length of a side, i.e. $\delta j / 2$.

Recall from eq. (4.1) that stress is the Force acting over an entire surface area and it can be written as $\sigma A = F$ in brief for infinitesimal considerations. Therefore the net $x$-component of force (i.e. that component of all of the forces, acting on all three faces, $x$, $y$ and $z$) is given by substituting $i$ and $j$ with $x$, $y$ and $z$ cyclically: $$ \sigma_x = \left( {\partial\tau_{xx} \over \partial x} + {\partial\tau_{xy} \over \partial y} + {\partial\tau_{xz} \over \partial z} \right) \delta x \delta y \delta z $$

Pay attention to the subscripts here. For the $y$ face the stress components, in order, are $\tau_{yx}$, $\tau_{yy}$ and $\tau_{yz}$ respectively. As has been our notation since chapter 4 the subscripts $ij$ refer to the $i$ component and the $j$ face. In the above formula we are interested in the $x$-component of force on the $x$, $y$ and $z$ faces, hence we keep the subscript $i$ fixed to $x$ and cycle $j$. This sort of term exists also for the $y$- and $z$-components.

Recall that in eq. (4.4) we equated surface forces and body forces acting on an infinitesimal cubic volume element of our fluid. We will renew that expression now writing the volume $\delta h \delta A$—for height $\delta h$ and area $\delta A$—as $\delta x \delta y \delta z$ instead, and writing the $\boldsymbol{\sigma} (\mathbf{r}, \hat{n}, t)$ as the $x$-component $\sigma_x$ alone: $$ \rho \delta x \delta y \delta z \left( {\d v_x \over \d t} \right) = \left( {\partial\tau_{xx} \over \partial x} + {\partial\tau_{xy} \over \partial y} + {\partial\tau_{xz} \over \partial z} \right) \delta x \delta y \delta z + \rho \delta x \delta y \delta z \left( {G_x \over m} \right)$$

In simpler form we have the following: $$\begin{equation} \rho {\d v_x \over \d t} =  \left( {\partial\tau_{xx} \over \partial x} + {\partial\tau_{xy} \over \partial y} + {\partial\tau_{xz} \over \partial z} \right)  + \rho {G_x \over m} \label{eq:eq-fluid-motion} \end{equation}$$or more compactly still, using Einstein notation, we write $$\begin{equation} \rho  \dot{v}_i  = \partial_j \tau_{ij} + \rho {G_x \over m} \tag{\ref{eq:eq-fluid-motion}} \end{equation}$$

This is known as the equation of motion for viscous fluids. It directly relates the $i$-directional acceleration of a mass (per unit volume) of a fluid to the surface forces (acting on a unit volume of that fluid) in the $i$ direction and the body forces (also per unit volume) in the same direction. While body forces are fairly straightforward to describe, e.g. the weight of an elemental volume of fluid is a result of the body force called 'gravity', the components of the stress tensor call for some thought. This will be our agenda in chapter 5.

  1. This being for a cube of length $\delta x$ we integrate from $-\delta x/2$ to $+\delta x/2$ as per convention and for ease of calculation without any loss of generality.

  2. These stresses act on the centre of the cube, or the centre of mass of whatever shape we consider our infinitesimal volume to be, and are present only in non-ideal fluids, which are the type of fluids we are discussing now.

  3. We are ignoring the so-called volumetric torque here, which is the torque resulting from a Body force, since such forces e.g. gravity generally act on the centre of mass and do not prompt any torque-induced rotation. However such an effect can be observed in case of magnetic fluids in which case we will simply introduce an appropriate torque (or torque per unit volume) term along the axis of rotation, the $z$-axis in our case.

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