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

[jeebs]

hi,
I'm having trouble understanding what the difference is between the B and H fields. What my understanding is, if you have a magnetic dipole on its own, its going to have some B field around it.
Then say you bring some material into the field, and the field within the boundaries of the material will be modified right? it'll increase or decrease or something? And this field is called the H field, am I right?

So why then do we have the expression B=\mu_0 H or B=\mu_0 (H + M) where M is the sample's magnetization?

Shouldn't the permeability, which has units, really just be some unitless number?
Why does H have units of magnetization, ie. the magnetic dipole moment per unit volume?

Thanks.


PS. if this is in the wrong section, please say so and I'll put it elsewhere.

 

[Bob S]

Based on Maxwell's equations, electric fields are generated by changing B fields, while H fields are generated by changing electric fields.
In dc fields, static electric E fields create currents (magnetization currents) I (when σ>0) which in turn produce static H fields. A static B field (actually d/dt ∫B·n dA = 0) cannot produce an electric field E.

Even if we used natural units where μ₀ = ε₀ = 1, this distinction between B and H remains. In magnetic materials, B is not linearly related to H due to the magnetization term M. If B = H in space, then B ≠ H in magnetic materials.

If voltage = d/dt ∫B·n dA, then how can curl H= σ·E = σ·voltage/length if B = H?

So even if μ₀ = ε₀ = 1, they have to have units. Going without units is about as rational as claiming that h-bar = c =1. (I may get FLAK on this).

Bob S

 

[Andy Resnick]

hi,
I'm having trouble understanding what the difference is between the B and H fields. <snip>

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.

 

[tiny-tim]

I'm having trouble understanding what the difference is between the B and H fields.

Hi jeebs!

It's all to do with the difference between free charge and bound charge (which together make total charge).

E and B are the total electric and magnetic fields.

D and H are the free electric and magnetic fields.

P and M are the bound electric and magnetic fields.​

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 ).

And
B = H + M (except that for the same historical reasons B is defined like E, so we need to divide it … why divide?? … by the permeability).

The latter equation says that the total magnetic field (divided by the permeability) equals the free magnetic field plus the bound magnetic field (the bound magnetic field is all those little loopy currents that make things magnetic).

When an electric current produces a magnetic field, it couldn't care what the field is going to be used for (ie for bold sweeping field lines or pokey little loops inside matter), so it produces a total field, which it's sensible to measure as B.

But once we put matter in the way, we can only measure the free field, H. As you say …

Then say you bring some material into the field, and the field within the boundaries of the material will be modified right? it'll increase or decrease or something? And this field is called the H field, am I right?

Basically, yes … (apart from the permeability factor, of course) B and H are the same away from matter, but in or near matter the matter soaks up some of the B, and all we measure is what's left, the H.

Finally, there's no reason in principle why B and H shouldn't be measured in the same units … but they're not!

 

[Gerenuk]

Only the B field is a real physical field and all of physics can be done with the B and E field alone. The H (and D) field is a mathematical trick to discard the consideration of induced magnetizations in materials.

Basically if you introduce magnetizable materials you have two choices.
1. Use B field only, but you have to integrate over all induced magnetic dipoles in materials and find a self-consistent solution
2. Use B and H field and find a self-consistent solution at the boundaries. No need to integrate over induced volume dipoles - you can ignore them.

The procedure for view (2) isn't usually as easy as saying "this object induces this field". You have to find self-consistent solutions of B and H field and which object created what you cannot tell in most cases.

In SI units B and H have different units.

Why does H have units of magnetization, ie. the magnetic dipole moment per unit volume?

This stems from the derivation of the "mathematical trick" which takes into account the induced dipoles. I can't remember which books have the derivation in them. Maybe
https://www.amazon.com/dp/0716703319/?tag=pfamazon01-20
In doubt, I can post it here.

 

[cabraham]

Only the B field is a real physical field and all of physics can be done with the B and E field alone. The H (and D) field is a mathematical trick to discard the consideration of induced magnetizations in materials.

Basically if you introduce magnetizable materials you have two choices.
1. Use B field only, but you have to integrate over all induced magnetic dipoles in materials and find a self-consistent solution
2. Use B and H field and find a self-consistent solution at the boundaries. No need to integrate over induced volume dipoles - you can ignore them.

The procedure for view (2) isn't usually as easy as saying "this object induces this field". You have to find self-consistent solutions of B and H field and which object created what you cannot tell in most cases.

In SI units B and H have different units.


This stems from the derivation of the "mathematical trick" which takes into account the induced dipoles. I can't remember which books have the derivation in them. Maybe
https://www.amazon.com/dp/0716703319/?tag=pfamazon01-20
In doubt, I can post it here.

Hmmm. You lost me with "only B is real, not H"! I'm having trouble with that one. Likewise for D & E. Two inductors, one air cored, and one ferromagnetic cored, are energized to the same B value, identical in all respects except for the core. The energies are vastly different as W = 0.5*B*H. One could argue that W = 0.5*(B^2)/mu, as well. Energy can be expressed only in terms of B and mu. But is mu "real"? It has to be or we have pure nonsense.

So here it is. Mu is real, and B is real. No argument so far? Since H = B/mu, then since B and mu are both real, how can their ratio, H, not be real as well? I'm just wondering.

D & H have to be every bit as real as E & B, or nothing makes sense. Since epsilon & mu are real, so are B, D, E, & H, as well. Am I missing something? Please enlighten me. You've engaged my curiosity. BR.

Claude

 

[Gerenuk]

So here it is. Mu is real, and B is real. No argument so far? Since H = B/mu, then since B and mu are both real, how can their ratio, H, not be real as well? I'm just wondering.

If course all vector fields exist. I meant real in the sense of "natural".

You can do physics with one simple law for the B field, but difficulties in the numerical solution.

Or you can do physics with complicating relations between B and H field, but easier numerics.

Nature doesn't care if the "numerical" solution is difficult or not. So I consider having only one simple law more "natural" and more "real".

However, for most practical reasons it's easier to do calculation with B and H fields despite their complicating interrelations.

I can try to work out these rules again. Only in the first case (B only) you can say that an object creates a particular field. With B and H field you have to make adjustments.

Btw, the most general law is
\vec{B}=\mu_0(\vec{H}+\vec{M})
Your relation with B=\mu H is only valid for special media (linear,...).

 

[tiny-tim]

magnetic energy density

Two inductors, one air cored, and one ferromagnetic cored, are energized to the same B value, identical in all respects except for the core. The energies are vastly different as W = 0.5*B*H. One could argue that W = 0.5*(B^2)/mu, as well.

Hi Claude!

0.5*B*H is the magnetic energy density, of course … J m^-3

 

[cabraham]

Hi Claude!

0.5*B*H is the magnetic energy density, of course … J m^-3

Yes, of course. My apologies. I got sloppy and forgot to multiply by the volume. The 2 inductors I used in my example were identical in all respects except for the cores, air vs. ferromagnetic. With equal volumes, the unequal energy densities result in unequal energies. I stand corrected. Thanks.

Claude

 

[Naty1]

A changing magnetic field creates an electric and 'a changing electric field creates a magnetic field.

Wikipedia has some relevant material here:
http://en.wikipedia.org/wiki/Electromagnetic_fields

For example, see:
http://en.wikipedia.org/wiki/Electromagnetic_fields#Dynamics_of_the_electromagnetic_field

In the past, electrically charged objects were thought to produce two types of field associated with their charge property. An electric field is produced when the charge is stationary with respect to an observer measuring the properties of the charge, and a magnetic field (as well as an electric field) is produced when the charge moves (creating an electric current) with respect to this observer. Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole — the electromagnetic field.

Once this electromagnetic field has been produced from a given charge distribution, other charged objects in this field will experience a force (in a similar way that planets experience a force in the gravitational field of the Sun). If these other charges and currents are comparable in size to the sources producing the above electromagnetic field, then a new net electromagnetic field will be produced. Thus, the electromagnetic field may be viewed as a dynamic entity that causes other charges and currents to move, and which is also affected by them

 

[kcdodd]

You can gain some insight from how using D and H have different properties than E and B.

(using natural units)

<br /> H = B - M<br />

<br /> D = E + P<br />Looking only at the static situation for a second, curl of H deals only with free current, because we took out the curl of the magnetic dipoles (bound currents). Looks prettier eh.

<br /> \nabla \times H = \nabla \times B - \nabla \times M = J_f<br />

Similar thing for electric field, only we want free charge instead of free current.

<br /> \nabla \cdot E = \rho_f - \nabla \cdot P<br />

<br /> \nabla \cdot D = \nabla \cdot E + \nabla \cdot P = \rho_f<br />

It's simpler because you don't have to take into account the currents and charges inside the the dipoles. Incidently, it's interesting that H can have non-zero divergence, like a magnetic source. And curl of D is non-zero, like magnetic current.

<br /> \nabla \cdot H = \nabla \cdot B - \nabla \cdot M = - \nabla \cdot M = \rho_m<br />

<br /> \nabla \times D = \nabla \times E + \nabla \times P = \nabla \times P = J_m<br />

 

[kinley]

When it comes to outside of a magnetic material both these fields are same. Inside a magnetic material they are completely different especially with regards to relative magnitude and direction. B field is dependent to a considerable extent on currents, both microscopic and macroscopic while the H field depends on microscopic currents. B field lines always form loops around the total current. In the case of H field the lines always loops around free current. They begin and end near magnetic poles.

 

[yungman]

Based on Maxwell's equations, electric fields are generated by changing B fields, while H fields are generated by changing electric fields.
In dc fields, static electric E fields create currents (magnetization currents) I (when σ>0) which in turn produce static H fields. A static B field (actually d/dt ∫B·n dA = 0) cannot produce an electric field E.

Even if we used natural units where μ₀ = ε₀ = 1, this distinction between B and H remains. In magnetic materials, B is not linearly related to H due to the magnetization term M. If B = H in space, then B ≠ H in magnetic materials.

If voltage = d/dt ∫B·n dA, then how can curl H= σ·E = σ·voltage/length if B = H?

So even if μ₀ = ε₀ = 1, they have to have units. Going without units is about as rational as claiming that h-bar = c =1. (I may get FLAK on this).

Bob S

My understanding is static electric field can cause FREE charges on conducting surface because of free electron movement. It also cause polarization of dielectric material which the charges on the surface consist of only BOUNDED charges.

Static electric field will not cause magnetic field and any surface current density. Am I getting this right?

 

[Bob S]

My understanding is static electric field can cause FREE charges on conducting surface because of free electron movement. It also cause polarization of dielectric material which the charges on the surface consist of only BOUNDED charges.

Static electric field will not cause magnetic field and any surface current density. Am I getting this right?

A static electric field on a conductor produces a dc current. You should think of the static (electric field) Maxwell equation for conductors as Curl H = σE = J, because ∂E/∂t = 0, where J is current density. The integral form of this in vacuum is Ampere's Law.

Wind a N= 100-turn coil around a solenoid coil form. It should be at least 12 ohms. Make it about L = 0.1 meter long. Find a dc (static) voltage source, like 12 volts. Make sure the coil resistance R is at least 12 ohms. Attach voltage source V to coil. The current I (=V/R) times N divided by L is H = NI/L ampere turns per meter, sometimes called the magnetic intensity.

I am using the word static to mean non-changing, like ∂E/∂t = 0.

Bob S

 

[yungman]

A static electric field on a conductor produces a dc current. You should think of the static (electric field) Maxwell equation for conductors as Curl H = σE = J, because ∂E/∂t = 0, where J is current density. The integral form of this in vacuum is Ampere's Law.

Wind a N= 100-turn coil around a solenoid coil form. It should be at least 12 ohms. Make it about L = 0.1 meter long. Find a dc (static) voltage source, like 12 volts. Make sure the coil resistance R is at least 12 ohms. Attach voltage source V to coil. The current I (=V/R) times N divided by L is H = NI/L ampere turns per meter, sometimes called the magnetic intensity.

I am using the word static to mean non-changing, like ∂E/∂t = 0.

Bob S

Sounds like we are talking about different things. Correct me if I mis understand this.

You are talking about putting a voltage across a conducting material where and electric field form inside the material and current flow even in DC. That I agree with you. As long as there is some dc resistance in the material that cause a voltage drop, an electric field will formed. And yes, when a current passing through, magnetic field will form.

What I am referring is a conducting material INSULATED in vacuum with electric field hitting it. Electric field only cause free charge to move ( eg, if the arrow of the electric field hitting the metal surface, electrons are drawn to the surface and leaving possitive charges on the opposite side of the metal.

 

[Bob S]

What I am referring is a conducting material INSULATED in vacuum with electric field hitting it. Electric field only cause free charge to move ( eg, if the arrow of the electric field hitting the metal surface, electrons are drawn to the surface and leaving possitive charges on the opposite side of the metal.

If "electric field cause free charge to move", and "electrons are drawn to the surface", then there are currents. If there are currents, then by Amperes Law, there are magnetic fields.

Bob S

 

[yungman]

If "electric field cause free charge to move", and "electrons are drawn to the surface", then there are currents. If there are currents, then by Amperes Law, there are magnetic fields.

Bob S

But that is at the initial condition when the electric field is first apply, the current flow as charge more to the surface. Yes I agree there will be magnetic field generated initially. But as everything settle down, as long as the electric field is constant, the free charge remain constant and nothing change, no more current moving. The whole thing is in a stand still and no current flows and the magnetic field disappeared. It just stay there frozen until the field intensity change. That is the reason I call it surface charge.

Yes in varying electric field, it become current and magnetic field is generated.

 

[Trifis]

The real question here is why the electric field (E) is weaker inside matter while the magnetic field (B) is stronger!

I'm surprised that nobody bothers to explain that. I mean if we think of magnetic properties of materials as a collection of microscopic magnetic dipoles then the application of an external magnetic field (H) should have had a similar effect with the polarization case in dielectrics, where the inner electric field (E) is weakened from the opposed electric field (P/ε) of the microscopic electric dipoles.

PS: In case of magnetism the dipoles align parallel with the external field while in electrostatics they align antiparallel...

 

[Trifis]

Moreover if the alignment of those dipoles was indeed analog (e.g. ferroelectrics vs ferromagnets), the analogon with repect to the formulas for D and H would still remain "unsettled", because their derivation does not depend on the above mentioned behaviour of the matter! Particularly as you already know, H field is the interior magnetic field MINUS the magnetization while D field is the interior electric field PLUS the polarization.

What's the point of those differences? Are they a direct consequence of the native difference between electricity and magnetism expressed through the maxwell equations? Please note that their derivation has NOTHING to do with the strength of the interior forces and still the two theories are not compatible...

 

[cabraham]

B is magnetic flux density, whereas H is magnetic field intensity. H has units of amp-turn/meter, whereas B has units of weber/turn-meter^2. In non-ferrous materials they have a simple inter-relation given by B = mu*H. In ferrous media it is a non-linear relation and numerical and/or graphical methods of computation are used.

One is an intensity, the other a density. A rough analogy exists for E & D as well, although I advise against trying to correlate D/E with B/H. D is the electric flux density, and E is the electric field intensity. All 4 quantities are equally important. I don't think we can define things using just 2 of them.

Claude

 

[Trifis]

Sorry but this is not a relevant answer!

Everyone is at least roughly aware of the nature of this fields... Both of my assertions deal with differing subjects.

One has to go back to how these quantities are derived through the macroscopic averaging process and note a fundamental difference both to the behaviour of dipoles as well as the maxwell equations between electrostatics and magnetostatics and if those two observations are somehow connected. I elaborated that already in my first posts.

 

[DrZoidberg]

Here is an interesting question:
Imagine there was a particle that has exactly the same properties as an electron except for it's charge. Instead of a negative charge this particle is a south monopole. Further imagine a particle nearly identical to a proton except that it is a north monopole. What happens if you replace every single electron and every single proton in the universe with those monopoles? Would the world come to an end?
-- disclaimer: the following statements are my personal opinions --
I say we wouldn't even notice the difference. The laws of physics and chemistry would remain the same. That is because the laws of physics that govern the behavior of electric and magnetic fields are 100% equivalent in every respect.
The different treatment that those two kinds of fields receive in maxwells equations is due to 2 reasons.
1. maxwell incorporated the fact that there are electric charges but no magnetic charges(monopoles) into his equations
2. different definitions are being used for things that are really equivalent and could be defined in an equivalent manner.

 

[Trifis]

@DrZoidberg What you are practically suggesting is that the existence of magnetic monopoles would make electicity and magnetism perfectly interchangeable when relativity is taken into account. E.g. a pure electric field could be "Lorentzly" transformed under this prerequisite into a pure magnetic field and vice versa.

That still doesn't explain the disanalogy of the H/D-field formulas! I've come to believe that the very reason we have given these forms in those general macroscopic quantities (namely D=E+ε_{0}P : weaker E-Field in matter --- H=B/μ_{0}-M : stronger B-Field in matter) is the fact that most of the electric materials are dielectric and not ferroelectric (which implies the weakening of the E-Field) and the opposite holding for magnetic materials. I do not see any other mathematical explanation since the definition ρ_{P}=-∇P used in the derivation of the D formula is totally arbitrary. Thus a similar assumption for the diamagnetic materials would yield H=B/μ_{0}+M instead!

PS: To make matters worse, polarization depends on the outer applied field M=χ_{m}B_{a}/μ_{0} while magnetization depends on the inner reduced electric field P=χ_{e}ε_{0}Ε_{i} ! I hate it when there is no symmetry where it should be...

 

[DrZoidberg]

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.
Before you complain, let me explain what I mean by "real field".
Imagine a magnetic monopole and a very thin magnet so that the monopole is able to just shoot through the magnet. It then goes around in circles, shooting through the magnet over and over again. If the B field would determine the force on the monopole, it would pick up more and more energy with each revolution. The H field however points from N to S inside a permanent magnet so there is no energy gain for the monopole particle. I know there are no monopoles but the laws of physics don't rule out their existence. Since the H field determines the force on the particle it is the real field.
Btw. in the same way you can use conservation of energy to show that the E field inside an electret has to point form + to -.
So I don't see an asymmetry there.

D=ε0E+P : stronger D-Field in matter
B=μ0*(H+M) : stronger B-Field in matter

The asymmetry between those two equations is just because P is not defined analogically to M. If you defined M as µ0*magnetization you would get B=μ0H+M. It's all just a matter of definition.

I wished someone with better math skills than me could post a complete proof for the equivalent bahavior of electric and magnetic fields. I'm sure it can all be derived from the known laws of electromagnetism.

 

[Trifis]

Hmmm that would clear up things if the E field was the real equivalent of H field instead of D field... BUT this is way too indistinguishable in the mainstream bibliography cause what is mostly stated is that D and H are the auxiliary fields introduced in order to simplify the maxwell equations in terms of FREE CHARGES/CURRENTS. And that too makes sense, because we tend to treat D and H field as the independent macroscopic fields which the respective E and B fields would be equal to in vacuum.

You are right to say that H field is the stable exterior exciter while B is the reduced interior. But think the case of electrostatics over. Consider a capacitor filled with dielectric material. The E field here is depended on the material and more often than not reduced due to the polarization. Thus it is the E field that is combined (by your definition) with P not the D field! D field on the other hand depends only on the free charges of the capacitor and we can thereby call it the exterior factor for the sake of argument... That again explains why we define ρ_{P}=-∇P and not ρ_{P}=+∇P as it should be! I think the founders have historically tangled up the concepts :P

Apropos the μ and ε assymetry is certainly not a problem for me!

 

[DrZoidberg]

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.

 

[Trifis]

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 possesses 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 referring 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.

 

[vanhees71]

On an elementary level, there is only one electromagnetic field, given by the Faraday tensor F_{\mu \nu} or, equivalently in the 3D formulation, \vec{E} and \vec{B}. 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

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

and the inhomogeneous equations

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

If you have macroscopic matter, it's way too complicated (and also not necessary!) to solve these fully microscopic Maxwell equations, since here \rho and \vec{j} 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 \vec{D} and \vec{H} 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

\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}}.

and the constitutive equations

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

Here, \vec{j}_{\text{free}} and \rho_{\text{free}} 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 \sigma is the conductivity and \epsilon and 1/\mu 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.

 

[DrZoidberg]

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 possesses 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

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

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

and the inhomogeneous equations

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

So that means when you have pure fields and no matter, j and p become 0 and you get

\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0
\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = 0
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?

 

[dipole]

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.

 

[DrZoidberg]

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.

 

[vanhees71]

So that means when you have pure fields and no matter, j and p become 0 and you get

\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0
\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = 0
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 \vec{E} and \vec{B} 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 \vec{E} and \vec{B}, while \vec{D} and \vec{H} 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.

 

[dipole]

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.

 

[Trifis]

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_{0}E+P , ∇D=ρ_{free}

H=B/μ_{0}-Μ , ∇H=j_{free}

There is an obvious asymmetry here... To my eyes at least!

 

[cabraham]

https://www.physicsforums.com/showpost.php?p=3930927&postcount=20

To all:

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

Claude

 

[dipole]

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_{0}E+P , ∇D=ρ_{free}

H=B/μ_{0}-Μ , ∇H=j_{free}

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.

 

[Trifis]

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.

 

[dipole]

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.

 

[Trifis]

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...

 

[dipole]

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.

 

[cabraham]

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

 

[DrZoidberg]

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.

 

[cabraham]

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

 

[vanhees71]

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_{0}E+P , ∇D=ρ_{free}

H=B/μ_{0}-Μ , ∇H=j_{free}

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 \vec{P} and magnetization \vec{M} 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 \vec{H} has to be grouped together with \vec{E}, which is not correct since relativity tells us that \vec{E} and \vec{B} have to be grouped together to the antisymmetric Faraday tensor and \vec{D} and \vec{H} 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!

 

[vanhees71]

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 (\vec{E},\vec{B}) and \vec{D} and \vec{H} are derived quantities from the linear-response approximation of the interaction between the electromagnetic field and (bulk) matter.

 

[DrDu]

This is just what I said, written in an explicit form in (imho ugly) SI units: The polarization \vec{P} and magnetization \vec{M} 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 \vec{j}=\partial \vec{P}/\partial t + \nabla \times \vec{M} and -\nabla \cdot \vec{P}=\rho.
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

 

[DrZoidberg]

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 \vec{H} has to be grouped together with \vec{E}, which is not correct since relativity tells us that \vec{E} and \vec{B} have to be grouped together to the antisymmetric Faraday tensor and \vec{D} and \vec{H} 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.

 

[Jano L.]

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 p and m, 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.

 

[DrDu]

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.

 

[Jano L.]

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:

<br /> \mathbf f = \rho_{free} \mathbf E + \mathbf j_{free} \times \mathbf B<br />

<br /> \mathbf f = \rho_{total} \mathbf E + \mathbf j_{total} \times \mathbf B<br />

<br /> \mathbf f = \rho_{free} \mathbf D + \mathbf j_{free} \times \mathbf H<br />

<br /> \mathbf f = \rho_{total} \mathbf D + \mathbf j_{total} \times \mathbf H<br />

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.