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Hidden momentum

  1. Sep 26, 2012 #1

    bcrowell

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    "hidden momentum"

    There's an interesting thing in E&M which is referred to as "hidden momentum" in this paper:

    Babson et al., Am. J. Phys. 77 (2009) 826
    http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2009_Griffiths.pdf

    The simplest example is a magnetic dipole immersed in an electric field, both of them at rest in a certain frame. The Poynting vector is the local density of momentum in the electromagnetic fields. The fields' total momentum doesn't vanish in general. But there's a theorem in SR that says that if a system's center of mass-energy is at rest, its total momentum must be zero. Babson resolves the problem by imagining the magnetic dipole as a current loop. Relativistic effects cause the flowing charged particles to have a total momentum (the "hidden momentum") that is not zero, and cancels the momentum of the fields. He does it for a couple of simple models of the charge carriers (balls rolling freely in a tube, and a perfect fluid).

    This all seems fine, but in other examples, it's not clear to me how the hidden momentum can come about.

    Suppose that the magnetic dipole is a fundamental particle such as a neutrino. How in the world does the hidden momentum come about? Does it have to be some kind of QFT effect, which we could imagine in terms of momentum of virtual particles?

    Suppose I have access to some magnetic monopoles. I put uniform electric charges +q and -q on opposite sides of a cube, and on two other opposite sides I put equal and opposite monopole charges. Where is the hidden momentum now?
     
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  3. Sep 27, 2012 #2
    Re: "hidden momentum"

    I think the problem here is too naive mixing of QM and SR. A static field with nonzero momentum is a quantum phenomenon without classical counterpart. This is a field configuration with maximum uncertainity of position (i.e. with no position at all).

    The analogy with charge carriers is good. You can imagine that the static field with nonzero momentum is some kind of a fluid (or a collection of tiny particles). This way, it can have momentum while no changing position at all. The thing that actually flows is the probability density.
     
  4. Sep 27, 2012 #3

    Bill_K

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    Re: "hidden momentum"

    If you mean a uniform electric field, I don't believe it. Consider first a magnetic monopole in an electric field. For any point where the magnetic field is B, there's a diametrically opposite point where the field is -B. Likewise the Poynting vectors at these two points will be equal and opposite, and so by symmetry the total field momentum vanishes.

    Now create a magnetic dipole from two magnetic monopoles. By superposition, the total field momentum still vanishes.
     
  5. Sep 27, 2012 #4

    mfb

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    Re: "hidden momentum"

    SM neutrino? I think even that could have some quantum mechanical analogon to a current loop, if it can have a dipole moment at all.
     
  6. Sep 27, 2012 #5

    bcrowell

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    Re: "hidden momentum"

    This argument totally makes sense to me, so there must be something wrong with my understanding of the Babson paper (which is freely available at the URL I gave in #1). Here's what they claim:

    So this is a nonuniform field, but notice how the result can be expressed in terms of the field, not its gradient. If I had an electric field created by some set of point charges, we could integrate this expression over them, and the claimed result would be equally valid. It therefore seems that they're implicitly claiming this for any field, even a uniform field.

    Later they analyze where the hidden momentum is:

    They then go on and show that the mechanical momentum of the charges is [itex]\mathbf{p}_{hid}=(1/c^2)(\mathbf{m}\times\mathbf{E})[/itex].

    The two examples are not exactly the same, but I think they're clearly saying that the differences are not important. In the first example they just refer to a dipole, in the second a specific model of a dipole, but that's because they need to examine where the hidden momentum is. In the first example they have a nonuniform field, in the second a uniform one, but I think that's irrelevant, as I argued above.

    So I'm perplexed. If Bill_K's analysis is right, then it seems that Babson's claimed result for [itex]\mathbf{p}_{em}[/itex] is wrong and should actually be zero. If that's the case, then the total momentum [itex]\mathbf{p}_{em}+\mathbf{p}_{hid}[/itex] is nonzero. We then have a special-relativistic system whose center of mass-energy remains stationary but whose total momentum is nonzero. That's impossible, since you can easily prove that in SR, the center of mass-energy is at rest if and only if the total momentum is nonzero. You simply write down the expressions for xcm and ptot in terms of the stress-energy tensor, do an integration by parts, and impose the divergence-free property of the stress-energy tensor. (Babson references a 1968 Phys Rev paper by Coleman, but anyway the theorem is easy to prove.)

    Since Bill_K's symmetry argument seems to contradict the expression for [itex]\mathbf{p}_{em}[/itex] given by Babson, it would be interesting to see how that result was calculated. Unfortunately Babson's reference 5 is to an Am J Phys article from 1969, which is paywalled: http://ajp.aapt.org/resource/1/ajpias/v37/i6/p621_s1?isAuthorized=no [Broken]
     
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  7. Sep 27, 2012 #6

    bcrowell

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    Re: "hidden momentum"

    I thought about this some more in the car on the way to work, doing so without hitting any lampposts or (as far as I remember) killing any pedestrians.

    Babson seems to be claiming that the result by Furry, pem=Exm (in units with c=1), is valid in the case of a point charge and a dipole, in the frame where they're at rest relative to one another. They seem to be portraying it as exact for a pointlike dipole. They then apply it to mechanical models of dipoles and get the result they want, which seems to imply that they're imagining their model dipoles as being sufficiently small for the result to hold.

    I don't think this can be right. If it were, then by integration pem=Exm would be valid for any electric field made by any static charge distribution, i.e., any static electric field. In fact, you could integrate over any static distribution of charges and dipoles. Bill's counterexample (two monopoles in a uniform electric field) can be obtained in this way, since you can chain together a bunch of dipoles and make them look like two monopoles separated by some distance.

    Maybe Furry's result is less general than Babson's description would make it appear. The question would then be under what circumstances it *is* valid.
     
  8. Sep 27, 2012 #7
    Re: "hidden momentum"

    Can a point particle be a dipole?
     
  9. Sep 27, 2012 #8

    bcrowell

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    Re: "hidden momentum"

    An electron has a magnetic dipole moment, and it's treated classically as a point particle. (I suggested a neutrino above, since it's electrically neutral and therefore will remain at rest in an electric field.)
     
  10. Sep 27, 2012 #9

    bcrowell

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    Re: "hidden momentum"

    I think I have a much more direct example where the Babson article just seems plain wrong. One of their examples is this: "A sphere of radius R carries a uniform polarization P and a uniform magnetization M (see Fig. 4). The momentum carried by the fields is [itex]p_{em}=(4/9)\pi \mu_o R^3 (M\times P)[/itex]."

    This is an example that can be constructed out of a surface layer of charges and a surface layer of magnetic monopoles. But for any static arrangement of electric charges and magnetic monopoles, the total momentum in the fields vanishes. To see this, write the fields at any given point as E1+E2+... and B1+B2+..., where the sum is over the contributions from all the charges or monopoles. The Poynting vector at this point is then (E1+E2+...)x(B1+B2+...). Distributing the cross product, you get a sum of terms which are the total momentum of a charge and a monopole. But the total momentum in the fields of such a charge-and-monopole pair is zero by symmetry.

    They also give the example of a capacitor in a magnetic field and say, "Naively, the electromagnetic momentum is ...," giving an expression for the momentum of the fields on the interior of the capacitor. But this is wrong by the argument in the preceding paragraph, and the reason it's wrong seems clear: they've neglected the fringing fields.
     
    Last edited: Sep 27, 2012
  11. Sep 27, 2012 #10

    bcrowell

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    Re: "hidden momentum"

    Hmm...the issue seems to be that you get different results with a point magnetic dipole than with a magnetic dipole composed of two magnetic monopoles separated by a finite distance: Milton and Meille, http://arxiv.org/abs/1208.4826 . Milton claims that there's an extra delta function contribution in the point-dipole case. I guess it's one of these issues where you can't interchange limits and integrals. I need to think about this some more and see if I believe Milton's argument.
     
  12. Sep 28, 2012 #11
    Re: "hidden momentum"

    A 'true' magnetic dipole composed of hypothetical actual monopoles would behave very differently to that of a 'real' current-loop model magnetic dipole - and basically for the reasons indicated in fig. 9 of the cited article by Babson et. al. There must be an excess momentum in the lower leg, simply by virtue of a speed differential owing to potential difference, continuity of flow-rate, and the dictates of SR. It is a dynamical effect totally absent for any static distribution of monopoles subject to any E field. As pointed out in some of the many other articles on this endlessly churned out topic in AJP, the effect is somewhat artificial in that the loop current cannot be an ordinary one composed of conducting wire, for which shielding surface charges eliminate any internally acting E field necessary to generate the needed velocity differential (or pressure differential in the 'incompressible fluid' alternative scenario) around the loop. One may seriously question whether an actual electron intrinsic magnetic moment behaves as per fig. 9 however. The fact that perfect rigidity is contrary to SR does not guarantee an intrinsic moment acts like a 'compressible' loop-current. Something imo needing experimental evidence to decide.

    Still, the real problem with that article and many similar ones is not in claiming hidden momentum is real - it certainly is, whether or not it applies to the case of intrinsic magnetic moments in E fields. It's quite easy to come up with mechanical analogues. E.g. - ball bearings circulating in a tube subject to gravity. Hidden momentum is real.

    The problem is in claiming HM as supporting evidence for the reality of SFM (stored field momentum) in static crossed fields, and that imho is extremely doubtful. Has no-one noticed there is a shell game going on with Babson's argument? In the end we are supposed to be relieved and happy that HM + SFM = 0. Sorry, but there are not two but three players in the game, third one being OMM ('overt' mechanical momentum) imparted in setting up and winding down such arrangements. OMM is acknowledged certainly, but I can see no proper accounting procedure in the end. Look at tables IV and V, and compare with conclusion statement in sect.VI. Notice something oddly out of place? Remember - there are three, not two players in this momentum game. Actually, it turns out afaik that always HM + OMM = 0 (linear momentum situations - does not apply to angular momentum scenario like Feynman disk where HM = 0). Which might make one ask where there is room for a physically real SFM. Just my crazy thoughts. Oh, one more thought. By mounting that tube-full-of-ball-bearings or equivalent onto a carousel, one can set up HM entirely in the absence of any imposed fields - a further question mark around the supposed 'balance' of SFM and HM. And, further thought two - get ready for a real shock when you look at the implications of having that loop of fig. 9 precess about the W axis! Poor old Emmy Noether might be turning in her grave.
     
    Last edited: Sep 28, 2012
  13. Sep 28, 2012 #12

    bcrowell

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    Re: "hidden momentum"

    It looks to me like some aspects of this phenomenon are much less interesting and much less important than they might have appeared. In cases involving a dipole where the momentum in the electromagnetic field is nonzero, the momentum is found within a region around the dipole that is approximately the size of the dipole itself. Since the "hidden" mechanical momentum is also located inside that region, these issues are of no interest unless you're specifically interested in modeling the internal structure of the dipole. The localization of the phenomenon seems to imply that if you're doing macroscopic physics, and the dipoles are microscopic, then you can't observe any macroscopic effects from these phenomena.

    This is completely different from the angular momentum of static fields, which is interesting and can be found far away from the sources.

    Babson does give an example that appears to have measurable real-world effects, which is a capacitor immersed in a magnetic field. When you discharge the capacitor, it receives a kick. But I think this one is also much ado about nothing. Suppose I supply the magnetic field using a permanent magnet. The permanent magnet consists of microscopic magnetic dipoles at some distance from the capacitor, and the total momentum in the electromagnetic field is equal to the momentum that is localized inside the dipoles. Any momentum in the fields between the plates of the capacitor is exactly canceled by the momentum of the fringing fields, so there's no need to calculate it explicitly.

    There is no mystery about how to avoid a contradiction with the theorem that says the total momentum must be zero iff the center of energy is at rest. The total momentum of the macroscopic fields vanishes. The momentum of each dipole also vanishes, since its mechanical momentum cancels the momentum in the local fields.

    In principle, the effects could be of interest in cases where the magnetic field is provided by macroscopic current loops. Suppose you have a solenoid immersed in an external electric field. Let A be the cross-sectional area of the wire, a the area of the solenoid in a plane perpendicular to the axis, l the linear dimension of the solenoid in the direction perpendicular to the electric field, w the correspnding dimension parallel to the field, and E the externally applied electric field. For a typical desktop current loop, l=0.1 m, A=5x10^-3 m^2, a=l^2, and I=10 A. Let the electric field be 10^6 V/m (about the point where air sparks). Then I get about 10^-12 kg.m/s for the hidden momentum phid=IaE/c^2. If the current is modeled by a perfect fluid, then the pressure difference as calculated by Babson between different parts of the wire is IaE/(c^2lA), which comes out to be about 10^-5 Pa. These effects seem much too small to measure. The momentum, which would be released when either field was discharged, would be masked by stray electric and magnetic forces many orders of magnitude greater. To make the pressure difference measurable, it seems like you'd need a super-gigantic wire loop.

    All the experimental observables seem to scale like surface effects, so to get an effect that's significant, you really want to go down to the microscopic scale, where surface-to-volume ratios get big. But at the microscopic scale, you need a quantum-mechanical model, not a classical one like this.
     
    Last edited: Sep 28, 2012
  14. Sep 28, 2012 #13
    Re: "hidden momentum"

    Hmm - when double-checking all the calc's done in Babson et al. article I have to concede their balance sheets do actually add correctly. So must withdraw the 'shell game' criticism as wrong - formally there is a balance between SFM and HM as claimed there. A stationary center of energy in the considered examples there is the result of four not three players in total. Net mechanical impulse owes to not just the electrical+magnetic forces Fem acting on the charges and currents, but also the mechanical reaction force Fhm that acts against the direction of rate-of-change of HM when being set up. It is the sum of these two mechanical impulses that cancel, ensuring a stationary center-of-energy. The other two - SFM and HM then cancel as shown there. Had missed seeing that they accounted for the purely mechanical reaction force Fhm owing to changing HM - it was there in 2nd para. following eq'n (44):
    OK so it all seems to work in examples they give, where HM is generated from actions of EM fields on non-precessing systems. I maintain though there is considerably more to it all. Standing by the other claims made in #11. Centripetal acceleration can replace an applied E or gravity field - as per carousel arrangement mentioned in #11. Suppose out in deep space a carousel having major axis z has mounted upon it's periphery an evenly spaced array of circular tubes each carrying frictionless circulating matter (fluid, ball-bearings etc.). Spin axes of such tubes oriented normal to both z and a radius vector r from axis z to the center of each tube, all such axes pointing say clockwise around the carousel periphery. Artificial 'g' from rotation of carousel about it's spin axis z will then set up in each tube an HM acting along z axis. Without any recourse to applied E or B or real g fields and thus without any connection to E×B type SFM. There are of course gyroscopic torques also acting on tubes, but such play no role in SFM and related considerations.

    Now for the enigmatic, difficult part to that. Suppose that before spin-up, carousel assembly initially had zero net motion in some rest frame S. We require total momentum be conserved. The only counter to HM subsequent to carousel spin-up must be an overt mechanical impulse of -|HM| along z axis imparted to the assembly as a whole. Implying it's center-of-mass will move along that axis at some steady drift velocity vz once HM in circulating tube matter has stabilized. In some other frame S' having motion say vy normal to z, there exists a steady rate-of-change of angular momentum dL'/dt' owing to the net system mass M motion along z given by dL'/dt' ~ -Mvy×vz (where both |vy,vz| << C so gamma factors can be ignored). An arbitrary figure, given there are an infinite number of possible frames S' in which L'(t') and dL'/dt' can be evaluated.

    There is some way to avoid this situation, apart from denying HM is possible? I can think of none. Strange but it seems inevitable consequence of a system having zero net linear momentum yet with an imparted moving center-of-energy. And then there is the other matter raised in #11 of precessing about a vertical axis of that loop in fig.9 of Babson et al. article. Opens up another can of worms.
     
  15. Jun 2, 2013 #14

    clem

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    I just came across this thread. There is no hidden momentum as described in the Babson paper. Details are in
    http://arxiv.org/pdf/1302.3880.pdf
    It is a bit complicated, but shows what they call hidden momentum just is not there.
     
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