2
$\begingroup$

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 $\pi$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – rob
    Commented Nov 24, 2023 at 4:20

1 Answer 1

0
$\begingroup$

Assumptions 1. and 2. are correct.

$\endgroup$
2
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jun 9 at 20:15
  • $\begingroup$ Please include at least some elaboration. $\endgroup$ Commented Jun 9 at 20:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.