# Why is the distortion in a non-homogeneous strain described as $\Delta u_i = \displaystyle \sum_j (e_{ij}- \omega_{ij})\Delta x_j$?

When we have a homogeneous strain - which may include both stretching and shear - all of the $e_{ij}$ are constants, and we can write $$u_x = e_{xx}\mathsf x + e_{xy}\mathsf y +e_{xz}\mathsf z\,.\tag{39.9}$$

[...] When the strains are not homogeneous, any piece of the jello may also get somewhat twisted - there will be a local relation. If the distortions are all small, we should have $$\Delta u_i = \sum_j (e_{ij}- \omega_{ij})\Delta x_j\,,\tag{39.10}$$ where $\omega_{ij}$ is an anti-symmetric tensor, $$\omega_{ij}= \tfrac12 (\partial u_j/\partial x_i- \partial u_i/\partial x_j)\,,\tag{39.11}$$ which describes the rotation.

This is excerpted from Lectures on Physics' Elastic Materials' The tensor of strain by Feynman. Here he describes the tensor of strains for both homogeneous and non-homogeneous substances.

I'm not comprehending how he formulated the definition of tensor of strain $e_{ij}$ for non-homogeneous substances.

Why is there $\Delta$ in $(39.10)$ in the distortion $\Delta u_i$ for non-homogeneous substances? It's not there in the distortion $u_x$ of homogeneous substances in $(39.9)$.

And why was it required to add the anti-symmetric tensor in the distortion of non-homogeneous substances viz. in $(39.10)\,?$ Is there no rotation in the distortion of homogeneous substances?

• for the people not familiar with Feynman's book/who don't have it lying around, can you provide a definition for the symbols he uses? I have a hunch as to what might be happening, but I'd be happy to be sure before writing an answer – Sanya Sep 23 '16 at 17:50

For the distortion in homogeneous strain, there is, as he writes, a linear relation between $u_j$ and $x_i$, so the derivatives $e_{ij}$ are actually constants and we can integrate $$\mathrm du_i = \sum_j e_{ij}~\mathrm dx_j$$ to give his formula $(39.9)$ $$u_i = \sum_j e_{ij} x_j$$ For non-homogeneous strain, the derivatives are actually position dependent and we have: $$\mathrm du_i = \sum_j \frac{\partial u_i}{\partial x_j}~\mathrm dx_j$$ or for small values of the distortions: $$\Delta u_i = \sum_j \frac{\partial u_i}{\partial x_j} \Delta x_j$$ If we look at $e_{ij} - \omega_{ij}$ we have the difference between symmetric and antisymmetric part of the gradient of the distortion field, i.e. the derivative, which we can see by evaluating: $$e_{ij} - \omega_{ij}=\frac{1}{2} \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{\partial u_j}{\partial x_i} + \frac{\partial u_i}{\partial x_j}\right)= \frac{\partial u_i}{\partial x_j}$$

And yes, I am pretty positive that a homogeneous strain, i.e. a constant derivative of the distortion w.r.t. the coordinates over all the body which is symmetric, is not possible with a rotation. This is, in the end, the polar decomposition theorem applied to the gradient matrix of the displacement field.