# Energy momentum tensor in polarizable media

I am trying to follow Soper's derivation of the energy-momentum tensor for polarizable media in his book 'Classical field theory'. Given a Lagrangian density $$\mathcal{L} = \mathcal{L}(\phi_A, \partial_\mu \phi_A)$$ the canonical energy-momentum tensor is $$$$T_\mu{}^\nu = \delta_\mu^\nu \mathcal{L} - (\partial_\mu \phi_A) \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_A)}$$$$ For the dipole-field interaction he gives the Lagrangian density $$$$\mathcal{L}_{E\mathcal{M}} = - \frac{1}{2} J_\lambda \epsilon^{\lambda \alpha \beta \sigma} F_{\alpha \beta} (\partial_\sigma R_a) \mathcal{M}_a.$$$$ Where \begin{align} J^\lambda & = n \epsilon_{abc} \epsilon^{\lambda\alpha\beta\gamma} \frac{1}{3!} (\partial_\alpha R_a) (\partial_\beta R_a) (\partial_\gamma R_a),\\ \mathcal{M}_a & = \frac{1}{\rho} \frac{\partial x_i}{\partial R_a} M_i, \\ \rho & = \sqrt{-J_\lambda J^\lambda}, \\ F_{\alpha \beta} & = \partial_\alpha A_\beta - \partial_\beta A_\alpha. \end{align} To calculate the canonical energy-momentum tensor I need the derivatives of the Lagrangian w.r.t. the derivatives of the fields $$\frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_A)}$$, where $$\phi_A = \{A_\alpha, R_a\}$$. Using \begin{align} \frac{\partial F_{\alpha \beta}}{\partial(\partial_\nu A_\gamma)} & = \delta_\alpha^\nu \delta_\beta^\gamma - \delta_\beta^\nu \delta_\alpha^\gamma, \\ \frac{\partial (\partial_\sigma R_a)}{\partial(\partial_\nu R_b)} & = \delta_\sigma^\nu \delta_{ab}, \\ (\partial_\mu R_b) \frac{\partial J^\lambda}{\partial(\partial_\nu R_b)} & = \delta_\mu^\nu J^\lambda - \delta_\mu^\lambda J^\nu, \end{align} whereby the last relation is given on page 42, I get the derivatives $$$$\begin{split} -(\partial_\mu A_\gamma) \frac{\partial \mathcal{L}}{\partial(\partial_\nu A_\gamma)} & = J_\lambda \epsilon^{\lambda \nu \gamma \sigma} (\partial_\mu A_\gamma) (\partial_\sigma R_a) \mathcal{M}_a,\\ -(\partial_\mu R_b) \frac{\partial \mathcal{L}}{\partial(\partial_\nu R_b)} & = - \delta_\mu^\nu \mathcal{L} - \frac{1}{2} J^\nu \epsilon_\mu{}^{\alpha \beta \sigma} F_{\alpha \beta} (\partial_\sigma R_a) \mathcal{M}_a \\ & \quad + \frac{1}{2} J_\lambda \epsilon^{\lambda \alpha \beta \nu} F_{\alpha \beta} (\partial_\mu R_a) \mathcal{M}_a. \end{split}$$$$ Therefore the canonical energy-momentum tensor would be $$$$\begin{split} (T_{E\mathcal{M}})^{\mu \nu} & = - \frac{1}{2} J^\nu \epsilon^{\mu \alpha \beta \sigma} F_{\alpha \beta} (\partial_\sigma R_a) \mathcal{M}_a \\ & \quad + \frac{1}{2} J_\lambda \epsilon^{\lambda \alpha \beta \nu} F_{\alpha \beta} (\partial^\mu R_a) \mathcal{M}_a \\ & \quad + J_\lambda \epsilon^{\lambda \nu \gamma \sigma} (\partial_\mu A_\gamma) (\partial_\sigma R_a) \mathcal{M}_a. \end{split}$$$$ Then adding a term $$(\partial_\gamma A^\mu) M^{\nu \gamma}$$ from the symmetrization procedure on page 148, where the relevant part for the dipole-field interaction is $$$$(\partial_\gamma A^\mu) J_\lambda \epsilon^{\lambda \gamma \nu \sigma}(\partial_\sigma R_a) \mathcal{M}_a$$$$ should give the symmetrized energy-momentum tensor given in Table 10.2 $$$$\begin{split} (\Theta_{E\mathcal{M}})^{\mu \nu} & = - \frac{1}{2} J^\mu \epsilon^{\nu \alpha \beta \gamma} F_{\alpha \beta} (\partial_\gamma R_a) \mathcal{M}_a \\ & \quad - \frac{1}{2} J^\nu \epsilon^{\mu \alpha \beta \gamma} F_{\alpha \beta} (\partial_\gamma R_a) \mathcal{M}_a \\ & \quad + \frac{1}{2} g^{\mu \nu} J_\gamma \epsilon^{\gamma \alpha \beta \gamma} F_{\alpha \beta} (\partial_\gamma R_a) \mathcal{M}_a \end{split}$$$$ When adding the term from the symmetrization procedure to the canonical energy-momentum I am not able to derive the same expression. Should $$\mathcal{M}_a$$ also be differentiating w.r.t. $$(\partial_\alpha R_a)$$ and if so how is the $$\frac{\partial x_i}{\partial R_a}$$ handled? What exactly does $$\mathcal{M}_a$$ represent if $$M_i$$ is the magnetization in the rest frame?