Question

Given that \(g\), the acceleration of gravity at the earth's surface, is \(\sim 9.8 \mathrm{~m} \mathrm{~s}^{-2}\), and that a year has \(\sim 3.2 \times 10^{7}\) seconds, verify that, in units of years and light years, \(g \approx 1\). A rocket moves from rest in an inertial frame \(S\) with constant proper acceleration \(g\) (thus giving maximum comfort to its passengers). Find its Lorentz factor relative to \(\mathrm{S}\) when its own clock indicates times \(\tau=1\) day, 1 year, 10 years. Find also the corresponding distances and times travelled in S. If the rocket accelerates for 10 years of its own time, then decelerates for 10 years, and then repeats the whole manoeuvre in the reverse direction, what is the total time elapsed in \(\mathrm{S}\) during the rocket's absence?

Short answer

#Answer#: For the three scenarios with 𝜏 = 1 day, 1 year, and 10 years, we have the following results: 1. 𝜏 = 1 day: - Lorentz factor: \(γ \approx 1.000002\) - Proper distance traveled: \(Δx' \approx 0.500002\) light years - Proper time elapsed: \(Δt \approx 0.500002\) years 2. 𝜏 = 1 year: - Lorentz factor: \(γ \approx 1.866\) - Proper distance traveled: \(Δx' \approx 0.866\) light years - Proper time elapsed: \(Δt \approx 1.623\) years 3. 𝜏 = 10 years: - Lorentz factor: \(γ \approx 2.29\) - Proper distance traveled: \(Δx' \approx 2.29\) light years - Proper time elapsed: \(Δt \approx 4.58\) years The total time elapsed for the round trip with 10 years of proper time in each acceleration/deceleration phase is approximately \(Δt_\text{total} = 36.64\) years in frame S.

Step-by-step solution

Verify statement about g

To verify that \(g \approx 1\) when expressed in units of years and light years, we have to convert the given g from meters per second squared to light years per year squared. We'll use the conversion factors: 1 year has about \(3.2 \times 10^7\) seconds and 1 light year is around \(9.461 \times 10^{15}\) meters. $$ g' = \frac{9.8 \frac{\text{m}}{\text{s}^2}}{1 \frac{\text{ly}}{\text{y}^2}} = \frac{9.8 \frac{\text{m}}{\text{s}^2}}{\frac{9.461 \times 10^{15} \text{m}}{(3.2 \times 10^7 \text{s})^2}} $$ Calculate the value of \(g'\): $$ g' \approx 1 $$ Thus, the statement is verified.

Find Lorentz factor, distances, and times

We have three scenarios where \(\tau\) = 1 day, 1 year, and 10 years. The proper distance traveled by the rocket is given by the equation: $$ Δx' = \frac{c^2}{g} (\sqrt{1 + (gτ/c)^2} - 1) $$ And the proper time elapsed is given by: $$ Δt = \frac{c}{g} (\sqrt{1 + (gτ/c)^2} - 1) $$ When calculating distances and times, we can make use of the fact that g is approximately equal to 1 in the required units. The Lorentz factor is given by the equation: $$ γ = \sqrt{1 + (gτ/c)^2} $$ Let's find out the values for each scenario: 1. 𝜏 = 1 day 2. 𝜏 = 1 year 3. 𝜏 = 10 years

Calculate total elapsed time for the round trip

The forward journey consists of two equal parts: accelerating for 10 years of proper time (t1) and decelerating for 10 years of proper time (t2). The same applies to the return journey (t3 and t4). To find the total time elapsed in S, we have to sum up the time elapsed in each part: $$ Δt_\text{total} = Δt_1 + Δt_2 + Δt_3 + Δt_4 $$ Using the previously calculated proper time formula (with the duration being 10 years) and considering that the time elapsed during acceleration and deceleration is equal: $$ Δt_\text{total} = 2 (Δt_{\text{forward}} + Δt_{\text{return}}) $$ After computing the total time elapsed, we can state the final answer.

Key concepts

Lorentz factor

In the realm of special relativity, the Lorentz factor, denoted as \( \gamma \), is a fundamental element used to describe how much time, length, and relativistic mass change for an object while it is moving at any speed. When an object moves close to the speed of light, it experiences changes predicted by relativity, which involve time dilation and length contraction.

The Lorentz factor is expressed as follows: \[\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\] where \( v \) is the velocity of the object, and \( c \) is the speed of light. In scenarios where velocities are a significant fraction of the speed of light, the Lorentz factor becomes considerably greater than 1.

This factor is crucial in the exercise, as it determines how time and distance are perceived when in a reference frame moving at a constant proper acceleration. It provides the bridge between the proper time, which is the time kept by a clock moving with the object, and the time observed from another inertial reference frame.

proper acceleration

Proper acceleration refers to the acceleration experienced by an object as measured in its own instantaneous rest frame. It's the acceleration that feels like a force acting on the object, such as the force you experience when a car accelerates or when you're standing on the Earth's surface due to gravity.

In the context of this exercise, proper acceleration is denoted by \( g \), similar to the gravitational acceleration on Earth, but it's constant for the entire motion of the rocket. This provides a sense of gravity to the passengers, making their experience akin to being on Earth's surface, which is why this kind of propulsion is described as offering "maximum comfort."

The interesting fact here is that, even if the rocket is traveling enormous distances through space over several years, passengers feel this constant push, reflecting a state of uniform acceleration. The computation of relative motions using proper acceleration results in the realization of effects like time dilation, examined in the following sections.

time dilation

Time dilation is an intriguing phenomenon in special relativity that occurs when two observers are in different frames of motion relative to each other. It results in one observer perceiving time to pass at a slower rate compared to another observer.

In this exercise, time dilation plays a fundamental role in determining how much time passes for the rocket in its own frame compared to the stationary frame \( S \). The relationship between proper time (\( \tau \)) and dilated time (\( t \)) is articulated through the Lorentz factor: \[t = \gamma \tau\] Here, \( \gamma \) is the Lorentz factor that stretches or compresses the time experienced by the moving object when viewed from the stationary frame.

This phenomenon implies that while a journey might last 10 years for the travelers on the rocket as per their proper time, the same journey could take much longer when viewed from an inertial frame of reference such as Earth, thanks to the effects of time dilation.

proper time

Proper time, denoted by \( \tau \), is the time interval measured by a clock that is at rest relative to the observer, meaning it moves along with the object experiencing the passage of this time. It's the most straightforward measurement of time, akin to what you would measure with a wristwatch traveling along with you.

In the framework of special relativity and this exercise, proper time is pivotal because it remains invariant for all observers. Whether in an accelerating rocket or inertial frame, proper time is what the traveler on the rocket perceives.

Over the course of the problem, the exercise refers to events such as the rocket's acceleration for 10 years of its proper time. It's crucial as it directly relates to what the passengers onboard would chart as the elapsed time, ensuring their world is consistent. It's also through proper time that the experiment links how far and for how long the rocket travels in the stationary frame compared to what the passengers tally, underlining the separation of experiences between stationary and moving frames in relativistic conditions.