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. I've already asked this on the physics stackexchange, but nobody answered me.

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    $\begingroup$ Crossposted from physics.stackexchange.com/q/144875/2451 $\endgroup$
    – Qmechanic
    Commented Nov 7, 2014 at 14:22
  • $\begingroup$ No response there @Qmechanic $\endgroup$ Commented Nov 30, 2014 at 2:00
  • $\begingroup$ @ArturodonJuan can you give sources (citations/links) of formulas given in this question? $\endgroup$ Commented Feb 21, 2020 at 8:55
  • $\begingroup$ @KamilKiełczewski For the definition of $\sigma$ I used the relevant Wikipedia article, but I believe I got most of the formulas from my head, i.e. they can be stated/derived from standard Newtonian mechanics + multivariable calculus. It's surprising that I asked this question a little over 5 years ago... $\endgroup$ Commented Feb 21, 2020 at 19:18
  • $\begingroup$ Thank you :) . Yes your question is still actual :) $\endgroup$ Commented Feb 21, 2020 at 20:14

1 Answer 1


Consider a small volume element

$$\{(s_1,s_2,s_3)\in \mathbb{R}^3: x_1 \leq s_1 \leq x_1 + \Delta x_1,x_2 \leq s_2 \leq x_2 + \Delta x_2,x_3 \leq s_3 \leq x_3 + \Delta x_3\}.$$

The surface (stress) force acting on that element in the $i$ direction consists of contributions from each of three pairs of faces where coordinates $s_1$, $s_2$ and $s_3$ are constant, respectively.

Consider one pair of where $s_1$ is fixed: $s_1 = x_1$ and $s_1 = x_1 + \Delta x_1$. The net force in the $i$ direction is

$$F_{i,1}\approx \sum_{j=1}^{3}\sigma_{ij}n_j|_{s_1=x_1 + \Delta x_1}\Delta x_2\Delta x_3+\sum_{j=1}^{3}\sigma_{ij}n_j|_{s_1=x_1}\Delta x_2\Delta x_3$$

This is approximate because we evaluate $\sigma_{ij}(s_1,s_2,s_3)$ at representative coordinates $s_2= x_2$ and $s_3=x_3$ . In general, we assume that the stress tensor is a continuously differentiable function of spatial position.

Here we use the fact that the surface force (in terms of the stress tensor and normal vector) is the product $\mathbb{\sigma}\cdot\mathbb{n}.$ On the face $s_1 = x_1 + \Delta x_1$ we have $\mathbb{n} = (1,0,0)$, and on the face $s_1 = x_1$ we have $\mathbb{n} = (-1,0,0).$


$$F_{i,1}\approx \sigma_{i1}|_{s_1=x_1 + \Delta x_1}\Delta x_2\Delta x_3-\sigma_{i1}|_{s_1=x_1}\Delta x_2\Delta x_3$$

The contribution to the force per unit volume is

$$\frac{F_{i,1}}{\Delta x_1 \Delta x_2 \Delta x_3}\approx \frac{\sigma_{i1}|_{s_1=x_1 + \Delta x_1}-\sigma_{i1}|_{s_1=x_1}}{\Delta x_1}.$$

Now consider the limit as the element shrinks to the point $(x_1,x_2,x_3)$ and $\Delta x_k \rightarrow 0.$

Then the contribution to the net surface force per unit volume at that point is

$$\hat{F_{i,1}}= \lim_{\Delta x_1 \rightarrow 0}\frac{\sigma_{i1}|_{s_1=x_1 + \Delta x_1}-\sigma_{i1}|_{s_1=x_1}}{\Delta x_1}=\frac{\partial \sigma_{i1}}{\partial x_1}.$$

Applying the same derivation to the other pairs of faces we obtain the total contribution to the net surface force per unit volume (in the $i$ direction) as

$$\hat{F_{i}}=\sum_{j=1}^{3}\frac{\partial \sigma_{ij}}{\partial x_j}$$

  • $\begingroup$ Oh my gosh thank you so much. That was so clear. I understand now. $\endgroup$ Commented Nov 6, 2014 at 23:41
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    $\begingroup$ I actually also learned that you can get to that answer by realizing that the total force in a particular direction would be given by the surface integral of those stresses contributing in that direction dotted with the normal component on the stress's respective face, and then applying the divergence theorem. However, I was looking for something more like this. $\endgroup$ Commented Nov 6, 2014 at 23:44
  • $\begingroup$ You're welcome. Its instructive to derive it both ways. $\endgroup$
    – RRL
    Commented Nov 7, 2014 at 2:14
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    $\begingroup$ @RRL - can you explain why in your bottom formula $\frac{\partial\sigma_{ij}}{\partial x_j}$ you switch indexes in sigma in comparison to OP question formulas where is $\frac{\partial\sigma_{ji}}{\partial x_j}$ (can you give some references/links). Becaouse this will give different result if stress tensor is non symetric (e.g. for some cases of non-Newtonian fluids)... As far I know the OP use right 'version' - e.g. here (MIT) eqation 2 and equation 6,7,8 for $\nabla\cdot\tau$ uses OP version (I can find much more examples) $\endgroup$ Commented Feb 21, 2020 at 9:12

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