# Lagrangian Dynamics of Fields¶

Note

Before step into this section, we suggest you review the Lagrangian dynamics of classical point particles, as we are not going to go through that.

The Lagrangian density of a field $$\phi(x)$$ is a map $$\mathcal L(\phi,\partial_\mu\phi)$$ of field $$\phi$$ and its derivative $$\partial_\mu\phi$$. Thus, the action can be constructed as

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

Now, perform the variation of the action

$\begin{split}\delta S &= \int \delta\mathcal L \,\mathrm d^4x\\ &= \int \left({\partial\mathcal L\over \partial \phi}\delta\phi + {\partial\mathcal L\over\partial\partial_\mu\phi}\delta\partial_\mu\phi\right)\,\mathrm d^4x\\ &= \int \left({\partial\mathcal L\over \partial \phi}\delta\phi + {\partial\mathcal L\over\partial\partial_\mu\phi}\partial_\mu\delta\phi\right)\,\mathrm d^4x\\ &= \int \left({\partial\mathcal L\over \partial \phi} - \partial_\mu {\partial\mathcal L\over\partial\partial_\mu\phi}\right)\delta\phi\,\mathrm d^4x + \int \partial_\mu \left({\partial\mathcal L\over\partial\partial_\mu\phi}\delta\phi\right)\,\mathrm d^4x\\ &= \int \left({\partial\mathcal L\over \partial \phi} - \partial_\mu {\partial\mathcal L\over\partial\partial_\mu\phi}\right)\delta\phi\,\mathrm d^4x = 0\end{split}$

The second equality is due to the fact that the variation of $$\phi$$ does not change the coordinate system; the fourth equality is because the second integral is a surface term under Stokes theorem, and the variation on the surface is zero.

Thus, to make the equality hold, the only possible way is

${\partial\mathcal L\over \partial \phi} - \partial_\mu {\partial\mathcal L\over\partial\partial_\mu\phi} = 0$

This is the Lagrangian equation for fields.