How Is Particle Distribution in a Solid Angle Derived in Mechanics?

[ranger281]

I'm reading Mechanics by Landau and Lifshitz, chapter IV, and trying to understand how in a (closed) center of mass system, with randomly distributed and oriented particles that disintegrate, "the fraction of particles entering a solid angle element $do_{0}$ is proportional to $do_{0}$, i.e. equal to $do_{0}/4π$". Then it is stated that "the distribution with respect to the angle $θ_{0}$ is obtained by putting $do_{0}=2π∗sin(θ_{0})dθ_{0}do_{0}=2π∗sin(θ_{0})dθ_{0}$".

What is the solid angle element $do_{0}$(and the equation for it)? How were those formulas obtained?

 

[vanhees71]

Take usual spherical coordinates for the unit sphere
$$\vec{x}=\begin{pmatrix}\cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta\end{pmatrix}.$$
The surface-element spanned by an infinitesimal rectangular-like shape given by the infinitesimal increments $\mathrm{d} \vartheta$ and $\mathrm{d} \varphi$ at the point $(\vartheta,\varphi)$ on the sphere is given by
$$\mathrm{d}^2 f = \mathrm{d} \vartheta \mathrm{d} \varphi \left | \frac{\partial \vec{x}}{\partial \vartheta} \times \frac{\partial \vec{x}}{\partial \varphi} \right|=\mathrm{d} \vartheta \mathrm{d} \varphi \sin \vartheta.$$

 

[ranger281]

Where does $do_{0}/4\pi$ come from?

Thank you for help.

 

[vanhees71]

The full solid angle is given by integrating over the entire spherical unit shell. Just draw the vector in dependence of $\vartheta$ and $\varphi$:

https://en.wikipedia.org/wiki/Spherical_coordinate_system#/media/File:3D_Spherical.svg

There you see that the full spherical shell is covered once for $\vartheta \in [0,\pi]$ and $\varphi \in [0,2 \pi[$. This gives
$$\Omega=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \sin \vartheta = 2 \pi \int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta = 4 \pi.$$
Now if something is spread uniformly across the sphere, the distribution must be
$$f(\varphi,\vartheta)=\frac{1}{4 \pi}.$$
Thus indeed the fraction entering the solid angle element $\mathrm{d}^2 \Omega$ is $\mathrm{d}^2 \Omega/(4 \pi)$.