I always thought that a change in electric field induces a magnetic field and vice-versa. Moreover, I imagined that any current distribution will give rise to a magnetic field. But then I wrote this down: Maxwell's equations in absence of magnetic field.

\begin{align} \nabla \cdot \mathbf{E}(\mathbf{x},t) &= \frac {\rho(\mathbf{x},t)} {\varepsilon_0}\\ \nabla \times \mathbf{E}(\mathbf{x},t) &= 0\\ \frac{\partial \mathbf{E}(\mathbf{x},t)} {\partial t}&= -\frac{\mathbf{j}(\mathbf{x},t)}{\varepsilon_0} \end{align}

the second equation gives \begin{align} \mathbf{E}(\mathbf{x},t)=-\nabla \phi(\mathbf{x},t) \end{align} so that the rest becomes

\begin{align} \nabla^2 \phi(\mathbf{x},t) &= -\frac {\rho(\mathbf{x},t)} {\varepsilon_0}\\ \nabla \frac{\partial\phi(\mathbf{x},t)} {\partial t}&= \frac{\mathbf{j}(\mathbf{x},t)}{\varepsilon_0} \end{align} (Note: already we can see charge conservation, i.e. $\partial_t \rho+\nabla\cdot\mathbf{j}=0$)

Then we get the usual solution of the first equation \begin{align} \phi(\mathbf{x},t) = \iiint \frac{\rho(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x', \end{align} which can be written as \begin{align} \partial_t\phi(\mathbf{x},t) = -\iiint \frac{\nabla_{\mathbf{x}'}\cdot\mathbf{j}(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x', \end{align} and the second equation, due to gradient theorem, becomes \begin{align} \partial_t\phi(\mathbf{x},t)&= \phi(\mathbf{0},t) + \frac{1}{\varepsilon_0}\int_0^1 \mathbf{j}(\lambda\mathbf{x},t)\cdot\mathbf{x}\,\mathrm d\lambda \end{align} so that \begin{align} \phi(\mathbf{0},t) + \frac{1}{\varepsilon_0}\int_0^1 \mathbf{j}(\lambda\mathbf{x},t)\cdot\mathbf{x}\,\mathrm d\lambda=-\iiint \frac{\nabla_{\mathbf{x}'}\cdot\mathbf{j}(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x' \end{align}

I guess what I'm trying to show here is $\mathbf{E}(\mathbf{x},t)=\mathbf{E}(\mathbf{x})$ so that $\mathbf{j}(\mathbf{x},t)=0$, but I might be wrong. Also I don't see an obvious way to continue with this derivation.


A current density doesn't always produce a magnetic field, so that configurations with $\mathbf{E}(\mathbf{r},t)$ and $\mathbf{B}=0$ do exist. The final question I'm wondering about is whether the last equation is some sort of a constraint on $\mathbf{j}(\mathbf{x},t)$ or is it an equality true in general?

  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Commented May 19, 2020 at 11:36

4 Answers 4


I think you are looking at this the right way but it is probably easier to think in terms of fields than potentials. Plugging $\newcommand{b}{\mathbf{B}}\renewcommand{e}{\mathbf{E}}\renewcommand{ed}{\dot{\e}}\newcommand{j}{\mathbf{j}}\renewcommand{z}{\mathbf{0}} \b=\z$ into$ \nabla \times \b = \ed + \j$ we get $\ed=-\j$. Then $\e = \e_0+\int_{t_0}^t -\j dt'$. We can now check if this definition of $\e$, together with $\b=\z$, satisfies Maxwell's equations. The ones concerning $\b$ are satisfied by construction. Checking Gauss's law, we find $$\nabla \cdot \e = \nabla \cdot \e_0 + \int_{t_0}^t -\nabla \cdot \j\, dt'=\rho_0 + \int_{t_0}^t \dot{\rho}\, dt'=\rho.$$

So Gauss's law checks out.

Now lets check the last equation. $$\z=-\dot{\b}=\nabla \times \e = \nabla \times \e_0 + \int_{t_0}^t -\nabla \times \j\, dt'.$$ If the rightmost side is to be zero for all $t$, then we must have that $\nabla \times \e_0=\z$ and then for all $t$, $\nabla \times \j=0$. The first equation tells us that $\e_0$ must be conservative, and the second tells us that $\j$ must be irrotational for all time.

In summary, we have found that $\b$ is zero then it is necessary to have $\ed = -\j$, so that $\e=-\int \j\, dt$, and then we found it is also necessary for $\j$ to be irrotational. Moreover, these two conditions are sufficient since you can construct a solution.

So in conclusion you can find a $\b=\z$ solution precisely when $\j$ is irrotational, in which case the solution is $\e=-\int \j\, dt$.

  • $\begingroup$ Another question. With your conditions of $$\mathbf E=\mathbf E_0-\int_{t_0}^t\mathbf j\,\text dt$$ $$\nabla\times\mathbf E_0=0$$ $$\nabla\times\mathbf j=0$$ Doesn't that just give us $$\nabla\cdot\mathbf B=0$$ and $$\nabla\times\mathbf B=0$$ What additional steps do we take to conclude that these conditions lead to $\mathbf B=0$? $\endgroup$ Commented May 19, 2020 at 15:22

As shown here section 18.2, it is possible to have configurations in which current density is non-zero but the magnetic field is zero. My understanding is that it is perfectly legitimate to have a time-varying Electric field and null magnetic field at all times. The simplest case is a variable current source $j(r,t)$ eminating radially from a source. Since $j(r,t)$ has spherical symmetry, $B=0$, however $E(r,t)$ varies in both space and time.


I think you have overlooked that a sentence like

a change in electric field induces a magnetic field and vice-versa.

is true in the vacuum, i.e. it is not valid in a region where non-zero charge density and current exist.


Magnetic fields are a consequence of special relativity as follows:

Given two charges A and B, and looking at the effect of A on B, you will get the correct result without any thought for magnetic fields unless both charges have a velocity. In any frame where one of the charges is motionless, magnetic field has no effect.

There are further restrictions. A must have a velocity that is perpendicular to the vector from A to B. B must have a velocity that is on the plane made by the AB line and A's velocity vector.

Magnetic fields provide the fudge factor required to correct for frames that display a relativity effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.