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Covariance equations of motion and symmetry

  • #1

Homework Statement


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


Homework 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]



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?
 
Last edited:

Answers and Replies

  • #2
Found it by using the action.

Thanks
 

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