Classical Lagrangian for an isotropic elastic solid

Question

Professor Fradkin’s course (CHAPTER 1. SECOND QUANTIZATION- equation 1.65) presents the Lagrangian density as:

$$L=\int d^3r\frac{\rho}{2}\left(\frac{\partial u}{\partial t}\right)^2-\frac{1}{2}\int d^3r \ \left[ K\nabla_i u_j\nabla_i u_j+\Gamma\nabla_i u_i\nabla_j u_j \right]$$

He says $K$ and $\Gamma$ are the elastic moduli.

Using the elastic energy $\frac{1}{2} \sigma_{ik}e_{ik}$, where the $e_{ik}$ is the infinitesimal strain tensor:

$$e_{ik}=\frac{1}{2}(u_{i,k}+u_{k,i})$$

I obtain for the potential energy:

$$\frac{1}{2}\left[\mu(\nabla_i u_j\nabla_i u_j+\nabla_i u_j\nabla_j u_i )+\lambda \nabla_i u_i\nabla_j u_j\right]$$

Where $\lambda$ and $\mu$ are the Lamé coefficients.

Setting $K=\mu$ and $\Gamma=\lambda$, I still have the extra term:

$$\nabla_i u_j\nabla_j u_i$$

Does anybody know what Professor Fradkin did?

Answer 1

$\nabla_i u_j \nabla_j u_i = \nabla_j u_j \nabla_iu_i+ spatial \: boundary \: terms$,

so the second term in your potential energy gives a contribution to third term. In other words $\Gamma \neq \lambda$ but $\Gamma$ gets a contribution also from the first Lamé parameter. The boundary terms can be omitted as they play no role in the variational prcedure to obtain the equations of motion.