Are the conducting electrons in a metal counted as free or bound charges? Caused by another question on the propagation of EM waves in a metal, I got a bit confused on the notion of free and bound charges in a metal (sometimes called induced or external). Are the free electrons in metals considered as free or as bound charges in a metal? By free or bounded I mean to indicate their role in the Maxwell equations: are they included in $\rho_{free}$ in $\nabla \cdot \vec{D} = 4\pi\rho_{free}$?
It seems to me that they should be counted as free charges, but then I encoutered the following sentence in Griffiths' book on electrodynamics : 

In the previous section I stipulated that the free charge density $\rho_f$ and the free current density $J_f$ are zero (...). Such a restriction is perfectly reasonable when you're talking about wave propagation through a vacuum or through insulating materials  (...) . But in the case of conductors we do not independently control the flow of charge, and in general $J_f$ is certainly not zero. In fact, according to Ohm's law, the (free) current density in a conductor is proportional to the electric field $J_f = \sigma E$.

This seems to imply that the conducting electrons lead to a free current. But that doesn't seem to make sense with the dissipation of free charge $\rho_f (t) = e^{-(\sigma/\epsilon)t}\rho(0)$ that is derived just below the stated sentence in Griffiths' book (if you take a neutral piece of metal, you don't expect the electrons to flow out spontaneously, you only expect charge to flow out if it wasn't neutral to begin with).
But, if we were to take the conducting electrons as bound charges, wouldn't em waves behave identically in a neutral conductor (with no external charges) as they do in an insulator?
 A: EDIT

By free or bounded I mean to indicate their role in the Maxwell equations: are they included in $\rho_{free}$ in $\nabla \cdot \vec{D} = 4\pi\rho_{free}$?

Displacement field is a concept that is in some cases useful for description of dielectrics. Its definition is
$$
\mathbf D = \epsilon_0 \mathbf E + \mathbf P
$$
where $\mathbf P$ is total electric dipole moment of neutral bodies (molecules) per unit volume of the medium. Because the bodies are neutral, divergence of $\mathbf P$ is minus bound charge density and divergence of $\mathbf D$ equals free charge density.
This makes it possible to use $\mathbf D$ to describe state of matter as opposed to $\mathbf E$ which describes the state of the force field that acts on the matter.
If we attempt to introduce $\mathbf P$ for metals, we need to be clear on what is electric dipole moment of neutral bodies in metal. However, according to microscopic models of metals, there are no such neutral bodies. The conduction particles move arbitrarily far from their starting point, making the stationary nuclei with the immobile electrons electrically charged. Such bodies cannot be accurately described merely by electric dipole moment. One would need to add electric charge of the bodies forming the stationary part of the metal as an independent quantity, and introduce a reference point to which their electric dipole moment is referring.
One could try to define $\mathbf P$ differently, for example as 
$$
\mathbf P(\mathbf x,t_1) = \int_{t_0}^{t_1}\mathbf j(\mathbf x, t')\,dt'.
$$
where $\mathbf j$ is total electric current density. Then one could retain the equations mentioned above also for metals, with the twist that total charge density $\rho$ would be involved instead of $\rho_{free}$. Then the answer to your question would be that $\nabla\cdot\mathbf D = \rho_{free}$ no longer is valid in metals so it is not possible to use this to distinguish bound and free charges.
But I think this description is very awkward and inappropriate use of dielectric concepts to describe metals. It would be like using displacement field instead of velocity field to describe incompressible fluid (not very useful).
Back to metals, I think it is much more clear and closer to microscopic models to use total electric charge density $\rho$, total electric charge current density $\mathbf j$ and electric field strength $\mathbf E$. $\mathbf P$ and $\mathbf D$ are not defined and there is also no dielectric constant, but there is electric conductivity instead, that relates electric strength $\mathbf E$ to electric charge current density $\mathbf j$ via the Ohm law:
$$
\mathbf j =\sigma \mathbf E.
$$
Now, the answer to your question is, the conduction charges are free charges, because they form electric current. But the density of these free charges $\rho_{free}$ is not equal to total charge density $\rho$. Most often, inside metal carrying electric current, total charge density vanishes while free charge density is negative due to conducting electrons.
A: Well, it has been four years since you posted this question, but I had the same question when I read Griffith's and actually asked it here and just received an answer in this post.
Before I continue I will answer that yes, electrons are free charge, but they are not the only free charge in a conductor. In a conductor free electrons leave the atom and leave a positive inmovable hole in the atom. Hence  $ \rho_f = \rho_{electron} +\rho_{inmovable-hole}$ where the electron density is negative. This is the $ \rho_f$ that you see in Gauss law.
Yet, the ohmic free current in a conductor is $$ \mathbf {J_f}= \rho_{electrons}\mathbf {V_{electrons}} + \rho_{inmovable-hole}\mathbf {V_{inmovable-hole}}$$
but $\mathbf {V_{inmovable-hole}}$ is zero since these holes dont move, hence:
$$ \mathbf {J_f}= \rho_{electrons}\mathbf {V_{electrons}}$$
You were asking how can there be a current if the relaxation equations indicate that $\rho_f $  goes to zero. Well according to the equations I just wrote, the fact that $\rho_f $ goes to zero just means that the electron and hole density just equal each other. Since the current is just conformed by electrons you can have zero $\rho_f $ and finite current.
Finally, neither electrons nor these holes are bounded charges, bounded charges are "paired" charges, that means they are basically dipoles distributed in the material. When a electric field is imposed in a dielectric these dipoles align with the electric field and cross the boundaries of your control volume creating a surface charge. They are accounted in the Displacement vector $\bf D$ and dont appear in the matter form of Gauss Law.
