Jackson writes in section 1.4 (third edition) that

\begin{align*} D_\alpha &= \epsilon_0 E_\alpha + \left(P_\alpha - \sum_\beta \frac{\partial Q'_{\alpha\beta}}{\partial x_\beta} + \ldots \right) \\ H_\alpha &= \frac{1}{\mu_0} B_\alpha - (M_\alpha + \ldots ) \end{align*}

and adds,

The quantities $\mathbf{P}$, $\mathbf{M}$, $Q'_{\alpha\beta}$, and similar higher order objects represent the macroscopically averaged electric dipole, magnetic dipole, and electric quadrupole, and higher moment densities of the material medium in the presence of applied fields.

In section 6.6 he derives these relations and states for D,

The third and higher terms are present in principle, but are almost invariably negligible.

and makes a similar statement about H.

So, is it true that the precise definitions of D and H really contain these higher-order terms, but that they're almost invariably dropped (e.g., by Griffiths, and all the lecture notes I could find online) because they're almost always negligible?

Or do we instead say that $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$ exactly and that $\mathbf{H} = \frac{1}{\mu_0} \mathbf{B} - \mathbf{M}$ exactly and that the macroscopic Maxwell's equations break down when the quadrupole and higher moment densities become significant?

  • $\begingroup$ Note also that for $\mathbf D=\epsilon_0 \mathbf E+\mathbf P$ to break down not only would the induced electric quadrupole density need to be large, but its divergence would also need to be large as well. You need a large induced moment which also changes quickly in space, and this is generally very difficult to arrange. $\endgroup$ May 4, 2015 at 13:40
  • $\begingroup$ @Emilio Pisanty , But it is possible to produce this kind of configuration in the context of optics. @ Brian Bi, I believe that what you are looking for is 'Spatial Dispersion' and 'Optical Activity' $\endgroup$
    – Hydro Guy
    May 4, 2015 at 13:47
  • $\begingroup$ Indeed, it is possible in the optical domain, but it does not stop being hard. I'm simply pointing out that there are additional barriers to be breached. $\endgroup$ May 4, 2015 at 13:56

1 Answer 1


We split $\rho=\rho_{\it ex}+\tilde\rho$ and $\jmath=\jmath_{\it ex}+\tilde\jmath$ into "explicit" and "medium" contributions. We can then define $P$ and $M$ by $\tilde\rho=-\nabla P$ and $\tilde\jmath=\nabla\times M$, and macroscopic fields by $D=E+P$ (I'm sloppy about units, there is a $4\pi$ in Gauss units) and $B=H+M$. The macroscopic equations are then exact, independent of the question where $\tilde\rho$ and $\tilde\jmath$ come from.

If moments beyond the dipole are important then what is lost is the realtion between $P$ and $M$ and the density of dipoles.


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