Do $\boldsymbol{\nabla}\cdot \mathbf E=0$ and $\boldsymbol{\nabla}\times\mathbf E=\boldsymbol 0$ imply uniform $\mathbf{E}$?

Do the equations $$\:\boldsymbol{\nabla}\cdot\mathbf{E}=0 \qquad \boldsymbol{\nabla}\times\mathbf{E}=\boldsymbol 0\:$$ imply that the vector field $$\mathbf{E}$$ is uniform? I think yes, since by Helmholtz Theorem the vector field is uniquely determined and only a uniform vector field satisfies these two conditions?

But then also: $$\operatorname{rot} \mathbf{E} = 0$$ implies $$\mathbf{E} = \operatorname{grad}\Phi$$ and $$\operatorname{div}(\operatorname{grad}\Phi) = 0$$ by Laplace equation. So the solution of the Laplace equation always yields a uniform electric field?

Your second line of thinking is closer to the truth.

It is correct that the gradient of any harmonic function (i.e., a function satisfying $$\nabla^2 \phi = 0$$) yields a divergence-free and curl-free vector field. Moreover, the potential for any uniform field $$\vec{E}$$ is $$\phi = - \vec{E} \cdot \vec{r}$$, which (it can be shown) is a harmonic function. So it is true that all uniform vector fields can be derived from a harmonic potential.

However, there are also non-uniform vector fields that derive from harmonic potentials. For example, $$\phi = \frac12 k(x^2 - y^2)$$ is a harmonic function, and it leads to a vector field of $$\vec{E} = k(-x \hat{\imath} + y \hat{\jmath})$$, which is not uniform.

The reason this seems to be at odds with the Helmholtz Theorem is that the Helmholtz Theorem requires a set of boundary conditions to uniquely determine a vector field, along with its divergence and curl. In particular, if we require that $$\vec{E} \to 0$$ as $$|\vec{r}| \to \infty$$, then the Helmholtz theorem tells us that the only divergence- and curl-free vector field that satisfies this is $$\vec{E} = 0$$ everywhere.1 As we have seen above, however, there are many other divergence- and curl-free vector fields which are possible if we do not require that $$\vec{E} \to 0$$ at infinity.

1 Note that Laplace's equation also requires a set of boundary conditions to uniquely determine a solution. In particular, if we require that $$\nabla^2 \phi = 0$$ and $$\phi \to \text{constant}$$ as $$|\vec{r}| \to \infty$$, then we must have $$\phi = \text{constant}$$ everywhere. This is consistent with the fact that the only divergence- and curl-free vector field that vanishes at infinity is the zero vector field.

• Thank you very much. Now everything makes sense. Commented Apr 7, 2022 at 13:53

Take $$\vec{E}(x,y,z) = (yz,xz,xy)^t\:.$$ It holds $$\nabla \cdot \vec{E}=0$$ together with $$\nabla \times \vec{E}=\vec{0}$$, but the field is not uniform.

The point is that $$\Delta \phi(x,y,z)=0$$ has infinitely many solutions depending on the boundary conditions.

In the case above $$\phi(x,y,z) = xyz$$ and $$\vec{E}= \nabla \phi$$ is the gradient of that $$\phi$$.

Not necessarily. Consider the vector field: $$$$\vec{E}(x,y,z) =\begin{pmatrix} x \\ -y \\ 0 \end{pmatrix}$$$$ with $$\vec\nabla\cdot\vec{E}=0$$ and $$\vec\nabla\times\vec{E}=\vec{0}$$.