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We know, that are materials that holds the Hooke Law under a specific range, this is, materials for which the deformation $dx$ is proportional to the force $dF$, $dF=-kdx$, which implies $E={1\over2}k{dx}^2$ under some range of $dF$.

Also, we have materials that are elastic under that range too, this is, their deformation is conservative, and in this sense, Hooke Law implies Elasticity.

We know a linear expression such as Hooke $dF=-kdx$ will imply a Conservative i.e. Elastic system, by calculus.

But there are "Non Hooke Elastic" materials, that is, that under some range of $dF$ are Elastic and does not hold Hooke Law, hence, the Energy should not be always $E={1\over2}k{dx}^2$ for these Elastic materials?. We could have dependence with temperature $T$ in the linear factor $k=k(T)$, but it is still Hooke, only with an adittional variable.

Or it is that elastic materials always accomplish Hooke law in some sense, but in some strict sense they are non Hooke?. How the conservative proof works under those cases?

Are there some specific simple examples for this?.

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This is correct, Hooke's law specifically means the range of deformations where the restoring force is linearly proportional to the deformation, which corresponds to the first term in Taylor expansion of force in deformation (or quadratic energy term near the elastic potential energy minimum). For greater deformations the force become nonlinear, then deformation becomes inelastic/plastic (i.e., the material does not anymore return to initial state), and eventually break/fructure occurs.

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  • $\begingroup$ This means that in an elastic range we could have additional terms? How then this force can be conservative, if we have terms in i.e. more powers of $x$?. We know that $1/x^2$ is conservative, but that $x^2$ is not conservative, so, how then we can have non hooke materials if $x^2$ is not conservative?. $\endgroup$
    – Brethlosze
    Commented Jun 3, 2022 at 17:02
  • $\begingroup$ Anything that can be described by a potential energy, i e., any force $f(x)=-\partial_x U(x)$ is conservative. The problem with quadratic force might be that it is breaking the inversion symmetry, since the negative and positive deformations result in the same force. $\endgroup$
    – Roger V.
    Commented Jun 3, 2022 at 17:16
  • $\begingroup$ So, a simple term $F=-kx^3$ will be conservative, symmetric, and non hooke, right? $\endgroup$
    – Brethlosze
    Commented Jun 3, 2022 at 17:39
  • $\begingroup$ Yes, this is correct. But such materials are rare. Same is true for the Ohm's law, btw. $\endgroup$
    – Roger V.
    Commented Jun 3, 2022 at 17:42

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