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The following quote is from the second volume of Feynman's lectures

The interesting theorem is that if the curl $\textbf{A}$ is zero, then $\textbf{A}$ is always the gradient of something.

Feynman didn't specify anything about that vector field $\textbf{A}$, yet, he asserted that once curl $\textbf{A}=\textbf{0}$, then it follows readily that $\textbf{A}$=grad $\psi$ for some scalar function $\psi$. What I know is that this isn't the case when $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$ or even when $\textbf{A}=\textbf{A}(\textbf{r})$, but $\textbf{A}$ isn't defined in a simply connected region.

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  • $\begingroup$ it is true locally in a sufficiently small neighborhood $\endgroup$
    – hyportnex
    Commented Oct 5, 2023 at 0:56
  • $\begingroup$ @hyportnex so it doesn't matter weather $\textbf{A}=\textbf{A}(\textbf{r})$ or $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$, we can always write $\textbf{A}$ as grad $\psi$ locally? $\endgroup$
    – Jack
    Commented Oct 5, 2023 at 1:01
  • $\begingroup$ @Jack Also, sorry to ask, but can you clarify what you mean when you write $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$? $\endgroup$
    – Maximal Ideal
    Commented Oct 5, 2023 at 1:03
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    $\begingroup$ A field is, by definition, a function a $\mathbf{r}$. For example, magnetic force is not a field, because it does not have this property, as it depends on the velocity of the charge experiencing the force, which is itself not something that is defined at arbitrary $\mathbf{r}$, but only at the location of the charge in question. $\endgroup$
    – Buzz ♦
    Commented Oct 5, 2023 at 1:15
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    $\begingroup$ even if you call the magnetic force on a charge a field, which is your right to do as you may call your breakfast a field, too, it is not $\mathbf A = \mathbf {f(\dot r)}$ but rather $\mathbf A = \mathbf {f(\dot r, r)}$ in which case its curl is not zero with respect to either variable. $\endgroup$
    – hyportnex
    Commented Oct 5, 2023 at 1:33

2 Answers 2

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From the blockquote you have provided, it seems that he is referring to the gradient theorem (fundamental theorem of calculus for line integrals).

Say we have a scalar field $\psi$. The theorem says that line integrals along a path $P$ from $C$ to $D$ through gradient fields $\mathbf{A}$ depend only on the endpoints of that path, not the particular route taken, i.e. $$ \int_P \mathbf{A} \cdot d \mathbf{r}=\psi(D)-\psi(C) . $$ For every closed path this means that $\nabla \times \mathbf{A} = 0$. This is the property of a conservative vector field and hence $\mathbf{A} = \nabla \psi$.

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    $\begingroup$ 1. Penultimate sentence should reach "since this holds for all closed paths, this means ...". 2. You have proved that if ${\bf A} = \nabla \psi$ then $\nabla \times {\bf A} = 0$ but not the converse. The question is whether $\nabla \times {\bf A}=0$ is enough to guarantee the existence of a $\psi$ of which $\bf A$ is the gradient. $\endgroup$
    – Andrew Steane
    Commented Oct 5, 2023 at 11:38
  • $\begingroup$ @AndrewSteane Thank you for the first comment. For the second bit, since we already see that $\nabla \times \mathbf{A} = 0$, then via Helmholtz decomposition (en.wikipedia.org/wiki/…) doesn't it then follow that $\mathbf{A} = \nabla \psi$? In particular, I have shown that if $\nabla \times \mathbf{A} = 0$ then $\mathbf{A} = \nabla \psi$. $\endgroup$
    – S.G
    Commented Oct 5, 2023 at 20:53
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I don't know of any case where the magnetic field or the $A$-potential is a function of velocity. The Lorentz force law $F = q\vec{v}\times\vec{B}$ depends on a given particle's velocity $\vec{v}$, but $\vec{B} = \vec{B}(t, x, y, z)$ itself is always a function of space (and time in dynamic cases).

I will only talk about vector fields $\vec{A} = \vec{A}(x, y, z)$ that are a function of space.

Anyways, Feynman's statement holds in a simply-connected region, which is usually taken for granted as a hidden assumption in introductory courses. Often we assume the vector fields are globally defined on $\mathbb{R}^{3}$.

Now I don't know what theorem Feynman had in mind, and I suspect the statement itself is the theorem Feynman was referring to, but the statement is a corollary of a number of other theorems. The simplest is to consider the Poincare lemma (after you translate the language of vector fields and differential forms and back). Another way is by the Helmholtz decomposition theorem assuming $\vec{A}$ is globally defined on $\mathbb{R}^{3}$ and satisfies a few other assumptions.

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  • $\begingroup$ The velocity can itself be a position dependant vector field, right? the same story as the velocity field of water, so can I say that, all in all, all physical fields are always position dependent? $\endgroup$
    – Jack
    Commented Oct 5, 2023 at 1:34

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