0

I try to calculate the intensity of a reflected signal from an ensamble of planar objects, randomly oriented in space.

With respect to the incoming beam, the orientation of the object is defined by two angles, $\phi$ and $\theta$, as sketched here: https://cloud.ill.fr/index.php/s/JjhSHxY3glwhyAY

For any given orientation, the signal $f(\phi, \theta)$ is as follows:

$$f(\phi, \theta) = \left\{ \begin{array}{l} 1 \text{ for } \theta<\theta_c \text{ and any } \phi \\ 0 \text{ for } \theta>\theta_c \text{ and any } \phi \end{array} \right.$$

In particular, I am interested in the result of the orientational average over $\phi$ (which I note as $\left<f(\phi, \theta)\right>$ and then calculating the limit for $\theta\rightarrow0$.

Edit: I have clarified the question better following the comment of David.

As I do not want to integrate over $\theta$, is the correct form of the average integral still $\int_0^{2\pi} f(\phi, \theta) d\phi$? i.e., the first part of: $$ \int_0^{2\pi}d\phi\int_0^{\pi} f(\phi, \theta)\sin\theta d\theta $$

thanks and best wishes, Leonardo

leonardo2887
  • 1
  • Since $f(\phi, \theta)$ does not vary with respect to $\phi$, only with respect to $\theta,$ you can say $\int_\alpha^\beta f(\phi, \theta),d\phi = (\beta - \alpha)f(\phi, \theta)$ for whatever range of angles $\alpha<\phi<\beta$ you are concerned with. But this seems to have little relationship to the physical problem. Your angles are the angles of spherical coordinates and there are standard formulas for integration; for example, see https://math.stackexchange.com/questions/131735/surface-element-in-spherical-coordinates – David K Aug 03 '21 at 12:45
  • Thanks David for the comment. Indeed, the question was ill-posed, and I hope to have it clarified a bit better. My question was not how to solve that integral, but which formula for integration I have to use, knowing that I want to integrate over $\phi$ alone and not over $\theta$. – leonardo2887 Aug 03 '21 at 15:25
  • It is still unclear what you think the "average over $\phi$" represents and why you think it is anything more than a trivial computation: $1$ if $\theta<\theta_c$, otherwise $0$. You have already implied that $f$ does not actually depend on $\phi$ in any way. – David K Aug 03 '21 at 16:31
  • I believe that your azimuth angles $\theta$ and $\theta_c$ only take values from $0$ to $\pi$, otherwise it makes no sense. The area of the spherical cap ${\theta\leqslant\theta_c}$ is $(1-\cos\theta_c)/2$, this should help you. But I still don't understand what you want to compute... – Tom-Tom Aug 10 '21 at 22:26

0 Answers0