# Ampères law: Getting $\nabla \times \vec H = \vec J_{free} + \frac{\partial \vec D}{\partial t}$ by taking the cross product

I've seen two different versions of Ampère's law and I'm having trouble connecting them:

$$\nabla \times \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t}$$

$$\nabla \times \vec H = \vec J_{free} + \frac{\partial \vec D}{\partial t}$$

Trying to "derive" the second version of it

Assumption: $$\vec H = \frac{\vec B}{\mu_0} - \vec M$$

$$\nabla \times \vec H = \nabla \times (\frac{\vec B}{\mu_0} - \vec M) = \vec J + \epsilon_0 \frac{\partial \vec E}{\partial t} - \vec J_m = \vec J_{free} + \epsilon_0 \frac{\partial \vec E}{\partial t} \stackrel{?}{\neq} \vec J_{free} + \frac{\partial \vec D}{\partial t}$$

Since according to my understanding $$\vec D = \epsilon \vec E \neq \epsilon_0 \vec E$$

Where am I going wrong?

The reason for this discrepancy is that your first equation is, in fact, incorrect. In a dielectric media, the field $$\boldsymbol{E}$$ is not the only one to affect the flux, but also $$\boldsymbol{P}$$. The latter give rise to bound-charge current density

$$\boldsymbol{J}_{\rm bound}=\mu_{0}\frac{\partial\boldsymbol{P}}{\partial t}$$

This contribution can be combined with $$\mu_{0}\varepsilon_{0}\frac{\partial\boldsymbol{E}}{\partial t}$$ to yield $$\mu_{0}\frac{\partial\boldsymbol{D}}{\partial t}$$, giving the correct term that you are missing.

Reference:

[1] Purcell, E. M., Morin, D. J., Electricity and Magnetism, 3rd Edition, pp. 505-507.

Igonring the non-local electromagnetic response (quadrupole and higher effects) one has

$$\mathbf{D}=\epsilon_0 \mathbf{E}+\mathbf{P}$$, where $$\mathbf{P}$$ is the polarization = density of electric dipoles per volume.

Also one has $$\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M}$$, where $$\mathbf{M}$$ is the magnetization = density of magnetic dipoles per volume.

The full current density is: $$\mathbf{J}=\mathbf{J}_{free}+\frac{\partial\mathbf{P}}{\partial t}+\mathbf{\nabla}\times\mathbf{M}$$

This shoud be sufficient to proove everything.Start with $$\mathbf{\nabla}\times\mathbf{H}=\mathbf{J}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$

The tricky bit is to show how the magnetization/polarization relate to current density. One can do it by considering point-like electric and magnetic dipoles. I don't have time to show it here (ask me if you are actually interested in it)