Lagrangian density for fieldsby spookyfish Tags: density, fields, lagrangian, lagrangian density, quantum field theory 

#1
Oct2413, 11:34 PM

P: 28

This is probably a minor point, but I have seen in some QFT texts the EulerLagrange equation for a scalar field,
[tex] \partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right)  \frac{\delta \cal L}{\delta \phi }=0 [/tex] i.e. [itex] \cal L [/itex] is treated like a functional (seen from the [itex] \delta [/itex] symbol). But why would it be a functional? Functonals map functions into numbers, and in our case [itex] \cal L [/itex] is a function of the fields (and their derivatives). 



#2
Oct2513, 12:01 AM

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If you want to be completely rigorous, the action is the true functional. The variational derivatives of the Lagrangian (density) should be considered distributions.




#3
Oct2513, 08:46 AM

P: 28

But why should there be a functional derivative of [itex] \cal L [/itex]? we have [itex] \cal L [/itex] which is a function of [itex] (\phi, \partial_\mu \phi) [/itex] and we differentiate (as a function) with respect to [itex] \partial_\mu \phi [/itex]




#4
Oct2513, 09:18 AM

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Thanks
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Lagrangian density for fields 



#5
Oct2513, 10:46 AM

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PF Gold
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