Covariance equations of motion and symmetry

by phoenixofflames
Tags: covariance, field theory, symmetry, variation
phoenixofflames is offline
Jan27-11, 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:
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)
L = L(F^a, \chi^{\alpha}, \chi^{\alpha}_{\mu})

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

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?
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phoenixofflames is offline
Jan29-11, 03:42 AM
P: 5
Found it by using the action.


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