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.
Best Answer
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