# I A magnetostatics problem of interest 2

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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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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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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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#### vanhees71

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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 ;-)).

#### Meir Achuz

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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.

#### vanhees71

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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.

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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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#### Meir Achuz

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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 $\bf M$.

#### Meir Achuz

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"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.

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@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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#### vanhees71

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Argh! Now I see, where my misunderstanding comes from! In Griffiths for the homogeneously magnetized spherical shell, he writes
$$\vec{j}_m=0, \vec{K}_m=\vec{M} \times \hat{n},$$
where $\hat{n}$ is the surface-unit normal vector along the boundary of the magnetized body.

In my interpretation it's included in $\vec{\nabla} \times \vec{M}$, because I take the surface current into account in terms of a Dirac-$\delta$-like contribution from generalized derivatives.

Both methods are equivalent, because the surface contribution can be also defined in terms of the socalled (in German) "Sprungrotation" (I don't know an expression in English for it, literally translated it means "jump curl"). The definition is hard to describe without a figure. So here it is

The blue area is the surface across which a vector field $\vec{V}$ may have singularities (in our case of a homogeneously magnetized body, the magnetization $\vec{M}$ has a jump from some finite value inside and 0 outside the body along its boundary).

Now you define the "Sprungrotation" $\text{Curl} \vec{V}$ along the surface by drawing the red curve with an arbitrary tangent-unit vector $\vec{t}$ along the surface and taking the limit of the red curve of contracting it to the point on the surface you want the operator $\text{Curl}$ define at, which goes as follows
$$\Delta h (\vec{t} \times \vec{n}) \cdot \text{Curl} \vec{V} = \Delta h \vec{t} \cdot (\vec{n} \times \text{Curl} \vec{V})= \Delta h \vec{t} \cdot (\vec{V}_2-\vec{V}_1).$$
Since this construction holds for all tangent unit vectors $\vec{t}$ along the boundary surface at the given point, you get
$$\vec{n} \times \text{Curl} \vec{V}=(\vec{V}_2-\vec{V}_1).$$
The two parts of the path along $\vec{n}$ don't contribute because they cancel in the limit, because there's only a jump across but not along the surface. Thus you have $\mathrm{Curl} \vec{V} \cdot{n}=0$ by definition and thus one gets
$$\vec{n} \times (\vec{n} \times \text{Curl} \vec{V})=-\text{curl} \vec{V} = \vec{n} \times (\vec{V}_2 -\vec{V}_1).$$
The surface current for our case of the homogeneously magnetized sphere (taking region 1 as the exterior, region 2 as the interior of this sphere) thus is
$$\vec{k}_m=\mathrm{Curl} \vec{M} = \vec{n} \times (\vec{M}_1-\vec{M}_2) = \frac{\vec{r}}{a} \times (-M \vec{e}_3)=M \cos \vartheta \vec{e}_{\varphi}.$$
So we can write the solution for the vector potential indeed as
$$\vec{A}(\vec{r})=\mu_0 \int_{\partial V} \mathrm{d}^2 f' \frac{\vec{M}(\vec{r}') \times \vec{n}'}{|4 \pi |\vec{r}-\vec{r}'|}.$$

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@vanhees71 I welcome your input on posts 33-35. $\\$ Also @Meir Achuz I welcome your input to the following: $\\$ Although $H=B-4 \pi M$ is used in some textbooks as the definition, I don't think they should cast it in stone like they do. If they define $H$ by this equation, then they need to show what other equations the entity $H$ obeys. $\\$ Other textbooks will define the $H=H_m +H_c$, but so many fail to tie the pieces together, and then simply assume $B=H+4 \pi M$ without proof. $\\$ Alternatively, the textbooks that define $H=B-4 \pi M$, then use the results (basically the alternative definition) of the other textbooks for $H=H_m+H_c$ without proof. $\\$ In my derivation of posts 26-28, I tie the pieces together. Defining $H_m$ and $H_c$ accordingly, I show/prove that the result $B=H+4 \pi M$ then follows from Biot-Savart along with $J_m=c \nabla \times M$.

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#### Meir Achuz

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${\bf B=H+}4\pi{\bf M}$ IS cast in stone. There is no place where it is not true.
Name one "Other textbooks".

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${\bf B=H+}4\pi{\bf M}$ IS cast in stone. There is no place where it is not true.
Name one "Other textbooks".
J.D. Jackson for one. In general, his Classical Electrodynamics is amazing, but IMO, he never proves, at least to my satisfaction, the magnetostatic equations that he employs. Other E&M textbooks I have worked extensively with are Pugh and Pugh, and Schwarz. $\\$ And I agree, $B=H+4 \pi M$ can be cast in stone, because I have independently proven it . Besides the proof in posts 26-28 that I did in 2012-2013, I also did another proof of it in 2009-2010, that I have previously linked on Physics Forums in other threads. See https://www.overleaf.com/read/kdhnbkpypxfk Until I succeeded in proving it with this proof in 2009-2010, I was very much unwilling to "cast this equation in stone", even though I had to accept it as gospel to pass the exams of the E&M courses that I had in 1975-1980. $\\$
The textbooks I have seen have not proven this equation in a manner that I was satisfied with. $\\$
I have to believe that most likely the physicists around 1880-1900 did a couple of very thorough proofs of the equation $B=H+4 \pi M$, and thereby it became cast in stone ever since, but somehow the modern day textbooks have omitted and/or lost some of the details.

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#### Meir Achuz

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Look at Eq. (5.81) in Jackson III.

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Look at Eq. (5.81) in Jackson III.
Unfortunately, I loaned out my copy of J.D. Jackson about 5 years ago, and never got it back. From what I remember though, J.D. Jackson uses a magnetic potential $\Omega$, and doesn't even introduce magnetic currents and magnetic surface currents. What I found lacking in his magnetic potential case is that although $\nabla^2 \Omega=0$, and $H=-\nabla \Omega$, (because $\nabla \cdot H=0$), there is no indication that any boundary conditions will be satisfied. That is why I took it upon myself to construct the proof in 2009-2010 that starts with a very simple case, and proves the boundary condition is satisfied. I then took it a couple of steps further, and proved $B=H+4 \pi M$ in general. $\\$ For example, if you take the simplest macroscopic case, which is a uniformly magnetized cylinder of semi-infinite length, (with the single end face at $z=0$ in the x-y plane), and you compute the potential $\Omega$ from the single pole end face, what guarantee is there that the $H=-\nabla \Omega$ that is computed will indeed be the correct one? i.e. that it gets the same answer as the surface current computation of the same scenario. Anyway, perhaps the textbooks successfully did this, but they did not do it to my satisfaction... I was skeptical until I actually rigorously proved it... $\\$ And to add to this, for this semi-infinite cylinder, $B_z =0$ by inspection by the pole model everywhere in the x-y plane except for $r<a$. When I did the calculation of this in 2010 with the surface currents, I was thinking I might even find it to be inconsistent with the pole model result. Instead, surprisingly, there was complete agreement !

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#### Meir Achuz

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1. Why do you keep 'proving' a definition?
2. There is 'complete agreement' because they are both correct calculations.
3. Your memory about Jackson is skewed. On page 196, just after using the scalar potential, he treats the vector potential with magnetic volume and surface currents. This comes after his definition of H=B/muzero-M on
page 192.

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1. Why do you keep 'proving' a definition?
2. There is 'complete agreement' because they are both correct calculations.
3. Your memory about Jackson is skewed. On page 196, just after using the scalar potential, he treats the vector potential with magnetic volume and surface currents. This comes after his definition of H=B/muzero-M on
page 192.
I don't have a good answer in responding to the above. Perhaps I had an older version of J.D. Jackson's book=the edition I had used cgs. And yes, they are correct calculations, but IMO the textbooks never seem to prove the pole formalism gives the same result for $B$ as the magnetic current formalism. Again, perhaps I missed something when I studied a couple of different E&M textbooks. I'd be interested in hearing @vanhees71 opinion, on whether the textbooks properly treat the equivalence of the $\vec{B}$ calculated by both methods. For the most part, the textbooks I have seen usually presented the pole method or the current method, but didn't tie the two together. Perhaps a later version of J.D. Jackson did precisely that. And again, I no longer have access to the edition that I had studied from=it is possible I overlooked portions of the discussion.

#### vanhees71

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I think any good textbook on CED has it. I learnt CED from Sommerfeld's textbook. Jackson has it for sure too.

I also don't think that there's a contradiction between my formulae and @Meir Achuz 's. It's only that I took the surface distribution into account by using derivatives in the sense of distributions, i.e., a Heaviside unitstep function gives a Dirac $\delta$ when its derivative is taken. You can of course also use the "Sprungrotation" (I'm still unsure how you call this operation in English, maybe "Surface Curl"?). Both methods are equivalent. The "Sprungrotation method", however, is more physical.

Of course, both magnetostatic methods are treated in the standard textbooks, i.e., a scalar potential for $\vec{H}$ in the absence of free currents and the vector potential for $\vec{B}$. Of course the latter approach is more natural since it's valid in the general case. That's why usually it is preferred compared to the magnetic-scalar-potential approach. Of course, both approaches are completely equivalent.

Right now I'm trying to continue to write on my SRT FAQ article for PF. It's amazing, how much knowledge about the relativistic formulation of classical electrodynamics has been lost with the decades since von Laue and Minkowski.

I also come from relativistic many-body QFT, where QED is the perfect playground to try out ideas. It's also a quite hot topic in relativistic heavy-ion collisions and related astro-nuclear physics (neutron stars/kilonovae and gravitational wave signals!). Now that for about 10-20 years relativistic viscous hydrodynamics is under control, also a renewed interest in relativistic magnetohydrodynamics has become a focus of current research. So I think, it's high time to review the macroscopic relativistic electrodynamics with modern tools, including relativistic kinetic theory derived from the Schwinger-Keldysh real-time-contour formulation. Particularly also vorticity in hydro- as well as magnetohydrodynamics still is a quite hot topic with a lot of new attention in hour heavy-ion community.

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Very good post above @vanhees71. Perhaps the E&M textbooks do, in fact, have a complete treatment of both the magnetic pole method and magnetic surface current methods. In that case, the extra calculations that I did that I mentioned in posts 26-28 and post 39 may have been unnecessary, and the results that I obtained may have been very predictable.$\\$ For me, though, I have to work with what I got out of reading and studying the textbooks=and I did put an enormous effort into the E&M studies= in two advanced undergraduate E&M courses as well as two graduate E&M courses=for me the results of my calculations, particularly the bunch in 2009-2010, came as a surprise=I did not expect that the magnetic surface current method of calculation would yield an identical result to that of the magnetic pole method. $\\$ It has been my experience that very few physics people really have a good handle on the magnetostatics subject. A couple years ago, I showed a couple of these calculations to a physics professor at a major university, and he told me that he had seen these types of calculations in graduate school, but he didn't completely understand the subject at that time, and by now he has forgotten most of it. $\\$ I may have taken some extra, perhaps unnecessary, steps to get a good handle on the subject, but for me, it took these extra steps to sort it all out.

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