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I'm studying the book Polarimetric Radar Imaging: From Basics To Applications and on page 31, there's this sentence:

In the following, we shall consider the propagation of an electromagnetic wave in a linear medium (free of saturation and hysteresis), free of sources. These hypotheses impose the conditions $\vec M(\vec r,t)=\vec P(\vec r,t)=\vec 0$ and $\vec J_a(\vec r,t)=\vec 0$

This implies that a linear medium is free of saturation. Then what's the meaning of saturation here? And how do we represent saturation or being free of saturation in mathematical sense and notation?

This also implies that a linear medium is free of hysteresis. What's the meaning of hysteresis? How do we represent hysteresis or being free of hysteresis in mathematical notation?

A linear medium is free of source and I think the mathematical notation for that will be $\vec J_a(\vec r,t)=\vec 0$. won't it? (Regarding the meaning of source here is the source to create electromagnetic field or electrical current)

In another Persian source, I have read that:

In a linear medium, if we have the value of one of the quantities, electromagnetic field intensity (H) or electromagnetic induction (B),the value of the unknown quantity can be computed in term of the value of known one

Would you explain the exact meaning of a linear medium in electromagnetism?

P.S.
In a ferromagnetic material saturation state occurs when all of the microscopic dipoles get the same direction and sense of the external electromagnetic field. So if the material or medium is free of saturation, that means these dipoles should neutralize each other and so $\vec P(\vec r,t)=\vec 0$, $\vec P$ is the polarization vector.

In a ferromagnetic material a portion of the microscopic dipoles remain in the same direction after removing the external electromagnetic field and cause the material to magnetize, the remaining magnetism in this way is called magnetic hysteresis. So if the material or the medium is free of hysteresis, that means that no magnetism remains after removing the external field and so $\vec M(\vec r,t)=\vec 0$, $\vec M$ is the magnetization vector.

That was an intuitive explanation for myself, please tell me what's the exact meaning and mathematical notation?

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  • $\begingroup$ It is not clear to me that the absence of source charges implies that $\vec{M} = \vec{P} = \vec{0}$. In fact, I would have thought that the electromagnetic wave would disturb the medium and cause a polarization to form. Similarly, a magnetic field would induce a magnetization. However, with no free sources I would expect $\vec{\nabla} \cdot \vec{P} = \vec{0}$. $\endgroup$
    – Brian Moths
    Commented Aug 26, 2015 at 20:06
  • $\begingroup$ $\nabla \cdot {\boldsymbol P} = \rho_b$ is for bound volume charges. Free charges are slightly different -- total field responds to $\nabla \cdot {\boldsymbol E} = \rho_f$ $\endgroup$
    – John M
    Commented Aug 26, 2015 at 20:08
  • $\begingroup$ @JohnM You got it backwards here and in a comment to your answer. $D$ gives you $\rho_f$ and $E$ of course gives you the total $\rho$. Since $\rho_f$ is zero, $\nabla \cdot D$ must be zero, but since we are linear, $\nabla \cdot P$ must be zero as well. $\endgroup$
    – Brian Moths
    Commented Aug 26, 2015 at 22:07

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In terms of a dielectric, it means there is a linear constitutive relation between the vectors.

$${\boldsymbol D} = \epsilon_0 {\boldsymbol E} + {\boldsymbol P}$$

Or

$${\boldsymbol D} = \epsilon_r \epsilon_0 {\boldsymbol E}$$

where $\epsilon_r \epsilon_0$ is a scalar relative and vacuum permitivitty. This way, there is now a linear relation between the applied field and the electric displacement. There are certainly situations where

$$D_j = \epsilon_{ij} E_i$$

for some rank two tensor $\epsilon_{ij}$, ie it is more favorable to apply a field along one direction and get a response along another -- re: anisotropy. If a dielectric material is free of saturation, then you can apply as much of a field as you want and get a corresponding electric displacement.

A hysteresis loop in a ferroelectric for example, does not allow this since there is a saturation for large field, so the mathematical way of expressing a dielectric material without a hysteresis is exactly the linear relation above for the electric displacement(and corresponding polarization). From Wiki:

enter image description here

In your first quote here, the author says the material is free of sources, which includes the bound charge source $\rho_b$ that arises $-\nabla \cdot{\boldsymbol P} = \rho_b \Rightarrow {\boldsymbol P} = 0$ as in the above comment. Think of a dielectric sphere in an external field problem that can be solved with separation of variables. There exists $\mathbf{no}$ bound charges within the material so the field inside is aligned with the external field and linearly increases with it.

You are also right that no polarization or magnetization should remain when the applied fields are removed $\mathbf{if}$ the material is free of hysteresis (see above hysteresis plot for zero ${\boldsymbol E} \Rightarrow {\boldsymbol D} \neq 0 $). Please let me know if I can clarify or expand anything else in the comments.

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  • $\begingroup$ I haven't studied electromagnetism a lot and just from Maxwell's equations, I know that $\nabla.D=\rho$ then how do we conclude that $\nabla.P=\rho$? Also tell me a little about ferromagnetic and ferroelectric materials. $\endgroup$
    – Sepideh Abadpour
    Commented Aug 26, 2015 at 21:25
  • $\begingroup$ Also, a ferroelectric at least, will have different charges.. $\nabla \cdot {\boldsymbol P}$ is from the bound charge that arises from the dipole on the lattice. Think of this as a fixed dipole($\pm$ charge) at some point in the crystal that generates an electric field that falls off like $1/r^3$ locally, it can rotate(anisotropy), it can be stretched but it can't be stretched to infinity (this contributes to hysteresis in my understanding). Free charges are different, these are charges that can float around, be moved by an applied field, ect. $\endgroup$
    – John M
    Commented Aug 26, 2015 at 21:30
  • $\begingroup$ Actually its $\nabla \cdot {\boldsymbol D} = \rho_f = \nabla \cdot \left(\epsilon_0 {\boldsymbol E} + {\boldsymbol P} \right) = \rho_\mathrm{total} - \rho_b $ $\endgroup$
    – John M
    Commented Aug 26, 2015 at 23:04
  • $\begingroup$ @SepidehAbadpour The difference is that (\nabla \cdot D = \rho_{free}) and (\nabla\cdot P = \rho_{bound}). Free charges are charges that are present in the material but not a part of the material. Bound charges are part of the material. Normally, they are zero as the material is electrically neutral but an electric field can cause a slight displacement between the electrons and nuclei inside a material creating an apparent charge which is called the bound charge. $\endgroup$
    – Poseidaan
    Commented Dec 4, 2022 at 19:07

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