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My lecture notes are confusing me here:

They state that if $phi(x,y,z)$ is a scalar function, then its gradient is a vector field $$F(x,y,z) := nabla phi$$

Why is the gradient a vector field?

Then it goes on to say

if $F(x,y,z)$ represents a gradient of some scalar function $phi(x,y,z)$
then $F(x,y,z)$ is a conservative or potential vector field
and $phi(x,y,z)$ is called the potential function of $F(x,y,z)$.

We just learned that if the curl is zero then it is a conservative vector field ?

I'm going to guess at what this means :

If you have a scalar function $phi(x,y,z)$ this means any old function of $x$, $y$, and $z$ then you calculate its gradient as a for a vector field $$frac{d}{dx}i + frac{d}{dy}j + frac{d}{dz}k$$

Then we know that the function $F(x,y,z)$ is a conservative vector field i.e., one with a curl of zero and the scalar function is called the potential function of $F(x,y,z)$

I also don't understand why this is the case? Does it have something to do with the initial vector function being a scalar and not a vector? I can do the calculations but I am just guessing at what is actually happening, would someone please be able to help with a clear explanation?

Thank you.

To begin, the reason the gradient of a scalar-valued function is a vector field is:
$$ nabla phi = langle frac{partial phi}{partial x}, frac{partial phi}{partial y},frac{partial phi}{partial z} rangle $$
which is the formula for a vector field. For each point $p$ in some region of $mathbb{R}^3$ we assign $nabla phi (p)$. This makes $nabla phi$ a vector field.

Now, if $vec{F} = nabla phi$ then $nabla times vec{F} = nabla times nabla phi = 0$. However, the converse is not necessarily true. It is possible to have $nabla times vec{F} =0$ throughout some domain $U subset mathbb{R}^3$ and yet there does not exist $phi$ on all of $U$ such that $nabla phi = vec{F}$. If you want the curl vanishing to imply the existence of the potential function $phi$ then you also need to have a topological condition on $U$. In particular, if $U$ is simply connected then we can apply Stokes' Theorem to arbitrary surfaces in $U$ and for each surface $S$:
$$ int_{partial S} vec{F} cdot dvec{r} = iint_S (nabla times vec{F}) cdot dvec{S} = 0.$$
Hence integrals of $vec{F}$ around loops in $U$ vanish hence $vec{F}$ is path-independent and then we can prove $displaystyle phi(vec{r}) = int_{p}^{vec{r}} vec{F} cdot dvec{r}$ is a valid formula for the construction of a potential in $U$.

In any event, if you are currently taking multivariate calculus and you were just introduced to conservative vector fields then relax. In a week or two these things should all gel together. There are a few moving pieces, but, once you see how they all fit it's really pretty. We have the following equivalence:
Suppose $U$ is an open connected subset of $mathbb{R}^n$ then the following are equivalent

1. $vec{F}$ is conservative; $vec{F}=nabla phi$ on all of $U$


2. $vec{F}$ is path-independent on $U$


3. $oint_C vec{F} cdot dvec{r} =0$ for all closed curves $C$ in $U$


4. (add condition $U$ be simply connected) $nabla times vec{F}=0$ on $U$.