We have completed a discussion of one of the photon interaction processes, the photoelectric effect. We will now begin a discussion of the process of

We have completed a discussion of one of the photon interaction processes, the photoelectric effect. We will now begin a discussion of the process of coherent scatter. 1

In this lecture we are going to identify and describe the process of coherent scatter and learn how to calculate cross sections for coherent scatter. Recall that the cross section is a measure of the probability of an interaction occurring. It is actually the attenuation coefficient per unit incident fluence density. 2

Let s first describe coherent scatter from a qualitative point of view. We start with an incident photon. We look upon the photon in this case as an electromagnetic wave. As the electromagnetic wave passes near an atom, it sets the electrons in the atom in motion. The electrons oscillate along with the electromagnetic wave. But what happens when electrons oscillate? They accelerate and decelerate so they emit energy in the form of electromagnetic radiation. Note that the frequency of oscillation, hence the energy of the scattered radiation, is identical to the frequency of oscillation of the incident radiation. The angle of scatter is very small, and most of these interactions are electronic interactions. There can be scatter from nuclei but it is relatively improbable. Scatter is primarily with electrons. We will be looking at the motion of the electrons caused by the electromagnetic wave that corresponds to the incident radiation. We are going to try to calculate how much radiation is emitted from these electrons when they oscillate. That s coherent scatter. 3

We will start by deriving an equation for Thomson scatter. Thomson scatter was really the first theoretical treatment of classical scatter. The problem with Thomson scatter is that it s wrong. The calculated cross section does not agree with experimental values. However, the cross section that we derive from Thomson scatter is included in all the scattering equations. So when we look at more sophisticated models for coherent scatter, we will see the Thomson scatter cross section as being the first approximation to the model; everything else will be a modification of it. So that is why we will calculate the effect of Thomson scatter first. It also gives us a good picture for a quantitative description of what is really going on. 4

First of all, what do we mean by the differential scatter cross section? Differential scatter cross section is defined as the probability that an incident photon produces a scattered photon at a specified solid angle. So the units of differential scatter cross section are a unit of area, per electron or per atom, depending on whether it s an electronic cross section or an atomic cross section, per unit solid angle. The unit of solid angle that we use is the steradian. There are 4π steradian in one sphere. We will see some applications of a differential cross section a little bit later on in this lecture. 5

What we really want to be able to do is to follow the energy. Rather than just calculating the differential cross section, we really want to determine the fraction of energy that s scattered in a particular direction. We obtain the fraction of energy scattered by multiplying the scattering cross section times the electron areal density times the angle subtended by the detector. Electron areal density is expressed as electrons per square meter, per square centimeter, etc., whatever unit. The angle subtended by the detector is expressed in steradian. The calculation of energy scattered in a particular direction is going to be our eventual goal in any of these treatments. So we derive the scattering cross section, multiply it by the areal density of electrons, and multiply that by the angle subtended by the detector. 6

A typical calculation is to determine the fraction of energy scattered per electron aerial density per solid angle subtended by the detector. Once again, we repeat the mantra, follow the energy. That s going to be our game plan for this lecture. 7

So how do we derive the scatter cross section using the model for Thomson scatter? Let s start with an electromagnetic wave. Recall from your E&M course that an electromagnetic wave has two components, an electric vector and magnetic vector. These are at right angles to each other and they are moving with a velocity c. Let s assume that the incident photon is unpolarized, so there are components of the electric field that are at right angles to each other. We ll designate the two components as E 1 and E 2. These two components are both perpendicular to the direction of propagation. This is your basic freshman year electrodynamics. This electric field comes in and interacts with an electron, causing the electron to accelerate. The acceleration of the electron induces an electric field. What we want to determine is the fraction of energy that s scattered in this induced electric field as related to the original incident electric field. 8

Let s take an electric field and have this electric field interact with a free electron. What is the force on the free electron due to the electric field? The force is proportional to the electron charge and the electric field vector. We have a component of force in each direction, direction 1 and direction 2, the two directions that are perpendicular to the direction of propagation of the electromagnetic wave. The force is a constant times the charge times the electric field. The constant is the same constant as that in the Coulomb force equation, 8.988 10 9 Nt m 2 /C 2. That s the force now on the free electron. What is now going to happen to the electron? 9

The electron s going to accelerate. Recall the basic equation F = ma. I hope you still remember that equation from your freshman physics course. We have determined what the force is. We can now relate the force to the acceleration of the electron. The acceleration is equal to the force divided by the mass of the electron m 0. So the acceleration is a constant times the electric charge times the electric field component divided by the mass of the electron. So far so good. Now from classical electricity and magnetism, the charge will radiate energy in the form of an electromagnetic wave. This is the scattered radiation whose intensity we are going to try to calculate. (I hope you are all with me so far in this derivation. The derivation so far is pretty straightforward, I think). 10

So we have an accelerated electron at the origin that s accelerating with components a 1 and a 2 parallel to the incident electric fields E 1 and E 2. This acceleration is going to produce an electric field at some point Q, a distance r from the accelerating electron. The component of the electric field that s produced in the E 1 direction is denoted as E 1 and the component of the electric field produced in the E 2 direction is denoted as E 2. Note that these two directions must be perpendicular to the vector r. 11

Let s figure out what this electric field vector is at the point Q. In the E 1 direction, the electric field is related to the acceleration, by this equation, which we obtain from basic electromagnetic theory. It is equal to the acceleration times the charge on the electron times the sine of the angle between E 1 and the line from the origin to Q, all this divided by c 2 times r. Let s plug in the value of a 1 into this equation. We find that the electric field is equal to the Coulomb constant times the square of the electron charge multiplied by the electric field component E 1 times the sine of the angle psi divided by the mass of the electron times c 2 r. (Is everyone with me so far?) 12

If we multiply the Coulomb force constant times the square of the charge of the electron and divide this by the mass of the electron times c 2, we obtain a constant that has the units of distance. This quantity is called the classical radius of the electron. It has the value of 2.818 times 10 15 m. This is a very small number. The diameter of the nucleus is 10 13 m, so this distance is about two orders of magnitude smaller than the nuclear diameter. If we plug this quantity in to our previous equation, we find that the electric field in the number 1 direction is proportional to the incident electric field in the same field times the ratio of the classical radius of the electron to the distance of interaction times the sine of the angle that the vector from the initial interaction point to our measurement point makes with the incident electric field direction. 13

So we have now computed the magnitude of one component of the electric field. It turns out that the second component gives us a similar equation but we find that E 2 is parallel to E 2 so we have that E 2 using the same argument we used to determine E 1 is r 0 over r times E 2 times the sine of 90 o. What we are saying here is that magnitude of the electric vector of the scattered radiation is going to be a very small fraction of the magnitude of the electric vector of the incident radiation because r 0 is so very much smaller than any distance we measure in this process. Let s stop our discussion at this point and let this sink in for a few days and we will continue discussing coherent scatter at our next lecture. 14