# Differential cross section

1. Jan 25, 2013

### Kontilera

Hello!
I´m confused about this concept.. It seems rather trivial, but my teacher is not that pedagogical and describes it as a rather diffcult concept so maybe I misunderstood it.

Given the definition in Sakurai and the scattering of only one particle it seem to be a kind of "denisty" per radians for the probability (or rather amplitude) that the particle will be scattered in this direction.

In other words for a small interval in our angle (lets say inbetween a and b) an estimation of the probability for our particle to come out in this direction should be given by:

$$P(\theta \in [a,b] )= |(b - a) \cdot \frac{d \sigma}{d \Omega}\big|_{\frac{a+b}{2}}\,\,|^2$$.

Is this a good intuitional picture to have in mind when going to the next lecture?

Last edited: Jan 25, 2013
2. Jan 25, 2013

### The_Duck

You're close, except that $\Omega$ is a solid angle and not just an angle. Also you shouldn't be squaring the differential cross section.

We need two angles to describe a 3D scattering process, usually $\theta$ (the angle the outgoing path makes to the incoming path) and $\phi$ (the azimuthal angle). The probability for a particle to scatter in a direction close to $(\theta, \phi)$--say, within $d\theta$ in the $\theta$ angle and and within $d\phi$ in the $\phi$ angle--is proportional to

$\cos \theta d\theta d\phi \frac{d \sigma}{d \Omega}(\theta, \phi)$

Here $\cos \theta d\theta d\phi$ is the amount of the solid angle we are looking at (which we might call $d \Omega$), and the analog of your (b-a) above.

Note that the expression above still isn't a probability. It has units of area; it's a cross section.

3. Jan 25, 2013

### Kontilera

Yeah, that makes sense. Looked up the definition again and it is actually a "probability density over the angles"... Thanks. :)

I know what I mean but "density" over angles? does it make sense? How would you say it when trying to explain the concept?

4. Jan 25, 2013

### Kontilera

This definition, pictorially describe in Sakurai, doesnt seem to obvious. I mean in the picture he paints the solid angle, $\Omega$ from a point in the potential area to an "area of observation", $\sigma$, but this solid angle does indeed depend on which point we choose inside our potential... Do we neglect these variantions our why dont they matter?

5. Jan 26, 2013

### andrien

In general theory of scattering,the source is composed of delta function sources(you treat it like that).So in case of generality you can not define any point inside the source.it will be considered as point source.