Is cross section is proportional

  • #1
If I show something that a cross section is proportional to:

[tex] 1 / (16p^2 sin^4( x) + 8mpsin^2 (x) + m^2) [/tex]

does it imply that the cross section is proportional to

[tex] 1 / sin^4 (x) [/tex] as well?
 
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Answers and Replies

  • #2
OlderDan
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alfredbester said:
If I show something that a cross section is proportional to:

[tex] 1 / (16p^2 sin^4( x) + 8mpsin^2 (x) + m^2) [/tex]

does it imply that the cross section is proportional to

[tex] 1 / sin^4 (x) [/tex] as well?
No. But can you factor the first denominator?
 
  • #3
[tex] 1 / (4psin^2(x) + m)(4psin^2(x) + m) [/tex]

Not sure what to do from there.
 
  • #4
OlderDan
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alfredbester said:
[tex] 1 / (4psin^2(x) + m)(4psin^4(x) + m) [/tex]

Not sure what to do from there.
The two factors in the denomnator are identical. What you have shown is that your thing is inversely proportional to the square of (4psin^2(x) + m). It is inversely proportiona to sin^4(x) only if m = 0. You could reach the same conclusion from your original expression, but it is a bit more evident when you factor it into a product of identical terms.
 
  • #5
That cause a problem then I was the scattering cross section is proportional to Mfi and

[tex] |M_{fi}|^2 /propto 1 / (q^2 + m^2)^2 [/tex]

Starting from the definition of q = [tex]p_{final} - p_{initial}.[/tex]

Show that the angular dependance of the scattering is then given simply by the Rutherford formula:CS= scattering cross secion

[tex] CS /propto 1 /(sin^4(\theta /2) [/tex]

I found [tex]q = 2psin(\theta /2)[/tex]
Since Mfi is proportional to the scattering CS I just tried sticking q into into the expression for [tex] |M_{fi}|^2 [/tex] as above but that clearly didn't work. I guess the wording angular dependance is key here.
 
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  • #6
OlderDan
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alfredbester said:
That cause a problem then I was the scattering cross section is proportional to Mfi and

[tex] |M_{fi}|^2 /propto 1 / (q^2 + m^2)^2 [/tex]

Starting from the definition of q = [tex]p_{final} - p_{initial}.[/tex]

Show that the angular dependance of the scattering is then given simply by the Rutherford formula:CS= scattering cross secion

[tex] CS /propto 1 /(sin^4(\theta /2) [/tex]

I found [tex]q = 2psin(\theta /2)[/tex]
Since Mfi is proportional to the scattering CS I just tried sticking q into into the expression for [tex] |M_{fi}|^2 [/tex] as above but that clearly didn't work. I guess the wording angular dependance is key here.
I had no context for your question, and I'm not completely following what you are trying to do. I have to log off now. See if this helps you any

http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html
 
  • #7
Thanks, I've got it now m=0 because the exchange boson is a photon :).
 

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