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Paradox regarding energy of dipole orientation

  1. Mar 4, 2007 #1
    "paradox" regarding energy of dipole orientation

    I've ran into a "paradox" concerning deriving the energy of a dipole's orientation in an external field. For example, the energy of a magnetic dipole m in an external field B is known to be:

    [tex]U= - \mathbf{m} \cdot \mathbf{B}[/tex]

    In Griffiths Intro to Electrodynamics, this is argued by looking at the torque on a small loop of current in an external magnetic field.

    The "paradox" arises instead when we try to derive it by looking at the energy in the magnetic field.

    [tex] U_{em} = \frac{1}{2} \int (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) d^3r [/tex]

    If we consider an external field B and the field due to a magnetic dipole B_dip, we have:

    [tex] U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 d^3r [/tex]
    [tex] U = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip})d^3r[/tex]

    the [tex]B^2[/tex] and [tex]B^2_{dip}[/tex] terms are independent of the orientation and so are just constants we will ignore. Which leaves us with:

    [tex] U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} d^3r [/tex]

    Now we have:
    [tex]\mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] + \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r})[/tex]

    If you work through the math, only the delta function term will contribute, which gives us:

    [tex]U=\frac{2}{3} \mathbf{m} \cdot \mathbf{B}[/tex]

    Which not only has the wrong magnitude, but the wrong sign. And thus the "paradox". Obviously, there is no paradox and I am just calculating something wrong, but after talking to several students and professors I have yet to figure out what is wrong here.

    One complaint has been that while doing the [tex]d\theta[/tex] and [tex]d\phi[/tex] integrals show that the non-delta function term doesn't contribute, it is unclear if this arguement holds for the point right at r=0. I believe it still cancels, but to alleviate that, let's look at a "real" dipole instead of an ideal one. A spinning spherical shell of uniform charge with a magnetic dipole m, has the magnetic field outside the sphere:

    for r>=R [tex]\mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] [/tex]

    So this is a good stand in for the idealized dipole, as it has a pure dipole field outside of the sphere. Inside the sphere the magnetic field is:

    for r<R [tex]\mathbf{B}_{dip} = \frac{2 \mu_0}{3} \frac{\mathbf{m}}{\frac{4}{3} \pi R^3}[/tex]

    which as you can see, reduces to the "ideal" case in the limit R -> 0.

    Here there is no funny business at r=0. The math again gives:

    [tex]U=\frac{2}{3} \mathbf{m} \cdot \mathbf{B}[/tex]

    What gives!?
    Who can help solve this "paradox"?
    Last edited: Mar 4, 2007
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  3. Mar 4, 2007 #2

    Hans de Vries

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    This reminds me of what is called the "4/3 problem" of classical electrodynamic
    energy/mass which has been studied by many physicist including Pointcaré
    and Feynman. See chapter 28 in Volume II of the Feynman lectures on physics.

    If the electrostatic energy of a point field with a cut off would be equivalent
    with a mass M then the momentum of the EM field of this particle moving at
    speed v corresponds to a momentum of a mass 4/3 M moving at v.


    Regards, Hans
    Last edited: Mar 4, 2007
  4. Mar 6, 2007 #3
    I've talked to some more grad students and still no one can figure this one out. I'm sure it is something really simple and we'll all feel like idiots afterward, but we just can't see it.

    Please, if anyone can help here it would be much appreciated. This problem has been nagging at the back of my head for a week now.
  5. Mar 6, 2007 #4
    Interesting paradox. I'll work it out and see what I can come up with.
  6. Mar 7, 2007 #5
    i don't know if this resolves the paradox, but the following can happen in the classical dipole model when applied to molecular systems:

    the charge distributions of two molecules approaching each other create and E field - this E field induces molecular dipoles. These molecular dipoles in turn contribute to the field, which enhances the dipoles. Dipole forces attract the molecules. At a certain distance, you reach a region of discontinuity where the dipole will grow unboundedly and the field will blow up - clearly a non-physical result. The reason this doesn't physically happen is due to the Pauli exclusion principle and electron-electron electrostatics, neither of which is captured in the classical dipole model. For this reason, computer simulations of molecules using explicit many-body polarization are done with "damping" regions that serve to remove the discontinuity.

    it may also turn out that in your case, the higher order multipole moments are important? what happens if you include quadrupolar terms?

    im just tossing these ideas out here, im not sure what the source of your seemingly paradoxical result actually is.
  7. Mar 10, 2007 #6
    Have you had any luck in figuring it out? Even if not, I'd be interested in hearing what results you got.

    I've been working on the electric dipole case now, which has a similar problem. Equivalently, the answer should still be U = - (p.E) However the electric dipole field has a different factor in front of the delta function term ... which confuses this even more.

    Please, if anyone has time to sit down and do a couple quick integrals and think through this, please do so. The math looks correct, so there must be some mistake in the logic and I can't find it.
  8. Jan 25, 2009 #7

    Jonathan Scott

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    Re: "paradox" regarding energy of dipole orientation

    [I saw the mention of this thread as an "unsolved paradox" in some other thread]

    The standard expressions for the Maxwell energy density and energy flow (Poynting vector) have been shown to be inconsistent under Lorentz transformations except when dealing with situations involving propagation of waves at c, with no local sources (charges or dipoles).

    The problems and some specific solutions were described quite clearly in an old paper by J W Butler which I don't have to hand right now, but from a Google search I'd guess it's probably called "A Proposed Electromagnetic Momentum-Energy 4-Vector for Charged Bodies".

    Butler's more general expression for electromagnetic energy density is described in Jackson "Classical Electrodynamics" (2nd edition) section 17.5 "Covariant Definitions of Electromagnetic Energy and Momentum". It should have a better chance of giving consistent results.
  9. Jan 25, 2009 #8


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    Re: "paradox" regarding energy of dipole orientation

    http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3421v3.pdf [Broken]
    Last edited by a moderator: May 3, 2017
  10. Jan 25, 2009 #9
    Re: "paradox" regarding energy of dipole orientation

    Oh wow! I didn't really expect this thread to be revived. Thanks guys.

    The only real updates since I last posted here is that a friend and I were confused why the [itex]\rho V + A \cdot j[/itex] worked for the electric dipole but the energy in the fields term did not. Looking at the derivation it quickly because obvious we neglected the surface term.

    We also studied this with finite source fields and finite dipoles to make us more sure of the math ... it gave the same answer.

    This also fixes the magnitude of the constant in front of [itex] m \cdot B[/itex] for the magnetic dipole case. However, the sign is still wrong. Using A.j of course gives the same answer (since they are mathematically equivalent here).

    I've asked professors and grad-students, but unfortunately, no one has an idea yet (although someone found a letter to a journal were Prof. David Griffith poses the same question. I could not find a response to his question though).

    I feel uncomfortable with some of their claims, and will have to read this closer later.
    Regardless, saying B.B is not an energy but A.j is does not solve this problem. It gives you the wrong sign just as the energy in the fields equation does (and as they must, since they are related by a vector calc identity).

    Thanks for the heads up.
    I'll give it a read through. This stupid sign error in my math/reasoning has always bugged me.
    Last edited by a moderator: May 3, 2017
  11. Jan 26, 2009 #10

    Hans de Vries

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    Re: "paradox" regarding energy of dipole orientation

    A very elementary example which requires Butler's expression to get the
    correct value for the energy flux is that of the (infinite) parallel plate
    capacitor moving perpendicular to the plane.

    Whatever the perpendicular velocity, the magnetic field B stays 0 always
    and so the energy flux contribution from the Poynting vector is 0 also.

    In Butler's expression the whole stress-energy tensor is used and the
    energy flux becomes:

    (energy density E2 at rest) TIMES (velocity) TIMES (gamma).

    Where the gamma term stems from the Lorentz contraction which results
    in a higher density of the energy flux. This is the result you would expect.

    One need Bloch's expression also to calculate the electromagnetic momentum
    of virtual photons from Klein Gordon transition currents.

    Regards, Hans
  12. Jan 26, 2009 #11


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    Re: "paradox" regarding energy of dipole orientation

    The + sign for mu.B is explained on page 10 of
    http://arxiv.org/PS_cache/arxiv/pdf/...707.3421v3.pdf [Broken]
    Last edited by a moderator: May 3, 2017
  13. Jan 26, 2009 #12
    Re: "paradox" regarding energy of dipole orientation

    I feel that fundamentally, you are somehow missing the energy of the current itself inside the loop. If you think about it, what you are calculating is just the energy stored in the fields outside of the loop. A dipole tends to point along the field lines, and when it does, the field it produces points along the magnetic field line. This of course strengthens the field and thus increases the energy contribution. However, the more important term would be the energy inside that loop of current, and it should be calculated in terms of ∫A·j dx (with an appropriate choice of j, then take the limit r->0). In your calculation, this infinite j term is completely neglected. The delta function in the B term does not take this into account, as it merely captures the B field at the center of the loop going to infinity. Indeed, the A·j term should capture the essential interaction of a twisting dipole.
    Last edited: Jan 26, 2009
  14. Jan 26, 2009 #13


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    Re: "paradox" regarding energy of dipole orientation

    Butler's paper has nothing to do with a magnetic moment at rest in a magnetic field.
  15. Jan 26, 2009 #14

    Jonathan Scott

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    Re: "paradox" regarding energy of dipole orientation

    That appears true to me. The main point which was relevant here is that it shows that the conventional energy density (both electrostatic and magnetic) only holds in the absence of sources, so it explains why it didn't work here. It also gives an expression which works for the energy of a single moving charge, but I don't know whether it can be integrated to cover those cases.

    The paper by J Franklin looks very interesting and useful, thanks, and as you point out it's very relevant to this particular case.
  16. Jan 28, 2009 #15
    Re: "paradox" regarding energy of dipole orientation

    No, I am including the energy inside the dipole as well. As mentioned above, I worked this out with a finite sized source and got the same answer as the point dipole (with the delta-function term ... which is indeed the 'inside' contribution). Also as mentioned, the A.j method is mathematically equivalent and sure enough, gives the same answer.

    Okay, I read up on the portion you mentioned in Jackson.
    That is not the problem here. What Jackson mentions is that the usual stress energy tensor doesn't transform correctly if there are singularities in the stress-energy tensor. So if there are point (or line or plane, etc.) sources, some steps can be taken to give a better definition.

    This doesn't apply to this problem for three reasons:
    1] The usual energy and momentum is still conserved. It can still be treated as an energy momentum as long as you are not transforming and mixing result from other coordinate systems.
    2] In a static situation (the sources do not depend on time), it appears that covariant defintion reduces to the usual one.
    3] I can easily choose a magnetic dipole source that is not a singularity (for example a sphere of uniform charge density spinning at a constant rate ... and super-impose another of opposite charge spinning the opposite way if you want to remove the electric field).

    As I already stated, I am very uncomfortable with this paper.
    In SR it does not make a difference if you use the usual energy density and surface term, or the A.j term. They keep complaining about the surface term as if that makes that form of the electromagnetic energy wrong. But it is not wrong. They are mathematically equivalent. And if the surface term bothers you that much, just note that the surface term only gives a contribution when you consider infinite sources (like they do, with an external B field everywhere or similarly with an external E field) or infinite time.

    Secondly, they do not explain the plus sign. They waive it away, and with what appears to be false logic.

    Third, while it doesn't matter in SR, it DOES matter in GR where the energy/momentum is located (in the fields or purely in the charges). The canonical method has always been to use the stress-energy tensor of the fields. This paper blatantly flies in the face of that.

    For all those reasons, let's please move on from that paper for this discussion. If you want to start another thread discussing their opinions in that paper, so be it.
    Last edited by a moderator: May 3, 2017
  17. Jan 29, 2009 #16
    Re: "paradox" regarding energy of dipole orientation

    So what if the math here is correct and there really is a paradox? What would this mean if the paradox is true? Can an experiment be conducted to verify the existence of such a paradox?
  18. Jan 29, 2009 #17
    Re: "paradox" regarding energy of dipole orientation

    With the energy having the opposite sign, it would mean that magnetic dipoles would try to anti-align with a magnetic field. We know experimentally that they allign with the field.

    There is no real paradox here (hence the use of "scare quotes").
    However, learning the resolution to this "paradox" will probably give me greater insight into the problem.
    The fact that the author of a popular electrodynamics textbook asked similar questions worries me that we might not be able to figure this out ourselves. I'd like to keep trying though. Any ideas people?
    Last edited: Jan 29, 2009
  19. Jan 30, 2009 #18

    Jonathan Scott

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    Re: "paradox" regarding energy of dipole orientation

    I don't remember the details that Jackson mentions right now, but I'm fairly sure Butler's paper shows that the standard expressions for the integrals of the assumed energy density (and Poynting vector flow) in space are only equal to the energy densities calculated by the other method (from the charges in the potentials) if there are no sources of any sort within the volume being considered; the equivalence is usually demonstrated by letting the volume extend to infinity and assuming the surface term tends to zero, but it doesn't if there are sources.

    I think he goes on to show that you do get a consistent result if you assume energy density related to E2/2 in the rest frame of a charge, in which case I think the energy-momentum density part of the tensor becomes (E2-B2)/2 times the four-velocity in some other frame, where the sign of the B term is the opposite of that obtained for waves travelling at c in the absence of sources.

    I first came across Butler's paper when I tentatively worked out the same result myself using four-vector algebra and thought that it contradicted the standard expressions but someone referred me to his paper. I still wonder about where the energy is "really" located, from the gravitational point of view, but until this new paper came along (which I haven't yet fully digested) I thought it could be consistently assumed to be in the field.
  20. Jan 30, 2009 #19
    Re: "paradox" regarding energy of dipole orientation

    Please reread the discussion in Jackson, for one of us is misunderstanding something, for I still disagree with you here. However, on a closer second reading I see that by divergent he didn't mean "singularity" but merely in the Del.whatever sense. So you are correct about that (any sources, singularities or not, ruins the general covariance).

    Yet, again, this doesn't mean the usual energy and field momentums cannot be used consistently within one inertial coordinate system.

    pg 756 in Jackson "Classical Electrodynamics"
    "the usual spatial integrals at a fixed time of the energy and momentum densities,
    [tex]u = \frac{1}{8\pi}(E^2 +B^2), \ \ \ \ g = \frac{1}{4\pi c} (E \times B)[/tex]
    may be used to discuss conservation of electromagnetic energy or momentum in a given inertial frame, but they do not transform as components of a 4-vector unless the fields are source-free."

    Also, later when working out the covariant definitions for the densities, he states that if the total momentum of the fields is zero, then the covariant definition just reduces to the usual definition. So in the cases considered here, (the total momentum in the fields is zero), we are already using the covariant definitions.
  21. Jan 31, 2009 #20

    Jonathan Scott

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    Re: "paradox" regarding energy of dipole orientation

    I can't identify which bit you mean by the last reference.

    As I see it, the first part of section 17.5 says that although the conventional expressions do not transform correctly, you can get an arbitrary but correctly transforming quantity by assuming that the conventional expression is correct in some frame then transforming that expression.

    Jackson then goes on to say (paragraph containing equation 17.44) that in the special case that there is a frame in which all the charges are at rest (so B is zero) then transforming to any other frame (17.44 and 17.45) gives an expression for the energy and momentum density which gives a physically plausible four-vector (and is the same as Butler's expression).
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