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)?

Please help me.


3 Answers 3


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


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:

enter image description here

Wikipedia's article con Cauchy stress.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.