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I need the general expression for the lagrangian density of a linear elastic solid. I haven't been able to find this anywhere. Thanks.

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the Lagrangian density is $L=T-U$, the difference between the kinetic and potential energy density, as we're used to in all of mechanics.

The kinetic energy density is $T = \rho v(x,y,z)^2/2$ where one has to calculate the density $\rho$ properly. The potential energy is more general and complicated,

$$U = \frac{1}{2} C_{ijkl}u_{ij}u_{kl}$$

where the tensor $C$ with four indices is called the elastic. The tensors $u$ are obtained from the displacement vectors as

$$ u_{ij} = \frac{1}{2} (\partial_i u_j + \partial_j u_i) $$

Note that the density $\rho$ in the kinetic energy also depends on the displacement vectors, if you want to re-express it via the mass density at rest (without displacement).

For isotropic materials,

$$ C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk} ) $$

where $\lambda$ and $\mu$ are known as Lamé constants. For more general (but linear) crystals, however, $C$ is a general tensor symmetric under the $ij$ exchange, $kl$ exchange, and the exchange of $ij$ and $kl$ as pairs.

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thanks. This is just what I needed. –  becko Jan 14 '11 at 2:54
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