Problem 18 - Fundamentals of Physics

Question

Inertial frame \(S^{\prime}\) moves at a speed of \(0.60 \mathrm{c}\) with respect to frame \(S\) (Fig. 37-9). Further, \(x=x^ {\prime}=0\) at \(t=t^{\prime}=0 .\) Two cvents are recorded. In frame \(S\), event 1 occurs at the origin at \(t=0\) and event 2 occurs on the \(x\) axis at \(x=3,0 \mathrm{~km}\) at \(t=4.0 \mathrm{prs}\). According to observer \(S^{\prime},\) what is the time of (a) event 1 and (b) event 27 (c) Do the two observers see the same sequence or the reverse?

Step-by-Step Solution

Answer

In frame \(S'\), Event 1 occurs at \(t' = 0\) and Event 2 occurs at \(t' = 2.75\). Both observers see the same event sequence.

Step 1: Identify Given Information

We have two inertial frames, \(S\) and \(S'\). The frame \(S'\) moves with a speed of \(0.60c\) relative to \(S\). Initially, \(x = x' = 0\) and \(t = t' = 0\). Event 1 happens at \(t = 0\), \(x = 0\), and Event 2 happens at \(t = 4.0\) years, \(x = 3.0\) km in frame \(S\). We need to find the time of these events in frame \(S'\).

Step 2: Use Lorentz Transformation for Time

The Lorentz transformation for time between frames \(S\) and \(S'\) is given by:\[t' = \gamma (t - \frac{vx}{c^2})\]where \(v = 0.60c\) is the relative velocity, and \(\gamma = \frac{1}{\sqrt{1-(v/c)^2}}\) is the Lorentz factor.

Step 3: Calculate Gamma Factor

Calculate \(\gamma\) with \(v = 0.60c\):\[\gamma = \frac{1}{\sqrt{1-(0.60)^2}} = \frac{1}{\sqrt{1-0.36}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25.\]

Step 4: Calculate Time of Event 1 in S'

For Event 1, \(t = 0\) and \(x = 0\). Using the time transformation:\[t' = 1.25 (0 - \frac{0.60c \times 0}{c^2}) = 0.\]Thus, the time of Event 1 in \(S'\) is \(t' = 0\).

Step 5: Calculate Time of Event 2 in S'

For Event 2, \(t = 4.0\) years and \(x = 3.0\) km (convert to light-years, recognize that \(1 \text{ year} = 9.46 \times 10^{12} \text{ km} = c \text{ light-years}\)), \(x = 3.0/9.46\). Now, use the time transformation formula:\[t' = 1.25 (4.0 - \frac{0.60c \times 3.0}{c^2}) = 1.25 (4.0 - 0.60 \times 3.0/(1 \text{ light-year}))\]\[t' = 1.25 (4.0 - 1.8) = 1.25 \times 2.2 = 2.75.\]Thus, the time of Event 2 in \(S'\) is \(t' = 2.75\) light-years.

Step 6: Evaluate Event Sequence for Both Observers

In frame \(S\), Event 1 occurs before Event 2 (\(t = 0\) and \(t = 4.0\) respectively). In frame \(S'\), Event 1 also occurs before Event 2 (\(t' = 0\) and \(t' = 2.75\) respectively). Thus, both observers see the same sequence of events.

Key Concepts

Inertial Frames

In physics, an inertial frame is often described as a reference frame in which an object not acted upon by forces moves at a constant velocity or is at rest. These frames of reference are crucial in classical and modern physics to observe and calculate the motion of objects.

In this context, the problem describes two inertial frames, \( S \) and \( S' \). Frame \( S' \) is moving at a velocity of \( 0.60c \), which means 60% the speed of light, relative to frame \( S \). This movement at such significant speed makes it essential to take into account the principles of Einstein's theory of relativity.

When analyzing events in different inertial frames, physicists use transformations to convert coordinates and measurements of one frame into another. In this scenario, the Lorentz Transformation is applied, which adjusts time and space readings between frames that have a constant velocity relative to each other. These transformations ensure the laws of physics hold equally in all inertial frames.

Relativity

The concept of relativity, introduced by Albert Einstein, revolutionizes how we perceive time and space. Einstein’s theory demolishes the idea of absolute time and space and asserts that the laws of physics are the same for all observers in any inertial frame.

Relativity becomes profound as speeds approach the speed of light. The equation used in the problem is derived from the Lorentz transformations, which mathematically describe how time and space are interconnected in relativity. According to these transformations, time and space are not independent but woven into a four-dimensional mesh known as "spacetime."

- The Lorentz factor \( \gamma \) plays a key role here. It's calculated as \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \), allowing for adjustments in measurements between different reference frames. In this problem, it was calculated using the given relative velocity \( 0.60c \), resulting in \( \gamma = 1.25 \).

Relativity also leads to intriguing phenomena like the changing sequence of events, although in this case, both observers agree on the order of events (Event 1 occurs before Event 2). This agreement is due to the relatively small time dilation present in this particular example.

Time Dilation

Time dilation is one of the most fascinating outcomes of Einstein's theory of relativity. It describes how time measured in one frame can differ from time measured in another moving at a significant fraction of the speed of light.

In the exercise, the time dilation effect is calculated using the Lorentz transformation formula: \[ t' = \gamma \left( t - \frac{vx}{c^2} \right) \] where \( t \) is the time in frame \( S \), \( t' \) is the time in frame \( S' \), \( v \) is the relative velocity, and \( c \) is the speed of light.

- Event 1 in frame \( S \) happens at \( t = 0 \). Since \( x = 0 \) for this event, it simplifies the calculation in \( S' \) to \( t' = 0 \).
- For Event 2, using the coordinates \( t = 4.0 \) years and \( x = 3.0 \) km, time in frame \( S' \) is calculated to be \( t' = 2.75 \) years using the Lorentz transformation.

Time dilation doesn't just affect the perception of time; it provides insight into the relative nature of simultaneous events across different frames. Here, even though both observers see events in the same order, the time intervals between events differ, reinforcing time's relative nature.