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Using the Maxwell equations and making no assumptions about the sources (charge and current densities), can one argue that $\vec{E}\perp \vec{B}$ at all times? Assuming that the EM field stems from a single source or set thereof.

I know that it is fairly easy to show that this is the case when the sources both vanish (i.e. when $\rho = 0$ and $\vec{J} = 0 $).

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_{0}}$$ $$\nabla \cdot \vec{B} = 0$$ $$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$$ $$\nabla \times \vec{B} = \mu_{0}\vec{J}+\frac{1}{c^2}\frac{\partial\vec{E}}{\partial t}$$

That is, if $\vec{E}$ and $\vec{B}$ both obey all of Maxwell's equations without exception,then does there always exist some vector $\vec{K}(\vec{x},t)$ such that

$$\vec{B}=\vec{K} \times \vec{E}?$$

If not, is there some way to specify a set of conditions under which this would be the case? One such condition would undoubtedly be the case of where there are no sources.

Or, at least, is there some explicit way to prove the counter point?

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  • $\begingroup$ You can only proof they're perpendicular for a single plane wave. The sources to vanish is not enough. See: physics.stackexchange.com/questions/61072/… $\endgroup$ – Jan Bos Jan 28 '18 at 14:01
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    $\begingroup$ Can you be clearer about the sources? You mention 3 seemingly incompatible things here: (1) "Making no assumptions about the sources", (2) "Single source", (3) "Single set of sources". Which of those are you really asking about? $\endgroup$ – Steve Byrnes Jan 28 '18 at 18:44
  • $\begingroup$ @SteveByrnes... To be honest, those three seem to me to be perfectly compatible with each other. All you need to assume is that the electric and magnetic fields both satisfy the Maxwell equations. If you have found an example where they do not, they you have more that a single set of sources. The "sources" in question are the current density and the charge desity, as demanded by the maxwell equations. I have made this very clear in the question. $\endgroup$ – Mahlomola Daniel Cwele Jan 29 '18 at 5:59
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No, this is not true. A simple example is a charged capacitor in the earth's magnetic field--unless you hold it just right, E and B are usually not orthogonal.

Even if there are no sources around, you can get E non-perpendicular to B (and even parallel to B) by superimposing multiple light waves traveling in different directions with different polarizations and wavelengths.

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  • $\begingroup$ The EM field in this example clearly violates the "single source" condition. The magnetic field of the earth and the charge in the capacitor have different sources. So, this ultimately does not answer the question. $\endgroup$ – Mahlomola Daniel Cwele Jan 28 '18 at 14:29
  • $\begingroup$ My "multiple light waves" example solves Maxwell's equations with no source terms whatsoever. $\endgroup$ – Steve Byrnes Jan 29 '18 at 13:10
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It depends what you mean by "a single source or set thereof". For example, may we take, as a set of sources, charges moving round a circular track at constant speed, equally spaced from one another? If so, consider a point, P, some way from the centre of the track and on its axis (line through the circle centre at right angles to the plane of the circle). At such a point the electric field and magnetic field due to the charges on the track are both directed along the axis and so parallel or antiparallel to each other.

This result is easily obtainable using mainly symmetry. It could be made quantitative using Coulomb's law and the Biot-Savart law and these, in turn, can (with care and time!) be deduced from Maxwell's equations.

Later addition The set-up in the first paragraph could be achieved in practice by spinning a disc (made of an insulator) with a charged edge.

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  • $\begingroup$ It seems that you are interested in finding an isolated example. As the question clearly shows, the interest is in whether this kind of conclusion would hold IN GENERAL, and for ALL SUCH SOURCES, not just for an isolated example. in fact, I have already provided an example under which the magnetic and electric field are known to be perpendicular, thereby stressing that this not the problem with which I need help. $\endgroup$ – Mahlomola Daniel Cwele Jan 28 '18 at 14:33
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    $\begingroup$ With respect, your question (as worded at the time I answered it) asked "can one argue that E⃗ ⊥B⃗ at all times?" A single exception to E⃗ ⊥B⃗ is enough to show that one $can't$ uphold the claim. $\endgroup$ – Philip Wood Jan 28 '18 at 14:38
  • $\begingroup$ The key in your quote is "at all times". But, please do not focus on a single part of the question, as I went to some length to make sure that the question is as clear as possible. $\endgroup$ – Mahlomola Daniel Cwele Jan 28 '18 at 14:52
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    $\begingroup$ I focused on what I could understand – the clear and unambiguous question in your first paragraph. [I note that the other answers do the same.] $\endgroup$ – Philip Wood Jan 28 '18 at 15:07
  • $\begingroup$ @MahlomolaDanielCwele - A single counterexample disproves your assertion that "$\vec{E}\perp \vec{B}$ at all times"! $\endgroup$ – freecharly Jan 28 '18 at 15:37
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The electric field is definitely not always perpendicular to the magnetic field. You can visualize this very easily by a counter-example:

Take a homogeneous magnetic field $\vec B$, e.g., due to a permanent magnet or inside a long wire coil. Then insert an electric point charge into this field. You can easily recognize that the electric $\vec E$ and the magnetic field $\vec B$ can have any angle between $0$ and $180$ degrees.

PS: The Coulomb field of the point charge is, of course, a solution to the Maxwell equation $$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_{0}}$$

Additional counter-example with a "single source":

Take a circular ring of positive charge rotating so that a stationary annullar electric current flows. On the axis on one side of the ring away from its center, you have always an electric field pointing away from the ring current. And you have also a magnetic field parallel or antiparallel to the electric field depending on which side of the ring you are looking at.

This disproves your claim that $\vec{E}\perp \vec{B}$ at all times for a "single source".

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  • $\begingroup$ But, can you guarantee that the magnetic field from the bar magnet and the electric field from the point charge will always obey all of maxwell's equations? for example, can you guarantee that $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ where $\vec{E}$ is the electric field of the point charge and $\vec{B}$ is from the bar magnet?? If you cant guarantee this, then this is not a counter example and it violates the condition of a "single source". $\endgroup$ – Mahlomola Daniel Cwele Jan 28 '18 at 14:43
  • $\begingroup$ @MahlomolaDanielCwele - You can be sure that all the other Maxwell laws are fulfilled in this counter-example. The electric field Coulomb field at $\vec r$ of the point charge $q$ at $\vec r_0$ : $$ \vec E=\frac {q (\vec r -\vec r_0)}{4 \pi \epsilon_0 |\vec r -\vec r_0|^3}$$ You can check that the curl of this is always zero. Also, the static magnetic field has always a zero time derivative. Thus $$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$$ is always fulfilled, because both the LHS and the RHS of this Maxwell equation are always zero. $\endgroup$ – freecharly Jan 28 '18 at 15:23
  • $\begingroup$ I realized that my added counter-example with a "single source" is in essence identical to the one given before by Philip Wood in his answer. $\endgroup$ – freecharly Jan 28 '18 at 20:07

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