A magnetostatics problem of interest 2

[vanhees71]

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
$$\vec{A}(\vec{r})=\mu_0 \int_{\partial V} \mathrm{d}^2 f' \frac{\vec{M}(\vec{r}') \times \vec{n}'}{|4 \pi |\vec{r}-\vec{r}'|}.$$

 

[Charles Link]

@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 $J_m=c \nabla \times M$.

 

[Meir Achuz]

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

 

[Charles Link]

${\bf B=H+}4\pi{\bf M}$ 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.

 

[Meir Achuz]

Look at Eq. (5.81) in Jackson III.

 

[Charles Link]

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 !

 

[Meir Achuz]

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.

 

[Charles Link]

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]

I think any good textbook on CED has it. I learned 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.

 

[Charles Link]

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.

 

[Charles Link]

I came up with an additional problem, a takeoff on what was posed in the OP, that is the kind of thing that anyone with a good understanding of the magnetic pole method and the magnetic surface current method should be able to readily work:

Consider a cylinder of length $L$ and radius $a$ that has uniform magnetization along its axis of magnitude $M_o$. Find the magnetic field strength $B$ at the center of this cylinder.

Show by both the pole and surface current methods that it has value $B=M_o \cos{\theta}$, (in units where $B=\mu_o H+M$), where $\theta=\arctan(a/(L/2))$.It's a good simple exercise with the result that the pole method and surface current method do give identical results for the magnetic field $\vec{B}$, as they always do.

 

[aclaret]

hello! I find this interesting problem, and I try to solve it :)

Lay down cylindrical coordinate chart with origin $\mathscr{O}$ centre of cylinder. Magnetic quantities $M, H, B \in R^3$ do satisfy $B = \mu_0 H + M$ from elementary electromagnetism theory. Denote with $x' \in R^3$ the position vector over the domain of integration. By theorem asserted by user charleslink in posting #1, write$$H = \frac{1}{4\pi \mu_0} \int_{\partial V} \frac{-\langle M, n \rangle x'}{|x'|^3} dA$$where $\langle x, y \rangle$ is standard inner product on $R^3$. Because the cylindrical symmetry, assume on physical basis that $H = h(\rho) e_z$ depend only on radial coordinate $\rho$, for some function $h: R \rightarrow R$ and where $e_z$ is upward pointing basis vector.

The integration trivial ;) ;), clearly you write $\partial V = S_1 \cup S_2 \cup S_3$ where

$S_1 = \{(x,y,z) \in R^3: z = \mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ is top face,
$S_2 = \{(x,y,z) \in R^3: z = -\mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ lower face,
$S_3 = \{(x,y,z) \in R^3: -\mathscr{l} < z < \mathscr{l}, \sqrt{x^2 + y^2} = a \}$ side face,

then define $\mathscr{l} := L/2$,
$$H_z = - 2\times \frac{M_0 \mathscr{l}}{4\pi \mu_0} \int_{S_1} \frac{\rho}{(\rho^2 + \mathscr{l}^2)^\frac{3}{2}} d\rho d\phi = - \frac{M_0 \mathscr{l}}{\mu_0} \left[ \frac{1}{\sqrt{\mathscr{l}^2 + \rho^2}} \right]_0^a = \frac{M_0 \mathscr{l}}{\mu_0} \left(\frac{1}{\sqrt{\mathscr{l}^2 + a^2}} - \frac{1}{\mathscr{l}}\right)
$$Obviously follows that$$\langle e_z, \mu_0 H + M \rangle = M_0 \left(\frac{\mathscr{l}}{\mathscr{l}^2 + a^2} - 1 \right) + M_0 = M_0 \cos{\theta}, \quad \cos{\theta} = \mathrm{arctan}(a/\mathscr{l})$$

 

[Charles Link]

hello! I find this interesting problem, and I try to solve it :)

Lay down cylindrical coordinate chart with origin $\mathscr{O}$ centre of cylinder. Magnetic quantities $M, H, B \in R^3$ do satisfy $B = \mu_0 H + M$ from elementary electromagnetism theory. Denote with $x' \in R^3$ the position vector over the domain of integration. By theorem asserted by user charleslink in posting #1, write$$H = \frac{1}{4\pi \mu_0} \int_{\partial V} \frac{-\langle M, n \rangle x'}{|x'|^3} dA$$where $\langle x, y \rangle$ is standard inner product on $R^3$. Because the cylindrical symmetry, assume on physical basis that $H = h(\rho) e_z$ depend only on radial coordinate $\rho$, for some function $h: R \rightarrow R$ and where $e_z$ is upward pointing basis vector.

The integration trivial ;) ;), clearly you write $\partial V = S_1 \cup S_2 \cup S_3$ where

$S_1 = \{(x,y,z) \in R^3: z = \mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ is top face,
$S_2 = \{(x,y,z) \in R^3: z = -\mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ lower face,
$S_3 = \{(x,y,z) \in R^3: -\mathscr{l} < z < \mathscr{l}, \sqrt{x^2 + y^2} = a \}$ side face,

then define $\mathscr{l} := L/2$,
$$H_z = - 2\times \frac{M_0 \mathscr{l}}{4\pi \mu_0} \int_{S_1} \frac{\rho}{(\rho^2 + \mathscr{l}^2)^\frac{3}{2}} d\rho d\phi = - \frac{M_0 \mathscr{l}}{\mu_0} \left[ \frac{1}{\sqrt{\mathscr{l}^2 + \rho^2}} \right]_0^a = \frac{M_0 \mathscr{l}}{\mu_0} \left(\frac{1}{\sqrt{\mathscr{l}^2 + a^2}} - \frac{1}{\mathscr{l}}\right)
$$Obviously follows that$$\langle e_z, \mu_0 H + M \rangle = M_0 \left(\frac{\mathscr{l}}{\mathscr{l}^2 + a^2} - 1 \right) + M_0 = M_0 \cos{\theta}, \quad \cos{\theta} = \mathrm{arctan}(a/\mathscr{l})$$

Very good @aclaret :) Next, do it with magnetic surface currents and Biot-Savart, where magnetic surface current density per unit length $\vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o}$.

 

[aclaret]

aww thank :), I'm happy that you like it! Ok, I try the second method... luckily, I don't think it too hard either ;)

Let me employ the same chart, origin $\mathscr{O}$. I effect cylindrical basis $\mathscr{B} = (e_{\rho}, e_{\phi}, e_z)$ of $R^3$ for analysis. Obvious that $K_m = (M_0/ \mu_0) e_{\phi} \in R^3$, so by theorem of charleslink in posting #1, obtain$$B = \frac{M_0}{4\pi} \int_{S_3} \frac{-e_{\phi} \times x'}{|x'|^3} dA$$Because $e_{\phi} \times x' = -a e_z + z e_{\rho}$, and clearly by the symmetry $e_{\rho}$ component integrate to zero, so now effecting the elementary surface integration$$B = \frac{M_0 a^2}{4\pi} e_z \int_{S_3} (a^2 + z^2)^{-\frac{3}{2}} d\phi dz = M_0 a^2 \left [ \frac{\mathscr{l}}{a^2(a^2 + \mathscr{l}^2)} \right] = M_0 \cos{\theta}$$I think that even quicker than the first method! Thank for teaching me, I always like find as many ways to solve the problem :)

 

[Charles Link]

@aclaret The two methods always give the same answer for $B$. I don't think that result is obvious, (there are a couple others who commented in the thread above who think a proof is unnecessary), but in any case I was able to prove that from first principles. You might also find this Insights article of interest: https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/

 

[Charles Link]

@aclaret See also https://www.overleaf.com/read/kdhnbkpypxfk

 

[Charles Link]

nice! my em knowledge is from course in applied linear analysis, so i don't understand so much about physical principles of the problem. if i may, let me ask, does there exist physical significance of quantity $j_{M}$ ("magnetic current"...)? i think, look like just mathematical trick ;)

The magnetization $M$ gives the number of microscopic/atomic current loops circulating per unit volume. When $M$ is uniform, there is zero magnetic current density $j_m$. The $j_m$ arises in a way that you can picture in two dimensions with a chess or checkerboard, where each square has a current loop circulating counterclockwise on its outer perimeter. The currents in adjacent squares precisely cancel, and the net effect is a current circulating on the outside of the board. That is basically what the magnetic current density $j_m$ or magnetic surface current per unit length $K_m$ is all about. $j_m=\nabla \times M/\mu_o$, and $K_m=M \times \hat{n}/\mu_o$, basically a $j_m$ with Stokes theorem applied at the surface boundary.

 

[Charles Link]

@aclaret See https://www.physicsforums.com/threa...perature-relationship-in-ferromagnets.923380/ for an interesting experiment with the temperature effects of permanent magnets that a couple of students did. See also post 21 where I measured the magnetic field from a permanent magnet using a boy scout compass.