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Chapter 44: Problem 56

Consider a collision in which a stationary particle with mass \(M\) is bombarded by a particle with mass \(m,\) speed \(t_{0}\). and total energy (including rest cnergy) \(E_{m}\). (a) Use the Lorentz transformation to write the velocities \(v_{m}\) and \(v_{M}\) of particles \(m\) and \(M\) in terms of the speed \(v_{\mathrm{cm}}\) of the center of momentum. (b) Use the fact that the total momentum in the center-of-momentum frame is zero to obtain an expression for \(v_{\mathrm{cm}}\) in terms of \(m, M,\) and \(v_{0}\). (c) Combine the results of parts (a) and (b) to obtain Eq. (44.9) for the total energy in the center-of- momentum frame.

Short Answer

Expert verified
The total energy in the center-of-momentum frame, as expressed by Eq. (44.9), is \(E_{cm} = \frac{(M+m)c^2}{\sqrt{1 - \left(\frac{mv_0}{(M+m)c}\right)^2}}\)

Step by step solution

01

- Lorentz transformation and velocities

According to the Lorentz transformation, the velocities of the particles in the center of momentum are given by \( v_m = \frac{v_0 + v_{cm}}{1 + \frac{v_0 v_{cm}}{C^2}}\) and \( v_M = \frac{v_{cm}}{1 + \frac{Mv_0}{mc^2}}\) in terms of the speed \(v_{cm}\) of the center of momentum, where \(v_0\) is the bombardment speed and \(C\) is the speed of light.
02

- Total momentum and center of momentum velocity

In the center of momentum frame, the total momentum is zero. Therefore, we have \(mcv_m + Mv_M = 0\), which gives the relationship \(v_{cm} = -\frac{mv_0}{M + m}\).
03

- Total energy in the center-of-momentum frame

Combining the results of steps 1 and 2, we can express the total energy in terms of the center of momentum velocity \(v_{cm}\) as \(E_{cm} = mc^2 \left(\frac{1}{\sqrt{1 - \frac{v_m^2}{c^2}}}\right) + Mc^2 \left(\frac{1}{\sqrt{1 - \frac{v_M^2}{c^2}}}\right) = mc^2 \left(\frac{1}{\sqrt{1 - \frac{v^2_{cm}}{c^2}}}\right) + Mc^2 \left(\frac{1}{\sqrt{1 - \frac{v^2_{cm}}{c^2}}}\right)\).
04

- Simplifying the total energy equation

By simplifying the equation, we obtain the total energy in the center of momentum frame as \(E_{cm} = \frac{(M+m)c^2}{\sqrt{1 - \left(\frac{mv_0}{(M+m)c}\right)^2}}\).

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Key Concepts

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

Center of Momentum Frame
In physics, the concept of the Center of Momentum Frame is crucial when analyzing collisions or interactions between particles. This frame is a special reference frame where the total momentum of the system is zero. When dealing with relativistic particles, which must comply with the principles of Einstein's theory of special relativity, working in this frame can drastically simplify the calculations. For example, in a collision between two particles, one stationary and the other moving, we can use the Lorentz transformation to determine the velocities of the particles within this frame. This transformation helps us translate the problem into a context where the combined momentum of both particles is precisely zero.
In the given exercise, determining the speed of the Center of Momentum (\(v_{cm}\)) involves using the principle that the total momentum in this frame is zero. This principle leads to the equation: \(v_{cm} = -\frac{mv_0}{M + m}\). With this, we can then accurately measure each particle's velocity relative to the center of momentum.
Relativistic Velocities
Understanding relativistic velocities is vital for studying high-speed particles鈥攖hose moving near the speed of light. The concept considers how different observers measure speeds relative to their reference frames. According to special relativity, velocities don't simply add up as they would in classical mechanics. Instead, the Lorentz transformation is used, offering an equation that gives the correct velocities when relativistic effects are significant.
In the exercise, the velocity of the bombarding particle (\(v_m\)) within the center of momentum frame is calculated using Lorentz transformation:
  • \(v_m = \frac{v_0 + v_{cm}}{1 + \frac{v_0 v_{cm}}{C^2}}\)
  • \(v_M = \frac{v_{cm}}{1 + \frac{Mv_0}{mc^2}}\)
Here, \(v_0\) is the particle's initial velocity, and \(C\) is the speed of light. These equations reveal how velocity measurements depend on the observer's reference frame, especially as speeds approach the speed of light. By accurately using these formulas, students can determine how the time dilation and length contraction affect observed velocities in different inertial frames.
Conservation of Momentum
The conservation of momentum is a key principle in physics, stating that the total momentum of an isolated system remains constant if there is no net external force acting on it. This is true regardless of whether the particles in the system are moving at classical speeds or relativistic velocities.
In the exercise, this principle is used in the center of momentum frame. The law assures that all momentum changes within the system result from internal forces alone. In mathematical terms, this implies:
  • $mcv_m + Mv_M = 0$
It's this conservation that simplifies the complex dynamics of a collision into a manageable problem. By recognizing that the momentum is conserved, students can confidently work through more sophisticated relativistic scenarios. When combined with the Lorentz transformation, conservation of momentum ensures that the analysis remains consistent with the principles of relativity. This helps derive expressions for energy and momentum that align with both Newtonian and Einsteinian views.

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Most popular questions from this chapter

About 10,000 cosmic-ray protons, each with hundreds of MeV of energy, impinge on each square meter of our upper atmosphere each second. They collide with atmospheric nitrogen and oxygen to produce secondary showers of newly created particles, including many muons. Muons have a mass of \(105.7 \mathrm{MeV} / c^{2}\) and an average lifetime of \(2.197 \mu \mathrm{s}\). Consider a secondary cosmic muon produced at an altitude of \(15.00 \mathrm{~km}\) aimed directly downward with an energy of \(6.000 \mathrm{GeV}\). With such high energy, the muon can travel a great distance into the earth without slowing down. (a) Determine the speed of this muon. (b) How far would this muon travel in one lifetime if there were no relativistic effects? (c) In the frame of the muon, what distance separates its creation position from the earth's surface? (d) If one lifetime passes in the frame of the muon, how much time passes in the frame of the earth? (e) How far does a muon travel in this time as seen from the earth? (f) Does the muon survive its trip to the surface? How far will it penetrate the earth in its lifetime?

In which of the following decays are the three lepton numbers conserved? In each case, explain your reasoning. (a) \(\mu^{-} \rightarrow \mathrm{e}^{-}+v_{e}+\bar{v}_{\mu}\) (b) \(\tau^{-} \rightarrow \mathrm{e}^{-}+\bar{\nu}_{e}+\nu_{\tau}\) (c) \(\pi^{+} \rightarrow \mathrm{e}^{+}+\gamma_{i}\) (d) \(\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\bar{\nu}_{e}\)

Determine the electric charge, baryon number, strangeness quantum number, and charm quantum number for the following quark combinations: (a) \(u d s ;(b) c \bar{u} ;(c) d d d ;\) and \((d) d \bar{c} .\) Explain your reasoning.

Which of the following reactions obey the conservation of baryon number? (a) \(p+p \rightarrow p+e^{+}\) (b) \(p+n \rightarrow 2 e^{+}+e^{-}\) (c) \(\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{-}+\bar{\nu}_{\mathrm{e}} ;\) (d) \(\mathrm{p}+\overline{\mathrm{p}} \rightarrow 2 \gamma\)

One proposed proton decay is \(\mathrm{p}^{+} \rightarrow \mathrm{e}^{+}+\pi^{0}\), which violates both baryon and lepton number conservation, so the proton lifetime is expected to be very long. Suppose the proton half-life were \(1.0 \times 10^{18} \mathrm{y} .\) (a) Calculate the energy deposited per kilogram of body tissue (in rad) due to the decay of the protons in your body in one year. Model your body as consisting entirely of water. Only the two protons in the hydrogen atoms in each \(\mathrm{H}_{2} \mathrm{O}\) molecule would decay in the manner shown; do you see why? Assume that the \(\pi^{0}\) decays to two \(\gamma\) rays, that the positron annihilates with an electron, and that all the encrgy produced in the primary decay and these secondary decays remains in your body. (b) Calculate the equivalent dose (in rem) assuming an RBE of 1.0 for all the radiation products, and compare with the 0.1 rem due to the natural background and the 5.0 rem guideline for industrial workers. Based on your calculation, can the proton lifetime be as short as \(1.0 \times 10^{18} \mathrm{y} ?\)
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