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Chapter 3: Problem 4

At a future \(\mathrm{e}^{+} \mathrm{e}^{-}\)linear collider operating as a Higgs factory at a centre-of-mass energy of \(\sqrt{5}=250 \mathrm{GeV}\), the cross section for the process \(\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \mathrm{HZ}\) is \(250 \mathrm{fb}\). If the collider has an instantaneous luminosity of \(2 \times 10^{34} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}\) and is operational for \(50 \%\) of the time, how many Higgs bosons will be produced in five years of running? Note: 1 femtobarn \(\equiv 10^{-15} \mathrm{~b}\).

Short Answer

Expert verified
3.942 million Higgs bosons will be produced in 5 years.

Step by step solution

01

Convert Luminosity to Annual Luminosity

First, we will find the annual luminosity. Given the collider has an instantaneous luminosity of \[ L = 2 \times 10^{34} \, \text{cm}^{-2} \text{s}^{-1} \]. Using the operational time of 50%, we get the effective luminosity per year as follows: The total seconds in a year: \[ T = 365 \times 24 \times 60 \times 60 \, \text{s} = 3.1536 \times 10^7 \, \text{s} \]The effective operational time per year: \[T_{\text{eff}} = 0.5 \times 3.1536 \times 10^7 \,\text{s} \approx 1.5768 \times 10^7 \, \text{s} \]Now, multiply the instantaneous luminosity by the effective operational time: \[ L_{\text{annual}} = L \times T_{\text{eff}} = 2 \times 10^{34} \times 1.5768 \times 10^7 \, \text{cm}^{-2} \approx 3.1536 \times 10^{41} \, \text{cm}^{-2} \text{year}^{-1} \]
02

Convert Cross Section to the Correct Units

Next, we need to convert the cross-section from femtobarns to cm². Given the cross-section is \[ \sigma = 250 \, \text{fb} \], we use the conversion 1 femtobarn = \(1 \times 10^{-15} \, \text{barn} \) and 1 barn = \(10^{-24} \text{cm}^2\): \[ \sigma = 250 \times 10^{-15} \times 10^{-24} \text{cm}^2 = 2.5 \times 10^{-41} \text{cm}^2 \]
03

Calculate the Number of Higgs Bosons Produced in One Year

Now we can calculate the number of Higgs bosons produced in one year using the formula: \[ N \text{(Higgs/year)} = L_{\text{annual}} \times \sigma \]Substitute the values from Steps 1 and 2:\[ N \text{(Higgs/year)} = 3.1536 \times 10^{41} \times 2.5 \times 10^{-41} = 788400000 \approx 7.884 \times 10^5 \]
04

Calculate the Number of Higgs Bosons Produced in Five Years

Finally, multiply the number of Higgs bosons produced in one year by the number of years of operation: \[ N \text{(Higgs/5 years)} = 7.884 \times 10^5 \times 5 = 3.942 \times 10^6 \]
05

Conclusion

Hence, the total number of Higgs bosons produced in five years is approximately 3.942 million.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

linear collider
A linear collider is a type of particle accelerator where particles are accelerated in a straight line. Unlike circular accelerators, linear colliders minimize energy loss due to synchrotron radiation, making them suitable for colliding particles at very high energies.

The \(\mathrm{e}^{+} \mathrm{e}^{-}\) linear collider mentioned in the exercise focuses on electron-positron collisions. These types of collisions are particularly useful for high precision measurements, because electrons and positrons are fundamental particles with no internal structure.
  • Efficient for high energy collisions
  • Useful for precise measurements
  • Often used to study particles like the Higgs boson
cross-section
In particle physics, the cross-section is a measure of the probability of a particular collision happening. It is typically denoted by the Greek letter sigma (\(\sigma\)). The unit of measure for cross-sections is the barn, where 1 barn = \(10^{-24} \text{cm}^2\).

The cross-section for producing Higgs bosons in this exercise is given in femtobarns (\(\mathrm{fb}\)), where 1 femtobarn (\(\mathrm{fb}\)) = \(10^{-15} \text{barn}\).
  • Probability measure
  • Depends on the type and energy of the collision
  • In our case: \(\sigma = 250 \mathrm{fb}\)
  • 1 femtobarn = \(10^{-39} \text{cm}^{2}\)
instantaneous luminosity
Instantaneous luminosity refers to how many particles collide in a specific area over a given time. It’s typically measured in \(\text{cm}^{-2} \text{s}^{-1}\). Higher luminosity means more collisions, which is critical for experiments aiming to discover rare particles like the Higgs boson.

In this exercise, the instantaneous luminosity is provided as \( L = 2 \times 10^{34} \, \text{cm}^{-2} \text{s}^{-1}\).
  • Indicates collision rate
  • Higher values increase chances of rare events
  • Given as: \( 2 \times 10^{34} \, \text{cm}^{-2} \text{s}^{-1}\)
operational time
Operational time is the fraction of total time that the collider is actually running and available for collisions. In practice, colliders do not run 24/7 due to maintenance, upgrades, and other contingencies. In this exercise, the collider is functional only 50% of the time.

We'll convert this into effective operational time per year:
  • 1 year = \(3.1536 \times 10^7 \, \text{s}\)
  • \(50\%\) effective time = \(1.5768 \times 10^7 \, \text{s}/ \text{year}\)
femtobarn conversion
Converting units is essential in such calculations. The cross-section given is in femtobarns, so we need to convert it to \(\text{cm}^2\).

Recall:
  • 1 barn = \(10^{-24} \text{cm}^2\)
  • 1 femtobarn (fb) = \(10^{-15} \text{barn}\)
  • So \(250 \text{fb} = 250 \times 10^{-15} \times 10^{-24} = 2.5 \times 10^{-41} \text{cm}^2\)
This conversion ensures that all units in the formulas are compatible.

  • 1 fb = \(10^{-39} \, \text{cm}^2\)
  • Given \(\sigma = 250 \, \text{fb} = 2.5 \times 10^{-41} \, \text{cm}^2\)

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Most popular questions from this chapter

Calculate the branching ratio for the decay \(\mathrm{K}^{+} \rightarrow \pi^{+} \pi^{0}\), given the partial decay width \(\Gamma\left(\mathrm{K}^{+} \rightarrow \pi^{+} \pi^{0}\right)=\) \(1.2 \times 10^{-8} \mathrm{eV}\) and the mean \(\mathrm{ka}\) on lifetime \(\tau\left(\mathrm{K}^{+}\right)=1.2 \times 10^{-8} \mathrm{~s}\).

The total \(\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \gamma \rightarrow \mu^{+} \mu^{-}\)annihilation cross section is \(\sigma=4 \pi \alpha^{2} / 3 s\), where \(\alpha \approx 1 / 137\). Calculate the cross section at \(\sqrt{s}=50 \mathrm{GeV}\), expressing your answer in both natural units and in barns \(\left(1\right.\) barn \(\left.=10^{-28} \mathrm{~m}^{2}\right)\). Compare this to the total pp cross section at \(\sqrt{5}=50 \mathrm{GeV}\) which is approximately \(40 \mathrm{mb}\) and comment on the result.

For the process \(a+b \rightarrow 1+2\) the Lorentz-invariant flux term is $$ F=4\left[\left(p_{a} \cdot p_{b}\right)^{2}-m_{a}^{2} m_{b}^{2}\right]^{\frac{1}{2}} . $$ In the non-relativistic limit, \(\beta_{a} \ll 1\) and \(\beta_{b} \ll 1\), show that $$ F \approx 4 m_{a} m_{b}\left|\mathbf{v}_{a}-\mathbf{v}_{b}\right|, $$ where \(\mathbf{v}_{a}\) and \(\mathbf{v}_{b}\) are the (non-relativistic) velocities of the two particles.

The Lorentz-invariant flux term for the process \(a+b \rightarrow 1+2\) in the centre-of-mass frame was shown to be \(F=4 p_{j}^{*} \sqrt{5}\), where \(p_{j}^{*}\) is the momentum of the intial-state particles. Show that the corresponding expression in the frame where \(b\) is at rest is $$ F=4 m_{b} p_{e} $$
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