What Is the Ratio of Rates for e+ e- -> mu+ mu- at sqrt(s)=5 GeV?

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In summary, at a center-of-mass energy of \sqrt{s}=5 GeV, the ratio of the rates for e^+ e^- collisions producing hadrons and \mu^+ \mu^- is primarily determined by the mass difference between the muon and available hadronic states. The matrix elements for both cases are similar and the rates have the same form when expressed in terms of momenta and masses for muons or hadrons in the final state.
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Homework Statement


For [tex]e^+ e^-[/tex] collisions at [tex]\sqrt{s}=5[/tex] GeV, estimate the ratio of the rates at which interactions produce hadrons and [tex]\mu^+ \mu^-[/tex]

Homework Equations


[tex]\sqrt{s} = 2E = E_{cm}[/tex]

[tex]\Gamma = \frac{S |p|}{8 \pi \hbar m_1^2 c} |M|^2[/tex] where M is the matrix element

[tex]\frac{d \sigma}{d \Omega} = \left(\frac{h c}{8 \pi} \right)^2 \frac{S |M|^2}{(E_1 + E_2)^2} \frac{|p_i|}{|p_f|}[/tex]

The Attempt at a Solution



So I know that at [tex]\sqrt{s}=5[/tex] GeV, the propogator has to be a [tex]\gamma[/tex] and the ratio of rates should favor the muon production as opposed to the hadrons, but I don't know how to calculate the rates. Likewise, I also know the propogator for [tex]e^+ e^- \rightarrow q \bar{q}[/tex] has to be a [tex]Z^0[/tex].Is the [tex]e^+ e^- \rightarrow \mu^+ \mu^-[/tex] considered a two-body scattering or what?

Basically, I'm torn as to calculating the rate. Do I need to explicitly find the matrix element for each case, or does that divide out?
 
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The electromagnetic couplings at the vertices are basically the same and the photon propagator in the middle has the same form, so the matrix elements are very similar. ([tex] e^+ e^- \rightarrow q \bar{q}[/tex] is dominated by intermediate photons, especially at c.o.m. energies far below the Z mass.) When you square the matrix element, you have to do sums over the incoming and outgoing spin states, but even this part is similar for the two cases. So the rates have the same form when expressed in terms of momenta and masses for muons or hadrons in the final state. You should check to make sure. The difference in rates is related to the large difference in masses between the muon and available hadronic states.
 

FAQ: What Is the Ratio of Rates for e+ e- -> mu+ mu- at sqrt(s)=5 GeV?

What is the ratio of rates for e+ e- -> mu+ mu-?

The ratio of rates for e+ e- -> mu+ mu- is a measure of the probability that an electron-positron pair (e+ e-) will produce a muon-antimuon pair (mu+ mu-) in a particle collision. It is typically denoted as R or Rμμ and is calculated by taking the number of muon-antimuon pairs produced divided by the number of electron-positron pairs produced.

How is the ratio of rates measured in experiments?

In particle physics experiments, the ratio of rates is typically measured by analyzing data from particle collisions and counting the number of electron-positron pairs and muon-antimuon pairs produced. This data is then used to calculate the ratio of rates.

What does the ratio of rates tell us about the particles involved in the collision?

The ratio of rates can give us information about the properties of the particles involved in the collision, such as their masses, interaction strengths, and decay probabilities. It can also provide insights into the underlying physical processes and interactions at play.

How does the ratio of rates for e+ e- -> mu+ mu- compare to other particle collision processes?

The ratio of rates for e+ e- -> mu+ mu- is just one of many possible particle collision processes that can occur. It is often compared to other collision processes, such as e+ e- -> hadrons or e+ e- -> tau+ tau-, to better understand the behavior and interactions of different particles.

Can the ratio of rates be used to test theories in particle physics?

Yes, the ratio of rates can be used to test various theories and models in particle physics. By comparing the measured ratio to theoretical predictions, scientists can validate or rule out different hypotheses and gain a deeper understanding of the fundamental building blocks of our universe.

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