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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.

Short Answer

Expert verified
The velocities of the two \(b\) particles in frame S are \(0.846c\) and \(-0.143c\).

Step by step solution

01

Analyze the Decay Process in Rest Frame of Particle a

In the rest frame of particle \(a\), the conservation of momentum must hold. Since particle \(a\) is at rest, the two \(b\) particles must have equal and opposite velocities to conserve momentum.Let the velocity of one \(b\) particle be \(v'_b\) in the rest frame of \(a\). Since the particles are identical and the total momentum is zero, \(v'_b = -v'_b\). We need to solve for \(v'_b\) using conservation of energy.
02

Apply Conservation of Energy in Rest Frame of Particle a

In the rest frame of particle \(a\), energy is conserved:\[ m_a c^2 = 2 \times \gamma_b m_b c^2 \]where \(\gamma_b\) is the Lorentz factor for particle \(b\):\[ \gamma_b = \frac{1}{\sqrt{1 - \left(\frac{v'_b}{c}\right)^2}} \]Given that \(m_a = 2.5 m_b\), we can substitute and solve:\[ 2.5 m_b c^2 = 2 \times \gamma_b m_b c^2 \]\[ \frac{2.5}{2} = \gamma_b \]\[ \gamma_b = 1.25 \]
03

Calculate the Velocity of b Particles in Rest Frame of a

Use \(\gamma_b = 1.25\) to find \(v'_b\):\[ \gamma_b = \frac{1}{\sqrt{1 - \left(\frac{v'_b}{c}\right)^2}} \]Solve for \(v'_b\):\[ 1.25 = \frac{1}{\sqrt{1 - \left(\frac{v'_b}{c}\right)^2}} \]\[ \sqrt{1 - \left(\frac{v'_b}{c}\right)^2} = \frac{1}{1.25} \]\[ 1 - \left(\frac{v'_b}{c}\right)^2 = \left(\frac{4}{5}\right)^2 \]\[ \left(\frac{v'_b}{c}\right)^2 = 1 - \frac{16}{25} \]\[ \left(\frac{v'_b}{c}\right)^2 = \frac{9}{25} \]\[ v'_b = 0.6c \]
04

Perform Lorentz Transformation to Find Velocities in Frame S

In frame \(S\), particle \(a\) has velocity \(u = 0.5c\) and the velocities of the \(b\) particles need to be calculated. Using velocity addition formula:\[ v_{b1} = \frac{v'_b + u}{1 + \frac{v'_b u}{c^2}} \]\[ v_{b2} = \frac{-v'_b + u}{1 - \frac{v'_b u}{c^2}} \]Substitute \(v'_b = 0.6c\):\[ v_{b1} = \frac{0.6c + 0.5c}{1 + \frac{0.6c \times 0.5c}{c^2}} = \frac{1.1c}{1.3} = 0.846c \]\[ v_{b2} = \frac{-0.6c + 0.5c}{1 - \frac{0.6c \times 0.5c}{c^2}} = \frac{-0.1c}{0.7} = -0.143c \]
05

Conclusion

The velocity of the \(b\) particle in the rest frame of \(a\) is \(0.6c\). In frame \(S\), the velocities of the \(b\) particles are \(0.846c\) and \(-0.143c\).

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Key Concepts

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

Lorentz Transformation
Understanding the Lorentz Transformation is essential in the realm of relativistic physics, where objects move at speeds close to the speed of light. This transformation allows us to convert the coordinates of an event from one inertial frame to another.

When transitioning between frames, the Lorentz transformation helps calculate how measurements of space and time vary. It accounts for the effects of time dilation and length contraction. The transformations are:
  • For time: \( t' = \gamma (t - \frac{vx}{c^2}) \)
  • For position: \( x' = \gamma (x - vt) \)
Where \( \gamma \) is the Lorentz factor, \( v \) is the relative speed between the frames, \( t \) is time in the stationary frame, and \( x \) is the position.

In practical problems like our exercise, using the Lorentz transformation simplifies finding the velocities in different frames, ensuring consistent results across varying perspectives.
Conservation of Momentum
In physics, momentum conservation is a fundamental principle that states the total momentum of a closed system remains constant over time, if it is not influenced by external forces. In the context of our exercise, momentum conservation is crucial when dealing with particle decays.

When particle \( a \) decays into two identical particles \( b \), the system's momentum before and after the decay must remain unchanged. In the rest frame of particle \( a \), which initially has zero momentum, the sum of the momenta of the two particles \( b \) must also be zero.

This leads to the condition:
\[ m_b v'_b + m_b (-v'_b) = 0 \]
This condition emphasizes that both particles travel at equal speeds but opposite directions, ensuring no net change in momentum.
Conservation of Energy
Within any closed physical system, the energy conservation principle maintains that the total energy remains constant over time. This principle is vital in analyzing high-speed particle interactions in relativistic mechanics.

In the given exercise, during the decay of particle \( a \) into two particles \( b \), the initial energy of particle \( a \) is entirely converted into the energies of the two \( b \) particles. This can be expressed as:
  • Initial energy: \( E_a = m_a c^2 \)
  • Final energy: \( 2 \times \gamma_b m_b c^2 \)
Equating these gives: \[ m_a c^2 = 2 \gamma_b m_b c^2 \] Substituting the given mass relationship, \( m_a = 2.5 m_b \), lets us solve for \( \gamma_b \): \[ \gamma_b = \frac{1.25}{1} \] This equation ensures that we respect the energy conservation rule, allowing us to find missing variables like velocity in the decay process.
Velocity Addition Formula
Velocity addition in Special Relativity differs from classical physics, due to the involvement of the speed of light. The relativistic velocity addition formula is an essential tool when dealing with moving reference frames.

In our scenario, to find particle \( b \)'s velocity in the original stationary frame \( S \), where particle \( a \) was initially traveling, we use the formula:
  • Velocity of particle \( b \) moving in direction of particle \( a \):\[ v_{b1} = \frac{v'_b + u}{1 + \frac{v'_b u}{c^2}} \]
  • Velocity of particle \( b \) moving opposite the direction of particle \( a \):\[ v_{b2} = \frac{-v'_b + u}{1 - \frac{v'_b u}{c^2}} \]
In this formula, \( v'_b \) is the velocity in the rest frame of particle \( a \), \( u \) is the velocity of the frame, and \( c \) is the speed of light.

By substituting known values, we ensure all velocities are calculated accurately, adhering to relativistic principles. The formula confirms that as velocities approach the speed of light, they never surpass it, honoring Einstein's principle of relativity.

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Most popular questions from this chapter

An excited state \(X^{*}\) of an atom at rest drops to its ground state \(X\) by emitting a photon. In atomic physics it is usual to assume that the energy \(E_{\gamma}\) of the photon is equal to the difference in energies of the two atomic states, \(\Delta E=\left(M^{*}-M\right) c^{2},\) where \(M\) and \(M^{*}\) are the rest masses of the ground and excited states of the atom. This cannot be exactly true, since the recoiling atom X must carry away some of the energy \(\Delta E .\) Show that in fact \(E_{\gamma}=\Delta E\left[1-\Delta E /\left(2 M^{*} c^{2}\right)\right] .\) Given that \(\Delta E\) is of order a few ev, while the lightest atom has \(M\) of order \(1 \mathrm{GeV} / c^{2},\) discuss the validity of the assumption that \(E_{\gamma}=\Delta E\)

(a) Show that if a body has speed \(v < c\) in one inertial frame, then \(v < c\) in all frames. [Hint: Consider the displacement four-vector \(d x=(d \mathbf{x}, c d t),\) where \(d \mathbf{x}\) is the three-dimensional displacement in a short time \(d t .]\) (b) Show similarly that if a signal (such as a pulse of light) has speed \(c\) in one frame, its speed is \(c\) in all frames.

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

Like time dilation, length contraction cannot be seen directly by a single observer. To explain this claim, imagine a rod of proper length \(l_{\mathrm{o}}\) moving along the \(x\) axis of frame \(\mathcal{S}\) and an observer standing away from the \(x\) axis and to the right of the whole rod. Careful measurements of the rod's length at any one instant in frame \(\mathcal{S}\) would, of course, give the result \(l=l_{\mathrm{o}} / \gamma\). (a) Explain clearly why the light which reaches the observer's eye at any one time must have left the two ends \(A\) and \(B\) of the rod at different times. (b) Show that the observer would see (and a camera would record) a length more than \(l\). [It helps to imagine that the \(x\) axis is marked with a graduated scale.] ( \(\mathbf{c}\) ) Show that if the observer is standing close beside the track, he will see a length that is actually more than \(l_{\mathrm{o}}\); that is, the length contraction is distorted into an expansion.

The muons created by cosmic rays in the upper atmosphere rain down more-or- less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about \(1.5 \mu\) s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of \(2000 \mathrm{m}\), and in the course of an hour detects 650 muons traveling at \(0.99 c\) toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive.) Needless to say, the relativistic answer agrees with experiment.

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