# Dielectrics vs. Diamagnetics

1. Jan 15, 2014

### pero2912

So I was thinking about similarities in the nature between these two which could justify similarity in the nomenclature.

So, diamagnetic materials are those whose atoms do not have permanent magnetic dipols, so when you apply magnetic force on them, they create local dipole oriented ANTIPARALLEL to the external magnetic field. That's why the magnetic susceptibility of diamagnets is negative.

Dielectric materials, similarly, are those whose atoms do not have permanent electric dipols, so when you apply electric force on them, they create local dipole oriented ANTIPARALLEL to the external electric field. But, for some reason electric susceptibility of dielectrics is always positive!

Second thing, when an atom has a permanent magnetic moment we call it a paramagnet, but when an atom has permanent electric moment (in.e. water) we call it again a dielectric.

I'd like to hear your opinion on this one.

Regards,

Pero

2. Jan 16, 2014

### Meir Achuz

That's a confusion caused by SI units.
In a dielectric material an electric field E polarizes molecules in the direction of the field (parallel not anti-parallel) whether they are permanent or induced electric dipoles.. In either case the polarization P is positive (and epsilon is greater than one, with D=epsilon E.
In a diamagnetic material, a B field polarizes molecules that do not have permanent magnetic moments in a direction opposite to the field. (This follows from Lenz's law.)
In SI units, even though it is B causing the magnetization, The relationship between B and H is written as B=mu H. That is why epsilon is greater than one for a dielectric material, and mu is less than one for a diamagnetic material.
A paramagnetic material has molecules with permanent magnetic moments, which the B field polarizes in the direction of the B field. Then the magnetization M is positive and mu is greater than one.

3. Jan 16, 2014

### DrDu

That is a convention which is not peculiar to the SI system. Up to some factors like 4π and c, this relationship looks alike in all other systems of units.
It may be somewhat unfortunate from a theoretical point of view, but it is easier to fix H in the usual geometry used for measurement, namely using long cylindrical coils.

4. Jan 16, 2014

### Meir Achuz

You are right that that is used in all known (to me) systems. It must go way back.

5. Jan 16, 2014

### Jano L.

Hey guys, you seem to think that theoretically there is some superiority in writing the linear relation as

$$\mathbf M = \chi \mathbf B$$

$$\mathbf M = \chi \mathbf H.$$

Could you give any reason? I see none, as both are macroscopic fields which are not in general equal to the field acting on the atoms and molecules composing the material and contributing to total magnetic moment. If you calculate magnetic force acting on a moving charge in a prolate hole parallel to the field, it is $q\mathbf v \times \mu_0 \mathbf H$. If you calculate the same for oblate hole, you get $q\mathbf v \times \mathbf B$. Actual molecules aren't in either kind of hole, but are packed close together which makes the acting field much more complicated. To express empiric linear relation, theoretically there does not seem reason to prefer any one of $\mathbf H,\mathbf B$.

Only experimentally, as DrDu says, there is advantage in using $\mathbf E$ in electrostatics and $\mathbf H$ in magnetostatics; they are both easily controllable and measurable, while $\mathbf D$ and $\mathbf B$ require some further steps.

6. Jan 17, 2014

### pero2912

Thanks for the help guys, you made thing have more sense. But one more question remains undiscussed:

7. Jan 17, 2014

### vanhees71

The definition of the dieelectric constant and permeablity of a medium in macroscopic electrodynamics is just historical, while only the choice of these quantities in the vacuum are subject to arbitrariness in the choice of the units. In the SI they are called $\epsilon_0$ and $\mu_0$ and are there only, because in the SI one wants to introduce a fourth base unit in addition to the three mechanical base units for time, length, and mass (i.e., s, m, kg in the SI).

The confusion goes back to a wrong understanding of the nature of the macroscopic field quantities $\vec{E}$, $\vec{B}$, $\vec{D}$, and $\vec{H}$. This would have been prevented only, if right in the very beginning the relativistic nature of the electromagnetic phenomena had been known. With the advent of special relativity (fully established concerning the physics by Einstein's famous paper of 1905 and mathematically analyzed fully by Minkowski in 1908) it's clear that $\vec{E}$ and $\vec{B}$ are one physical object and s oare $\vec{D}$ and $\vec{H}$. The latter are derived quantities, containing the medium response to perturbations by an external electromagnetic field. In general they are functionals of the electromagnetic field. In linear-response approximation for a homogeneous and isotropic medium one get's the known simplifying constitutive relations. Unfortunately by the historical misunderstanding of the electromagnetic quantities one writes
$$\vec{D}=\epsilon \vec{E}, \quad \vec{H}=\frac{1}{\mu} \vec{B},$$
which are valid for the Fourier components wrt. to time and $\epsilon=\epsilon(\omega)$ and $\mu=\mu(\omega)$, as long as spatial dispersion can be neglected.

8. Jan 17, 2014

### Meir Achuz

Look at the textbook derivation of H. It is defined as B-4\pi M as a mathematical convenience.

9. Jan 17, 2014

### Meir Achuz

F always equals qvXB.
H_parallel is continuous so H in a prolate hole equals H in the matter, and then B in the hole equals H in the matter.
For an oblate hole, B is continuous.

10. Jan 17, 2014

### Jano L.

Perhaps I did not say it clearly enough. Given two vectors describing magnetic field in a body of matter at point $\mathbf x$ ($\mathbf H,\mathbf B$), and making a hole around $\mathbf x$ afterwards, the magnetic force on a test charge should behave as I wrote above. The two fields give different value for the magnetic force for different kind of a hole. Molecule at $\mathbf x$ in the $original$ body without hole will feel microscopic field $\mathbf b(\mathbf x)$ that is most probably much more complicated than either $\mathbf H(\mathbf x)$ or $\mathbf B(\mathbf x)$ since it reflects the presence of myriads of microscopic molecules around. So there is no reason to say $\mathbf H$ or $\mathbf B$ is "responsible" or "causes" magnetization and the other one does not; it is the microscopic field $\mathbf e, \mathbf b$ that should play this role.

11. Jan 17, 2014

### Meir Achuz

What I meant was that H and B are equal inside a prolate hole, and each are equal to H inside the surrounding material. You are correct about needing e or b if there is no hole, but you want the field a molecule feels.

12. Jan 18, 2014

### DrDu

I think these theoretical constructs of drilling holes are not admissible and least so on a microscopic level. Permeability refers to homogeneous media and a hole severely violates homogeneity.
There is also no need to distinguish between microscopic and macroscopic fields.
Rather, the equations are best formulated in Fourier space:
$B(\omega,k)=\mu(\omega,k) H(\omega,k)$.
The value of $\mu$ usually tabulated is then $\mu(0,0)$.

13. Jan 18, 2014

### Jano L.

DrDu, I mentioned holes just to point out the meaning of the macroscopic fields: you need macroscopic hole (much smaller than the char. length of variation of the field, of course, the field has to be uniform across the region) to place and measure motion of the test charge and then the $\mathbf H,\mathbf B$ fields give magnetic force inside it for two extreme cases: long prolate hole, and wide oblate hole.

This shows that neither of the two fields is any better in the role of "cause" of the magnetizing force on atom/molecule, since it is clear the atom/molecule $is~not$ inside such a hole, but is surrounded by other molecules and acted upon by more complicated magnetic field.

There is need if we are interested in calculating force acting on the charges in individual molecule at $\mathbf x$. In classical theory, this is not just equal to
$$q\mathbf E(\mathbf x,t) + q\mathbf v/c \times \mathbf B(\mathbf x,t),$$
with macroscopic fields, or in quantum theory the Hamiltonian is not just minimal coupling Hamiltonian with $\mathbf A(\mathbf x,t)$ giving the above macroscopic fields. Such macroscopic fields do not obey Maxwell's equations when the grainy nature of the electric charge density of the medium is taken into account, and thus are not reliable for the calculation of what happens to individual molecule.

To remedy this, one has to describe the field in a more refined way, for example by microscopic field $\mathbf e(\mathbf x),\mathbf b(\mathbf x)$ with high-resolution in space. The two kinds of fields then have in general $different~value$ at the position of the molecule, although on average they have to have the same phase at #\mathbf x## of molecule and be proportional in magnitude (not equal) to preserve macroscopic dispersion theory.

14. Jan 20, 2014

### tiny-tim

maybe i'm making a fuss about (literally! ) nothing, but why do people refer to prolate and oblate holes?

most people (myself certainly) aren't sure of the difference between prolate and oblate …

why aren't they just called rod holes and disc holes ?​