# Why does the pure shear term / strain deviator tensor have non-zero entries on the main diagonal?

In a textbook of mine an operation is performed, of which I think the goal is to get zeros on the main diagonal of a matrix (the matrix represents strain). But im not sure that is the goal and Im also not sure it is accomplished.

Say vector field $u$ represents displacement, then I think my textbook says the pure shear term would be:

$$\epsilon = \frac{1}{2} [\nabla u + (\nabla u)^T ] - \frac{1}{3} \nabla \cdot u I$$

I thought the pure shear term would be:

$$\pmatrix{0,\frac{1}{2}(\frac{du_x}{dy}+\frac{du_y}{dx}) ,\frac{1}{2}(\frac{du_x}{dz}+\frac{du_z}{dx}) \cr\frac{1}{2}(\frac{du_x}{dy}+\frac{du_y}{dx}) , 0,\frac{1}{2}(\frac{du_y}{dz}+\frac{du_z}{dy}) \cr\frac{1}{2}(\frac{du_x}{dz}+\frac{du_z}{dx}) ,\frac{1}{2}(\frac{du_y}{dz}+\frac{du_z}{dy}) , 0}$$

So I thought the main diagonal terms (volumetric strain) of $\frac{1}{2} [\nabla u + (\nabla u)^T ]$:

$$\pmatrix{ \frac{du_x}{dx}, 0, 0 \cr 0, \frac{du_y}{dy}, 0 \cr 0, 0, \frac{du_z}{dz} }$$

Would be cancelled out by $- \frac{1}{3} \nabla \cdot u I$. However When I manually fill out $- \frac{1}{3} \nabla \cdot u I$ they dont seem to completely cancel out the volumetric strain because I get:

$$\nabla \cdot u = \frac{du_x}{dx}+\frac{du_y}{dy}+\frac{du_z}{dz}$$

Which is a scalar so I do:

$$- \frac{1}{3} \nabla \cdot u I = - \frac{1}{3} \pmatrix{ \frac{du_x}{dx}+\frac{du_y}{dy}+\frac{du_z}{dz}, 0, 0 \cr 0, \frac{du_x}{dx}+\frac{du_y}{dy}+\frac{du_z}{dz}, 0 \cr 0, 0, \frac{du_x}{dx}+\frac{du_y}{dy}+\frac{du_z}{dz} }$$

Either Im doing this wrong, or I dont interpret the action of this first equation $\epsilon = \frac{1}{2} [\nabla u + (\nabla u)^T ] - \frac{1}{3} \nabla \cdot u I$ correctly. So my question is: what am I doing wrong, or why is the volumetric strain not cancelled out by this?

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The tensor you think is the pure shear is not rotationally invariant. Physics, instead, does not depend on the reference frame you are adopting to describe objects, so the notion of pure shear should be this way invariant. If you rotate the axes $x,y,z \to x',y',z'$, the form $$\pmatrix{0,\frac{1}{2}(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{ \partial x}) ,\frac{1}{2}(\frac{\partial u_x}{\partial z}+\frac{\partial u_z}{\partial x}) \cr\frac{1}{2}(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{ \partial x}) , 0,\frac{1}{2}(\frac{\partial u_y}{\partial z}+\frac{\partial u_z}{\partial y}) \cr\frac{1}{2}(\frac{\partial u_x}{\partial z}+\frac{\partial u_z}{\partial x}) ,\frac{1}{2}(\frac{\partial u_y}{\partial z}+\frac{\partial u_z}{\partial y}) , 0}$$ would not be preserved, but it could get some diagonal terms depending on the rotation, still remaining traceless ($A+B+C=0$): $$\pmatrix{A,\frac{1}{2}(\frac{\partial u'_x}{\partial y'}+\frac{\partial u'_y}{ \partial x'}) ,\frac{1}{2}(\frac{\partial u'_x}{\partial z'}+\frac{\partial u'_z}{\partial x'}) \cr\frac{1}{2}(\frac{\partial u'_x}{\partial y'}+\frac{\partial u'_y}{ \partial x'}) , B,\frac{1}{2}(\frac{\partial u'_y}{\partial z'}+\frac{\partial u'_z}{\partial y'}) \cr\frac{1}{2}(\frac{\partial u'_x}{\partial z'}+\frac{\partial u'_z}{\partial x'}) ,\frac{1}{2}(\frac{\partial u'_y}{\partial z'}+\frac{\partial u'_z}{\partial y'}) , C} \:.$$ I go to illustrate you how this form preservation is actually obtained with the correct definitions.
Let us start form the linearized "deformation" tensor (actually the linearized deformation tensor $\epsilon$ is the symmetric part of $L$ for the reason I will write shortly): $$L:= \nabla u\:.$$ The rotation group, $SO(3)$, acts on it as follows: $L \to L' := RLR^t$.
The idea is to decompose it with respect to the orthogonal decomposition of the space $D$ of double Cartesian tensors as a direct sum of irreducible and invariant subspaces under the above mentioned action of $SO(3)$. From the physical side, the fact that this decomposition is rotationally invariant means that it does not depend on the reference frame (assuming that the reference frames are connected by the group of isometries of the Euclidean space). One has: $$D = A \oplus S \oplus T\:,$$ where $A$ is the space of antisymmetric Cartesian tensors, $S$ is the space of traceless symmetric Cartesian tensors, $T$ is the space of Cartesian tensors proportional to $I$. So, if $t\in D$, its decomposition into $SO(3)$ invariant components is: $$t = \frac{1}{2}(t-t^t)\:\:\oplus \:\: \frac{1}{2}(t+t^t) - \frac{tr (t)}{3}I \:\:\oplus \:\: \frac{tr (t)}{3}I\:.$$ Referring to $L$, the first part $\frac{1}{2}(L-L^t)$ has nothing to do with deformations as it describes infinitesimal rotations of the continuous body you are studying (they could be cancelled simply changing reference frame). The second and third parts have to do with true deformations, in fact their sum is the linearized deformation tensor $\epsilon$. The second one, it being traceless, does not involve volume deformations. For this reason it is called pure shear. The last one is completely isotropic and describes pure volume deformations. As said this decomposition does not depend on the reference frame.
Thank you for this very thorough and insightful answer. This is much more insightful than wikipedia just mentioning the strain deviator tensor. With your answer I get an inkling as to why it works this way! I have not had grouptheory, after your introduction of $L'$ I dont completely follow any more: What are double cartesian tensors and how do you decompose their space under the action of $SO(3)$ (why is it appropriate to use $A$, $S$ and $T$)? –  Leo Jan 27 '14 at 10:31
You can think of a double Cartesian tensor as an assignment of a $3\times 3$ matrices $A$ to every orthogonal reference frame, with the constraint that changing reference frame with $\vec{x}' = R\vec{x}+ \vec{c}$, the corresponding matrices are related as $A'= RAR^t$. Physics is described form the class $\{A\}$ and not by a single element in that class. –  Valter Moretti Jan 27 '14 at 10:37
So, every physically meaningful property of a tensor has to be invariant under the action of the rotations $R\in SO(3)$. The decomposition I mentioned is invariant under that action, e.g., if a matrix $A$ is antisymmetric $RAR^t$ is such. So antisymmetry is a property of the tensor represented by $A$, not only of $A$. This property could have physical meaning. –  Valter Moretti Jan 27 '14 at 10:42
The fact that a matrix is symmetric and traceless is $SO(3)$ invariant, so it is good to describe physics, whereas the fact that a matrix is symmetric with vanishing elements on the diagonal is not $SO(3)$ invariant, so it depends on arbitrary choices (your choice of the reference frame): there is no good physics inside! –  Valter Moretti Jan 27 '14 at 10:46