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Frame \(\mathcal{S}^{\prime}\) travels at speed \(V_{1}\) along the \(x\) axis of frame \(\mathcal{S}\) (in the standard configuration). Frame \(\mathcal{S}^{\prime \prime}\) travels at speed \(V_{2}\) along the \(x^{\prime}\) axis of frame \(\mathcal{S}^{\prime}\) (also in the standard configuration). By applying the standard Lorentz transformation twice find the coordinates \(x^{\prime \prime}, y^{\prime \prime}, z^{\prime \prime}, t^{\prime \prime}\) of any event in terms of \(x, y, z, t .\) Show that this transformation is in fact the standard Lorentz transformation with velocity \(V\) given by the relativistic "sum" of \(V_{1}\) and \(V_{2}\)

Short Answer

Expert verified
The transformation demonstrates the standard Lorentz transformation with composite velocity \( V = \frac{V_1 + V_2}{1 + V_1 V_2} \).

Step by step solution

01

Lorentz Transformation for Frame S to S'

The Lorentz transformation from frame \( \mathcal{S} \) to frame \( \mathcal{S}' \), which moves with velocity \( V_1 \) along the \( x \)-axis (assuming speed of light \( c = 1 \), for simplicity), is given by:\[ x' = \gamma_1 (x - V_1 t) \]\[ t' = \gamma_1 (t - \frac{V_1 x}{c^2}) = \gamma_1 (t - V_1 x) \]\[ y' = y \]\[ z' = z \]where \( \gamma_1 = \frac{1}{\sqrt{1 - V_1^2}} \) is the Lorentz factor for velocity \( V_1 \).
02

Lorentz Transformation for Frame S' to S''

For the second transformation, from frame \( \mathcal{S}' \) to frame \( \mathcal{S}'' \), which moves with velocity \( V_2 \) along the \( x' \)-axis, is given by:\[ x'' = \gamma_2 (x' - V_2 t') \]\[ t'' = \gamma_2 (t' - \frac{V_2 x'}{c^2}) = \gamma_2 (t' - V_2 x') \]\[ y'' = y' \]\[ z'' = z' \]where \( \gamma_2 = \frac{1}{\sqrt{1 - V_2^2}} \) is the Lorentz factor for velocity \( V_2 \).
03

Substitute to Find S to S'' Transformation

Substitute the transformations from Step 1 into Step 2:\[ x'' = \gamma_2 (\gamma_1 (x - V_1 t) - V_2 \gamma_1 (t - V_1 x)) \]\[ = \gamma_2 \gamma_1 ((1 + V_1 V_2) x - (V_1 + V_2) t) \]\[ t'' = \gamma_2 (\gamma_1 (t - V_1 x) - V_2 \gamma_1 (x - V_1 t)) \]\[ = \gamma_2 \gamma_1 ((1 + V_1 V_2) t - (V_1 + V_2) x) \]\[ y'' = y \]\[ z'' = z \]
04

Simplify to Recognize Composite Velocity

Using the simplifications \((1 - V_1^2)(1 - V_2^2) = (1 - (V_1 + V_2)^2 + V_1^2V_2^2)\), the expression simplifies to;\[ V = \frac{V_1 + V_2}{1 + V_1 V_2} \]Verify that this is the relativistic velocity addition formula. The transformations confirm this composite velocity \( V \) so that final coordinate system \( \mathcal{S}'' \) is moving with velocity \( V \) relative to \( \mathcal{S} \) as expected.

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Key Concepts

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

Relativistic Velocity Addition
In classical physics, when two velocities are added, they simply sum up, like how cars on a road would. But, when velocities approach the speed of light, things start to behave differently. This is where the relativistic velocity addition formula comes in handy. It helps us understand how velocities combine when speeds are close to the speed of light, ensuring that nothing exceeds this ultimate speed limit.

Imagine two reference frames: Frame \(\mathcal{S}^\prime\) is moving with velocity \(V_1\) relative to Frame \(\mathcal{S}\), and Frame \(\mathcal{S}^{\prime \prime}\) is moving with velocity \(V_2\) relative to \(\mathcal{S}^\prime\). You might think you could just add \(V_1\) and \(V_2\), but relativity introduces a unique twist. The formula for combining these velocities is:
  • \(V = \frac{V_1 + V_2}{1 + V_1 V_2}\)
This equation ensures that the sum never exceeds the speed of light. The denominator adjusts the total velocity, showcasing how space and time interact near light speeds.
Lorentz Factor
The Lorentz factor is a crucial part of understanding relativistic effects. It plays an essential role in calculating how time, length, and relativistic mass change for objects in motion relative to an observer.

Mathematically, the Lorentz factor, denoted as \(\gamma\), is expressed as:
  • \(\gamma = \frac{1}{\sqrt{1 - v^2}}\)
Here, \(v\) represents the velocity of the moving object, often a fraction of the speed of light. The Lorentz factor increases as velocity approaches the speed of light, indicating more pronounced relativistic effects.

For each frame under consideration, when calculating transformations, we define:
  • \(\gamma_1 = \frac{1}{\sqrt{1 - V_1^2}}\) for the first transformation into Frame \(\mathcal{S}^\prime\)
  • \(\gamma_2 = \frac{1}{\sqrt{1 - V_2^2}}\) for the second transformation into Frame \(\mathcal{S}^{\prime\prime}\)
The Lorentz factor impacts how time slows (time dilation) and how lengths contract (length contraction), essential phenomena in special relativity.
Reference Frames
Reference frames are crucial for understanding motion and transformations in relativity. A reference frame is a perspective from which an observer measures and describes physical phenomena. In the theory of relativity, different observers might be in motion relative to each other, resulting in distinct frames of reference.

In our problem, we dealt with three reference frames:
  • \(\mathcal{S}\) - the initial stationary frame where observations begin.
  • \(\mathcal{S}^\prime\) - moving along the \(x\)-axis of \(\mathcal{S}\) with velocity \(V_1\).
  • \(\mathcal{S}^{\prime\prime}\) - moving along the \(x^\prime\)-axis of \(\mathcal{S}^\prime\) with velocity \(V_2\).
Choosing the right reference frame is important because it determines how we perceive time, distance, and velocity. Relativity tells us that measurements can change depending on these frames. When transitioning from \(\mathcal{S}\) to \(\mathcal{S}^\prime\), and then to \(\mathcal{S}^{\prime\prime}\), we use Lorentz transformations to account for these variations.
Relativity Theory
At the heart of Einstein's special relativity lies the assertion that the laws of physics are consistent for all observers, regardless of their relative motion. This revolutionary theory turned classical mechanics on its head, showing that space and time are intertwined.

The two pillars of relativity are:
  • The speed of light is constant and insurmountable for all observers.
  • The laws of physics are the same in all inertial frames, those which are either at rest or moving at a constant velocity.
With relativity, concepts like past, present, and future are not absolute. Instead, they depend on the observer’s speed and position. This drastically affects our understanding of time and space.

Relativity affects everyday technology, from GPS systems, which account for time dilation due to satellite speeds, to improving our understanding of cosmic events. The implications of relativity spread far and wide, inviting us to see the universe through a new lens.

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Most popular questions from this chapter

Show that any two zero-mass particles have a CM frame, provided their three- momenta are not parallel. [Hint: As you should explain, this is equivalent to showing that the sum of two forward light-like vectors is forward time-like, unless the spatial parts are parallel.]

Newton's first law can be stated: If an object is isolated (subject to no forces), then it moves with constant velocity. We know that this is invariant under the Galilean transformation. Prove that it is also invariant under the Lorentz transformation. [Assume that it is true in an inertial frame \(\mathcal{S}\), and use the relativistic velocity-addition formula to show that it is also true in any other \(\mathcal{S}^{\prime} .\) ]

A positive pion decays at rest into a muon and neutrino, \(\pi^{+} \rightarrow \mu^{+}+\nu .\) The masses involved are \(m_{\pi}=140 \mathrm{MeV} / c^{2}, m_{\mu}=106 \mathrm{MeV} / c^{2},\) and \(m_{\nu}=0 .\) (There is now convincing evidence that \(m_{\nu}\) is not exactly zero, but it is small enough that you can take it to be zero for this problem.) Show that the speed of the outgoing muon has \(\beta=\left(m_{\pi}^{2}-m_{\mu}^{2}\right) /\left(m_{\pi}^{2}+m_{\mu}^{2}\right) .\) Evaluate this numerically. Do the same for the much rarer decay mode \(\pi^{+} \rightarrow \mathrm{e}^{+}+\nu,\left(m_{\mathrm{e}}=0.5 \mathrm{MeV} / c^{2}\right)\)

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.

A rocket is traveling at speed \(0.9 c\) along the \(x\) axis of frame \(\delta\). It shoots a bullet whose velocity \(\mathbf{v}^{\prime}\){ measured in the rocket's rest frame } \(\mathcal{S}^{\prime} )\) is \(0.9 c\) along the \(y^{\prime}\) axis of \(\mathcal{S}^{\prime} .\) What is the bullet's velocity (magnitude and direction) as measured in \(\mathcal{S}\) ?

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