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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 pion \(\left(\pi^{+} \text {or } \pi^{-}\right)\) is an unstable particle that decays with a proper half-life of \(1.8 \times 10^{-8}\) s. (This is the half-life measured in the pion's rest frame.) (a) What is the pion's half-life measured in a frame \(\mathcal{S}\) where it is traveling at \(0.8 c ?\) (b). If 32,000 pions are created at the same place, all traveling at this same speed, how many will remain after they have traveled down an evacuated pipe of length \(d=36 \mathrm{m} ?\) Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive. (c) What would the answer have been if you had ignored time dilation? (Naturally it is the answer (b) that agrees with experiment.)

Prove that if \(T\) and \(a\) are respectively a four-tensor and a four-vector, then \(b=T \cdot a=T G a\) is a four-vector; that is, it transforms according to the rule \(b^{\prime}=\Lambda b\)

A particle of mass 12 MeV/c \(^{2}\) has a kinetic energy of 1 \(\mathrm{MeV}\). What are its momentum (in MeV/c) and its speed (in units of \(c\) )?

A robber's getaway vehicle, which can travel at an impressive 0.8c, is pursued by a cop, whose vehicle can travel at a mere 0.4c. Realizing that he cannot catch up with the robber, the cop tries to shoot him with bullets that travel at \(0.5 c\) (relative to the cop). Can the cop's bullets hit the robber?

A neutral pion (Problem 15.86) is traveling with speed \(v\) when it decays into two photons, which are seen to emerge at equal angles \(\theta\) on either side of the original velocity. Show that \(v=c \cos \theta\)

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