# Inverse-Square vector fields have both a divergence and curl of $0$?

Consider an inverse- square vector field

$$\vec{F} = \frac{x}{r^3}\hat{x} + \frac{y}{r^3}\hat{y} + \frac{z}{r^3}\hat{z} = \frac{\hat{r}}{r^2}$$

where $r = \sqrt{x^2 + y^2 + z^2}$. The curl $\nabla \times \vec{F} = \vec{0}$, therefore we might go looking for a potential $V$. I find that $V = -1/r$ works and therefore one can say that $\vec{F} = \nabla V$ is derivable from a potential function $V$. I'll point out right now that $\vec{F}$ is undefined at the origin.

The divergence $\nabla \cdot \vec{F} = 0$. Therefore, we might go looking for a vector potential $\vec{A}$ such that $\vec{F} = \nabla \times \vec{A}$. One would say that $\vec{F}$ is derivable from a vector potential $\vec{A}$. But I'm having trouble seeing that an inverse-square vector field is derivable from both a vector potential and a scalar potential. So I know we have a trouble point at the origin. Yet this trouble point doesn't really seem to affect the "conservativeness" or path-independence of the vector field. But this trouble point does seem to affect the surface-independence of the vector field. As long as the surface doesn't wrap around the origin, I'd expect the inverse-square vector field to be surface-independent for a given boundary curve.

Can an inverse-square vector field be derivable from both a scalar potential and a separate a vector potential? (Helmholtz theorem comes to mind. But the question I'm asking involves two separate equations. One $\nabla V$ gives $\vec{F}$ and another $\nabla \times \vec{A}$ gives $\vec{F}$ as well).

• In a region $V$ bounded by $S$ for which both $\nabla\cdot \vec A(\vec r)=0$ and $\nabla\times\vec A(\vec r)=0$, Helmholtz's Theorem reveals $$\vec A(\vec r)=\nabla \oint_S \frac{\hat n'\cdot \vec A(\vec r')}{|\vec r-\vec r'|}\,dS'-\nabla \times\oint_S \frac{\hat n'\times \vec A(\vec r')}{|\vec r-\vec r'|}\,dS'$$ If $\vec A(\vec r)=\frac{\hat r}{r^2}$ for $\vec r\ne0$, then we find $$\vec A(\vec r)=\nabla \oint_S \frac{\hat n'\cdot \hat r'}{r'^2|\vec r-\vec r'|}\,dS'-\nabla \times\oint_S \frac{\hat n'\times \hat r'}{r'^2|\vec r-\vec r'|}\,dS'$$ for $\vec r\ne 0$ and $\vec r\in V$ – Mark Viola May 18 '18 at 18:48
• @MarkViola Don't those surface integrals go to zero since $\vec{F}$ falls faster than $1/r$? – DWade64 May 18 '18 at 19:13
• The boundary $S$ cannot enclose the origin (And therefore, the domain $V$ cannot be $\mathbb{R^3}$ ). And the surface integrals sum to $\displaystyle \frac{\hat r}{r^2}\ne0$ for any $\vec r\ne 0$.. – Mark Viola May 18 '18 at 19:21
• @MarkViola Thank you for your comments. Since we already know that $V = -1/r$ is the scalar potential which gives the inverse-square vector field, shouldn't that first surface integral (in the last line) be $-1/r$ so that $\nabla$ of said surface integral gives $\hat{r}/r^2$? In which case, this leaves no room for having the 2nd term (the curl of the 2nd surface integral). This term must be 0 somehow? Therefore we find that the surface integral (the vector potential) is either 1) a constant or 2) the dirac string vector potential shown below. Aren't we led to a contradiction? – DWade64 May 22 '18 at 15:34
• @MarkViola if $\nabla \times \vec{F} = 0$, then all we need is a scalar potential, and that's it. If $\nabla \cdot \vec{F} = 0$, then all we need is a vector potential, and that's it. Both of these previous statements were made under the assumption of domains not containing the origin. But Helmholtz's theorem is say something different? I already know that $V = -1/r$ is the scalar potential? I'm confused how Helmholtz either supports or contradicts these previous statements – DWade64 May 22 '18 at 15:38

If all you want is an inverse-square vector field in a topologically trivial region, then the Dirac string vector potential does the trick: $$\vec{A} = \frac{1 - \cos \theta}{r \sin \theta} \hat{\phi} = \frac{\tan (\theta/2)}{r} \hat{\phi}.$$ It is not too hard to show that the curl of this vector field is $$\vec{\nabla} \times \vec{A} = \frac{1}{r^2} \hat{r}.$$ However, $\vec{A}$ is ill-defined when $\theta = \pi$, which means that it can't be extended to all of space. As noted in my previous answer, there are topological obstructions to extending any such vector field over all of space.
• I'm going to have to think some more about this, but since $\hat{r}/r^2$ is undefined at the origin, why is $V$ defined over all of space (except the origin)? Meaning, when we circulate around an $E$ field, the trouble point at the origin should cause trouble, but it doesn't? That is, we can circulate around the unit circle, and when we get back to our starting point, the potential energy is the same value? So although the trouble point doesn't really hurt the conservativeness of $E$, it does hurt $A$? The trouble point causes a $4\pi$ in fluxes through closed surfaces containing the origin – DWade64 May 21 '18 at 14:39
• @DWade64: I'm not quite sure what you're getting at, but note that $\vec{A}$ can't possibly be conservative. A necessary condition for a conservative force is that $\vec{\nabla} \times \vec{F} = 0$; but $\vec{A}$ has a non-zero curl by construction. – Michael Seifert May 21 '18 at 14:45
• Do you know how this result fits in with Helmholtz's theorem? $\Phi$ and $\vec{A}$ are made up of 2 integrals each. Now I try to apply Helmholtz's theorem to a ball of space not including the origin (say something in the 1st quadrant). The two volume integrals are $0$. We are left with the 2 surface integrals. Somehow, one of the surface integral in Helmholtz's theorem should yield the $\vec{A}$ that you give? – DWade64 May 27 '18 at 19:23
• I'm trying to understand how this above example also fits with the fact that $\nabla (-1/r)$ fits the vector field. How does $V = -1/r$, the $\vec{A}$ that you give, and Helmholtz all support one another. It might have to be with the uniqueness of the decomposition. For instance, over the ball of space that I give, is there a unique decomposition? If there is, I feel like we get into contradictions. But if there isn't, then maybe a decomposition by those two surface integrals gives an answer that can be made to give just your $\vec{A}$, or just $V = -1/r$, or some weird combo summing to field – DWade64 May 27 '18 at 20:32