Trying to understand the idea behind path independence for line integrals I’m trying to grasp the idea behind the line integral for conservative vector fields. If I have some vector valued function $F$ which can be represented as $\nabla f$ then I have a conservative field. Now what I’m not fully comprehending is that why the integral $$\int_C F \ dr$$ is path independent. I’ve looked several explanations for this and the analogy for FTC is quite clear so this makes sense in ”symbolic” form, but all the explanations I’ve seen usually explain this via some physical experiement such as traversing to the top of a mountain and since the gravitational field is conservative the work done to get to the top would be independent of the path I’m taking. I’m not a physicist and even this doesn’t quite make sense, I would assume that the work done from the gravitational field to me would be much greater if I start to go for example up sideways instead of directly walking up the mountain? If work is defined as $W=f \cdot d$ wouldn’t the $d$ be much greater if I’m taking a longer path and thus $W$ would get larger?
Is there some other kind of intuition for this that wouldn’t rely so much on physical concepts such as the gravitational, electric or magnetic fields?
Edit: Ah perhaps if I’m taking a path sideways the field vectors would be directed towards me at some angle and thus wouldn’t do as much work as if they would be directed right towards me?
 A: There are two main definitions for the definition of a conservative vector field $\mathbf{F}:\mathbb{R}^d\to\mathbb{R}^d$:

*

*$\exists \Phi:\mathbb{R}^d\to\mathbb{R} \text{ such that }\mathbf{F}=\nabla\Phi$


*In the special case that $d=2$ or $d=3$, we can also say $\nabla\times\mathbf{F}=\mathbf{0}.$


*And of course often given as a definition is $\oint\mathbf{F}(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=0$, which is what you're asking about.
I'll show that the first and third definitions imply the second. Let's start with the first. Letting $O$ be a closed smooth curve in $\mathbb{R}^d$,
$$\oint_{O} \mathbf{F}\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\oint_{O} \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}$$
Seeing as $O$ is smooth, we can parameterize it as $O=\{\mathbf{x}(t)~|~t\in(0,T)\}$, with of course $\mathbf{x}(0)=\mathbf{x}(T)$. With this in mind, we can represent the line element as
$$\mathrm{d}\mathbf{x}=\mathbf{x}'(t)\mathrm{d}t$$
So then
$$\oint_{O} \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\int_0^T \nabla\Phi(\mathbf{x}(t))\boldsymbol{\cdotp}\mathbf{x}'(t)\mathrm{d}t$$
However, $\nabla\Phi(\mathbf{x}(t))=\left(\nabla\Phi\circ\mathbf{x}\right)(t)$, and so via the chain rule,
$$\mathrm{D}_t\left(\Phi\circ\mathbf{x}\right)(t)=\left(\nabla\Phi\circ\mathbf{x}\right)(t)\boldsymbol{\cdotp}\mathbf{x}'(t)=\nabla\Phi(\mathbf{x}(t))\boldsymbol{\cdotp}\mathbf{x}'(t)$$
Hence,
$$\oint_{O} \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\int_0^T \mathrm{D}_t\left(\Phi\circ\mathbf{x}\right)(t)\mathrm{d}t$$
So by FTC,
$$=\left(\Phi\circ\mathbf{x}\right)(T)-\left(\Phi\circ\mathbf{x}\right)(0)$$
$$=\Phi(\mathbf{x}(T))-\Phi(\mathbf{x}(0))=0.$$
Which as per our parameterization, is precisely zero. One can take this idea further to say that for any smooth curve $C$, not necessarily closed, parameterized as $C=\{\mathbf{x}(t)~|~t\in[a,b]\}$ that
$$\int_C \nabla\Phi(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\Phi(\mathbf{x}(b))-\Phi(\mathbf{x}(a))$$
That is to say, it only depends on the endpoints.
In the special case of two or three dimensions, showing (3) implies (2) is practically instant. Let $\Omega$ be the region enclosed by $O$, that is to say, $\partial\Omega=O$. Using Stokes's theorem, one can say that
$$\oint_O \mathbf{F}(\mathbf{x})\boldsymbol{\cdotp}\mathrm{d}\mathbf{x}=\int_{\Omega}\left(\nabla\times\mathbf{F}\right)\boldsymbol{\cdotp}\mathrm{d}\mathbf{n}$$
But, the RHS is trivially zero since by assumption $\nabla\times\mathbf{F}=\mathbf{0}$. Therefore the LHS is zero as well.
