Divergence of $H$ I have read in D.J. Griffith's electrodynamics text that if one finds the field $H$ by Ampère's law, that means that the divergence of $H$ is zero, because the free current alone determines the answer. Why is that?
 A: In general, you need to know both the $\textrm{curl}$ and $\textrm{div}$ of a vector field to determine the field. Now you always have $\textrm{div}\textbf{B}=\textrm{div}(\textbf{H+M})=0$ hence $\textrm{div}\textbf{H}=-\textrm{div}\textbf{M}$ always, so if you can find $\textbf{H}$ from symmetry arguments using Ampere's law $\textrm{curl}\textbf{H}=\textbf{J}$ or equivalently $\oint \textbf{H}\cdot d\textbf{s} = I$ then you implicitly assume that $\textbf{M}=0$, and thus $\textrm{div}\textbf{H}=0$. 
A: The definition of the H field is given as follows:
$$\mathbf{H}=\dfrac{\mathbf{B}}{\mu_0}-\mathbf{M}$$
where $\mathbf{B}$ is the total magnetic field in all space and $\mathbf{M}$ is the magnetization vector. Taking the line integral of $\mathbf{H}$ on a closed loop and using Ampere's law, we get,
$$\oint \mathbf{H}\cdot d\mathbf{l}=\oint\dfrac{\mathbf{B}}{\mu_0}\cdot d\mathbf{l}- \oint \mathbf{M}\cdot d\mathbf{l}=\mathbf{I}_{\text{total}}-\mathbf{I}_{\text{bound}}$$
This is beacuse the magnetic field $\mathbf{B}$ takes into account all the currents in the region, whereas the magnetiztion vector $\mathbf{M}$ only depends on the bound currents produced in the material by the external field. 
Thus we can say that,
$$\oint \mathbf{H}\cdot d\mathbf{l}= \mathbf{I}_{\text{free}}$$
Now, the free current cannot have a divergence since this would mean that there is an accumulation of charge somewhere in space and that there is a diffrence between the charge entering a closed region and the charge leaving the same closed region. Thus if $\mathbf{H}$ were to be found using Ampere's law, 
$$\nabla \cdot \mathbf{H}=0$$
A: The divergence of $\mathbf{H}$ is not, necessarily, zero. Similarly, the curl of $\mathbf{D}$ isn't necessarily zero, either. What this boils down to is an exercise in Helmholtz decomposition, where you break the field down in to parts that have:

*

*non-zero divergence (divergenceful) and zero curl (irrotational),

*zero divergence (divergenceless) and non-zero curl (solenoidal), and

*divergenceless and irrotational.

As you can see from vector calculus, we have
\begin{align}
\nabla \times (\nabla \Phi) &= 0 \text{ and}\\
\nabla \cdot \left(\nabla\times \mathbf{A}\right) & =0
\end{align}
for any reasonably smooth functions $\Phi$ and $\mathbf{A}$. This can be proven by brute force, or more elegantly by showing that interchanging the order of the two derivatives means the expressions must equal themselves times $-1$, and so must be zero.
It turns out that you can show that the three parts of the field can be found from the field using the formulas:

*

*divergenceful part of $\mathbf{F} = -\nabla \frac{1}{4\pi} \int_V \frac{\nabla' \cdot\mathbf{F}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} \operatorname{d}V'$

*solenoidal part of $\mathbf{F} = \nabla \times \frac{1}{4\pi} \int_V \frac{\nabla' \times\mathbf{F}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} \operatorname{d}V'$ and

*divergenceless and irrotational part of $\mathbf{F} = \nabla \frac{1}{4\pi} \oint_{\partial V}  \frac{\hat{n}' \cdot \mathbf{F}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\operatorname{d}V' - \nabla \times \frac{1}{4\pi} \int_{\partial V} \frac{\hat{n}' \times\mathbf{F}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} \operatorname{d}V'$.

Now, compare these equations with Maxwell's equations. The first is the combination of the integral and differential forms of Coulomb's law/Gauss's law, the second is the same for the Biot-Savart law/Ampere's law, and the third can be seen as Coulomb's law for surface charges on the edge of our region of interest $V$ and the Biot-Savart law for surface currents in the same place.
Crucially, if we start with the curl of the field, e.g. $\nabla \times \mathbf{H} = \mathbf{J}_{\text{free}}$, then you'll only get the solenoidal part of the field. Its divergence will, therefore, vanish.
So, where does the divergenceful part of $\mathbf{H}$ come from?
Look at the defining equations for the auxiliary fields:
\begin{align}
    \mathbf{D} &= \epsilon_0\mathbf{E} + \mathbf{P}\ \mathrm{and} \\
    \mathbf{H} & = \frac{1}{\mu_0}\mathbf{B} - \mathbf{M}.
\end{align}
Now, take the curl of both sides of the first, and the divergence of both sides of the second. Because $\nabla\times \mathbf{E} = 0$ and $\nabla \cdot \mathbf{B} = 0$ in static problems you will find
\begin{align}
    \nabla \times \mathbf{D} & = \nabla \times \mathbf{P} \ \mathrm{and}\\
    \nabla \cdot \mathbf{H} & = - \nabla \cdot \mathbf{M}.
\end{align}
What this implies in terms of  is that the solenoidal part of the displacement field, $\mathbf{D}$, is fixed by the solenoidal part of the polarization density, $\mathbf{P}$, and the divergenceful part of the magnetizing field, $\mathbf{H}$, is fixed by the divergenceful part of $\mathbf{M}$.
The simplest problems where the divergenceful part of $\mathbf{H}$ and the solenoidal part of $\mathbf{D}$ contribute are in infinite planes of finite thickness, $h$, (for concreteness, say the surface normal is $\hat{k}$) where $\mathbf{P}$ is uniform and in the plane (say, $\hat{i}$ direction), and where $\mathbf{M}$ is uniform and perpendicular to it (say $\hat{k}$ direction). In those cases, the curl and divergence of the dipole densities do not vanish at the surfaces, respectively, affecting their corresponding macroscopic fields (for a more realistic problem, just make it a disc of finite radius).
It is, therefore, more complete to say that the free charge density fixes the divergenceful part of $\mathbf{D}$ and the free current density fixes the solenoidal part of $\mathbf{H}$. This statement even applies in electrodynamics. The statement about the divergenceful part of $\mathbf{H}$ also applies in electrodynamics. The only modification that needs to be made for electrodynamics would be to the $\nabla\times \mathbf{D}$ equation, but I've never seen it actually used.
A: The differential Ampere's Law is $$\textrm{curl}\textbf{H}=\textbf{J}$$ Taking the divergence of this gives $$\textrm{div}(\textrm{curl}\textbf{H})=\textrm{div}\textbf{J}=0$$because the divergence of any curl field is zero. $$\textrm{div}\textbf{J}=0$$ means that the current is divergence free (solenoid), i.e., it has no sources or sinks so that what flows into a certain volume also has to flow out of it. This expresses the conservation of charge for time independent (stationary) currents and charges.
A: The only place I can find this is in Griffiths (4th ed.) p.283. Griffiths does not say that the divergence of the H-field is zero if you find the H-field using Ampere's law and the free current. This is evidently not true since $\nabla \cdot {\bf H} = -\nabla \cdot {\bf M}$.
What Griffiths appears to say is that you can assume that $\nabla \cdot {\bf M}=0$ and hence that the H-field is divergence-free "if the problem exhibits cylindrical, plane, solenoidal or toroidal symmetry".
I find this statement quite difficult to interpret. It is easier just to say that you can only assume that the divergence of H is zero if the divergence of M is zero. That is automatically satisfied in non-magnetic media; it is also true in a completely homogeneous magnetic medium with uniform magnetisation or where you can assume ${\bf B} = \mu {\bf H}$; but it may not be true in all magnetic media and certainly isn't true where there are boundaries between (magnetic) media.
Maybe what Griffiths means is that if you are able to use
$$ \oint {\bf H}\cdot d{\bf l} = I_{\rm free}$$
to get the H-field and you are able to argue in terms of symmetries that the magnitude of the field is constant and there are no components perpendicular to the path you've integrated over, then that means the field is purely solenoidal and has no divergence. Or to put it another way, if the divergence of the field is non-zero, Ampere's law doesn't tell you about the divergent part.
