Using displacement field vs. electric field to calculate curl of magnetic field So let's say we have a medium with polarization $\vec{P} = \gamma \nabla \times E$, with no free currents or charges.
So we know that $H = \frac{1}{\mu_0} B - M$ , $D = \epsilon_0 E+P$ and $\nabla \ \times H =\mu_0 J_{free}+\frac{\partial D}{\partial t} $ here reduces to $\nabla \ \times H =\frac{\partial D}{\partial t} $
So since there is no free current I mistakenly thought that $\nabla \times B = \mu_0 J + \frac{\partial E}{\partial t}$ reducing to here $\nabla \times B = \frac{\partial E}{\partial t}$ but if I say that
$$\nabla \times B = \mu_0 \nabla \times H = \mu_0 \frac{\partial D}{\partial t} = \mu_0 \epsilon_0 \frac{\partial E}{\partial t} + \mu_0 \frac{\partial P}{\partial t} \not= \frac{\partial E}{\partial t}$$
so why do I need to use the displacement field here? this is like saying there is a current due to the polarization but what is this current? what am I missing here?
 A: Of course there is current due to changes in polarization. For polarization to change, charged particles have to change positions, and this motion means there is electric current.
In a dielectric with no magnetization, total current can be expressed as
$$
\mathbf J = \frac{\partial \mathbf P}{\partial t}.
$$
In a magnetic medium with no electric polarization, total current can be expressed as
$$
\mathbf J = \nabla \times \mathbf M.
$$
There is no universal formula for total current, it depends on the medium. In magnetic conductor in ohmic regime, total current is
$$
\mathbf J = \nabla \times \mathbf M + \sigma \mathbf E.
$$
A: By your account, you have the following constitutive relations:
$$ = ε_0 + γ∇×, \hspace 1em  = μ_0( + ),$$
and I'm assuming $γ$ is a constant. From the Maxwell equation
$$\frac{∂}{∂t} + ∇× =  ⇒ ∇× = -\frac{∂}{∂t},$$
so we can write
$$ = ε_0 - γ\frac{∂}{∂t}.$$
You're assuming $ = $, from which it follows:
$$∇× - \frac{∂}{∂t} =  =  ⇒ ∇× = \frac{∂}{∂t}.$$
Thus, upon substitution of $$, we have:
$$∇× = μ_0 ∇× + μ_0 ∇× = μ_0 \frac{∂}{∂t} + μ_0 ∇×,$$
and upon further substitution of $$, we have (assuming $γ$ is constant):
$$\frac{∂}{∂t} = ε_0 \frac{∂}{∂t} - γ \frac{∂^2}{∂t^2},$$
which, after applying $μ_0 ε_0 = (1/c)^2$, results in the formula:
$$∇× = μ_0 ∇× + μ_0 ∇× = \frac{1}{c^2} \frac{∂}{∂t} - μ_0 γ \frac{∂^2}{∂t^2} + μ_0 ∇×.$$
With the constitutive relations, you can eliminate one field from each pair $(,)$ and $(,)$.
The final equation can be arranged as:
$$∇× - \frac{1}{c^2} \frac{∂}{∂t} = μ_0\left(∇× - γ \frac{∂^2}{∂t^2}\right),$$
so, there's an effective current given in the parentheses on the right, arising ultimately from the constitutive relations. Bringing back the $-$ Maxwell equation, we also have
$$\frac{∂^2}{∂t^2} = \frac{∂}{∂t}\left(\frac{∂}{∂t}\right) = \frac{∂}{∂t}\left(-∇×\right) = -∇×\left(\frac{∂}{∂t}\right).$$
Assuming, again, that $γ$ is constant,
$$∇× - \frac{1}{c^2} \frac{∂}{∂t} = μ_0 ∇×\left( + γ \frac{∂}{∂t}\right).$$
So, the effective current, by virtue of being a curl, is divergence-free.
