# In magnetism, what is the difference between the B and H fields?

Jano L.
Gold Member
I just can cite my former solid state theory teacher in that context: "Never average!
Jackson is not a solid state theorist but from particle physics, he does not know that."
In solid state theory you don't calculate the polarization from the moments of the atoms or molecules but you determine epsilon from e.g. linear response.
If you do not average dipole moments, you are calculating different thing than macroscopic polarization.

But as I understood it, the procedure Adler is describing is just a particular way to evaluate the macroscopic polarization, using the formalism of epsilon in Fourier domain. This epsilon then gives macroscopic polarization, whose meaning is exactly average bulk dipole moment. It may be hidden in the flood of formulae, but he uses density matrix to calculate averages.

DrDu
If you do not average dipole moments, you are calculating different thing than macroscopic polarization.
True, but the point I want to make is that averaging is something you can do if you like and quite trivially so but it is not some essential ingredient needed to define magnetisation or polarisation as many introductory e-dynamics books try to make you believe. Also I don't think that macroscopic H and D fields are any better defined than their microscopic counterparts. The freedom of the choice of gauge as you call it is also present here. E.g. in optics one sets B=H by definition and only considers polarisation P. The price one has to pay is that P becomes non-local, which is unavoidable at higher frequencies anyhow. See Landau Lifgarbagez, Electrodynamics of continua.
Only at zero frequency this is inconvenient as P then diverges in this scheme.
Another example are metals whose response may either be described in terms of a complex dielectric constant (leading to polarisation) or of a real conductivity (giving rise to a current density).

Jano L.
Gold Member
it is not some essential ingredient needed to define magnetisation or polarisation
So then, how would you define polarization without averaging? As the polarization potential $\mathbf p$ of microscopic charge and current densities? This is highly oscillating in space and is not related to the equally highly oscillating microscopic field $\mathbf e$ in any simple way. Such epsilon would be very irregular function determined by the exact position of the atoms, and hence describing particular body instead of general character of the medium.

The linear response relation, if forced upon these microscopic quantities, would make the corresponding $\epsilon(\mathbf q, \omega)$ depend very violently on both spatial coordinates and q.

In short, this all would mean complex microscopic theory. But in microscopic domain the polarization and linear response description is not very useful - one works directly with charged particles and equations of motion.

E.g. in optics one sets B=H by definition and only considers polarization P.
I do not think B=H is an arbitrary definition. It is an approximation, valid because the magnetic moments are weak and negligible - Landau comments this. We could be exact and keep the weak magnetization in H.

Another example are metals whose response may either be described in terms of a complex dielectric constant (leading to polarisation) or of a real conductivity (giving rise to a current density).
Hmm, conduction can be $described$ by Hertz polarization potential, current being its time derivative. It has mathematical advantages, that's right, but I do not think this is enough to call it polarization. In case the current is direct, the polarization potential grows linearly in time. Calling it polarization leads to misleading picture. There is nothing in metal that increases in time, the metal does not undergo any change, unlike dielectric, where growing polarization means growing displacement of bound charge.

DrDu
I do not think B=H is an arbitrary definition. It is an approximation, valid because the magnetic moments are weak and negligible - Landau comments this. We could be exact and keep the weak magnetization in H.

Hmm, conduction can be $described$ by Hertz polarization potential, current being its time derivative. It has mathematical advantages, that's right, but I do not think this is enough to call it polarization. In case the current is direct, the polarization potential grows linearly in time. Calling it polarization leads to misleading picture. There is nothing in metal that increases in time, the metal does not undergo any change, unlike dielectric, where growing polarization means growing displacement of bound charge.
For the understanding of the approach with B=H, I find the following article by Agranovich and Gartstein very useful
http://siba.unipv.it/fisica/articoli/P/PhysicsUspekhi2006_49_1029.pdf [Broken]
The prototype of a dielectric function of a metal is the Lindhard (or refinements like Sjolander Singwi) dielectric function which is treated in every text book on many body physics see e.g. here:
www.itp.phys.ethz.ch/education/lectures_fs10/Solid/Notes.pdf [Broken]
Evidently, it includes the effect of conductivity.

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So
E = D + P (except that for historical reasons E is defined differently, so we need to multiply it by the permittivity, and for some reason P is multiplied by minus-one ).
Maybe P is multiplied by -1 because the dipole moment points from the negative to the positive charge, so a positive polarization actually reduces the field due to the displacement field.

Jano L.
Gold Member
DrDu,
thank you for the links. I think the ambiguity of M arises only if one stays within macroscopic theory. There one cannot define P and M solely by

rho = -div P
j = dP/dt + ∇×M,

as there is infinitely many solutions of these equations.

So if one requires only the above equations, it is possible to define P an M in many ways - P and M are not unique. They do not even have to be zero in the vacuum.

However, P and M are usually introduced with help of a microscopic model, as average electric dipole moment of neutral molecules and average magnetic moment of those molecules, multiplied by density of those molecules. As long as the medium is thought to consist of small neutral aggregates of bound charged particles , P and M are unique and so are D and H, so there is no additional freedom in their choice.

On the other hand, metals and other conducting media probably cannot be modelled by localized aggregates of bound charged particles, and the quantities P and M cannot have the same meaning as in the above theory. I think, when one uses p and m for conducting media, one really uses polarization potentials with the only defining requirement - the above equations for j and rho.

The confusion arises because for oscillating fields, even the free charges behave as if they were bound - they oscillate around temporary equilibrium positions. The polarization potentials p, m then look almost as if they corresponded to the polarization and magnetization M from the theory of dielectrics/magnets.

DrDu
However, P and M are usually introduced with help of a microscopic model, as average electric dipole moment of neutral molecules and average magnetic moment of those molecules, multiplied by density of those molecules. As long as the medium is thought to consist of small neutral aggregates of bound charged particles , P and M are unique and so are D and H, so there is no additional freedom in their choice.
That's an important point. I don't think it is rigorously feasible any more to interpret P and M as dipole moment densities once you are using quantum mechanics.

Jano L.
Gold Member
I am not sure of that. In molecular spectroscopy, Schroedinger calculated expected value of dipole moment of hydrogen atom using his time-dependent equation. This was one of triumphs of his wave mechanics. The same equation is used today in orthodox quantum mechanics and the calculation itself remains the same. Similar calculations are made for molecules today - one calculates induced oscillating dipole moment as linear combination of matrix elements of dipole operator. I think as long as the molecule is neutral (most of spectroscopy), the definition of P as

density * <dipole moment>

makes good sense. I agree though that in conducting media, things are different and polarization in this sense most probably cannot be introduced.

DrDu
Yes, but in extended media as opposed to isolated molecules, you cannot say exactly where one atom or molecule ends and the other one begins. Computationally, it is easier to calculate the polarisation directly.
The difference in polarisation of metals and insulators reflects itself in the spatial dispersion (k dependence) of the dielectric function in metals while this dispersion is often negligible in insulators.

DrDu
I think as long as the molecule is neutral (most of spectroscopy), the definition of P as

density * <dipole moment>

makes good sense.
This definition does not work even for materials which show spontaneous static polarisation (ferroelectrics or piezoelectrics). In fact one needs some quite advanced topological methods to calculate this polarisation:
http://www.physics.rutgers.edu/~dhv/pubs/local_preprint/dv_piezo.html

You can express the H-field as the gradient of a scalar potential when all currents are steady (have no time dependence). The more general case of time-varying currents means you can no longer do that.

Jano L.
Gold Member
DrDu.,
thanks, the connection with the Berry phase is surprising.

Muphrid,
that cannot be true globally, since in vacuum H is proportional to B which can have closed lines of force when currents are stationary (circles around straight wire)

Yep, you're right, I was trying to rederive it a bit too fast. Per wiki, H can be a gradient of a scalar potential only when there is no free current.

Can I ask one further question

With all these fields I find it useful to think of a logical cause-effect sequence to understand what happens

First we have a B field, say

then we put in this B-field a paramagnet

The interaction between the dipole moments in the paramagnet and the B-field leads to some degree of alignment, and a nonzero magnetisation M vector arises. M = f(B)

This then adds to the B-field inside the paramagnet, giving it a boost.

But we don't say B = B + M(B). We instead use H. Why is a THIRD necessary?

so H = B/mu0 + M

This third quantity then has entirely different units, so if I want to know how many Webers are passing through some current loop for a practical calculation, I can't use H, because the units are no longer Wb/m^2.

Can someone frame this apparent overkill of vector fields in terms of a useful, logical calculation? Why can't we just have a single B-field with a corrective term due to bound magnetic dipoles in the same unit system?

Jano L.
Gold Member
The resulting field ##B## is not a sum of the initial field ##B_{ext}## and magnetization ##M##, but a sum ##B_{ext} + \delta B##, where ##\delta B## is the field of the paramagnet. This contribution due to the paramagnet depends also on the shape of the paramagnet and there is no general relation between it and the magnetization. The field ##H## is just another convenient quantity to describe the field in magnetostatics; again, there is no general relation between it and the magnetic field ##B##.

isn't a BH curve a general relation?

Philip Wood
Gold Member
I do think it ugly and confusing that, in the SI, H and M are defined so as to have different units from B, owing to the factor of $\mu_0$. Similarly for D, P and E. Can someone persuade me that I'm wrong to think this?

Jano L.
Gold Member
There is the relation

$$\mathbf B = \mu\mathbf H$$

but this is only approximation, and ##\mu## depends on the kind of material, so I would say, there is no general relation between the two. B, H are independent variables with different meaning in general.

1 person
Jano L.
Gold Member
I do think it ugly and confusing that, in the SI, H and M are defined so as to have different units from B, owing to the factor of μ0. Similarly for D, P and E. Can someone persuade me that I'm wrong to think this?
The reason they have different units is a good one.

##\mathbf M## is so defined so that ##\nabla \times\mathbf M## gives magnetization current density ##\mathbf j##, so the unit of ##\mathbf M## is ##\text{A.m}^{-1}##. ##\mathbf B## is defined so that ##q\mathbf v \times \mathbf B## gives force, so the unit of B is ##\text{N.A}^{-1}\text{.m}^{-1}##.

Philip Wood
Gold Member
Jano L. Agreed, but I want to go back a stage... In a vacuum, for steady current, we have
$$\nabla\times\vec{B}=\mu_0 \vec{J}$$
in which $\vec J$ is the free current density.
I just wish that $\vec M$ had been defined similarly by $\nabla\times\vec{M}=\mu_0 \vec{j}$.

Jano L.
Gold Member

Philip Wood
Gold Member
Why have the $\mu_0$ in one equation, but not in the other (post 70)? Why give the field vector due to bound currents a different unit from the field vector due to free currents?

Jano L.
Gold Member
I think one good answer is that I've given in #69. In other words, magnetic field and magnetization are different quantities with different meaning, use and therefore different units.

Suppose we defined magnetization in the way you suggest. What benefit is there in calling magnetization ##\mathbf M## a quantity which does not give average magnetic moment of element, but gives ##\mu_0## times that magnetic moment?

Philip Wood
Gold Member
Yes, I agree that you'd lose the interpretation of M as mean magnetic moment per unit volume - but not if the magnitude of magnetic moment for a current loop had been defined as $\mu_{0} I \Delta S$ rather than $I \Delta S$ in the first place (Georgi?)! At this stage, I hear howls of protest!

Jano L.
Gold Member
So you see it, such definition puts the awkward constant ##\mu_0## into another equation. It is equally bad as the SI convention.

I think the best thing to do is to stick to SI when talking to general audience, and to use whatever system suits you in your own research. I like the convention where only ##c## appears in the Maxwell equations and no crazy ##\epsilon_0,\mu_0## appears at all. But when communicating with others, the ugly SI convention is beneficial because it is widely known and accepted.