Asymmetry in e+e- anihilation

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In summary: I don't understand what you're asking. In summary, the asymmetry in the Z exchange contribution gives a measure of the Z exchange contribution.
  • #1

neu

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I understand the following:

The cross section [tex]\frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \propto 1 + cos^2 \theta [/tex] for a purely vectorial (electromagnetic) interaction. Hence [tex]\sigma[/tex] is expected to be symetric in [tex]cos \theta[/tex].

The axial vector (weak) coupling of the Z boson violates parity and give an asymmetric contribution to the [tex]\sigma[/tex] distribution.

So obviously the asymmetry give a measure of the Z exchange contribution.

OK, so Q: I don't understand why this reasoning doesn't apply when you consider [tex]e^+ e^- \rightarrow e^+ e^- [/tex]
Surely this behaves the same, since the photon and Z exchange is equally probable?
 
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  • #2
OK, well is it because
[tex]
e^+ e^- \rightarrow e^+ e^-
[/tex]

can occur via t-channel as well, since there's no species change. If so how does this affect angular distribution?
 
  • #3
well, first off, the photon and Z boson exchanges are not equally probable: they involve different coupling constants, and the Z-boson is heavy while the photon is not (more relevant at lower energies).

I'm not quite sure the answer to your question, but the fact that there is a t-channel process might bury the relatively small asymmetry coming from the axial Z coupling. That's just a first-blush guess.
 
  • #4
Yes, I realized I'd written that badly. I meant to say the photon and Z couplings are just as they are for muon pair production as electron pair production/scatter.

So Why is it that

[tex] \frac{d \sigma (e^+e^- \rightarrow e^+e^-)}{d\Omega} \neq \frac{d \sigma (e^+e^- \rightarrow \mu^+\mu^-)}{d\Omega} [/tex]

My thinking is that [tex] \frac{d \sigma}{d\Omega} \propto \mid M_{z} +M_{\gamma}\mid ^2[/tex]

where there is s, t and u channel scatter for [tex] (e^+e^- \rightarrow e^+e^-) [/tex] and only s channel for [tex] (e^+e^- \rightarrow \mu^+\mu^-) [/tex]

And so : [tex] \frac{d \sigma (e^+e^- \rightarrow e^+e^-)}{d\Omega} > \frac{d \sigma (e^+e^- \rightarrow \mu^+\mu^-)}{d\Omega} [/tex]


How does that sound?
 
  • #5
neu said:
How does that sound?

strange! I'm a little confused what you're asking. are you only asking why Bhabha scattering is not the same as muon production? That's straightforward - the processes are not topologically the same (that is, as you say, there is a t-channel diagram in bhabha but not mu-production), so there is no reason why they are the same. that one is "larger" than the other: that's not immediately obvious since (in principle) there could be DEstructive interference making bhabha scattering smaller.

but none of this seems to have anything to do with your original question about why you can't measure the Z-contribution to Bhabha scattering but you can for mu-production. I think it depends on precisely what you are trying to measure.

I'm not doing a good job answering your question because I'm confused. help me out: what precisely do you want to do with bhabha scattering that you claim you cannot do?
 
  • #6
OK, I'm not being clear, thanks for sticking with me. I'll re-state my question.

Question:
Ignoring mass effects/ phase space arguments, why at leading order do we expect:

[tex]
\frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \neq \frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow e^+e^-)
[/tex]

And which should be larger.

My thoughts:
-only s-channel for muon production, but s, t and u-channel for bhabha,
-so bhabha scatter/anihillation cross section should be larger (aside from phase space favouring low final state mass).

Another question:
For photon exchange we have
[tex]
\frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \propto 1 + cos^2 \theta
[/tex]

This is symmetric in cos, but measurement shows asymmetry due to the axial vector coupling of the Z boson. Why doesn't this reasoning apply (correct me if I'm wrong) for bhabha scatter/anihillation.My thoughts:
Even though, as you say, the two processes are topologically different, they can still have Z/photon exchange for all processes.

I'm thinking that u and t-channel processes don't posses the same symmetric (in cos) properties as the s-channel process so they wash out this method of measuring the Z contribution.
 
  • #7
neu said:
only s-channel for muon production, but s, t and u-channel for bhabha,

Correct.

neu said:
so bhabha scatter/anihillation cross section should be larger (aside from phase space favouring low final state mass).

Incorrect. You don't know if it is larger or smaller without doing the calculation, because you don't know the relative phase of the s and t channel contributions. The interference could be either constructive or destructive. (In QM you add amplitudes and then square)



neu said:
Another question:

This has me totally confused. If you write down the angular distribution using photon exchange, you don't see the parity violating effect of the Z. That's true for muon exchange and it's true for Bhabbas. If you include the Z, both processes show parity violation.
 
  • #8
neu said:
My thoughts:
-only s-channel for muon production, but s, t and u-channel for bhabha,
-so bhabha scatter/anihillation cross section should be larger (aside from phase space favouring low final state mass).

see vanadium 50's comments.

Another question:
For photon exchange we have
[tex]
\frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \propto 1 + cos^2 \theta
[/tex]

This is symmetric in cos, but measurement shows asymmetry due to the axial vector coupling of the Z boson. Why doesn't this reasoning apply (correct me if I'm wrong) for bhabha scatter/anihillation.

you're saying it doesn't?! Naively I'd expect both processes should be sensitive to the axial Z-boson coupling. The only thing I can think of is that there are some experimental barriers. Can you give me a reference to this conflict?

My thoughts:
Even though, as you say, the two processes are topologically different, they can still have Z/photon exchange for all processes.

I'm thinking that u and t-channel processes don't posses the same symmetric (in cos) properties as the s-channel process so they wash out this method of measuring the Z contribution.

t-channel processes most assuredly do NOT have the same angular dependence as s-channel processes, that is true; you can see this by explicitly doing the calculation, or by expanding in partial waves and noticing that the t-channel has no s-wave component, and is therefore not isotropic. In any event, this might be true, but please clarify my previous question about where you heard Bhabha is such a bad boy.
 
  • #9
Thanks for your responses, things are clear now.

It seems I've misinterpreted things regarding the second question I asked. It's a question of how the amplitudes differ. Of course both have an axial vector contribution.

It was to do with a conversation I had with a lecturer.

Thanks again for your help.
 

1. What is asymmetry in e+e- annihilation?

Asymmetry in e+e- annihilation refers to the phenomenon in which the production of particles and antiparticles in a high energy collision is not exactly equal. This can be caused by differences in the initial state, such as the polarization of the beams, or by differences in the interaction between particles and antiparticles.

2. Why is asymmetry in e+e- annihilation important?

Asymmetry in e+e- annihilation is important because it provides valuable information about the underlying physics of the collision. By measuring the degree of asymmetry, scientists can gain insight into the fundamental interactions between particles and antiparticles and potentially discover new particles or phenomena.

3. How is asymmetry in e+e- annihilation measured?

Asymmetry in e+e- annihilation is typically measured by comparing the number of particles produced in the forward direction (in the same direction as the initial beams) to the number produced in the backward direction (in the opposite direction). The difference between these two numbers is known as the asymmetry.

4. What is the significance of a large asymmetry in e+e- annihilation?

A large asymmetry in e+e- annihilation can indicate the presence of new particles or interactions that are not predicted by current theories. This can lead to further research and potentially new discoveries in the field of particle physics.

5. How does asymmetry in e+e- annihilation differ from other types of asymmetry?

Asymmetry in e+e- annihilation is a specific type of asymmetry that occurs in high energy collisions of particles and antiparticles. It is different from other types of asymmetry, such as spatial or temporal asymmetry, which refer to differences in space or time, respectively.

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