# Not physical result from the presence of surface charge density between dielectrics

Suppose a non-zero a surface charge density $\sigma$ to be present on the interface between two different dielectrics materials (indicated by the subscripts 1 and 2).

The boundary condition $D_{2_{\perp}}-D_{1_{\perp}}=\sigma$ for the electric displacement field (let's suppose $D=\varepsilon E$ with $\varepsilon \in \mathbb{R}$) says that for an incident electromagnetic wave on the surface:

$$E_{\perp}e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}+ E_{\perp}''e^{i(\mathbf{k}''\cdot \mathbf{x}-\omega'' t +\varphi')}=\frac{\sigma}{\varepsilon_2} +\frac{\varepsilon_1}{\varepsilon_2}E_{\perp}'e^{i(\mathbf{k}'\cdot \mathbf{x}-\omega' t+\varphi'')}$$

where $E$ is the incident electric field, $E''$ the reflected and $E'$ the transmitted.

I can write $\sigma=\sigma e^{0}$. Because the exponential are linearly indipendent the following equations must be true:

$$(\mathbf{k}\cdot \mathbf{x})_{z=0}-\omega t=(\mathbf{k}'\cdot \mathbf{x})_{z=0}-\omega' t+\varphi'=(\mathbf{k}''\cdot \mathbf{x})_{z=0}-\omega'' t+\varphi''=0$$ where the condition $z=0$ means that the phase is evaluated on the incident plane.

Since this conditions must be true for every position $(x,y)$ and for every time $t$, follows that:

$$0=\varphi'=\varphi''$$ $$\omega=\omega'=\omega''=0$$ $$\mathbf{k}=\mathbf{k}'=\mathbf{k}''=\mathbf{0}$$

i.e. after the wave has encountered the interface, it has become a constant and no longer a wave. This result is obviously unphysical, so there is a problem. Can someone please show me how to get out of this problem?

• I haven't worked it out entirely but there appears to be something very fishy with your logic, which appears to depend on the idea that if $a + b = c + d$ then all of these 4 terms have the exact same complex phase. That's simply not true, for example $(1) + (i) = (2 - 2i) + (-1+3i)$ relates 4 very different complex phases. In addition these conditions should be true for every time $t$ choosing a specific position $\vec x$ where the wave is incident; I am not sure if that helps you or not but it suggests $\omega = \omega' = \omega''$, so "wavefronts don't go out any faster than they come in." Feb 25 '17 at 23:08
• The idea is that the exponentials are linearly indipendent, so a sum of exponential with different exponent is zero if and only if the coefficients are zero. Since this is not the case (the coefficients are non-zero electric fields) then the exponent must be the same. Am I wrong? about the position: the only constraint is that $\mathbf{x}$ must be evaluated on the interface, so I can just write $z=0$, but the $x$ and $y$ coordinates are free. I edited my question to be more clear. Thank you for your precious help. Feb 26 '17 at 0:45
• Yes, you are wrong. For example any sum of the $N^\text{th}$ roots of unity is 0, $\sum_{k=0}^{N-1} e^{2\pi ik/N}=0.$ I would not try to do fancy footwork in mathematics like that without very firm understanding first. Feb 26 '17 at 2:10
• Of course you're right. I guess exponentials must be linearly indipendent only if the exponent is real. The reason why I asked this question is beacuse I received another answer that mislead me. Could you please give a look to it? physics.stackexchange.com/questions/314612/… Feb 26 '17 at 2:18
• @CRDrost OP is essentially correct in the mathematical manipulations: exponentials of any sort are indeed linearly independent, which means that if $\sum_{k=1}^N c_k e^{\alpha_k x}\equiv 0$ for all $x$ in a real interval, and all the $\alpha_k$ are distinct, then all the $c_k$ must vanish. OP loses his footing on more physical grounds. Feb 26 '17 at 9:57

You have gotten yourself in trouble by inserting a constant surface charge density into a field that was not designed to describe it. There is nothing that forces the field to look like that combination of incident-reflected-transmitted plane waves; rather, it is one possible solution of the Maxwell equations which is interesting and which is useful to model certain situations - namely, the transmission of a plane wave at the interface between two linear dielectrics. If you change the situation, then there is no guarantee the fields will remain the same.

In your case, you do obtain a contradiction, but the lesson from it is "you have chosen the wrong Ansatz to describe your field". Luckily, it's pretty easy to describe the fields, using the superposition principle: if you want to describe the field caused by an incoming plane wave plus a static surface charge density, simply solve separately for both and then add the solutions. This will add constant fields pointing away from the boundary, which will solve all your concerns:

The Gauss' Theorem applied to the charged plane grants that:

$$D_{\sigma_1}+D_{\sigma_2}=\sigma \tag{1}$$

Where $D_{\sigma_i}$ is the electric displacement field in the $i$th dielectric generated by the charged plane itself. Using the boundary condition $D_{2_{\perp}}-D_{1_{\perp}}=\sigma$ for the total field (static + wave) you get:

$$D_{\perp}e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}+ D_{\perp}''e^{i(\mathbf{k}''\cdot \mathbf{x}-\omega'' t +\varphi')}+D_{\sigma_2}=\sigma -D_{\sigma_1}+D_{\perp}'e^{i(\mathbf{k}'\cdot \mathbf{x}-\omega' t+\varphi'')} \tag{2}$$ where $D_{\sigma_1}$ is subtracted because is pointing in the opposite direction of $D_{\sigma_2}$.

Putting $(1)$ in $(2)$ you get:

$$D_{\perp}e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}+ D_{\perp}''e^{i(\mathbf{k}''\cdot \mathbf{x}-\omega'' t +\varphi')}=D_{\perp}'e^{i(\mathbf{k}'\cdot \mathbf{x}-\omega' t+\varphi'')}$$

from that follow the know phase relationships:

$$0=\varphi'=\varphi''$$ $$\omega=\omega'=\omega''$$ $$\mathbf{k}=\mathbf{k}'=\mathbf{k}''$$

...unless, that is, the incoming plane wave were to mess with the surface charge and make it non-constant, but you haven't specified enough of the physics to tell what would happen. In the simplest case, you would simply get an oscillatory component to the surface charge, and it then follows from the linear dependence argument (as well as from the physical consideration that the dynamics of the charge are driven by the plane wave) that it will oscillate with the same exponential as the fields. In fact, if you look carefully, this is already happening in the standard configuration, if you rephrase the Gauss law in terms of the $\mathbf E$ fields instead of $\mathbf D$.

• I've not specified enough just because it's not a real physical situation, but just me trying to figure out what happens if the interface is charged. Is it physical to suppose that the charge won't oscillate with the wave and remain constant? Let's suppose that for now. Let's also say that I want to ignore the electric field generated by the surface charge density, I just want to find out what happens to the plane wave after it has encountered the interface. How can I do the math? Do I have to ignore the presence of $\sigma$? Feb 26 '17 at 11:00
• As I said, it depends, but if you just want to ignore the effects of the surface charge, why include it in the first place? For this sort of free-form discussion, though, go to the chat room. Feb 26 '17 at 11:13
• I don't want to ignore all the effects of $\sigma$, I'm just not interested in the constant electric field. I'm interested only in the oscillating one. I can rephrase my question in this way: "Is Snell's law still valid if the interface is charged?". Sorry if the question might be stupid. p.s. I'm quite new to this site and I don't know how to move this discussion into a chat before the 8 comments limit is reached. Sorry about that and thank you again. Feb 26 '17 at 11:26
• I think I solved my problem doing explicitly the calculations. I suggested an edit for your answer. Let me know if you agree. Feb 26 '17 at 14:13
• @Alessandro for the record, edits of this magnitude are usually more of a change of the author's intent than we normally consider acceptable on this site. Feb 26 '17 at 14:13