Paradox regarding energy of dipole orientation

[Hans de Vries]

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.

There are many well know examples where this goes wrong...

1) In radiation the (E^2+B^2) energy grows indefinitely from an oscillating charge
with constant average energy momentum according to the (\rho V + \mathbf{A} \cdot \mathbf{j}) method.

2) The infamous 4/3 problem for the electromagnetic mass of the electron.

3) maybe you want to add electric amd magnetic dipoles as examples where it fails...


The latter example is still somewhat surprising as most static examples do OK.
Maybe if you consider the (\rho V + \mathbf{A} \cdot \mathbf{j}) change in the source of the fixed B field
due to a rotation of the magnetic dipole, that you get the equivalency back.
You can't ignore this contribution to the change of the total energy of the
system.

The total change of (E^2+B^2) should then be equal to the change in
(-m\cdot B) plus the change in (\rho V + \mathbf{A} \cdot \mathbf{j}) of the source of the fixed B field.


Regards, Hans

 

[JustinLevy]

3) maybe you want to add electric amd magnetic dipoles as examples where it fails...

You can't argue against a proof in textbooks by noting things unrelated to this situation, and then stating "maybe" the textbook proof is therefore wrong. Work out this problem using A.j if you don't believe me. You will get the same answer.

I'm sorry if I am getting snappy, but I am getting frustrated by your insistence that the textbooks are wrong here without showing any math. I've reread the derivation in two textbooks now. The two methods are equivalent for this situation. If you are going to continue to disagree with the textbooks, show your A.j calculation.

 

[Hans de Vries]

You can't argue against a proof in textbooks by noting things unrelated to this situation, and then stating "maybe" the textbook proof is therefore wrong. Work out this problem using A.j if you don't believe me. You will get the same answer.

I'm sorry if I am getting snappy, but I am getting frustrated by your insistence that the textbooks are wrong here without showing any math. I've reread the derivation in two textbooks now. The two methods are equivalent for this situation. If you are going to continue to disagree with the textbooks, show your A.j calculation.

I'm not saying that "textbooks are wrong". These are your words. You must be referring
to undergraduate textbooks which make statements valid in a limited context only.

Certainly the finite propagation speed of the electromagnetic field is ignored here.
The E/B field energy components aren't necessarily at all related to the interaction
terms at all at the same time defined at a certain reference frame.

In quantum field theory the Interaction term 4-momenta and the EM field 4-momenta
are different quantities all together and it's very odd to say that "both are equivalent"

If this is all relevant for your particular (static) situation is something else. For so far
you haven't responded to the suggestions I made about this in my last post. Regards, Hans

 

[Hans de Vries]

3) maybe you want to add electric amd magnetic dipoles as examples where it fails...

Work out this problem using A.j if you don't believe me. You will get the same answer.

In both cases the E^2+B^2 method gives a different result as the A.j method
applied on the dipole in isolation. The A.j method is correct. This is particular
easy to see in the case of the electric dipole.

The dipole is defined by p=q.r where q is the charge of the positive/negative poles
and r is the distance between the two charges. Rotating the dipole from up to down
moves both charges over a distance of r which means a change of energy of 2qrE=2pE
as it should be.

In the general case the energy is -p.E which varies from -pE to +pE.

If you try to calculate the energy of the dipole by calculating the energy of the
E field instead (with the delta defined as in Jackson chapter 4) then you will get
the right sign but the wrong magnitude, as you did mention yourself also if I
remember well.Regards, Hans.

 

[Sam Park]

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

You may be dismissing the textbook explanation of your conundrum too quickly. The interaction between two (or more) current loops is more complicated than one might think. In a set of N current loops, the magnetic flux through the ith loop has a contribution to it arising from the current in each of the other loops. The magnetic flux through the ith loop is the integral of B*n over any surface bounded by the loop (where "n" is the unit vector normal to the surface), and this depends not just on the current through the ith loop, but on the current through each of the loops, i.e., the flux through the ith loop is written as the sum of (L_ij)(I_j) for j = 1 to N where the constants L_ij are the coefficients of inductance. You see, there is mutual inductance, so it isn't just superimposing two independent B fields, which I think is what your analysis assumes. The coefficients of inductance are given by the double line integral of (1/|ri - rj|) dl1 dl1 around the ith and jth loops, and hence L_ij = L_ji. Then the overall magnetostatic energy of the system of loops is the sum of (1/2) L_ij I_i I_j for i,j = 1 to N. For example, with N=2 loops the energy is

U_B = (1/2) (L11 I1 I1 + L12 I1 I2 + L21 I2 I1 + L22 I2 I2)

Now, for an incremental displacement of these loops, holding all the currents constant, only the L_ij coefficients change, while the I terms are constant, so we have

d(U_B)I = (1/2) SUM d(Lij) Ii Ij

On the other hand, making the displacement at constant flux... well, you can work it out yourself... it comes out d(U_B)Q = -d(U_B)I. This is exactly analogous to how, in electrostatics, the change in energy when displacing a set of conductors at constant charge equals the negative of the change for the same displacement at constant potential. In the electrostatic case, the relevant process is displacement at constant charge when determining the forces acting on the conductors, which corresponds in the magnetic case to displacement at constant flux when determining the torques acting on the current loops.

A "perfect" dipole made from monopoles has a magnetic field different from a "perfect" dipole made from a current density.

True, and sort of interesting, when we consider the limiting case of a point-like magnetic dipole, for which the field is identical except at r = 0.

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.

Hmmm... if you're talking about an individual electron, which is usually regarded as a point-like particle, I don't think its magnetic moment can necessarily be modeled as physically spinning charge, like a current loop, although it has a lot in common with that. How would we know, for a point-like charge, whether the "field" at r=0 was like the limit of a current loop, or like the limit of two monopoles? Presumably the only way to tell is how the particle behaves. It behaves like U = -m.B, but this just begs the question. Anyway, I think the magnetic interaction between current loops can't be represeted by treating each loop as an isolated dipole field and superimposing them.

 

[JustinLevy]

I'm not saying that "textbooks are wrong". These are your words.

You are disagreeing with the textbooks. You can't take the stance that the textbooks are correct, and that you are correct, without contradicting yourself.

You must be referring to undergraduate textbooks which make statements valid in a limited context only.

The only requirement is that you are asking about the energy of a system in which the fields are currently constant in time in the system.

Here, let me reproduce the equivalence proof here real quick.

Starting point:
U = \frac{1}{2} \int \left[ \rho V + \mathbf{A}\cdot\mathbf{J} \right]d\tau
Where U is the electromagnetic energy of the system, \rho is the charge density, V is the scalar potential, A is the vector potential, J is the current density, and d\tau is the volume element with the integral over the volume you wish to know the energy of.
NOTE: This starting point already assumes you are asking about the energy of a system in which the fields are currently constant in time in the system. However, I will still make clear any assumptions used in the derivation regardless if they are already assumed, in order to make it clear I am aware the derivation assumes that as well.


Now, we go through the steps:
1) use maxwell's equation (assumption, the E fields are constant in time for the region of integration)
U = \frac{1}{2} \int \left[ \epsilon_0 (\nabla\cdot\mathbf{E}) V + \mathbf{A}\cdot \frac{1}{\mu_0}(\nabla \times\mathbf{B}) \right] d\tau


2) Use some vector calc identities:
\mathbf{A}\cdot (\nabla \times\mathbf{B}) = \mathbf{B} \cdot \mathbf{B} - \nabla\cdot (\mathbf{A}\times\mathbf{B})
(\nabla\cdot\mathbf{E}) V = \nabla \cdot (V\mathbf{E}) - \mathbf{E}\cdot(\nabla V)
so we have
U = \frac{1}{2} \int \left[ \epsilon_0 (\nabla \cdot (V\mathbf{E}) - \mathbf{E}\cdot(\nabla V)) + \frac{1}{\mu_0}(\mathbf{B} \cdot \mathbf{B} - \nabla\cdot (\mathbf{A}\times\mathbf{B})) \right] d\tau


3) Using the definition of the potentials (assuming that the magnetic field is constant in time for the region of integration)
U = \frac{1}{2} \int ( \epsilon_0 E^2 + \frac{1}{\mu_0}B^2) d\tau <br /> + \frac{1}{2} \int \left[ \nabla \cdot (\epsilon_0 V\mathbf{E} - \frac{1}{\mu_0}\mathbf{A}\times\mathbf{B}) \right] d\tau


4) Using a vector calc identity
U = \frac{1}{2} \int ( \epsilon_0 E^2 + \frac{1}{\mu_0}B^2) d\tau + U_{\mathrm{surface \ term}}
where
U_{\mathrm{surface \ term}} = \frac{1}{2} \oint (\epsilon_0 V\mathbf{E} - \frac{1}{\mu_0}\mathbf{A}\times\mathbf{B}) \cdot d\mathbf{a}



If you are integrating over all space, and your sources are finite (i.e. A and V go to zero at infinity) the result is simply:
U = \frac{1}{2} \int ( \epsilon_0 E^2 + \frac{1}{\mu_0}B^2) d\tau
If you are looking at a finite region, or your sources are infinite, then you must include the surface term.

If this is all relevant for your particular (static) situation is something else. For so far you haven't responded to the suggestions I made about this in my last post.

Because, if you are disagreeing with a textbook, it should be your onus to prove your point. Instead, here I am, wasting time reproducing a proof that you should have just gone and looked up if you still disagreed with the textbooks. It is very frustrating.

In both cases the E^2+B^2 method gives a different result as the A.j method applied on the dipole in isolation. The A.j method is correct. This is particular easy to see in the case of the electric dipole.

No, that is just flat out wrong. Again, unless you wish to disagree with textbooks and disagree with maxwell's equations and vector calc identities.

The dipole is defined by p=q.r where q is the charge of the positive/negative poles
and r is the distance between the two charges. Rotating the dipole from up to down
moves both charges over a distance of r which means a change of energy of 2qrE=2pE
as it should be.

In the general case the energy is -p.E which varies from -pE to +pE.

If you try to calculate the energy of the dipole by calculating the energy of the
E field instead (with the delta defined as in Jackson chapter 4) then you will get
the right sign but the wrong magnitude, as you did mention yourself also if I
remember well.

I got the wrong magnitude initially because I was using an infinite source and was not aware of the surface term that must be included in these cases.

As I mentioned, and at least one other poster mentioned as well, the (E^2+B^2) method works fine for the electric dipole. If you use finite sources, there is no problem. If you use infinite sources, and remember to use the surface term as the derivation requires, then you get the correct answer here as well.


Many of the last few posts could have been avoided if you checked for yourself when I first asked you to, instead of continuing on insisting the textbooks were wrong without even working it out yourself. Now, please work out the A.j case yourself if you still believe this somehow magically fixes the situation.


Sam,
on a skim of your post I see that there could be something wrong with that square loop argument. I took to much time on the A.j stuff to have time to really think over your post. Thank you for responding; I am not ignoring it, and I will get to it eventually.

 

[Hans de Vries]

You are disagreeing with the textbooks. You can't take the stance that the textbooks are correct, and that you are correct, without contradicting yourself.

Nonsense, You're aggressive "claim to textbook authority" discussion style is inappropriate
since you are producing results that, as you say yourself, conflict with experimental data.

The only requirement is that you are asking about the energy of a system in which the fields are currently constant in time in the system.

Wrong, we are talking about point dipoles here and experimental results concerning the
magnetic moments of elementary particles. Your equivalence proof assumes that the
fields of the particle act on the particle itself which is not the case with the experimental
data.

Have you ever studied the numerous attempts to model an elementary particle as
a classical charge distributions interacting with itself?

Maybe you are talking entirely classical here. Don't assume that other people do
when you are talking about point sources! The fact that other people are aware of
these issues does mean in your eyes that they are disagreeing with your textbooks?

The correct result corresponding with the experimental data is obtained when only
the external field interacts with the circular point current. Look at any textbook
which derives the magnetic dipole's energy in an external field. They calculate the
torque produced by the external field acting on the circular current.Regards, Hans.

 

[JustinLevy]

Nonsense, You're aggressive "claim to textbook authority" discussion style is inappropriate
since you are producing results that, as you say yourself, conflict with experimental data.

The difference is I admit that because the final result I get is different from the textbooks (albeit derived a different way), that something in my calculation is wrong. You instead, disagree with the textbooks and claim you are correct.

If you are going to disagree with multiple textbooks, and I even took the time to reproduce the proof here, then it is your onus to prove the textbooks are wrong.

Instead, you have replied once again without any math backing up your statements that the textbooks are wrong. I am getting incredibly frustrated with this, as I am sure has become apparrent. If you still disagree, again, please work out the A.j case to prove it is different.

Wrong, we are talking about point dipoles here and experimental results concerning the magnetic moments of elementary particles.

Damn it! I posted a long calculation showing that this problem persists even with finite sized dipoles. And we don't have to go to experimental results of elementary particles to check these calculations, as this can be seen just with a current loop.

Is U=+m.B incorrect to predict how a dipole will orient in a magnetic field? You betcha. We all agree on this. But that does NOT automatically make your argument against the textbook proof regarding A.j and B^2 correct.

Your equivalence proof assumes that the fields of the particle act on the particle itself which is not the case with the experimental data.

I started from the energy in A.j, which you said is correct.
Are you now saying that it is NOT correct?

Second, if you bothered to look at my calculation, the B^2 energy term is taken as (B_external^2 + 2B_external . B_dipole + B_dipole^2). Because the B_dipole^2 term is independent of orientation, it is not used to calculate the orientation energy. So again, your arguments do not apply.

This is getting very frustrating.
If you don't feel like working out the math, fine. But at least give the equation you feel should be solved to get the correct answer. How is that for a compromise?

 

[Hans de Vries]

This is getting very frustrating.
If you don't feel like working out the math, fine. But at least give the equation you feel should be solved to get the correct answer. How is that for a compromise?

The math is what we should talk about isn't it? So far I see only angry personal attacks.
How about stopping that as a start and concentrate on the math? Your thread on this
subject and your math work IS appreciated. Don't let your temper spoil it.

OK, I'm interested in your 1/3 surface term, I'll look at that and I'll post some simplifications
to do the E^2+B^2 integral in the mean time.Regards, Hans.

 

[Hans de Vries]

I'll post some simplifications to do the E^2+B^2 integral in the mean time.

One should be able to derive the point dipole fields in a much simpler way.
This is what I get: We start with a static point charge \delta(\vec{r}) which obtains
over time a potential field given by.

<br /> \mbox{field}\Big\{\,\delta(\vec{r})\,\Big\} ~~=~~ \frac{1}{4\pi r}<br />

The reversed operator which derives the source point charge from the field
is just the (minus) Laplacian.

<br /> -\nabla^2\Big\{\,\frac{1}{4\pi r}\,\Big\} ~~=~~ \delta(\vec{r})<br />

Integrating the delta function over space shows equal contributions from the
three spatial components.

<br /> \int \delta(\vec{r})~d\vec{r} ~=~<br /> -\int\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right]<br /> \frac{d\vec{r}}{4\pi r} ~=~ \frac13+\frac13+\frac13 ~=~ 1<br />These 1/3 fractions will lead to the delta functions in the dipole fields as we will see.

We can define vector dipole and axial dipole sources by using differential operators
on the monopole \delta(\vec{r}) and derive their potential and electromagnetic fields.<br /> \begin{array}{lcll}<br /> j^o &amp;=&amp; -~\mbox{div}\,\left(~\vec{\mu} \,\delta(\vec{r})~\right) &amp;<br /> \mbox{vector dipole point charge density} \\<br /> \vec{j} &amp;=&amp; +~\mbox{curl}\left(~\vec{\mu}\, \delta(\vec{r})~\right) &amp;<br /> \mbox{axial dipole point current density} \\<br /> \\<br /> A^o &amp;=&amp; -~\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) &amp;<br /> \mbox{vector dipole electric potential} \\<br /> \vec{A} &amp;=&amp; +~\mbox{curl}\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) &amp;<br /> \mbox{axial dipole magnetic potential} \\<br /> \\<br /> \vec{E} &amp;=&amp; \mbox{grad}\left(\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right)\right)&amp;<br /> \mbox{vector dipole electric field} \\<br /> \vec{B} &amp;=&amp; \,\mbox{curl}\left(\mbox{curl} \left(~\vec{\mu}\,\frac{1}{4\pi r}\,\right)\right)&amp;<br /> \mbox{axial dipole magnetic field} \\<br /> \end{array}<br />The expressions for the electromagnetic fields do implicitly contain the delta
functions at the center with the right magnitude. The E and B fields are
related to each other by the standard vector identity.

\mbox{curl}(\mbox{curl}\vec{A}) ~=~ \mbox{grad}(\mbox{div}\vec{A})-\nabla^2\vec{A}

We have seen that the last term (the Laplacian) yields \delta(\vec{r}). The two therefor
differ only at the center by a delta function in the \vec{\mu} direction. We can see
this explicitly if we align the dipoles with the z-axis so we can write.

<br /> \begin{aligned}<br /> \vec{E} &amp;= &amp;\bigg[~<br /> \mathbf{\hat{x}}\, \frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~<br /> \mathbf{\hat{y}}\, \frac{\partial}{\partial y}\frac{\partial}{\partial z} ~+~~~<br /> \mathbf{\hat{z}}\, \frac{\partial^2}{\partial z^2} ~~~~~~~~~~~~~ <br /> &amp;\bigg] ~ \frac{1}{4\pi r}\\ \\<br /> \vec{B} &amp;= &amp;\bigg[~<br /> \mathbf{\hat{x}}\,\frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~<br /> \mathbf{\hat{y}}\,\frac{\partial}{\partial y}\frac{\partial}{\partial z} ~-~<br /> \mathbf{\hat{z}}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)~ <br /> &amp;\bigg] ~ \frac{1}{4\pi r}\\<br /> \end{aligned}<br />The only difference is in the z-components. The total difference between the
two is the Laplacian and thus \delta(\vec{r}). The vector dipole gets -1/3 while the
axial dipole gets +2/3. Upon integration over space only the even functions
(the z-components) survive and we get for the mixed energy component which
depend on the orientation of the dipoles:

<br /> U_B ~=~ \frac{2}{3}\,\vec{\mu}\cdot\vec{B} <br /> ~~~~\mbox{and} ~~~~<br /> U_E ~=~ -\frac{1}{3}\,\vec{\mu}\cdot\vec{E}<br />

This should then be adjusted with the surface terms to get the j.A values.Regards, Hans

 

[Hans de Vries]

I made some modifications to the post above which further simplifies things.

Regards, Hans

 

[Sam Park]

In case anyone is still interested in the original question, I found another reference (Becker's Electromagnetic Fields and Interactions) that talks specifically about this, and it agrees with what I posted earlier based on Schwartz's treatment. In a nutshell, we can't treat displacements of current loops by considering just the magnetic field, if the currents are held constant, because in that case the effects of mutual induction must be taken into account, and the electric field does work on the system. Becker shows explicitly that, holding the currents constant, a current loop will do work equal to the INCREASE in the magnetic field energy, and this double expenditure is supplied by the electromotive force that must be applied during the displacement in order to maintain the constancy of the currents. On the other hand, if no emf is applied to keep the currents constant, they will change in response to a displacement, but the flux will be constant. For such a displacement, the work done equals the reduction in the magnetic field energy.

 

[Meir Achuz]

Thank you Sam. Of course, Becker, Schwartz, or any other textbook that considers this has the same answer. It also was given in post #11 of this thread, but people must not have looked at the link, so I copy it here:

"We note that the sign of the magnetic dipole energy is opposite that for the electric dipole. That is because the current producing the magnetic dipole is kept constant by a constant current source. The magnetic dipole would rotate so as to increase its energy, thus tending to align with the magnetic field. For this case, with the current kept constant by a constant current source, the force on the point dipole will be the positive gradient of the energy:
{\bf F=\nabla(\mu\cdot B)}.
The magnetic dipole force thus has the same form and sign as the electric dipole force."

 

[JustinLevy]

It also was given in post #11 of this thread, but people must not have looked at the link

I did not ignore the link. I commented on the paper. It makes judgements which are not justified, and when it comes to the quote which applies to this situation, they just state an answer without any math to justify it.

Considering that it is an unpublished source that comes to a conclusion that disagrees with the canonical choice for treating E&M in GR, this paper should not be given much weight.

I finished up by saying "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."

This is not to say that Sam's points and references are wrong, only that I do not want to consider that archiv paper clem linked to in post #8.

the force on the point dipole will be the positive gradient of the energy

How can the force EVER be the positive gradient of the energy?

In case anyone is still interested in the original question, I found another reference (Becker's Electromagnetic Fields and Interactions) that talks specifically about this

Great!
I'll see if I can find a copy to read.

It seems though that you guys are suggesting that a magnetized object would anti-align to an external field, where as a loop with frozen in charges sent spinning would align with an external field. This doesn't quite make sense to me yet. Hopefully your great find will clear this all up. Thanks!

 

[Meir Achuz]

It seems though that you guys are suggesting that a magnetized object would anti-align to an external field, where as a loop with frozen in charges sent spinning would align with an external field. This doesn't quite make sense to me yet. Hopefully your great find will clear this all up. Thanks!

You still have it backwards. The torques and force are the same for each. That is why the positive gradient of U is necessary in the constant current case. It is just the same as for electrostatic energy where the force is the positive gradient of U at constant voltage.
This is in every textbook. Why do you insist on fighting when you just need to learn?

 

[Hans de Vries]

In case anyone is still interested in the original question, I found another reference (Becker's Electromagnetic Fields and Interactions) that talks specifically about this, and it agrees with what I posted earlier based on Schwartz's treatment. In a nutshell, we can't treat displacements of current loops by considering just the magnetic field, if the currents are held constant, because in that case the effects of mutual induction must be taken into account, and the electric field does work on the system. Becker shows explicitly that, holding the currents constant, a current loop will do work equal to the INCREASE in the magnetic field energy, and this double expenditure is supplied by the electromotive force that must be applied during the displacement in order to maintain the constancy of the currents. On the other hand, if no emf is applied to keep the currents constant, they will change in response to a displacement, but the flux will be constant. For such a displacement, the work done equals the reduction in the magnetic field energy.

This is (of course) the correct explanation.

The A field of the background field will add an extra EM induced momentum to
the loop current when aligned, or subtract it when it is anti aligned. This is what
happens in the calculation.

But a B field can not increase the (total) momentum of a moving charge. The
momentum is the sum of the inertial PLUS the induced momentum. This means
that the inertial momentum has to go down by the same amount as the
inductive momentum goes up.


Regards, Hans

 

[JustinLevy]

In case anyone is still interested in the original question, I found another reference (Becker's Electromagnetic Fields and Interactions) that talks specifically about this, and it agrees with what I posted earlier based on Schwartz's treatment. In a nutshell, we can't treat displacements of current loops by considering just the magnetic field, if the currents are held constant, because in that case the effects of mutual induction must be taken into account, and the electric field does work on the system. Becker shows explicitly that, holding the currents constant, a current loop will do work equal to the INCREASE in the magnetic field energy, and this double expenditure is supplied by the electromotive force that must be applied during the displacement in order to maintain the constancy of the currents. On the other hand, if no emf is applied to keep the currents constant, they will change in response to a displacement, but the flux will be constant. For such a displacement, the work done equals the reduction in the magnetic field energy.

I was not able to find anyone with Becker's book so I could read that section.
Since you have it there, would you mind answering some questions for me?

You say "in that case the effects of mutual induction must be taken into account, and the electric field does work on the system." I am confused by this statement, because that is how the magnetic energy is derived in the first place (at least in some textbooks), by considering the work required against the induced EMF that needs to be done to setup that magnetic field configuration. So it seems like that is already included in the calculation.

Reading through your arguments again, I can see now at least some of what you mean. For example:
1) Take a charged ring (charges unable to move with respect to the ring) and spin the ring to make a magnetic dipole. This will align with an external magnetic field. And in doing so, the induced emf will slow the ring down (the dipole will change)... so while U=+m.B, the energy could still decrease?
2) Consider now a loop of perfectly conducting wire, with a fixed current (a battery is attached to provide energy to maintain the current if necessary). Again, this will align with an external magnetic field. We can compare this to #1, by letting it align with the field, then applying the energy to increase the magnetic dipole back to the starting value. It is clear from this, that the energy could decrease as in #1, then energy is transferred from some other source (in this case a battery) into electromagnetic energy to increase the dipole. The transfer from the "battery"/internal energy source, will not change the energy of the system ... it will only change the energy in the EM fields. So the energy of the system still decreases when aligning to the field.

However, even taking those two situations as solved, as I mentioned before,
"It seems though that you guys are suggesting that a magnetized object would anti-align to an external field, where as a loop with frozen in charges sent spinning would align with an external field."

This is because there is no internal energy being used up to maintain the magnetic dipole of a magnetized object (is there?). So for this case where the dipole remains constant without expenditure of energy as with the two loop cases given above, U=+m.B seems to require it to anti-align. So there is still something I am missing here.

Any ideas to fill in the last bit here would be great.

You still have it backwards. The torques and force are the same for each. That is why the positive gradient of U is necessary in the constant current case. It is just the same as for electrostatic energy where the force is the positive gradient of U at constant voltage.
This is in every textbook. Why do you insist on fighting when you just need to learn?

Ugh. I am trying to learn. Just because I don't like that unpublished source you gave does not mean I am unreceptive. Two things I've "fought" against were the notions that the 'solution' to this problem was that the equations might not be covariant, or that the A.j method is somehow different than the B^2 method, and I "fought" against those because those two ideas are not correct or not applicable here (as I hope we can all agree on now).

I understand that the "torques and force are the same" for those two situations. I understand that somehow in the end the energy should be represented as U = -m.B for describing the alignment. I've never argued that these calculations prove a real paradox or anything (in fact I've consistently said the opposite). The point of this is that the calculations don't seem to match up, so how do we correct/account for this?. So you telling me that a magnetized object aligns with the field because the torques are the same is not helpful for I already know that. That is avoiding the question.

Basically, a magnetized object seems to break the described method to get around this problem because it doesn't require a "battery" or any internal energy to maintain the dipole as far as I know. So going by the energy calculations, U=+m.B, it should anti-align. So there is still more to this that we need to figure out.

The A field of the background field will add an extra EM induced momentum to
the loop current when aligned, or subtract it when it is anti aligned. This is what
happens in the calculation.

But a B field can not increase the (total) momentum of a moving charge. The
momentum is the sum of the inertial PLUS the induced momentum. This means
that the inertial momentum has to go down by the same amount as the
inductive momentum goes up.

I thought we resolved this. The A.j method is equivalent to the B^2 method. A simple proof from a textbook was reproduced here for you to read. Please stop claiming these methods are different without math to back up your statements.

Here, let me do the calculation using A.j real quick:
Now for a constant external magnetic field,
\mathbf{A} = -\frac{1}{2} \mathbf{r} \times \mathbf{B}_0
where the chosen gauge is \nabla \cdot \mathbf{A} = 0.
For a quick verification:
\nabla \cdot \mathbf{A} = \frac{-1}{2} \nabla \cdot ( \mathbf{r} \times \mathbf{B}_0) = \frac{-1}{2} \nabla \cdot \left[ \hat{x}(y B_z - z B_y) + \hat{y} (z B_x - x B_z) + \hat{z} (x B_y - y B_x) \right] = 0

\begin{align*}<br /> \nabla \times \mathbf{A} &amp;= \frac{-1}{2} \nabla \times (\mathbf{r} \times \mathbf{B}_0) \\<br /> &amp;= \frac{-1}{2} \nabla \times \left[ \hat{x}(y B_z - z B_y) + \hat{y} (z B_x - x B_z) + \hat{z} (x B_y - y B_x) \right] \\<br /> &amp;= \frac{-1}{2} \left[ \ \hat{x}(\partial_y (x B_y - y B_x) - \partial_z (z B_x - x B_z)) + \hat{y}(\partial_z (y B_z - z B_y) - \partial_x (x B_y - y B_x)) \right. \\<br /> &amp; \hspace{ 1 in} + \left. \hat{z}(\partial_x (z B_x - x B_z) - \partial_y (y B_z - z B_y))\ \right] \\<br /> &amp;= \frac{-1}{2} \left[ \hat{x}( -B_x - B_x) + \hat{y}(- B_y -B_y) + \hat{z}(- B_z - B_z) \right ] = \mathbf{B}_0<br /> \end{align*}<br />

Now, starting with the A.j energy density, for currents j in an external field B_0
U = \int \mathbf{A}\cdot\mathbf{j} \ d^3r = \int \frac{-1}{2}(\mathbf{r}\times\mathbf{B}_0)\cdot\mathbf{j} \ d^3r = \int \frac{-1}{2}(\mathbf{j}\times\mathbf{r})\cdot\mathbf{B}_0 \ d^3r
= \mathbf{B}_0 \cdot \int \frac{1}{2}(\mathbf{r}\times\mathbf{j}) \ d^3r
which, since the second term is just the definition of the magnetic moment, gives
U = + \mathbf{B}_0 \cdot \mathbf{m}

The A.j and B^2 methods are equivalent.

 

[atyy]

I don't know if this is related, but you may like to take a look the Feynman lectures discussion on this. Whether it's +/-m.B depends on whether you count the energy to set up the external B field or something like that (I'm garbling it).

Edit: Volume 2, Chapter 15

 

[Hans de Vries]

Justin, You are continuing to misinterpret and misrepresent my statements. I'm not
talking about the A.j versus B^2 method.

If you would had simply read the link clem gave you 3 weeks ago then you would have
spared yourself a lot of time. The mathematical contents of the paper is perfectly
mainstream and perfectly relevant for this issue despite the author's interpretation
of DE+BH as a probability density only.

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

Vq+A.j is DE+BH only under certain conditions since (see eq.49) of the link above.

Vq+A.j = DE+BH + surface term + time dependent term

The paper calculates the surface term in two ways see eq.39 and eq.42.The real issue is the constant current assumed in the calculations and the energy
that has to be put in the loop to keep the current constant.Regards, Hans

 

[Meir Achuz]

Justin:
It might help you to read the derivation of Eq. (4.102) in Jackson (2nd Ed.) (or another text) where he derives the positive gradient for the electric case at constant V. The same procedure applies in the magnetic case at constant current. You could also read the section in that preprint on a permanent moment which does have -mu.B.

 

[Meir Achuz]

Justin:
All you have shown is that A.j gives the right answer.
If you dot B_0 with B from a dipole, you will get the same wrong answer as in your first post.

"Please stop claiming these methods are different without math to back up your statements." shows you have not taken my advice.
You are now on your own.

 

[Sam Park]

However, even taking those two situations as solved, as I mentioned before, "It seems though that you guys are suggesting that a magnetized object would anti-align to an external field, where as a loop with frozen in charges sent spinning would align with an external field." This is because there is no internal energy being used up to maintain the magnetic dipole of a magnetized object (is there?). So for this case where the dipole remains constant without expenditure of energy as with the two loop cases given above, U=+m.B seems to require it to anti-align. So there is still something I am missing here.

There is an important difference between the energy calculations for a magnetic dipole produced by “free current” and one produced by a magnetized subtance. The distinction is captured by the difference between the B and the H fields. Outside any magnetic material, B and H are strictly proportional to each other, but inside magnetic material they are quite different. The potential energy density of a magnetic field is really B.H/2, and reduces to B^2/(2 mu) only outside of any magnetic material.

Take for example a spherical shell of uniformly magnitized material. Outside the shell the B and H fields are essentially identical, being that of a perfect dipole, for which the field energy is indifferent to orientation. Inside the sphere, there is a uniform field, but the direction of the H field is opposite the direction of the B field in this interior region. This gets back to my first post in this thread, asking if you were really sure the components of the cross-term B_f and B_d really had the same sign. I should have said that, in the region that counts (inside the magnetized sphere), the field that counts is the H_d field, which points in the opposite direction of the B_d field.

The reason we need to use H is that a magnetic field B is caused by two different types of currents, which are sometimes called “free current” and “bound current”. The free currents are the ones you would usually identify as electrical currents in a system, produced by hooking up a battery to a wire, or some similar method. In contrast, the “bound currents” are the ones we infer must be present in atoms comprising a permanently magnetized substance. Unlike free currents, bound currents don’t do any work, so when we perform calculations involving the energy of the magnetic field we need to distinguish between the fields produced by these two kinds of current. This is why we need to use the field energy expression based on the H field when considering magnetized objects.

 

[JustinLevy]

I don't know if this is related, but you may like to take a look the Feynman lectures discussion on this. Whether it's +/-m.B depends on whether you count the energy to set up the external B field or something like that (I'm garbling it).

Edit: Volume 2, Chapter 15

I think I have that around here somewhere. Thanks, I'll check it out.

Justin, You are continuing to misinterpret and misrepresent my statements. I'm not talking about the A.j versus B^2 method.

I'm sorry if there is a miscommunication. Your post seemed to be claiming that using the A field gave the correct answer and the B field did not. Even now, going back and rereading your post, I get the same impression ... that you are saying that using the A field gives the correct answer and the B field does not. Can you please reread your previous post and clarify?

If you would had simply read the link clem gave you 3 weeks ago then you would have spared yourself a lot of time. The mathematical contents of the paper is perfectly mainstream and perfectly relevant for this issue despite the author's interpretation of DE+BH as a probability density only.

As stated before, I READ THAT PAPER. Please everyone stop bringing up that paper. It is not useful for this situation. Here's what they say about the dipole problem:
"In fact, [ F =\nabla (\mu \cdot B) ] should be considered a separate force law for permanent magnetic dipoles on a par with the Lorentz force on charge or the j×B force on currents, since it cannot be derived from those force laws."

Their answer is a non-answer. Their "solution" to the dipole problem is to postulate a new force. If you want to take that as an answer, fine, but you don't need to reference that paper, for stating it outhand yourself is giving as much justification as they gave it.

Secondly, while I agree much of their math is straight-forward and correct, their paper is very non-mainstream. For some reason they have some serious philosophical problems with EM radiation carry energy and go through all kinds of manipulations to try to justify it, but fall very short. In doing so they take a classical theory and turn it into a theory with manifestly non-local interactions, come to the "conclusion" that EM radiation doesn't carry energy or momentum, and completely disregard the stress energy tensor as physical despite its use in GR (instead opting for a "energy in the charges" approach which would make GR tortuous).

I have reproduced a proof from a textbook in this thread already showing the relation between the "energy in the fields" and the "energy in the charges" methods. No further discussion about the relation between these methods should be needed. No further discussion of that paper should be needed (or wanted) either.

Vq+A.j is DE+BH only under certain conditions since (see eq.49) of the link above.

Vq+A.j = DE+BH + surface term + time dependent term

The paper calculates the surface term in two ways see eq.39 and eq.42.

You complain that I keep interpretting your posts as saying only one of the two energy calculations methods is allowed here. Yet you keep bringing it up in every post.

I showed the relation between the two. I've even explicitly done the calculation for both cases and got the same answer. These are the same here. If we are in agreement, why do you keep bringing it up? It is clear that the "solution" to this problem does not lie in our choice of calculating with the "energy in the fields" or the "energy in the charges" methods. So let's please move on.

The real issue is the constant current assumed in the calculations and the energy that has to be put in the loop to keep the current constant.

Okay, let's consider an example that doesn't have a constant current so we don't need to worry about the energy necessary to keep that going.

Consider a spinning ring of charge, initially anti-aligned to a magnetic field of magnitude B supplied by an external source. Initially the magnetic moment of the ring has a magnitude m1. Thus the energy of orientation is U = -m1 B. Let it align with the field, and the magnitude of the dipole moment is now m2, with the energy U=+m2 B. This energy must have come from somewhere, and with no current source to blaim it must have come from the Kinetic energy or from the self energy in the dipole field. The kinetic energy is proportional to M (the mass of the ring) and the self energy of the dipole field is proportional to m1^2. Each of these can be adjusted separately as well as the external field. So what prevents us from making a situation where m1.B > the kinetic energy + self energy of the dipole field ? In that case, it is clear that the total energy would be minimized with the dipole anti-aligned. If it goes to torque here, there isn't even enough energy to bring this to rest, let alone align the dipole with the field ... so what happens?

It might help you to read the derivation of Eq. (4.102) in Jackson (2nd Ed.) (or another text) where he derives the positive gradient for the electric case at constant V. The same procedure applies in the magnetic case at constant current.

With some source of energy providing the necessary work to keep the current the same, I think I understand that case now. It is the magnetized material and non-current source cases that still seem strange to me.

You could also read the section in that preprint on a permanent moment which does have -mu.B.

As noted several times already, I have read that paper. It is unpublished, non-mainstream, and when it gets to the situations relavent here, just states answers (and even postulates new forces) instead of working anything out. Let us please drop that paper.

Justin:
All you have shown is that A.j gives the right answer.
If you dot B_0 with B from a dipole, you will get the same wrong answer as in your first post.

No, I have shown that BOTH give the same answer. Not only have I worked this out explicitly, I have even reproduced the textbook proof that shows the relation between the "energy in the fields" and the "energy in the charges" methods. The methods give the same results here.

If you have some philosophical issue with the surface terms like those preprint authors, then just don't consider infinite sources. The answer to this problem has nothing to do with A.j versus B^2 methods. This has been shown many times, so please please let us move on from that.

"Please stop claiming these methods are different without math to back up your statements." shows you have not taken my advice.
You are now on your own.

Those methods are equivalent for this problem. This has been shown explicitly by working it out both ways for this problem, as well as a general proof from a textbook. If you still wish to claim otherwise, then you are wrong. Saying so does not make me unwilling to learn. It just makes me unwilling to ignore textbooks and explicit mathematical examples, and instead believe otherwise because someone says so without any math to back it up.

;-------------------------------------------

Sam,
I thought you could work out problems with that involve M in two ways:

Method 1:
Given M and the free current:
- Calculate the bound current
- Calculate the magnetic field using these current densities
- Calculate whatever you need using the magnetic field now

method 2:
Given M and the free current:
- Calculate H
- Calculate whatever you need using B and H now

I thought these two methods (use the bound current, or use H) were equivalent. In particular, when I had to take an EM course, I remember them requiring us to do many problems using both methods to apparently drive home that either way worked. So shouldn't I get the same energy either way?

What you brought up seems to clearly answer no. It seems obvious now, but I will need to read more to jive these two thoughts in my head. This however seems to raise another issue since using the relations already derived, and given that H is defined as H= \frac{1}{\mu_0} B - M we arrive at the relation between B_external and M inside the sphere as H= \frac{1}{\mu_0} B_{ext} + (\frac{2}{3} M) - M = \frac{1}{\mu_0} B_{ext} - (\frac{1}{3} M). This seems to raise another "magnitude of the energy" issue.

Thanks for the input. I think you've sketched together enough that I can see all the answers are in your suggestions and examples there ... I just need to sort through this some more. At least now it is abundantly clear how the magnetized and non-magnetized situations arrive at different signs.

 

[Hans de Vries]

I'm sorry if there is a miscommunication. Your post seemed to be claiming that using the A field gave the correct answer and the B field did not.

Justin, I think everybody here agrees about the following numbers:

jA = 1 (mechanical energy transformed into electromagnetic field energy)
B^2 = 2/3 (electromagnetic field energy inside the volume)
S = 1/3 (surface term of the electromagnetic field energy)
T = 0 (time dependent term)

The miscommunication here is that you equate the "B^2 method" with B^2+S+T while
others simply mean B^2=2/3. Read for instance clem's last post and your response to it.


Regards, Hans

 

[Hans de Vries]

As stated before, I READ THAT PAPER. Here's what they say about the dipole problem:

"In fact, [ F =\nabla (\mu \cdot B) ] should be considered a separate force law for permanent magnetic dipoles on a par with the Lorentz force on charge or the j×B force on currents, since it cannot be derived from those force laws."

Their answer is a non-answer. Their "solution" to the dipole problem is to postulate a new force.

This the whole point. F =\nabla (\mu \cdot B) is the correct force (with the right sign)
on the dipole when the Lorentz force is applied on a circular current !

The author assumes a mysterious sign flip (stemming from elementary
particle physics). There are however no sign flips. The sign is always like this.
There are no mysterious sign flips in the force.

Consider a spinning ring of charge

An increase in flux (increase in B) through the ring induces an electric field
along the loop \partial B/\partial t = -\nabla\times E which opposes the current in the loop and
slows down the spinning ring.

The mechanical energy from the angular momentum of the ring decreases
by a factor \mu\cdot B while the total field energy increases by the same factor.

This will reduce the current in the loop, but the amount by which the current
decreases depends on the mass (-> angular momentum) of the spinning ring.
A very heavy spinning ring will provide the kinetic energy without hardly slowing
down and the current remains almost the same.

A static non-homogeneous magnetic field B can not increase the total
energy of the dipole (it is a conservative field), but it can transform the
kinetic energy from the angular momentum into kinetic energy from linear
motion.

Note that none of the energy calculations done or mentioned this tread
contains such a "kinetic energy from linear momentum" because a static
situation is always assumed. The calculation of the various energies of
an accelerating dipole is much more involved.

Without doing such calculations all we can tell is that apparently all the
kinetic energy that becomes available from the slowing down of the
spinning ring (\mu\cdot B) would be available for an increase of the linear
momentum via F =\nabla (\mu \cdot B)


Regards, Hans

 

[gabbagabbahey]

Sam,
I thought you could work out problems with that involve M in two ways:

Method 1:
Given M and the free current:
- Calculate the bound current
- Calculate the magnetic field using these current densities
- Calculate whatever you need using the magnetic field now

method 2:
Given M and the free current:
- Calculate H
- Calculate whatever you need using B and H now

I thought these two methods (use the bound current, or use H) were equivalent. In particular, when I had to take an EM course, I remember them requiring us to do many problems using both methods to apparently drive home that either way worked. So shouldn't I get the same energy either way?

What you brought up seems to clearly answer no. It seems obvious now, but I will need to read more to jive these two thoughts in my head. This however seems to raise another issue since using the relations already derived, and given that H is defined as H= \frac{1}{\mu_0} B - M we arrive at the relation between B_external and M inside the sphere as H= \frac{1}{\mu_0} B_{ext} + (\frac{2}{3} M) - M = \frac{1}{\mu_0} B_{ext} - (\frac{1}{3} M). This seems to raise another "magnitude of the energy" issue.

Thanks for the input. I think you've sketched together enough that I can see all the answers are in your suggestions and examples there ... I just need to sort through this some more. At least now it is abundantly clear how the magnetized and non-magnetized situations arrive at different signs.

My previous post was in error (I hadn't fully read all of the posts to see that you were in fact having a very different issue); Sam's resolution of the "paradox" is correct. For a detailed discussion of it, see R.H. Young, Am. J. Phys 66, 1043 (1998) --- In which Young answers Griffiths' question on why the Hamiltonian is -\vec{\mu}\cdot\vec{B}--- and references cited therein.

 

[Meir Achuz]

This the whole point. F =\nabla (\mu \cdot B) is the correct force (with the right sign)
on the dipole when the Lorentz force is applied on a circular current !

The author assumes a mysterious sign flip (stemming from elementary
particle physics). There are however no sign flips. The sign is always like this.
There are no mysterious sign flips in the force.

Read it again.
The force on a magnetic dipole is always F =\nabla (\mu \cdot B).
But the energy on a current loop where the current is kept fixed is given by
U=\mu \cdot B, with the force F=\nabla U. The energy of a permanent magnetic dipole in a magnetic field is given by U=-\mu \cdot B, with the force F=-\nabla U.
The force F =\nabla (\mu \cdot B) for a permanent magnet cannot be derived from the Lorentz force because a permanent magnet is not due to current loops but to the electron's magnetic moment, which is unrelated to current loops.

 

[Hans de Vries]

Read it again.
The force on a magnetic dipole is always F =\nabla (\mu \cdot B).
But the energy on a current loop where the current is kept fixed is given by
U=\mu \cdot B, with the force F=\nabla U.

The force is per definition F=-\nabla U and there is no sign change which would
violate the conservation of energy. A static magnetic field is conservative and
can only transform one kind of kinetic energy into another, in this case rotational
kinetic energy into translational kinitic energy as described in post #75.

The energy of a permanent magnetic dipole in a magnetic field is given by U=-\mu \cdot B, with the force F=-\nabla U.
The force F =\nabla (\mu \cdot B) for a permanent magnet cannot be derived from the Lorentz force because a permanent magnet is not due to current loops but to the electron's magnetic moment, which is unrelated to current loops.

According to the Dirac equation there IS a current loop associated with the electron's
magnetic moment. See for instance: Sakurai, Advanced Quantum Mechanics eq(3.205)
and further in the section on the Gordon Decomposition.

A paper which discusses this in the context see:
http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf"

According to the prevailing belief, the spin of the electron or some other
particle is a mysterious internal angular momentum for which no concrete
physical picture is available, and for which there is no classical analog.
However, on the basis of an old calculation by Belinfante [Physica 6 887
(1939)], it can be shown that the spin may be regarded as an angular
momentum generated by a circulating flow of energy in the wave field
of the electron. Likewise, the magnetic moment may be regarded as
generated by a circulating flow of charge in the wave field. This provides
an intuitivelyl appealing picture and establishes that neither the spin nor
the magnetic moment are “internal” — they are not associated with the
internal structure of the electron, but rather with the structure of the field.
Furthermore, a comparison between calculations of angular momentum in
the Dirac and electromagnetic fields shows that the spin of the electrons
is entirely analogous to the angular momentum carried by a classical
circularly polarized wave.

see equation (22)Regards, Hans

 

[Hans de Vries]

According to the Dirac equation there IS a current loop associated with the electron's
magnetic moment. See for instance: Sakurai, Advanced Quantum Mechanics eq(3.205)
and further in the section on the Gordon Decomposition.

A paper which discusses this in the context see: http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf"

Regards, Hans

For an intuitive picture see the introductory chapter of my book (section 1.9)
http://physics-quest.org/Book_Chapter_EM_basic.pdf

Or the chapter on the Gordon decomposition itself (section 18.2 and 18.3)
http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdfRegards, Hans
(P.S. Some work in progress here regarding the decomposition of the axial current I see now)

 

[Meir Achuz]

The force is per definition F=-\nabla U and there is no sign change which would
violate the conservation of energy.

I am a bit surprised that you say that so late in this thread. The thread started with the fairly simple demonstration that U=+\mu\cdot{\bf B} for a loop with a constant current in a B field. This required {\bf F}=+\nabla U to get the correct form that
{\bf F}=+\nabla(\mu\cdot{\bf B}). F also equals \nabla U for the force of one capacitor plate on the other, when the capacitor is kept at constant voltage. This is shown in most EM texts. In every case, {\bf F}=+\nabla(\mu\cdot{\bf B}), but the energy can have either sign depending on whether or not a constant voltage or constant current source is providing energy to keep the voltage or current constant.

For an electron, U=-\mu\cdot{\bf B} because there is no external source feeding energy to the electron. This is independent of any speculation about the source of the moment.

 

[Hans de Vries]

I am a bit surprised that you say that so late in this thread. The thread started with the fairly simple demonstration that U=+\mu\cdot{\bf B} for a loop with a constant current in a B field. This required {\bf F}=+\nabla U to get the correct form that {\bf F}=+\nabla(\mu\cdot{\bf B}). F also equals \nabla U for the force of one capacitor plate on the other, when the capacitor is kept at constant voltage. This is shown in most EM texts. In every case, {\bf F}=+\nabla(\mu\cdot{\bf B}), but the energy can have either sign depending on whether or not a constant voltage or constant current source is providing energy to keep the voltage or current constant.

For an electron, U=-\mu\cdot{\bf B} because there is no external source feeding energy to the electron. This is independent of any speculation about the source of the moment.

This is all correct Meir,
My original response was to Justin Levi's attack on section 4 of this paper:

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

Which discusses the magnetic moment of elementary particles. Which have, as you
say, U=-\mu\cdot{\bf B} because they are passive.

As stated before, I READ THAT PAPER. Please everyone stop bringing up that paper. It is not useful for this situation. Here's what they say about the dipole problem:

"In fact, [ F =\nabla (\mu \cdot B) ] should be considered a separate force law for permanent magnetic dipoles on a par with the Lorentz force on charge or the j×B force on currents, since it cannot be derived from those force laws."

Their answer is a non-answer. Their "solution" to the dipole problem is to postulate a new force. If you want to take that as an answer, fine

From "Here's what they say about the dipole problem" I concluded that the
author says that Justin's "paradox": U=+\mu\cdot{\bf B} was explained by telling that
"elementary particles obey different laws".

Re-reading section 4 shows that the author correctly states that U=-\mu\cdot{\bf B}
So the author of the paper is correct here, besides the other independent
discussion about the nature of the magnetic moment of elementary particles.Regards, Hans.