Can a static non-conservative vector field have scalar potential?

Question

STATEMENT#1: A vector field can be considered as conservative if the field can have its scalar potential.
STATEMENT#2 If we can have non-zero line integral of any vector field along with a single loop then the field can be considered as non-conservative.

STATEMENT#3 If a static vector field F is defined everywhere, then if we get curl(F)=0 then we can say that 𝐅 is a static conservative vector field.

STATEMENT#4 (I think, this is the wrong statement) for a static vector field A, if we get curl(A)=0, then we can make a conclusion about curl(A) will be zero only for the region where A is defined.

STATEMENT#4" (Correction in that statement) If a vector field 𝐀 is not defined everywhere then we can't conclude anything about the field being rotational or irrotational just based on the curl of that field is zero or non-zero.

Please make corrections in these statements.

Answer 1

You cannot define $\mathbf F$ as the gradient of a scalar potential unless it is conservative. If the restriction of $\mathbf F$ to a particular domain is irrotational (i.e. $\nabla\times\mathbf F = 0$ everywhere in that domain) and the domain is simply connected, then the restriction of $\mathbf F$ to that domain is conservative and can be expressed as the gradient of a scalar potential. However, you can still define a scalar potential even in the case where $\mathbf F$ is non-conservative. By the Helmholtz Theorem, you can say $\mathbf F = -\nabla \Phi + \nabla \times \mathbf A$, where $\Phi$ is the scalar potential and $\mathbf A$ is the vector potential. The basic idea of this is that $\mathbf F$ can be decomposed into an irrotational part $(-\nabla\Phi)$ and a divergence-free part $(\nabla\times\mathbf A)$, where is conservative field refers to the case where the divergence-free part is zero.

In the example you have given, $\nabla\times\mathbf H$ is zero outside the wire, but non-zero inside the wire. The Maxwell-Ampère Law is true everywhere, not just on the surface of the wire, and if you use its integral form to evaluate the line integral, it is easy to see that it is non-zero. As $\mathbf H$ is non-conservative, it cannot be expressed as the gradient of a scalar potential, because that does not include the divergence-free part. This is because, while $\mathbf H$ is irrotational outside the wire, the set of points outside the wire is not simply connected. Another way of looking at this is that while the current density is zero outside the wire, the magnetic vector potential is not, so it is the $\nabla\times\mathbf A$ term that gives rise to the magnetic field outside the wire. (Note: $\mathbf A$ is traditionally used to refer to the vector potential corresponding to the $\mathbf B$ field, but here I use it to refer to the vector potential of $\mathbf H$.)

Answer 2

STATEMENT#3 If a static vector field $\mathbf F$ is defined everywhere, then if we get $\operatorname{curl}(\mathbf F)=0$ then we can say that 𝐅 is a static conservative vector field.

I assume that when you say $\mathbf F$ is defined everywhere, you mean that its domain is all of $\mathbb R^n$ for some $n$ (usually 2 or 3). Assuming that $\operatorname{curl}(\mathbf F)$ is also defined on all of $\mathbb R^n$ and vanishes everywhere, then this statement is correct because $\mathbb R^n$ is simply connected. However, note that this is a fairly limited case - we often model magnetic fields on restricted domains which do not satisfy this property.

STATEMENT#4 (I think, this is the wrong statement) for a static vector field $\mathbf A$, if we get $\operatorname{curl}(\mathbf A)=0$, then we can make a conclusion about $\operatorname{curl}(\mathbf A)$ will be zero only for the region where $\mathbf A$ is defined.

This statement is incorrect because it doesn't make sense. What does it mean for $\operatorname{curl}(\mathbf A)$ to be zero or not in a region where $\mathbf A$ is not defined? The domain of definition of $\operatorname{curl}(\mathbf A)$ is a subset of the domain of definition of $\mathbf A$, and it only makes sense to talk about a vector field and its properties on the domain on which it is defined.

STATEMENT#4" (Correction in that statement) If a vector field 𝐀 is not defined everywhere then we can't conclude anything about the field being rotational or irrotational just based on the curl of that field is zero or non-zero.

This is incorrect. A continuously-differentiable vector field defined on some open domain $U$ is called irrotational if its curl vanishes at all points in $U$. The fact that $U\neq \mathbb R^n$ in general is irrelevant.

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Does this mean that 𝐅 will be a static conservative vector field for the region where 𝐅 is defined and curl(𝐅)=0?

No. This is true only if the region $U$ on which $\mathbf F$ is defined is simply connected. If $U$ is not simply connected, then you will generally be able to write it as the gradient of a scalar potential in patches, but you cannot do it for all of $U$.

As an example, consider the vector field

$$\mathbf F = -\frac{y}{x^2+y^2} \hat x + \frac{x}{x^2+y^2} \hat y$$

defined on $U = \left\{(x,y)\in\mathbb R^2 : x^2+y^2>1 \right\}$. This is what the magnetic field outside of a wire of radius $r=1$ looks like.

Because $U$ is not simply connected, we cannot find a scalar field $\phi$ defined on all of $U$ such that $\mathbf F = \nabla \phi$, despite the fact that $\operatorname{curl(\mathbf F)}$ is zero on all of $U$ (that is, $\mathbf F$ is irrotational).

That being said, consider the domain $V=\left\{(x,y)\in\mathbb R^2:x^2+y^2>1\text{ and }y>0\right\}$, which is a subset of $U$.

This domain is simply connected, because any closed path in $V$ can be shrunk to a point without leaving $V$, which means that $\mathbf F$ restricted to $\mathbf V$ is conservative. Therefore, we can find some scalar field $\phi$ defined on $V$ such that $\mathbf F\big|_{V} = \nabla \phi$. You can verify that

$$\phi(x,y) = -\tan^{-1}\left(\frac{x}{y}\right)$$

gives us the field we're looking for. We also note (as of course we must) that this field cannot be defined on all of $U$, because $U$ includes points for which $y=0$.

Furthermore, even if we were to use the more general $\varphi=-\tan^{-1}(x,y)$ which simply assigns an angle $\theta\in(-\pi,\pi]$ to each point $(x,y)\neq (0,0)$, this still doesn't work on all of $U$ because it is not continuous (and therefore not differentiable) along the negative $x$-axis.

Answer 3

Your question is too long for me to read through the whole thing. But it is useful to treat a magnetic field, whose curl is not zero, as the gradient of a magnetic scalar potential. You just have to be careful to use the potential only in a region where the curl is zero.