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So, what they do is that they assume linear elasticity and Cartesian coordinates. Starting from the volumetric integral of the stress tensor times a variation in the displacement vector, they are able to show that the work related to this small change is, roughly speaking, minus the stress tensor times a variation of the strain tensor.

Ok, so what I'd like to know is if this result applies to nonlinear elasticity as well. I've seen claims that it does, citing that the proof is more or less elementary. I can't quite immediately see how the L&L approach could be generalized, though. Help anyone? Ideally I'd like to see a result without the assumption of Cartesian coordinates, but any pointers would be nice.

I've written below the version found in L&L:

[tex]\int_V \delta R dV = \int_S \sigma_{ik} \delta u_i df_k - \int_V \sigma_{ik}\frac{\partial\delta u_i}{\partial x_k} dV[/tex]

Take V infinite and assume that stress vanishes at infinity. Noticing that the stress tensor is symmetric we may write

[tex]\int_V \delta R dV = -\frac{1}{2}\int_V \sigma_{ik} \delta \left(\frac{\partial u_i}{\partial x_k} + \frac{\partial u_k}{\partial x_i}\right) dV = -\int_V \sigma_{ik}\delta u_{ik} dV[/tex]

Thus, they conclude, [tex]\delta R = - \sigma_{ik}\delta u_{ik}[/tex].

This I don't think would work for the nonlinear case, as it's missing the term

[tex]\frac{\partial u_l}{\partial x_i}\frac{\partial u_l}{\partial x_k}[/tex]

Any help would be much appreciated.