# Fresnel Equations Boundary Conditions

1. Nov 11, 2004

### ObsessiveMathsFreak

I'm having trouble with the meaning of the boundary condition in the derivation of fresnels quations, namely that the component of E tangental to the surface is continous across the boundary.

My trouble is, what physically does this corresspond to. Is it something to do with the divergence of E. What quantity is being conserved and why?

2. Nov 11, 2004

### da_willem

When there is no changing magnetic field present the curl of the electric field is zero. This is one of the famous Maxwell equations . Mathematically this condition: $\nabla \times \vec{E}=0$ can be written in integral form by using the 'curl theorem':

$$\int \vec{E} \cdot d \vec{l}=0$$

Where the integral should be evaluated along a closed loop. This is because the electrical force is conservative. The above integral maybe rings a bell, it is related to the definition of the electrical potential. It says if you move around a closed loop, the electrical potential will be the same on arrival at your starting point. A conservative force preserves energy, so the boundry condition follows in the end from the conservation of energy.

If you have a certain boundry on it you can draw a thin rectangular loop around it. If you let the sides normal to the surface go to zero their contribution to the integral will be zero. So when the length of the loop parallel to the surface is l the intgegral becomes: $E^{//}_{above}l-E^{//}_{below} l=0$ where // indicates the parallel component. So the parallel components of the electric field must be the same above and below a boundry in absence of a changing magnetic field.

Last edited: Nov 12, 2004
3. Nov 11, 2004

### ObsessiveMathsFreak

But wouldn't your argument imply that the Fields perpendicualr to the surface are also the same?

4. Nov 12, 2004

### da_willem

No. Lets again draw a thin rectangular loop but now let the sides parallel to the boundry go to zero. If you evaluate the integral again the only contributions come from the sides perpendicular to the boundry. So it just says:

$$\int E^{perp.}_{left} dl = \int E^{perp.}_{right} dl$$

But the electric field may change as a function of position so we can't evaluate the intgerals without knowing the field. Notice that the 'left electric field' is infinitesimally close to the 'right electric field' because we let the sides of the rectanglular loop parallel to the boundry go to zero. This argument does no prediction about the perpendicular component of the electric field...

If you would like to evaluate the perpendicular component you would have to draw a 'Gaussian pillbox' and use Gauss's law. This has to do with the divergence of the electric field.

5. Nov 17, 2004

### ObsessiveMathsFreak

I'm still not getting this. The papers I've read say that the part of the E field parallel to the surface is continous across the surface, but the part perpendicular to the surface is not.

I see where the argument with the loop is going, but nowhere in your argument did you mention the surfaces themselves.
I'm sorry to sound so ignorent after two replies, but I still haven't a clue why this is the case.
$$\mathbf{E_{||}}$$ is continous across the boundary, but $$\mathbf{E_{perp}}$$ is not.

Is this because when the || loop sides are brought to zero, the perp sides are in two media, but when the perp sides are brought to zero, the || sides are each only in one medium.
In the former case, what is the justification for saying that the electric field is different in different media.(I know this is logically the case, but I'm wondering what constants, or which of Maxwells equations, gives rise to this change)

In case anyone is wondering, I need this for a study on optical traps/tweezers.

6. Nov 18, 2004

### da_willem

Ok, well imagine an interface between two different media 1 and 2. These will have a different electrical permittivity $\epsilon_1$ and $\epsilon_2$. Now when these two media are dielectrics they will be polarized by an electric field: there will be a dipole moment induced (crudely said a shift of the electron cloud around the nucleus). If these induced dipole moments are inhomogeneuous the result will be a net volumecharge in the dielectric. This is called a 'bound charge'. As a consequence there will be an additional electric field from the dipoles aside the externally applied field.

You will have to take into account this bound charge in Gauss's law (http://en.wikipedia.org/wiki/Gauss's_law). If you would like to use some sort of Gauss's law for only the 'free charge', this is the charge not resulting from the induced dipoles you will have to use a different quantity than the electric field in the Gauss integral. This quantity is called the 'electric displacement', and often you will encounter Guass' law with the electric displacement instead of the electric field.

If there is no charge present the perpendicular component of the electric displacement will have to be the same under and above the interface because of Gauss's law. As the electric displacement is not the same as the electric field this means this does not mean the perpendicular component of the electric field will have to be continuous across the boundry! In the case of linear dielectrics (the polarisation is linear in the applied electric field) there is a simple relation between the electric displacement and the electric field:

$$\vec{D}=\epsilon \vec{E}$$

So the condition that the perpendicular component of the electric displacement will have to be the same under and above the interface (from the modification of Gauss's law):

$$D^{perp.}_{1}-D^{perp.}_{2}=0$$

yields $$\epsilon_1 E^{perp.}_{1}=\epsilon_2 E^{perp.}_{2}$$

Notice that the argument (following from a vanishing curl of the electric field)for the parallel component of the electric field still stands.
Note also that this is all in the absence of a surface charge. If there would be a surface charge present it would follow from Gauss's law (or Coulombs law if you like) that perpendicular component of the electric field is not continuous but differs on both sides.

This is a far too short explanation. I hope this clarifies things but I think it has only raised some more questions, so I advise you to read a book on it, maybe do a course on electrostatics