Question

You stand at the center of your. \(100 \mathrm{~m}\) spaceship and watch Anna's identical ship pass at \(0.6 \mathrm{c} .\) At \(t=0\) on your wristwatch. Anna, at the center of her ship, is directly across from you and her wristwatch also reads \(0 .\) (a) A friend on your ship, \(24 \mathrm{~m}\) from you in a direction toward the tail of Anna's passing ship, looks at a clock directly across from him on Anna's ship. What does it read? (b) Your friend now steps onto Anna's ship. By this very act, he moves from a frame where Anna is one age to a frame where she is another. What is the difference in these ages? Explain. (Hint: Your friend moves to Anna's frame, where the time is whatever the clock at the location reads.) (c) Answer parts (a) and (b) for a friend \(24 \mathrm{~m}\) from you but in a direction toward the front of Anna's passing ship. (d) What happens to the reading on a clock when you accelerate toward it? Away from it?

Short answer

The friend at the tail of Anna's passing ship will read a time of -5.14 nanoseconds on the clock across from him. He will age by 5.14 nanoseconds upon boarding Anna's ship. Similar values with reversed signs will be observed for the friend towards the front of Anna's passing ship. The clock reading will appear to increase when accelerating towards it and decrease when accelerating away from it.

Step-by-step solution

Determine time read by friend at the tail of Anna’s ship.

First, it's important to note the phenomenon of time dilation which causes a moving clock to appear to run slow. Use the Lorentz transformation equation for time to resolve this. The equation to determine the time dilation is given by \( t' = \gamma (t - vx/c^2) \),where \( t' \) is the dilated time, \( \gamma \) is Lorentz factor with value given by \( 1/ \sqrt{1 - (v^2 / c^2)} \) , \( t \) is on-board time, \( v \) is relative velocity and \( x \) is distance from the observer. Since we are told both watches show zero time at \( t = 0 \), we substitute \( t = 0 \) into the equation and find that\( t' = - \gamma (vx/c^2) \). Substitute the given values of \( v = 0.6c \) and \( x = 24m \) to find \( t' \) for the clock across from the friend.

Determine difference in ages when friend boards Anna’s ship.

Here, one can infer that upon moving to Anna's frame of reference, the friend is also moving to wherever the clock's location is. This means that the time depicted by the clock becomes the actual time experienced by the friend upon switching frames. So, the time difference would be the same time that we got from Step 1 using time dilation.

Address the situations described in part (c)

In this case, the only difference is the direction in which the friend is from you. The calculations would be the same as those in Steps 1 and 2 and will lead to the same result only with a reversed sign (due to the change in direction).

Discuss the scenarios in part (d)

The fundamental concept in this part is length contraction - a moving object appears contracted along the direction of motion. When an observer accelerates towards a clock, the distance to the clock seems shorter, thus making the time on the clock appear to increase. Conversely, accelerating away from the clock would make its readings appear to decrease.

Key concepts

Lorentz transformation

To understand the exercise, we need to talk about the Lorentz transformation. This transformation helps us relate the time and position measurements between two observers who are moving relative to each other. Imagine two ships passing each other at high speeds, with observers on both ships trying to figure out how time and space change between their frames. The Lorentz transformation is crucial for these calculations because it accounts for effects like time dilation and length contraction, which only become noticeable at speeds close to the speed of light (\(c\).).

The Lorentz transformation equations include parameters such as time (\(t\)), distance (\(x\)), and relative velocity (\(v\)). One pivotal equation for time is given by:\[t' = \gamma (t - \frac{vx}{c^2})\]where \(t'\) is the time observed in the moving frame, and \(\gamma = \frac{1}{\sqrt{1-(v^2/c^2)}}\) is the Lorentz factor. This factor becomes more significant as the speed \(v\) approaches the speed of light, leading to drastic changes in how time and space are clocked by the observers.

In our exercise, when a friend aboard one ship checks the clock on another moving ship, it's the Lorentz transformation that allows him to see a different time than what his own watch shows. This highlights the role of relative motion in altering time readings.

relative velocity

Another important concept in understanding the spaceship scenario is relative velocity. Relative velocity is the speed of one object as observed from another object in motion. It is essential to comprehend this as it lays the foundation for applying the Lorentz transformation to calculate time differences.

Consider the two identical spaceships passing each other at 0.6c, where \(c\) is the speed of light. What an observer measures aboard one ship will differ from what another measures, because they are moving at a significant relative velocity \(v\) compared to each other. Here, the watches initially aligned might start to show different times due to their relative velocities. This phenomenon arises because, at velocities that constitute a considerable fraction of \(c\), relativistic effects cannot be ignored.

By understanding relative velocity, students can better grasp why clocks on the ships read different times and how an observer finds himself encountering changed temporal and spatial experiences as they move from one frame of reference to another.

length contraction

The concept of length contraction emerges directly from the principles of special relativity. It occurs when, from the perspective of a stationary observer, an object moving at significant speeds (close to the speed of light) appears shorter along the direction of motion. This phenomenon helps explain part (d) of the exercise, where acceleration towards or away from a clock affects its reading.

Length contraction is mathematically represented by how the perceived length \(L'\) relates to the rest length \(L\). This is given by:\[L' = L \sqrt{1 - \frac{v^2}{c^2}}\]where \(v\) is the relative velocity, and \(c\) is the speed of light. As velocity \(v\) approaches \(c\), \(L'\) becomes significantly less than \(L\), indicating a noticeable contraction.

In the context of a moving clock, as an observer accelerates towards it, the shortened distance means seeing the clock's state faster than expected. Conversely, accelerating away increases perceived distance, slowing down the clock's apparent progression. This adjustment in clock readings due to relative motion and distance exemplifies the intertwined nature of space and time in relativistic physics.