Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $F$?

share|improve this question
    
Related –  Vijay Murthy Apr 5 '12 at 9:51
add comment

1 Answer

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.


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 $m>1$.

share|improve this answer
    
I understand how this implies that we don't see higher powers of terms like $u_{i,j}^5$ (because this is small compared to, say $u_{i,j}^4$), but I don't understand how this implies that we don't see higher derivatives, like $u_{i,jk}$. Can you elaborate on this? –  Jules Apr 5 '12 at 10:20
    
It's pretty much equivalent: one way to define the derivatives is to consider the closest-matching polynomial function, and interpret this as a Taylor series. –  leftaroundabout Apr 5 '12 at 10:46
    
Right! But the reason that we can ignore the later terms in a Taylor expansion is that each derivative gets multiplied by a power of $\Delta x$, which is small. It's not that the derivatives themselves are small. What is the reason that $u_{i,jk} \ll u_{i,j}$? –  Jules Apr 5 '12 at 10:51
    
Well, there is no reason for this, because it isn't true! It's actually not even possible to compare those derivatives directly, since they have different physical dimensions. Only together with the coefficients this is meaningful. –  leftaroundabout Apr 5 '12 at 11:23
    
Sure, but you could say exactly the same about comparing $\Delta x^5$ with $\Delta x^4$. The point is that the coefficient in $F$ is just a constant. So asymptotically you can still compare them. As far as I can see there can be two reasons why $u_{i,jk}$ does not appear in $F$: (1) it is asymptotically smaller than the other derivatives because of some assumption or (2) its coefficient in $F$ is zero for some reason. My question is: which is it, and why? –  Jules Apr 5 '12 at 11:38
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.