fields action principle

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:

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

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

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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Euler-Lagrange equations – Field Theory