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One way to set up the system of synchronized clocks in a frame \(\mathcal{S}\), as described at the beginning of Section 15.4, would be for the chief observer to summon all her helpers to the origin \(O\) and synchronize their clocks there, and then have them travel to their assigned positions very slowly. Prove this claim as follows: Suppose a certain observer is assigned to a position \(P\) at a distance \(d\) from the origin. If he travels at constant speed \(V\), when he reaches \(P\) how much will his clock differ from the chief's clock at \(O\) ? Show that this difference approaches 0 as \(V \rightarrow 0\).

Short Answer

Expert verified
The clock difference approaches zero as speed \(V\) approaches zero.

Step by step solution

01

Understanding Time Dilation

When an observer moves at a speed relative to a frame, their clock as measured in the stationary frame experiences time dilation. According to the theory of relativity, if an observer travels at a constant speed \(V\), their time \(t'\) is related to the stationary frame time \(t\) by the equation: \( t' = \frac{t}{\sqrt{1 - \frac{V^2}{c^2}}} \), where \(c\) is the speed of light.
02

Calculate time taken at speed V

To find the time it takes for the observer to reach position \(P\), we use the formula \(t = \frac{d}{V}\), where \(d\) is the distance from origin \(O\) to \(P\) and \(V\) is the constant speed of the observer.
03

Calculate elapsed time on observer's clock

Substitute \(t\) from Step 2 into the time dilation formula to calculate the time \(t'\) on the observer's clock: \[ t' = \frac{\frac{d}{V}}{\sqrt{1 - \frac{V^2}{c^2}}} \]
04

Calculate clock difference

The difference in time between the observer's clock and the stationary clock at origin \(O\), \(\Delta t\), is given by \(\Delta t = t - t'\). Simplify the expression to find: \[ \Delta t = \frac{d}{V} - \frac{\frac{d}{V}}{\sqrt{1 - \frac{V^2}{c^2}}} \]
05

Analyze limit as V approaches 0

Analyze the expression \(\Delta t\) as the speed \(V\) approaches zero. Since the term \(\frac{V^2}{c^2}\) becomes negligible, the expression for \(\Delta t\) simplifies to 0. This is because the Lorentz factor \(\sqrt{1 - \frac{V^2}{c^2}}\) approaches 1 as \(V\) approaches 0, thus making \(\Delta t\) approach 0.

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Key Concepts

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

Synchronized Clocks
Synchronized clocks play a crucial role in understanding time dilation and the theory of relativity. Imagine a group of observers, each tasked with carrying a clock to a different position in space. To ensure accurate measurement across different locations, these clocks must be synchronized. This is often done by initially having all observers meet at a single point—like the origin of a coordinate system—to synchronize their clocks.
Once synchronized, the observers can then move to their designated positions at a very slow velocity. The necessity of slow travel arises from ensuring that the time on their clocks remains as close as possible to the synchronized time at the origin. The slower the observers move, the less the impact time dilation will have on their clocks, maintaining the synchronization between clocks effectively.
  • Initial synchronization at the origin
  • Slow travel to minimize time discrepancies
  • Maintaining synchronization ensures accurate measurements
Constant Speed
Traveling at a constant speed is a key aspect in analyzing time dilation. In the scenario where an observer is moving from an origin point to another point at a constant speed, it allows us to predictably calculate the effects of time dilation. The observer covers the distance relying on the formula for speed, which is distance divided by time, or simply expressed as \( V = \frac{d}{t} \).
However, due to the effects of relativity, the time as experienced by the observer, known as proper time, differs from the time experienced in the stationary frame. This difference arises when the observer speed is significant compared to the speed of light. Yet, if the observer travels extremely slowly, the influence of relativity shrinks. Thus, maintaining a constant, slow speed ensures that time measurements remain consistent with negligible relativity effects, allowing the clocks to stay in sync.
  • Allows predictable measurement of time
  • Enables analysis using simple equations
  • Reduces relativistic effects when speed is low
Theory of Relativity
The theory of relativity, proposed by Albert Einstein, fundamentally changes how we understand space and time. It essentially states that the laws of physics are the same for all observers in uniform motion and that the speed of light is constant, regardless of the relative motion of an observer. These foundational principles lead to effects like time dilation.
In the context of synchronized clocks moving at constant speed, time dilation becomes crucial. For an observer in motion relative to a stationary frame, their clock ticks slower compared to the clocks at rest in the stationary frame. This requires us to use the mathematical framework provided by the theory of relativity to calculate the difference in time experienced by moving observers, ensuring we can accurately deal with high-speed travel scenarios.
  • Alters core understanding of space and time
  • Explains time dilation and length contraction
  • Ensures consistent physical laws across different inertial frames
Lorentz Factor
The Lorentz factor is a fundamental component of calculations involving time dilation and the theory of relativity. It is denoted by \( \gamma \) and is defined as \( \gamma = \frac{1}{\sqrt{1 - \frac{V^2}{c^2}}} \). This factor comes into play when analyzing how time and space are perceived by observers in relative motion.
As an observer's speed \( V \) approaches zero, the Lorentz factor approaches 1, which means time as perceived by the moving observer becomes identical to the time in the stationary frame. However, as the speed increases and approaches the speed of light \( c \), the Lorentz factor increases drastically, reflecting substantial time dilation and resulting in significant differences in time reporting between frames.
  • Fundamental for adjusting time and space calculations
  • Affects the perceived passage of time relative to motion
  • Approaches 1 as speed decreases, reducing relativistic effects

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Most popular questions from this chapter

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.

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 rocket traveling at speed \(\frac{1}{2} c\) relative to frame \(\mathcal{S}\) shoots forward bullets traveling at speed \(\frac{3}{4} c\) relative to the rocket. What is the speed of the bullets relative to \(\mathcal{S} ?\)

(a) Find the 3 \(\times 3\) matrix \(\mathbf{R}(\theta)\) that rotates three- dimensional space about the \(x_{3}\) axis, so that \(\mathbf{e}_{1}\) rotates through angle \(\theta\) toward \(\mathbf{e}_{2}\). (b) Show that \([\mathbf{R}(\theta)]^{2}=\mathbf{R}(2 \theta),\) and interpret this result.

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 \(\pi^{+} \rightarrow \mathrm{e}^{+}+\nu,\left(m_{\mathrm{e}}=0.5 \mathrm{MeV} / c^{2}\right)\)

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