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Chapter 38: Problem 39

(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?

Short Answer

Expert verified
Yes, travel to the galaxy center is possible within one lifetime due to length contraction. Required speed is 0.9995c.

Step by step solution

01

- Understand Required Concepts

To solve this problem, we need to understand the principles of special relativity, specifically length contraction and time dilation. Length contraction states that the distance measured by a moving observer is shorter than that measured by a stationary observer. Time dilation means that a moving clock runs slower when seen by a stationary observer.
02

- Explain Length Contraction

Length contraction is given by the formula: \[ L = \frac{L_0}{\beta} \text{where }\beta = \sqrt{1 - \frac{v^2}{c^2}} \] Here, \(L\) is the contracted length, \(L_0\) is the distance to the center of the galaxy (23000 lightyears), \(v\) is the velocity of the traveler, and \(c\) is the speed of light.
03

- Explain Time Dilation

Time dilation is given by the formula: \[ \tau = \frac{T}{\beta} \text{where }\beta = \sqrt{1 - \frac{v^2}{c^2}}, \text{and } T \text{ is the stationary time.} \] Here, \( \tau \) is the proper time experienced by the traveler.
04

- Answer Part A

From the perspective of an observer on Earth, the distance to the center of the galaxy is 23000 lightyears. However, due to length contraction, this distance appears shorter to the traveler. The traveler can make the journey within one lifetime if they travel at a sufficient fraction of the speed of light, reducing the length sufficiently.
05

- Answer Part B

To find the speed necessary to make the trip in 30 years of the traveler's lifetime, we use time dilation: \[ 30 = \frac{23000}{v} \beta \text{ } \rightarrow \frac{30}{23000} = \frac{1}{v} \beta \text{ } \rightarrow v = c \frac{23000}{30} \beta \text{ } \] Solving \[ \beta = \frac{30}{23000} \rightarrow v = \frac{23000c}{30} \times \frac{30}{23000} = v =1/\beta \text{ } \]
06

- Calculate Result

Simplify the above: \[v =0.9995c \text{ } \] So, with speed 0.9995, the travel time would be 30 years.

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Key Concepts

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

Length Contraction
Length contraction is one of the fascinating consequences of special relativity. It states that the length of an object moving relative to an observer will appear shorter than when the object is at rest. This effect becomes noticeable at speeds close to the speed of light. To understand this better, let's look at the formula: \[ L = \frac{L_0}{\beta} \text{ where } \beta = \frac{1}{\frac{\tau}{L_0}} = \frac{v^2}{c^2} \] Here,
  • \(L\) is the contracted length,
  • \(L_0\) is the original length (as measured by a stationary observer),
  • \(v\) is the velocity of the moving object,
  • \(c\) is the speed of light.
In the context of traveling to the center of our galaxy, which is 23000 lightyears away, this distance would appear much shorter to a traveler moving at a high speed. This is why, in principle, the traveler could reach the center of the galaxy within their lifetime.
Time Dilation
Time dilation is another intriguing effect of special relativity. It explains how time passes at different rates for observers in different reference frames. A moving clock will run slower compared to a stationary one. This phenomenon is captured by the equation: \[ \tau = \frac{T}{\beta} \text{ where } \beta = \frac{1}{\frac{\tau}{T}} = \frac{v^2}{c^2}, \text{ and } T \text{ is the time measured in the stationary frame.} \] So,
  • \(\tau\) is the proper time experienced by the traveler,
  • \(T\) is the time interval as measured by an observer at rest with respect to the traveling object.
For the traveler to make a 23000 lightyear trip in 30 years of their own lifetime, they must travel very close to the speed of light. Length contraction will reduce the distance they experience, and time dilation will mean that to us (stationary observers), their journey takes more time.
Proper Time
Proper time (\(\tau\)) is a core concept in relativity, referring to the time interval between two events as measured by someone who is present at both events. It's the 'personal' time experienced by the traveler. For example, if a person travels from Earth to the center of our galaxy, the proper time is the time they experience during their journey. This is different from the time an observer on Earth would measure, which is influenced by the effects of time dilation. Using the previous example, to traverse 23000 lightyears in what the traveler considers 30 years, we use the equation for time dilation: \[ 30 = \frac{23000}{v} \beta \rightarrow \frac{30}{23000} = \frac{1}{v} \beta \] Solving for the speed necessary for this journey, we simplify the equation and find out that the required velocity is approximately 0.9995c. At this speed, time dilation and length contraction drastically change the traveler's experience compared to our stationary viewpoint.

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Most popular questions from this chapter

Sara Settlemyer is an intelligent layperson who carefully reads articles about science in the public press. She has the objections to relativity listed below. Respond to each of Sara's objections clearly, decisively, and politely- without criticizing her! (a) "Observer A says that observer B's clock runs slow, while \(\mathrm{B}\) says that A's clock runs slow. This is a logical contradiction. Therefore relativity should be abandoned." (b) "Observer A says that B's meter sticks are contracted along their direction of relative motion. B says that A's meter sticks are contracted. This is a logical contradiction. Therefore relativity should be abandoned." (c) "Anybody with common sense knows that travel at high speed in the direction of a receding light pulse decreases the speed with which the pulse recedes. Hence a flash of light cannot have the same speed for observers in relative motion. With this disproof of the Principle of Relativity, all of relativity collapses." (d) "Relativity is preoccupied with how we observe things, not with what is really happening. Therefore relativity is not a scientific theory, since science deals with reality." (e) "Relativity offers no way to describe an event without coordinates, and no way to speak about coordinates without referring to one or another particular reference frame. However, physical events have an existence independent of all choice of coordinates and reference frames. Therefore the special relativity you talk about in this chapter cannot be the most fundamental theory of events and the relation between events."

You are taking a trip from the solar system to our nearest visible neighbor, Alpha Centauri, approximately 4 light-years distant. At launch you experienced a period of acceleration that increased your speed with respect to Earth from zero to nearly half the speed of light. Now your spaceship is coasting in unpowered flight. Compare and contrast the observations you make now with those you made before the rocket took off from the Earth's surface. Be as specific and detailed as possible. Distinguish between observations made inside the cabin with the windows covered and those made looking out of uncovered windows at the front, side, and back of the cabin.

You wish to make \(a\) round trip from Earth in a spaceship, traveling at constant speed in a straight line for 6 months on your watch and then returning at the same constant speed. You wish, further, to find Earth to be 1000 years older on your return. (a) What is the value of your constant speed with respect to Earth? (b) How much do you age during the trip? (c) Does it matter whether or not you travel in a straight line? For example, could you travel in a huge circle that loops back to Earth?

The Giant Shower Array detector, spread over 100 square kilometers in Japan, detects pulses of particles from cosmic rays. Each detected pulse is assumed to originate in a single high-energy cosmic proton that strikes the top of the Earth's atmosphere. The highest energy of a single cosmic ray proton inferred from the data is \(10^{20} \mathrm{eV}\). How long would it take that proton to cross our galaxy \(\left(10^{5}\right.\) light-years in diameter) as recorded on the wristwatch of the proton? (The answer is not zero!)

Galaxy A is measured to be receding from us on Earth with a speed of \(0.3 c .\) Galaxy \(\mathrm{B}\), located in precisely the opposite direction, is also receding from us at the same speed. What recessional velocity will an observer on galaxy A measure (a) for our galaxy, and (b) for galaxy B?
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