Why does elastic energy only depend on first derivatives?

Question

Say we have an elastic material that is deformed with displacement function $u : \mathbb{R}^n \rightarrow \mathbb{R}^n$. It is reasonable to assume that the energy required for such a displacement is an integral over local properties of $u$:

$$E = \int_S F(u_i, u_{i,j}, u_{i,jk}, \cdots)$$

Where $u_{i,jk\cdots z} = \frac{\partial^m u_i}{\partial x_j \partial x_k\cdots \partial x_z}$. It is clear that $F$ cannot depend on $u_i$, because the energy is translation invariant. As it turns out, for small deformations $F$ depends on all $u_{i,j} = \frac{\partial u_i}{\partial x_j}$ and does not depend on higher derivatives. Furthermore, $F$ is a quadratic in $u_{i,j}$, namely $F = \sum T_{ijkl} u_{i,j} u_{k,l}$ where $T_{ijkl}$ depends on the material. Is there a reason for this, or is this just an empirical fact? Is there a set of intuitive assumptions that lead to this form of $F$?

Answer 1

It's not so much that $F$ doesn't depend on higher-order derivatives, it's just that, on a sufficiently small scale (which is what you deal with when performing such an integral), the first-order term is always dominant. That's obvious enough: if you simply Taylor-expand in the displacement, all terms approach zero as $\mathcal{O}(\Delta x^m)$, so for $\Delta x\ll\chi$ (where $\chi$ is some kind of characteristic length, of the order of magnitude of the inverse of the derivatives' coefficients in $F$), the contribution of the higher-order terms is always neglectable.

This argument fails when the coefficients are not in the same order of magnitude. That can happen if you zoom down to the scale of the material's internal structure: the integral is actually an approximation of a sum over some kind of discrete "building units". An example would be a metal spring squeezed down so that the windings just touch each other. In this case, squeezing a bit more requires much more energy than releasing the spring to the same absolute displacement yields, so the contribution is of higher-than-first order.

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All right, my argument is sort of circular.

Again, the integral is really just an approximation of a sum over displacements $\mathbf{u}_i$ of discrete particles. We're modeling these displacements by a displacement field, which is a smooth function $\mathbf{u}(\mathbf{x})$ such that $\mathbf{u}(\mathbf{x}_i)=\mathbf{u}_i$. This function should not introduce any extra information, i.e. its Fourier transform should vanish for wave numbers above the reciprocal lattice grid dimension $\kappa$, and in fact much earlier because we're not dealing with microscopic displacements. In $\mathbf{k}$-space, the derivatives are just multiplication by $\mathrm{i}\mathbf{k}$, so "$|\partial^m\mathbf{u}|\ll\kappa^m$". The interactions between the particles are assumed to be predominantly between direct neighbours, so $F$ should evaluate $\mathbf{u}$ only with $\Delta x\leq\frac1\kappa$. But then, the actual contributions is $F$ are terms $\Delta x^m\cdot \partial^m_{\mathbf{x}}\mathbf{u}\approx \eta^m$, with $\eta\ll \frac\kappa\kappa=1$. So the terms appearing in $F$ quickly vanish for $m>1$.