Covariance equations of motion and symmetryby phoenixofflames Tags: covariance, field theory, symmetry, variation 

#1
Jan2711, 05:38 PM

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: [tex] E_i = \left(\frac{\partial L}{\partial \chi^i}\right)  \partial_{\mu} \left(\frac{\partial L}{\partial \chi^i_{\mu}}\right) [/tex] Symmetry transformation [tex] \delta \chi^i = \xi^{\alpha} (\chi) [/tex] Lagrangian [tex] L = L(F^a, \chi^{\alpha}, \chi^{\alpha}_{\mu}) [/tex] [tex] \chi^{\alpha}_{\mu} = \partial_{\mu} \chi^{\alpha} [/tex] 3. The attempt at a solution [tex]E_i &= \left(\frac{\partial L}{\partial \chi^i}\right)  \partial_{\mu} \left(\frac{\partial L}{\partial \chi^i_{\mu}}\right) [/tex] [tex]= \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] [/tex] [tex]= \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] [/tex] [tex]= E^{'}_i + \left(\frac{\partial \xi^{\alpha}}{\partial \chi^i} \right) E_{\alpha}[/tex] at first order in xi. The answer is [tex] \delta E_i =  \left(\frac{\partial \xi^{\alpha}}{\partial \chi^i} \right) E_{\alpha}[/tex] 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 [tex]\delta L[/tex] is not zero and doesn't need to be a complete derivative. What does this covariance exactly mean? 



#2
Jan2911, 03:42 AM

P: 5

Found it by using the action.
Thanks 


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