Covariance equations of motion and symmetry

by phoenixofflames
Tags: covariance, field theory, symmetry, variation
 P: 5 1. The problem statement, all variables and given/known data Hi, I need to proof the covariance of the equations of motion under an infinitesimal symmetry transformation. 2. Relevant equations Equations of motion: $$E_i = \left(\frac{\partial L}{\partial \chi^i}\right) - \partial_{\mu} \left(\frac{\partial L}{\partial \chi^i_{\mu}}\right)$$ Symmetry transformation $$\delta \chi^i = \xi^{\alpha} (\chi)$$ Lagrangian $$L = L(F^a, \chi^{\alpha}, \chi^{\alpha}_{\mu})$$ $$\chi^{\alpha}_{\mu} = \partial_{\mu} \chi^{\alpha}$$ 3. The attempt at a solution $$E_i &= \left(\frac{\partial L}{\partial \chi^i}\right) - \partial_{\mu} \left(\frac{\partial L}{\partial \chi^i_{\mu}}\right)$$ $$= \left(\frac{\partial L}{\partial \chi^{'\alpha}}\right) \left(\frac{\partial \chi^{'\alpha}}{\partial \chi^i}\right) - \partial_{\mu} \left[\left(\frac{\partial L}{\partial \chi^{' \alpha}_{\beta}}\right) \left(\frac{\partial \chi^{' \alpha}_{\beta}}{\partial \chi^i_{\mu}} \right) \right]$$ $$= \left(\frac{\partial L}{\partial \chi^{'i}}\right) + \left(\frac{\partial L}{\partial \chi^{'\alpha}}\right)\left(\frac{\partial \xi^{\alpha}}{\partial \chi^i}\right) - \partial_{\mu} \left[\left(\frac{\partial L}{\partial \chi^{'i}_{\mu}}\right) + \left(\frac{\partial L}{\partial \chi^{' \alpha}_{\mu}}\right) \left(\frac{\partial \xi^{\alpha}}{\partial \chi^i} \right)\right]$$ $$= E^{'}_i + \left(\frac{\partial \xi^{\alpha}}{\partial \chi^i} \right) E_{\alpha}$$ at first order in xi. The answer is $$\delta E_i = - \left(\frac{\partial \xi^{\alpha}}{\partial \chi^i} \right) E_{\alpha}$$ I have no clue actually how to do this... because L is a function of Chi, but I take the partial derivative towards chi' ,... Actually I have no clue how to do it mathematically correct.. Is it completely wrong or... Is there another way,.. Note that $$\delta L$$ is not zero and doesn't need to be a complete derivative. What does this covariance exactly mean?