Sam sees two events as simultaneous: (i) Event \(A\) occurs at the point (0,0,0) at the instant 0: 00: 00 universal time; (ii) Event \(B\) occurs at the point \((500, \mathrm{~m}, 0,0)\) at the same moment. Tim, moving past Sam with a velocity of \(0.999 c \hat{x}\), also observes the two events. a) Which event occurred first in Tim's reference frame? b) How long after the first event does the second event happen in Tim's reference frame?

Short Answer

Expert verified
Answer: In Tim's reference frame, Event B occurred first, and Event A happened approximately 7.475 x 10^-6 s after Event B.

Step by step solution

01

Calculate the relative velocity and gamma factor

Given that Tim is moving past Sam with a velocity of \(0.999c\hat{x}\), so the relative velocity \(v = 0.999c\). The gamma factor is calculated using the formula: $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$. Substitute \(v = 0.999c\) and \(c = 3 \times 10^8 \mathrm{~m/s}\), we get: $$\gamma = \frac{1}{\sqrt{1 - \left(\frac{0.999(3\times 10^8 \mathrm{~m/s})}{3\times 10^8 \mathrm{~m/s}}\right)^2}} \approx 22.366$$.
02

Apply Lorentz transformation for Event A

In Sam's frame, the coordinates of Event A are \((t_A, x_A, y_A, z_A) = (0, 0, 0, 0)\). Now, let's apply the Lorentz transformation to find the coordinates of Event A in Tim's frame \((t_A', x_A', y_A', z_A')\): $$t_A' = \gamma(t_A - \frac{vx_A}{c^2}) = 22.366(0 - \frac{0.999c \cdot 0}{c^2}) = 0$$ $$x_A' = \gamma(x_A - vt_A) = 22.366(0 - 0.999c \cdot 0) = 0$$ $$(y_A', z_A') = (y_A, z_A) = (0, 0)$$. Thus, Event A coordinates in Tim's reference frame are \((t_A', x_A', y_A', z_A') = (0, 0, 0, 0)\).
03

Apply Lorentz transformation for Event B

In Sam's frame, the coordinates of Event B are \((t_B, x_B, y_B, z_B) = (0, 500\mathrm{~m}, 0, 0)\). Now, let's apply the Lorentz transformation to find the coordinates of Event B in Tim's frame \((t_B', x_B', y_B', z_B')\): $$t_B' = \gamma(t_B - \frac{vx_B}{c^2}) = 22.366(0 - \frac{0.999c \cdot 500\mathrm{~m}}{c^2}) \approx -7.475 \times 10^{-6}\mathrm{~s}$$ $$x_B' = \gamma(x_B - vt_B) = 22.366(500\mathrm{~m} - 0.999c \cdot 0) \approx 11,183\mathrm{~m}$$ $$(y_B', z_B') = (y_B, z_B) = (0, 0)$$. Thus, Event B coordinates in Tim's reference frame are \((t_B', x_B', y_B', z_B') \approx (-7.475 \times 10^{-6}\mathrm{~s}, 11,183\mathrm{~m}, 0, 0)\).
04

a) Identify the first event in Tim's reference frame

By comparing the time coordinates of Event A and Event B in Tim's reference frame, we can determine which event occurred first: $$t_A' = 0\mathrm{~s}$$ $$t_B' \approx -7.475 \times 10^{-6}\mathrm{~s}$$. Since \(t_B'\) is negative, it means that Event B occurred before Event A in Tim's reference frame.
05

b) Calculate the time difference between the two events in Tim's reference frame

To find the time interval between the two events in Tim's reference frame, we can simply take the difference between the time coordinates: $$\Delta t' = t_A' - t_B' = 0\mathrm{~s} - (-7.475 \times 10^{-6}\mathrm{~s}) \approx 7.475 \times 10^{-6}\mathrm{~s}$$. So, in Tim's reference frame, the second event (Event A) happens approximately \(7.475 \times 10^{-6}\mathrm{~s}\) after the first event (Event B).

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Key Concepts

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

Lorentz Transformation
Imagine you are standing still while a train zips by at nearly the speed of light. Would you and a passenger on the train agree on what's happening and when? According to Einstein's theory of relativity, the answer is 'no'. This is where the Lorentz transformation comes into play.The Lorentz transformation equations are mathematical formulas that address how measurements of time and space by two observers moving relative to each other are related. They show that time and space coordinates in one frame can be converted into those in another moving frame. When Sam and Tim observe events A and B, they are in different reference frames, with Tim moving at a significant fraction of the speed of light relative to Sam. By applying the Lorentz transformations, we can calculate where and when Tim perceives these events to occur.The equations take into account the relative speed and apply a factor called gamma (1/ 1 - frac{v^n2}{c^n2})) which increases as the speed approaches the speed of light. Essentially, as Tim zooms past Sam, the Lorentz transformation reveals how the measurements of these events differ for Tim compared to those made by Sam.
Simultaneity
When two events are understood to occur at the same time, they are said to be simultaneous. However, simultaneity is not a universal experience. According to Einstein's theory of relativity, two events that are simultaneous in one frame of reference may not be in another if the observers are moving relative to one another.In the given exercise, Sam sees events A and B as simultaneous - they both occur at the instant 00:00:00 universal time. However, for Tim, who is in motion relative to Sam, the timing is skewed due to the high velocity (0.999 c). The outcome of this peculiarity is that simultaneity is relative - what’s simultaneous for one observer may not be for another. Highlighting the profound implication that there is no absolute universal clock ticking the same way for everyone across the cosmos.
Time Dilation
Now let's talk about the concept of time stretching or time dilation. It sounds like science fiction, but it's a very real aspect of the universe. Time dilation occurs because, according to the theory of relativity, the speed of light is constant in all reference frames. Let’s say an astronaut travels at a high speed in a spaceship; she will age more slowly compared to someone on Earth.In the exercise, Tim is moving so fast that time for him is moving slower compared to Sam. This can be quantified using the gamma factor calculated in the problem, which alters the perceived passage of time. Time dilation predicts that Tim will find that event B happened before event A, revealing that what is a simultaneous occurrence for Sam, stretches apart for Tim. This leads to the astonishing realization that observers in motion relative to one another will measure different time intervals between the same two events.
Reference Frames
The concept of reference frames is foundational in understanding relativity. A reference frame is essentially a viewpoint from which an observer makes measurements of space and time. It could be thought of as the 'stage' where physics happens for that observer.In classical physics, reference frames were considered to be largely equivalent—Newton's laws of motion work the same regardless of whether you're on a moving train or standing still on Earth. However, when we enter the realm of relativity, things change dramatically. The motion of the reference frame becomes very important when dealing with high speeds that are fractions of the speed of light, as in the case of Sam (stationary reference frame) and Tim (moving reference frame). This affects our understanding of simultaneous events, time dilation, and spatial measurements - all because we are looking at the universe from different 'stages'.

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Most popular questions from this chapter

Which quantity is invariant-that is, has the same value-in all reference frames? a) time interval, \(\Delta t\) d) space-time interval, b) space interval, \(\Delta x\) \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\) c) velocity, \(v\)

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