I'm confused about some notations in electromagnetism. Please help me to solve these questions.

I first studied electromagnetism with David cheng's book, but

From the book : Lukas Novotny, Bert Hecht - Principles of nano-optics-,

The Maxwell's equation (for linear, isotropic case) is stated as bellows :

$$\begin{align} \nabla \times H&=J+\cfrac{\partial D}{\partial t} \tag{1}\\ \nabla \times E&=-\cfrac{\partial B}{\partial t} \tag{2}\\ \nabla \cdot B&=0 \tag{3}\\ \nabla \cdot D&=\rho \tag{4} \end{align}$$

Here, $J=J_s+J_c$,

where $J_s$ is the source current, $J_c$ is the conduction current

of course, for the continuity eqauation, $ρ$ must satisfy


then, from constitutive relations(* means the convolution) $D=ε_0ε_r*E=ε_0E+P,\ H= \cfrac{B}{μ_0}-M, J_c=σ*E$ (D : electric displacement, P : Polarization vector, M : Magnetization vector)

the net current density becomes $J_{net}=J_s+J_c+J_p+J_m$,

where $J_p=\cfrac{∂P}{∂t},\ J_m=∇×M$

then the net charge density must satisfy $ρ_{net}=ρ_s+ρ_c+ρ_p+ρ_m $

we also have these two continuity equations :

$\cfrac{∂ρ_{net}}{∂t}+∇\cdot J_{net}=0, \cfrac{∂ρ}{∂t}+∇\cdot J=0$

[Question 1]

From the two continuity equations, we also have $\cfrac{∂(ρ_p+ρ_m)}{∂t}+∇\cdot (J_p+J_m)=\cfrac{∂(ρ_p+ρ_m)}{∂t}+∇\cdot (J_p)=0$

since $∇\cdot(J_m)=∇\cdot(∇×M)=0$

is there a case $ρ_m≠0$ ? i.e. Does the equation $\cfrac{∂ρ_p}{∂t}+∇\cdot J_p=0$ hold in any situation?

[Question 2-1]

By applying the Fourier transform to all these equations, the book says, we can define ( $\hat X$ : Fourier transformed quantity, j=\sqrt{-1})

$\hat ε_c:=\hat ε_r+\cfrac{j \hat σ}{ωε_0}$

If we set $\hat J_i:=\hat J_c+\hat J_p$, then we may define $\hat σ_c$ so that $J_i=(\hat σ+(-jωε_0)(\hat ε_r-1))\hat E=\hat σ_c \hat E$

These two quantities are related by $\hat ε_c=1+\cfrac{j\hat σ_c}{ωε_0}$.

The problem happens in other books, especially in plasmonics

they use $\hat ε=1+\cfrac{j\hat σ}{ωε_0}\ldots (5)$

I cannot tell whether the pair {ε, σ} in (5) is {$ε_c, σ_c$} or {$ε_r, σ$}. Which is true? and Why?

[Question 2-2]

Derivation of the Drude-Sommerfield dielectric function uses the 'Fourier transformed Polarization vector', but I think it is $\hat P=ε_0(\hat ε_r -1)\hat E$, not $\hat P_c=ε_0(\hat ε_c -1)\hat E$, ($P_c$ is used, since it's obviously not the P) (* Using phasors/plane wave case is equivalent with the Fourier transform). Which is true? and Why?

Moreover, when we say 'complex permittivity'(or (complex) 'dielectric function') and 'complex conductivity', are they {$ε_0\hat ε_r$(or $\hat ε_r$), $\hat σ$}? or {$ε_0\hat ε_c$(or $\hat ε_c$), $\hat σ_c$}?

Please help me!

If these questions appear the exam in the electromagnetism class, a great panic would happen.

  • $\begingroup$ Too many questions. Lets look at the first one for now. What is $\rho_m$? You did not define it anywhere. What are its units? $\endgroup$ – Cryo Apr 26 at 21:34
  • $\begingroup$ Oh, It's the electric charge density due to the current density $J_m=▽×M$ (not the magnetic charge density, it's obviously 0) $\endgroup$ – radodhae_Mol Apr 27 at 1:06
  • $\begingroup$ I also used SI units, and the total continuity equation can be derived by $curl(H)=curl(\cfrac{B}{μ_0}-M)=(J_s+J_c)+\cfrac{\partial D}{\partial t}=(J_s+J_c)+\cfrac{\partial (ε_0E+P)}{\partial t}$ $▽\cdot curl(B)=0=μ_0(▽\cdot(J_s+J_c+\cfrac {\partial P}{\partial t} +curl(M))+\cfrac{\partial ▽\cdot ε_0E}{\partial t})$ $\endgroup$ – radodhae_Mol Apr 27 at 1:47
  • $\begingroup$ You do not need the charge density for magnetization. Magnetization current density is the current density due to presence of magnetization, magnetization is the density of magnetic dipoles per volume. The charge density of the magnetic dipole is zero. So there is no need to introduce $\rho_m$. You will have to set it to zero later on anyway. I think this answers question 1 $\endgroup$ – Cryo Apr 27 at 21:45
  • $\begingroup$ Question 2.1. All are true. In the case of linear, isotropic, non-magnetic materials the response is usually described with two real numbers. I have seen people use real dielectric constant and real conductivity, complex dielectric constant (no conductivity), complex conductivity (no dielectric constant). Choose the one you like best, but make sure to stick with it throughout the computation. This is a bit like units, you can measure distance in meters or feet, but avoid using both within the same calculation. $\endgroup$ – Cryo Apr 27 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.