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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.

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    $\begingroup$ $circlewithalinethroughit$? $\endgroup$ – Avitus Nov 5 '14 at 11:02
  • $\begingroup$ I assume OP means $\phi$. I've changed the post accordingly, but please correct me if any of my guesses are wrong. Also, OP, the symbol $\nabla$ is spelled nabla, "nambla" is something else entirely... $\endgroup$ – Travis Nov 5 '14 at 11:07
  • $\begingroup$ "Circle with "that" line in it" ...the empty set $\;\emptyset\;$ ? The greek letter phi $\;\Phi\;,\;\;or\;\;\phi\;$ ? Do you really don't know this and you're studying mathematics? And even if we know (of course, the empty set cannot be), what is the definition of that circlewiththatlineinit, which most probably is a function? $\endgroup$ – Timbuc Nov 5 '14 at 11:08
  • $\begingroup$ @Timbuc Often students haven't encountered many Greek letters by the time they start multivariable calculus, except perhaps $\alpha$, $\theta$, and $\pi$, and maybe a few more if they've taken a little physics. $\endgroup$ – Travis Nov 5 '14 at 11:09
  • $\begingroup$ Thanks so much for letting me know this about the Greek letters. I didn't know that but now I do ! Next time I don't know what a symbol is I can google Greek Alphabet ! Travis you are correct in your guesses. Thanks. $\endgroup$ – user3528592 Nov 5 '14 at 11:15
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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 $S$: $$ \int_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = 0.$$ Hence integrals of $\vec{F}$ around loops in $U$ vanish hence $\vec{F}$ is path-independent and then we can prove $\displaystyle \phi(\vec{r}) = \int_{p}^{\vec{r}} \vec{F} \cdot d\vec{r}$ 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. $\oint_C \vec{F} \cdot d\vec{r} =0$ for all closed curves $C$ in $U$
  4. (add condition $U$ be simply connected) $\nabla \times \vec{F}=0$ on $U$.
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