A magnetostatics problem of interest 2

In summary, the conversation discusses the important problem of a uniformly magnetized cylinder of finite length in magnetostatics. Two methods, the magnetic pole method and the magnetic surface current method, are used to calculate the magnetic field inside and outside the cylinder. Both methods yield identical results for the vector B. The conversation also mentions the simpler case of a homogeneously magnetized sphere and provides a link to the solution. While a closed-form solution for the magnetized cylinder of finite length may not exist, the agreement between the two methods is still remarkable.
  • #1
Charles Link
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A very important problem in magnetostatics is the uniformly magnetized cylinder of finite length. Permanent cylindrical magnets can be modeled as having approximately uniform magnetization, and it is of much interest, given such a uniformly magnetized cylinder, to be able to calculate the magnetic field both outside and inside the magnetized cylinder.
The solution of the magnetic field ## \vec{B} ## can be computed by both the magnetic pole method and by the magnetic surface current method. Both methods yield identical results for the vector ## \vec{B} ##. We will summarize the two methods below:
1) Magnetic pole method: Magnetic pole density (fictitious) is given by ## \rho_m=-\nabla \cdot \vec{M} ##. The result for uniform magnetization ## \vec{M}=M_o \hat{z} ## is simply a magnetic surface charge density at the end faces of the cylinder given by ## \sigma_m=\vec{M} \cdot \hat{n} ## resulting in + and - poles at the end faces of the cylinder. The vector ## \vec{H}## is then computed everywhere as ## \vec{H}(\vec{x})=\int \frac{ \sigma_m (\vec{x}-\vec{x}')}{4 \pi \mu_o |\vec{x}-\vec{x}'|^3} \, dA ## and ## \vec{B}=\mu_o \vec{H}+\vec{M} ##. This last equation takes some work to prove in detail, but we will simply use it in the solution for ## \vec{B} ##. It has in fact been proven.
2) Magnetic surface current method: Magnetic surface current density ## \vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o} ## . For a uniform ## \vec{M}=M_o \hat{z} ## the result is a surface current density per unit length ## \vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o} ## on the outer surface of the cylinder, in the same geometry as a solenoid. The magnetic field ## \vec{B} ## is then found everywhere inside and outside the cylinder using Biot-Savart as ## \vec{B}(\vec{x})=\frac{\mu_o}{4 \pi} \int \frac{ \vec{K}_m \times (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3 } \, dA ##, where this ## dA ## is over the cylindrical surface, unlike the previous ## dA ##, where the integration is over the end faces. ## \\ ##
This is an important problem in magnetostatics, and it is hoped that the E&M (electricity an magnetism) students at the upper undergraduate level and higher find it of interest.
 
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  • #3
Working the case of the magnetic field ## \vec{B} ## inside the uniformly magnetized cylinder of infinite length is perhaps the easiest one to compute by both methods. ## \\ ## For the pole method case, there are no poles, so ## \vec{H}=0 ##, and ## \vec{B}=\mu_o \vec{H}+\vec{M}=\vec{M} ##. ## \\ ## For the surface current method, Ampere's law (in place of Biot-Savart) applies to get the same answer. ## \oint \vec{B} \cdot dl=\mu_o I ## gives ## \vec{B}=\vec{M} ## because ## I=K_m L =M L/\mu_o ##.
 
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  • #4
Note for the above: The case of the magnetic surface currents for the uniformly magnetized cylinder of infinite length has the same geometry as that of a solenoid of infinite length with surface current per unit length ## K_m=\frac{M_o}{\mu_o} ## replacing ## nI ## in the formula for the magnetic field ## \vec{B}=\mu_o n I \hat{z} ##, where ##n ## is the number of turns per unit length. Thereby ## \vec{B}=M_o \hat{z} ##.
 
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  • #5
A minor correction to post 1, but I don't want to attempt to edit it, the ## dA ## in both cases should be ## dA' ##.
 
  • #6
A full derivation of this insight for finite length is given in the answer (given in the book) to problem 8.3 in Franklin, "Classical Electromagnetism", Vol. II.
 
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  • #7
Meir Achuz said:
A full derivation of this insight for finite length is given in the answer (given in the book) to problem 8.3 in Franklin, "Classical Electromagnetism", Vol. II.
I'm glad to see there is at least one E&M textbook that has it. IMO this should be an important part of the undergraduate physics curriculum.
 
  • #8
Hm, I have it for the simpler case of a homogeneously magnetized (hard ferromagnet) sphere using both the scalar potential for ##\vec{H}## and the vector potential for ##\vec{B}## and accordingly both the method using the magnetization directly and the surface-current formulation.

I wonder, whether there's a closed solution for the homogneously magnetized cylinder of finite length. I haven't tried yet. It seems to be more complicated, because the magnetic potential depends on both ##\rho## (radial cylinder coordinate) and ##z##.
 
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  • #9
@vanhees71 The homogeneously magnetized sphere is also another magnetostatics problem that I think the upper level undergraduate physics students would do well to have studied, and if they really want to challenge themselves, to be able to work it routinely. I do think the universities would do well to place increased emphasis on the advanced E&M courses for the undergraduate physics majors. ## \\ ## And could you perhaps post a "link" to your solution(s) of the magnetized sphere. It would help to make this thread more complete. Thanks. ## \\ ## And I don't think a closed-form solution exists for the magnetized cylinder of finite length, but the very interesting thing is that both the "magnetic pole" method and "magnetic surface current" method give the exact same result for ## \vec{B} ##. They are two very different methods, and the agreement, although shown by mathematical proof to be the same, is quite remarkable.
 
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  • #10
The homogeneously magnetized sphere is in my German lecture notes on E&M for high school-teacher students. This manuscript is matching the needs of this special audience. Unfortunately we have had the Bologna reform in Europe and thus "module plans". For the standard lectures thus we are forced to teach them in a given order, i.e., I'm not free to do "relativity first", because that's part of another module. That's why these lecture notes follow the old-fashioned 19th-century approach to E&M. Otherwise I think they are not too bad:

https://th.physik.uni-frankfurt.de/~hees/publ/theo2-l3.pdf
The homogeneously magnetized sphere can be found in Sect. 3.3.3 (worked out in both ways discussed here).
 
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  • #11
@vanhees71 I worked through a good part of your section 3.3.3 including equations 3.3.18 and 3.3.19 leading up to some of the calculations of section 3.3.3. My German is just good enough that I am able to work through your calculations.
I haven't gone through your entire solution yet, but it really appears to be an E&M work of art. Thank you very much.:smile:
Edit: The problem is precisely analogous to the dielectric sphere with uniform polarization. Originally, our physics class in college on 1975 was shown the Legendre polynomial solution, but since then Griffith's E&M has a solution to the problem, and the surface polarization charge problem was also done in closed form. Let me see if I can find a couple of "links"... See post 13 of https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5514160 And see post 9 of "Discuss in the Community" of https://www.physicsforums.com/threads/the-homopolar-generator-an-analytical-example-comments.932758/ And see post 2 of https://www.physicsforums.com/threa...ormly-polarised-cylinder.942173/#post-5959482 Griffiths works the problem for a cylinder oriented sideways, but no doubt this technique would also work for the uniformly polarized sphere.
 
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  • #12
And a follow-on: For those physics students with a keen interest in E&M, the solutions in the "links" of post 10 and post 11 might be of much interest. ## \\ ## IMO, for the magnetostatic problem, the magnetic surface current method really provides the better solution from a physics sense. The magnetic pole solutions are mathematically quite elegant, and can be shown to give the exact same answer for ## \vec{B} ##, but IMO, they hide and even confuse the underlying physics.
 
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  • #13
The underlying physics is that ferromagnetism comes from the polarization of the intrinsic magnetic moments of electrons, and not from electric currents.
 
  • #14
Meir Achuz said:
The underlying physics is that ferromagnetism comes from the polarization of the intrinsic magnetic moments of electrons, and not from electric currents.
I disagree with this interpretation. Many who were taught the magnetic "pole model" subscribe to this. The pole model uses the equation ## B=\mu_o H+M ##, and assigns a "local" contribution to ## B ## that results from the magnetization ## M ##. It also treats ## H ## and ## B ## as two different types of magnetic fields. The magnetic surface current method is able to avoid these (magnetic pole) explanations, which I believe are incorrect. In the magnetic surface current explanation, the ## H ## is shown to be a mathematical construction, and does not represent an actual magnetic field. In addition, the ## M ## in the equation ## B=\mu_o H+M ## is shown to come from the contribution of the surface currents that result from any finite distribution of magnetization ## M ##. In the absence of currents in conductors, for the magnetic field ## B ## in the material, the ## H ## is simply a correction that appears for shapes other than cylinders of infinite length. ## \\ ## Outside the material, the ## H ## from the magnetic poles does give the magnetic field strength ## B ##, (depending on the system of units there may be a proportionality constant), but the reasons why this happens to work are open to debate. For any uniformly magnetized cylinder of finite length, (outside the material), the ## H ## from the poles (using inverse square law) completely agrees with the ## B ## from the magnetic surface currents (using Biot-Savart)... Inside the material (inside the cylinder), a local ## M ## needs to be added artificially to the ## B ##. This is where the pole model, IMO, fails to explain the underlying physics. In the magnetic surface current explanation, the contribution to ## B ## in the amount of ## M ## actually comes from the surface currents. ## \\ ## The uniformly magnetized cylinder of finite length can be considered to be the building block for any arbitrary distribution of magnetization. If we look at the building block, we can say the ## B ## needs to have added to the ## H ## an amount of value ## M ## to give the ## B ## its proper continuity etc., and that there may, in fact, be a "local" contribution of ## M ## to ## B ##. This certainly could be a possibility. The alternative would be that the agreement between magnetic pole model and magnetic surface current model is simply mathematical coincidence.
 
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  • #15
No, @Meir Achuz is right. There's not only fundamental electric charge but also fundamental magnetic dipole moments. Ferromagnetism on the microscopic level is a quantum phenomenon (exchange interaction of electrons and spontaneous symmetry breaking).
 
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  • #16
vanhees71 said:
No, @Meir Achuz is right. There's not only fundamental electric charge but also fundamental magnetic dipole moments. Ferromagnetism on the microscopic level is a quantum phenomenon (exchange interaction of electrons and spontaneous symmetry breaking).
I agree that the magnetic dipole is fundamental. My consideration here is with solving the macroscopic problem. In Grififith's E&M textbook, he begins with an arbitrary distribution of microscopic magnetic dipoles, and he derives an equation for the vector potential ## A ## that shows there is an "equivalent" magnetic surface current that can be considered as the source of this vector potential. ## \\ ## Even though I know that it works, and I have even proven that it works, the complete prescription for the magnetic pole method, with the equation ## B=\mu_oH+M ##, is not completely intuitive for me.
 
  • #17
Of course, concerning the macroscopic theory you are right. There we average out all quantum effects and have no clue about the inner workings of the microscopic consistuents.

Here, we obviously consider a "hard ferromagnet", i.e., we describe it as something with a constant magnetization ##\vec{M}(\vec{x})## (with the magnet at rest of course). We have a case of magnetostatic then, i.e.,
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{H}=\vec{j}_{text{free}}=0.$$
The constitutive equation in this case simply is
$$\vec{B}=\mu_0 (\vec{H}+\vec{M}),$$
where I use the standard definition of ##\vec{M}## concerning factoring out ##\mu_0##, used in the SI (which is always very confusing, but that's due to historical misunderstandings and the inertia of physicists to make consistent definitions after gaining new knowledge, but that's a minor obstacle).

There are now two ways to proceed to solve the above equations. The mathematical more convenient one starts from the curl-freeness of ##\vec{H}##, which allows to introduce a magnetic potential,
$$\vec{H}=-\vec{\nabla} \phi_m.$$
From ##\vec{\nabla} \cdot \vec{B}=0## we find
$$\vec{\nabla} \cdot \vec{H}=-\Delta \phi_m=-\vec{\nabla} \cdot \vec{M}=\rho_m.$$
Here ##\rho_m## is an effective "magnetic charge". If ##\vec{M} \simeq \text{const}## what you get in fact is a magnetic "surface charge", but that's a bit cumbersome to treat in general (the div on the rhs. of the previous equation gives a ##\delta##-distribution like singularity with support along the surface, but that's indeed not very convenient to evaluate for general shapes). That's why we use a trick since we solve our equations anyway by integrals.

Using the Green's function of the D'Alembert operator we have
$$\phi_m(\vec{x})= \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho_m(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}. \qquad (1)$$
Now we can write
$$\frac{\rho_m(\vec{x}')}{|\vec{x}-\vec{x}'|}=-\frac{\vec{\nabla}' \cdot \vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|} = \vec{M}(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}-\vec{\nabla}' \frac{\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Plugging this into the previous equation, the integral over the total divergence vanishes due to Gauss's theorem (with the boundary surface at inrinity and a magnet of finite extent). This leads to
$$\phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}') \cdot (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}=-\vec{\nabla} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
In the final step we have used
$$\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|}=-\vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}$$
and than carried the nabla wrt. ##\vec{x}## out of the integral.

I think this is what you call the "magnetic-pole method". As the derivation shows there are of course no real magnetic (mono)poles, but ##\vec{M}## occurs as a magnetic-dipole distribution with its potential, as can also immediately seen from the 1st expression for ##\phi_m##.

The other method is to use ##\vec{\nabla} \cdot \vec{B}##, which is generally valid (in contradistinction to the pole method which is restricted to the here discussed static case with vanishing free currents) and introduce the vector potential
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Since there's gauge freedom we can make life easier by imposing the Coulomb-gauge condition (almost always most convenient in the static case),
$$\vec{\nabla} \cdot \vec{A}=0.$$
Then
$$\vec{\nabla} \times \vec{B} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A}.$$
On the other hand you have
$$\vec{\nabla} \times \vec{B} = \mu_0 \vec{\nabla} \times(\vec{H}+\vec{M}) = \mu_0 \vec{\nabla} \times \vec{M}=\mu_0 \vec{j}_m.$$
Here for ##\vec{M} \simeq \text{const}## you get an effective surface current.

[Note in proof:]
It's related to the boundary condition
$$\vec{k}_m=\frac{1}{\mu_0} \vec{n} \times (\vec{B}_>-\vec{B}_<),$$
i.e., the jump of the magnetic field across the boundary of the magnet.
 
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  • #18
vanhees71 said:
The constitutive equation in this case simply is

→B=μ0(→H+→M),B→=μ0(H→+M→),​

\vec{B}=\mu_0 (\vec{H}+\vec{M}),
I don't know that I am willing to accept this as a constitutive equation, without proof. Biot-Savart's equation follows from Maxwell's ## \nabla \times B=\mu_o J_{total} ##. I do of course accept Maxwell's equations as constitutive.## \\ ## The equation ## B=\mu_o (H+M) ## can be shown to be a result of Maxwell's equations and Biot-Savart, (as I myself have done independently). I do think they would do well to derive ## B=\mu_o (H+M) ## in the textbooks from Biot-Savart. ## \\ ## Some textbooks base the equation ## B=\mu_o(H+M) ## on the analogous ## D=\epsilon_o E+P ##, (there are different definitions for ## M ##: Some books use ## B=\mu_o H+M ## ), but I think this is a little bit of a stretch. ## \\ ## I should add here what ## H ## represents is this equation: it is a combination of what is basically the ## B ## (divided by ## \mu_o ##), from currents in conductors, along with a magnetic pole contribution=fictitious magnetic charge density ## \rho_m=-\nabla \cdot M ##, with ## H ## computed from the inverse square law. The "local" contribution to ## B ## in the amount of ## \mu_o M ##, when derived from Biot-Savart and the surface currents, turns out to be a non-local contribution, with ## H ## from the poles as a correction factor for geometries other than long cylinder. ## \\ ## One correction in @vanhees71 's last line above : Surface current per unit length ## \vec{K}_m=\vec{M} \times \hat{n} ##.
 
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  • #19
"In addition, the MM in the equation B=μoH+MB=μoH+M is shown to come from the contribution of the [surface currents] that result from any finite distribution of magnetization M."
That is NOT CORRECT in a permanent ferromagnet.
There is a term (in derivations for ##{\bf A}##) that reads
$${\bf A(r')=\int} d^3r\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$
That looks and acts like a surface current, but IS NOT an actual surface current for a permanent ferromagnet.
Ferromagnetism comes from the intrinsic dipole moments of electrons in the iron atom.
It is this misunderstanding that has you confused.
Actually neither method is 'wrong'. Each is a useful mathematical method of finding B and H for a permanent ferromagnet. I find the dipole method easier if it is not confounded by SI.
 
  • #20
Meir Achuz said:
"In addition, the MM in the equation B=μoH+MB=μoH+M is shown to come from the contribution of the [surface currents] that result from any finite distribution of magnetization M."
That is NOT CORRECT in a permanent ferromagnet.
There is a term (in derivations for ##{\bf A}##) that reads
$${\bf A(r')=\int} d^3r\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$
That looks and acts like a surface current, but IS NOT an actual surface current for a permanent ferromagnet.
Ferromagnetism comes from the intrinsic dipole moments of electrons in the iron atom.
It is this misunderstanding that has you confused.
Actually neither method is 'wrong'. Each is a useful mathematical method of finding B and H for a permanent ferromagnet. I find the dipole method easier if it is not confounded by SI.
I think we all (@vanhees71 included here) have a pretty good understanding that the magnetic surface currents are more of a mathematical resultant of a distribution of microscopic magnetic dipoles than an actual current that can be measured with an ammeter. We have previously discussed this in detail in other threads.
 
  • #21
I apologize for restating the obvious. I guess I was still replying to your first three posts without looking in detail at later posts.
 
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  • #22
Meir Achuz said:
I apologize for restating the obvious. I guess I was still replying to your first three posts without looking in detail at later posts.
@vanhees71 is our in-house E&M expert. I like to bounce ideas off of him=sometimes I get it right, and other times he makes corrections to my inputs. In any case, I do think they would do well to place more emphasis on some of the fundamental magnetostatics problems in the present physics curriculum for the undergraduate students. It is only once in a great while these days that I see a homework post on Physics Forums that addresses some of these very fundamental magnetostatic calculations, such as the one in this thread.
 
  • #23
Charles Link said:
I don't know that I am willing to accept this as a constitutive equation, without proof. Biot-Savart's equation follows from Maxwell's ## \nabla \times B=\mu_o J_{total} ##. I do of course accept Maxwell's equations as constitutive.## \\ ## The equation ## B=\mu_o (H+M) ## can be shown to be a result of Maxwell's equations and Biot-Savart, (as I myself have done independently). I do think they would do well to derive ## B=\mu_o (H+M) ## in the textbooks from Biot-Savart. ## \\ ## Some textbooks base the equation ## B=\mu_o(H+M) ## on the analogous ## D=\epsilon_o E+P ##, (there are different definitions for ## M ##: Some books use ## B=\mu_o H+M ## ), but I think this is a little bit of a stretch. ## \\ ## I should add here what ## H ## represents is this equation: it is a combination of what is basically the ## B ## (divided by ## \mu_o ##), from currents in conductors, along with a magnetic pole contribution=fictitious magnetic charge density ## \rho_m=-\nabla \cdot M ##, with ## H ## computed from the inverse square law. The "local" contribution to ## B ## in the amount of ## \mu_o M ##, when derived from Biot-Savart and the surface currents, turns out to be a non-local contribution, with ## H ## from the poles as a correction factor for geometries other than long cylinder. ## \\ ## One correction in @vanhees71 's last line above : Surface current per unit length ## \vec{K}_m=\vec{M} \times \hat{n} ##.
These are constitutive equations. They follow from microscopic electrodynamics (many-body QFT) in terms of linear-response (Green-Kubo) theory, and they are very much simplified, but amazingly successful to describe a lot of phenomena (at least) qualitatively right.

To get the systematics straight you usually put all "free charges, currents and magnetizations" in the equations for the field components ##\vec{D}## and ##\vec{H}##. The only idiosyncrazy of the common convention is how the susceptibilities (or in the most simple approximation with just constant permittivities and permeabilities) are distributed. It's in fact a convention, but systematically was to flip ##\mu## to the other side, but this would confuse people more than anything, and that's why one leaves it as is.

The most "physical system of units" are Heaviside-Lorentz units. There you have the Maxwell equations

$$\vec{\nabla} \times \vec{B} +\frac{1}{c} \partial_t \vec{B}=0,\\
\vec{\nabla} \cdot \vec{B}=0 \\
\vec{\nabla} \times \vec{H} -\frac{1}{c} \partial_t \vec{D} = \frac{1}{c} \vec{j}_{\text{free}},\\
\vec{\nabla} \cdot \vec{D}=\rho.$$
The constitutive equations in the most simple form then read
$$\vec{D}=\epsilon \vec{E}, \quad \vec{B}=\mu \vec{H}.$$

Then I'm a bit puzzled about your remarks concerning the magnetic current. For the discussed case you have (in SI units) always
$$\vec{j}_m=\vec{\nabla} \times \vec{M}.$$
If you have a body in a region ##V## and you assume a homogeneous isotropic ##\vec{M}=\text{const}## across ##V## (it's ##\vec{M}(\vec{x})=0## for ##\vec{x} \notin V## of course). Then ##\vec{j}## becomes a Dirac ##\delta##-distribution with support on the boundary ##\partial V## of the body. That's effectively a surface current.

Take the example of a sphere of radius ##a##. Then you have
$$\vec{M}(\vec{x})=M \Theta(a-r) \vec{e}_3, \quad r=|\vec{x}|.$$
The curl is (using spherical coordinates when convenient)
$$\vec{j}_m=\vec{\nabla} \times \vec{M} = -M \vec{e}_3 \times \vec{\nabla} \Theta(a-r)=M \vec{e}_3 \times \vec{e}_r \delta(a-r)=M \sin \vartheta \vec{e}_{\varphi} \delta(r-a).$$
This means there's a magnetic surface current
$$\vec{k}_m=M \sin \vartheta \vec{e}_{\varphi}$$
along the surface of the sphere.

From
$$\Delta \vec{A}=-\mu_0 \vec{j}_m$$
you get
$$\vec{A}(r)=\int_{0}^{\pi} \mathrm{d} \vartheta' \int_{0}^{2 \pi} \mathrm{d} \varphi' a^2 \sin \vartheta \frac{\vec{k}_m(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Here $$\vec{x}'=a (\cos \varphi' \sin \vartheta',\sin \varphi' \sin \vartheta ',\cos \vartheta').$$
Of course it's more convenient to solve the differential equation in spherical coordinates using the obvious ansatz
$$\vec{A}(\vec{x})=\vec{e}_{\varphi} A_{\varphi}(r,\vartheta),$$
which automatically fulfills the Coulomb gauge condition. Also be careful to use
$$-\Delta \vec{A}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A}),$$
because you must not naively calculation the Laplace operator of a vector field in non-Cartesian coordinates.

For the rest of the calculation see my (German) lecture notes, pp. 91ff

https://itp.uni-frankfurt.de/~hees/publ/theo2-l3.pdf
 
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  • #24
Meir Achuz said:
"In addition, the MM in the equation B=μoH+MB=μoH+M is shown to come from the contribution of the [surface currents] that result from any finite distribution of magnetization M."
That is NOT CORRECT in a permanent ferromagnet.
There is a term (in derivations for ##{\bf A}##) that reads
$${\bf A(r')=\int} d^3r\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$
That looks and acts like a surface current, but IS NOT an actual surface current for a permanent ferromagnet.
Ferromagnetism comes from the intrinsic dipole moments of electrons in the iron atom.
It is this misunderstanding that has you confused.
Actually neither method is 'wrong'. Each is a useful mathematical method of finding B and H for a permanent ferromagnet. I find the dipole method easier if it is not confounded by SI.
I'm not sure what you mean by this formula. I guess it's the following idea: From
$$\Delta \vec{A}=-\mu_0 \vec{\nabla} \times \vec{M}$$
you get
\begin{equation*}
\begin{split}
\vec{A}(\vec{x}) &=\mu_0 \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \times \vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|} \\
&=+ \mu_0 \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{M}(\vec{x}') \times \vec{\nabla}' \frac{1}{4 \pi |\vec{x}-\vec{x}'|} \\
&=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{M}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.
\end{split}
\end{equation*}
 
  • #25
It may be worthwhile for me to present a derivation of the equation ## B=\mu_o (H+M) ## that I did in 2012-2013 using Biot-Savart and ## \nabla \times M=J_m ## along with some vector identities. It is done for the case where there are no currents in conductors. The contribution from currents in conductors is simply an add-on to both sides of the equation. For the case at hand, ## \vec{H} ## comes from contributions from the magnetic poles with magnetic charge density ## \rho_m=-\nabla \cdot \vec{M} ##. The proof is somewhat lengthy, so let me put it in post 26. It might take me couple of tries to get the Latex right, because there are quite a few equations.
 
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  • #26
I'm going to use c.g.s. units for the derivation, and I will show that ## B=H+4 \pi M ##.
Let ## a=-\nabla \frac{1}{|x-x'|}=\frac{x-x'}{|x-x'|^3} ##.
Let ## b=M(x') ##. Then ## \nabla' \times b=\nabla' \times M(x')=\frac{J_m(x')}{c} ##.
Biot-Savart gives ## B(x)=\int (\nabla' \times b) \times a \, d^3x'=-\int a \times (\nabla' \times b) ##.
By a vector identity ## B=-\int [\nabla'(a \cdot b)-(b \cdot \nabla') a-(a \cdot \nabla') b-b \times \nabla' \times a] \, d^3 x' ##.
The first term integrates to zero because of a vector identity similar to Gauss' law that ## \int \nabla \psi \, d \tau=\int \psi \, dA ##, and for a finite sample size, ## b ## vanishes on the large outer surface that surrounds the magnetized sample.
The 4th term is zero because it is the curl of a gradient. This leaves the two middle terms.
The derivation is a little more lengthy than I anticipated, so let me just add a couple of key steps here, and I will try to fill in the pieces at a later time:
## \nabla' \cdot a=-4 \pi \delta(x-x') ##, and ## \nabla \cdot a=-\nabla' \cdot a=4 \pi \delta(x-x') ##.
Now ## b \cdot \nabla' a=\nabla' \times (a \times b)+b(\nabla' \cdot a)-a(\nabla' \cdot b)+(a \cdot \nabla')b ##.
Again, I hope to fill in all the steps=there are quite a number of them=but in any case, the result of all of this vector calculus is that ## B=H+4 \pi M ##.
It is worth mentioning one more step in this derivation: ## (a \cdot \nabla') b=(b \cdot \nabla')a-\nabla' \times (a \times b) -b (\nabla' \cdot a)+a (\nabla' \cdot b) ##.
This last substitution seems to be necessary, because in a later step ## (b \cdot \nabla')a=-(b \cdot \nabla)a ## (without a prime on the last gradient operator ) is used to evaluate this term, while evaluation of ## (a \cdot \nabla')b ## as is would be problematic.
Much of the rest is just evaluating the terms, and also determining that certain integrals equal to zero, because they equate to surface integrals, etc.
The editor is giving me problems again, so I think I will go to post 27 to add additional steps.
 
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  • #27
To add to post 26 (I already put this in post 26, but it lost it):
##\int (b \cdot \nabla)a \, d^3x'=\int [\nabla(a \cdot b)-(a \cdot \nabla)b-a \times \nabla \times b -b \times \nabla \times a] \, d^3x'##
##=\nabla \int(a \cdot b) \, d^3x' ##.
The last result follows because the gradient operation here is unprimed and operating on a term (##b ##) with a primed coordinate in the second and third terms. Meanwhile, the 4th term is the curl of a gradient, so that it is zero. Since the gradient operator on the first term is unprimed, and the integration is over the primed coordinate, the unprimed gradient can be pulled out in front of the integral.
The remaining steps that I haven't shown in this derivation are, for the most part simpler. I will show one more, and that is in evaluating ## \int (b \cdot a) \, d^3x' ##, the terms become
## M \cdot (\nabla' \frac{1}{|x-x'|})=\nabla' \cdot (\frac{M}{|x-x'|})-\frac{\nabla' \cdot M}{|x-x'|} ##, and the first term integrates to zero using Gauss' law.
Much of the rest is relatively straightforward. I would have liked to show it all, but it is quite a lot harder writing it out in Latex, than it is to write it out by hand.
 
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  • #28
To summarize posts 26 and 27, (I'm working in cgs there, but now going back to SI) starting with Biot-Savart, and ## \nabla \times M=J_m ##, and using a bunch of vector identities, the result follows that ## \vec{B}=\mu_o (\vec{H}+\vec{M}) ##.
Here ## \vec{H} ## is defined (cgs) as ## \vec{H}(x)=\int \frac{\rho_m(x')(x-x')}{|x-x'|^3} \, d^3x' ##, where ## \rho_m=-\nabla \cdot \vec{M} ##.
This derivation doesn't consider the ## H ## that is from currents in conductors, which simply gets tacked onto the above result (## B=\mu_o(H+M) ##) by redefining ## H ## to include the contribution from currents in conductors using Biot-Savart. ## \\ ##
From these results, it is apparent that, what is basically the result of the magnetic pole model, that ## B=\mu_o(H+M) ##, goes hand-in-hand with magnetic field calculations that are based on magnetic currents and Biot-Savart. It is open to debate which one is the more fundamental. ## \\ ## Additional item: The pole model equation ## B=\mu_o(H+M) ## follows from Biot-Savart and ## \nabla \times M=J_m ##.
I don't believe the converse would be the case. ## \\ ## And an additional comment: In grouping large numbers of the fundamental building block, which is the microscopic magnetic dipole, and making a macroscopic magnetized object, it would appear that in the mathematics, it can be treated as having magnetic charges, (regions of ## \rho_m=- \nabla \cdot M ##), or regions of magnetic currents, (where ## J_m=\nabla \times M ##). Perhaps in a sense, both are equally fictitious, but it seems necessary to employ one or the other, and to essentially treat it as real, in order to compute the magnetic field ## B ##.
@vanhees71 I think you might find the above algebraic derivation of interest.
 
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  • #29
Of course, the full Maxwell equations clearly say, it's "currents" rather than "monopoles". Of course, it's no problem to add also magnetic monopoles to the electromagnetic theory (see Dirac's famous paper), but it seems as if Nature doesn't realize this idea ;-)).
 
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  • #30
The posts are coming so fast, I can't read or respond to them, but I will reply to this
from Vanhees71:

"I'm not sure what you mean by this formula. I guess it's the following idea:
##\Delta \vec{A}=-\mu_0 \vec{\nabla} \times \vec{M}...##"

We are doing things in opposite directions. I develop the potential integral for magnetization, ##\bf M##, in the same way as the Coulomb integral is developed from Coulomb's law for a point charge q.

I start with the vector potential for a chunk, ##\Delta V,## of magnetized matter,
$$\Delta{\bf A}=\frac{{\bf M}\Delta V\times {\bf r}}{r^3},$$
where ##{\bf M}##, the magnetization is defined as the magnetic moment per unit volume.
I don't see any other way to give a physical definition of ##\bf M##.
Going to the integral limit gives
$${\bf A(r)}=\int\frac{{\bf M(r')\times r'}d^3r'}{\bf|r-r'|^3}.$$
Then, some vector manipulation leads to

$${\bf A(r)}=\int\frac{{\bf\nabla'\times M(r')}d^3r'}{\bf|r-r'|}+
\int\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$

This suggests that ##\bf\nabla\times M## behaves like, but IS NOT, a current; and that
##\bf M\times dS## behaves like, but IS NOT, a differential surface current.
 
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  • #31
I see, now your formulae make sense to me. The second integral, however is over the boundary of the integration volume. If it contains the whole body in its interior this vanishes. The only exception is if ##\vec{M}## itself contains a singular surface-magnetization piece. As your derivation shows, ##\vec{M}## is a volume density, namely magnetization per volume.
 
  • #32
What @Meir Achuz has is the same derivation that Griffith's shows in his E&M book=I believe it is in section 6.2. The numerator of the second term should be recognized as the surface current per unit length ## \vec{K}_m=\vec{M} \times \hat{n} ##. ## \\ ##
And I think it is rather fascinating that both a "magnetic charge" description and a "magnetic current" description are able to generate the exact same result for the magnetic field ## \vec{B} ##. The "magnetic charge" at the end faces of the macroscopic cylinder, and the "magnetic surface current" on the outer surface of the macroscopic cylinder are both rather fictitious, and are both the result of some vector calculus, but the magnetic field ## \vec{B} ## can be calculated in either manner.
 
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  • #33
I will try to correct some misinterpretations.
##{\bf H=B-}4\pi{\bf M}## is the definition of ##\bf H##.
A definition cannot be derived. ##\bf B## ALWAYS equals ##{\bf H+}4\pi{\bf M}##, for any possible model.

Biot and Savart proposed the "law of Biot-Savart", which had nothing to do with ##\bf M##.
 
  • #34
"What @Meir Achuz has is the same derivation that Griffith's shows in his E&M book."
That is why I was surprised when you seemed to contradict it, and Vanhees71 questioned my integral from Griffiths.
 
  • #35
@Meir Achuz I think if you read through the derivation in posts 26-28, you would find it of interest. What I am basically showing with this is that the macroscopic "pole" model of magnetism, and a magnetic current description, go hand-in-hand, even though they are so different mathematically. ## \\ ## In the pole (magnetic charge) model, often the equation ## B=H+4 \pi M ## is assumed set in stone because it is analogous to ## D=E+4 \pi P ##. ## \\ ## Alternatively, if you begin with a magnetic current description, ## \nabla \times M=\frac{J_m}{c} ## (cgs) and Biot-Savart, ## B(x)=\int \frac{J_m(x') \times (x-x')}{c|x-x'|^3} \, d^3x' ##, then, ## \\ ## defining ## H=B-4 \pi M ##, ## \\ ## the result is ## H =H_m+H_c## where ## \\ ## ## H_m(x)=\int \frac{\rho_m(x')(x-x')}{|x-x'|^3} \, d^3x' ##, with ## \rho_m=-\nabla \cdot M ##,
along with the contribution ## H_c ## from currents in conductors using Biot-Savart. ## \\ ##
That's why I actually prefer to derive the result ## B=H+4 \pi M ##, using the above as definitions for ## H_m ## and ## H_c ##, with ## H=H_m +H_c ##.
 
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