# Differential Cross Section and Factor of $\pi$

I hope this is not a double-post, but the other threads couldn't help me:

In my calculations of the differential cross section $$\frac{d\sigma}{d\Omega}$$, I am always a factor $$\pi$$ lower than the reference data. This is driving me NUTS. As I have checked my code multiple times and found no mistake, my best guess is that I am lacking the understanding of some basic definitions.

Here is my understanding:

In experimental physics, the differential cross section is defined as follows:

$$\frac{d\sigma}{d\Omega} = \frac{N}{F \cdot \rho \cdot \epsilon \cdot \Delta\Omega}$$

Where $$N$$ is the count of desired interactions, $$F$$ is the number of incoming particles, $$\rho$$ is the target area density in inverse microbarns, $$\epsilon$$ is the reconstruction efficiency and $$\Delta\Omega$$ is the solid angle element in which the particles are detected.

I assume that I am making a mistake related to $$\Delta\Omega$$. I assume that following equation holds:

$$\Delta\Omega = 2 \cdot \pi \cdot \Delta \cos(\theta_{CM}).$$

If my detector spans from $$\cos(\theta_{CM})=0.9$$ to $$\cos(\theta_{CM})=1.0$$, I think that $$\Delta \cos(\theta_{CM})=0.1$$, meaning it is the angle covered by the detector. That would imply $$\Delta\Omega = 0.63$$.

1. Are these assumptions correct?

2. When data is given as $$\frac{d\sigma}{d\cos(\theta)}$$, I assume I could just convert this to $$\frac{d\sigma}{d\Omega}$$ by multiplying with $$\frac{1}{2\pi}$$. Correct?

I am going crazy over this missing factor of $$\pi$$.

• Beware that $\mathrm d\Omega$ refers to a small patch of solid angle, but $\mathrm d(\cos\theta)$ refers to a complete annulus at constant scattering angle. If you imagine a detector in your downstream beam at $\cos\theta>0.9$, a small detector might cover the entire solid angle. But around $\theta=90^\circ$, you need an entire ring of detectors to get the entire angular slice.
– rob
Commented Nov 24, 2023 at 4:20