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I Does a rotating magnetic field possess angular momentum?

  1. Mar 14, 2017 #1
    Does a rotating magnetic field possess angular momentum in the direction of rotation?

    I suppose this comes down to a broad question about the physical nature of fields in general. I love the Einstein-de Haas effect, where an iron core spins in the opposite direction of the induced spin angular momentum of the electrons producing the field. That’s all very clear and intuitive.

    And that got me wondering about the case of a rotating magnetic field. Let’s say we have an iron toroidal core with a series of coils around it (let’s say that there’s an air gap between the coils and the toroid, and the toroid is somehow floating freely inside them), as depicted in Fig. 16-13 of this Feynman Lecture:
    http://www.feynmanlectures.caltech.edu/II_16.html

    Now we induce a rotating magnetic field. Will the iron core rotate in the opposite direction?

    Maybe I’m mistakenly thinking of a magnetic field as a form of “stuff,” which has its own independent physical properties like angular momentum when rotated. It’s perplexing because this page discusses the momentum of fields stored in space, but the directions of the flow are still counter-intuitive to me:
    http://www.feynmanlectures.caltech.edu/II_27.html

    If anyone can shed some light on this I’d really appreciate it; I can’t find any good references to help me figure out how to think about this question.
     
  2. jcsd
  3. Mar 14, 2017 #2

    mfb

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    The magnetic field has an energy density, and the electromagnetic field has a momentum given by the Poynting vector. The time-variable magnetic field will induce an electric field, you can use this plus the magnetic field to calculate the Poynting vector, and calculate angular momentum based on that.

    I can imagine that it is non-zero, although it would be tricky to trace back where it came from. The coils could carry away angular momentum as well.
     
  4. Mar 14, 2017 #3

    Vanadium 50

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    A magnetic field cannot rotate. It can have a curl, but it exists at each point in space and time. You can't detach and rotate it.
     
  5. Mar 14, 2017 #4
    Thanks for Poynting me in the right direction on this ; I found this excellent paper, which indicates that the Poynting vector in a transformer, which is exactly what we’re looking at in that Feynman diagram Fig.16-13, is directed from one coil through the center to the other (although in this paper they use a rectangular iron core rather than a circle):

    “The Poynting vector field and the energy flow within a transformer”
    http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Poynting_transformer.pdf

    So the Poynting vector would rotate as the magnetic field rotates.

    I’m not sure what to make of this since the Poynting vector represents the momentum/energy flow, and in this case it's rotating perpendicular to that flow. And I can’t find anything online about a “rotating Poynting vector.” Is that equivalent to angular momentum?
     
    Last edited: Mar 14, 2017
  6. Mar 14, 2017 #5
    But Richard Feynman seems to disagree:

    “A field just like that of a rotating magnet can be made with an arrangement of coils such as is shown in Fig. 16–13.”
    http://www.feynmanlectures.caltech.edu/II_16.html

    It’s my understanding that in all physical ways, including inductive effects, a rotating magnetic field is equivalent to a rotating magnet without the rotating matter. That strikes me as bizarre and counterintutitive, but that frequently seems to be the nature of physics.
     
  7. Mar 14, 2017 #6

    mfb

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    No, the Poynting vector is already something connected to angular momentum.
    "A field like that from a rotating field" doesn't mean there is any physical rotation of a field, although that might be a nice description in some cases. But not if you want to look at angular momentum.
     
  8. Mar 14, 2017 #7

    Vanadium 50

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    Then why did he put in in scare quotes?

    I'm done here.
     
  9. Mar 15, 2017 #8
    It’s startling just how difficult it is to find an in-depth authoritative description of the physics of a rotating magnetic field online. You’d think an idea this old would be discussed comprehensively somewhere.

    I did find a huge and useful old book called “Magnetic Fields: A Comprehensive Theoretical Treatise for Practical Use” (Heinz E. Knoepfel, c.1931, 2000 edition) that addresses the question of whether or not a rotating magnetic field actually rotates or not. On .pdf page 163 he writes:

    “Consider a conducting bar centered on the z-axis: The magnetic field diffusion here is identical (in a coordinate system rotating with H) to that described in figure 4.3-6, where instead the bar rotates in a constant field H.”
    http://ed.quantum-bg.org/Knoepfel.pdf

    Figure 4.3-6 is way down on .pdf page 215, and it shows how stationary magnetic field lines twist within and outside of a conductive rod that’s rotating within the field. The earlier quote indicates that the exact same distortion happens to the field lines of a rotating magnetic field spinning around a stationary conductive rod. That seems pretty definitive to me: the same field line distortions are manifest regardless of whether you choose the reference frame of the stationary magnetic field with a rotating rod, or a stationary rod and a rotating magnetic field.

    But that doesn’t answer the question. I’ve scoured the internet for hours and I can’t find anything that addresses question of momentum in a rotating magnetic field (angular or otherwise), or the physical meaning of a rotating Poynting vector, or how I can find the angular momentum now that I know what the Poynting vector is doing.

    If anyone's ever seen a good treatment of the physics of a rotating magnetic field that gets into these details I'd be grateful for a title.

    Thank you.
     
  10. Mar 15, 2017 #9

    jtbell

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    For Poynting vector (energy flow density) ##\vec S = \frac 1 {\mu_0} (\vec E \times \vec B)##, the (linear) momentum density is ##\mathcal{\vec P} = \mu_0 \epsilon_0 \vec S = \epsilon_0 (\vec E \times \vec B)## and the angular momentum density is ##\mathcal{\vec L} = \vec r \times \mathcal{\vec P} = \epsilon_0 [\vec r \times (\vec E \times \vec B)]##, where ##\vec r## is the position vector from the point around which the angular momentum is being calculated.

    (reference: Griffiths, Introduction to Electrodynamics, 3rd ed., section 8.2.4)
     
    Last edited: Mar 15, 2017
  11. Mar 15, 2017 #10
    Thank you jtbell, but I need to find a treatment specifically pertaining to rotating magnetic fields, because it's an exceptional case.

    For example, you might assume that you could just analyze the Poynting vector of the inductors in the array at various times in the cycle to get the time-averaged angular momentum. But this paper points out http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Poynting_transformer.pdf (toward the end of Section II) that the magnetic field within each inductive winding in a similar transformer circuit, is actually insignificant (totally contrary to what I've been taught about inductors) - the magnetic vector field *outside of and between* the inductors is actually much larger (Eq. 4 ), and it's that exterior field that's responsible for the energy transport. So I'm not sure where the position vector begins - inside one inductor, on the outside of an inductor, or in the center of this field in-between, which is where the energy transport is happening. And what's the physical significance, if any, of the rotating Poynting vector on the question of angular momentum?

    I can't find any academic treatment of the rotating magnetic field that explains all of this, and after seeing how weird and counterintuitive it seems to be, I have no confidence in my ability to figure it out from first principles. If anyone else does, then they may well prove to be an unsung God of Physics, because this is much more complicated than I thought it would be.
     
  12. Mar 16, 2017 #11

    mfb

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    You don't need that. You just have to find the Poynting vector everywhere. Which is not necessarily easy.
     
  13. Mar 17, 2017 #12
    I asked this question to a physics professor friend who got his PhD in magnetism, and he got back to me yesterday without an answer. So apparently I should’ve ranked this question at a higher level than “Intermediate.”

    Perhaps there’s an easier way to approach this. I keep thinking that it should come down to the angular momentum of the electrons, like in the Einstein-de Haas effect. If we look at it that way, it seems analogous to a mechanical system consisting of a series of gyroscopes at every point of a circular ring, where they’ll all precessing in the plane of the ring in tandem, so the axes align through the center point in a rotating manner around the ring. Since the angular momentum pseudo-vector points along the axis of spin, which is rotating in place at every point of the ring, then the time-averaged momentum at every point would sum to zero.

    And if we look at the rotating magnetic field itself as if it’s generated by a single circular perpendicular current (kinda like one big electron), then we’d get the same result – rotating angular momentum averaging to zero with each cycle.

    I’m not sure if the mechanical analogy is entirely applicable here, but it seems to be consistent with the description of the Einstein-de Haas experiment.
     
    Last edited: Mar 17, 2017
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