Derivative of a Wilson Line
It's my first post on this website so please excuse any breaches of protocol that I'm unaware of.
I've come across a formula for taking derivatives of Wilson Lines with respect to points on the path, but I'm having trouble deriving it. I found it in Equation 12 of https://arxiv.org/pdf/1210.2581.pdf , where it is presented without proof or citation. I therefore expect that it is an elementary proof that is very familiar to people. My attempt to prove it goes as follows:
$W(x,y) \equiv \hat{P} \exp[ig\int A_\mu(s) d s^\mu] $
Defines the path ordered <PERSON> line with starting point $y^\mu$ and endpoint $x^\mu$. Given a particular field $A_\mu$ I think of this as a functional of the path. Instead, if we can parametrize the path via $s^\mu(t)$, then the <PERSON> line can instead be though of a function of the endpoints:
$W(x,y) = \hat{P} \exp[ig\int_{t_y}^{t_x} A_\mu(s(t))\frac{ ds^\mu}{dt} dt]$
Where we've implied $s^\mu(t_y) = y^\mu$ and $s^\mu(t_x) = x^\mu$. I would like to get $\frac{\partial W(x,y)}{\partial x^\mu}$, but I only have $t_x$ as an accessible variable in the expression. 818
So then I take $\frac{ \partial W(x,y)}{\partial t_x}$ instead:
$\frac{ \partial W(x,y)}{\partial t_x} = igA_\mu(x) \frac{ ds^\mu}{d t}|_{t=t_x}W(x,y) $
Then I think to myself that since $x^\mu$ is a function of $t_x$, maybe I can invert the function and use the chain rule:
$\frac{ d}{dx^\mu} =^? \frac{dt_x}{dx^\mu}\frac{d}{dt_x}$
(By this logic, it should really be $\frac{\partial t_x}{\partial x^\mu}$, but I'm thinking we can write $\frac{ d t_x}{d x^\mu} = (\frac{ d x^\mu}{d t_x})^{-1}$). Then, this $\frac{dt_x}{d x^\mu}$ looks like it will cancel the $\frac{ ds^\mu}{dt}|_{t=t_x}$ in the above expression since $s^\mu(t_x) = x^\mu$. So I write
$\frac{ \partial W(x,y)}{\partial x^\mu} = igA_\mu(x) W(x,y).$
Note that if I took the derivative with respect to the start point $y^\mu$ instead, I would get a similar result but with an extra minus sign from the fundamental theorem of calculus:
$\frac{ \partial W(x,y)}{\partial y^\mu} = -ig W(x,y) A_\mu(y).$
(Throughout I have assumed that the chain rule maintains the ordering) Then, if I want to extend this to taking derivatives with respect to any point on the path, not just an endpoint, I can use the fact that:
$W(x,y) = W(x,z)W(z,y)$
Straightforward application of the product rule and our above expressions then gives:
$\frac{ \partial W(x,y)}{\partial z^\mu} = 0.$
Which seems obvious because we've only allowed $z^\mu$ to move along the pre-specified path for the <PERSON> line. So the <PERSON> line itself does not change. The only way to get a nonzero derivative is to use the start point or the endpoint, because changing them changes the length of the entire path. However, in my understanding, this is actually incorrect. The result is not zero! Equation 12 of https://arxiv.org/pdf/1210.2581. 818