# Why div B = 0 but div H not necessary 0?

yungman
I am referring specificly to page 273 6.3.2 A Deceptive Parallel in the "Introduction to Electrodynamics" 3rd edition by David Griffith that I am sure most of you have. The book said:

1) $\nabla \cdot \vec B = 0 \hbox { does not imply } \nabla \cdot \vec H = 0$

2) $\nabla \cdot \vec H = \nabla \cdot \frac{1}{\mu_0} \vec B - \nabla \cdot \vec M = -\nabla \cdot \vec M$.

$\hbox { 3) Only } \nabla \cdot \vec M = 0 \Rightarrow \vec B = \mu_0 \vec H$.

My questions are:

1) When is $\nabla \cdot \vec M \hbox { not equal to } 0$ ?

2) Is it true that all diamagnetic and paramagnetic material, $\vec M \hbox { is parallel to } \vec B_{ext}$ ? Or the material has to be linear and isotropic on top of dia and paramagnetic?

3) Sounds like to me only the ferromagnetic material that $\vec M$ is not parallel to $\vec B_{ext}$ until all domains are lined up with the external magnetic field?

4) what are the example of non-isotropic material?

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Mentor
1) When is $\nabla \cdot \vec M \hbox { not equal to } 0$ ?

Consider a bar magnet which is uniformly magnetized parallel to its length. $\nabla \cdot \vec M \ne 0$ at the ends of the magnet.

yungman
Consider a bar magnet which is uniformly magnetized parallel to its length. $\nabla \cdot \vec M \ne 0$ at the ends of the magnet.

That is also the part I am confuse.

$$\vec M = \frac{\sum^n_{k=1} \vec m_k}{\Delta v} \hbox { and } \vec m = \hat n IS$$

Where n is number of $\vec m$ in $\Delta v$. and $\hat n$ is unit normal of the oriented surface of $\vec m$ and I is the current generate by the electrol circulating in the dipole. S is the area of the dipole.

In reality, the total flux coming out of the end of the magnet is not straight out forever, it is like magnetic field that bend and circulate back to the end of the magnet which result of

$$\nabla \cdot \vec B = 0$$.

But we just treat $\vec M$ as vector pointing straight out.

Yes if you treat $\vec M$ just point straight out and never curl back like the magnetic field, then divergence of M is not zero. But this is not magnetic field anymore. If that is your argument, any magnetic material in the presence of external magnetic field $\nabla \cdot \vec H$ will never be zero. When is $\nabla \cdot \vec M$ ever be zero?

Gold Member
M represents the magnetization of a material. The divergence of M could be non-zero when the magnetization is non-uniform. So if we have a problem where the permeabiltiy is homogeneous, then we can always have zero divergence of the magnetization. The trivial example would be a non-magnetic material in free-space but this is trivial because M is zero. But of course if we have materials that have the same permeabiltiy then there is no spatial dependence on the permeability and we then know that the divergence of M must be zero since we can pull out the \mu from Gauss' Law.

In terms of magnetization this just means that the combined strength, number and orientation of the polarizable magnetic dipoles in the material remains constant.

EDIT: M does not curl back for the same reasons that a magnetic field does. M is not directly a magnetic field but it is the polarization of the magnetic moments inside of a material. Disregarding any inherent magnetic properties of a material, normally the microscopic moments are randomly oriented. Only when we apply an external field do some of these moments become polarized. When they are polarized, the applied field rotates the moments or induces an additional magnetic moment. Thus, the net induced moments are aligned along the direction of the applied field. So M only curls if the material is inhomogeneous or if the applied field is curling.

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dgOnPhys
...
My questions are:

1) When is $\nabla \cdot \vec M \hbox{ not equal to } 0$ ?

2) Is it true that all diamagnetic and paramagnetic material, $\vec M \hbox { is parallel to } \vec B_{ext}$ ? Or the material has to be linear and isotropic on top of dia and paramagnetic?

3) Sounds like to me only the ferromagnetic material that $\vec M$ is not parallel to $\vec B_{ext}$ until all domains are lined up with the external magnetic field?

4) what are the example of non-isotropic material?

4. Crystal in general have anisotropic properties (depending on their symmetries). In general for a non-ferromagnetic crystal, the relation between field H and magnetization is linear but not scalar: susceptibility is not a number but a (symmetric) matrix.

Not sure about your other questions (it has been too long since I studied this stuff). If it helps you can think of $-\nabla \cdot \vec M = \rho_{b}$, as density of bound magnetic charge... so in most cases it is hard to have a piece of volume looking like it has an effective magnetic (monopole) charge. It gets easier at the surface where bound magnetic charge is $\vec M \cdot \hat{\vec n}$ (which is basically the divergence on a 2D boundary) so a possible answer to your first question is: at the surface of a magnetized material where the magnetization field is not parallel to the surface.

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granpa
Why div B = 0 but div H not necessary 0?
I am referring specificly to page 273 6.3.2 A Deceptive Parallel in the "Introduction to Electrodynamics" 3rd edition by David Griffith that I am sure most of you have. The book said:
$\nabla \cdot \vec B = 0 \hbox { does not imply } \nabla \cdot \vec H = 0$

its treating magnetization as though it were the result of dipoles. (pairs of magnetic monopoles)

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dgOnPhys
its treating magnetization as though it were the result of dipoles. (pairs of magnetic monopoles)

Mentor
Electrons have an intrinsic ("spin") magnetic dipole moment of about $-9.3 \times 10^{-24}$ J/T. No magnetic monopoles are involved.

dgOnPhys
Electrons have an intrinsic ("spin") magnetic dipole moment of about $-9.3 \times 10^{-24}$ J/T. No magnetic monopoles are involved.
Guys, cheer up, nobody is talking about free monopoles, I am talking about equivalent bound sources. It is common practice in (good) text-books to point out how magnetization, the volumetric density of magnetic dipoles, is equivalent to
a distribution of magnetic charge of volumetric density $-\nabla \cdot \vec M$ and surface density $\hat{\vec n} \cdot \vec M$,
OR alternatively (in most textbooks)
a distribution of electric currents of volumetric density $\nabla \times \vec M$ and surface current density $\vec M \times \hat{\vec n}$

(Same is true for polarization and electric fields, one can use magnetic currents as sources)

E.g. take a cylinder with homogeneous magnetization parallel to its z-axis, you can think of the sources of magnetization either as superficial currents along the azimuthal direction or as two distribution of magnetic charge (of opposite sign) on the two cylinder bases.

Better this way?

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granpa
It is common practice in (good) text-books to point out how
magnetization, the volumetric density of magnetic dipoles,
is equivalent to
a distribution of magnetic charge​
of volumetric density $-\nabla \cdot \vec M$ and surface density $\hat{\vec n} \cdot \vec M$,​

OR alternatively (in most textbooks)

a distribution of electric currents
of volumetric density $-\nabla \times \vec M$ and surface current density $\vec M \times \hat{\vec n}$​

True, but they are only the same for the external field.

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yungman
Guys, cheer up, nobody is talking about free monopoles, I am talking about equivalent bound sources. It is common practice in (good) text-books to point out how magnetization, the volumetric density of magnetic dipoles, is equivalent to
a distribution of magnetic charge of volumetric density $-\nabla \cdot \vec M$ and surface density $\hat{\vec n} \cdot \vec M$,
OR alternatively (in most textbooks)
a distribution of electric currents of volumetric density $-\nabla \times \vec M$ and surface current density $\vec M \times \hat{\vec n}$

(Same is true for polarization and electric fields, one can use magnetic currents as sources)

E.g. take a cylinder with homogeneous magnetization parallel to its z-axis, you can think of the sources of magnetization either as superficial currents along the azimuthal direction or as two distribution of magnetic charge (of opposite sign) on the two cylinder bases.

Better this way?

You have to forward slash the itex.

yungman
Are dia and paramagnetic materials consider homogeneous because $\vec M$ is so weak that

$$\vec B = \mu_0 \vec H$$

We only have to concern with ferromagnetic material?

dgOnPhys
You have to forward slash the itex.

Thx I had not checked the posting

dgOnPhys
True, but they are only the same for the external field.

Well, it has been a while but are you sure about that? can you provide an example?

granpa
a uniformly magnetized infinitely long bar magnet

Guys, cheer up, nobody is talking about free monopoles, I am talking about equivalent bound sources. It is common practice in (good) text-books to point out how magnetization, the volumetric density of magnetic dipoles, is equivalent to

The magnetization cannot be defined as a density of magnetic dipoles in general. The problem is that in quantum mechanics, one cannot separate unambiguously one dipole from the other. In older classical electrostatics on the other hand, it was not unusual to define the strength of a dipole via the strength of two monopoles of opposite charge at unit distance.
Nowadays magnetization is simply a restatement of charge conservation as $$\nabla j_\mathrm{int}=0$$ in magnetostatics, it follows that j _int(the current density due to the atomic currents inside the medium) can be expressed as $$-\nabla \times M$$, the minus sign being convention. This equation defines M only up to the gradient of a scalar field $$\grad \xi$$. So it would be possible to define M in such a way that $$\nabla M=0$$. However, in this way not much is won over a description in terms of j_int as the dependence of M on B would be very non-local (that is, of the form $$M(r)=\int dr'^3 \chi(r,r')B(r')$$, with $$\chi(r,r')$$ falling off slowly with the distance of r from r') even in cases where a description in terms of well separated atomic dipoles is possible. A description, which on the macroscopic scale is localized, i.e., $$M(r)=\chi(r) B(r)$$ necessarily means that M(r) has to have a longitudinal component.

dgOnPhys
a uniformly magnetized infinitely long bar magnet

Can you elaborate? how is the description in term of surface current failing for the interior field?

dgOnPhys
The magnetization cannot be defined as a density of magnetic dipoles in general. The problem is that in quantum mechanics, one cannot separate unambiguously one dipole from the other. In older classical electrostatics on the other hand, it was not unusual to define the strength of a dipole via the strength of two monopoles of opposite charge at unit distance.
Nowadays magnetization is simply a restatement of charge conservation as $$\nabla j_\mathrm{int}=0$$ in magnetostatics, it follows that j _int(the current density due to the atomic currents inside the medium) can be expressed as $$-\nabla \times M$$, the minus sign being convention. This equation defines M only up to the gradient of a scalar field $$\grad \xi$$. So it would be possible to define M in such a way that $$\nabla M=0$$. However, in this way not much is won over a description in terms of j_int as the dependence of M on B would be very non-local (that is, of the form $$M(r)=\int dr'^3 \chi(r,r')B(r')$$, with $$\chi(r,r')$$ falling off slowly with the distance of r from r') even in cases where a description in terms of well separated atomic dipoles is possible. A description, which on the macroscopic scale is localized, i.e., $$M(r)=\chi(r) B(r)$$ necessarily means that M(r) has to have a longitudinal component.

1. we are talking about CLASSICAL physics
2. magnetization is defined through a CONTINUOUS distribution of dipoles not a discrete one; an inaccurate model at microscopic scale but still the model
3. elementary dipoles are defined as opposite monopole charges at INFINITESIMAL distance preserving dipole moment in the limit
4. With $$\nabla M=0$$ I guess you mean div M =0 not grad M =0

granpa
Can you elaborate? how is the description in term of surface current failing for the interior field?

Huh? The surface current gives the correct result.
The dipole distribution reduces to finite charges at the ends of the infinite bar and therefore results in zero field at any finite point. (which is only correct for the exterior field)

I thought it was pretty obvious.

yungman
Are dia and paramagnetic materials consider homogeneous because $\vec M$ is so weak that

$$\vec B = \mu_0 \vec H$$

We only have to concern with ferromagnetic material?

Also

$$\nabla X \vec H = \vec J_{free}$$

I want to confirm that in the boundary condition, even if there is no free current density, H is not zero. It just mean

$$\int_C \vec H \cdot d\vec l = 0 \Rightarrow \vec H_{1t} = \vec H_{2t}$$

Where 1t and 2t are for the tangential component of H in top layer and bottom layer.

1. we are talking about CLASSICAL physics
2. magnetization is defined through a CONTINUOUS distribution of dipoles not a discrete one; an inaccurate model at microscopic scale but still the model
3. elementary dipoles are defined as opposite monopole charges at INFINITESIMAL distance preserving dipole moment in the limit
4. With $$\nabla M=0$$ I guess you mean div M =0 not grad M =0

To 1. I only wanted to say that we are talking about continuous charge and current distributions, whether classical or not I want to leave open. For these, it is not clear how to represent them unambiguously in terms of a dipole density.
To 4. Yes, sorry, I wrote div and chaged it to nabla, that's why the dot is missing.
5. What I really wanted to say is the following: One can write down various forms of the macroscopic Maxwell equations. E.g., instead of introducing polarization and magnetization, one can deal directly with the charge and current distribution of the medium instead. Whatever one uses, the Maxwell equations have to be supplemented by a set of material equations which describe the dependence of the magnetization or of the currents on the magnetic field B. The current is a very non-local function of the magnetic field. The reason for the utility of the magnetization lies in the fact that it can be chosen in such a way as to depend (approximately) only on the local value of the magnetic field. Now by the Helmholtz theorem, every field can be decomposed into it's transverse and longitudinal part, only the latter having a non-vanishing divergence.
In principle, for the Maxwell equations, it would be sufficient to consider only the transverse part of the magnetization, however, this would be a very non-local function of B. So the non-vanishing of div M in the usual formulation is due precisely to the demand for a local relation between M and B.

dgOnPhys
Huh? The surface current gives the correct result.
The dipole distribution reduces to finite charges at the ends of the infinite bar and therefore results in zero field at any finite point. (which is only correct for the exterior field)

I thought it was pretty obvious.

Let me see if I am following: you are saying
- equivalent current distributions describe fields inside and outside matter
- equivalent charge distributions work only outside

How does this work out for polarization? Same formal equations but I don't recall many examples using bound magnetic currents in places of bound charges to solve polarization problems...

I guess I will have to go and dig my old books out of storage...

yungman
How about my question? Seems like this thread has be hijacked and turn into classical vs quantum physics!

granpa
How about my question? Seems like this thread has be hijacked and turn into classical vs quantum physics!

Let me see if I am following: you are saying
- equivalent current distributions describe fields inside and outside matter
- equivalent charge distributions work only outside

How does this work out for polarization? Same formal equations but I don't recall many examples using bound magnetic currents in places of bound charges to solve polarization problems...

I guess I will have to go and dig my old books out of storage...

bound electric charges are fine since they actually exist.

bound magnetic charges (monopoles) only work for the external field because they don't actually exist.

I don't understand why you are having so much trouble understanding this.

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No, this has nothing to do with quantum mechanics. The divergence of H does not vanish when the relation between B and H is assumed to be local. If $$B(r)=\mu(r) H(r)$$ and $$\nabla \cdot B=0$$ then $$\nabla\cdot H(r)=-\frac{1}{\mu(r)}H(r)\cdot \nabla \mu(r)$$ as the resutl of a simple vector analytical identity, namely $$\nabla \cdot a A =(\nabla a)\cdot A+a \nabla\cdot A$$ with a a function and A a vector.

yungman

bound electric charges are fine since they actually exist.

bound magnetic charges (monopoles) only work for the external field because they don't actually exist.

I don't understand why you are having so much trouble understanding this.

.......

dgOnPhys
.......

Sorry yungman, apparently the side topic originating from your initial question seems to be annoying you... I will open a new thread and see if we can get this sorted out. Please join in if you feel like it

yungman
Sorry yungman, apparently the side topic originating from your initial question seems to be annoying you... I will open a new thread and see if we can get this sorted out. Please join in if you feel like it

It's no problem, just don't forget my question! lalbatros
1. we are talking about CLASSICAL physics

Might that not be the crux of the matter?

I think indeed that in strict classical physics: $\hbox \nabla \cdot \vec M = 0$

To add magnetisation in classical electrodynamics, you need a additional empirical model.
Based on classical charges and their motions, you can't get any magnetisation.

granpa
Based on classical charges and their motions, you can't get any magnetisation.
I don't know what you mean unless you are saying that classically electrons should stop circling the nucleus and shouldn't have spin.

http://en.wikipedia.org/wiki/Magnetization
The origin of the magnetic moments responsible for magnetization can be either microscopic electric currents resulting from the motion of electrons in atoms, or the spin of the electrons or the nuclei

dgOnPhys
Might that not be the crux of the matter?

I think indeed that in strict classical physics: $\hbox \nabla \cdot \vec M = 0$

To add magnetisation in classical electrodynamics, you need a additional empirical model.
Based on classical charges and their motions, you can't get any magnetisation.

All it takes to get classical magnetization is a distribution of classical magnetic dipoles. If you want to make a magnetic dipole through motion just spin a charged sphere on its axis. The only problem with this classical model is the wrong gyromagnetic ratio.

All it takes to have $\hbox \nabla \cdot \vec M \neq 0$ is an inhomogeneous material e.g. think of the interface between different materials

lalbatros
All it takes to get classical magnetization is a distribution of classical magnetic dipoles. If you want to make a magnetic dipole through motion just spin a charged sphere on its axis. The only problem with this classical model is the wrong gyromagnetic ratio. ...

The magnetic field associated to a (spinning) charged particle or to a "magnetic dipole" is divergence-free.
How do you explain then that the macroscopic sum of the microscopic magnetic fields could be non-divergence-free?

lablatros, I do not exactly understand what you want to say. Do you claim that M is somehow a macroscopic sum of microscopic magnetic fields? That would be wrong.

dgOnPhys
The magnetic field associated to a (spinning) charged particle or to a "magnetic dipole" is divergence-free.
How do you explain then that the macroscopic sum of the microscopic magnetic fields could be non-divergence-free?

The thing with point-like charged particle and point-like dipoles (magnetic or not) is that they are described by a source distribution which is zero almost everywhere:
$\rho(\vec r)=q \delta(\vec r - \vec r_0)$
$\rho(\vec r)=- \vec p \cdot \nabla \delta(\vec r - \vec r_0)$

still when you are allowed to integrate on them (rather than summing) you can create any CONTINUOUS distribution of charge or dipole density.

If you just sum over a discrete set and measure the average resulting field, I am not sure things would be as peachy, perhaps this is what DrDu is saying in the post right before mine... if one tries to make a classical model out of a finite number of charges and dipoles not all average magnetization fields can be represented...

DrDu?

lalbatros
lablatros, I do not exactly understand what you want to say. Do you claim that M is somehow a macroscopic sum of microscopic magnetic fields? That would be wrong.

This is indeed my claim, and I don't think it wrong.
How else would you define the magnetisation?

By definition, B = H + M ,
where H is the magnetic field caused by "external currents"
end wher M is the magnetic field caused by "microscopic currents".

Actually, the decomposition between H and M is arbirary in principle.
It is the purpose of the "theory of magnetism" to separate "external currents" as cause and "microscopic currents" as effects and to analyze the physics behind it.

One could imagine other decompositions for other purposes.
Coupled coils could be an example.
The H field would be the field caused by the "controlled loop" and M the field from the other loop.
In this case, div H = 0 as well as div M = 0.

The question remains then:

when could we have div M <> 0 and what is the physical meaning ?

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