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Chapter 2: Problem 1

When expressed in natural units the lifetime of the \(\mathrm{W}\) boson is approximately \(\tau \approx 0.5 \mathrm{GeV}^{-1} .\) What is the corresponding value in S.I. units?

Short Answer

Expert verified
The lifetime \(\tau\) is approximately \(3.29 \times 10^{-25} \, \mathrm{s}\).

Step by step solution

01

- Understand Natural Units

In natural units, certain physical constants such as the speed of light ( c) and Planck's constant (鈩�) are set to 1. Here, the lifetime \(蟿\) is given as \(0.5 \, \mathrm{GeV}^{-1}\).
02

- Convert to Standard Units

Convert \(蟿\) from \(\mathrm{GeV}^{-1}\) to seconds. Use the relation \(1 \, \mathrm{GeV} = 1.60218 \, \times 10^{-10} \, \mathrm{J}\) and \(h = 6.626 \, \times 10^{-34} \, \mathrm{Js}, c = 3 \, \times 10^8 \, \mathrm{m/s}\).
03

- Apply Conversion Factor

Remember \(\mathrm{J} = \frac{\mathrm{kg} \cdot \mathrm{m}^2}{\mathrm{s}^2}\), to convert 1 GeV to J: \(1\, \mathrm{GeV} = 1.60218 \, \times 10^{-10} \, \mathrm{J}\). To convert \(\mathrm{GeV}^{-1}\) to \(\mathrm{s}\): \(\tau = 0.5\, \mathrm{GeV}^{-1} = 0.5 \times\frac{6.582119 \times 10^{-25}\, \mathrm{s}}{1 \, \mathrm{GeV}} \).
04

- Calculate the Lifetime

Calculate, \(\tau \approx 0.5 \times 6.582119 \times 10^{-25} \, \mathrm{s} \). This yields: \(\tau \approx 3.29106 \times 10^{-25} \, \mathrm{s}\).

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Key Concepts

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

natural units
In particle physics, natural units simplify many equations by setting fundamental constants to 1. For example, the speed of light \(c\) and Planck's constant \((\hbar)\) are both set to 1. This means that time and distance can be expressed without needing to account for these constants directly. Natural units make calculations more straightforward because you avoid carrying around these constants within equations. It's helpful to know these units when dealing with high-energy phenomena, making complex interactions easier to analyze. Understanding natural units allows you to move seamlessly between energy, length, and time scales in several domains of physics.
W boson lifetime
The W boson is a fundamental particle with a very short lifetime. In natural units, its lifetime is approximately \(\tau \approx 0.5 \, \mathrm{GeV}^{-1}\). The W boson mediates weak nuclear force, one of the four fundamental forces of nature. Lifetime refers to the average time before a particle decays into other particles. Understanding these lifetimes helps scientists predict how particles behave in collisions and decays, which is crucial for experiments in large particle accelerators like the LHC. Converting this lifetime into S.I. units 鈥� as shown in the original solution 鈥� gives a more intuitive grasp of how fleeting such particles are, typically in the order of \(10^{-25} \mathrm{s}\) for the W boson.
unit conversion in physics
Unit conversion is vital for comparing results from different experiments and ensuring consistency across scientific studies. To convert the lifetime of the W boson from \(\mathrm{GeV}^{-1}\) to seconds, follow some key steps:
  • Understand the equivalence of energy units: \(1 \, \mathrm{GeV} = 1.60218 \times 10^{-10} \, \mathrm{J}\).
  • Use Planck's constant: For conversion, the fact that \(h = 6.626 \times 10^{-34} \, \mathrm{Js}\) and knowing that \(c = 3 \, \times \ 10^8 \, \mathrm{m/s}\) are placeholders.
These conversions allow physicists to translate measurements in particle physics into more familiar terms. This step ensures comprehension and maintains continuity in research. Properly converting units ensures both accuracy and precision in scientific communication.
Planck's constant
Planck's constant (denoted as h) is a fundamental quantity in quantum mechanics. It describes the quantization of energy levels. The constant has a value: \(h = 6.626 \times 10^{-34} \mathrm{Js}\). In many calculations, its reduced form \((\hbar = h/2\pi)\) is used and is often set to 1 in natural units. Planck鈥檚 constant plays a critical role in defining the scale at which quantum effects become significant. It bridges the concepts of energy and frequency (\

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Most popular questions from this chapter

For the process \(1+2 \rightarrow 3+4\), the Mandelstam variables \(s, t\) and \(u\) are defined as \(s=\left(p_{1}+p_{2}\right)^{2}\), \(t=\left(p_{1}-p_{3}\right)^{2}\) and \(u=\left(p_{1}-p_{4}\right)^{2}\). Show that $$ s+u+t=m_{1}^{2}+m_{2}^{2}+m_{3}^{2}+m_{4}^{2} $$

A particle of mass \(3 \mathrm{GeV}\) is travelling in the positive \(z\)-direction with momentum \(4 \mathrm{GeV}\); what are its energy and velocity?

At the HERA collider, \(27.5 \mathrm{GeV}\) electrons were collided head-on with \(820 \mathrm{GeV}\) protons. Calculate the centre-ofmass energy.

In a collider experiment, \(\Lambda\) baryons can be identified from the decay \(\Lambda \rightarrow \pi^{-} p\), which gives rise to a displaced vertex in a tracking detector. In a particular decay, the momenta of the \(\pi^{+}\)and \(p\) are measured to be \(0.75 \mathrm{GeV}\) and \(4.25 \mathrm{GeV}\) respectively, and the opening angle between the tracks is \(9^{\circ}\). The masses of the pion and proton are \(139.6 \mathrm{MeV}\) and \(938.3 \mathrm{MeV}\). (a) Calculate the mass of the \(\Lambda\) baryon. (b) On average, \(\Lambda\) baryons of this energy are observed to decay at a distance of \(0.35 \mathrm{~m}\) from the point of production. Calculate the lifetime of the \(\Lambda\).

In the laboratory frame, denoted \(\Sigma\), a particle travelling in the z-direction has momentum \(\mathbf{p}=p_{z} \hat{\mathbf{z}}\) and energy \(E\). (a) Use the Lorentz transformation to find expressions for the momentum \(p_{z}^{\prime}\) and energy \(E^{\prime}\) of the particle in a frame \(\Sigma^{\prime}\), which is moving in a velocity \(\mathbf{v}=+v \hat{z}\) relative to \(\Sigma\), and show that \(E^{2}-p_{2}^{2}=\left(E^{\prime}\right)^{2}-\left(p_{2}^{\prime}\right)^{2}\). (b) For a system of particles, prove that the total four-momentum squared, $$ p^{\mu} p_{\mu} \equiv\left(\sum_{i} E_{i}\right)^{2}-\left(\sum_{i} \mathbf{p}_{i}\right)^{2} $$ is invariant under Lorentz transformations.
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