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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 mad physicist claims to have observed the decay of a particle of mass \(M\) into two identical particles of mass \(m,\) with \(M<2 m .\) In response to the objections that this violates conservation of energy, he replies that if \(M\) was traveling fast enough it could easily have energy greater than \(2 m c^{2}\) and hence could decay into the two particles of mass \(m\). Show that he is wrong. [He has forgotten that both energy and momentum are conserved. You can analyse this problem in terms of these two conservation laws, but it is much simpler to go to the rest frame of \(M .]\)

A traveler in a rocket of proper length 2d sets up a coordinate system \(\mathcal{S}^{\prime}\) with its origin \(O^{\prime}\) anchored at the exact middle of the rocket and the \(x^{\prime}\) axis along the rocket's length. At \(t^{\prime}=0\) she ignites a flashbulb at \(O^{\prime} .\) (a) Write down the coordinates \(x_{\mathrm{F}}^{\prime}, t_{\mathrm{F}}^{\prime}\) and \(x_{\mathrm{B}}^{\prime}, t_{\mathrm{B}}^{\prime}\) for the arrival of the light at the front and back of the rocket. (b) Now consider the same experiment as observed from a frame \(\delta\) relative to which the rocket is traveling with speed \(V\) (with \(\delta\) and \(S\) ' in the standard configuration). Use the inverse Lorentz transformation to find the coordinates \(x_{\mathrm{F}}, t_{\mathrm{F}}\) and \(x_{\mathrm{B}}, t_{\mathrm{B}}\) for the arrival of the two signals. Explain clearly in words why the two arrivals are simultaneous in \(\mathcal{S}^{\prime}\) but not in \(\mathcal{S} .\) This phenomenon is called the relativity of simultaneity.

(a) A meter stick is at rest in frame \(\mathcal{S}_{\mathrm{o}}\), which is traveling with speed \(V=0.8 c\) in the standard configuration relative to frame \(\mathcal{S}\). (a) The stick lies in the \(x_{\mathrm{o}} y_{\mathrm{o}}\) plane and makes an angle \(\theta_{\mathrm{o}}=60^{\circ}\) with the \(x_{\mathrm{o}}\) axis (as measured in \(\mathcal{S}_{\mathrm{o}}\) ). What is its length \(l\) as measured in \(\mathcal{S}\), and what is its angle \(\theta\) with the \(x\) axis? [Hint: It may help to think of the stick as the hypotenuse of a \(30-60-90\) triangle of plywood.] (b) What is \(l\) if \(\theta=60^{\circ} ?\) What is \(\theta_{\mathrm{o}}\) in this case?

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 \(c \text { is always less than } c . \text { See Problem } 15.43 .]\)

A particle of mass 12 MeV/c \(^{2}\) has a kinetic energy of 1 \(\mathrm{MeV}\). What are its momentum (in MeV/c) and its speed (in units of \(c\) )?

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