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Scattering Cross Section - At high energies

  • #1
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Homework Statement



Why does ## \frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow \tau^+ + \tau^- } \rightarrow 1## at high energies?

Would it be the same if it was ## \frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow e^+ + e^- }##?

Homework Equations




The Attempt at a Solution


[/B]
For the first reaction, propagator factor for the photon is ##\frac{1}{(2E)^2 - m_\gamma^2} = \frac{1}{4E^2}##. Two vertex factors are ##g_{EM}^2##. Thus ## |M_{fi}|^2 = |\frac{g_{EM}^2}{4E_e^2}|^2##.

Fermi's golden rule is given by:
[tex]\Gamma = 2\pi |M_{fi}|^2 \frac{dN}{dE_f} [/tex]

Differential cross section is number of particles scattered per unit solid angle per unit time divided by flux:

[tex]\frac{d\sigma}{d\Omega} = \frac{1}{2\nu_e} \Gamma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{dN}{dE_f} [/tex]

The density of states is given by: ##dN = \frac{p^2_\mu}{(2\pi)^3} d\Omega dp_\mu## and substituting gives:

[tex] d\sigma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{p^2}{(2\pi)^3} \frac{dp}{dE_0} d\Omega [/tex]

At high energies, ##E_\mu \approx \frac{1}{2} E_0##, so ## \frac{dp_\mu}{dE_0} = \frac{1}{2} \frac{dp_\mu}{dE_\mu} = \frac{E_\mu}{p_\mu}##.
 
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Answers and Replies

  • #2
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At high energies, ##v_\mu \approx c##, so ##\frac{dp_\mu}{dE_\mu} = \frac{1}{2} \frac{E_\mu}{p_\mu} = \frac{1}{2c} = \frac{1}{2}##, letting c=1. This gives:

[tex] d\sigma = \frac{1}{2\nu_e} 2\pi \left( \frac{g_{EM}^2}{4E_e^2} \right)^2 \frac{p^2}{(2\pi)^3} \frac{1}{2} d\Omega [/tex]

[tex] \sigma = \frac{1}{2\nu_e} 2\pi \left( \frac{g_{EM}^2}{4E_e^2} \right)^2 \frac{p^2}{(2\pi)^3} \frac{1}{2} 4\pi [/tex]

[tex] = \frac{1}{64} \frac{1}{\nu_e} \frac{1}{\pi} \frac{g_{EM}^2}{E_e^4} p_\mu^2 [/tex]

Using ##\frac{g_{EM}^2}{4\pi} = \alpha##,

[tex]\sigma = \frac{\pi}{4} \frac{1}{\nu_e} \frac{p_\mu^2}{E_e^4} \alpha^2 [/tex]

For highly relativistic case, ##p_\mu \approx E_\mu = E_e## and ##\nu_e \approx c = 1##.

[tex]\sigma = \frac{\pi}{4} \frac{1}{E^2} = \pi \frac{\alpha^2}{(2E)^2} = \pi \frac{\alpha^2}{s^2} [/tex]
where ##s=2E## is the centre of mass energy.
 
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  • #3
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Since the scattering cross section only depends on:

a) Centre of mass energy
b) ##g_{EM}##

Wouldn't both ##\rightarrow 1## at high energies (##p >> m##)?
 
  • #4
ChrisVer
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Wouldn't both →1\rightarrow 1 at high energies (p>>mp >> m)?
I find the 2nd question a little tricky...
For the first you dealt only for one possible channel : the s-channel ..
For the second case however, the electrons can also "not interact" at all... apart from annihilating and giving again e^+/- pair..
 
  • #5
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I find the 2nd question a little tricky...
For the first you dealt only for one possible channel : the s-channel ..
For the second case however, the electrons can also "not interact" at all... apart from annihilating and giving again e^+/- pair..
I think I got it.

Part (a)
The cross section is proportional to ##|M_{fi}|^2##.

First process is given by:

[tex]e^+ + e^- \rightarrow \gamma^* \rightarrow \mu^+ + \mu^-[/tex]

The propagator factor for this process is ## \frac{1}{P \cdot P - m_\gamma^2} = \frac{1}{P \cdot P}## since ##m_\gamma = 0 ##.

Second process is given by:

[tex]e^+ + e^- \rightarrow Z^0 \rightarrow \tau^+ + \tau^-[/tex]

The propagator factor for this process is ## \frac{1}{P \cdot P - m_Z^2} ##.

At high energies, this propagator factor ##\approx \frac{1}{P \cdot P}##

Thus they approach 1 at high energies.

Part (b)

Wouldn't this always be true? Given that the propagator factor for both are equal to ##\frac{1}{P \cdot P}##.
 
  • #6
ChrisVer
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The part a is clear for you I think... Even for photon interactions alone, the ratio you give approaches 1 for high energies...it can also be done for the Z0.

For part b, you don't have to think of propagators alone...
you can have:
[itex]e^- e^+ \rightarrow \gamma^* \rightarrow e^- e^+[/itex] (s-channel)
as well as
[itex] e^- e^+ \rightarrow e^- e^+[/itex] ...by t-channel...they don't have to annihilate to photon and produce the pairs again as you have to do for the case of muon products.
 
  • #7
ChrisVer
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a1.jpg
 
  • #8
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  • #9
ChrisVer
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For high energies, the propagators of the weak and electromagnetic interaction are the same. It goes as [itex]\frac{1}{s}[/itex]
The difference comes by the 2nd diagram that is possible for the electron/positron products interaction and doesn't exist for the muons.
 
  • #10
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For high energies, the propagators of the weak and electromagnetic interaction are the same. It goes as [itex]\frac{1}{s}[/itex]
The difference comes by the 2nd diagram that is possible for the electron/positron products interaction.
Which process dominates? Which is the "s-channel" and which is the "t-channel"?
 
  • #11
ChrisVer
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Which process dominates? Which is the "s-channel" and which is the "t-channel"?
In my figure the diagram with muons and the same one with electrons are the s-channel processes.
The extra diagram for the [itex]e^- \rightarrow e^-[/itex] and [itex] e^+ \rightarrow e^+[/itex] which exists only for the [itex]e^-e^+ \rightarrow e^+ e^-[/itex] process is the t-channel....

I don't understand what you mean by "dominates". I am explainging why:
[itex] \frac{e^+e^- \rightarrow \mu^+\mu^-}{e^+e^- \rightarrow e^+e^-}[/itex] shouldn't be 1.
 
  • #12
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In my figure the diagram with muons and the same one with electrons are the s-channel processes.
The extra diagram for the [itex]e^- \rightarrow e^-[/itex] and [itex] e^+ \rightarrow e^+[/itex] which exists only for the [itex]e^-e^+ \rightarrow e^+ e^-[/itex] process is the t-channel....

I don't understand what you mean by "dominates". I am explainging why:
[itex] \frac{e^+e^- \rightarrow \mu^+\mu^-}{e^+e^- \rightarrow e^+e^-}[/itex] shouldn't be 1.
All three are in the same picture.

By "dominate" I mean which is most frequent/most likely?
 
  • #13
ChrisVer
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a1.jpg
 
  • #14
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  • #15
ChrisVer
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Both are the same (I think), since the interaction is symmetric under the interchange [itex]t \leftrightarrow s[/itex]
But nevertheless, how can this "help" you?
 

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