# If $\vec{\nabla} \cdot \vec{V} \neq 0$ at only one point, will this prevent us from saying that $\vec{V}=\vec{\nabla} \times \vec{U}$?

This question has an answer in the language of high level mathematics. Can somebody explain this in the language of vector calculus.

Part I: Let us consider Cartesian coordinate system with origin $$O$$ and axes $$x,y,z$$. Let:

$$r=\sqrt{x^2+y^2+z^2}$$

$$\text{and }\vec{V}=0\ (\hat{i}) +\dfrac{\partial}{\partial z} \left( \dfrac{1}{r} \right) (\hat{j}) -\dfrac{\partial}{\partial y} \left( \dfrac{1}{r} \right) (\hat{k})$$

It is obvious that $$\dfrac{1}{r}$$ is defined everywhere except at the origin.

Now let us take the divergence of $$\vec{V}$$:

$$\vec{\nabla} \cdot \vec{V}=0$$

Since $$\dfrac{1}{r}$$ is not defined at the origin, $$\vec{\nabla} \cdot \vec{V}=0$$ is true everywhere except at the origin.

Since $$\vec{\nabla} \cdot \vec{V} \neq 0$$ at a point $$P (0,0,0)$$, will this prevent us from concluding $$\vec{V}=\vec{\nabla} \times \vec{U}$$ at points other than $$P$$? Why? Why not?

Part II: If $$\vec{\nabla} \cdot \vec{V} \neq 0$$ at points on a one dimensional arbitrary curve in space, will this prevent us from concluding $$\vec{V}=\vec{\nabla} \times \vec{U}$$ at other points not on the curve? Why? Why not?

NOTE - For both Part I and Part II:

If (Why/Why not) is beyond the scope of vector (multivariable) calculus, just reply as yes/no. However please try your best to explain (Why/Why not) in the language of vector (multivariable) calculus.

SEMI ANSWER: Please point out the limitations

I have stumbled upon a derivation in the language of elementary vector calculus. Please point out if there are any limitations in my derivation. In the context of advanced mathematics (de Rham cohomology or Poincare lemma), it seems to me that there are limitations.

Derivation:

To prove: At all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$), if $$\vec{\nabla} \cdot \vec{B}=0$$, then $$\vec{B}=\vec{\nabla} \times \vec{A}$$

Proof:

At all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$): \begin{align} \vec{B} &= B_x\ (\hat{i}) + B_y\ (\hat{j}) + B_z\ (\hat{k})\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j}) + \int^{(x,y,z)}_{(x,y,\infty)} \dfrac{\partial B_z}{\partial z}\ dz\ (\hat{k})\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j}) - \int^{(x,y,z)}_{(x,y,\infty)} \left( \vec{\nabla} \cdot \vec{B} - \dfrac{\partial B_z}{\partial z}\ \right) dz\ (\hat{k})\\ &\text{{Since \vec{\nabla} \cdot \vec{B}=0}}\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j}) - \int^{(x,y,z)}_{(x,y,\infty)} \left( \dfrac{\partial B_x}{\partial x} + \dfrac{\partial B_y}{\partial y}\ \right) dz\ (\hat{k})\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j})\ + \left[ \dfrac{\partial}{\partial x} \left(- \int^{(x,y,z)}_{(x,y,\infty)}B_x\ dz \right) -\dfrac{\partial}{\partial y} \left( \int^{(x,y,z)}_{(x,y,\infty)}B_y\ dz \right) \right] (\hat{k})\\ &\text{{By changing the order of integration and differentiation}}\\ \end{align}

At all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$), let's define:

$$\displaystyle \vec{A}=\int^{(x,y,z)}_{(x,y,\infty)}B_y\ dz\ (\hat{i}) - \int^{(x,y,z)}_{(x,y,\infty)}B_x\ dz\ (\hat{j}) + 0\ (\hat{k}) + \vec{\nabla}f$$

where $$f$$ is an arbitrary function of $$(x,y,z)$$

Therefore at all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$):

\begin{align} \vec{B} &= \left( \dfrac{\partial A_z}{\partial y}-\dfrac{\partial A_y}{\partial z} \right) (\hat{i}) +\left( \dfrac{\partial A_x}{\partial z}-\dfrac{\partial A_z}{\partial x} \right) (\hat{j}) +\left( \dfrac{\partial A_y}{\partial x}-\dfrac{\partial A_x}{\partial y} \right) (\hat{k})\\ &= \vec{\nabla} \times \vec{A} \end{align}

• Does the statement say something about how it is defined $\vec U$? – manooooh Nov 27 '18 at 7:54
• $\vec{U} \rightarrow \vec{U} +\vec{\nabla}f$ is the vector whose curl gives $\vec{V}$......$f$ is an arbitrary function of $x,y,z$ – Joe Nov 27 '18 at 9:03
• The origin isn't in the domain of $\bf V$, so it doesn't even make sense to ask whether $\nabla \cdot {\bf V} = 0$. It's true that $\nabla \cdot {\bf V} = 0$ everywhere on the domain of $V$. – Travis Nov 27 '18 at 10:43
• How do we know that the origin isn't in the domain of $\bf {V}$? – Joe Nov 27 '18 at 11:00
• Because the quantity $\frac{1}{r}$ that appears in the definition of $\bf V$ isn't defined at the origin. – Travis Nov 27 '18 at 12:42

## 2 Answers

You seem to know that a divergence free field $$\vec V$$ can be regarded as curl of some other field: There is a field $$\vec U$$ such that $$\vec V={\rm curl}(\vec U)$$. This is a consequence of the so-called Poincaré Lemma.

But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $$\vec U$$ only if the domain of $$\vec V$$ is, e.g., a ball or star shaped. For your field $$\vec V$$ this is not the case. Therefore we only can say the following: Each point $${\bf p}$$ in the punctured space $$\dot{\mathbb R}^3:={\mathbb R}^3\setminus\{{\bf 0}\}$$ is the center of an open ball $$B_r({\bf p})\subset \dot{\mathbb R}^3$$ such that within $$B_r({\bf p})$$ the field $$\vec V$$ can be written in the form $$\vec V={\rm curl}(\vec U)$$ for some $$\vec U$$ defined in $$B_r({\bf p})$$ only. These local fields $$\vec U$$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $$\vec U_*$$, which then would be defined on all of $$\dot{\mathbb R}^3$$.

Of course, it could be that "by coincidence" the $$\vec V$$ in your example possesses a global vector potential $$\vec U_*$$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $$\vec V$$ across a sphere around $${\bf 0}$$ should be $$=0$$) is fulfilled. Consider as an analogue the functions $$z\mapsto{1\over z}$$ and $$z\mapsto{1\over z^2}$$ in the punctured complex plane $$\dot{\mathbb C}$$. Both have local primitives. But one of them does not have a global primitive in $$\dot{\mathbb C}$$, the other does, the reason being that $$\int_{\partial D}{1\over z}\>dz=2\pi i\ne0\>, \qquad \int_{\partial D}{1\over z^2}\>dz=0\ .$$

Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.

• Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand. – Joe Nov 27 '18 at 11:59
• You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand. – Christian Blatter Nov 27 '18 at 14:22
• "The flux of $\vec{V}$ across a sphere around $\bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere.... – Joe Nov 27 '18 at 16:37
• Full answer. What is the meaning of $\dot{\Bbb R}^3$? – manooooh Nov 27 '18 at 16:51
• I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${\mathbb R}^3$ or balls. – Christian Blatter Nov 27 '18 at 16:52

It is a consequence of the Kelvin-Stokes theorem:

https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem

If you take a potential $$U=\frac{1}{r}$$ with associated "electric field" $$E:=\nabla U$$ then you know that the flux of $$E$$ through a unit sphere $$\Sigma$$ centered around $$(0,0,0)$$ is 1.

However by Kelvin-Stokes if you could write $$E=curl(B)$$ for a certain vector field $$B$$ then the flux of $$E$$ through $$\Sigma$$ would be equal to 0 since $$\Sigma$$ is closed surface.