Say I have a point charge somewhere of charge $Q$ and I want to calculate the work it'd take to bring another charge, of charge $q$ this time, into a distance of $R$ from it.
I know the final results I'm supposed to expect: $k\frac{Qq}{R}$, and most of the reasoning makes sense, but when I attempted to derive it myself, I arrived at the following conundrum:
To my understanding, when performing a change in integration coordinates, I am supposed to multiply the integrand by the Jacobian determinant. In this case, the coordinates that I see as "natural" (why?) are $(x,y)$, and I wanted to use spherical coordinates to easily express the distance in the electrical force formula, so I'd theoreitcally have to multiply my integrand by $|J|=r$.
This is wrong since I would get:
$\int_\infty^R \left( E(r) \cdot r \right) dr$
And I'm supposed to receive the above without the multiplicative $r$ to obtain the known result.
That brings me to a more general question: why do I choose to consider $(x,y)$ as the "natural" coordinates? And why it is with respect to them that I would consider coordinates "changed"? And why is it that in the above case I do not need a Jacobian determinant? And if I chose to calculate this integral in $(x,y)$ coordinates, would I need to multiply by a determinant to change from $(r,\theta)$ to $(x,y)$, in this case (or in other words: are the "natural" coordinates here actually spherical?)?
I feel like I'm missing something obvious and would appreciate assistance.
So the general rule when you have $n$ variables, and you are integrating some function over $\mathbb R^n$ is as follows: $$\int_{\phi(V)}Fdy^1\cdots y^n=\int_V|D\phi|F\circ \phi dx^1\cdots dx^n$$ where $\phi$ is the coordinate transformation $(y_1,\dots, y_n)=\phi(x_1,\dots, x_n)$ and $|D\phi|$ is my notation for the absolute value of the determinant of the Jacobian $D\phi$.
In one variable, the determinant is just the derivative, so if $y=\phi(x)$, then the integration formula just states that: $$\int_{\phi(a)}^{\phi(b)}f(y)dy=\int_a^b\frac{d\phi}{dx}f\circ \phi(x)dx$$ This should just look like $u$-substitution, because it is.
Now, a line integral of a vector field, is an integral a long a path. If $\gamma:(a,b)\longrightarrow\mathbb R^n$, given by $t\mapsto (x^1(t),\dots, x^n(t))$ is a path connecting $\gamma(a)$ and $\gamma(b)$, then we have that: $$\int_{C}\vec{F}\cdot d\gamma=\int_a^b(\vec{F}\circ \gamma(t))\cdot \dot{\gamma}(t)dt$$ where $C$ is the path sketched out by $\gamma$. Now for line integrals of vector fields, the rule is much different, because when you change coordinates you get different basis vectors, but we still have an equality when changing coordinates.
In other words, if $\vec{F}$ is given in standard basis coordinates we have that $E=F_x\hat{x}+F_y\hat{y}$. In polar coordinates, we have that: $$F=(F_x\cos\theta+F_y\sin\theta)\hat{r}+(F_y\cos\theta-F_x\sin\theta)\hat{\theta}$$ The new curve is given by $$\phi\circ \gamma=\left(\sqrt{x^2(t)+y^2(t)},\arctan^{-1}(y,x)\right)$$ Luckily for us, if our curve is given by a straight line, then we have that $\gamma(t)=(t,\theta)$ (after reparameterizing by arc length), where $\theta$ is the angle that the straight line makes, and moreover, the electric field only has components in the $\hat{r}$ direction. So, while we could integrate in $x$ and $y$ coordinates, the electric field is given by: $$E=|E(r)|\hat{r}$$ for all in these coordinates, and $d\gamma(t)=\hat{r}dt$, our integral becomes: \begin{align} \int_C\vec{E}\cdot d\gamma=\int_{t_0}^{t_1}|E(t)|dt \end{align} since the value of $t$ at each point is the distance $r$, we have that: \begin{align} \int_C\vec{E}\cdot d\gamma=\int_{r_0}^{r_1}|E(r)dr \end{align}
The crux of this story is that when you integrate vector fields over paths, you can change coordinates, but then you also need to change not only your line element, but also the basis vectors in which you write your vector fields. I think Griffith's Electrodynamics should probably talk about this in more depth in it's first chapter.
ETA: The Jacobian matrix is precisely the matrix which rewrites the vector fields in a new basis, so that’s where it comes in.