0

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.

Can anyone please give a good derivation for this equation and explain all the terms in the equation and give an intuitive idea about this equation? Your small efforts are so precious for me please help me to move more

Thomas Fritsch
  • 39,317
ROBIN RAJ
  • 545
  • 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

1 Answers1

2

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
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$/\mathrm{s}$).

Thomas Fritsch
  • 39,317
  • "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
  • @ROBINRAJ Consider for example "Rutherford scattering" (google for that). All particles passing the target with large impact parameter $b$ are scattered very little, hence $\theta\approx 0$. Only those with small $b$ result in a large $\theta$. That means $dN(\theta,\phi)/d\Omega$ and $d\sigma(\theta,\phi)/d\Omega$ have a very pronounced maximum at $\theta=0$. – Thomas Fritsch Aug 31 '19 at 14:21
  • Is it scattering crosssection is the square of the maximum value of impact parameter multiplied by $\pi$ – ROBIN RAJ Sep 01 '19 at 12:20
  • Is that statement right?@Thomas Fritsch – ROBIN RAJ Sep 02 '19 at 07:12
  • @ROBINRAJ The impact parameter $b$ can be arbitrarily large. Therefore, in my opinion, your statement doesn't make much sense. – Thomas Fritsch Sep 02 '19 at 07:40
  • @ROBINRAJ By the way, please note that the comments section is not for long-winded discussions. If you have a new dedicated problem, then ask a new question (may be linking back to this old question for context). – Thomas Fritsch Sep 02 '19 at 08:23