Paradox regarding energy of dipole orientation

[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:

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

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

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

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

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.

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

The limit $R \rightarrow 0$ yields an ideal magnetic dipole:
$$\mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] + \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r})$$

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

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:

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

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,
$$U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} \ d^3r = \frac{1}{\mu_0} \int B \hat{\mathbf{z}} \cdot \left[ \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] + \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r}) \right] d^3r$$

The first term does not contribute, for
$$\begin{align*} \int & \hat{\mathbf{z}} \cdot [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] d^3r \\ & = \int [ 3(m_x \sin\theta\cos\phi + m_y \sin\theta\sin\phi + m_z \cos\theta)\cos\theta - m_z ] r^2 \sin\theta \ d\phi d\theta dr \end{align*}$$
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
$$\int [ 3\cos^2\theta - 1] \sin\theta \ d\theta = 0.$$

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

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

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 $r<R$, the result is identical to attempt 1, so the energy is:

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

This new term, where $r > R$, 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.

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

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

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

Performing the $\phi$ integration yields

[tex]
\begin{align*}
U_2 &= \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 \\
&= \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.
\end{align*}
[/itex]

Do a change of variables $u=\cos\theta$ and complete the integral.
$$\begin{align*} U_2 &= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \int_R^\infty \int_{-1}^1 \frac{1}{r^4} u^2 \ du dr \\ &= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \frac{2}{3} \int_R^\infty \frac{1}{r^4} \ dr \\ &= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \frac{2}{3} \frac{1}{3R^3} \\ &= \frac{1}{3} (\frac{2 \mu_0}{3} \frac{\mathbf{m_1}}{\frac{4}{3}\pi R^3} \cdot \mathbf{m_2}) \end{align*}$$

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
$$\begin{align*} U_2 = \frac{1}{3} \mathbf{m_2} \cdot \mathbf{B} \end{align*}$$

The final result is therefore
$$\begin{align*} U = + \mathbf{m} \cdot \mathbf{B} \end{align*}$$

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 unstable, and the minimum energy is with the dipole aligned opposing the field.

While it is hard to do such integral in your head, one can see that it seems intuitively reasonable that a magnetic dipole alligned against a field would reduce the total magnetic field, while one alligned with the field would increase the magnetic field ... therefore from the "energy in the fields" 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.

... Hence the "paradox".

 

[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
$$U= \int_{\pi/2}^\theta \mathbf{T} \cdot d\mathbf{\theta'} = \int_{\pi/2}^\theta m B \sin\theta' \ d\theta' = - m B \cos\theta$$

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

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.

 

[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)
$$U = \frac{1}{2} \int ( \epsilon_0 E^2 + \frac{1}{\mu_0}B^2) d\tau + \frac{1}{2} \int \left[ \nabla \cdot (\epsilon_0 V\mathbf{E} - \frac{1}{\mu_0}\mathbf{A}\times\mathbf{B}) \right] d\tau$$


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.

$$\mbox{field}\Big\{\,\delta(\vec{r})\,\Big\} ~~=~~ \frac{1}{4\pi r}$$

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

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

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

$$\int \delta(\vec{r})~d\vec{r} ~=~ -\int\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right] \frac{d\vec{r}}{4\pi r} ~=~ \frac13+\frac13+\frac13 ~=~ 1$$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.$$\begin{array}{lcll} j^o &=& -~\mbox{div}\,\left(~\vec{\mu} \,\delta(\vec{r})~\right) & \mbox{vector dipole point charge density} \\ \vec{j} &=& +~\mbox{curl}\left(~\vec{\mu}\, \delta(\vec{r})~\right) & \mbox{axial dipole point current density} \\ \\ A^o &=& -~\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) & \mbox{vector dipole electric potential} \\ \vec{A} &=& +~\mbox{curl}\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) & \mbox{axial dipole magnetic potential} \\ \\ \vec{E} &=& \mbox{grad}\left(\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right)\right)& \mbox{vector dipole electric field} \\ \vec{B} &=& \,\mbox{curl}\left(\mbox{curl} \left(~\vec{\mu}\,\frac{1}{4\pi r}\,\right)\right)& \mbox{axial dipole magnetic field} \\ \end{array}$$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.

$$\begin{aligned} \vec{E} &= &\bigg[~ \mathbf{\hat{x}}\, \frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~ \mathbf{\hat{y}}\, \frac{\partial}{\partial y}\frac{\partial}{\partial z} ~+~~~ \mathbf{\hat{z}}\, \frac{\partial^2}{\partial z^2} ~~~~~~~~~~~~~ &\bigg] ~ \frac{1}{4\pi r}\\ \\ \vec{B} &= &\bigg[~ \mathbf{\hat{x}}\,\frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~ \mathbf{\hat{y}}\,\frac{\partial}{\partial y}\frac{\partial}{\partial z} ~-~ \mathbf{\hat{z}}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)~ &\bigg] ~ \frac{1}{4\pi r}\\ \end{aligned}$$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:

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

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:
$$\nabla \cdot \mathbf{A} = \frac{-1}{2} \nabla \cdot ( \mathbf{r} \times \mathbf{B}_0) = \frac{-1}{2} \nabla \cdot \left[ \hat{x}(y B_z - z B_y) + \hat{y} (z B_x - x B_z) + \hat{z} (x B_y - y B_x) \right] = 0$$

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

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.