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

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  • #26
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Yes, books on electromagnetism often state that D and H are the auxiliary fields. But that is not the same as saying that they are equivalent.

If you keep the charge on a capacitor the same than E depends on the material. If you keep the voltage constant then D depends on the material. But a capacitor is not a good example because there is no good magnetic equivalent to a capacitor due to the lack of monopoles.
Let's instead look at a ring. If you have an iron ring and there is a changing electric field going through its center perpendicular to the plane of the ring, then there is a magnetic field induced that magnetizes the iron. The H field in this case depends on the rate at which the electric field changes and is independent of the ring. The B field depends on the properties of the iron. Without the iron the B field would be much weaker.
Now take a ring made from a good dielectric, e.g. ceramic. If there is a changing magnetic field going through its center an E field is induced that polarizes the ceramic. Here the E field is independent of the material of the ring but the D field depends on it's material properties. It's much higher with the ring in place.
 
  • #27
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Yes, books on electromagnetism often state that D and H are the auxiliary fields. But that is not the same as saying that they are equivalent.
How do you explain the fact then that those two possess the free charge/current dependency? Wouldn't be plausible that ∇E=ρ ? I was hoping that you/somebody would go into that, since I highlighted it...

If you keep the voltage constant then D depends on the material.
Can you prove that? I kept refering to our four fields with the qualifiers "interior" (E and B) and "exterior" (D and H), having implied an exterior field is produced by an outer "free" source and is thereby always stable.

But a capacitor is not a good example because there is no good magnetic equivalent to a capacitor due to the lack of monopoles.
I don't see why the monopoles play a role in this case... Why wouldn't a local homogenous H field be an equivalent?

Not to mention that I enjoyed the ring example.
 
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  • #28
vanhees71
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On an elementary level, there is only one electromagnetic field, given by the Faraday tensor [itex]F_{\mu \nu}[/itex] or, equivalently in the 3D formulation, [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex]. This relativistic point of view clearly shows that these two fields belong together since they build the one covariant antisymmetric Faraday tensor. They obey Maxwell's equations (in Heaviside-Lorentz units that are most natural), i.e., the homogeneous equations

[tex]\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0[/tex]

and the inhomogeneous equations

[tex]\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E} = \rho.[/tex]

If you have macroscopic matter, it's way too complicated (and also not necessary!) to solve these fully microscopic Maxwell equations, since here [itex]\rho[/itex] and [itex]\vec{j}[/itex] denote all the many charges of all atoms building the matter. It is impossible to solve these complicated equations, including all binding effects of the matter (which requires a fully quantum theoretical description by the way).

Instead one can solve approximate equations for macroscopic electromagnetic fields that are small compared to the microscopic electromagnetic fields between atomic nuclei and electrons forming, atoms, molecules, solids, etc. In this case one can describe the reaction of the medium in an averaged way by using the socalled linear-response theory. For a homogeneous isotropic medium, this yields to (induced and proper) electric polarization and magnetization. The microscopic charge and current distributions are "coarse grained" by taking the average over volume elements that are small compared to the macroscopic scales of change but large to the microscopic (atomic) scales (implying that for electromagnetic waves the wavelength of the macroscopic fields is large compared to atomic distance scales of the medium).

In this case one introduces the auxilliary fields [itex]\vec{D}[/itex] and [itex]\vec{H}[/itex] that form in relativistic notation also an antisymmetric 2nd-rank tensor. In non-relativistic approximation, the inhomogeneous Maxwell equations are then written in the form

[tex]\vec{\nabla} \times \vec{H}-\frac{1}{c} \partial_t \vec{D}=\sigma \vec{E}+\vec{j}_{\text{free}}, \quad \vec{\nabla} \cdot \vec{D}=\rho_{\text{free}}.[/tex]

and the constitutive equations

[tex]\vec{D}=\epsilon \vec{E}, \quad \vec{H}=\frac{1}{\mu} \vec{B}.[/tex]

Here, [itex]\vec{j}_{\text{free}}[/itex] and [itex]\rho_{\text{free}}[/itex] are the free electric current and charge densities, which are given by the coarse-graining procedure (averaging out the microscopic charge distributions) explained above. Further [itex]\sigma[/itex] is the conductivity and [itex]\epsilon[/itex] and [itex]1/\mu[/itex] the the electric and magnetic permeabilities. That the latter appear in the numerator and denominator of the constitutive equations, respectively, is an unfortunate historical heritage of the pre-relativity era of electromagnetic theory, we have to live with.
 
  • #29
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Trifis - The magnetic equivalent of the E field is (if I understand the physics correctly) the H field and not the B field. H and E are real fields, B and D are combinations of field and magnetization/polarization - a mathematical trick so to speak to simplify calculations.
I think I was a little confused here. If you just look at the math it can seem as if E is equivalent to H because that assumption can make part of the math more symmetric but looking at the physics of how magnetic dipoles differ from electric dipoles it's clear that E is equivalent to B.

How do you explain the fact then that those two possess the free charge/current dependency? Wouldn't be plausible that ∇E=ρ ? I was hoping that you/somebody would go into that, since I highlighted it...
I'm not sure what you mean. ∇E=ρ is basically Gauss's law. Just without the ε0 but ε0 is just a unit conversion factor in this case.

http://en.wikipedia.org/wiki/Maxwell's_equations
I looked at the wikipedia article on maxwell's equations and found this
Wikipedia said:
For one reason, Maxwell's equations can be made fully symmetric under interchange of electric and magnetic fields by allowing for the possibility of magnetic charges with magnetic charge density ρm and currents with magnetic current density Jm.
So apparently the only reason maxwell's equations are asymmetric is because matter contains no monopoles.

the homogeneous equations

[tex]\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0[/tex]

and the inhomogeneous equations

[tex]\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E} = \rho.[/tex]
So that means when you have pure fields and no matter, j and p become 0 and you get

[tex]\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0[/tex]
[tex]\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = 0[/tex]
So it is only the properties of matter (e.g. the presence of electric charges) that create asymmetry? For pure fields the equations are completely symmetric?
 
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  • #30
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The H field (the magnetic field) is the field in vacuum. This field can induce a magnetization of ponderable matter, and the total field (vacuum plus induced field) is the B field (magnetic induction). It is sometimes confusing that the conventions of E and D are different than B and H.

Constitutive relations are used to relate the H and B (as well as E and D) fields.
In vacuum the H and B field are equivalent. You write as if the H field is somehow more fundamental, it certainly is not.

My understanding is that the H field is simply a mathematical tool, it depends only on the free currents and not on bound currents like the B field does, and so simplifies calculations. You can calculate the H fields from the known free currents, and then from that calculate the B field. At the end of the day though, it's the B field that we measure and is a "physical" thing, not the H field.

Plus, relativistic transforms show that in vacuum, an electric field transforms to the B field, not the H field.
 
  • #31
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Plus, relativistic transforms show that in vacuum, an electric field transforms to the B field, not the H field.
In vacuum B is not just equivalent to H it is identical. B=H (if you measure both in the same units e.g. gaussian units) so your statement makes no sense. Otherwise you are correct I think.
 
  • #32
vanhees71
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So that means when you have pure fields and no matter, j and p become 0 and you get

[tex]\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0[/tex]
[tex]\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = 0[/tex]
So it is only the properties of matter (e.g. the presence of electric charges) that create asymmetry? For pure fields the equations are completely symmetric?
Of course, if there is no matter, you have a free electromagnetic field, but Maxwell's equations for [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] are always valid, also at presence of matter. You only have to make sure that the charge and current densities denote the complete set of charges und currents of the elementary particles building up this matter (including the magnetic moments of all particles!).

As I wrote earlier, this complicated set of equations cannot be solved for macroscopic matter, and fortunately that's not necessary, but you can use approximations, of which the simplest and often applicable in everyday life is linear response theory. Then you get macroscopic classical electromagnetics as you find it in almost all textbooks on the subject, although these very often omit to stress the basic fact that on a fundamental level, there is only one electromagnetic field with components [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex], while [itex]\vec{D}[/itex] and [itex]\vec{H}[/itex] are auxiliary quantities denoting the average fields over macroscopic small, microscopic large regions, containing the effective average charge and current distributions of matter. In these macroscopic Maxwell equations only the external charges and currents brought into the system from outside and not taken care of by the average charge and current distributions of the matter.
 
  • #33
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In vacuum B is not just equivalent to H it is identical. B=H (if you measure both in the same units e.g. gaussian units) so your statement makes no sense. Otherwise you are correct I think.
Yes well this is just a bit of semantics, but "equal" is more precise than "equivalent" I suppose.
 
  • #34
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It is good to know that there are some people willing to let us know what we already know...
BUT the primary question remains emphatically unanswered, so may I state it again:

D=ε[itex]_{0}[/itex]E+P , ∇D=ρ[itex]_{free}[/itex]

H=B/μ[itex]_{0}[/itex]-Μ , ∇H=j[itex]_{free}[/itex]

There is an obvious asymmetry here... To my eyes at least!
 
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  • #36
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It is good to know that the are some people willing to let us know what we already know...
BUT the primary question remains emphatically unanswered, so may I state it again:

D=ε[itex]_{0}[/itex]E+P , ∇D=ρ[itex]_{free}[/itex]

H=B/μ[itex]_{0}[/itex]-Μ , ∇H=j[itex]_{free}[/itex]

There is an obvious asymmetry here... To my eyes at least!
That's your question, not the OP's. I think the short answer is that the H field is a just a mathematical tool, and shouldn't be thought of too literally. In CGS units they have the same units, the fact that they have different units in SI is one of the shortcomings of that rather outdated system. Most advanced and newer E&M textbooks use CGS.
 
  • #37
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That's your question, not the OP's. I think the short answer is that the H field is a just a mathematical tool, and shouldn't be thought of too literally. In CGS units they have the same units, the fact that they have different units in SI is one of the shortcomings of that rather outdated system. Most advanced and newer E&M textbooks use CGS.
This thread was abandoned for more than two years, so actually it is my question which is to be ventilated. I didn't open a new thread, because I felt it belonged here...

Once again the confusion is not down to either the historically ε/μ conventions or the units! That'd be way too straightforward to be inquired.

I am not going to repeat the previous discussion about outer/inner fields, their amplification/weakening in matter and how the derivations should be analogous and symmetrical in both electrostatics and magnetostatics. You could easily look it up, I think.
I believe my point sums up to the fact that in the first formula there is a + between E and P whereas in the second there is clearly a - between B and M.
 
  • #38
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This thread was abandoned for more than two years, so actually it is my question which is to be ventilated. I didn't open a new thread, because I felt it belonged here...

Once again the confusion is not down to either the historically ε/μ conventions or the units! That'd be way too straightforward to be inquired.

I am not going to repeat the previous discussion about outer/inner fields, their amplification/weakening in matter and how the derivations should be analogous and symmetrical in both electrostatics and magnetostatics. You could easily look it up, I think.
I believe my point sums up to the fact that in the first formula there is a + between E and P whereas in the second there is clearly a - between B and M.
I see, I didn't realize this was an old thread. I would have made a new one to avoid such confusion.

The answer to your question is that it's because of the way the bound charges and bound currents in the material interact. The polarization always acts "against" the field in a linear material. Meaning the bound charges in the material try to separate in such a way as to cancel the external field, and so the electric field inside a dialectic is smaller than in vacuum.

In a magnetic material, one can think of the electrons bound to the atoms as lots of little magnetic dipoles, in the presence of an external magnetic field, these dipoles tend to align, and the effect is that their fields add with the external field.

Again, the D and H fields are just mathematical tools, don't think of them as real fields.
 
  • #39
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The answer to your question is that it's because of the way the bound charges and bound currents in the material interact. The polarization always acts "against" the field in a linear material. Meaning the bound charges in the material try to separate in such a way as to cancel the external field, and so the electric field inside a dialectic is smaller than in vacuum.

In a magnetic material, one can think of the electrons bound to the atoms as lots of little magnetic dipoles, in the presence of an external magnetic field, these dipoles tend to align, and the effect is that their fields add with the external field.
Exactly what I inferred too! The "creator" of these equations has determined the algebraic signs so as to satisfy the behaviour of the majority of materials! But what if we consider the case of ferroelectics where the polarization amplifies the electric field or respectively the case of diamagnets where the tiny dipols align antiparallel with the external exciter?

The elusion of a more general expression have caused plenty of misunderstandings to the later generations. For instance some have already tried to parallel D to B field due to this infelicitous notation...
 
  • #40
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Well in the case of dialectics, in general the Polarization is related to the E-field by (in CGS units)

P = χE

where χ is a rank 2 tensor, not just a scalar. So those equations are generally correct, I believe, but it is not generally correct to assume a linear relationship between P and E. I know next to nothing about nonlinear dialectics though so I can't really say more.
 
  • #41
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I see, I didn't realize this was an old thread. I would have made a new one to avoid such confusion.

The answer to your question is that it's because of the way the bound charges and bound currents in the material interact. The polarization always acts "against" the field in a linear material. Meaning the bound charges in the material try to separate in such a way as to cancel the external field, and so the electric field inside a dialectic is smaller than in vacuum.

In a magnetic material, one can think of the electrons bound to the atoms as lots of little magnetic dipoles, in the presence of an external magnetic field, these dipoles tend to align, and the effect is that their fields add with the external field.

Again, the D and H fields are just mathematical tools, don't think of them as real fields.
Sorry, but please review my post earlier in this thread. Since mu & epsilon are real, as are E & B, then D & H cannot be anything less than real. Take a capacitor using a dielectric with substantial absorption factor. We apply an E field from an external source. We get polarization. We then reduce the external E field to zero, but the dielectric does not return to its original un-energized state. It retains energy due to dielectric absorption.

Anyone who has worked in the lab with large caps knows the danger involved. D is retained after E goes to zero. The D-E curve of a dielectric has a hysteresis just like magnetic materials do. Zero E can co-exist with non-zero D. How can D not be "real"? It makes no sense. I am not an expert on this matter, but please enlighten me. What am I missing?

Claude
 
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  • #42
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Anyone who has worked in the lab with large caps knows the danger involved. D is retained after E goes to zero. The D-E curve of a dielectric has a hysteresis just like magnetic materials do. Zero E can co-exist with non-zero D. How can D not be "real"? It makes no sense. I am not an expert on this matter, but please enlighten me. What am I missing?
In a dielectric, D represents the displacement of charges. The positive charges in the material move a little bit to one side and the negative charges move a little to the other. If that dielectric is in between two connected metal plates it will induce opposite charges in the metal. Then there will be no E inside the dielectric but still a D since the charges are still displaced. So D is certainly something real, a real displacement of charges, but it is not really an electric field.
 
  • #43
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In a dielectric, D represents the displacement of charges. The positive charges in the material move a little bit to one side and the negative charges move a little to the other. If that dielectric is in between two connected metal plates it will induce opposite charges in the metal. Then there will be no E inside the dielectric but still a D since the charges are still displaced. So D is certainly something real, a real displacement of charges, but it is not really an electric field.
I agree completely but for 1 exception, ref bold highlight of your quote. D is indeed real, and is an electric flux density, as opposed to E which is an electric field intensity. Some texts call D the electric displacement. These are all quite real, albeit D does indeed differ from E. I would agree that D is not a true "intensity". But I also state that E is not a real "density".

A similar scenario exists for H & B. When a ferrous bar is not magnetized, then a coil is wrapped around it then current applied, the magnetic domains become aligned in the bar. A B-H curve describes the relation between B (density) and H (intensity). If we remove the external current, H goes to zero. But B generally does not, a residual magnetism is held, known as remnance.

Similar to the dielectric case, we send H to zero, and still have a non-zero B. In the absence of intensity, we still retain some density. B is not the same as H, again, density is different from intensity. But I don't regard either as virtual or non-real. Then again, one can regard all fields as nothing but mere math constructs, i.e. B, D, E, & H all represent something that cannot be explained by action at a distance. All incurred forces require time to propagate, & fields describe this phenomenon.

Anyway, this stuff is interesting to ponder. Thanks for your input.

Claude
 
  • #44
vanhees71
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It is good to know that there are some people willing to let us know what we already know...
BUT the primary question remains emphatically unanswered, so may I state it again:

D=ε[itex]_{0}[/itex]E+P , ∇D=ρ[itex]_{free}[/itex]

H=B/μ[itex]_{0}[/itex]-Μ , ∇H=j[itex]_{free}[/itex]

There is an obvious asymmetry here... To my eyes at least!
This is just what I said, written in an explicit form in (imho ugly) SI units: The polarization [itex]\vec{P}[/itex] and magnetization [itex]\vec{M}[/itex] are the (linear) response of the medium to the electromagnetic field, i.e., it's the part of the field which comes from the average over the charge and current distributions of the matter over macroscopically small but microscopically large regions in space.

There is no asymmetry in these equations except for the man-made convention to define magnetization with an opposite sign than the analogous polarization. The reason for this odd convention is that before the discovery of relativity people thought that [itex]\vec{H}[/itex] has to be grouped together with [itex]\vec{E}[/itex], which is not correct since relativity tells us that [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] have to be grouped together to the antisymmetric Faraday tensor and [itex]\vec{D}[/itex] and [itex]\vec{H}[/itex] to the antisymmetric tensor, I'd call Minkowski tensor since Minkowski has been the first to write down the macroscopic Maxwell equations and the constitutive equations in a manifestly covariant way.

Maxwell's theory is only "asymmetric" in the fact that it assumes, in accordance with empirical facts, that there do not exist elementary magnetic charges (magnetic monopoles). There's no problem in principle to extend Maxwell theory in this respect as has been shown by Dirac in the 1930ies. A very good description of this you can find in

J. Schwinger, Classcial Electrodynamics

which is anyway a very good book on E&M!
 
  • #45
vanhees71
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https://www.physicsforums.com/showpost.php?p=3930927&postcount=20

To all:

What was wrong with my answer in post #20 linked above?

Claude
Nothing is wrong with that, but I never understood why introduce new names and confusing oneself by distinguishing "quantity" and "intensity" quantities. This goes back to Sommerfeld's famous lectures, which I still consider the best textbooks ever written on classical physics, but I never understood this wording.

In a modern view, there is only the electromagnetic field [itex](\vec{E},\vec{B})[/itex] and [itex]\vec{D}[/itex] and [itex]\vec{H}[/itex] are derived quantities from the linear-response approximation of the interaction between the electromagnetic field and (bulk) matter.
 
  • #46
DrDu
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This is just what I said, written in an explicit form in (imho ugly) SI units: The polarization [itex]\vec{P}[/itex] and magnetization [itex]\vec{M}[/itex] are the (linear) response of the medium to the electromagnetic field, i.e., it's the part of the field which comes from the average over the charge and current distributions of the matter over macroscopically small but microscopically large regions in space.
No, you don't have to average nor do you need (linear) response. P and M perfectly well (although not uniquely) defined microscopic quantities.
Namely they are solutions of the equations [itex]\vec{j}=\partial \vec{P}/\partial t + \nabla \times \vec{M}[/itex] and [itex]-\nabla \cdot \vec{P}=\rho[/itex].
These equations are completely general and can be used to replace either the full current and charge density or only part of it in the microscopic Maxwell equations. P and M are only specified up to a term which vanishes on taking the derivatives. It (namely the magnetization) is usually determined by the condition that P remains finite in the limit of zero frequency.

There is a transformation of the Hamiltonian, known as the Power Zienau Woolley transformation which allows to remove the appearance of the magnetic vector potential introducing the polarization instead.
A good deal of molecular electrodynamics is based on this approach
See e.g.
http://onlinelibrary.wiley.com/doi/...X(1999)74:5<531::AID-QUA9>3.0.CO;2-H/abstract
http://books.google.de/books/about/...ctrodynamics.html?id=rpbdozIZt3sC&redir_esc=y
 
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  • #47
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There is no asymmetry in these equations except for the man-made convention to define magnetization with an opposite sign than the analogous polarization. The reason for this odd convention is that before the discovery of relativity people thought that [itex]\vec{H}[/itex] has to be grouped together with [itex]\vec{E}[/itex], which is not correct since relativity tells us that [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] have to be grouped together to the antisymmetric Faraday tensor and [itex]\vec{D}[/itex] and [itex]\vec{H}[/itex] to the antisymmetric tensor
There is a physical reason for why M is defined that way. If P increases in a dielectric the change in P induces a magnetic field according to the right hand rule. That means the magnetic field that a changing P induces is identical in direction to the field that a changing E is inducing.
A changing M induces an electric field according to the left hand rule, so the direction of the E field is the same as that of a changing B field. If M was defined as H - B then you'd have to use the right hand rule for a changing M.
btw. I don't think that you can derive which field is real from relativity. Sure relativity says that in a vacuum an electric field transforms into a magnetic field. But since H=B in vacuum, the math doesn't tell you if it transforms into H or B. Except of course if you define B as the field that E transforms into. Then B is by definition the real field.
B was originally defined through the E field it induces. Basically B was defined as that property of space or materials (e.g iron) that induces an E field when it changes. If iron contained no electric charges but instead dipoles made out of magnetic monopoles, it could still be used as a transfomer core. It would behave just like normal iron, all the math used in tranformer design would still seem to give the correct results but now it's not the real magnetic field anymore that determines the induced E field. Just like in a dielectric it is changes in the D field that determines the induced B field, in a monopole diamagnetic it's the change in the magnetic auxilary field that determines the induced E field. So if you define B as the field that induces the E field you get a problem there.
Anyway, that B is the real field and not H could only be derived from the physics of how magnetic dipoles in iron work (i.e. they are like tiny current loops and not like monopole dipoles) since the properties of those dipoles determine if it's the real field or the auxilary field that induces an E field and B was originally defined as the field that induces an E field when it changes.
 
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  • #48
Jano L.
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No, you don't have to average nor do you need (linear) response. P and M perfectly well (although not uniquely) defined microscopic quantities.
It is true that we do not need linear response. In ferromagnetism we use M and H with no problem.

However, the general understanding of P and M is that these give electric and magnetic moment per unit volume, in macroscopic theory. Because in theory the matter consists of molecules with localized and moving charge distributions, P and M have to be defined by averaging of the moments of the neutral molecules over some small domain. Both P and M are therefore smooth and unique physical quantities.


The theory in Thirunamachandran and Craig is a different matter. It concerns microscopic theory. It is quite an interesting attempt to describe all microscopic charges in terms of auxiliary functions [itex]p[/itex] and [itex]m[/itex], using the same formulae that hold for bound charges and currents in the macroscopic theory.

But as you say, these p, m are not defined uniquely - much like the electromagnetic potentials. There are many choices for p and m, so I am reluctant to call these polarization/magnetization. But perhaps one can derive macroscopic P and M from the microscopic p, m - if so, in which gauge? Do you have some more accessible references on this? I would be interested.
 
  • #49
DrDu
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It is true that we do not need linear response. In ferromagnetism we use M and H with no problem.

However, the general understanding of P and M is that these give electric and magnetic moment per unit volume, in macroscopic theory. Because in theory the matter consists of molecules with localized and moving charge distributions, P and M have to be defined by averaging of the moments of the neutral molecules over some small domain. Both P and M are therefore smooth and unique physical quantities.
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.
The standard reference is
S. Adler, Quantum theory of the dielectric constant in real solids, Phys. Rev. 126:413-420 (1962)
The hard part is to find the material equation which links j or P to E. Whether to use j or P and how to take the Fourier component with k=0 (= averaging) is relatively trivial.
 
  • #50
Jano L.
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Comment on this reality issue:

If one of the fields B/H was not unique, like the vector potential, one could say it is not the real field.

But, in macroscopic theory, both B and H have unambiguous values almost everywhere (except surfaces of discontinuity). So there is no point saying that one is more real than the other. Both are equally real.

Perhaps the motivation behind the question is, how should we calculate the density of force in the medium? Formally, many possibilities suggest themselves:

[tex]
\mathbf f = \rho_{free} \mathbf E + \mathbf j_{free} \times \mathbf B
[/tex]

[tex]
\mathbf f = \rho_{total} \mathbf E + \mathbf j_{total} \times \mathbf B
[/tex]

[tex]
\mathbf f = \rho_{free} \mathbf D + \mathbf j_{free} \times \mathbf H
[/tex]

[tex]
\mathbf f = \rho_{total} \mathbf D + \mathbf j_{total} \times \mathbf H
[/tex]

or some other combination?

The answer is, the force density in a medium cannot be expressed in such a simple way. The correct way to find the force density is to study carefully the microscopic forces acting on the charged particles composing the medium. For linear media, energy considerations can be used to derive the force density, but then it can be more complicated than the above formulae. See, e.g. Panofsky & Phillips.
 

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