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Chapter 15: Problem 61

A particle of mass 3 MeV/c \(^{2}\) has momentum 4 MeV/c. What are its energy (in MeV) and speed (in units of \(c\) )?

Short Answer

Expert verified
The energy is 5 MeV and the speed is 0.8c.

Step by step solution

01

Define Relativistic Energy-Momentum Relationship

The relativistic energy-momentum relationship is given by \[E^2 = (pc)^2 + (m_0c^2)^2\]where:- \(E\) is the total energy- \(p\) is the momentum- \(m_0\) is the rest mass of the particle- \(c\) is the speed of light.
02

Substitute Known Values into the Formula

Given:- mass \(m_0 = 3 \, \text{MeV}/c^2\)- momentum \(p = 4 \, \text{MeV}/c\)Substituting into the formula:\[(E)^2 = (4 \, \text{MeV})^2 + (3 \, \text{MeV})^2\].
03

Calculate the Energy

Calculate \(E^2\):\[(E)^2 = 16 \, \text{MeV}^2 + 9 \, \text{MeV}^2 = 25 \, \text{MeV}^2\].Taking the square root gives:\[E = \sqrt{25 \, \text{MeV}^2} = 5 \, \text{MeV}\].
04

Use the Energy-Momentum Relationship for Speed

The speed \(v\) of the particle is given by the relationship \[v = \frac{pc^2}{E}\].
05

Substitute Values to Solve for Speed

Substitute the known values into the speed formula:\[v = \frac{(4 \, \text{MeV}/c) \cdot c^2}{5 \, \text{MeV}}\].Simplify to get:\[v = \frac{4}{5}c = 0.8c\].

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Key Concepts

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

Energy-Momentum Relationship
The energy-momentum relationship is a fundamental principle in relativistic mechanics, allowing us to understand the link between a particle's energy, momentum, and rest mass. It is expressed mathematically as:\[E^2 = (pc)^2 + (m_0c^2)^2\]Here, \(E\) represents the total energy of the particle, \(p\) is its momentum, \(m_0\) is the rest mass, and \(c\) is the speed of light (approximately \(3 \times 10^8 m/s\)). This formula is crucial as it helps predict and calculate the total energy of particles moving at relativistic speeds—speeds close to the speed of light, where classical physics no longer applies.
  • \((pc)^2\) is the energy due to momentum alone.
  • \((m_0c^2)^2\) is the contribution due to the rest mass energy.
When particles are accelerated to high speeds, their momentum and energy change according to the principles of relativity, and the energy-momentum relationship becomes much more significant than at lower speeds. This equation reminds us that energy and momentum are intertwined in the relativistic realm, emphasizing that energy is not merely an isolated property but relates intimately with how fast and how massive an object is.
Relativistic Energy
Relativistic energy takes into account the total energy possessed by an object moving at speeds that are substantial fractions of the speed of light. Conventional Newtonian physics, which considers energy solely based on mass and velocity, fails at such high velocities. Here, relativistic effects become significant.The rest energy, given by \(E_0 = m_0c^2\), represents the inherent energy an object possesses due solely to its mass. When a particle moves at relativistic speeds, its total energy is a combination of this rest energy and additional energy from its momentum. This additional energy is a non-negligible part of the total energy, contrary to classical scenarios.In our example, where a particle has a rest mass of 3 MeV/c\(^2\) and momentum of 4 MeV/c, substitution yields \[(E)^2 = (4 \, \text{MeV})^2 + (3 \, \text{MeV})^2 = 25 \, \text{MeV}^2\]Solving this provides a total energy of 5 MeV, demonstrating the combination of kinetic and rest mass energies.
Particle Speed Calculation
Determining the speed of a particle in relativistic mechanics involves not just simple velocity calculations, but understanding its energy and momentum relationships. The speed \(v\) can be computed using the equation:\[v = \frac{pc^2}{E}\]In the example exercise, with a momentum of 4 MeV/c and energy of 5 MeV, substituting these values gives:\[v = \frac{(4 \, \text{MeV}/c) \cdot c^2}{5 \, \text{MeV}}\]This simplifies to:\[v = \frac{4}{5}c = 0.8c\]Because the speed of light \(c\) is the ultimate speed limit, relativistic speed computations often yield results as fractions of \(c\). Here, the calculated speed of the particle is 0.8c, highlighting significant relativistic effects, as the speed is a substantial fraction of the speed of light. Such considerations are essential for accurately determining how particles behave at high velocities, affording insight into the dynamics of fast-moving particles often seen in high-energy physics and particle accelerators.

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Most popular questions from this chapter

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 space explorer \(A\) sets off at a steady \(0.95 c\) to a distant star. After exploring the star for a short time, he returns at the same speed and gets home after a total absence of 80 years (as measured by earth-bound observers). How long do \(A\) 's clocks say that he was gone, and by how much has he aged as compared to his twin \(B\) who stayed behind on earth? [Note: This is the famous "twin paradox." It is fairly easy to get the right answer by judicious insertion of a factor of \(\gamma\) in the right place, but to understand it, you need to recognize that it involves three inertial frames: the earth-bound frame \(\mathcal{S}\), the frame \(\mathcal{S}^{\prime}\) of the outbound rocket, and the frame \(\mathcal{S}^{\prime \prime}\) of the returning rocket. Write down the time dilation formula for the two halves of the journey and then add. Notice that the experiment is not symmetrical between the two twins: \(A\) stays at rest in the single inertial frame \(\delta\), but \(B\) occupies at least two different frames. This is what allows the result to be unsymmetrical.]

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.

Verify directly that \(x^{\prime} \cdot y^{\prime}=x \cdot y\) for any two four- vectors \(x\) and \(y,\) where \(x^{\prime}\) and \(y^{\prime}\) are related to \(x\) and \(y\) by the standard Lorentz boost along the \(x_{1}\) axis.

A useful form of the quotient rule for three-dimensional vectors is this: Suppose that a and b are known to be three-vectors and suppose that for every orthogonal set of axes there is a \(3 \times 3\) matrix T with the property that \(\mathbf{b}=\mathbf{T}\) a for every choice of \(\mathbf{a},\) then \(\mathbf{T}\) is a tensor. (a) Prove this. (b) State and prove the corresponding rule for four-vectors and four-tensors.
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