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(Note $\vec{F}$ and $\vec{G}$ are arbitrary 3D vector fields)

So I have been messing with some PDE recently. Some expressions came to mind include

$$\nabla \cdot \vec{F}=0 \hspace{12mm}[1]$$ $$\nabla \times \vec{F}=\vec{0}\hspace{12mm}[2]$$ $$\nabla \phi=0\hspace{12mm}[3]$$

I often seen in some workings in physics textbooks that $[2]$ usually implies

$$\vec{F}=\nabla\phi\hspace{12mm}[4]$$

In physics we are taught about scalar potentials and vector potentials which satisfy the following because of vector calculus identities

$$\nabla\phi=\vec{F},\nabla \times \vec{F}=\vec{0}\hspace{12mm}[5]$$

$$\nabla\times\vec{G}=\vec{F}, \nabla \cdot \vec{F}=0\hspace{12mm}[6]$$

However I noticed for $[1]$ it does not necessary means $[6]$ because there exist irrotational vector fields e.g. $\vec{K}$ where the divergence vanishes everywhere as shown in this link because of how the net inflow/outflow of vectors in an infinitesimal volume is zero

$$\vec{K}=\begin{pmatrix} \frac{x}{r^3} \\\frac{y}{r^3}\\\frac{z}{r^3}\end{pmatrix}$$

where $r^2=x^2+y^2+z^2$

Q1 But can we also have an analogous case for $[2]$, where there exists a vector field $\vec{S}$ without singularities such that it satisfy $[2]$ but not $[4]$ i.e. there is no $\phi$ such that

$$\vec{S}=\nabla\phi$$

is true?

Q2

If no such $\vec{S}$ exists, how to explain why we cannot have a notion of net microscopic circulation analogous to net inflow/outflow of vectors in divergence?

Attempt at imagining what it might look like based on the intuition that $\vec{S}$ has no net microscopic circulation seemed to give me singularities... enter image description here

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  • $\begingroup$ What do you mean by "without singularities"? $K$ has no singularities on its domain (which has a "hole" at $0$). Are you looking for a vector field $\bf F$ on, e.g., all of $\mathbb{R}^n$ for which $\nabla \times {\bf F} = 0$ but which has no potential? There is no such field; one can show this directly, but this is also the statement that a particular cohomology group of $\mathbb{R}^3$ vanishes. $\endgroup$ Feb 4, 2015 at 7:22
  • $\begingroup$ I am not sure whether I have used the correct term. What I am trying to say is that the vector field is smooth so that mixed derivative rules apply (no sudden changes or cusp or similar thing in the functions that form the vector components) I have not learnt cohomology before thus that might be why I am not aware about the proof. I will try to read more about it $\endgroup$
    – Secret
    Feb 4, 2015 at 7:31
  • $\begingroup$ Well, $K$ is also smooth everywhere it is defined, i.e., everywhere except the origin. $\endgroup$ Feb 4, 2015 at 7:32

1 Answer 1

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No. The underlying reason is topological: There's an object called de Rham cohomology that detects whether such functions exist. Unfortunately, it's hard to go into any details without introducing a lot of topological machinery. For $\mathbb{R}^3$, we can explicitly construct $\phi$ with $F = \nabla \phi$ by integrating: $\phi(x) = \int_0^x F.ds$, where the integration is taken along a path from $0$ to $x$. By Stokes' theorem, this integral is path-independent, and so defines a nicely behaved function $\phi$.

On the other hand, if $\phi$ has a singularity, then there's no guarantee that the path won't go through it, and the resulting integral may not be path-independent even when the path does avoid it. The classic example is the function $F$ you give in your post. It's easier to consider the $2$-dimensional version. There, $\int_C F.ds = 2\pi$, where $C$ is the unit circle. If $F$ were a gradient, that integral would have to vanish by path-independence. (Think of the circle as two paths from $1$ to $-1$ joined at their endpoints.) It turns out that in many situations (specifying the exact conditions requires a bit of topology), that's the only obstruction to being the gradient of a field. Specifically, there are integrals over finitely many spaces that vanish iff a irrotational field $F$ has $F = \nabla \phi$ for some $\phi$.

I realize that the above is vague and possibly not very useful, but the answer to your question is 'no' in general: Irrotational fields on $\mathbb{R}^3$ with singularities, or more generally fields on other spaces, are not necessarily of the form $F = \nabla \phi$. To go further into the matter, the next place to look is de Rham cohomology.

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