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

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?

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.

• So writing $F_\mu$ in the form of (1) gives $F_\mu(x) = \int \partial_\mu\phi(y) \delta(x-y) dy$. Attempting to integrate by parts, I get $F_\mu(x) = \phi - \int \phi(y) \delta'(x-y) dy$. Am I to assume the first term vanishes when evaluated over the interval (I am not sure how to argue this is true) or is this not correct? Commented Mar 5, 2022 at 4:12
• @Hannah The surface term will be $\phi(y)\delta(x-y)$ evaluated at infinity, away from the point $x$. Since $\delta$ vanishes away from $y=x$, it will vanish and the surface term vanishes as well. As for the remaining term, just beware that you should get a four-derivative, so it should look like $\partial_\mu \delta$ instead of simply $\delta'$ (recall that $x$ and $y$ are points in a four-dimensional spacetime) Commented Mar 5, 2022 at 4:50
• I assumed $d=4$ in the previous comment, but I think you get the idea Commented Mar 5, 2022 at 4:51
• OK, that makes sense. Now, I am still not confident about applying the functional derivative, although the form is more familiar. My instinct is to write $- \frac{\delta}{\delta \phi(x)} \int\phi(y)\partial_\mu(\delta(x-y))dy = -\int [ \frac{\delta \phi(y)}{\delta \phi(x)}\partial_\mu(\delta(x-y))+0]dy = -\int \delta(y-x)\partial_\mu(\delta(x-y))dy$. This appears to end up being the 4-derivative of unity which is zero. Is that right? Commented Mar 5, 2022 at 5:17
• @Hannah in that expression, you are computing the functional derivative of $F_\mu(x)$ with respect of $\phi(x)$ (the arguments are the same), that's why it's weird. Try differentiating $F_\mu(x)$ wrt $\phi(z)$. You are on the right path though! Commented Mar 5, 2022 at 7:16