The principle of stationary action derivation

Start with the action$$S[\phi]=\int d^4x \, \mathcal{L}(\phi,\partial_\mu \phi)$$S[ϕ]=∫d4xL(ϕ,∂μ​ϕ)

The physical field configuration is the one for which a small variation of the field,$$\phi(x)\rightarrow \phi(x)+\delta\phi(x)$$ϕ(x)→ϕ(x)+δϕ(x)

does not change the action to first order:$$\delta S=0$$δS=0

Now vary the action:$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \delta(\partial_\mu\phi) \right]$$δS=∫d4x[∂ϕ∂L​δϕ+∂(∂μ​ϕ)∂L​δ(∂μ​ϕ)]

Since variation and differentiation commute,$$\delta(\partial_\mu\phi) = \partial_\mu(\delta\phi)$$δ(∂μ​ϕ)=∂μ​(δϕ)

so$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \partial_\mu(\delta\phi) \right]$$δS=∫d4x[∂ϕ∂L​δϕ+∂(∂μ​ϕ)∂L​∂μ​(δϕ)]

Now integrate the second term by parts:$$\int d^4x\, \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \partial_\mu(\delta\phi) = – \int d^4x\, \partial_\mu \left[ \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right] \delta\phi$$∫d4x∂(∂μ​ϕ)∂L​∂μ​(δϕ)=−∫d4x∂μ​[∂(∂μ​ϕ)∂L​]δϕ

The boundary term vanishes because we assume$$\delta\phi=0$$δϕ=0

on the boundary of spacetime.

Therefore,$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi} – \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right) \right]\delta\phi$$δS=∫d4x[∂ϕ∂L​−∂μ​(∂(∂μ​ϕ)∂L​)]δϕ

For the action to be stationary for arbitrary $\delta\phi$δϕ, the bracket must vanish:$$\boxed{ \frac{\partial \mathcal{L}}{\partial \phi} – \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right) =0 }$$∂ϕ∂L​−∂μ​(∂(∂μ​ϕ)∂L​)=0​

This is the Euler–Lagrange equation for a field.

For the scalar field,$$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi – \frac{1}{2}m^2\phi^2$$L=21​∂μ​ϕ∂μϕ−21​m2ϕ2

we get$$\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi$$∂ϕ∂L​=−m2ϕ

and$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi$$∂(∂μ​ϕ)∂L​=∂μϕ

So the Euler–Lagrange equation becomes$$-m^2\phi-\partial_\mu\partial^\mu\phi=0$$−m2ϕ−∂μ​∂μϕ=0

or$$\boxed{ (\partial_\mu\partial^\mu+m^2)\phi=0 }$$(∂μ​∂μ+m2)ϕ=0​

That is the Klein–Gordon equation.

In plain English:$$\boxed{ \delta S=0 \quad\Rightarrow\quad \text{field chooses stationary action} \quad\Rightarrow\quad \text{Euler–Lagrange equation} \quad\Rightarrow\quad \text{Klein–Gordon equation} }$$δS=0⇒field chooses stationary action⇒Euler–Lagrange equation⇒Klein–Gordon equation