Covariance equations of motion and symmetry

[phoenixofflames]

Homework Statement

Hi, I need to proof the covariance of the equations of motion under an infinitesimal symmetry transformation.

Homework 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
$$L = L(F^a, \chi^{\alpha}, \chi^{\alpha}_{\mu})$$

$$\chi^{\alpha}_{\mu} = \partial_{\mu} \chi^{\alpha}$$

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?

 

[phoenixofflames]

Found it by using the action.

Thanks