Let's have an ideal conductor with current density $\mathbf{J}$. The ideal conductor takes up the entire space, effectively resulting in the entire space $\mathbb{R}^3$ being permeated with $\mathbf J$.

In case of uniform current density $\mathbf{J}$ taking up the entire space, $\mathbf{E}$ being $0$ everywhere, what's $\mathbf{B}$ everywhere in space? I thought it would be $0$ (because the contributions from all elementary currents should cancel each other out), but that seems to contradict Ampère's circuital law

$\nabla\times \mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}$, since


Is, in that case, $\mathbf{B}=0$ everywhere, is the entire situation incompatible with Maxwell's equations, or some third possibility?


Sorry for my poor english !

I think we can go into steady state and forget about the electric field.

The problem is well known for infinite volume charges distributions. In particular for a volume distribution of uniform charge for which "by symmetry" the electric field should be zero, non compatible with the Maxwell Gauss equation.

In the case of the uniform current density, I think that the Biot and Savart integral is not really convergent. It depends on how to stretch the limits to infinity. We only find $0$ if we stretch the limits to infinity symmetrically. For example, if the current distribution is cylindrically symmetrical, of radius $R$, the field at the distance from the axis is ${{B}_{\theta }}=\frac{1}{2}{{\mu }_{0}}jr$. If we stretch $R$ to infinity, this field does not tend to $0$.

The symmetry arguments are not usable either because the solution of the Maxwell equations is not unique.

In the end, one should expect difficulties with non-physical charges or currents distributions. We can consider ourselves lucky to have to have so few problems with infinite surface and linear distributions!

I can try to clarify a bit (it's too long for a simple comment) :

You will notice that when we study Gauss' theorem we consider the infinite straight wire, the infinite plane but never the "infinite cube".

I think that the passage to the limit of an infinite volume distribution is not consistent with the equations of electromagnetism. In particular, there is no longer a single solution.

I also think that the arguments that give a zero integral are not convincing because the limit of the integral depends totally on how one makes the boundaries to infinity. A bit like someone who would prove that $\sum\limits_{0}^{+\infty }{{{(-1)}^{k}}}=0$ that because at each 1 we can associate a -1 !

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  • $\begingroup$ Thanks. So Maxwell's equations aren't compatible with $\varrho=const.\neq0$ everywhere in space, not only Ampère's law, but e.g. Gauss' law too, because it predicts that $\boldsymbol{\nabla}\cdot\mathbf{E}=\frac{\varrho}{\varepsilon_0}$, which would result in $0 = \frac{\varrho}{\varepsilon_0}\neq0$. The general answer, therefore, is that the situation is impossible to begin with because Maxwell's equations are incompatible with $\varrho$ being a nonzero constant everywhere in space. Is that right? $\endgroup$ – Golden Gleam Jan 21 '19 at 23:22
  • $\begingroup$ Even though nobody confirmed that my interpretation of your answer was correct, I'm sufficiently confident that I can accept your answer and rely on my interpretation of it. Thanks! $\endgroup$ – Golden Gleam Jan 25 '19 at 0:25
  • $\begingroup$ I am also surprised that so few people have answered your question. I tried to complete the answer. $\endgroup$ – Vincent Fraticelli Jan 25 '19 at 5:37
  • $\begingroup$ Thanks, I already upvoted your answer before, but it didn't show because I don't have enough reputation yet. $\endgroup$ – Golden Gleam Jan 27 '19 at 8:38

Congrats! What you've discovered, more or less, is that the case of an infinite uniform current density pervading all space in magnetostatics is the complementary notion to the case of an infinite uniform charge density pervading all space in electrostatics. And it is well-known that in this case, such conditions do not uniquely specify the electric field; likewise, here, they do not uniquely specify the magnetic field. That is, this is a place where the laws of EM theory are non-predictive. The reason this is not a problem is quite simply that the Universe is not either a uniform charge or current distribution, and thus this pathology is not encountered.

Many physical theories have these kinds of pathologies - another example is that Newtonian mechanics admits a structure called "Norton's dome" in which that many possible future evolutions are consistent with the equations of motion from the same initial conditions, however this structure is not one we can physically realize because it requires a perfectly ideal mathematically pointed surface and point particle resting thereupon, neither of which are possible in the "real" Universe thanks to atoms and to quantum mechanics, the former of which prevents the pointed surfaces, and the latter of which prevents a state of simultaneous complete rest and fully-determined location of a point particle from ever happening. Moreover, it may be that "point" particles like electrons aren't even truly points at all, as in string theory, though this is purely hypothetical as of now. No evidence for string theory has ever been found. Yet another example is the idea of a "naked singularity" in general relativity, which was actually discussed in a recent thread here concerning black holes.

Generically, the "solution" to these is to note that the conditions they require turn out to be "absurd" in some way, as in the infinite uniform density, or the notion of the need for a surface with mathematically ideal properties. The naked singularity one is a bit more troublesome because it is actually still a wide open problem as to whether or not that a realistic physical process might result in one (this is the so-called "cosmic censorship problem"), or what general relativity would say is one.

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While it's true that the electric field inside an ideal conductor with uniform $J$ is zero, that doesn't imply the magnetic field inside it of it will be zero.

In fact, you can easily calculate the magnetic field inside an ideal conductor with uniform current density $\vec{J}$ using Ampere's law integral version:

$$\oint \vec{B}\cdot d\vec{l}=\mu_0 I$$

In general just because $E$ or $B$ is zero in a region of space doesn't necessarily mean the other should also be zero. So there's no contradiction here.

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  • 1
    $\begingroup$ Isn't the Ampere Law $\int Bd l =\mu I$? $\endgroup$ – Karthik Jan 21 '19 at 13:44
  • $\begingroup$ Thanks, I know $\mathbf{E}=0$ doesn't imply $\mathbf{B}=0$, my question was about the ideal conductor being spatially infinite, not just any ideal conductor. $\endgroup$ – Golden Gleam Jan 21 '19 at 23:32
  • $\begingroup$ @KarthikV Yes, it is. (As long as there are no other currents aside from $I$, or as long as we include the other currents in $I$.) $\endgroup$ – Golden Gleam Jan 21 '19 at 23:40
  • $\begingroup$ Right. So does $dS$ mean the circumference element in your formula? $\endgroup$ – Karthik Jan 22 '19 at 0:35
  • $\begingroup$ @KarthikV I'm the author of the question, I didn't write the answer. 🙂 $\endgroup$ – Golden Gleam Jan 23 '19 at 2:45

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