Evaluating Scattering Integral

In summary, the person is trying to integrate over the Dirac delta and is getting lost. They ask for help and are told to change to spherical coordinates and to only include the terms affecting the integral over r. They are also told that the Dirac delta will only affect the integral over r.
  • #1
Waxterzz
82
0
Hi,

I really don't have a clue to solve this.

1SR45bL.png
I tried something like the dirac function identity:

utoxnMx.png


But then I saw it's dk' not dk' and couldn't got it straight.

Can someone help me with this?
 
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  • #2
As it is said, you have to change to spherical coordinates. The Dirac delta will then only affect the integral over r.
 
  • #3
DrClaude said:
As it is said, you have to change to spherical coordinates. The Dirac delta will then only affect the integral over r.

I don't see it. dk' is not a volume element?
 
  • #4
1a83c429226b2c6a2da143ae35a00751.png


dk' is a vector. Then I have an integral comprising of 3 terms each involving unit vectors.

I really don't have a clue.
 
  • #5
Wait wait, I am extremely confused.

I just have to integrate over the k' first?
 
  • #6
Mr
DrClaude said:
As it is said, you have to change to spherical coordinates. The Dirac delta will then only affect the integral over r.
6 hours and I still have no clue, can you please hold my hand and solve it with me. I mean, just say what I have to do, I will solve it, but give me instructions.

I mean what's the deal with the dirac function and the dk', it's supposed to be a volume element and the dirac term, why isn't it delta(k-k')
 
  • #7
##d\vec{k}## is a volume element. Write it as a volume element in spherical coordinate. To make the integrand even more transparent, use the cosine law to express ##|\vec{k}-\vec{k'}|^2## in terms of ##k##, ##k'##, and ##\theta##.
 
  • #8
blue_leaf77 said:
##d\vec{k}## is a volume element. Write it as a volume element in spherical coordinate. To make the integrand even more transparent, use the cosine law to express ##|\vec{k}-\vec{k'}|^2## in terms of ##k##, ##k'##, and ##\theta##.
mW21BmR.png


How come dk' is a volume element, it's the derivative of a vector. Most textbooks, the volume element is called a d tau or a dV

My head is a mess. It's like I completely forgot how to calculus.

Good news: I know I'm wrong.

Really don't see it
 
  • #9
What happened to the Dirac delta? You should do the integral over k' first.
 
  • #10
Waxterzz said:
How come dk' is a volume element, it's the derivative of a vector. Most textbooks, the volume element is called a d tau or a dV
It's not a derivative, it is an infinitesimal vector element.
 
  • #11
DrClaude said:
It's not a derivative, it is an infinitesimal vector element.

It's been this the whole time?, with r being k'

eqn_vol_e.gif


Ps: After this I'm going to finally learn LaTeX
 
  • #12
Why did the Dirac delta disappear in the second equation from the last one?
 
  • #13
TEAEU8n.png
sv97Qux.png


This is what I got uptil now, but I have to leave.

Thanks for help, give me feedback if you want to, and I will post update when I'm back, probably tomorrow. Thanks for the patience anyway!
 
  • #14
DrClaude said:
It's not a derivative, it is an infinitesimal vector element.

blue_leaf77 said:
Why did the Dirac delta disappear in the second equation from the last one?

Hi,

Sorry for the late reply.

Hope this is somewhat more clear, because last post was a bit messy. Still haven't learned LaTex.

34XvfUc.png


Is this valid?

I couldn't evaluatie the last integral, because of the square.

Do I need to use partial fractions?

Thanks
 
  • #15
Hi myself,

I found out I need to include something like 1- ( stuff going in the z direction after k scattering) / (stuff that would go in z direction if there was no scattering at all), so 1 - k' projected on z axis /k = 1 - (k cos theta) / k

Yes, I'm a noob.

So I'll start over. :')
 

1. What is a scattering integral and how is it used in scientific research?

A scattering integral is a mathematical equation that is used to describe the behavior of scattered particles or waves in a given system. It is commonly used in a variety of scientific fields, such as physics, chemistry, and engineering, to study the interactions between particles or waves and their surrounding environment.

2. How is a scattering integral different from other types of integrals?

A scattering integral is unique in that it takes into account the interactions between scattered particles or waves and their surrounding environment. This makes it more complex than other types of integrals, which typically only consider the behavior of individual particles or waves.

3. What factors are considered when evaluating a scattering integral?

When evaluating a scattering integral, factors such as the properties of the scattering medium, the energy and direction of the scattered particles or waves, and the interactions between these particles or waves and the medium are taken into consideration. Other factors, such as the size and shape of the scattering objects, may also be considered depending on the specific research being conducted.

4. How is the accuracy of a scattering integral determined?

The accuracy of a scattering integral is determined by comparing its results to experimental data or other known solutions. If the results closely match the expected behavior of the scattered particles or waves, then the integral is considered to be accurate.

5. What are some applications of scattering integrals in scientific research?

Scattering integrals have numerous applications in scientific research. They are commonly used in fields such as optics, acoustics, and radar to study the behavior of light, sound, and electromagnetic waves as they interact with different materials and structures. They are also used in particle physics to study the behavior of subatomic particles and in meteorology to model the scattering of light by atmospheric particles.

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