# Understanding derivation of scattering from Maxwells eqns

1. May 7, 2014

### Steve Drake

I am trying to follow a Maxwell's equations derivation for light scattering but don't understand 'why' the authors do the steps they do at this start bit. Help would be greatly appreciated...

It starts with the incident electric field equation.
$$\textbf{E}_{0}(\textbf{r},t) = \textbf{E}_0 \exp \left[i \left(\textbf{k}_0 \cdot r-\omega t\right)\right] \qquad [1]$$
Then they show
$$\nabla \times \textbf{E}(\textbf{r},t) = -\frac{\partial}{\partial t}\textbf{B}(\textbf{r},t) \qquad [2] \\ \nabla \times \textbf{H}(\textbf{r},t) = \frac{\partial}{\partial t}\textbf{D}(\textbf{r},t) \qquad [3]$$
The scatterer is described by
$$\textbf{D}(\textbf{r},t) = \varepsilon(\textbf{r}) \cdot \textbf{E}(\textbf{r},t) \qquad [4] \\ \textbf{B}(\textbf{r},t) = \mu_{0} \textbf{H}(\textbf{r},t) \qquad [5]$$
Now this is where I get lost

"Taking the curl of eq.[2] using that
$$\nabla \times ( \nabla \times \textbf{E}) = \nabla ( \nabla \cdot \textbf{E}) - \nabla^{2} \textbf{E}$$ and substitutions of Eqns [3,4,5] yields a single equation for the total electric field strength,
$$\nabla(\nabla \cdot \textbf{E}(\textbf{r},t)) - \nabla^{2}\textbf{E}(\textbf{r},t) = -\mu_{0} \varepsilon(\textbf{r}) \cdot \frac{\partial^{2}}{\partial t^{2}} \textbf{E}(\textbf{r},t)$$
So why do they just decide to take the curl of equation 2 using that identity? And then subbing all those equations in? What does it mean physically? I really need everything spelled out in very basic ways to help me understand.

Thanks

2. May 7, 2014

### Fredrik

Staff Emeritus
This type of question is usually impossible to answer. Maybe the first person who did this didn't have a reason, and just decided to try as many things as possible to see if anything interesting comes out.

In this case, I suppose we can start with the observation that (3)-(5) make it obvious that the curl of B can be expressed using only the E field. So by taking the curl of (2), we will get an equation that involves only the E field.

Last edited: May 7, 2014
3. May 7, 2014

### Steve Drake

Thanks for your reply. This is actually from a modern book treating light scattering. I am not wondering about the original derivations Maxwell did, I am after the physical reason why the book takes this path to (eventually after 4 pages) derive an expression for the scattered field.

Can you go into a bit more detail here please? I cant understand why we would want this?

Thanks

4. May 7, 2014

### Fredrik

Staff Emeritus
I'm not convinced that there is a physical reason (other than the fact that the fields satisfy Maxwell's equations). The main reason why they take that path is that they have found that it leads to an interesting result. The fact that they're not telling you about any other reasons could mean that they're not aware of any.

Because it's useful in the next step? I can't tell you more than this, because I don't know what the next step looks like, or what the final result looks like. The final result should be a clue. Perhaps it was first written down as an intelligent guess based on results of experiments. In that case it makes sense to ask "How can we get that from Maxwell's equations?".

Last edited: May 7, 2014
5. May 7, 2014

### UltrafastPED

Equation (1) is the definition of the incident light; equations (2) and (3) are Maxwell's equations - the later for the case with no currents. Equations (4) and (5) are definitions of the fields for the case of linear response - the simplest case.

So starting with these you work the pieces - since (3) has curl E but not B, and (4) has curl B, it makes sense to take the curl of (3) so that (5) can be substituted. And of course, if you are using vector equations you need to have the vector identities handy.

This gets you to the final result.

This is a case where the physical inputs are Maxwell's equations under special conditions; the resulting scattering equations are only valid for the given conditions. This is the physical content.

These relations were originally worked out by physicists with excellent to outstanding mathematical skills.

6. May 7, 2014

### dauto

They take the curl of equation 2 because they want to use equation 3 to eliminate the magnetic field. Equation 3 has a curl of the magnetic field, so naturally taking the curl of equation 1 is the first step of the solution. It's not a case of trying things blindly and hopping to hit gold.

7. May 7, 2014

### Steve Drake

Thank you very much, that is excellent. It is details like these that books just do not say and I find them incredibly necessary for my understanding. So eqn [4] and [5] are put in as they describe how the material 'responds' to the incoming field. The one thing I still cant wrap my head around is what is the produced expression telling us? Something to do with the curl of the E field equals the rate of change..? Edit: I mean I know what it is saying mathematically but what is it saying in a physical sense?

8. May 7, 2014

### Steve Drake

This kind of derivation is in a lot of light scattering books and I have never understood it because they never seem to go into enough detail for me. I am beginning to 'get' it now though.

9. May 7, 2014

### UltrafastPED

The scattering equation will typically be further simplified so that it matches specific physical conditions - the one given is very general.

So you will look at the results "far away" after scattering off of a dipole field.

Eventually you will get to do a homework problem which shows why the sky is blue during the day ...