B Angular distribution to energy distribution

AI Thread Summary
To transform angular distributions into energy distributions for elastic nuclear reactions, one can integrate the differential angular cross sections over all angles. This method is similar to techniques used in photoionization, where energy distributions are derived from angular distributions. The discussion highlights that while the photoelectric effect does not directly involve energy distribution, the integration approach remains applicable. Understanding these relationships is crucial for accurate modeling in nuclear physics. The conversation emphasizes the importance of these transformations in analyzing nuclear reactions.
Ado
Messages
26
Reaction score
4
Hi!

I wonder, in the case of elastic reactions with nuclear potential, how go from an angular distribution to an energy distribution?

I have relationships on differential angular cross sections for neutron and proton elastic nuclear reations and I would like to transform them into differential energy cross sections.

Do you know how to proceed ?

Thank you in advance.
 
Physics news on Phys.org
At least in the field of photoionization, the energy distribution can be obtained from angular distribution by integrating the latter over all angles.
 
Exact :) the Photoelectric effect does not involve energy distribution, it is convenient =)
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top