I was looking at the equation for the differential force $dF$ caused from a stress tensor $\sigma$ acting on a differential surface $dS$. In mathematical terms,

$dF=\sigma \cdot dS$

Taking the integral of both sides yields.

$$F=\int \sigma \cdot dS$$

What is the nature of this integral? How does on go about calculating it? I would like to see an example given a non-trivial $\sigma$.

Can this be generalized to $F=\int \sigma \cdot dV$?

My idea was to integrate with respect to the variables $x$ and $y$ and represent $dS$ as a normal vector multiplied by the $dx$ and $dy$ differentials. Taking the product of the right hand side, you then integrate with respect to the differentials.

  • $\begingroup$ In this case, the dot product helps because it gets rid of one rank of the tensor. But in general, I think the opposite of $\frac{\partial}{\partial x^\mu}$ is the obvious $\int\,dx^\mu$, which would be approached the normal tensor way. $\endgroup$ – Jim Aug 19 '15 at 16:41
  • $\begingroup$ @Jim I meant matrix product by $\sigma \cdot dS$. Still, it does yield a vector, I just don't know what to do with it :/ $\endgroup$ – Zach466920 Aug 19 '15 at 16:43
  • $\begingroup$ I'd double check with math.se, but I think you just do the integral for each component of the vector and the result will still be a vector. Honestly, I'm more used to tensor form than matrix form $\endgroup$ – Jim Aug 19 '15 at 16:45
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    $\begingroup$ You probably do know what to do with it and just don't recognize the fact. Here's a problem that you know involving integrals and vectors: given a particle with initial position $\vec{x}_0$, initial velocity $\vec{v}_0$ and constant acceleration $\vec{a}$, find the position as a function of time. The extension to higher rank tensors is similar. $\endgroup$ – dmckee --- ex-moderator kitten Aug 19 '15 at 16:47
  • $\begingroup$ @dmckee yes, but $dS$ has to be a vector. I don't actually know how to pick $dS$ so integration can be done under something like $dx \ dy$. With your example the equation already comes with $dt$ in a nice form. $\endgroup$ – Zach466920 Aug 19 '15 at 16:53

You can imagine three equations one for each component of the force and then the tensor can be thought of as three vectors.

For instance $F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$ where $d\vec S=\hat n dA.$ So now it is a regular flux surface integral. You just have three of them:

$F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$, $F_y=\int \sigma_{yx}n_x+\sigma_{yy}n_y+\sigma_{yz}n_zdA,$ and $F_z=\int \sigma_{zx}n_x+\sigma_{zy}n_y+\sigma_{zz}n_zdA.$

See how it almost looks like a vector on the left and a matrix times a vector on the right? It's not more complicated than that.

$$F=\int \sigma dS=\int \sigma n dA.$$


Seems as though my idea is the correct one. Here's an example.

Since, $dS$ is normal to the surface and has a magnitude equal to the area of a differential. $dS$ can be represented by a cross product of the differentials of the relevant surface.


$$\sigma= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} $$

We want to find the force over the surface of a square parallel to the x-y plane. The cross product of the differentials yields $[0,0,dx \cdot dy]$. Taking the transpose of this we get $dS$. Putting this into the formula yields,

$$\sigma \cdot dS=[dx \cdot dy,0, dx \ dy]^t$$

Taking the integral over the square gives,


Where $s$ is the side length of the square. This method can be generalized for other surfaces of course.


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