# Why does $E=\nabla\phi$ follow from $\nabla\times E=0$?

I understand that using one of Maxwell's equations, $$\vec{\nabla} \times \vec{E}(\vec{x})=0,$$ it can be said that $$\vec{E}(\vec{x})=-\vec \nabla \phi(\vec{x}).$$

However, I can't find or understand why. Can anyone point me in the right direction?

• en.wikipedia.org/wiki/… - click "show" in the gray box to see a proof Commented Jan 29, 2014 at 13:22

## 4 Answers

This is just a consequence of Helmholtz's theorem - Any vector field can be written as: $$\mathbf{F}=-\boldsymbol{\nabla}\Phi+\boldsymbol{\nabla}\times\mathbf{A}$$ Now take the curl of both sides - since the curl of $\mathbf{F}$ is zero by assumption, and the curl of a divergence is also zero (write it out), we must have $\boldsymbol{\nabla}\times\boldsymbol{\nabla}\times\mathbf{A}=0$. The curl of a curl is in general not zero, so for this to hold for all points $x$, we demend that $\mathbf{A}=0$. Thus, $$\boldsymbol{\nabla}\times\mathbf{F}=0\ \ \longrightarrow\ \ \mathbf{F} =-\boldsymbol{\nabla}\Phi$$

• Just out of curiosity, this looks very similar to the Lorentz force. Is there any noteworthy connection? Commented Jan 29, 2014 at 11:06
• @user2538235 - not quite, but maxwell's law's become a lot more "connected" when you look at them through the eyes of special relativity: en.wikipedia.org/wiki/… Commented Jan 29, 2014 at 11:12
• @nbubis "The curl of a curl is in general not zero, so for this to hold for all points 𝑥, we demand that 𝐀=0". But then A could be a constant, non-zero vector field as well isn't it?. Commented Feb 11, 2021 at 1:43

The easiest way to see this connection, I think, is not through the differential form of that equation but through its integral form. It is one of the standard exercises of vector calculus to prove that $$\nabla\times \mathbf{E}=0\ \text{on all space}\ \Longleftrightarrow \oint_C\mathbf{E}\cdot\text d \mathbf{l}=0 \ \text{for all closed curves }C. \tag1$$

Once you have that, it is fairly easy to prove the existence of the potential, which you can do by simply constructing it directly: $$\phi(\mathbf{r})=-\int^\mathbf{r}_C\mathbf{E}\cdot\text d \mathbf{l},$$ where the integral is over any curve $C$ which starts at some reference point $\mathbf{r}_0$ and ends at $\mathbf{r}$. The fact that the integrals over closed loops are zero means that no matter what path you take to $\mathbf{r}$, you will have the same value of the potential. It is then a trivial matter to show that the gradient of this function is indeed the initial electric field $\mathbf{E}$.

That second step is very intuitive, but it still looks disconnected from your initial statement unless you can put some meaning behind the equivalence in (1). That is, in fact, not really an exercise: it is a straightforward application of Stokes' theorem, which is the correct generalization of the fundamental theorem of calculus to more than one dimension. This means that you can read statement (1) as saying something like

if the (correct) derivative of the field is zero, then the (appropriately generalized) signed sum of its values at a 'boundary' is zero.

There are many ways to make this more obvious. You can, for example, consider a very small loop, and show that the contour integral is essentially the curl there. You can also split a given loop into a grid of many, smaller loops, and show that the circulation integral must vanish in all of them if the curl does, so that the whole integral must vanish. Both of these are typically explained in depth in any electromagnetism textbook; I would refer you to Purcell and Griffiths if you're still having trouble.

You can probably find information about this in any textbook on electrodynamics. Try Jackson's "Classical Electrodynamics".

A direct way is to Fourier transform your equations. Then ${\bf{k}}{\times}{\bf{E}}{(\bf{k})}=0$ implies that ${\bf{E}}{(\bf{k})}$ is along $\bf{k}$, which is your second equation.

I know that Maxwell's equations are for $\mathbb{R}^3$ space, but let me give you an example for $\mathbb{R}^2$ since it is easier to understand and you were just asking for a general direction/idea:

Assume we have a vector field $\vec{v}: \mathbb{R}^2 \mapsto \mathbb{R}^2$ given by:

$v(x,y) = Q(x,y)\vec{e}_x + P(x,y)\vec{e}_y$

where $Q, P: \mathbb{R}^2 \mapsto \mathbb{R}$. Recall that the curl ($\nabla \times \vec{v}$) on $\mathbb{R}^2$ is defined as:

$\nabla \times \vec{v} = \partial_xP - \partial_yQ$.

Let now assume that $\nabla \times \vec{v} = 0$ then we have:

$0 = \partial_xP - \partial_yQ \Leftrightarrow \partial_xP = \partial_yQ$.

If we now assume that there exists a potential $f: \mathbb{R}^2 \mapsto \mathbb{R}$ for this vector field then

$\vec{v} = \nabla f = \vec{e}_x\partial_x f + \vec{e}_y\partial_y f = Q(x,y)\vec{e}_x + P(x,y)\vec{e}_y$

must hold as well as:

$Q(x,y) = \partial_x f$ and $P(x,y) = \partial_y f$.

The question is now wether we can really find a function $f$ such that the above conditions hold. Let $f \equiv \int P dy$ then $\partial_y f = P$ and we further find:

$\partial_x f = \frac{\partial}{\partial x} \int P dy = \int \frac{\partial P}{\partial x} dy = \int \frac{\partial Q}{\partial y} dy = Q + C(x)$

But since the constant only depends only on $x$ the curl respectively $\partial_y C(x)$ still vanishes and thus it is not a contradiction to the statement above.

So we have showed that we can find such a potential function. I don't want to go into more detail here. This proof was more to give you a general idea since there are more subtle things to it such that the space where the vector field is defined must be simply connected etc.

• Urgje gave a way more elegant answer but I wanted to give you a "straight" forward idea which might make it even easier for you to understand it. Commented Jan 29, 2014 at 10:23