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My system consist of a cylinder with axis Z that can contract and dilate along this axis. It obeys microscopically Hooke's law of elasticity: $${\cal{L}}=\frac12\rho\dot{u_z}^2-\frac12C_{zzzz}(\partial_z u_z)^2$$ with $\cal{L}$ the Lagrangian density, $u_z(z,t)$ the local displacement to the rest position, $C_{zzzz}$ the unique component of the stiffness tensor (equal to the Young Modulus $Y$) relevant here considering the unique degree of freedom of my system, and $\rho$ of course for density. Supposing that the stretch is uniform in my cylinder $u_z(z,t)$ obeys the following equation: $$u_z(z,t)=\frac{z}{l}\delta z(t)$$ with $l$ the rest length of the cylinder, $\delta z$ the total stretch at its top, and the base being being static by exterior constraint and defined as $z=0$. My goal is to rederive from there the global Hooke law, by integrating the Lagrangian density over x,y,z; But from there I arrive at: $$L=\frac12m\delta\dot{z}^2-\frac12(\frac{3C_{zzzz} S}{l})\delta z^2$$ My problem is that I get a wrong value for the global stiffness of my bar: $K=3\frac{YS}{l}$, the factor 3 shouldn't be there. Is it purely geometrical or is there a misunderstanding in my reasoning? It may seem dull at first to derive this result, but I'll be way more confident in seeing the right result emerging from this calculus.

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Plus, if I had added a dissipative term to my first Lagrangian density, would it be possible to derive a damping term for my global oscillations. Should I need to work it out from the wave equations? (because dissipative terms and Lagrangian description don't seem quite compatible for me).

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Your erroneous assumption is that $C_{zzzz}$ corresponds directly to the Young's modulus. In fact, it does not, but in the case of your example is 1/3 of the Young's modulus. Therefore, the global stiffness is, in fact, $\frac{Y S}{l}$, as expected. To understand the relationship between the Young's modulus and the coefficient $C_{zzzz}$, take a look at Landau and Lifshitz, paying particular attention to equations (4.1) through (4.3), (5.3) and (5.5). (If you aren't convinced these equations apply to your problem, take a look at equation (10.3)... if it still doesn't make sense, look at equation (1.3).)

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  • $\begingroup$ Thanks for the clarification, indeed it explains the difference in the results. Do you have any idea regarding the damping? What would the term to add at the microscopic equations to recover a global $F_{\text{dissipation}}= - m \Gamma \delta \dot{z}$? $\endgroup$
    – Learning is a mess
    Feb 25, 2013 at 16:33
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    $\begingroup$ Take a look at L&L $\S 34$... $\endgroup$
    – KDN
    Feb 25, 2013 at 16:45
  • $\begingroup$ My edition stops at §31 =) I'll look for a more recent one ;) $\endgroup$
    – Learning is a mess
    Feb 25, 2013 at 16:48
  • $\begingroup$ Ok I did find a recent scan on the internet, the paragraphs you were pointing at was included in my book copy, they must have added supplements in between; I'll try deriving the damping from the viscosity ;) $\endgroup$
    – Learning is a mess
    Feb 25, 2013 at 16:51
  • $\begingroup$ I posted a link to an online archive of the 2nd edition in your reference question... it is mostly complete with fairly minor revisions, I think. $\endgroup$
    – KDN
    Feb 25, 2013 at 16:57

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