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I am reading the electrodynamics book by Griffith and am puzzled by a derivation step there. At some point, he shows that ${\rm curl} \vec{A} = \vec{0}$ and then says that since $\vec{A}$ is curl-free, $\vec{A}$ must be a gradient of some function. I know that if $\vec{A}$ is a gradient of some function, then ${\rm curl} \vec{A} = \vec{0}$ as can be shown by explicitly writing this down in the coordinate form, but Griffith goes in the opposite logical direction. How does he do that? I guess the Helmholtz decomposition theorem (aka the fundamental theorem of vector calculus) has to be used, but I do not see how. Can you shed some light on this?

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As Griffiths notes in Appendix B on the Helmholtz theorem,

Any (differentiable) vector function $\mathbf{F}(\mathbf{r})$ that goes to zero faster than $1/r$ as $r \to \infty$ can be expressed as the gradient of a scalar plus the curl of a vector: $$ \mathbf{F}(\mathbf{r}) = \mathrm{grad} \left[ - \frac{1}{4 \pi} \int \frac{\mathrm{div}(\mathbf{F})}{|\mathcal{R}|} \, d \tau' \right] + \mathrm{curl} \left[ \frac{1}{4 \pi} \int \frac{\mathrm{curl}(\mathbf{F})}{|\mathcal{R}|} \, d \tau' \right] $$

This is Eq. B.10 of Griffiths, but I have rewritten it slightly to make the significance of the terms clearer. Note that the divergence and curl inside the integrals are evaluated at the point $\mathbf{r}'$, and that $\mathcal{R} \equiv \mathbf{r} - \mathbf{r}'$.

In this form it should be obvious that if $\mathrm{curl}(\mathbf{F}) = 0$, then the second term is zero. This then implies that $\mathbf{F}(\mathbf{r})$ is the gradient of a scalar function, namely the first quantity in square brackets. Similarly, if $\mathrm{div}(\mathbf{F}) = 0$, then $\mathbf{F}$ is the curl of some vector function.

As other answers have noted, this construction is somewhat specific to $\mathbb{R}^3$; if you ever have to deal with a space that is topologically non-trivial, then you can't use this technique. But for the purposes of learning electrodynamics from Griffiths, assuming that you're in $\mathbb{R}^3$ is almost always a good assumption.

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Hint: If $A$ is a 1-form/covector field, then the implication $$\mathrm{d}A=0\quad\Rightarrow\quad\exists f \text{ locally}:~A=\mathrm{d}f$$ is a special case of Poincare lemma. A proof can be found in any good textbook on differential forms. Note that there can in principle be topological obstructions that hinter a globally defined 0-form $f$.

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    $\begingroup$ Of course, Griffiths never discusses de Rham cohomology classes in his E&M book (more's the pity.) $\endgroup$ – Michael Seifert Jun 14 at 14:53
  • $\begingroup$ A pretty solid (and constructive!) proof is also on the linked Wikipedia page! $\endgroup$ – Bence Racskó Jun 14 at 22:12
  • $\begingroup$ Oh yeah that deceptively simple proof! $\endgroup$ – lineage Jun 15 at 4:57
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First of all: the result as stated is not universally true. If you know that $\nabla\times\mathbf F=0$ throughout some region $V$, then you can conclude that $\mathbf F$ is the gradient of some scalar function if you also know that the region $V$ is simply connected. If $V$ is multiply-connected, then the result does not hold.

That said, if you do know that you're working in a simply-connected region, then the result does follow from the Helmholtz decomposition theorem. As mentioned in the Wikipedia page you have linked already, the decomposition implies that you can write $\mathbf F = -\nabla\Phi + \nabla\times \mathbf A$ for scalar and vector potentials given by $$ {\displaystyle {\begin{aligned}\Phi (\mathbf {r} )&={\frac {1}{4\pi }}\int _{V}{\frac {\nabla '\cdot \mathbf {F} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,\mathrm {d} V'-{\frac {1}{4\pi }}\oint _{S}\mathbf {\hat {n}} '\cdot {\frac {\mathbf {F} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,\mathrm {d} S' \\ \mathbf {A} (\mathbf {r} )&={\frac {1}{4\pi }}\int _{V}{\frac {\nabla '\times \mathbf {F} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,\mathrm {d} V'-{\frac {1}{4\pi }}\oint _{S}\mathbf {\hat {n}} '\times {\frac {\mathbf {F} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,\mathrm {d} S'\end{aligned}}} . $$ From the latter, you can simply read out that the volume-integral contribution to $\mathbf A$, given by an integral of a term proportional to the curl $\nabla\times\mathbf F$, vanishes under your conditions. There is also a surface term integrated over $S=\partial V$, and here you need to supplement your assumptions by either

  • assuming that $V$ covers all of space and $\mathbf F$ falls off to zero fast enough that the term vanishes, or
  • assuming that the term vanishes identically, say, by requiring that $\hat{\mathbf n}\times\mathbf F \equiv 0$ over the surface.

Under those (reasonable) extensions, then you get $\mathbf A=0$, which completes the proof.

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Here's a very special case of Poincare's Lemma, specialized to provide you with an explicit formula for a gradient. Suppose $U\subset\Bbb{R}^n$ is a star-shaped open subset with respect to the origin (for example, $U$ could be an open ball around the origin, or an open ellipsoid, or for simplicity assume $U=\Bbb{R}^n$) and $F:U\to \Bbb{R}^n$ is a vector field of class $\mathcal{C}^1$ (meaning all the partial derivatives $\frac{\partial F_i}{\partial x^j}$ exist and are continuous on $U$). If for all $1\leq i,j\leq n$ we have that $\frac{\partial F_i}{\partial x^j}=\frac{\partial F_j}{\partial x^i}$ on $U$ (when $n=3$, this is equivalent to $\text{curl}(F)=0$ on $U$), then the function $f:U\to\Bbb{R}$ defined as \begin{align} f(x)&:=\int_0^1\sum_{i=1}^nF_i(tx)\cdot x^i\,dt\tag{$*$} \end{align} is a scalar potential (i.e $\text{grad}(f)=F$). The proof of this is a standard exercise in differentiating under the integral sign.

Some remarks:

  • The reason we made the star-shaped assumption on $U$ is because in defining the integral, we need all the values $F_i(tx)$, for $0\leq t\leq 1$, to be defined, i.e $F$ has to be defined on the line segment joining the origin and the point $x\in U$ in question.

  • The motivation for this formula for $f$ is also pretty straight forward. We're just trying to apply the fundamental theorem of calculus in reverse. Note that if $f$ is sufficiently smooth (eg $\mathcal{C}^1$), then to calculate the value $f(x)$ in terms of the partial derivatives $\frac{\partial f}{\partial x^i}$, we just do the following: \begin{align} f(x)&=f(0)+\int_0^1\frac{d}{dt}[f(tx)]\,dt \tag{FTC}\\ &=f(0)+\int_0^1\sum_{i=1}^n\frac{\partial f}{\partial x^i}(tx)\cdot x^i\,dt \tag{chain rule} \end{align} Therefore, we have a formula for $f(x)$ in terms of the derivatives; such representation formulas are very important in several areas of mathematics, and this is a very simple example. Now in dealing with scalar potentials, of course a constant offset doesn't matter, so let us forget about the $f(0)$ term. Next, remember our original goal: we wanted to start with $F$ and end up with some formula for $f$. So, of course, if we did have $F=\text{grad}(f)$, then it means $\frac{\partial f}{\partial x^i}=F_i$ for all $i$.

This is why $(*)$ is the way it is. We just take a simple fundamental theorem of calculus and chain rule application, and try to turn the argument around. At this point, we have to check that everything works out, and in this case it does (and this is a good exercise in differentiation for you).

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Excellent mathematical derivations have been posed in other answers. You should definitely look them up.

You shouldn't have been surprised to find that a curl free field is expressible as a scalar gradient. This is exactly$^1$ the same as saying - conservative fields have potentials.

This is since for a conservative field, we want its integral over a loop to be zero for the displacement is zero and the field shouldn't have done any work. If this occurs for a field, a function may be described at each point that gives the$^3$ value of this work done in bringing to that point a test particle from some reference point. In other words, the field is said to have a potential energy function. The field then becomes the gradient of this potential. The reason curl comes into the picture is because curl can be thought of as the 'limiting line integral of a field per unit area'$^2$. So to make the work done by the field in a loop zero, the integral of the curl in the enclosing area must vanish, and it will for something curlless.


$^1$ within the constraints of the theorem.

$^2$ there's a lot to unpack here, you may find reveiwing a derivation of the expression for calculating curl useful.

$^3$ negative of the

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