I am currently studying the textbook Surface Enhanced Raman Spectroscopy -- Analytical, Biophysical, and Life Science Applications by Sebastian Schlücker, Wolfgang Kiefer. Chapter 1.2.2 Planar Surfaces says the following:

Once the complex dielectric function $\epsilon(\lambda)$ is known, all the electromagnetic properties of the material can be calculated in different geometries. The normal reflectance $R$ (in the direction perpendicular to the surface) arises as a natural consequence of matching the boundary conditions of the fields at the interface. $^{2)}$

$^{2)}$ The standard boundary conditions for all electromagnetic problems require that the components of the electric field parallel to the surface (on both sides of the surface) are equal, as well as the perpendicular components of the displacement vector $\mathbf{D} = \epsilon(\lambda) \mathbf{E}$. In standard notation for an interface between medium 1 and 2: $E^\parallel_1 = E^\parallel_2$, and $\epsilon_1(\lambda) E^\perp_1 = \epsilon_2(\lambda)E^\perp_2$. The normal reflectance at a planar surface between the two media is given by $R = \left\vert \dfrac{n_2 - n_1}{n_2 + n_1} \right\vert^2$, with $n_1 = \sqrt{\epsilon_1}$ and $n_2 = \sqrt{\epsilon_2}$.

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Given this explanation, the concept is quite vague in my mind. I was wondering if someone would please take the time to clarify this, showing the relevant mathematics to make the concept more concrete. In particular, I'm curious about the differential equations and boundary conditions that it is referring to. I would greatly appreciate it if people would please take the time to carefully explain this for a novice such as myself.

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    $\begingroup$ Can you help by telling us a little about where you are in your studies? For example, it would help us decide where to start if you can tell us if you are familiar with Gauss' Law. $\endgroup$
    – garyp
    Commented May 13, 2020 at 1:31
  • $\begingroup$ @garyp I would say that I'm a novice. I am familiar with the integral form of Gauss's law: $$\oint_S \vec{E} \cdot \hat{n} \ da = \dfrac{q_{\text{enc}}}{\epsilon_0},$$ where the left side is the electric flux, and the right side is the total amount of charge contained within that surface, divided by the permittivity of free space. I am not experienced with these concepts, so a careful explanation, aimed at the novice, would be appreciated. I also have knowledge of basic ODEs/PDEs. $\endgroup$ Commented May 13, 2020 at 10:35

1 Answer 1


Let me take you through deriving the boundary conditions. I had a fantastic Prof, who explained this very well. We can derive them by evaluating some of Maxwell's equations at the interface.

Boundary Condition 1

First, let's assume there is a surface charge density $\rho_{sf}$. Let's begin with Gauss' Law: $$ \oint \vec{D}\cdot \mathrm{d}\vec{S} = Q_{encl} $$

We used the displacement field here, defined as $\vec{D} = \varepsilon \vec{E}$, and $Q_{encl}$ is the charge enclosed by a Gaussian box we're about to draw. Let's make it a cylinder around the interface, of height $h$.

Gauss Law Diagram

In words, Gauss' law states that the net flux of an electric (displacement) field in a closed surface is directly proportional to the enclosed electric charge. Our cylinder is the closed surface - let's work the integral out!

First, there are no constraints on our choice of the Gaussian surface's height, so we can take the limit $h \rightarrow 0 $. This means the surface integral for the curved surface goes to zero, and we're left with the ends:

$$ \int_{top} \vec{D}\cdot \mathrm{d}\vec{S} = (\vec{D}_1\cdot \hat{n})A $$ $$ \int_{bottom} \vec{D}\cdot \mathrm{d}\vec{S} = -(\vec{D}_2\cdot \hat{n})A $$

where A is the surface area of the top and bottom end of the cylinder, the subscript number indicates the medium and $\hat{n}$ is the surface normal. Note that for the bottom end, the integral is negative as the surface normal and the displacement field point in opposite directions.

Not forgetting about charge enclosed $Q_{end}$: in the limit of $h \rightarrow 0$, this becomes equal to the surface charge density multiplied by the area of the top/bottom surface! So, Gauss' law becomes:

$$ (\vec{D}_1\cdot \hat{n} - \vec{D}_2\cdot \hat{n})A = A\rho_{surf} $$

Cancelling $A$ and evaluating the dot products, we end up with the first boundary condition for the displacement field vector component perpendicular to the interface:

$$ \boxed{D_{1\perp} - D_{2\perp} = \rho_{surf}} $$

We could substitute $\vec{D} = \varepsilon \vec{E}$ here. Also, the presence of a surface charge will depend on the type of interface.

Boundary Condition 2

For the next boundary condition, let's evaluate the integral form of Faraday's Law:

$$ \oint \vec{E}\cdot\mathrm{d}\vec{l} = - \frac{\mathrm{d}}{\mathrm{d}t} \int \vec{B}\cdot\mathrm{d}\vec{S} $$

In words, this states that for any closed loop path, the sum of the length elements times the electric field in the direction of the length element is equal to the time derivative of the magnetic flux through the surface created by the closed loop path.

Faraday's Law Diagram

So, let's draw a rectangular loop of height $h$ and length $l$ around the interface. We can split the path integral of Faraday's Law up into a sum of 4 parts, for each side of the loop. Again, we can take the limit $h\rightarrow0$, which means we are only left with the path integrals parallel to the surface:

$$ \int_{top} \vec{E}\cdot\mathrm{d}\vec{l} = (\vec{E}_1 \cdot \hat{n})l $$ $$ \int_{bottom} \vec{E}\cdot\mathrm{d}\vec{l} = -(\vec{E}_2 \cdot \hat{n})l $$

where $\hat{n}$ is the unit vector in direction of the loop. Not forgetting about $\int \vec{B}\cdot\mathrm{d}\vec{S}$: in the limit of $h\rightarrow 0$, as the surface area of the enclosed path goes to 0, $\int \vec{B}\cdot\mathrm{d}\vec{S} = 0$. This leaves us with Faraday's law in the following form:

$$ (\vec{E}_1 \cdot \hat{n} - \vec{E}_2 \cdot \hat{n})l = 0 $$

If we cancel $l$ and evaluate the dot products, we're left with the second boundary condition, for the components of the electric field parallel to the interface:

$$ \boxed{E_{1\parallel} - E_{2\parallel} = 0} $$

I hope you now understand where the boundary conditions come from. I won't tackle the derivation of the reflectivity now, but I recommend you look up the derivation of the Fresnel equations if you're interested: here a possible source: https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/lecture13_0.pdf

If you would like me to, I could derive the reflectivity tomorrow.

  • $\begingroup$ Thanks for taking the time to post such an elaborate answer. So the enclosed charge $Q_{\text{encl}}$ is the cylinder? $\endgroup$ Commented May 13, 2020 at 23:32
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    $\begingroup$ The enclosed charge Q is any net-electrical charge within the cylinder. If we make the cylinder infinitesimally small, it's just the charge forming part of our surface. Drawing Gaussian shapes, like spheres, or cubes, are a common way of analysing a problem like this - it's just a neat way of taking advantage of the mathematical formalism and flux. Though in your example I see there is no surface charge density or charge enclosed - it will undoubtedly depend on the materials. $\endgroup$
    – Samalama
    Commented May 13, 2020 at 23:47
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    $\begingroup$ Can you please elaborate on why there are no constraints on our choice of the Gaussian surface's height? $\endgroup$ Commented May 14, 2020 at 12:16
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    $\begingroup$ The way we draw a Gaussian closed surface is our choice, depending on which part of space we wish to investigate. Maybe I phrased it poorly, let me try another way: we choose $h \rightarrow 0$ because we're interested in the boundary conditions, i.e. how do the fields behave at that particular point in space where the two media meet. Therefore, we don't wish to include contributions from the fields in other parts of space (such as parts of medium 1 or 2). Note also that it's $h \rightarrow 0$ not $h=0$, as the latter would no longer be a closed surface. $\endgroup$
    – Samalama
    Commented May 14, 2020 at 12:32

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