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In the 24 th century the fastest available interstellar rocket moves at \(v=0.75 c .\) Mya Allen is sent in this rocket at full (constant) speed to Sirius, the Dog Star, the brightest star in the heavens as seen from Earth, which is a distance \(8.7\) ly as measured in the Earth frame. Assume Sirius is at rest with respect to Earth. Mya stays near Sirius, slowly orbiting around that Dog Star, for 7 years as recorded on her wristwatch while making observations and recording data, then returns to Earth with the same speed \(v=0.75 \mathrm{c}\). According to Earth-linked observers: (a) When does Mya arrive at Sirius? (b) When does Mya leave Sirius? (c) When does Mya arrive back at Earth? According to Mya's wristwatch: (d) When does she arrive at Sirius? (e) When does she leave Sirius? (f) When does she arrive back on Earth?

Short Answer

Expert verified
(a) 11.6 years (b) 18.6 years (c) 30.2 years (d) 7.65 years (e) 14.65 years (f) 22.3 years.

Step by step solution

01

Calculate travel time to Sirius (Earth frame)

Use the formula for time: \[ t = \frac{d}{v} \]where \( d = 8.7 \) ly and \( v = 0.75c \). The time \( t \) is therefore:\[ t = \frac{8.7 \text{ ly}}{0.75c} = \frac{8.7}{0.75} \text{ years} = 11.6 \text{ years}. \]Mya arrives at Sirius after 11.6 years in the Earth frame.
02

Calculate time at Sirius (Earth frame)

According to Earth-linked observers, Mya stays at Sirius for 7 years. Hence, her stay duration of 7 years is recorded in the Earth frame as well.
03

Calculate total time at Sirius and leaving Sirius (Earth frame)

Adding the travel time and the stay duration:\[ t_{leave} = 11.6 + 7 = 18.6 \text{ years}. \]Mya leaves Sirius 18.6 years after leaving Earth.
04

Calculate return travel time (Earth frame)

The return travel time is identical to the outbound travel time:\[ t_{return} = 11.6 \text{ years}. \]Mya arrives back on Earth after:\[ t_{total} = 18.6 + 11.6 = 30.2 \text{ years}. \]Mya arrives back at Earth 30.2 years after she originally left.
05

Calculate travel time to Sirius (Mya's frame - time dilation)

Use the time dilation formula:\[ t' = t \sqrt{1 - \frac{v^2}{c^2}} \]For the trip to Sirius, substitute \( t = 11.6 \text{ years} \) and \( v = 0.75c \):\[ t' = 11.6 \sqrt{1 - (0.75)^2} = 11.6 \sqrt{1 - 0.5625} = 11.6 \sqrt{0.4375} \approx 7.65 \text{ years}. \]Mya's wristwatch records 7.65 years for the trip to Sirius.
06

Calculate total time (Mya's frame)

The stay time of 7 years is not affected by time dilation (since it is at rest). For the return trip, the time dilation calculation is the same and also equals 7.65 years. To find the total time on Mya's wristwatch:\[ t'_{total} = 7.65 + 7 + 7.65 = 22.3 \text{ years}. \]Mya's wristwatch records a total of 22.3 years. Thus, the return trip is 22.3 years as per her wristwatch.

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Key Concepts

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

Time Dilation
In the realm of relativity, time dilation refers to the stretching or contracting of time experienced by an object in motion relative to an observer at rest. To understand time dilation, let's consider Mya's interstellar journey to Sirius in the 24th century. Mya's rocket travels at 0.75 times the speed of light, denoted as \(0.75c\). According to Earth-based observers, the journey to Sirius, 8.7 light-years away, takes \(t = \frac{d}{v}\), calculated as 11.6 years. However, due to time dilation, Mya's experiences differ. Time dilation is governed by the formula:
\[ t' = t \sqrt{1 - \frac{v^2}{c^2}} \]
where \(t\) is the time in the Earth frame, \(v\) is the velocity of the rocket, and \(c\) is the speed of light. Substituting the values:
\[ t' = 11.6 \sqrt{1 - (0.75)^2} = 11.6 \sqrt{1 - 0.5625} = 11.6 \sqrt{0.4375} \approx 7.65 \text{ years} \]
Hence, Mya's wristwatch records 7.65 years for the trip to Sirius. The same dilation applies for the return trip, making her total travel time 15.3 years, plus the 7 years spent near Sirius. In total, Mya experiences her journey as 22.3 years, reflecting significant time dilation compared to 30.2 years for Earth observers.
Lorentz Transformation
The Lorentz transformation equations bridge the differences between measurements in different inertial frames moving at constant velocities relative to each other. For Mya's journey to Sirius, Lorentz transformations allow us to compute the length and time measurements from Mya’s perspective. The key equations are:
\[ x' = \gamma (x - vt) \]
and
\[ t' = \gamma (t - \frac{vx}{c^2}) \]
where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\). Here, \(\gamma \approx 1.511\) when \(v = 0.75c\). Utilizing Lorentz transformation helps understand how distances and times contract in Mya's frame. For example, the distance Mya measures to Sirius would be
\[ x' = \frac{x}{\gamma} = \frac{8.7 \text{ ly}}{1.511} \approx 5.76 \text{ ly} \]
Hence, while Earth observers see the distance as 8.7 light-years, Mya perceives it as 5.76 light-years due to Lorentz contraction. This contraction combined with time dilation allows Mya’s journey and activities to take lesser time in her frame compared to Earth’s frame.
Interstellar Travel
Interstellar travel, the concept of journeying between stars within a galaxy, brings fascinating implications within the framework of relativity. One of the core challenges is accounting for the effects of traveling at relativistic speeds. Mya's journey to Sirius at 0.75c offers a clear example. Key points to consider:
  • Distance: In interstellar terms, distances are vast. Sirius is 8.7 light-years from Earth, making the travel feasible only at high speeds closer to light speed.
  • Relativistic Speed: High velocities like 0.75c involve significant relativistic effects, altering time and space perceptions significantly due to time dilation and length contraction.
  • Journey Time: For Earth observers, Mya's journey takes 30.2 years (11.6 years each way plus 7 years stationed). For Mya, time dilation reduces her perceived travel time to 22.3 years.
These differences highlight the necessity of understanding relativity for realistic interstellar travel scenarios. The impact of relativity causes Earth-based observers and travelers to experience time and distance differently, posing unique challenges and fascinating possibilities for future space exploration and human travel across the cosmos.

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Most popular questions from this chapter

Astrophysicists describe the redshift of receding astronomical objects using the redshift factor \(z\), defined implicitly in the following equation: $$ \lambda_{\text {observed }} \equiv(1+z) \lambda_{\text {emitted }} $$ Here \(\lambda_{\text {observed }}\) is the wavelength of light observed from Earth, while \(\lambda_{\text {emitted }}\) is the wavelength of the light emitted from the source as measured in the rest frame of the source. The emitted wavelength is known if one knows the emitting atom, identified from the pattern of different wavelengths characteristic of that atom. Astrophysicists measuring the redshifts of light from extremely remote quasars calculate a \(z\) -factor in the neighborhood of \(z \approx 6 .\) Use the Dopplershift equations of special relativity to determine how fast such quasars are moving away from Earth. Note: Actually, for such distant objects the unmodified Doppler shift formula of special relativity does not apply. Instead, one thinks of the space between Earth and the source expanding as the universe expands; the wavelength of the light expands with this expansion of the universe as it travels from the source quasar to us.

A train moves at \(10 \mathrm{~km} / \mathrm{h}\) along the track. A passenger sprints toward the rear of the train at \(10 \mathrm{~km} / \mathrm{h}\) with respect to the train. Our knee-jerk motto says that the train clocks "run slow" with respect to clocks on the track, and the runner's watch "runs slow" with respect to train clocks. Therefore the runner's watch should "run doubly slow" with respect to clocks on the track. But the runner is at rest with respect to the track. What gives? (This example illustrates the danger of the simple knee-jerk motto "Moving clocks run slow.")

(a) Can a person, in principle, travel from Earth to the center of our galaxy, which is 23000 ly distant, in one lifetime? Explain using either length contraction or time dilation arguments. (b) What constant speed with respect to the galaxy is required to make the trip in \(30 \mathrm{y}\) of the traveler's lifetime?

A flash of light is emitted at an angle \(\phi^{\prime}\) with respect to the \(x^{\prime}\) axis of the rocket frame. (a) Show that the angle \(\phi\) the direction of motion of this flash makes with respect to the \(x\) axis of the laboratory frame is given by the equation $$ \cos \phi=\frac{\cos \phi^{\prime}+v^{\mathrm{rel}} / c}{1+\left(v^{\mathrm{rel}} / c\right) \cos \phi^{\prime}} . $$ Optional: Show that your answer to Problem 34 gives the same result when the velocity \(v^{\prime}\) is given the value \(c .\) (b) A light source at rest in the rocket frame emits light uniformly in all directions. In the rocket frame \(50 \%\) of this light goes into the forward hemisphere of a sphere surrounding the source. Show that in the laboratory frame this \(50 \%\) of the light is concentrated in a narrow forward cone of half-angle \(\phi_{0}\) whose axis lies along the direction of motion of the particle. Derive the following expression for the halfangle \(\phi_{0}\) : $$ \cos \phi_{0}=v^{\mathrm{rel}} / c $$ This result is called the headlight effect. (c) What is the half-angle \(\phi_{0}\) in degrees for a light source moving at \(99 \%\) of the speed of light?

Quite apart from effects due to the Earth's rotational and orbital motion, a laboratory reference frame on the Earth is not an inertial frame, as required by a strict interpretation of special relativity. It is not inertial because a particle released from rest at the Earth's surface does not remain at rest; it falls! Often, however, the events in an experiment for which one needs special relativity happen so quickly that we can ignore effects duc to gravitational accclcration. Considcr, for cxamplc, a proton moving horizontally at speed \(v=0.992 c\) through a 10 -m-wide detector in a laboratory test chamber. (a) How long will the transit through that detector take? (b) How far does the proton fall vertically during this time lapse? (c) What do you conclude about the suitability of the laboratory as an inertial frame in this case?

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