1
$\begingroup$

In electricity and magnetism, we use the scalar potential to derive the electric field and the vector potential to derive the magnetic field because $\nabla\cdot B=0$ and $\nabla\times E=0$.

I was just thinking, there are regions in space where $\nabla \times B=0$ as well. For these regions can we define a magnetic scalar potential? To what extent is this potential useful? (It is obviously not universally applicable. But is it useful in any ways?)

$\endgroup$
  • 3
    $\begingroup$ Isn't it because of $\nabla\cdot\mathbf B=0$ that we define a vector $\mathbf A$ such that $\mathbf B=\nabla\times\mathbf A$? $\endgroup$ – Kyle Kanos Sep 23 '15 at 20:13
  • $\begingroup$ I didn't see your question until just now. The magnetic scalar potential can be very useful, and is too often dismissed witout much consideration. I treat and use it in some detail in my textbook <amazon.com/Classical-Electromagnetism-Jerrold-Franklin/dp/…> See all of Section 7.10 and Section 7.11.2. $\endgroup$ – JERROLD FRANKLIN Jul 18 '16 at 21:41
  • $\begingroup$ @JERROLD FRANKLIN : I bought your book. Only a used copy I'm afraid, but still. I have a special interest in the way electromagnetism is taught. $\endgroup$ – John Duffield Jul 20 '16 at 6:37
4
$\begingroup$

Yes, we can define a magnetic scalar potential in some problems, specifically if the current density vanishes in some places. Note that the condition is not $\nabla \cdot B = 0$ since this is always true. To define the magnetic scalar potential requires that there be a quantity whose curl is zero (curls of gradients are zero), which is to say $\nabla \times H = 0$. This can often be very handy, as (obviously) it's easier to work with than the vector potential. As you note, however, it isn't generally useful, and for most "interesting" problems you will not be able to use it. The Wikipedia page is here, and the relevant section of Jackson's Classical Electrodynamics is 5.9.

$\endgroup$
-4
$\begingroup$

In electricity and magnetism, we use the scalar potential to derive the electric field and the vector potential to derive the magnetic field because ∇⋅B=0 and ∇×E=0.

IMHO there's a deeper reason than the mathematical expressions: the field concerned is the electromagnetic field. See section 11.10 of Jackson's Classical Electrodynamics where he says "one should properly speak of the electromagnetic field Fµv rather than E or B separately". Also see Wikipedia and note this: "Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole - the electromagnetic field". Or look at electromagnetic radiation on Wikipedia and see this: "the curl operator on one side of these equations results in first-order spatial derivatives of the wave solution, while the time-derivative on the other side of the equations, which gives the other field, is first order in time". What you might think of as two fields are actually two different aspects of the same thing. I think the best way to really appreciate this is via an analogy: imagine you're in a canoe and the electromagnetic wave is a big ocean wave coming at you. Up you go, and as you do the tilt of your canoe denotes E, and the rate of change of tilt denotes B. But there's only one wave there. In similar vein when it's the electron's electromagnetic field, there's only one field there. The electromagnetic field. That's why $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$$ Not because changing one field creates the other, or because one circulates round the other. The equals is an "is". It's because they're two aspects of the same thing.

I was just thinking, there are regions in space where ∇×B=0 as well. For these regions can we define a magnetic scalar potential? To what extent is this potential useful? (It is obviously not universally applicable. But is it useful in any ways?)

You can find mathematical explanations such as this one, but IMHO they don't really get to the bottom of it. IMHO to do that you have to appreciate that it's it's the electromagnetic field, and that it isn't totally unlike the gravitomagnetic field. See this NASA article where you can read things like twist the dimple and if space is twisted and the twists of space-time:

enter image description hereImage courtesy of NASA

The electromagnetic field isn't a million miles away from the gravitomagnetic field, which is why Heaviside was able to come up with gravitomagnetism as a gravitational and electromagnetic analogy in a journal called The Electrician. Think of the electromagnetic field as a "twist" field akin to the gravitomagnetic field. Let's depict it using a fairly common vector-field image:

enter image description herePublic domain image by Fibonacci, see Wikipedia.

OK, here's the important thing: if you were flying into this but thought you weren't moving, you might not realise it was a "twist" field. You might think of it as a "turn" field instead. You will be aware that curl is also called rot which is short for rotor. It's for good reason. A magnetic field causes electrons to go round in circles, it really is a rotor field, a "turn" field. And most crucially, we have frame dragging in space. We do not have discontinuities. There are no regions where space is rotating freely like a roller bearing. It might look that way when you're moving through the electromagnetic field or vice versa, but space is connected, and the "twist" diminishes away from the centre. It's like the non-irrotational field show on the right here. The curl is not zero. For it to be zero you need a lump of space rotating freely, but space just isn't like that, and that's why we have no magnetic monopoles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.