# Conservative electric field must be static?

My question means, by Maxwell equations: $$\nabla\times \vec{E}=0\stackrel{?}{\implies} \frac{\partial \vec{E}}{\partial t}=0$$

I think that is right, this is my explanation,

Intuitive explanation: A conservative electric field must be generated by static system of charges (unproven), which directly satisfy the Coulomb's law, then the electric field doesn't change with time explicitly.

Mathematical explantion: We know, $$\nabla\times \vec{E}=0$$ and $$\nabla\cdot \vec{E}=\rho/\epsilon_0$$, while the boundary value of full space shouldn't change with time (unproven). Then by Helmholtz decomposition, we could find the unique electric field $$\vec{E}$$, which doesn't change with time explicitly, for time doesn't appear in equations.

Is my explanation correct? And how to prove the unproven part mathematically and strictly in explanations? Please help!

• Are you considering currents and magnetic fields? If yes, then I think your conjecture is false. Feb 24, 2022 at 9:42

No, $$\nabla\times\vec{E}=0\implies\frac{\partial \vec{B}}{\partial t}=0\implies\partial_t(\nabla\times\vec{B})=0\implies\frac{\partial^2\vec{E}}{\partial t^2}=0$$ but not that $$\frac{\partial \vec{E}}{\partial t}=0$$.

There is no additional constraint that would be imposed on $$\frac{\partial \vec{E}}{\partial t}$$ by requiring that $$\vec{E}$$ is curl-free because we have already worked out the constraint coming on $$\frac{\partial \vec{E}}{\partial t}$$ from the two dynamical Maxwell equations. The two source Maxwell equations don't affect $$\frac{\partial \vec{E}}{\partial t}$$.

For example, consider the vacuum solution:

$$\vec{E}=t\hat{z},\vec{B}=\frac{1}{2}(-y\hat{x}+x\hat{y})$$

• Thanks for your answer, but I don't agree with your example. Your example neglected the source of $\vec{E}$ and $\vec{B}$, although it satisfied Maxwell equations when $\rho=0$ and $\vec{J}=0$. But I think electromagnetic field could "exist" without source, but it must be "generated" by source, like EM wave. No such $\rho$ and $\vec{J}$ in real world can generate this kind of $\vec{E}$ and $\vec{B}$, so I don't think it disproved the question. Feb 24, 2022 at 15:04
• @VictorZhang I didn't "neglect" the source, it's an exact solution to Maxwell equations. :) Your claim that an electromagnetic field must be generated by a source is not supported by Maxwell equations. Several vacuum solutions exist for $t\in(-\infty, \infty)$ and their existence shows that electromagnetic fields can exist on their own just fine. In any case, my argument preceding the example is general and holds regardless of whether you choose vacuum solutions or non-vacuum solutions. I'll try to construct a non-vacuum example in the evening :)
– user87745
Feb 24, 2022 at 15:32
• I don't think every mathematical solution of Maxwell equations will lead to a "real" situation. For example, I can also construct a solution: $\vec{E}=(x \hat{x}+y \hat{y}+z \hat{z})\cos{\omega t}$. At a same time, every location in the space has a same $\rho$ by Gauss' law. However, every location has a different $\vec{E}$, that's a contradiction of translational symmetry. This kind of situation will not happen in physical world. Feb 26, 2022 at 5:50
• @VictorZhang Obviously, the real world has a specific initial condition. For example, $\vec{E}=\frac{kq}{r^2}\hat{r}$ is also not a real situation because the real world does not contain only a single spherical charge. Regarding my promised non-vacuum example, I realized that it's notoriously hard to come up with a non-vacuum solution for any electromagnetic situation if you consider also the Lorentz force acting on the sources in the Maxwell field. If I am allowed to ignore this backreaction, there are many trivial examples I can give. [...]
– user87745
Feb 26, 2022 at 6:48
• [...] For example, using the linearity of the Maxwell equations, just superimpose a Columb solution with my vacuum solution and you've got a non-vacuum solution that violates your conjecture.
– user87745
Feb 26, 2022 at 6:48