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I am new to electromagnetism, and have a question regarding the possibility to leave out the $B$ component entirely for picturing electromagnetism, as follows please.

The local relationship between any point $x$ and its immediate neighbors that will naturally give rise to a global wave "pattern" can be specified by:

$$ \frac{\partial^2 y}{\partial t^2} = c \frac{\partial^2 y}{\partial x^2} $$

When time is iterated step by step, and at each time-step we perform this iteration locally for all $x$ and its immediate neighbors using this specified relationship (with finite-difference method for example), a "pattern" emerges that we are familiar with, resembling that of water waves. The "speed" at which this wave travels outward from a point source is dependent of the constant c in this local relationship, which we can then interpret as the "speed" of the wave pattern as we time-evolve it.

In electromagnetism, the two starting Maxwell equations $$ \frac{\partial E}{\partial t} = - k_1\frac{\partial B}{\partial x}$$ and $$ \frac{\partial B}{\partial t} = - k_2\frac{\partial E}{\partial x}$$

usually just ends up being combined to give a single self-referencing relationship

$$ \frac{\partial^2 E}{\partial t^2} = c\frac{\partial^2 E}{\partial x^2}$$

which is the exact local relationship that we have known will give rise to a wave like pattern spreading out at the speed of c.

My question is, knowing with hindsight that B can be explained in other ways (using special relativity), is anything lost in our understanding of E if we just picture E exactly like a single-component water waves, and leave out B altogether in the picture?

That engineers who need to calculate effects of B are of course free to unearth the concept to fit their specific needs, but that it is not strictly essential to include B in our simpler understanding of E?

Please advise. Thank you.

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    $\begingroup$ There's more things for electromagnetics to explain than just how an EM wave propagates. $\endgroup$
    – The Photon
    Oct 23, 2021 at 18:19
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    $\begingroup$ Without the magnetic field, how would you explain why a compass needle points north? $\endgroup$
    – d_b
    Oct 23, 2021 at 18:53
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    $\begingroup$ you can have no $\vec E$ and still have a $\vec B$.. $\endgroup$ Oct 23, 2021 at 20:26
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    $\begingroup$ It's not true that you can always find a frame with zero magnetic field, so it's not true that "special relativity + electric fields" is enough to fully explain magnetism. It is true that a Lorentz boost will convert some electric field into some magnetic field, and vice versa, so you can conclude that magnetic fields are necessary from electric fields + special relativity, but you can't conclude that all magnetic fields would be purely electric in some frame... this point has been lost on many people on the internet. $\endgroup$
    – Andrew
    Oct 23, 2021 at 21:31
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    $\begingroup$ @Andrew I would upvote that comment ten times if I could. It blows my mind how the majority of people who have learned these arguments seem to have taken the exact opposite of the intended lesson from them. $\endgroup$
    – knzhou
    Oct 24, 2021 at 4:22

3 Answers 3

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is anything lost in our understanding of E if we just picture E exactly like a single-component water waves, and leave out B altogether in the picture?

Yes, you lose quite a bit, actually. In particular, the first equation is only valid in the absence of currents, and the solutions for the wave equation do not include electrostatic fields. The wave equation, while important, is not the sum total of all electromagnetic behavior. The magnetic field is necessary as are currents and charges.

Without them you lose:

Circuits

Energy transfer

Near field effects

Electrostatic effects

Magnetic dipoles

Etc.

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    $\begingroup$ @ Dale thank you for answering! I think I realize now there are other aspects to B that cannot be explained away. $\endgroup$
    – user315366
    Oct 24, 2021 at 8:01
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Magnetic field due to motion of charges is just an electric field viewed from a different frame of reference. But the magnetic field due to the spin of particles is an intrinsic effect that does not depend on the reference frame. Therefore, it is wrong to assume that all magnetic fields are just electric fields from a different frame of reference.

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  • $\begingroup$ @ Curiouser thank you! Is the magnetic field due to spin of particles the same/indistinguishable experimentally from those macroscopic current in a wire variety of B? $\endgroup$
    – user315366
    Oct 24, 2021 at 8:04
  • $\begingroup$ What you mean is that only in special cases there is a reference frame in which the magnetic field vanishes. $\endgroup$
    – my2cts
    Oct 24, 2021 at 10:28
  • $\begingroup$ What if you have a loop of current carrying wire? Can you show me a frame where the magnetic field is zero everywhere? $\endgroup$
    – Andrew
    Oct 24, 2021 at 17:53
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What you are looking for is the four potential $(V,\vec A)$, where $V$ is the potential energy and $\vec A$ the potential momentum.

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  • $\begingroup$ @ my2cts thank you. i looked it up and found this survey article for learning electromagnetism through gauge theory arxiv.org/pdf/physics/9803023.pdf $\endgroup$
    – user315366
    Oct 24, 2021 at 19:21

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