# Lagrangian density for fields

 P: 48 This is probably a minor point, but I have seen in some QFT texts the Euler-Lagrange equation for a scalar field, $$\partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right) - \frac{\delta \cal L}{\delta \phi }=0$$ i.e. $\cal L$ is treated like a functional (seen from the $\delta$ symbol). But why would it be a functional? Functonals map functions into numbers, and in our case $\cal L$ is a function of the fields (and their derivatives).
 Sci Advisor HW Helper PF Gold P: 2,603 If you want to be completely rigorous, the action is the true functional. The variational derivatives of the Lagrangian (density) should be considered distributions.
 P: 48 But why should there be a functional derivative of $\cal L$? we have $\cal L$ which is a function of $(\phi, \partial_\mu \phi)$ and we differentiate (as a function) with respect to $\partial_\mu \phi$
Thanks
P: 4,160
Lagrangian density for fields

 Quote by spookyfish This is probably a minor point, but I have seen in some QFT texts the Euler-Lagrange equation for a scalar field, $$\partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right) - \frac{\delta \cal L}{\delta \phi }=0$$ i.e. $\cal L$ is treated like a functional (seen from the $\delta$ symbol). But why would it be a functional? Functonals map functions into numbers, and in our case $\cal L$ is a function of the fields (and their derivatives).
If they wrote it that way it's a misprint. The derivatives should be ∂'s, not δ's.
 Quote by spookyfish But why should there be a functional derivative of $\cal L$? we have $\cal L$ which is a function of $(\phi, \partial_\mu \phi)$ and we differentiate (as a function) with respect to $\partial_\mu \phi$