Differential cross section/simple integral

In summary: CSci203/LectureNotes/LectureNotes-LaTeX.pdfIn summary, the student is trying to solve an equation for sigma, but is having trouble.
  • #1
jonwell
11
0
At least it should be a simple integral...

Homework Statement



The whole text is here- http://i35.tinypic.com/2nisnp.jpg

Basically (I think) I need to integrate the differential over all angles theta and phi, and get sigma(naught) back out.


Homework Equations



given in pic

The Attempt at a Solution



If I use the given substitution and integrate I get log(tan(pi/2)) which is log(0) which is broken... I don't really know where to go other than that.

Thanks! Been reading the forum for a long time but this is my first question :)
 
Physics news on Phys.org
  • #2
You shouldn't be getting any Log terms when you integrate; if you show me your work, I can tell you where you are going wrong.
 
  • #3
I substitute sin(theta) for d(omega), and the integral for 1/sin(theta) is log(tan(theta/2)) at least as given by mathematica and the back of my book.

I kind of figured I shouldn't be getting that, which is what makes me think I'm approaching the problem all out of whack.
 
  • #4
You should be substituting [itex]sin(\theta) d \theta d \phi[/itex] for [itex]d \Omega[/itex] not just [itex]sin(\theta)[/itex]...[itex]d \Omega[/itex] is a differential, [itex]sin(\theta)[/itex] is just a function...you should have:

[tex]\frac{d \sigma}{d \Omega}=\frac{d \sigma}{sin(\theta) d \theta d \phi}=\frac{\sigma_0}{4 \pi}[/tex]

[tex]\Rightarrow d \sigma=\frac{\sigma_0}{4 \pi} sin(\theta) d \theta d \phi[/tex]

...do you follow?

Then just integrate both sides of the equation.
 
  • #5
gabbagabbahey said:
You should be substituting [itex]sin(\theta) d \theta d \phi[/itex] for [itex]d \Omega[/itex] not just [itex]sin(\theta)[/itex]...[itex]d \Omega[/itex] is a differential, [itex]sin(\theta)[/itex] is just a function...you should have:

[tex]\frac{d \sigma}{d \Omega}=\frac{d \sigma}{sin(\theta) d \theta d \phi}=\frac{\sigma_0}{4 \pi}[/tex]

[tex]\Rightarrow d \sigma=\frac{\sigma_0}{4 \pi} sin(\theta) d \theta d \phi[/tex]

...do you follow?

Then just integrate both sides of the equation.


I figured it was something simple. I follow perfectly, thank you very much :)

[...I need to figure out that LaTeX code stuff, that's pretty cool.]
 
  • #6
Your welcome. There's an introduction to LaTeX here :smile:
 

1. What is a differential cross section?

A differential cross section is a mathematical concept used in physics to describe the probability of a particle scattering or interacting with another particle. It is typically denoted by σ and is expressed in units of area, such as square meters or barns.

2. What is the difference between differential and total cross section?

The differential cross section measures the probability of particles scattering into specific angles or energies, while the total cross section measures the overall probability of particle scattering in any direction. The total cross section is obtained by integrating the differential cross section over all possible angles or energies.

3. How is the differential cross section calculated?

The differential cross section is calculated using the quantum mechanical scattering amplitude, which takes into account the properties and interactions of the particles involved. It also depends on factors such as the relative speed and angle of the particles before and after the interaction.

4. What is a simple integral in relation to differential cross section?

A simple integral is a mathematical operation used to calculate the total cross section from the differential cross section. It involves summing up all the individual probabilities of scattering at different angles or energies, and is represented by the integral symbol "∫".

5. How is the differential cross section used in particle physics experiments?

The differential cross section is a fundamental concept in particle physics and is used to study the properties and interactions of particles. It is often measured in experiments, and the data obtained can be compared to theoretical predictions to test and refine our understanding of the underlying physics principles.

Similar threads

  • Introductory Physics Homework Help
Replies
25
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Nuclear Engineering
Replies
0
Views
616
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
794
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
3K
Back
Top