Paradox regarding energy of dipole orientation

[JustinLevy]

"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:

U= - \mathbf{m} \cdot \mathbf{B}

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.

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

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

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

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

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

Now we have:
\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})

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

U=\frac{2}{3} \mathbf{m} \cdot \mathbf{B}

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 d\theta and d\phi integrals show that the non-delta function term doesn't contribute, it is unclear if this argument 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 \mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ]

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 \mathbf{B}_{dip} = \frac{2 \mu_0}{3} \frac{\mathbf{m}}{\frac{4}{3} \pi R^3}

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:

U=\frac{2}{3} \mathbf{m} \cdot \mathbf{B}


What gives!?
Who can help solve this "paradox"?

 

[Hans de Vries]

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.

http://www.google.com/search?num=100&hl=en&safe=off&q=“4/3+problem”+energy&btnG=SearchRegards, Hans

 

[JustinLevy]

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.

 

[arunma]

Interesting paradox. I'll work it out and see what I can come up with.

 

[quetzalcoatl9]

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, I am not sure what the source of your seemingly paradoxical result actually is.

 

[JustinLevy]

Interesting paradox. I'll work it out and see what I can come up with.

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.

 

[Jonathan Scott]

[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.

 

[Meir Achuz]

http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3421v3.pdf

 

[JustinLevy]

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 \rho V + A \cdot j 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 m \cdot B 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).

http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3421v3.pdf

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).

[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.

Thanks for the heads up.
I'll give it a read through. This stupid sign error in my math/reasoning has always bugged me.

 

[Hans de Vries]

[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.

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

 

[Meir Achuz]

However, the sign is still wrong.
This stupid sign error in my math/reasoning has always bugged me.

The + sign for mu.B is explained on page 10 of
http://arxiv.org/PS_cache/arxiv/pdf/...707.3421v3.pdf

 

[tim_lou]

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.

 

[Meir Achuz]

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.

Butler's paper has nothing to do with a magnetic moment at rest in a magnetic field.

 

[Jonathan Scott]

Butler's paper has nothing to do with a magnetic moment at rest in a magnetic field.

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.

 

[JustinLevy]

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.

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.

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.

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).

The + sign for mu.B is explained on page 10 of
http://arxiv.org/PS_cache/arxiv/pdf/...707.3421v3.pdf

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.

 

[Charlie_V]

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?

 

[JustinLevy]

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?

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?

 

[Jonathan Scott]

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).

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 traveling 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.

 

[JustinLevy]

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,
u = \frac{1}{8\pi}(E^2 +B^2), \ \ \ \ g = \frac{1}{4\pi c} (E \times B)
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.

 

[Jonathan Scott]

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,
u = \frac{1}{8\pi}(E^2 +B^2), \ \ \ \ g = \frac{1}{4\pi c} (E \times B)
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.

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).

 

[gabbagabbahey]

This might not be a very satisfactory answer for you; but I think you are making this way too complicated.

When you calculate the energy by finding the amount of work required to move the dipole in from infinity and rotating it into position (as per Griffiths) you get -m.B. That is the amount of work the field does; the amount of energy that is stored in the field as a result is -(-m.B)=+m.B.

 

[JustinLevy]

Jonathan Scott,
I feel we are going in circles.

Do you agree with these two points:
1] The usual energy and field momentums can be used consistently within one inertial coordinate system.
2] In this case with static fields and E=0, the covariant definition just reduces to the usual definition.

If you disagree with either of these, please show the math explaining specifically what you mean ... for I feel I am misunderstanding the specifics of your complaint.

This might not be a very satisfactory answer for you; but I think you are making this way too complicated.

When you calculate the energy by finding the amount of work required to move the dipole in from infinity and rotating it into position (as per Griffiths) you get -m.B. That is the amount of work the field does; the amount of energy that is stored in the field as a result is -(-m.B)=+m.B.

Please work the math out for yourself, for you appear to be post-dicting an argument to fix the sign to match what you feel is correct, instead of actually doing the calculation. In creating such a "post-dicting correction" you are actually creating new problems instead of solving the main problem.

For example, consider an electric dipole. The energy is -p.E , analogous to the magnetic dipole case. If you calculate the energy in the fields, you get the correct answer (-p.E). If your argument was correct, then now the energy in the fields for an electric dipole is wrong!

 

[Jonathan Scott]

Jonathan Scott,
I feel we are going in circles.

Do you agree with these two points:
1] The usual energy and field momentums can be used consistently within one inertial coordinate system.
2] In this case with static fields and E=0, the covariant definition just reduces to the usual definition.

If you disagree with either of these, please show the math explaining specifically what you mean ... for I feel I am misunderstanding the specifics of your complaint.

"1] The usual energy and field momentums can be used consistently within one inertial coordinate system."

I'd agree that this is what Jackson says (and so do others). However, it really depends what you use it for. Basically, the "stuff" described by the conventional expressions obviously obeys a conservation law by Poynting's theorem, and has the right dimensions, but it doesn't really physically match the expected distribution of energy and momentum in many cases.

"2] In this case with static fields and E=0, the covariant definition just reduces to the usual definition."

The case in Jackson where the covariant definition reduces to the usual definition (around equation 17.44 in my 2nd edition copy) is quite specifically the one where there is a frame in which all charges are at rest and B is zero, not the other way round. There may well be an analogous result for a static B field, but I don't know.

 

[JustinLevy]

"2] In this case with static fields and E=0, the covariant definition just reduces to the usual definition."

The case in Jackson where the covariant definition reduces to the usual definition (around equation 17.44 in my 2nd edition copy) is quite specifically the one where there is a frame in which all charges are at rest and B is zero, not the other way round. There may well be an analogous result for a static B field, but I don't know.

Maybe I'm missing something, but I don't understand why we can't agree on that point.

Following Jackson, the covariant definition reduces to the usual definition in a frame he denotes as K'. Later he defines K' as the frame in which the total momentum in the fields (using the usual definition) is zero. Since I have chosen a frame where E=0 everywhere, this is indeed that frame. Yes, if there were no moving charges, that would also be a specific example of such a frame ... but in this case there is no such inertial frame where all the charges are at rest.

So again, I feel we are completely justified in using the usual energy and momentum densities for the electromagnetic fields. This caveat you brought up is interesting, but is ultimately unrelated to this problem as it does not require changing any of our calculations.

 

[Jonathan Scott]

Maybe I'm missing something, but I don't understand why we can't agree on that point.

Following Jackson, the covariant definition reduces to the usual definition in a frame he denotes as K'. Later he defines K' as the frame in which the total momentum in the fields (using the usual definition) is zero. Since I have chosen a frame where E=0 everywhere, this is indeed that frame. Yes, if there were no moving charges, that would also be a specific example of such a frame ... but in this case there is no such inertial frame where all the charges are at rest.

So again, I feel we are completely justified in using the usual energy and momentum densities for the electromagnetic fields. This caveat you brought up is interesting, but is ultimately unrelated to this problem as it does not require changing any of our calculations.

I have very little more I can say, as I've already stated what I believe he is saying, but I'll have another go at expressing it in a different way.

In the earlier part (starting at the paragraph containing equation 17.37 end ending at the paragraph containing equation 17.43) he says that the "correct four-vector character" can be obtained by taking the conventional expression in an arbitrary frame, of which the "natural choice" is a "rest" frame where ExB is zero, and transforming that to the actual frame. This doesn't mean it's the "correct" energy and momentum, but rather that whatever it is, its integral transforms correctly as a four-vector. (I'm not even entirely sure that he has considered the case where ExB is zero due to E being zero but B being non-zero, as the rest of the discussion is about the motion of charges.)

In the next section, starting at the paragraph containing equation 17.44, he restricts discussion to the specific case where there is a frame in which all charges are at rest and B is therefore zero (which does NOT include your case). In that case, we get Butler's four-vector expression for the energy and momentum density, which has the opposite sign for the B² term compared with the conventional expression. However, this obviously reduces to the conventional expression in the rest frame because B is zero.

I do not know whether there is some closely related result which can be obtained for the case where there are static magnetic fields and no unbalanced charges, but I do know that no such result is described in that section of my copy of Jackson and I don't think that one can extrapolate to one without providing specific reasoning.

 

[Sam Park]

I've been working on the electric dipole case now, which has a similar problem. Equivalently, the answer should still be U = - (p.E) ...

Do you encounter the same "paradox" in the more general problem of determining the energy of a arbitrary assembly of charges? It's straightforward to determine the energy both in terms of the work required to bring the charges into the final configuration, and in terms of the integral of the field energy. Naturally they both give the same result. I would think the question about a dipole is just a special case of this, so it might help if you started with the more general case (where you know you can get the same answer both ways) and then specialize to the case of a simple dipole.

 

[JustinLevy]

Do you encounter the same "paradox" in the more general problem of determining the energy of a arbitrary assembly of charges?

That post you referred to is quite a ways back in the discussion. At that point I was still using infinite sources. Using a finite source (or including a surface term at infinity) solves this problem. So the electric dipole case works fine.

However, even with finite sources, the energy of a magnetic dipole in a magnetic field has the wrong sign. This happens when calculating using A.j as the energy density as well.

I have very little more I can say, as I've already stated what I believe he is saying, but I'll have another go at expressing it in a different way.

In the earlier part (starting at the paragraph containing equation 17.37 end ending at the paragraph containing equation 17.43) he says that the "correct four-vector character" can be obtained by taking the conventional expression in an arbitrary frame, of which the "natural choice" is a "rest" frame where ExB is zero, and transforming that to the actual frame.

Good, so we agree on that. So using his choice laid out there, the covariant expression he presents reduces to the usual equations we've been using so far.

This doesn't mean it's the "correct" energy and momentum, but rather that whatever it is, its integral transforms correctly as a four-vector.

Oh come on!
Previously you agreed that the usual energy and momentum terms can be used as invariants of time and spatial translations respectively for a given inertial frame.
NOW, you even agree that the covariant expression (which is equal to the equations we're using here) transform correctly as a four-vector.

So, it fits the definition of the energy, and has the properties you'd expect from an energy. What more do you need!?


[EDIT: You seem to be complaining that I am generalizing a specific example, and want to know if that is justified. Jackson actually says the opposite. When he gets to the part where he discusses the static charges he says "For electromagnetic configurations in which all the charges are at rest in some frame (the Abraham-Lorentz model of a charged particle is one example), the general formulas can be reduced to more attractive and transparent forms." (emphasis added is mine)

If you want to maintain your objection, explain why that one specific case of the general formula makes my calculations wrong despite the fact that my calculations agree with the general formula and is not of the subset included in that "specific case" the reduced equations were for? ]

 

[Sam Park]

That post you referred to is quite a ways back in the discussion. At that point I was still using infinite sources. Using a finite source (or including a surface term at infinity) solves this problem. So the electric dipole case works fine.

However, even with finite sources, the energy of a magnetic dipole in a magnetic field has the wrong sign. This happens when calculating using A.j as the energy density as well.

Hmmm... So you're saying that, with your original approach, the sign came out wrong for both the electric dipole and magnetic dipole, but now with your new approach the sign problem has disappeared from the electric dipole case, but remains unchanged for the magnetic dipole case? So there must have been two completely independent errors in your original approaches, one reversing the sign of the electric dipole energy, and another reversing the sign of the magnetic dipole energy, and you've corrected one of them but not the other. This reminds me of a story about a student of Dirac who asked for help because he got the wrong sign for some quantity, so he knew he must have made a mistake with the sign somewhere, and Dirac said "Yes, or an odd number of them".

The only post in this thread that presented actual equations to explain the sign problem you're talking about was the very first post, but I gather your thinking has changed since then, so it might be helpful for you to present an update to that original post, showing whatever calculation(s) you currently believe give the wrong sign.

 

[JustinLevy]

Originally I was considering infinite sources (as you can see in the original posting). Working out the magnetic dipole in an external magnetic field and using just a volume integral of the (E^2+B^2) energy density term resulted in:
U = + (2/3) m.B
Doing likewise for the electric dipole gave:
U = -(1/3) p.E

So there were two problems originally, the magnitude was wrong for both, and the sign was wrong for the magnetic dipole case (not the electric dipole case).


What has changed since then:
It turns out that the energy density equations are not correct if there are infinite sources, unless one includes a surface term at infinity. Including the surface term, or using finite sources fixed the magnitude problem. Now we have:
U = + m.B
U = - p.E


Once writing the equations down, I am confident of my ability to solve them as I have checked many times, and multiple grad students have worked it out themselves, and no one here objects to the math. As mentioned, Prof. David Griffiths (who wrote a popular E&M textbook) even wrote to a journal asking a similar question (unfortunately I didn't see a followup answer article or anything anywhere). So I don't expect to find a sign error in my math.

The problem must lie in the application of the equations ... maybe I'm mis-applying the equations to this situation. Jonathan Scott brought up one such idea. It brought up some interesting facts that I hadn't heard before, and was an interesting read, but it appears to me to be a dead end. If you have any other ideas, please do let us know.

 

[cup]

I didn't read trough this thread very carefully, but are you sure that you are allowed to do the manipulations

<br /> U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 d^3r <br />

<br /> U = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip})d^3r<br />

On \mathbf{B}_{dip} which contains the dirac delta function?

I think it might work out if you consider \mathbf{B}_{dip} as a weird function defined in terms of the dirac delta function from the beginning, instead of plugging it in at the end of the calculation, having assumed it to be ordinary.

 

[cup]

To clarify: you should look around to find out if there is any formalism for squaring the dirac delta function.

If there isn't then the starting point of the derivation is unfounded.

 

[Sam Park]

I didn't read trough this thread very carefully, but are you sure that you are allowed to do the manipulations

<br /> U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 d^3r <br />

<br /> U = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip})d^3r<br />

On \mathbf{B}_{dip} which contains the dirac delta function?

I think it might work out if you consider \mathbf{B}_{dip} as a weird function defined in terms of the dirac delta function from the beginning, instead of plugging it in at the end of the calculation, having assumed it to be ordinary.

From just a simplistic point of view, the quantity U is being defined as the integral of a squared magnitude, (Bf + Bd)^2, which is obviously positive-definite, but of course the self-terms Bf^2 and Bd^2 are then subtracted from this, leaving just the cross term 2Bf*Bd. The sign of this is positive if Bf and Bd both have the same sign, so the only way for the argument of the integral to ever be negative is for Bd to have the opposite sign of Bf somewhere. Depending on how you define those separate quantities, you could write the original argument as (Bf - Bd)^2, in which case, after subtracting the self-terms, the argument is -2Bf*Bd, which is to say, the sign of the overall answer is reversed. So it's crucial for Bf and Bd to have their signs defined on a consistent basis. Neither of their signs matters individually, but the product of their signs matters, so it's important to make an even number of errors when assigning their signs!

If Bd and Bf really always have the same sign, then clearly the integral of Bf*Bd must be positive, so I'd suggest focusing on the question of whether they really always have the same sign. Remember it's a dot product of two vectors. Which way do those two vectors point?

 

[JustinLevy]

To clarify: you should look around to find out if there is any formalism for squaring the dirac delta function.

If there isn't then the starting point of the derivation is unfounded.

To make sure that is not a problem, this was worked out with finite sources (a spinning sphere of charge has a non-zero magnetic dipole moment outside, a constant field inside, and all other multipole moments are zero). The answer is exactly the same (you don't even need to take the limit the size goes to zero because it is a perfect dipole ... but you could if you want, and of course it reproduces the "point" dipole).

So that is not the problem here.

If Bd and Bf really always have the same sign, then clearly the integral of Bf*Bd must be positive, so I'd suggest focusing on the question of whether they really always have the same sign. Remember it's a dot product of two vectors. Which way do those two vectors point?

Sam, I don't mean to be disrespectful, but can you please work out the problem yourself to check? Your comments are not making any sense, and I think this is because you haven't worked out the problem yourself.

Bd and Bf are vector fields. Their direction and magnitude are defined at every point. Regardless of how I orient the dipole, there are places where the fields are parallel, and places where the fields are anti-parallel. Only after doing the whole integral can I find out if the result is positive or negative. Furthermore, the fields are completely defined by the sources. Are you saying I calculated the fields from the sources wrong? Because not only have I checked those equations myself, but at least two textbooks agree with those equations for the fields of a dipole. I didn't even bother deriving that in the openning post; I just gave them verbatum from a textbook.

Please, if you insist the math in my calculation is faulty, please try working it out yourself so you can be convinced and we can move beyond that. For I am convinced the error is not in the calculation itself, but in the assumptions we've made in writing down / applying the equations to the physics.

 

[cup]

A formula such as

<br /> U_{em} = \frac{1}{2} \int (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) d^3r <br />

comes with certain requirements on the objects represented by the symbols in the formula.
The integrand has to be a function.

If you go ahead and (implicitly) assume that it's a function. This allows you to use the familiar algebraic rules that you've learned, and make some derivations, like:

<br /> U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 d^3r <br />
<br /> U = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip})d^3r<br />
<br /> U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} d^3r <br />

...then that's totally fine and dandy.

But when you're done with your derivations, you can't just change your mind about the symbols and say:

"
well, the thing that I (implicitly) said was a function above wasn't really a function, it was this thing:

<br /> \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})<br />

"

which is not a function, i.e. does not obide by all of the algebraic rules that you used in the initial derivation.

As soon as you changed the meaning of \mathbf{B}_{dip}, the algebraic manipulations you performed initially are not valid anymore: the string of equations don't follow anymore.

To sum up:

The final integral incidentally DOES make sense mathematically... but you lost the physics (coming from the very first formula) when you changed the meaning of \mathbf{B}_{dip}, because the string of equations in the initial derivation does not hold anymore, because the algebraic manipulations are unjustified, because the thing is not a function.

 

[Sam Park]

Sam, I don't mean to be disrespectful, but can you please work out the problem yourself to check? Your comments are not making any sense, and I think this is because you haven't worked out the problem yourself. Bd and Bf are vector fields. Their direction and magnitude are defined at every point.

Yes, I know B is a vector field that varies from place to place. That’s why I said, in order for the integral to give a negative value, the vectors must be pointing in opposite directions somewhere. My intent was to suggest that you check the directions of your vectors at some key points near the dipole where the biggest contributions to the integral will be, and convince yourself that the two vectors are indeed pointing in negative directions relative to each other (i.e., the dot product is negative) at those points, as a sanity check on your assignment of signs to the two components.

Please, if you insist the math in my calculation is faulty, please try working it out yourself so you can be convinced and we can move beyond that. For I am convinced the error is not in the calculation itself, but in the assumptions we've made in writing down / applying the equations to the physics.

I entirely agree that the problem is not in the arithmetic, it’s in the assignment of physical meanings to the symbols. That has been the point of my messages. Sorry if I didn’t make that clear.

Representing magnetic dipoles as two equal and opposite “magnetic charges”, we can carry through the derivation of the potential energy of a given dipole in a given magnetic field. We regard the given field as produced by a suitable distribution of magnetic charges, and then we compute the work required to bring another pair of magnetic charges to a certain configuration in that field. Now, we can also compute the corresponding change in the magnetic field energy, and these come out to be the same, both equal –u*B. The derivation is essentially identical to the case of electric dipoles. Are you saying this derivation is inapplicable in the magnetic case?

 

[JustinLevy]

cup,
PLEASE read what I wrote in the previous post.
This was also worked out with finite sources for all the fields, and the answer is still the same (before even taking the limit that the dipole goes to a point dipole, which of course you can take if you want and you get the same result).

I'm tired of repeating myself. A professor who wrote a popular textbook on E&M even came up with the same result, and wrote into a journal asking about it. The math is not faulty here, if you don't believe me, please do it yourself until you are convinced of this as well. The problem must lie in the assumptions we've made in writing down / applying the equations to the physics, not in the mechanics of doing the calculation itself.

 

[JustinLevy]

Representing magnetic dipoles as two equal and opposite “magnetic charges”, we can carry through the derivation of the potential energy of a given dipole in a given magnetic field.

This is inherently different, for now Del.B is not zero. In particular, the field between the "charges" now is in the opposite direction. (A "point" magnetic dipole made this way would have a different constant in front of the term that contributes at the origin.)

The derivation is essentially identical to the case of electric dipoles. Are you saying this derivation is inapplicable in the magnetic case?

Yes.

 

[Sam Park]

This is inherently different, for now Del.B is not zero. In particular, the field between the "charges" now is in the opposite direction. (A "point" magnetic dipole made this way would have a different constant in front of the term that contributes at the origin.)

Well, there are multiple distinct field configurations that possesses the same effective dipole moment u, but all of them have potential energy -u*B, so the particular choice of configuration doesn't matter.

Since the field configuration I described explicitly has a magnetic dipole moment of u, and since it gives consistent results for the field energy, the "paradox" has at least been reduced. What would help now is for you to specify precisely what field configurations (with dipole moment u) have internal energy opposite to the internal energy of the configuration I described. This might help to isolate the source of the problem. First, show that your field configuration has a dipole moment of u, and second, show the calculation that implies it has potential energy +u*B. I don't think you've ever shown the first part of this for your example.

 

[Hans de Vries]

Well, there are multiple distinct field configurations that possesses the same effective dipole moment u, but all of them have potential energy -u*B, so the particular choice of configuration doesn't matter.

Since the field configuration I described explicitly has a magnetic dipole moment of u, and since it gives consistent results for the field energy, the "paradox" has at least been reduced. What would help now is for you to specify precisely what field configurations (with dipole moment u) have internal energy opposite to the internal energy of the configuration I described. This might help to isolate the source of the problem. First, show that your field configuration has a dipole moment of u, and second, show the calculation that implies it has potential energy +u*B. I don't think you've ever shown the first part of this for your example.

The form of the delta is quite well established in the one photon Breit-Fermi interaction between
the proton and the electron which need to be right in order to predict the 21 cm hydrogen line. Regards, Hans

 

[Sam Park]

The form of the delta is quite well established in the one photon Breit-Fermi interaction between the proton and the electron which need to be right in order to predict the 21 cm hydrogen line.

Yes, I think we all agree that the accepted expression for the energy of a magnetic dipole is correct. The task is to reconcile it with the calculations. I think maybe I can (finally) shed some light on this. There's a discussion of this in Schwartz's "Principles of Electrodynamics".

When displacing a current loop, in the presence of a magnetic field produced by one or more other current loops, we need to distinguish two different ways of carrying out the displacement, and the two different physical consequences. We might think the right way is to displace our tiny current loop while holding the currents (in all the loops) constant, but this will necessarily result in a change in the flux in the loops (like tiny generators). While the flux is changing, the integral of E*dl around the loop will not be zero, and hence the electric field will do work on the currents in the loops, and this work is in addition to the purely magnetic effect we are trying to evaluate.

Another way of carrying out the displacement is with the fluxes (not the currents) held constant. Using this method, the integral of E*dl around each loop is automatically zero, so we do only the magnetic work. Therefore, the relevant quantity is the change in energy for a spatial displacement while holding constant flux (not constant current).

Unfortunately, it's much more straightforward to evaluate the change in energy at constant current, so this is usually what is done, but then we make use of a remarkable theorem, which says that for a given displacement of a set of current loops, the change in magnetostatic internal energy due to carrying out that displacement at constant current is the negative of the change in energy due to carrying out the same displacement at constant flux.

So, I suspect the original poster is evaluating the change in energy at constant current, and therefore getting the negative of the relevant value, which is the change in energy at constant flux.

 

[Jonathan Scott]

Sorry I've not been keeping up with this discussion; various minor illnesses have been depriving me from sleep which doesn't help my concentration.

I've just gone back and revisited the original post, and this has reminded me of something which might just possibly be related that I noticed quite recently when looking at stress-energy tensors in GR (relating to the Kumar mass).

If you consider a field with multiple sources, and draw some arbitrary plane somewhere between the sources, you can then consider the "pressure" (force per area, energy per volume) through that plane, so for example if the sources are charges and the fields are electrostatic, then it should be possible to find some integral over that plane which adds up to the total force across that plane. I then expected to be able to move that plane between sources to find out whether the integral over a volume matched any conventional expression for the total energy.

I had expected that integral to be related to the difference between the sum of the squares of the separate fields and the square of the sum of the fields, in the same way as the difference between the total energy of some configuration of sources and the sum of the energies of the individual sources. That is, if the total field is F = (F₁+F₂) then I expected the force term to arise from the 2F₁.F₂ term in the square. However, that gave a factor of 2/3 compared with the expected value.

However, if I take each individual field, take its projection perpendicular to the plane (in either direction - it doesn't matter which as it gets squared) and then take the products with itself and every other field, I get a very similar expression which however integrates to give the full expected force between the sources. This probably works over any surface between the sources too, not just a flat plane. I'm sorry I'm too hazy at the moment to state this mathematically; I presumably have it written down somewhere but not easily to hand.

When I looked at the difference between these two approaches, I found that it's basically related to using a scalar approach to a tensor situation which results in loss of data. For example, the flow of x-momentum in the x-direction may seem to behave like a scalar, even though it should be treated like the 11 component of a tensor.

Basically, this result means that there is a subtle distinction between the effect of two separate fields and the effect of a single combined field. Although the effective field adds up linearly, the resulting distribution of pressure and energy in space does not necessarily do so.

Sorry to be so vague about this; I just hope this gives some sort of clue for now. When I'm more awake I will try to dig out something a bit more mathematical.

 

[cup]

cup,
PLEASE read what I wrote in the previous post.
This was also worked out with finite sources for all the fields, and the answer is still the same (before even taking the limit that the dipole goes to a point dipole, which of course you can take if you want and you get the same result).
...

Sorry about that.

Well, please show me the finite derivation you speak of.

This is what I get when I attempt it:

<br /> \left\{<br /> \begin{array}{c c}<br /> \mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] &amp; r &gt; R \\<br /> \mathbf{B}_{dip} = \frac{2 \mu_0}{3} \frac{\mathbf{m}}{\frac{4}{3} \pi R^3} &amp; r \leq R \\<br /> \end{array}<br /> \right.<br />

<br /> U<br /> =<br /> \frac{1}{\mu_0}<br /> \int_{B(0, R)}<br /> \mathbf{B} \cdot \mathbf{B}_{dip}<br /> d^3r<br /> +<br /> \frac{1}{\mu_0}<br /> \int_{\mathbf{R}^3 - B(0, R)}<br /> \mathbf{B} \cdot \mathbf{B}_{dip}<br /> d^3r<br /> =<br /> U_{inside}<br /> +<br /> U_{outside}<br />

<br /> U_{inside}<br /> =<br /> \frac{2}{3}<br /> \frac{1}{\frac{4}{3} \pi R^3}<br /> \int_{B(0, R)}<br /> \mathbf{B} \cdot \mathbf{m}<br /> d^3r<br /> =<br /> \frac{2}{3}<br /> \mathbf{B} \cdot \mathbf{m}<br /> \frac{1}{\frac{4}{3} \pi R^3}<br /> \int_{B(0, R)}<br /> d^3r<br /> =<br /> \frac{2}{3}<br /> \mathbf{B} \cdot \mathbf{m} <br />

<br /> U_{outside}<br /> =<br /> \frac{1}{4 \pi}<br /> \int_{\mathbf{R}^3 - B(0, R)}<br /> \frac{3(\mathbf{m} \cdot \hat{\mathbf{r}}) (\mathbf{B} \cdot \hat{\mathbf{r}}) - \mathbf{B} \cdot \mathbf{m}}{r^3}<br /> d^3r<br /> =<br /> ?<br />

The nominator of course only depends on direction, so we can try to do the integral in spherical shells. But the the radial integral then becomes of 1/r from R to infinity. What do you do with that?

Why do you ignore U_outside?

 

[JustinLevy]

Well, please show me the finite derivation you speak of.

Okay, I'll try to type something up.

Why do you ignore U_outside?

That integral is zero. It is a multiplication of two different spherical harmonics (legendre polynomials) which are orthogonal, so the result is zero.

Also I notice you are using an infinite external source, if you do this you have left out one term (a surface term at infinity). You can either include this term, or just use a finite source.


Okay, let me try to write out the calculations.
----------------------

Consider a spherical shell of radius R and of uniform charge spinning such that it has a magnetic dipole of \mathbf{m}. The field outside the sphere is that of a perfect magnetic dipole, and inside the magnetic field is a constant.

<br /> \mathbf{B}_{dip} = <br /> \begin{cases}<br /> \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] &amp; \text{if } r \geq R \\<br /> \frac{2 \mu_0}{3} \frac{\mathbf{m}}{\frac{4}{3} \pi R^3} &amp; \text{if } r &lt; R<br /> \end{cases}<br />

The limit R \rightarrow 0 yields an ideal magnetic dipole:
<br /> \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})<br />

The energy in electromagnetic fields is given by:
<br /> U_{em} = \frac{1}{2} \int (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) d^3r<br />

where d^3r is the volume element and the integration is over all space.

Using an infinite external source:

If we consider an external field \mathbf{B} and the field due to a magnetic dipole \mathbf{B}_{dip}, we have:

<br /> U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 \ d^3r = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip}) \ d^3r<br />

the B^2 and B^2_{dip} terms are independent of the orientation and so are just constants we will ignore.

Choosing axes such that the external field is in the z direction,
<br /> U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} \ d^3r =<br /> \frac{1}{\mu_0} \int B \hat{\mathbf{z}} \cdot \left[ \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}) \right] d^3r<br />

The first term does not contribute, for
<br /> \begin{align*}<br /> \int &amp; \hat{\mathbf{z}} \cdot [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] d^3r \\<br /> &amp; = \int [ 3(m_x \sin\theta\cos\phi + m_y \sin\theta\sin\phi + m_z \cos\theta)\cos\theta - m_z ] r^2 \sin\theta \ d\phi d\theta dr <br /> \end{align*}<br />
of which the m_x and m_y terms vanish after the \phi integration, and the m_z term will vanish after the \theta integration for it contains
<br /> \int [ 3\cos^2\theta - 1] \sin\theta \ d\theta = 0.<br />

Therefore only the delta function term will contribute to the interaction energy.
<br /> U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r}) \ d^3r = (\mathbf{B} \cdot \mathbf{m}) \int \frac{2}{3} \delta^3(\mathbf{r}) d^3r<br />

<br /> U = + \frac{2}{3} \mathbf{m} \cdot \mathbf{B}<br />

This has the wrong magnitude and the wrong sign. The reason the magnitude is wrong is because we used an infinite source. When doing so, the energy in the electromagnetic fields also requires a surface term
\frac{1}{2\mu_0} \int (\mathbf{A}\times\mathbf{B}) \cdot d\mathbf{a}

If you work this out, it gives another \frac{1}{3} \mathbf{m} \cdot \mathbf{B}. Or you can just use a finite source, which I'll do here as it is more satisfying for we avoid any limits to infinity.

Another possible complaint is that while doing the \theta and \phi integrals show that the non-delta function terms don't contribute, it is unclear if this argument holds for the point right at r=0 where this coordinate system is undefined. Also the delta function term can be unsettling. However, since the dipole is in an external field, and the finite sized dipole just has a constant field inside (where the external field is also constant) it is trivial to do this with the finite dipole and get the same result (before even taking the limit it approaches an ideal 'point' dipole). The math results in the same answer, so that is not where the problem resides.

Using a finite external source:

In the last attempt, even though the integration was over "all space", the source of the external field was not including anywhere in space. While this satisfies Maxwell's equations, a finite source would be preferred. Since it has already been noted that a spinning shell of uniform charge produces a constant magnetic field inside, this can be used as the source and the limit R \rightarrow \infty be taken if the equivalent of a constant external magnetic field "everywhere" is needed.

For r&lt;R, the result is identical to attempt 1, so the energy is:

<br /> U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} \ d^3r = \frac{2}{3} (\mathbf{m} \cdot \mathbf{B}) + U_2<br />
where
<br /> U_2 = \frac{1}{\mu_0} \int_R^\infty \int_0^\pi \int_0^{2\pi} \mathbf{B} \cdot \mathbf{B}_{dip} \ r^2 \sin\theta \ d\phi d\theta dr<br />

This new term, where r &gt; R, both the source and the dipole under investigation have perfect dipole fields. The magnetic dipole of the source will be referred to as \mathbf{m_1} while that of the perfect dipole is \mathbf{m_2}. The axes are chosen such that the external field (and therefore \mathbf{m_1}) is in the z direction.

<br /> U_2 = \frac{1}{\mu_0} \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{\mu_0}{4 \pi r^3} \left[ 3 m_1 \cos\theta \hat{\mathbf{r}} - m_1 \hat{\mathbf{z}} \right] \cdot \mathbf{B}_{dip} \ r^2 \sin\theta \ d\phi d\theta dr<br />

Using the same logic as in the last attempt, the term in the z direction with no angular dependence will not contribute to the integral here. Expanding out the remaining terms results in

<br /> \begin{align*}<br /> U_2 = &amp; \frac{1}{\mu_0} \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{\mu_0}{4 \pi r^3} 3 m_1 \cos\theta \ \hat{\mathbf{r}} \\<br /> &amp; \cdot \ \frac{\mu_0}{4 \pi r^3} [ 3(m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta)\hat{\mathbf{r}} - \mathbf{m_2} ] <br /> r^2 \sin\theta \ d\phi d\theta dr \\<br /> U_2 = &amp; \frac{3 \mu_0}{16 \pi^2} m_1 \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{r^4} \cos\theta [ 3(m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta) \\<br /> &amp; - (m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta)] \sin\theta \ d\phi d\theta dr \\<br /> U_2 = &amp; \frac{3 \mu_0}{8 \pi^2} m_1 \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{r^4} \cos\theta (m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta) \sin\theta \ d\phi d\theta dr<br /> \end{align*}<br />

Performing the \phi integration yields

<br /> \begin{align*}<br /> U_2 &amp;= \frac{3 \mu_0}{4 \pi} m_1 \int_R^\infty \int_0^\pi \frac{1}{r^4} \cos\theta m_{2z} \cos\theta \sin\theta \ d\theta dr \\<br /> &amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \int_R^\infty \int_0^\pi \frac{1}{r^4} \cos^2 \theta \ \sin\theta \ d\theta dr.<br /> \end{align*}<br /> [/itex]<br /> <br /> Do a change of variables u=\cos\theta and complete the integral.<br /> &lt;br /&gt; \begin{align*}&lt;br /&gt; U_2 &amp;amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \int_R^\infty \int_{-1}^1 \frac{1}{r^4} u^2 \ du dr \\&lt;br /&gt; &amp;amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \frac{2}{3} \int_R^\infty \frac{1}{r^4} \ dr \\&lt;br /&gt; &amp;amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \frac{2}{3} \frac{1}{3R^3} \\&lt;br /&gt; &amp;amp;= \frac{1}{3} (\frac{2 \mu_0}{3} \frac{\mathbf{m_1}}{\frac{4}{3}\pi R^3} \cdot \mathbf{m_2})&lt;br /&gt; \end{align*}&lt;br /&gt;<br /> <br /> Referring to the source equations, the relationship between \mathbf{m_1} and the constant external field \mathbf{B} inside the source simplifies the result to<br /> &lt;br /&gt; \begin{align*}&lt;br /&gt; U_2 = \frac{1}{3} \mathbf{m_2} \cdot \mathbf{B}&lt;br /&gt; \end{align*}&lt;br /&gt;<br /> <br /> The final result is therefore<br /> &lt;br /&gt; \begin{align*}&lt;br /&gt; U = + \mathbf{m} \cdot \mathbf{B}&lt;br /&gt; \end{align*}&lt;br /&gt;<br /> <br /> While the magnitude is now correct, the sign is still wrong. This calculation seems to indicate the state in which the dipole is aligned with the external field is <i>unstable</i>, and the minimum energy is with the dipole aligned <i>opposing</i> the field.<br /> <br /> While it is hard to do such integral in your head, one can see that it seems intuitively reasonable that a magnetic dipole alligned <i>against</i> a field would reduce the total magnetic field, while one alligned <i>with</i> the field would increase the magnetic field ... therefore from the &quot;energy in the fields&quot; standpoint, this result seems to make sense. However, we know from experiment that this must be wrong as a dipole left to rotate on its own will try to allign with the field. <br /> <br /> ... Hence the &quot;paradox&quot;.

 

[Hans de Vries]

As far as I can say:

The correct solution is using the interaction term I\cdot A instead of the EM field energy.
This is after all what the energy expression describes for infinitesimal circular currents
in the a vector potential field with curl, for instance:

-\mu\cdot B ~~\propto~~ - \left(\nabla\times I\right)\cdot \left(\nabla\times A\right)

The energy is experimentally determined by looking at the behavior of the object
possessing the magnetic moment, for instance the electron, not by looking at what
the EM field does.

The physics can be sketched quantum mechanically as follows: The circular A field
tries to impose a phase change rate along the circle on the circular current. This is
not allowed because of the quantization. The phase change rate must be compensated
by an equal (negative) phase change rate of the inertial momentum along the circle.

That is: The total (canonical) momentum along the circle stays the same but the
inertial momentum becomes:

1) Less if A runs in the same clockwise direction as I. (less energy)
2) More if A runs anti-clockwise to I. (more energy)The magnetic field B will impose a torque on the magnetic dipole which will start
to precess with the Lamor frequency. The torque can be derived from the angle
depended energy. The Lamor precession frequency can be derived quantum
mechanically from the Dirac Spinor.

If a Dirac spinor with an arbirary spinning direction is split in a "spin-up" component
and a "spin-down" component relative to the magnetic field, and if the two separate
components are given a different energy with a delta +/- the magnetic energy.

Then the addition of the two components represents a precessing spinor and the
precessing frequency equals the Lamor frequency.

I'm in the process of working this out for my book.Regards, Hans.

 

[JustinLevy]

Well, there are multiple distinct field configurations that possesses the same effective dipole moment u, but all of them have potential energy -u*B, so the particular choice of configuration doesn't matter.

It does matter. I just gave an example where it is clear that it matters.
If you insist on using magnetic monopoles, please note then that you can't even define a vector potential anymore via B = Del x A if you do this. Furthermore, as already explained, a "perfect" dipole made from monopoles has a magnetic field different from a "perfect" dipole made from a current density. What you are presenting is not analogous to the problem at hand.

So, I suspect the original poster is evaluating the change in energy at constant current, and therefore getting the negative of the relevant value, which is the change in energy at constant flux.

I don't agree with that.
Here's a quick derivation of the energy to orient a dipole with constant current:

- consider a square loop with sides length L, and current I (magnetic dipole moment m = I L^2)
- there is a constant external magnetic field B
- have the dipole initially perpendicular to the magnetic field

Let's calculate the work required to orient the dipole (keeping current constant, and hence the dipole constant) in the external field, we find:
force on wire segments due to external field, F = I L B
with dipole at angle theta to magnetic field, torque T = (I L B) L sin(theta) = IL^2 B sin(theta) = m B sin(theta)

which finially gives us the energy to put it into an orientation as
U= \int_{\pi/2}^\theta \mathbf{T} \cdot d\mathbf{\theta&#039;} = \int_{\pi/2}^\theta m B \sin\theta&#039; \ d\theta&#039; = - m B \cos\theta

U= - \mathbf{m} \cdot \mathbf{B}

So, contrary to your claims, deriving with a constant current (constant dipole) condition does NOT lead to the wrong sign. It yields the correct sign.

Also, as already mentioned, an electron or proton have a constant dipole moment ... (so this would be like the "constant current" method you claim is the invalid one), yet we know experimentally the energy should be described as U = - m.B for these particles.

Anyway you cut it, it seems like the answer should be U = -m.B

The problem only arises when trying to calculate the energy by using the energy in the fields.

 

[Hans de Vries]

The correct solution is using the interaction term I\cdot A instead of the EM field energy.

In fact this is just like defining the energy of a charge q in a potential field \Phi
as q\Phi instead of calculating the change in total energy of the charge's E field
due to the extra E field, E = -\mbox{grad}(\Phi).Regards, Hans

 

[JustinLevy]

The correct solution is using the interaction term I\cdot A instead of the EM field energy.

That is mathematically equivalent to the energy in the fields. They are related by vector calc identities (and definitions of maxwell's equations and the potentials). This is derived in most textbooks, so I hope we can agree the A.j and the (E^2+B^2) method is identical.

If you don't agree, please point out where the textbooks made an unstated (and possibly incorrect?) assumption. For I don't currently see how your statements provide a solution to the problem here.

 

[Hans de Vries]

That is mathematically equivalent to the energy in the fields. They are related by vector calc identities (and definitions of maxwell's equations and the potentials). This is derived in most textbooks, so I hope we can agree the A.j and the (E^2+B^2) method is identical.

If you don't agree, please point out where the textbooks made an unstated (and possibly incorrect?) assumption. For I don't currently see how your statements provide a solution to the problem here.

Indeed, I don't agree, given for instance the (unfortunately) failed atemps to express
the energy-momentum of a charged particle in terms of its electromagnetic field energy-
momentum.

Also, the interaction terms are relativistically covariant, while the EM energy-momentum
as generally presented in the textbook is not. A more elaborate approach is needed for
instance along the lines of Butler's approach as described by Jackson in his last chapter.

Regards, Hans.

 

[cup]

Justin,

In the finite case, can you state clearly what \mathbf{B} and \mathbf{B}_{dip} are chosen to be, and why this should represent the desired physical situation as R \to \infty?

 

[JustinLevy]

Hans,
The (\rho V + \mathbf{A} \cdot \mathbf{j}) and (E^2+B^2) methods are mathematically equivalent as shown by several textbooks. If you want to argue against these textbooks, please show us mathematically why they are NOT the same despite their proofs.

Second, the section in Jackson regarding Butler already was brought up ... in the case we are considering here, the covariant definition reduces to the usual energy density definition we are using here. So that argument does not effect the validity of the calculations.

Justin,

In the finite case, can you state clearly what \mathbf{B} and \mathbf{B}_{dip} are chosen to be, and why this should represent the desired physical situation as R \to \infty?

B_dip is the field of the dipole we are orienting with respect to a source field B which is a constant field everywhere the current of B_dip is. In fact, the source field B is constant in an arbitrarily large region around the current of B_dip. You can take the limit of R -> infinity if you want the region in which the source field is constant to be infinite (ie. a constant external field everywhere).

However, as you can see, the answer doesn't depend on R (as long as R is large enough to enclose the dipole current with a constant magnetic field). This makes sense since the physics is local, and should only depend on the magnetic field at the dipole we are orienting.