1

I know that if the curl of a vector field is zero, then the vector field must be the gradient of a scalar field. But if the divergence of a vector field is zero, is it a must that the vector field be the curl of some vector field? If so please provide mathematical proof.

Integrand
  • 8,457
Sai Krishna Garlapati
  • 49
  • It is not clear what would be the gradient of a vector field. In $\mathbb{R}^3$ the gradient of a scalar field is a vector – Miguel Sep 07 '20 at 19:01
  • The gradient of a vector field would be a tensor encoding the change in the field. Therefore if the field is of the form $\sum f_i(x,y,z) e_i$ then the gradient is made of all $\partial_i f_j$ – Rachid Atmai Sep 07 '20 at 19:04
  • @Miguel sorry, I meant the divergence. – Sai Krishna Garlapati Sep 07 '20 at 19:05
  • It's not always true. If you assume the vector field is defined on a contractible subset then it's true (for example if it's defined on an open ball, or for example a star-shaped open set or for example on the whole of $\Bbb{R}^n$). In this case, the theorem you're looking for is a special case of Poincare's lemma – peek-a-boo Sep 07 '20 at 19:10
  • 1
    @peek-a-boo Exactly. I'm confused by the question. There are lots of examples of curl-free vector fields which are not gradients of any scalar-valued functions. – Rachid Atmai Sep 07 '20 at 19:17
  • I strongly recommend this book: https://g.co/kgs/wkfCBt – Miguel Sep 07 '20 at 19:46
  • 1
    This is an important question which, strangely enough, does not have many good answers on math.stackexchange. I don’t think it should be closed due to lack of context; this is clearly a fundamental question that many people (myself included) will want to understand. In particular, is there a way to answer without either invoking Poincare’s lemma or reverse engineering Poincaré’s lemma? If so, I can’t find it on math.stackexchange. – littleO Sep 30 '23 at 12:20
  • I’m voting to close this question because it has a very good answer. If @littleO finds that dropping the Poincare lemma adds anything to better understanding this should perhaps be asked as a new question. – Kurt G. Oct 01 '23 at 04:08
  • 1
    Is having a good answer a reason to close a question? We typically leave questions open after an answer has been accepted in hopes that other illuminating perspectives will appear. – littleO Oct 01 '23 at 04:13
  • 1
    @littleO indeed, while I think my answer for finding $f$ is obvious (it’s just FTC in reverse to get the candidate integral, then a simple computation to check it actually works), I’m less ‘satisfied’ by my answer for the vector potential $\mathbf{A}$, because although Poincare’s lemma is rather intuitive (it’s a dressed up version of the FTC argument above, by taking the interior product of the radial vector field with the differential form in question), it feels like cheating. It would indeed be nice to see a less differential from-ish motivation for the specific case of vector potential. – peek-a-boo Oct 02 '23 at 02:34
  • Right now, the best I have is “vector field to scalar potential” decreases the degree of the object involved, so the integral for $f$ has no $t$ factor. However, going from a field to a vector potential is a ‘higher order’ object (vector potential vs scalar potential before). Next, the $F_i(tx)\cdot x^i$ which was initially simply scalar multiplication, should now somehow involve cross products since curl involves cross products. i.e by lots of handwaving and hindsight and symbolic matching/guessing, we can arrive at the integral formula for $\mathbf{A}(x)$, but it’s not satisfying of course. – peek-a-boo Oct 02 '23 at 02:40

1 Answers1

6

Without further assumptions, neither of the statements you made are true. It is possible to have a vector field with $0$ curl, yet it not be the gradient of some function, and it is also possible to have a divergence-free vector field yet it not be the curl of some vector field. You need to impose certain topological restrictions on the domain of the vector fields.

If you assume the vector field is defined on a contractible subset then both statements are true (for example if it's defined on an open ball, or for example a star-shaped open set or for example on the whole of $\Bbb{R}^3$). In this case, the theorems you're looking for are special cases of Poincare's lemma.

For the first statement, if you have a vector field $\mathbf{F}:\Bbb{R}^n \to \Bbb{R}^n$ of class $\mathcal{C}^1$ such that for all $i,j\in\{1,\dots, n\}$, $\frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i}$ (in the case $n=3$, this is equivalent to saying the curl is zero), then the function $f:\Bbb{R}^n\to \Bbb{R}$ given by \begin{align} f(x) &= \int_0^1 \sum_{i=1}^n F_i(tx)\cdot x^i \, dt. \end{align} is a scalar potential: $\text{grad}(f) = \mathbf{F}$ (proof is a direct calculation).

For simplicity, let's say your vector field $\mathbf{F}:\Bbb{R}^3\to \Bbb{R}^3$ is defined everywhere, is of class $\mathcal{C}^1$, and is divergence free. Then, the vector field $\mathbf{A}:\Bbb{R}^3\to \Bbb{R}^3$ defined as \begin{align} \mathbf{A}(\mathbf{x}) &:= \int_0^1 t\cdot [\mathbf{F}(t\mathbf{x}) \times \mathbf{x}]\, dt \end{align}, where $\times$ is the cross product in $\Bbb{R}^3$, will satisfy $\text{curl}(\mathbf{A}) = \mathbf{F}$. The proof is a tedious, but direct calculation. Note that we need $\mathbf{F}$ to be defined everywhere so that the integral above makes sense (or at the very least, if we want this specific proof to work, we need the domain to be star-shaped with respect to the origin).

peek-a-boo
  • 55,725
  • 2
  • 45
  • 89
  • I think you should comment the first sentence of the OP is not true either, being exactly the same case of Poincaré's lemma. – Miguel Sep 07 '20 at 22:06
  • @Miguel Thanks for the suggestion. I have edited my answer slightly and addressed that. – peek-a-boo Sep 07 '20 at 22:18
  • 1
    Where did your $\mathbf{A}(\mathbf{x})$ come from? – Descartes Before the Horse May 26 '21 at 04:48
  • 1
    @DescartesBeforetheHorse I knew the proof of Poincare's lemma in the general case (and there the formula is very memorable to me because it very easily generalizes what I wrote for the scalar potential), and from there I just reverse-engineered this particular case of curl to deduce what $\mathbf{A}$ ought to be. – peek-a-boo May 26 '21 at 05:00
  • @peek-a-boo Are you sure there is a multiplication with $t$ in the integrator in the formula for $\mathbf{A}$ ? – Numa Apr 10 '22 at 05:08
  • @Numa yes I'm sure – peek-a-boo Apr 10 '22 at 05:53
  • @peek-a-boo Thank a lot you for the quick reply, as this is already an old thread here! Then I will start to do the tedious calculations to see what happens. I asked just because I calculated the following 2d version: $A(\mathbf{x}) = \int_0^1 \mathbf{F}(t\mathbf{x}) \wedge \mathbf{x} , dt$ which appears to work very nicely. But if I would add a multiplication with $t$ then it would not work. – Numa Apr 10 '22 at 09:39
  • @Numa I would recheck your calculations. You need the $t$ there. The general calculation of Poincare's lemma involves a $k$-form $\omega$ on $\Bbb{R}^n$ (or a star-shaped open set in $\Bbb{R}^n$, say WLOG containing the origin), and a $k-1$ form $P(\omega)$ defined as $P(\omega):=\sum_I\sum_{\alpha=1}^k(-1)^{\alpha-1}\left(\int_0^1t^{k-1}x^{i_{\alpha}}\omega_I(tx)\right),dx^{i_1}(x)\wedge \cdots \widehat{dx^{i_{\alpha}}(x)}\wedge\cdots \wedge dx^{i_k}(x)$, where the sum is over all multindices $I=(i_1,\dots, i_k)$ with $1\leq i_1<\cdots <i_k\leq n$. – peek-a-boo Apr 10 '22 at 09:40
  • Then, one can show that $d(P(\omega))+P(d\omega)=\omega$, so that if $d\omega=0$ (the generalization of $\text{grad}(f)=0$, or $\text{div}(\mathbf{F})=0$) then $d(P(\omega))=\omega$, i.e $\omega$ is exact. Edit: the integral above should of course have a $dt$ at the end. – peek-a-boo Apr 10 '22 at 09:46
  • @peek-a-boo Thank you for the detailed reply! I tripple-checked my calculations and they are correct without $t$ and multiplying $t$ will lead to an incorrect formula. Thanks to your detailed reply, even though I do not know much about differential forms, I think I get now where the difference between $2$d and $3$d is: In the $2$d case (a divergence free $2$d vector field is the "curl" of a scalar function) which I calculated by hand, one has $P(\omega)$ is a zero form and thus $\omega$ would be a one form(?) and thus $k-1 = 0$. – Numa Apr 10 '22 at 10:53
  • @Numa oh right in 2D the curl is a little weird; rather than thinking of it as a vector one can think of it as a scalar, so you're right. – peek-a-boo Apr 10 '22 at 21:28
  • 1
    @peek-a-boo Thanks for the reply! I just realized ( and I wonder why I haven't realized this before, and just want to share it for the record) that in 2d, the problem div$(\mathbf{F}) = 0 \Rightarrow \mathbf{F} = $"curl"$(f)$ is equivalent to curl$(\mathbf{F}) = 0 \Rightarrow \mathbf{F} = \nabla f$, thus the formula you posted for $f$ does the job, after applying the rotation $R = \begin{bmatrix}0&1\-1&0\end{bmatrix}$. Indeed div$(\mathbf{F}) = 0 \iff $curl$(R\mathbf{F}) = 0$. In a nutshell, what I was calculating was a rotation of your formula for the scalar potential in $2d$ – Numa Apr 11 '22 at 14:18