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Imagine the following (moving) charge distribution in a one-dimensional space

$$f(x) = \begin{cases} 1 & |x - vt| \leq 1 \\ 0 & else \end{cases}$$

Given the ends of this distribution two to Amperes & Faradays law there should be an infinitely strong electric field.

How would an expirement (moving some charged charge carrier) work out?

Niclas
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  • I'm not sure what you mean, ampere and faradays law work for isolated electrons that are in motion. – Triatticus Feb 29 '24 at 21:11
  • Basically a moving charge carrier would be equivalent to a current, but this current would abruptely end and thus the change in B relative to time would infinite by amperes law (current ends immediately), thus an infintely high E field rotation would be created by faradays law – Niclas Feb 29 '24 at 21:19
  • In general $\mathbf J = \rho \mathbf v$ but for a point particle at location $\mathbf r_q$ the charge density is $\rho(\mathbf r) = q \delta(\mathbf r-\mathbf r_q)$ with velocity $\mathbf v = \frac{d \mathbf r_q}{dt}$ – hyportnex Feb 29 '24 at 22:40
  • "Given the ends of this distribution two to..." Can you please clarify what you mean by this fragment? Writing "...two to..." does not make sense. Is there a typo? Or multiple typos? – hft Feb 29 '24 at 23:30
  • "How would an expirement [sic]... work out?" This seems too vague to answer... What experiment? What does "work out" mean? If "work out" means to ask about what would actually result in the real world then that is hard to say since the real world is not one dimensional. – hft Feb 29 '24 at 23:31
  • The fact that the Leinárd-Weichert potentials/fields exist is already a testament to that the fields are not infinite for a point charge. – Triatticus Mar 01 '24 at 00:55

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In one-dimensional space EM theory with $E$ and $B$ fields does not work. You can only have a static E-field, it is not very interesting (but if you wish, see ref1 ref2).

You could, however, keep your defined $f(x)$ exactly as it is and use it in 3 space dimensions, where it will describe an infinite slab of uniform charge parallel to the $yz$-plane and moving in the $x$ direction.

To solve it we can start with $v=0$, which gives just a static E-field, finite everywhere. Then transform it to a frame of reference where $v$ is nonzero, see ref3. Nothing infinite will come out of that.

You can then also try to solve it directly. It seems your reasoning would still apply and give an infinite $E$-field if you use Ampere's and Faraday's laws. So you basically prove that Maxwell was right when he added the $\partial{\bf E}/\partial t$ to Ampere's law to complete it. You will see that here it cancels the ${\bf J}$ term! (Note that ${\bf E}$ inside the charged slab is not zero, it points from the "middle plane" outwards with increasing strength if you move from this middle plane to the slab's surfaces.)

From ref3 we see that even for relativistic speed, the moving slab still has the same ${\bf E}$ as in the static case and that ${\bf B}$ remains zero (from ${\bf E}_{\|'}= {\bf E}_{\|}$ and the three equations following it). So this is actually just as uninteresting as EM in one space dimension where you would have this result by definition.

Jos Bergervoet
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