# Why is 3D stress tensor acting only on three surfaces?

I'm trying to learn about the stress tensor (in 3D)

The tensors are said to have directions (the first subindex $i$ in $\sigma_{ij}$) and specify the surface upon which they act (the second subindex $j$ in $\sigma_{ij}$).

What confuses me is why is it defined only to act on three surfaces, even when the cube has six surfaces?

• That imaginary volume is infinitesimally small. Commented Aug 17, 2016 at 10:30
• @lemon No, that doesn't do it - the forces on the back faces are still present, they're just omitted from that diagram for clarity. Commented Aug 17, 2016 at 10:34
• @EmilioPisanty Yes, the forces are present on opposing sides but the stress on opposing sides is the same as a consequence of the infinitesimal separation... Commented Aug 17, 2016 at 10:40
• @lemon no, as it happens it's not. The force on an infinitesimal element is $\mathrm df_i=\sigma_{ij}n_j \mathrm d A$, and it flips sign for the back faces because the normal vector $\vec n$ has opposite direction. Commented Aug 17, 2016 at 10:54

That diagram shows only that set of three arrows for clarity: it is the smallest set that displays all the information necessary, but adding arrows to all six faces would make the diagram too crowded.

In general, if you have a material with a given stress tensor $\sigma_{ij}$, and you look at one small internal surface patch of area $A$ and normal vector $\vec n$, then the force $\vec F$ transmitted by the material's stress across that surface patch is given by $$F_i=\sum_j \sigma_{ij}n_j A.$$ In particular, for the three faces shown with $\vec n=\hat e_k, \ k=x,y,z$, you get a force with $i$th component $\sigma_{ik}$ acting on the face with normal $\hat e_k$. For the back faces, $\hat n=-\hat e_k$ changes direction, so the forces change direction, but they're obviously still there.

(It's a good exercise at this point to draw those arrows for one face and its opposite. You should get normal arrows pointing away from each other, either pushing in (pressure) or pulling out (tension). The tangential arrows ($\sigma_{yx}$ and $\sigma_{zx}$, say) should point away from each other in a shearing motion. The net result should be deformations of the unit cube, but no net force.)

• Does acting on surface, say x , mean that the force acts on both surfaces that have normals in the x direction? Like that, say, fluid passes through the cube through both of the faces in that direction? Commented Aug 17, 2016 at 10:52
• Or is it meant that the fluid acts inside the cube? So if the direction of the tensor is reversed, then it acts on the opposing face? Commented Aug 17, 2016 at 10:52
• The tensor has no direction: it is a linear transformation that takes the outward normal of a surface to the force acting through it. Note that the direction of the normal really matters: the force on opposite surfaces is opposite, because even if both faces have normals in the $x$ direction, one of them points towards $+x$ and the other towards $-x$. Commented Aug 17, 2016 at 10:58

We do consider other 3 faces, as we take into account (Sigma x), it means other opposite face of it is also taken but written only once which means the magnitude is same but direction is opposite, i.e. you will see 9 components but still there are 18 components, we only write 9 since other 9 will have same magnitude. So, think twice is it worthy to write twice ? answer is NO as if you know the magnitude, you can apply it on the element, if you are still confuse then think we take element under equilbrium condition.