# Clarification for the formulae for Differential Cross Section in Scattering theory [duplicate]

I am trying to study scattering theory using "Quantum Mechanics Concepts and Applications" by "Nouredine Zettili" . He starts from the formula

$${d \sigma( \theta,\phi) \over d \Omega } = {1 \over J_{inc}}{d N( \theta,\phi) \over d \Omega }$$

where $${d \sigma( \theta,\phi) \over d \Omega }$$ is called differential cross-section. $$J_ {inc}$$ is the incident flux.

But the book does not try to explain all the terms in the equation, so I don't get an intuitive idea about this equation.

• That answer is not sufficient, because it only explains what is differential cross-section is, Not try to derive above formulae and not try to explain the RHS of the above formulae. Can you please help me @ThomasFritsch – ROBIN RAJ Aug 29 '19 at 14:22
• So the only missing thing to understand is $dN = J_{\text{inc}} d\sigma$. For that we need to know how these are defined in your book. – Thomas Fritsch Aug 29 '19 at 15:09
• It does not provide much more information about the equation, Does not explain the terms in the equation, I think My question already provides all the details given by the textbook. @ThomasFritsch Fritsch – ROBIN RAJ Aug 29 '19 at 16:21

You understand already the meaning of differential cross-section $$\frac{d\sigma}{d\Omega}$$.
It is the ratio of an incoming cross-section area $$d\sigma$$ (measurable in $$\mathrm{m^2}$$) to the corresponding outgoing solid angle $$d\Omega$$ (measurable in steradian) as shown in the following image.

image from Wikipedia: Scattering cross-section

Now Zettili's book (page 617) says:

The differential cross section, which is denoted by $$d\sigma(\theta,\phi)/d\Omega$$, is defined as the number of particles scattered into an element of solid angled $$d\Omega$$ in the direction $$(\theta,\phi)$$ per unit time and incident flux: $$\frac{d\sigma(\theta,\phi)}{d\Omega} = \frac{1}{J_{inc}} \frac{dN(\theta,\phi)}{d\Omega}, \tag{11.2}$$ where $$J_{inc}$$ is the incident flux (or incident current density); it is equal to the number of incident particles per area per unit time.

So in the last sentence he says, the incoming flux $$J_{inc}$$ can be measured in particles$$/(\mathrm{m^2 \cdot s})$$.

You also know that $$\frac{d\sigma}{d\Omega}$$ can be measured in $$\mathrm{m^2 / steradian}$$.

From these two facts together with equation (11.2) you can conclude,
$$\frac{dN}{d\Omega} = J_{inc}\frac{d\sigma}{d\Omega}$$ can be measured in particles$$/(\mathrm{steradian \cdot s})$$.

So $$dN$$ is the number of particles per unit time going out into solid angle $$d\Omega$$. Thus $$dN$$ can be measured in particles$$/\mathrm{s}$$).

• "So dN is the number of particles per unit time going out into solid angle dΩ". I am confused with this statement because of $J_{inc}$ is incident flux not the outgoing flux from the target , Can you give a clarification @Thomas Fritsch – ROBIN RAJ Aug 29 '19 at 19:45
• @ROBINRAJ $dN$ is proportional to incoming flux $J_{inc}$. When you use more incoming particles per time, then of course you get more outgoing particles per time. – Thomas Fritsch Aug 29 '19 at 20:00
• Actually dN number of particle come from dΩ solid angle is due to d$\sigma$ change in cross-section, I think this equation and condition is only satisfied if and only if the target has some uniformity or Some continuous symmetry, Otherwise d$\sigma$ change does not give dN number of particles comes out from corresponding dΩ change is it right @Thomas Fritsch – ROBIN RAJ Aug 29 '19 at 20:33
• @ROBINRAJ $d\sigma(\theta,\phi)$ and $dN(\theta,\phi)$ don't need to be uniform. They may depend on $\theta$ and $\phi$. I have omitted to write $(\theta,\phi)$ only for simplicity. – Thomas Fritsch Aug 29 '19 at 20:40
• this implies cross-section $\sigma$ depends on $\theta$ and $\phi$, can you explain intuitively this dependence.@Thomas Fritsch – ROBIN RAJ Aug 31 '19 at 13:58