Calculating the functional derivative of $\partial_\mu\phi$ with respect to $\phi$

Question

Given $F_\mu=\partial_\mu\phi$, I need to find the functional derivative $\frac{\delta F}{\delta \phi}$. I am not familiar with the treatment of functional derivatives outside the context of finding Euler-Lagrange equations from the action and I'm struggling to apply various definitions of the functional derivative to this situation. What is the best way to proceed?

Answer 1

I trust you know that, in $d$ dimensions, $$\frac{\delta \phi(x)}{\delta \phi(y)} = \delta^{(d)}(x-y).$$

This comes from the fact that given a function $f(x)$, we can treat it as if it were a functional by noticing that $$f(x) = \int f(y) \delta^{(d)}(x-y) \ \mathrm{d}^dy \tag{1}$$ and proceeding with the usual rules.

Since most of the remaining work has already been done by OP in the comments, I'll complete the answer. Notice we can write $$F_\mu(x) = \int \partial^y_\mu \phi(y) \delta^{(d)}(x-y) \ \mathrm{d}^d y,$$ where $\partial^y_\mu$ denote the derivative with respect to $y$. Integrating by parts, we see that \begin{align} F_\mu(x) &= \int \partial^y_\mu \phi(y) \delta^{(d)}(x-y) \ \mathrm{d}^d y, \\ &= \oint_{\text{surface}} \phi(y) \delta^{(d)}(x-y) \ \mathrm{d}S_\mu - \int \phi(y) \partial^y_\mu\delta^{(d)}(x-y) \ \mathrm{d}^d y, \end{align} where we wrote the volume integral of the gradient as a surface integral. Since the original integral was in the entire space, the surface over which we are integrating is at infinity, away from $y=x$. Since the delta will vanish everywhere apart from $y=x$, we get to

\begin{align} F_\mu(x) &= - \int \phi(y) \partial^y_\mu\delta^{(d)}(x-y) \ \mathrm{d}^d y, \\ &= \int \phi(y) \partial^x_\mu\delta^{(d)}(x-y) \ \mathrm{d}^d y, \end{align} where we used the antisymmetry in $x \leftrightarrow y$ of $\delta^{(d)}(x-y)$ to write $\partial^x_\mu\delta^{(d)}(x-y) = - \partial^y_\mu\delta^{(d)}(x-y)$. Now we see that \begin{align} \frac{\delta F_\mu(x)}{\delta \phi(z)} &= \int \frac{\delta \phi(y)}{\delta \phi(z)} \partial^x_\mu\delta^{(d)}(x-y) \ \mathrm{d}^d y, \\ &= \int \delta^{(d)}(y - z) \partial^x_\mu\delta^{(d)}(x-y) \ \mathrm{d}^d y, \\ &= \partial^x_\mu\delta^{(d)}(x-z). \end{align}

The choice of using different arguments in the functional derivative is for generality: we could consider $x=z$ just as we can consider $\delta^{(d)}(0)$ instead of the more general $\delta^{(d)}(x-z)$. However, this is less general and might lead to divergences if the functional derivative turns out to be a distribution, as is often the case.