# Stress Force - Understanding Cauchy Stress Tensor

I've been trying to understand the derivation for the Cauchy Momentum Equation for so long now, and there is one part that every derivation glides over very quickly with practically no explanation (I'm guessing they assume the reader knows it already).

The part I'm stuck with is how they relate stress tensor, $\sigma_{ji}$, to the sum of the forces' on a infinitesimal block of volume $dV$. I'll give you a bit of the context of the situation I'm in. Here's how this part of every derivation goes.

Suppose you have a differential/infinitesimal [rectangular prism] volume of fluid $dV$, side-lengths $dx_j$, density $\rho$, and acceleration in the $i^{\textrm{th}}$ direction $a_i$. Applying Newton's second law per unit volume in the $i^{\textrm{th}}$ direction gives us

$$\rho\,a_i = \sum F_i, \,\,\,\,\textrm{\sum F_i is the net body force in the i^{\textrm{th}} direction}$$

Now for the net body force in the $i^{\textrm{th}}$ direction, we have the external body forces $f$ and the stress [surface really] forces, which is given as the rate of the stress variation in the $i^{\textrm{th}}$ direction, $\sum \limits_{j} \frac{\partial \sigma_{ji}}{\partial x_j}$. Thus, we have by Newton's second law per unit volume of fluid in the $i^{\textrm{th}}$ direction,

$$\rho\,a_i=f_i + \sum \limits_{j} \frac{\partial \sigma_{ji}}{\partial x_j}$$

And so the total force (all of the previous equation multiplied by the volume of the unit of fluid which it pertained to) is given as

$$\rho\,a_i\,dV = f_i\,dV + \sum \limits_{j} \frac{\partial \sigma_{ji}}{\partial x_j}dV$$

How on earth is the total stress force in a particular direction (used as a body force) given as the rate of the stress variation in that direction?! I see how the units work out, but I can't see any logic behind it. I thought that $\sigma_{ji}$ represented the stress (force per unit area) on the $dx_j$ side pointing in the $i$ direction. If that's the case, how do the relate that surface force to being a body force, especially in the way that is stated above (saying it's the rate of stress variation in that direction)?

how on earth.....? A possible way to look at it goes like this, let's consider a little cube length $$l$$, then the stress force in the $$i$$th direction acting on $$j$$th surface element is $$-\sigma_{ji}(x_i,x_j,x_k)*dA$$ where $$dA=l^2$$. The force in the $$i$$th direction acting on the other surface element parallel to the first is $$\sigma_{ji}(x_i+dx_i,x_j,x_k)*dA$$, so the total stress force acting in the $$i$$th direction is $$(\sigma_{ji}(x_i+dx_i,x_j,x_k)-\sigma_{ji}(x_i,x_j,x_k))*dA$$ Now divide by the volume element $$dV=dx_idx_jdx_k$$ and consider $$dA=dx_idx_k$$ to obtain $$f_i^{(j)}=\frac{\sigma_{ji}(x_i+dx_i,x_j,x_k)-\sigma_{ji}(x_i,x_j,x_k)}{dx_j}=\frac{\partial\sigma_{ji}}{\partial x_j}$$ Then total force per unit volume in the $$i$$th direction is given by your desired equation

I'm 7 years late, but I'll answer anyway hoping to be useful to someone else.

Note: I will use no bar for scalars, one bar for vectors (hat for versors) and two bars for tensors (for brevity, when I will say "tensor" I will always mean "rank 2 tensor", but of course scalars and vectors too are tensors).

## Introduction

The momentum conservation of a generic fluid (no matter if it is compressible, viscous, etc.) is expressed the Cauchy equation: $$\begin{equation} \nabla \cdot \bar{\bar{\sigma}} + \rho \bar{f} = \frac{\partial (\rho \bar{v})}{\partial t} + \nabla \cdot \bar{\bar{F}} \tag{1} \end{equation}$$ A little scary! But if you want to invest time and energy in studying what I write, I will explain to you the meaning of each term well, and why this equation represents the momentum conservation in a continuum. It is worth studying the Cauchy equation, because it is the starting point for reaching the Euler equation (don't ask to me how is that Euler died before Cauchy was born) and the Navier-Stokes equation (this one is defined by Cengel-Cimbala as the cornerstone of fluid mechanics, and the study of its solutions is one of the seven so called "millennium problems"). $$\rho$$ and $$\bar{v}$$ are density and velocity, others terms will be defined in the next sections.

## Definition 1: stress tensor $$\bar{\bar{\sigma}}$$

The first problem is to find a way to describe mathematically forces exerted within the continuum. The idea is to suppose that for each infinitesimal imaginary surface $$d\bar{A}$$ inside the continuum exists a rank 2 tensor field $$\bar{\bar{\sigma}}$$ such that $$\bar{\bar{\sigma}} \cdot d\bar{A}$$ is the force $$d\bar{F}$$ that the "upper" side of the continuum (the one containing the tiny vector $$d\bar{A}$$) exerts on the other side. If this is the first time you have to do with stress tensor you may find a bit repelled by this abstract object: after all we usually speak about forces exerted between different bodies, while here we have a tiny surface in the continuum, and this surface does not divide the body into two parts. But if you think for a while you will see this is not a problem: you can't cut inside the pulp a tiny surface and measure the forces acting between the two sides, but this doesn't mean that forces inside the continuum aren't present, they exists and this is the reasonable way to describe them. But really it is? I mean, another perplexity could be this one: who ensures that the law $$\begin{equation} d\bar{F} = \bar{\bar{\sigma}} \cdot d\bar{A} \tag{2} \end{equation}$$ describe reality correctly? Well... as often happens in physics, we try with the simpler hypothesis (and often nature helps us because the simpler hypothesis works). $$d\bar{F}$$ is a function of $$d\bar{A}$$ (and we suppose that it is proportional to $$dA=|d\bar{A}|$$), and a dependence like $$d\bar{F} = k d\bar{A}$$ is excluded because, as you can guess (think to a torsion) and as we will see later, in general $$d\bar{F}$$ and $$d\bar{A}$$ can have different directions. So a linear law as (2) is the simplest that we can use. I can't prove the existence of stress tensor, I'll simply assume that (2) works. A posteriori, we can justify (2) by observing that, with other hypotheses, it leads to Navier-Stokes equation, which has some experimental confirmations.

## Symmetry of stress tensor

We will consider $$\bar{\bar{\sigma}}$$ symmetric because we will exploit the Cauchy equation in finding Euler equation and Navier-Stokes equation, and in these contexts stress tensors are symmetric by construction. So we have no problem and we can exploit the extended divergence theorem (see below). Anyway in books I read that the symmetry has a deeper origin and that it can be done a general proof. To be honest, I didn't understand these proofs because they work exploiting rotational equilibrium, without justifying it, but we have no need to go deep here if our ultimate goals are Cauchy equation, Euler equation and Navier-Stokes equation.

In short, symmetry of stress tensor is essential to write Cauchy equation and in its application (in which too extended divergence theorem is exploited), but this is not a big problem for us: we actually will always handle symmetric stress tensors, so our considerations about these problems stop here.

## Definition 2: vector $$\bar{f}$$

$$\bar{f}$$ is a vector such that multiplied by density $$\rho$$ it gives the density of body forces (forces that acts through the volume of the body, in contrast with contact forces) $$\begin{equation} \rho \bar{f} = \frac{d\bar{F}_{body}}{dV} \tag{3} \end{equation}$$ Note that $$\bar{g}\rho = \bar{g}\frac{dm}{dV}=\frac{\bar{g}dm}{dV}=\frac{d\bar{F}_{grav}}{dV}$$ where $$d\bar{F}_{grav}$$ is the force exerted by gravity on the portion of the fluid considered: typically body force is due to gravity and we can identify $$\bar{f}$$ with gravitational field. This is the best thing to do now, to not be distracted by things that are not essential here (anyway I'll continue to use a more general $$\bar{f}$$ symbol: body forces can also have electromagnetic origin, or be due to the fact that we aren't in an inertial system).

## Definition 3: mass flux and momentum flux $$\bar{\bar{F}}$$

As we can define a vector field $$\bar{J}$$ (probably the reader is already familiar with it) that describes the flux of mass, i.e. defined in a way that $$\begin{equation} \bar{J} \cdot d\bar{A} = \frac{dm}{dt} \qquad \textrm{(dm = mass through d\bar{A} in its direction during t\to t+dt)} \tag{4} \end{equation}$$ similarly we can define a tensor field $$\bar{\bar{F}}$$ (less know) that describes the flux of momentum, i.e. defined in a way that $$\begin{equation} \bar{\bar{F}} \cdot d\bar{A} = \frac{d\bar{p}}{dt} \qquad \textrm{(d\bar{p} = momentum through d\bar{A} in its direction during t\to t+dt)} \tag{5} \end{equation}$$ Unlike what was done with the stress tensor, I can prove the existence of $$\bar{\bar{F}}$$ finding it, I will do it later.

## Proof that $$\bar{J}=\rho\bar{v}$$

The fluid volume that goes through $$d\bar{A}$$ in time $$dt$$ is (if $$d\bar{A}$$ is small and ignoring higher order infinitesimal volumes) the product of $$v dt$$ times the shadowed surface in figure, that is $$dA \cos \theta$$ where $$\theta$$ is the angle between $$\bar{v}$$ and $$d\bar{A}$$. We conclude that mass through $$d\bar{A}$$ in time $$dt$$ is, $$dm = \rho v dA \cos \theta dt = \rho \bar{v} \cdot d\bar{A} dt$$. Observing (4) the proof is ended.

## Intermission: some conventions

To go on, it is better to introduce some conventions and notations. In cartesian coordinates the outer product between two vectors is by definition the tensor $$\begin{equation}\tag{6} \bar{A} \otimes \bar{B} = \begin{pmatrix} A_x B_x & A_x B_y & A_x B_z \\ A_y B_x & A_y B_y & A_y B_z \\ A_z B_x & A_z B_y & A_z B_z \end{pmatrix} \end{equation}$$ I add that it is convenient, in doing calculations, writing vectors their "natural" way as columns, and that inner (the usual dot product) and outer product can be profitably be seen as matrix product such that

• in the inner product we transpose term first term

• in the outer product we transpose the second term

You cannot protest, these are definitions, this is the grammar with which I will write the equations. Note that (6) can be seen as $$\begin{pmatrix} A_x \\ A_y \\ A_z \end{pmatrix} \begin{pmatrix} B_x & B_y & B_z \end{pmatrix}$$, and you can easily see why the convention works with inner (i.e. dot) product. We will always manage symmetric tensors (their representation will be a symmetric matrix) so transposition of tensors can be ignored in what follows (but transposition of vectors is important).

## Proof that $$\bar{\bar{F}} = \rho \bar{v} \otimes \bar{v}$$

Let's consider the $$\bar{J}=\rho\bar{v}$$ proof. As seen, the mass through $$d\bar{A}$$ in time $$dt$$ is $$\rho \bar{v} \cdot d\bar{A} dt$$. By multiplying by $$\bar{v}$$ I find momentum $$d\bar{p} = \rho (\bar{v}\cdot d\bar{A}) dt \bar{v}$$. Observing (5) and our thesis $$\bar{\bar{F}} = \rho \bar{v} \otimes \bar{v}$$, we see that to end the proof we have to proof that $$\begin{equation} (\bar{v} \cdot d \bar{A}) \bar{v} = (\bar{v} \otimes \bar{v}) \cdot d\bar{A} \end{equation}$$ i.e $$\begin{equation} \left[ \begin{pmatrix} v_x & v_y & v_z \end{pmatrix} \begin{pmatrix} dA_x \\ dA_y \\ dA_z \end{pmatrix} \right] \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \begin{pmatrix} v_x v_x & v_x v_y & v_x v_z \\ v_y v_x & v_y v_y & v_y v_z \\ v_z v_x & v_z v_y & v_z v_z \end{pmatrix} \begin{pmatrix} dA_x \\ dA_y \\ dA_z \end{pmatrix} \end{equation}$$ You can focus for example on the $$x$$ component and check that this is an identity.

## Extended divergence theorem

Everywhere in books, sites, YouTube, etc. they speak ad nauseam about ordinary divergence theorem, but almost nobody speaks about the extended one, which is almost equally important. I found it this summer in Cengel-Cimbala book (to whom I stole the name, I used to call it "alternative divergence theorem"), and I find it strange that it is not given the proper prominence it should have in literature. If you are interested, you will find a proof in my Stack Exchange answer to "How do the electric or magnetic field contain momentum"? The theorem states that given a symmetric tensor field $$\bar{\bar{M}}$$ defined inside a volume $$V$$ delimited by a surface $$S$$, we have (note that in both sides we have vectors) $$\begin{equation} \int_V (\nabla \cdot \bar{\bar{M}}) d V = \oint_S \bar{\bar{M}} \cdot d \bar{A} \tag{7} \end{equation}$$

## Cauchy equation proof

Let's consider a portion of fluid $$V$$ (not necessarily small). The net force on it by the fluid around is $$\bar{F}_{net} = \oint_S d \bar{F}_{sup}$$ where $$S$$ is the surface delimiting $$V$$ and $$d\bar{F}_{sup}$$ are force acting on an infinitesimal surface that make $$S$$. Exploiting definition of stress tensor given before, we write $$\bar{F}_{net} = \oint_S \bar{\bar{\sigma}} \cdot d \bar{A} = \int_V \nabla \cdot \bar{\bar{\sigma}} dV$$, where I exploited (7) (as said, I'll use only symmetric stress tensor, so I can use it). Please note that $$\nabla \cdot \bar{\bar{\sigma}}$$ is a short notation for $$\begin{equation} \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{pmatrix} \begin{pmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{pmatrix} \end{equation}$$ $$\rho \bar{f} dV$$ is the body (not contact) force acting on elementary volume so the sum of body forces is given by $$\int_V \rho \bar{f} dV$$. We conclude that total force on $$V$$ can be written in this way: $$\begin{equation} \int_V ( \nabla \cdot \bar{\bar{\sigma}} + \rho \bar{f}) dV \end{equation}$$ It must be equal to $$\frac{d\bar{p}}{dt}$$, that has two terms. On one hand, evidently $$\rho \bar{v}$$ is the density of momentum and so we have a term $$\frac{d}{dt} \int_V \rho \bar{v} dV = \int_V \frac{\partial(\rho \bar{v})}{\partial t} dV$$. On the other hand, we have to take into account the momentum that, in interval $$t \to t + dt$$, goes through the surface that envelops the volume. We have seen that $$\frac{d\bar{p}}{dt} =\bar{\bar{F}} \cdot d\bar{A}$$ (by definition con momentum flux $$\bar{\bar{F}}$$), so $$\frac{d\bar{p}}{dt} = \int_S \bar{\bar{F}} \cdot d \bar{A} = \int_V \nabla \cdot \bar{\bar{F}} dV$$, where we exploited again (7) (the tensor $$\rho \bar{v} \otimes \bar{v}$$ is evidently symmetric, look to the definition of outer product above). Putting all together we can write $$\int_V \left( \nabla \cdot \bar{\bar{\sigma}} + \rho \bar{f} - \frac{\partial (\rho \bar{v})}{\partial t} - \nabla \cdot \bar{\bar{F}} \right) dV=0$$. The equation must be true for every $$V$$ so the integrand is zero and the Cauchy equation is proved.

## Material derivative

Let's define the material derivative as the operator (the symbol $$\otimes$$ is explained above: don't worry too much about it, now you simply have to manage it as a well defined operator to have shorter equations, learn what it does) $$\begin{equation} \frac{D}{Dt} = \frac{\partial }{\partial t} + \bar{v} \cdot \nabla \otimes \tag{8} \end{equation}$$ Usually symbol $$\otimes$$ is omitted in material derivatives, but I find it is better writing it explicitly to be consistent with convention introduced before and to emphasize that in $$\nabla \otimes \bar{v}$$ the column $$\bar{v}$$ must be transposed, to get a rank 2 tensor. There would be many things to say about material derivative, that is a bridge between lagrangian view and eulerian view in description of fluids, and that can be exploited to proof Cauchy equation using directly second Newton law, but I can't transform an answer in a book, and if your purpose is to write Cauchy equation in a more compact way, you can take material derivative as a convenient definition to write shorter formulas.

Please note that with rules about outer product written above, and remembering that $$\nabla = \left( \frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z} \right)$$, we have that $$\begin{equation} \bar{v} \cdot \nabla \otimes \bar{v} = \begin{pmatrix} v_x & v_y & v_z \end{pmatrix} \begin{pmatrix} \frac{ \partial v_x}{ \partial x} & \frac{ \partial v_y}{ \partial x} & \frac{ \partial v_z}{ \partial x} \\ \frac{ \partial v_x}{ \partial y} & \frac{ \partial v_y}{ \partial y} & \frac{ \partial v_z}{ \partial y} \\ \frac{ \partial v_x}{ \partial z} & \frac{ \partial v_y}{ \partial z} & \frac{ \partial v_z}{ \partial z} \end{pmatrix} \end{equation}$$ i.e. each component of this vector is the sum of three terms.

## A shorter version of Cauchy equation

An alternative shorter way to write Cauchy equation is $$\begin{equation} \frac{D \bar{v}}{Dt} =\frac{1}{\rho} \nabla \cdot \bar{\bar{\sigma}} + \bar{f} \tag{9} \end{equation}$$ where I exploited the material derivative written above. Using (8) and comparing with (1), we see that to prove that (1) and (9) are the same we must prove that the following equation is an identity $$\begin{equation} \dot{\rho} \bar{v} + \nabla \cdot \bar{\bar{F}} = \rho \bar{v} \cdot \nabla \otimes \bar{v} \tag{10} \end{equation}$$ Exploiting continuity equation $$\dot{\rho} = - \nabla \cdot (\rho \bar{v})$$ we can write (10) in this way $$\begin{equation} \nabla \cdot \bar{\bar{F}} = \rho \bar{v} \cdot \nabla \otimes \bar{v} + (\nabla \cdot (\rho \bar{v}) ) \bar{v} \end{equation}$$ Now consider that $$\bar{\bar{F}}$$ can be seen as the outer product of vector $$\rho \bar{v}$$ with vector $$\bar{v}$$, so to end the prof we must to show that it is true the following identity $$\begin{equation} \nabla \cdot (\bar{A} \otimes \bar{B}) = \bar{A} \cdot \nabla \otimes \bar{B} + ( \nabla \cdot \bar{A} ) \bar{B} \end{equation}$$ If you have learned the operator definitions in the "intermission" above, checking this identity is long but not difficult as it may seem looking at it (note that in the middle term there aren't parenthesis but it turns out that the order is indifferent), so (9) is proved.

Let's do it in 1D for simplicity: you consider a portion of thread of length $dL$ and section $S$, with a net body force density $f$, say $f=\rho S g$ where $\rho S$ is the lineic mass density. On $dL$, you also have stress from the rest of the thread, which are $T_+ = \sigma_{zz} (z+dL) S$ at the $z+dL$ end and $T_- = -\sigma_{zz}(z) S$ at the other end.

So: $\rho S a_z dL = T_+ + T_- + f dL$. Divide by $dL$ and make it go to 0 to recover the $\partial_z \sigma_{zz}$ term.

Recall the Gauss' divergence theorem by which:

$$\sum_j \frac{\partial \sigma_{ij}}{\partial x_j}dV = \sigma_{ij}dS_j$$

Therefore, the equilibrium equation is no longer mysterious, and the sum of forces is the internal forces $$F_{int,i} = f_i\text{d}V$$ plus the forces applied on the surface $$F_{sup,i} = \sigma_{ij}\text{d}S_j$$:

$$F_{int,i} + F_{sup,i} = \rho a_i\text{d}V = \text{d}m\ a_i$$

To understand why the components of the stress tensor can be understood as the forces on the surface, it is only necessary to reflect on this representation: 