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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

Two particles \(a\) and \(b\) with masses \(m_{a}=0\) and \(m_{b}>0\) approach one another. Prove that they have a CM frame (that is, a frame in which their total three-momentum is zero). [Hint: As you should explain, this is equivalent to showing that the sum of two four-vectors, one of which is forward light-like and one forward time-like, is itself forward time-like.]

(a) By exchanging \(x_{1}\) and \(x_{2}\), write down the Lorentz transformation for a boost of velocity \(V\) along the \(x_{2}\) axis and the corresponding \(4 \times 4\) matrix \(\Lambda_{\mathrm{B} 2}\). ( \(\mathbf{b}\) ) Write down the \(4 \times 4\) matrices \(\Lambda_{\mathrm{R}+}\) and \(\Lambda_{\mathrm{R}-}\) that represent rotations of the \(x_{1} x_{2}\) plane through \(\pm \pi / 2,\) with the angle of rotation measured counterclockwise. (c) Verify that \(\Lambda_{\mathrm{B} 2}=\Lambda_{\mathrm{R}-} \Lambda_{\mathrm{B} 1} \Lambda_{\mathrm{R}+},\) where \(\Lambda_{\mathrm{B} 1}\) is the standard boost along the \(x_{1}\) axis, and interpret this result.

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.

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.

As a meter stick rushes past me (with velocity v parallel to the stick), I measure its length to be \(80 \mathrm{cm} .\) What is \(v ?\)

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