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In the theory of elasticity there is an important object, the displacement increment vector $u_i$. The derivative of such an object can be decomposed into symmetric and antisymmetric parts:

$$ \partial_i u_j = \underbrace{\frac{1}{2} (\partial_i u_j + \partial_j u_i)}_{u_{ij}} + \frac{1}{2} (\partial_i u_j - \partial_j u_i) $$

The antisymmetric part describes rotation:

It is obvious that whole body rotation doesn't change energy. But local rotations can lead to changes of energy. But usually energy of a deformed body is described by:

$$ E = \lambda^{ijkl} u_{ij} u_{kl} $$

Why do we not include contributions of the antisymmetric part? Local rotations will change energy of system.

G. Smith
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Nikita
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1 Answers1

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The antisymmetric term $\frac{1}{2}(\partial_j u_i - \partial_i u_j)$ does not correspond to an internal deformation of the elastic continua, but rather a rigid rotation of the system.

This means that the antisymmetric term $O_{ij}$ (following Fetter and Walecka's naming convention) does not produce a change in the dimensions of the body (IE: does not change the length, area, or volume of the body as a whole), and thus does not contribute to the elastic energy of the system.

We have more reasons to believe that the elastic energy sees no contribution from a rigid rotation (or at least no contribution to first order). We know this since we derive the energy from the Stress Tensor $T_{ij}$, which is symmetric. The terms we can build with $O_{ij}$ that are symmetric are beyond our first order approximation (and are thus very small).

Kyle Perez
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  • But local rotations (not rotations of rigid body) must give contributions to energy. Or why not? – Nikita Oct 16 '20 at 19:32
  • @Nikita, Fetter and Walecka call $O_{ij}$ a local rigid rotation, and it corresponds to rotating the object in question. However, if you were to have something like a circle, and pick a radial cut, and stretch it around the circle, deforming it, there would clearly be an energy cost to do this (as you compress one part of the material and stretch another part). The energy cost to do this would be described by the energy equation you presented. Importantly, this behaviour isn't due to a local rigid rotation and is instead described by the symmetric $\epsilon_{ij}$ matrix (the symmetric matrix). – Kyle Perez Oct 16 '20 at 20:16