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

[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.

 

[Jano L.]

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.]

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
The prototype of a dielectric function of a metal is the Lindhard (or refinements like Sjolander Singwi) dielectric function which is treated in every textbook on many body physics see e.g. here:
www.itp.phys.ethz.ch/education/lectures_fs10/Solid/Notes.pdf
Evidently, it includes the effect of conductivity.

 

[vin300]

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.]

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 modeled 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.]

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