Show that the real K-G equation, $(\Box + m^2)\phi=0$ is the EOM for the action $S=\frac12\int d^4x(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2)$

Question

This question concerns a real scalar field.

Show that the real Klein-Gordon equation, $(\Box + m^2)\phi=0$ is the equation of motion, $\delta S[\phi(x)]/\delta\phi(x)=0$, for the action

$$S=\frac12\int d^4x\left(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2\right)$$ by performing a functional variation. Here $\mu = 0,1,2,3$, ($3+1$ space-time dimensions).

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The Lagrangian density is
$$\mathcal{L}=\mathcal{L}\left(\phi,\ \partial^\mu{\phi},\ \partial_\mu{\phi}\right)=\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2\tag{A}$$

Now since the change in functional variation $$\frac{\delta S[\phi(x)]}{\delta\phi(x)}=\lim_{\delta \phi\to 0}\left(\frac{S(\phi + \delta\phi)-S(\phi)}{\delta\phi}\right)=0$$ It follows that $S(\phi + \delta\phi)-S(\phi)$ must be zero for any $\delta\phi$.

The variation in the action, $S$ is therefore $$\delta{S[\phi]}=\frac12\int d^4x\left[\phi+\delta\phi,\ \partial^\mu{\phi}+\delta\left(\partial^\mu{\phi}\right),\ \partial_\mu{\phi}+\delta(\partial_\mu{\phi})\right]$$ and after Taylor expanding to first order in $\delta\phi$ becomes $$\delta{S[\phi(x)]}$$ $$=\int \frac{d^4x}{2} \left[\mathcal{L}\left(\phi, \partial^\mu{\phi}, \partial_\mu{\phi}\right)+\frac{\partial\mathcal{L}}{\partial \phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial\left(\partial^\mu\phi\right)}\delta\left(\partial^\mu\phi\right)+\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi\right)}\delta\left(\partial_\mu\phi\right)-\mathcal{L}\left(\phi, \partial^\mu{\phi}, \partial_\mu{\phi}\right)\right]$$

$$=\frac12\int d^4x\left[-2m^2\phi\delta\phi+\partial_\mu{\phi}\delta\left(\partial^\mu\phi\right)+\partial^\mu{\phi}\delta\left(\partial_\mu\phi\right)\right]\tag{B}$$

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This is as far as I can get and matches the first line of the solution. I will typeset this solution in exactly the same way as the author did to illustrate my confusion:

Take the action

$$S=\frac12\int d^4x\left(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2\right)$$

Now we vary it; $$\delta{S} = \frac12\int d^4x\left[\left(\delta\partial^\mu\phi\right)\partial_\mu{\phi}+\partial^\mu{\phi}\delta\left(\partial_\mu\phi\right)-2m^2\phi\delta\phi\right]\tag{1}$$ $$= \frac12\int d^4x\left[\color{#085}{\left(\partial^\mu\delta\phi\right)\partial_\mu{\phi}}+\partial^\mu{\phi}\left(\partial_\mu\delta\phi\right)-2m^2\phi\delta\phi\right]\tag{2}$$ $$= \int d^4x\left[\partial^\mu\phi\partial_\mu{\delta\phi}-m^2\phi\delta\phi\right]\tag{3}$$ $$=\int d^4x \left[-\partial_\mu (\partial^\mu\phi)\delta\phi-m^2\phi\delta\phi\right]+\int d^4x \partial_{\mu}\left(\delta\phi\partial^\mu\phi\right)\tag{4}$$ and we drop the last term, the total divergence due to boundary conditions, so, $$\delta{S}=-\int d^4x \left(\partial^2\phi+m^2\phi\right)\delta\phi\tag{5}$$

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Was it correct to have 3 arguments for the Lagrangian density in $(\mathrm{A})$, namely to distinguish between the contravariant and covariant derivatives?

What is the justification for commuting the $\delta$ past the derivative in going from $(1)$ to $(2)$?

How did the author get from $(2)$ to $(3)$? It's almost as if the first term in the integrand of $(2)$ (marked green) has been forgotten about.

To go from eqn. $(3)$ to $(4)$ I think integration by parts has been used to factor the $\delta\phi$ out, so integrating the first term of $(3)$ by parts gives $$\int\partial^\mu{\phi}\left(\partial_\mu\delta\phi\right)d^4x=\color{blue}{\left[\partial^\mu{\phi}\int\partial_\mu\left(\delta\phi\right) d^4x \right]}-\int\bigg(\partial_\mu(\partial^\mu\phi)\delta\phi\bigg) d^4x$$ This explains the negative sign for the first term in $(4)$, but why does the boundary term (marked blue) not match the final term of equation $(4)$?

I don't really understand why the final term of $(4)$ can be dropped, but a more pressing question I have is how $(5)$ was deduced from $(4)$.

Put another way, it was my understanding that the covariant derivative is such that $$\partial_\mu\equiv\frac{\partial}{\partial x^\mu}\tag{C}$$
and the contravariant derivative is defined as $$\partial^\mu\equiv\frac{\partial}{\partial x_\mu}\tag{D}$$ but the way it's written in going from $(4)$ to $(5)$ suggests that $$\partial_\mu(\partial^\mu\phi)\stackrel{\color{red}{\mathrm{?}}}{=}\partial^2\phi$$ But how can this possibly be true? Okay, so the $\mu$ index is summed over in accordance with the Einstein summation convention since it is a repeated (dummy) index, but by virtue of $(\mathrm{C})$ and $(\mathrm{D})$ without the $\mu$ index $\partial^2\phi$ does not tell me which variable the field, $\phi$ is being differentiated with respect to.

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Update:

I've been given a good answer that addresses most of my questions nicely. The only part that still puzzles me is how the author was able to immediately write down eqn. $(1)$ in the solution. Is it blatantly obvious that the functional variation, $\delta{S} = \frac12\int d^4x\left[\left(\delta\partial^\mu\phi\right)\partial_\mu{\phi}+\partial^\mu{\phi}\delta\left(\partial_\mu\phi\right)-2m^2\phi\delta\phi\right]$? I had to go through several lines of logical reasoning to justify that equation, including a Taylor expansion.

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Closing remarks

With the bounty time reaching its conclusion and my subsequent comments below one of the answers I know there may not be time to address these questions before the bounty ends. So I'll award the bounty regardless of whether I get a reply to these comments.

I would just like to say a massive thanks to all those that took time and effort to write such great answers. I know I haven't been very good at keeping on top of this question, I just wish I had more time.

Answer 1

We will use the following definition of functional differentiation: \begin{equation} \frac{\delta S}{\delta\phi(y)}=\lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(S[\phi+\epsilon\delta^{(4)}(x-y)]-S[\phi]\right), \end{equation} where $\delta^{(4)}(x-y)$ is the Dirac delta function.

Since \begin{equation} S[\phi]=\frac{1}{2}\int d^4x\left(\partial^\mu{\phi}\partial_\mu{\phi}-m^2\phi^2\right), \end{equation} we have \begin{align} S[\phi+\epsilon\delta^{(4)}(x-y)]&=\frac{1}{2}\int d^4x[\partial^\mu\left(\phi+\epsilon\delta^{(4)}(x-y)\right)\partial_\mu\left(\phi+\epsilon\delta^{(4)}(x-y)\right)\\ &-m^2\left(\phi+\epsilon\delta^{(4)}(x-y)\right)^2]\\ &=\frac{1}{2}\int d^4x[\left(\partial^\mu\phi+\epsilon\partial^\mu\delta^{(4)}(x-y)\right)\left(\partial_\mu\phi+\epsilon\partial_\mu\delta^{(4)}(x-y)\right)\\ &-m^2\left(\phi+\epsilon\delta^{(4)}(x-y)\right)^2]\\ &=S[\phi]+\frac{\epsilon}{2}\int d^4x[(\partial^\mu\phi)\partial_\mu\delta^{(4)}(x-y)+(\partial^\mu\delta^{(4)}(x-y))\partial_\mu\phi\\ &-2m^2\phi\delta^{(4)}(x-y)]+\mathcal{O}(\epsilon^2) \end{align} Since $X^\mu Y_\mu=Y^\mu X_\mu$, we have $(\partial^\mu\delta^{(4)}(x-y))\partial_\mu\phi=(\partial^\mu\phi)\partial_\mu\delta^{(4)}(x-y)$. Therefore, \begin{align} S[\phi+\epsilon\delta^{(4)}(x-y)]&=S[\phi]+\epsilon\int d^4x[(\partial^\mu\phi)\partial_\mu\delta^{(4)}(x-y)-m^2\phi\delta^{(4)}(x-y)]+\mathcal{O}(\epsilon^2). \end{align} Because of the integration by parts, we know that \begin{align} \int d^4x(\partial^\mu\phi)\partial_\mu\delta^{(4)}(x-y)=\int d^4x\partial_\mu((\partial^\mu\phi)\delta^{(4)}(x-y))-\int d^4x(\Box\phi)\delta^{(4)}(x-y), \end{align} where we have used the definition $\partial_\mu\partial^\mu=\Box$.

Discarding boundary terms, we find \begin{align} S[\phi+\epsilon\delta^{(4)}(x-y)]&=S[\phi]-\epsilon\int d^4x(\Box\phi+m^2\phi)\delta^{(4)}(x-y)+\mathcal{O}(\epsilon^2)\\ &=S[\phi]-\epsilon(\Box\phi(y)+m^2\phi(y))+\mathcal{O}(\epsilon^2). \end{align} Substituting this into the definition of functional differentiation, we get \begin{equation} \frac{\delta S}{\delta\phi(y)}=\lim_{\epsilon\to 0}\left(-(\Box\phi(y)+m^2\phi(y))+\mathcal{O}(\epsilon)\right)=-(\Box\phi(y)+m^2\phi(y)). \end{equation} Finally, by using the condition \begin{equation} \frac{\delta S}{\delta\phi(y)}=0, \end{equation} we obtain the following equation of motion \begin{equation} \Box\phi+m^2\phi=0. \end{equation}

Addressing the question in the update of your post.

My answer is no, it is not blatantly obvious. For this reason, let me do the following identification [See section 2.3 Functional Derivatives in Greiner & Reinhardt, Field Quantization] $$\delta\phi(x)=\epsilon\delta^{(4)}(x-y).$$ Having done that, note that $$\delta\left(\partial^\mu\phi(x)\right)=\partial^\mu\phi^\prime(x)-\partial^\mu\phi(x)=\partial^\mu\left(\phi^\prime(x)-\phi(x)\right)=\partial^\mu\left(\delta\phi(x)\right)$$. Therefore, $$\delta\left(\partial^\mu\phi(x)\right)=\epsilon\partial^\mu\delta^{(4)}(x-y).$$ Now, we have all it is necessary to do the following demonstration.$$\delta S[\phi]=S[\phi+\delta\phi]-S[\phi]=S[\phi+\epsilon\delta^{(4)}(x-y)]-S[\phi].$$ We have already done the calculation of $S[\phi(x)+\epsilon\delta^{(4)}(x-y)]-S[\phi]$ before. Therefore, we get \begin{align} \delta S[\phi]&=\frac{\epsilon}{2}\int d^4x[(\partial^\mu\phi)\partial_\mu\delta^{(4)}(x-y)+(\partial^\mu\delta^{(4)}(x-y))\partial_\mu\phi\\ &-2m^2\phi\delta^{(4)}(x-y)]+\mathcal{O}(\epsilon^2)\\ \Rightarrow \delta S[\phi]&=\frac{1}{2}\int d^4x[(\partial^\mu\phi)\delta(\partial_\mu\phi)+\delta(\partial^\mu\phi)\partial_\mu\phi\\ &-2m^2\phi\delta\phi]+\mathcal{O}\left((\delta\phi)^2\right), \end{align} where we have used $\delta\phi(x)=\epsilon\delta^{(4)}(x-y)$ and $\delta\left(\partial^\mu\phi(x)\right)=\epsilon\partial^\mu\delta^{(4)}(x-y)$.

Answer 2

Step $(A)$. Actually, the action is a functional, that is a linear form defined on a functional vector space. Practically, it means that $S[\phi]$ takes a function $\phi$ as input and nothing else is needed, because the functional derivative will include automatically the contributions coming from all its derivatives, being covariant, contravariant or even higher-order derivatives of $\phi$. Nevertheless, the concrete dependency of $S$ on the derivatives of $\phi$ are usually listed within the Lagrangian, as you did in $(A)$, in order to specify of the characteristics of the considered system (e.g. : explicit time dependency, second-order terms, etc.), but it is only a matter of conventions.

Step $(1) \rightarrow (2)$. The functional differential commutes with the partial derivative because they don't act on the same objects; more precisely, they both act on $\phi$ separate points of view, where $\phi$ plays different roles. Indeed, $\delta\phi$ is an infinitesimal variation of the field $\phi$ as an independent input of the action functional - it is the functional analog of $\mathrm{d}x$ in standard calculus, while $\partial^\mu\phi$ represents the usual derivative of $\phi$, that its variation as a dependent function with respect to the independent variable $x_\mu$. Those two limits (since derivatives are basically limits) don't depend on each other.

Step $(2) \rightarrow (3)$. In fact, the green term and the following one are joined together, since they are equal. Indeed, one has : $$ \partial_\mu\phi \, \partial^\mu\delta\phi = g_{\mu\nu}\partial^\nu\phi \, g^{\mu\rho}\partial_\rho\delta\phi = \delta_\nu^\rho \, \partial^\nu\phi \, \partial_\rho\delta\phi = \partial^\mu\phi \, \partial_\mu\delta\phi $$

Step $(3) \rightarrow (4)$. You are right when saying that an integration by parts took place. However, the boundary term generated by the kinetic term $\int\mathrm{d}^4x\; \partial^\mu\phi \, \color{red}{\partial_\mu}\delta\phi$, with the integration by parts being done with respect to the red derivative, is given by $\int\mathrm{d}^\color{red}{3}x\; \partial^\mu\phi \, \delta\phi$ and it is assumed to vanish for a sufficiently smooth field $\phi$.

Step $(4) \rightarrow (5)$. The last integrand is a derivative, so that it counters the integral and adds a constant contribution to the action, which can be ignored, since we are only interested in the extrema (and thus the derivatives) of the functional. In a more standard calculus context, it is equivalent to the argument that the quantity $\int_a^b f'(x) \mathrm{d}x = f(b) - f(a)$ is constant with respect to $f(x)$ as a functional, even if it is a function of $a$ and $b$.

Step $(5)$. The (not very common) symbol $\partial^2$ represents the d'Alembertian : $\partial^2\phi = \partial_\mu\partial^\mu\phi = \square\phi$. I guess it has been chosen in analogy with the Laplacian $\Delta$ as the formal "norm" of the nabla $\vec{\nabla}$, because the d'Alembertian operator is the relativistic norm of the gradient $\partial^\mu$. It also recalls that it plays the role of a second-order derivative in this 4D Lorentzian space.

Final remark. I feel that you are a bit confused by the notation $-$ and relativistic computations with all the mess of indices and the tensorial notation are easy for nobody $-$, that is why I would recommend you to compare the present system with a mere oscillator harmonic, since the real Klein-Gordon equation is nothing else than a harmonic oscillator of frequency $m$ in a four-dimensional space with a Lorentzian metric.

Answer 3

Let ${M}$ be the spacetime manifold. Let there be an action functional \begin{align*} {S}[\Phi]=\frac{1}{2}\int_{M}\mathrm{d}^{4}{x}{\,}\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}\displaystyle{\,}\bigg({g}^{\mu\nu}({x})\frac{\partial\Phi({x})}{\partial{x}^{\mu}}\frac{\partial\Phi({x})}{\partial{x}^{\nu}}-{m}^{2}\Phi({x})^{2}\bigg){\,}{.} \end{align*} For the variational derivative of the action functional follows \begin{align*} \frac{\delta{S}[\Phi]}{\delta\Phi({y})}&=\frac{1}{2}\int_{M}\mathrm{d}^{4}{x}{\,}\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}\displaystyle{\,}\frac{\delta}{\delta\Phi({y})}\bigg({g}^{\mu\nu}({x})\frac{\partial\Phi({x})}{\partial{x}^{\mu}}\frac{\partial\Phi({x})}{\partial{x}^{\nu}}-{m}^{2}\Phi({x})^{2}\bigg)\\[0.5em] &=\frac{1}{2}\int_{M}\mathrm{d}^{4}{x}{\,}\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}\displaystyle{\,}\bigg({2}{g}^{\mu\nu}({x})\frac{\partial\Phi({x})}{\partial{x}^{\nu}}\frac{\delta}{\delta\Phi({y})}\frac{\partial\Phi({x})}{\partial{x}^{\mu}}-2{m}^{2}\Phi({x})\frac{\delta\Phi({x})}{\delta\Phi({y})}\bigg){\,}{.} \end{align*} It is known that \begin{align*} \frac{\delta\Phi({x})}{\delta\Phi({y})}=\delta^{4}({x}-{y}){\quad}\mathrm{and}{\quad}\frac{\delta}{\delta\Phi({y})}\frac{\partial\Phi({x})}{\partial{x}^{\mu}}=\frac{\partial}{\partial{x}^{\mu}}\frac{\delta\Phi({x})}{\delta\Phi({y})}=\frac{\partial\delta^{4}({x}-{y})}{\partial{x}^{\mu}}{\,}{.} \end{align*} Substituting our knowledge into the equation yields \begin{align*} \frac{\delta{S}[\Phi]}{\delta\Phi({y})}&=\int_{M}\mathrm{d}^{4}{x}{\,}\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}\displaystyle{\,}\bigg({g}^{\mu\nu}({x})\frac{\partial\Phi({x})}{\partial{x}^{\nu}}\frac{\partial\delta^{4}({x}-{y})}{\partial{x}^{\mu}}-{m}^{2}\Phi({x})\delta^{4}({x}-{y})\bigg)\\[0.5em] &=\int_{M}\mathrm{d}^{4}{x}{\,}\delta^{4}({x}-{y}){\,}\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}\displaystyle{\,}\bigg(-\frac{1}{\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}}\displaystyle\frac{\partial}{\partial{x}^{\mu}}\bigg(\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({x})]}\displaystyle{\,}{g}^{\mu\nu}({x})\frac{\partial\Phi({x})}{\partial{x}^{\nu}}\bigg)-{m}^{2}\Phi({x})\bigg)\\[0.5em] &=-\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({y})]}\displaystyle{\,}\bigg(\frac{1}{\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({y})]}}\displaystyle\frac{\partial}{\partial{y}^{\mu}}\bigg(\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}({y})]}\displaystyle{\,}{g}^{\mu\nu}({y})\frac{\partial\Phi({y})}{\partial{y}^{\nu}}\bigg)+{m}^{2}\Phi({y})\bigg){\,}{.} \end{align*} The variation of the action ${S}[\Phi]$ with respect to the field $\Phi$ is \begin{align*} \frac{\delta{S}[\Phi]}{\delta\Phi}=-\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}\displaystyle{\,}\bigg(\frac{1}{\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}}\displaystyle\frac{\partial}{\partial{x}^{\mu}}\bigg(\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}\displaystyle{\,}{g}^{\mu\nu}\frac{\partial\Phi}{\partial{x}^{\nu}}\bigg)+{m}^{2}\Phi\bigg){\,}{.} \end{align*} For the corresponding equation of motion $\frac{\delta{S}[\Phi]}{\delta\Phi}={0}$ follows \begin{align*} \frac{1}{\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}}\displaystyle\frac{\partial}{\partial{x}^{\mu}}\bigg(\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}\displaystyle{\,}{g}^{\mu\nu}\frac{\partial\Phi}{\partial{x}^{\nu}}\bigg)+{m}^{2}\Phi={0}{\,}{.} \end{align*} The d'Alembert operator is defined as \begin{align*} \square\Phi={g}^{\mu\nu}\frac{\mathcal{D}}{\mathcal{D}{x}^{\mu}}\frac{\mathcal{D}\Phi}{\mathcal{D}{x}^{\nu}}=\frac{1}{\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}}\displaystyle\frac{\partial}{\partial{x}^{\mu}}\bigg(\textstyle\sqrt{-\mathrm{det}[{g}_{\mu\nu}]}\displaystyle{\,}{g}^{\mu\nu}\frac{\partial\Phi}{\partial{x}^{\nu}}\bigg){\,}{.} \end{align*} The notation $\frac{\mathcal{D}}{\mathcal{D}{x}^{\mu}}$ representing the covariant derivative. Rewriting for the d'Alembert operator finally gives the Klein-Gordon equation \begin{align*} \square\Phi+{m}^{2}\Phi={0}{\,}{.} \end{align*} For a flat spacetime simply substitute ${g}_{\mu\nu}=\eta_{\mu\nu}$ and $\frac{\mathcal{D}}{\mathcal{D}{x}^{\mu}}=\frac{\partial}{\partial{x}^{\mu}}$. Also, the notation for the partial derivatives in special relativity is just $\partial_{\mu}:=\frac{\partial}{\partial{x}^{\mu}}$ and $\partial^{\mu}:=\eta^{\mu\nu}\frac{\partial}{\partial{x}^{\nu}}$, or more specific \begin{align*} \partial_{\mu}=\bigg(\frac{1}{c}\frac{\partial}{\partial{t}},\frac{\partial}{\partial{x}},\frac{\partial}{\partial{y}},\frac{\partial}{\partial{z}}\bigg){\quad}\mathrm{and}{\quad}\partial^{\mu}=\mp\bigg(\frac{1}{c}\frac{\partial}{\partial{t}},-\frac{\partial}{\partial{x}},-\frac{\partial}{\partial{y}},-\frac{\partial}{\partial{z}}\bigg){\,}{.} \end{align*}

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The variational derivative $\frac{\delta}{\delta\Phi({x})}$ is an operator on functionals $\mathcal{L}[\Phi({x})]:=\mathcal{L}({x},\Phi({x}),\partial\Phi({x}),\partial^{2}\Phi({x}),\cdots,\partial^{w}\Phi({x}))$. \begin{align*} \frac{\delta\mathcal{L}[\Phi({x})]}{\delta\Phi({y})}&=\sum_{{k}={0}}^{w}\frac{\partial\mathcal{L}[\Phi({x})]}{\partial\partial^{k}\Phi({x})/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{k}}}\frac{\delta}{\delta\Phi({y})}\frac{\partial^{k}\Phi({x})}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{k}}}\\[0.5em] &=\sum_{{k}={0}}^{w}\frac{\partial\mathcal{L}[\Phi({x})]}{\partial\partial^{k}\Phi({x})/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{k}}}\frac{\partial^{k}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{k}}}\frac{\delta\Phi({x})}{\delta\Phi({y})}\\[0.5em] &=\delta^{4}({x}-{y})\sum_{{k}={0}}^{w}(-{1})^{k}\frac{\partial^{k}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{k}}}\frac{\partial\mathcal{L}[\Phi({x})]}{\partial\partial^{k}\Phi({x})/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{k}}}\\[0.5em] &+\sum_{{k}={1}}^{w}\sum_{{m}+{n}={k},{m}\geqslant{n}}\frac{\partial^{m}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg(\frac{\partial\mathcal{L}[\Phi({x})]}{\partial\partial^{{m}+{n}}\Phi({x})/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}\partial{x}^{\beta_{1}}\cdots\partial{x}^{\beta_{n}}}\frac{\partial^{n}\delta({x}-{y})}{\partial{x}^{\beta_{1}}\cdots\partial{x}^{\beta_{n}}}\bigg) \end{align*} An action functional is then defined through \begin{align*} {S}[\Phi]=\int_{M}\mathrm{d}^{4}{x}{\,}\mathcal{L}[\Phi({x})]{\,}{.} \end{align*} The variational derivative of the action functional is \begin{align*} \frac{\delta{S}[\Phi]}{\delta\Phi({y})}=\int_{M}\mathrm{d}^{4}{x}{\,}\frac{\delta\mathcal{L}[\Phi({x})]}{\delta\Phi({y})}=\sum_{{k}={0}}^{w}(-{1})^{k}\frac{\partial^{k}}{\partial{y}^{\alpha_{1}}\cdots\partial{y}^{\alpha_{k}}}\frac{\partial\mathcal{L}[\Phi({y})]}{\partial\partial^{k}\Phi({y})/\partial{y}^{\alpha_{1}}\cdots\partial{y}^{\alpha_{k}}}{\,}{.} \end{align*} Here, the Lagrangian is a scalar density of weight ${1}$. One can then show that the variational derivative satisfies the following conditions: \begin{align*} (\mathrm{i})&{\quad}\frac{\delta({F}({x})\mathcal{L}_{A}[\Phi({x})]+{G}({x})\mathcal{L}_{B}[\Phi({x})])}{\delta\Phi({y})}={F}({x})\frac{\delta\mathcal{L}_{A}[\Phi({x})]}{\delta\Phi({y})}+{G}({x})\frac{\delta\mathcal{L}_{B}[\Phi({x})]}{\delta\Phi({y})}\\[0.5em] (\mathrm{ii})&{\quad}\frac{\delta(\mathcal{L}_{A}[\Phi({x})]\mathcal{L}_{B}[\Phi({x})])}{\delta\Phi({y})}=\mathcal{L}_{A}[\Phi({x})]\frac{\delta\mathcal{L}_{B}[\Phi({x})]}{\delta\Phi({y})}+\mathcal{L}_{B}[\Phi({x})]\frac{\delta\mathcal{L}_{A}[\Phi({x})]}{\delta\Phi({y})} \end{align*} Applying these rules to our action functional ${S}[\Phi]$ gives \begin{align*} \frac{\delta}{\delta\Phi({y})}\bigg(\frac{\partial\Phi({x})}{\partial{x}^{\mu}}\frac{\partial\Phi({x})}{\partial{x}^{\nu}}\bigg)&=\frac{\partial\Phi({x})}{\partial{x}^{\nu}}\frac{\delta}{\delta\Phi({y})}\frac{\partial\Phi({x})}{\partial{x}^{\mu}}+\frac{\partial\Phi({x})}{\partial{x}^{\mu}}\frac{\delta}{\delta\Phi({y})}\frac{\partial\Phi({x})}{\partial{x}^{\nu}}{\quad}\mathrm{and}\\[0.5em] \frac{\delta\Phi({x})^{2}}{\delta\Phi({y})}&={2}\Phi({x})\frac{\delta\Phi({x})}{\delta\Phi({y})}={2}\Phi({x})\delta^{4}({x}-{y}){\,}{.} \end{align*}