# What exactly does the quantity $\sigma(\theta)$ represent in scattering problems? What is the interpretation of the differential cross section?

So I don’t understand intuitively what the scattering cross section is. The equation also seems quite random to me. I have looked at derivations but I cannot wrap my head around. I am looking for some guidance.

What exactly does the quantity $$\sigma(\theta)$$ represent? It is not a differential and it is not a cross section, so why is it called differential cross section?

If I understand the units correctly it is $$\left[L\right]^2$$ or ‘area per angle squared’/ (maybe area per solid angle?)

$$\sigma\left(\theta\right)=-\frac{p}{\sin{\theta}}\frac{dp}{d\theta}$$ What exactly is in words. Why is it useful?

And what does the angle represent? If I understand correctly the term $$\frac{dp}{d\theta}$$represents how if you change the scattering angle $$\theta$$ how the impact parameter changes. But my guess would be that the impact parameter $$p$$ is the independent variable (since you change this in the experiment)

My questions:

1. What is the interpretation of the quantity $$\sigma(\theta)$$?
2. Why is it useful?
3. What is the interpretation of the quantity $$\frac{dp}{d\theta}$$?

There's a lot of different terminology and notation surrounding cross sections, which can make it difficult to learn the subject. However, fundamentally what is going on is not difficult -- we are trying to construct a mapping from the "input" of the scattering experiment (a beam of particles flying toward a scatterer) to the "output" (particles emanating in all directions outward from the scatterer).

I will follow the notation on wikipedia.

## What is the differential cross section?

The differential cross section $$\frac{d \sigma}{d\Omega}$$ has units of $${\rm m^2 sr^{-2}}$$, where $${\rm sr}$$ is the unit of solid angle, a steridian.

This figure from wikipedia helps explain what it represents Imagine a beam of incoming particles. For now, assume those particles are classical. Now, imagine a small area of that beam, represented by $$d\sigma$$. The fact that wikipedia has chosen to call this quantity the "differential cross section" on the figure is very annoying -- I would call it an "infinitesimal cross section of the incoming beam." It is a small part of the whole beam we are going to focus on. Our question is: where will this part of the beam end up after the scattering experiment?

Then can follow the trajectory as our collection of particles pass by the "scattering center" and fly off to infinity. Now imagine a large sphere centered on the scattering center. The infintesimal beam of particles we are following will now fly through an infintestimal solid angle of this sphere, $$d\Omega$$.

The differential cross section $$\frac{d\sigma}{d\Omega}$$ is the ratio of the incoming area, to the outgoing solid angle. In general, it is a function of (a) the "input" parameters (the impact parameter $$b$$ as well as the azimuthal angle $$\varphi_{\rm in}$$), and (b) the "output" parameters (the scattering angle $$\theta$$ and azimuthal angle $$\varphi_{\rm out}$$). However, for a classical particle, we only really need to specify one of the input or output variables, since classical mechanics is deterministic. A particle with a given impact parameter and input azimuthal angle will always end up with the same scattering angle and output azimuthal angle. In practice, we don't really control the impact parameter of any individual particle, but we do measure the output scattering angle and azimuthal angle, so we think of the differential cross section as a function of the output variables $$\frac{d\sigma}{d\Omega}(\theta, \varphi_{\rm out})$$ Incidentally, in quantum mechanics we consider the differential cross section to be a function of these variables as well. In quantum mechanics, we typically imagine the incoming particle is in a superposition of input parameters corresponding to a plane wave, and then measure the probability of the particle to scatter at different scattering angles and output azimuthal angles.

Finally, note that we actually only want to look at the solid angle of the scattered beam. Classically, any particle that has no interaction with the scatter does not contribute to the differential cross section. Quantum mechanically, we remove the "free" part of the wavefunction (typically a plane wave) that does not interact with the scatterer, and are only interested in the part of the wave function that is outgoing and has been affected by the scatterer.

## Azimuthally symmetric scatterer

If the scatterer is azimuthally symmetric about the axis of the beam, then the differential cross section does not depend on the azimuthal angle $$\varphi_{\rm out}$$. Then we can simply integrate over the azimuthal angle, and obtain $$\frac{d\sigma}{d(\cos \theta)}(\theta) = \int_0^{2\pi} d\varphi_{\rm out} \frac{d\sigma}{d\Omega}(\theta) = 2\pi \frac{d\sigma}{d\Omega}$$

## Total cross section

Finally, sometimes you want to compute the total cross section, $$\sigma$$. Loosely, you can interpret this as the total area of the original beam of particles that interacted with the scatterer. In general, this is given by integrating the differential cross section over all solid angles $$\sigma = \int d\Omega \frac{d\sigma}{d\Omega} = \int_0^\pi \sin \theta d\theta \int_0^{2\pi} d\varphi_{\rm out} \frac{d\sigma}{d\Omega}$$ For azimuthally symmetric scatterers, this simplifies $$\sigma = \int d\cos\theta \frac{d\sigma}{d\cos\theta} = \int_0^\pi \sin \theta d\theta \frac{d\sigma}{d\cos\theta}$$ The cross section gives you a rough idea of the "size" of the scatterer (at least in the directions perpendicular to the beam). A larger cross section means the target is easier to hit. A smaller cross section means it is harder to hit.

## Miscellaneous notes

Finally, some notes.

• Alternative notation: There are lots of different alternative notations. Sometimes, people write $$\frac{d\sigma}{d(\cos \theta)}$$ as $$\frac{d\sigma}{d\theta}$$. I find this confusing because in order to get to the total cross section $$\sigma$$, you need to integrate against $$\sin\theta d\theta=d\cos \theta$$, not $$d\theta$$. Second, you seem to have used $$\sigma(\theta)$$ to refer to $$\frac{d\sigma}{d\cos(\theta)}$$. You can do that, so long as you are clear on what you mean.
• What does this business about the differential cross section $$\frac{d\sigma}{d\Omega}$$ not being a differential mean? Well, first, the reason we use the notation $$\frac{d\sigma}{d\Omega}$$ for the differential cross section is that we can get to the total cross section by integrating the differential cross section against the solid angle. However, if we think about going the other way, from the total cross section to the differential cross section, we have to keep in mind that the total cross section is not a function of solid angle, so it does not really make sense to differentiate the total cross section with respect to angles. In a probability analogy, imagine that the total probability that a bang goes off during the day $$p$$. We can also ask if the probability that the bang goes off depends on the time of day. We might refer to the probability that the bang goes off between the time interval $$t$$ and $$t+dt$$ as $$\frac{dp}{dt}$$. Then the total probability that the bang goes off sometime during the day is $$p=\int_0^{\rm 1\ day} dt \frac{dp}{dt}$$. However, strictly speaking, if we interpret $$\frac{dp}{dt}$$ as an actual derivative, then $$\frac{dp}{dt}=0$$ since the total probability $$p$$ does not depend on $$t$$. Therefore, we have to remember $$\frac{dp}{dt}$$ is just notation for the probability per unit time and not an actual derivative of $$p$$.

So I don’t understand intuitively what the scattering cross section is.

Intuitively, the cross section is just that--it is the cross-sectional area of the object that is scattering the probe particles in the beam. I will call the object that is scattering the probe particles the "scatterer."

We are only interested in a cross-sectional area since the dimension along the beam of probe particles does not matter. (Or rather, the beam just "sees" the cross sectional area, not the volume.)

If we are probing a fixed scatterer, the cross section should be the same whether we are scattering x-rays or electrons or alpha particles, or whatever. (This is not entirely true, because some particles might not interact at all, e.g., if the scatterer is scattering via electromagnetic forces, then neutral particles can't "see" it.) (And, of course, the full description of scattering will account for probe differences, but the properties of the scatterer will remain the same.)

Consider the example of a hard sphere of radius $$R$$. If we shoot a beam of particles at the hard sphere then particles that are within $$R$$ of the center of the sphere will scatter and particles that are outside of $$R$$ will not. Therefore the total scattering cross section of the hard sphere is $$\pi R^2$$. In other words, we know that when we integrate $$\int d\Omega \frac{d\sigma}{d\Omega}$$ we better end up with $$\pi R^2$$. We just have to arrive at this result in a funny way because we are trying to back out a result regarding the scatterer based only on a measurement of the scattered probe particles.

Another way to this about this is that we are trying to figure out the "shape" of the scatterer, but we only have information about the scattered probe particles. (We can't "see" the scatterer directly, we can only measure the scattered probe particles.)

Suppose we are scattering off of a cube with sides of length $$L$$. If the beam is oriented directly along one of the cube edge directions the cross section will be $$L^2$$. (Of course, in real life it is hard to perfectly align a scatterer with the beam, so in real life will will likely have to average over all the cube's orientations).

Suppose we are scattering off of a Coulomb potential. Of course, this will only scatter charged particles, but it will scatter any charged particles regardless of how big their impact parameter is (assuming it is a completely unscreened Coulomb potential). Therefore the scattering cross section is infinite in this case.

Why is it useful?

Because we usually can not just go a "look" at the scatterer. The scatterer is usually something like a molecule, or a collection of atoms in a solid, which is/are invisible to the naked eye.

The only way we can "see" these things is by shooting in a beam of probe particles and looking at the angular distribution of scattered particles.

We have to back out information about the "shape" (or other properties) of the scatterer using only the information obtained from the scattered probe particles.

Image you are asked to figure out the shape of an object, but you can't look at it. Image that the only thing you can do is to close your eyes and throw ping-pong balls at it. You throw lots of ping-pong balls at the object and some of them bounce directly back at your face, some of them make glancing blows and bounce off part of the walls, and some pass by untouched. You can glean info about the shape of the unseen object by carefully recording which ping-pong balls bounce off which ways for which impact parameters. If you throw billions of ping-pong balls you can map out the shape (or rather the cross-sectional shape) of the unseen object.

Scattering is one of our only tools by which we can study small objects like atoms, molecules, and so on.