$E,H,\rho,\vec{j}$ intensive? $B,D,\Phi,\vec{A}$ extensive? Why not $ DdE,BdH ,\Phi\delta\rho,\vec{A}\cdot \delta \vec{j}$ as infinitesimal work?

Question

$u$ is the e&m field energy

$\frac{\partial u}{\partial t}=H\cdot \frac{\partial B}{\partial t}+E\cdot \frac{\partial D}{\partial t}$

Compared with thermo physics: $PdV,TdS,\mu dN$

${H,E,P,T,\mu}$ are intensive variables

${B,D,V,S,N}$ are extensive variables

But in microscopic level, there is no $H$ field. $B$ should be more meaningful.

My question is, why are $H$ and $E$ intensive variable. Why are $D$ and $B$ extensive variables?

I think all of them $E,D,H,B$ should be intensive, because energy density $u$ itself is an intensive variable.

Answer 1

Think of $D$ and $B$ as $D=\epsilon _0 E + P$ and $B=\mu_0 H + M$, resp. Now the electric and magnetic polarization densities $P$ and $M$ are volumetric densities of extensive quantities, because the total polarization of the material is proportional to the amount of matter present in the field, hence $P$ and $M$ correspond to "extensive" variables when energy exchange is considered. Fields that induce the polarizations are naturally intensive parameters, and indeed $EdP$ or $HdM$ is the electric/magnetic work expended to which one may add the field energy $\frac{1}{2} \epsilon_0 E^2$ and $\frac{1}{2} \mu_0 H^2$ in the absence of the ponderable matter and obtain the total differential work $\epsilon_0 EdE + EdP = EdD$ or $\mu_0 HdH + HdM = HdB$ but neither $D$ nor $B$ is "extensive" in a conventional sense instead its respective polarization is extensive.

Answer 2

Are you referring to intensive and extensive properties? This makes little sense to me for field variables, but it does make a little, so I'll roll with it.

E is the electric field, including the permittivity of the material it is passing through. D does not include permittivity but only reflects the charge contained in a region of space, and the flux sent through an enclosing surface by that charge. If I put one coulomb of electric charge on a metal ball, Gauss's law applied over any region including the ball will account for exactly one coulomb. On the other hand, if I integrate E without adjustment, the result will depend on the permittivity of other objects on the surface of the region.

Likewise with B and Ampere's law: to recover the amount of current from H you need to know permeability along the line of integration.

Note the section "Generality of classification" at the end of the above Wikipedia link. Extension of the intensive/extensive construct to field variables is not condoned by IUPAC, since the principle of dividing the system doesn't hold. Such properties only apply to a discrete subsystem, not to a homogeneous material.