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A particle of unknown mass \(M\) decays into two particles of known masses \(m_{a}=0.5 \mathrm{GeV} / c^{2}\) and \(m_{b}=1.0 \mathrm{GeV} / c^{2},\) whose momenta are measured to be \(\mathbf{p}_{a}=2.0 \mathrm{GeV} / \mathrm{c}\) along the \(x_{2}\) axis and \(\mathbf{p}_{b}=1.5 \mathrm{GeV} / c\) along the \(x_{1}\) axis. \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV} .\right)\) Find the unknown mass \(M\) and its speed.

Short Answer

Expert verified
\(M \approx 2.94 \, \text{GeV}/c^2\), \(v \approx 0.660c\).

Step by step solution

01

Calculate Energy of Particle a

Use the energy-momentum relation for particle \(a\):\[ E_a = \sqrt{(m_a c^2)^2 + (p_a c)^2} \]Given \(m_a = 0.5 \, \text{GeV}/c^2\) and \(p_a = 2.0 \, \text{GeV}/c\), calculate \(E_a\):\[ E_a = \sqrt{(0.5)^2 + (2.0)^2} = \sqrt{0.25 + 4} = \sqrt{4.25} \, \text{GeV} \]\[ E_a \approx 2.06 \, \text{GeV} \]
02

Calculate Energy of Particle b

Use the energy-momentum relation for particle \(b\):\[ E_b = \sqrt{(m_b c^2)^2 + (p_b c)^2} \]Given \(m_b = 1.0 \, \text{GeV}/c^2\) and \(p_b = 1.5 \, \text{GeV}/c\), calculate \(E_b\):\[ E_b = \sqrt{(1.0)^2 + (1.5)^2} = \sqrt{1 + 2.25} = \sqrt{3.25} \, \text{GeV} \]\[ E_b \approx 1.80 \, \text{GeV} \]
03

Calculate Total Energy and Momentum of Initial Particle

The total energy \(E\) of the original particle is the sum of the energies of particles \(a\) and \(b\):\[ E = E_a + E_b = 2.06 + 1.80 = 3.86 \, \text{GeV} \]Calculate the total momentum vector \(\mathbf{p}\):\[ \mathbf{p} = (p_{b(x_1)}, p_{a(x_2)}) = (1.5 \, \text{GeV}/c, 2.0 \, \text{GeV}/c) \]Magnitude of \(\mathbf{p}\):\[ p = \sqrt{1.5^2 + 2.0^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5 \, \text{GeV}/c \]
04

Find Unknown Mass M of Initial Particle

Use the invariant mass equation:\[ M c^2 = \sqrt{E^2 - (pc)^2} \]Substitute values:\[ M = \sqrt{(3.86)^2 - (2.5)^2} = \sqrt{14.8996 - 6.25} = \sqrt{8.6496} \, \text{GeV}/c^2 \]\[ M \approx 2.94 \, \text{GeV}/c^2 \]
05

Calculate Speed of Initial Particle

Use the relation \( E = \gamma M c^2 \) and \( p = \gamma M v \):First, find \( \gamma \):\[ \gamma = \frac{E}{M c^2} = \frac{3.86}{2.94} \approx 1.31 \]Find \( v \) using \( p = \gamma M v \):\[ v = \frac{p}{\gamma M} = \frac{2.5}{1.31 \times 2.94} c \]\[ v \approx 0.660 c \]
06

Conclusion: Unknown Mass and Speed

The unknown mass \(M\) is approximately \(2.94 \, \text{GeV}/c^2\), and its speed is approximately \(0.660c\).

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Key Concepts

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

Energy-Momentum Relation
The energy-momentum relation is a crucial concept in relativistic mechanics. It links the energy of a particle with its momentum and mass, allowing us to derive quantities like energy when momentum and mass are known. Given as:
  • \( E = \sqrt{(m c^2)^2 + (pc)^2} \)
This equation stems from Einstein’s theory of relativity. It shows how energy is not merely dependent on mass but also momentum, especially significant at high speeds.
For our case with particle \( a \) and \( b \), using this relation helps us find their respective energies knowing their masses and momenta.
Knowing the energy of the final state particles is key as it helps find the initial state quantities such as total energy and momentum, later used to find the invariant mass.
Invariant Mass
Invariant mass, often denoted as \( M \), is a scalar quantity that remains the same, irrespective of the reference frame. It's sometimes called rest mass and provides a direct measure of a particle system's "weight".
The invariant mass can be determined using the total energy and momentum of the system through:
  • \( M c^2 = \sqrt{E^2 - (pc)^2} \)
Here, \( E \) is the total energy, and \( p \) is the total momentum.
This formula tells us that even though the energy and momentum can vary based on the observer's frame of reference, the invariant mass is constant.
In our particle decay example, knowing the decay products' energies and momenta helps compute the original particle's mass.
Particle Decay
Particle decay refers to a process where a particle transforms into other particles. It's a common phenomenon in high-energy physics and is governed by conservation laws.
In the decay process considered in the problem, a particle of unknown mass decays into two particles with known masses and measurable momenta.
Important points to remember about particle decay:
  • Conservation of Energy: Total energy before decay equals total energy after.
  • Conservation of Momentum: Total momentum before decay equals total momentum after.
By applying these conservation laws, you can find the energy and momentum of the decayed particles and use them to compute unknown quantities, such as the original particle's invariant mass and its speed, using relativistic equations.
Relativistic Speed
Relativistic speed occurs when a particle moves at speeds close to the speed of light, \( c \). At these speeds, traditional velocity calculations using Newtonian mechanics break down, as relativistic effects become significant.
The speed of a particle can be found using the relation:
  • \( E = \gamma M c^2 \)
  • \( p = \gamma M v \)
Here, \( \gamma \) is the Lorentz factor given by \( \gamma = \frac{1}{\sqrt{1 - (v/c)^2}} \), reflecting how time and length distort at high velocities.
In our calculation, the unknown particle's speed approximates to 0.660c, indicating significant relativistic effects due to its high velocity.
Understanding relativistic speed ensures accurate analysis of a particle's behavior at speeds nearing that of light, crucial in high-energy physics experiments.

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Most popular questions from this chapter

(a) What is a particle's speed if its kinetic energy \(T\) is equal to its rest energy? ( \(\mathbf{b}\) ) What if its energy \(E\) is equal to \(n\) times its rest energy?

One way to create exotic heavy particles is to arrange a collision between two lighter particles $$a+b \rightarrow d+e+\cdots+g$$ where \(d\) is the heavy particle of interest and \(e, \cdots, g\) are other possible particles produced in the reaction. (A good example of such a process is the production of the \(\psi\) particle in the process \(\left.e^{+}+e^{-} \rightarrow \psi, \text { in which there are no other particles } e, \cdots, g .\right)\) (a) Assuming that \(m_{d}\) is much heavier that any of the other particles, show that the minimum (or threshold) energy to produce this reaction in the CM frame is \(E_{\mathrm{cm}} \approx m_{d} c^{2} .\) (b) Show that the threshold energy to produce the same reaction in the lab frame, where the particle \(b\) is initially at rest, is \(E_{\mathrm{lab}} \approx m_{d}^{2} c^{2} / 2 m_{b} .\) (c) Calculate these two energies for the process \(e^{+}+e^{-} \rightarrow \psi,\) with \(m_{e} \approx 0.5 \mathrm{MeV} / c^{2}\) and \(m_{\psi} \approx 3100 \mathrm{MeV} / c^{2} .\) Your answers should explain why particle physicists go to the trouble and expense of building colliding-beam experiments.

A particle \(a\) traveling along the positive \(x\) axis of frame \(\mathcal{S}\) with speed 0.5c decays into two identical particles, \(a \rightarrow b+b,\) both of which continue to travel on the \(x\) axis. (a) Given that \(m_{a}=2.5 m_{b},\) find the speed of either \(b\) particle in the rest frame of particle \(a .\) (b) By making the necessary transformation on the result of part (a), find the velocities of the two \(b\) particles in the original frame S.

As seen in frame \(\mathcal{S}\), two rockets are approaching one another along the \(x\) axis traveling with equal and opposite velocities of \(0.9 c .\) What is the velocity of the rocket on the right as measured by observers in the one on the left? [This and the previous two problems illustrate the general result that in relativity the "sum" of two velocities that are less than \(c \text { is always less than } c . \text { See Problem } 15.43 .]\)

When a radioactive nucleus of astatine 215 decays at rest, the whole atom is torn into two in the reaction $$^{215} \mathrm{At} \rightarrow^{211} \mathrm{Bi}+^{4} \mathrm{He}$$ The masses of the three atoms are (in order) \(214.9986,210.9873,\) and \(4.0026,\) all in atomic mass units. (1 atomic mass unit \(=1.66 \times 10^{-27} \mathrm{kg}=931.5 \mathrm{MeV} / c^{2}\).) What is the total kinetic energy of the two out coming atoms, in joules and in MeV?

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