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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.

EDIT:

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?

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  • $\begingroup$ 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 May 19 '20 at 11:36
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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 $\e=-\int \j\, dt$.

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  • $\begingroup$ 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 $\mathbf B=0$? $\endgroup$ – BioPhysicist May 19 '20 at 15:22
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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.

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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.

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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.

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