Euler-Lagrange equations – Field Theory

Consider a function \phi(x) where x is a point in spacetime. If we assume that the Lagrangian is dependent on \phi(x) and its derivative \partial_{\mu} \phi(x) . The action is then given by,

S = \int d^{4}x  \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) 

According to the principle of least action, we have:

\delta S = 0 = \int d^{4}x   \delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x))

\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi  + \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta(\partial_{\mu} \phi)

\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi  + \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \partial_{\mu} (\delta\phi)

According to product rule,

\frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \partial_{\mu} (\delta\phi) + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \delta\phi = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right) 

Therefore,

\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \left( \frac{\partial \mathcal{L}}{\partial \phi} -  \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right)  \right)\delta \phi  + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right) 

\delta S =  \int d^{4}x  \left[  \left( \frac{\partial \mathcal{L}}{\partial \phi} -  \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right)  \right)\delta \phi  + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right) \right] 

The second term in that equation is zero because the end points of the path are fixed i.e

20181118_210223

This then leads to the Euler-Lagrange equation for a field as :

\frac{\partial \mathcal{L}}{\partial \phi}  = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right)  

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