March 17, 2003
How does relativity theory resolve the Twin Paradox?
Ronald C. Lasky, a lecturer at Dartmouth College's Thayer School of Engineering, explains. Time must never be thought of as pre-existing in any sense; it is a manufactured quantity. --Hermann Bondi
Paul Davies's recent article "How to Build a Time Machine" has rekindled interest in the Twin Paradox, arguably the most famous thought experiment in relativity theory. In this supposed paradox, one of two twins travels at near the speed of light to a distant star and returns to the earth. Relativity dictates that when he comes back, he is younger than his identical twin brother.
The paradox lies in the question "Why is the traveling brother younger?" Special relativity tells us that an observed clock, traveling at a high speed past an observer, appears to run more slowly. (Many of us solved this problem in sophomore physics, to demonstrate one effect of the absolute nature of the speed of light.) Since relativity says that there is no absolute motion, wouldn?t the brother traveling to the star also see his brother?s clock on the earth move more slowly? If this were the case, wouldn?t they both be the same age? This paradox is discussed in many books but solved in very few. When the paradox is addressed, it is usually done so only briefly, by saying that the one who feels the acceleration is the one who is younger at the end of the trip. Hence, the brother who travels to the star is younger. While the result is correct, the explanation is misleading. Because of these types of incomplete explanations, to many partially informed people, the accelerations appear to be the issue. Therefore, it is believed that the general theory of relativity is required to explain the paradox. Of course, this conclusion is based on yet another mistake, since we don't need general relativity to handle accelerations. The paradox can be unraveled by special relativity alone, and the accelerations incurred by the traveler are incidental. An explanation follows.
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Let us assume that the two brothers, nicknamed the traveler and the homebody, live in Hanover, N.H. They differ in their wanderlust but share a common desire to build a spacecraft that can achieve 0.6 times the speed of light (0.6c). After working on the spacecraft for years, they are ready to launch it, manned by the traveler, toward a star six light-years away. His craft will quickly accelerate to 0.6c. For those who are interested, it would take a little more than 100 days to reach 0.6c at an acceleration of 2g's. Two g's is two times the acceleration of gravity, about what one experiences on a sharp loop on roller coaster. However, if the traveler were an electron, he could be accelerated to 0.6c in a tiny fraction of a second. Hence, the time to reach 0.6c is not central to the argument. The traveler uses the length-contraction equation of special relativity to measure distance. So the star six light-years away to the homebody appears to be only 4.8 light-years away to the traveler at a speed of 0.6c. Therefore, to the traveler, the trip to the star takes only eight years (4.8/0.6), whereas the homebody calculates it taking 10 years (6.0/0.6). It is instructive to discuss how each would view his and the other?s clocks during the trip. Let?s assume that each has a very powerful telescope that enables such observation. Surprisingly, with careful use of the time it takes light to travel between the two we can explain the paradox.
Both the traveler and homebody set their clocks at zero when the traveler leaves the earth for the star (event 1). When the traveler reaches the star (event 2) his clock reads eight years. ( Click here for graph.) However, when the homebody sees the traveler reach the star, the homebody?s clock reads 16 years. Why 16 years? Because, to the homebody, the craft takes 10 years to make it to the star and the light six additional years to come back to the earth showing the traveler at the star. So to the homebody, the traveler?s clock appears to be running at half the speed of his clock (8/16.)?
As the traveler reaches the star he reads his clock at eight years as mentioned, but he sees the homebody?s clock as it was six years ago (the amount of time it takes for the light from the earth to reach him), or at four years (10-6). So the traveler also views the homebody?s clock as running half the speed of his clock (4/8).
On the trip back, the homebody views the traveler?s clock going from eight years to 16 years in only four years' time, since his clock was at 16 years when he saw the traveler leave the star and will be at 20 years when the traveler arrives back home (event 3). So the homebody now sees the traveler's clock advance eight years in four years of his time; it is now twice as fast as his clock. On the trip back, the traveler sees the homebody?s clock advance from four to 20 years in eight years of his time. Therefore, he also sees his brother?s clock advancing at twice the speed of his. They both agree, however, that at the end of the trip the traveler?s clock reads 16 years and the homebody?s 20 years. So the traveler is four years younger. The asymmetry in the paradox is that the traveler leaves the earth?s reference frame and comes back, whereas the homebody never leaves the earth. It is also an asymmetry that the traveler and the homebody agree with the reading on the traveler?s clock at each event, but not vice versa. The traveler?s actions define the events.
The Doppler effect and relativity together explain this effect mathematically at any instant. The interested reader will find the combination of these effects discussed in The Fundamentals of Physics, by David Halliday et al. (John Wiley and Sons, 1996). Paul Davies also does a nice job explaining the Twin Paradox in his book About Time (Touchstone 1995, ppf 59.) My explanation follows Davies?s closely; I hope my graph adds further clarity. The reader should also note that the speed that an observed clock appears to run depends on whether it is traveling away from or toward the observer. The sophomore physics problem, mentioned earlier, is a special case as it applies only when the motion of the traveler passes the observer?s reference frame with no separating distance in the direction of motion.
For those with a little more formal physics background, a spacetime diagram also explains the paradox nicely. It is shown with the supporting calculations for the Doppler effect on the observed time. Proper time is time in the frame of the observer.?
Twin A makes a round trip at 0.6c to a star 18 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals does B send during the trip
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Q: Twin A makes a round trip at 0.6c to a star 18 light-years away, while twin B stays on the earth.…
Twin A makes a round trip at 0.6c to a star 18 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals
does B send during the trip
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concept of modern physic 6 edition beiser chapter 1 problem 22.Twin A makes a round trip at 0.6c to a star 12 light-years away,while twin B stays on the eart...
Science. Physics. Physics questions and answers. Twin A makes a round trip at 0.6c to a star 12 light years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. (a) How many signals does A send during the trip?
The velocity (v) of twin A is 0.6c. 2. Twin A makes a round trip at 0.6c to a star 12 light years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning.
Step 1/3 1. First, we need to determine the time it takes for Twin A to travel to the star and back to Earth. Since the star is 12 light-years away and Twin A is traveling at 0.6c (60% the speed of light), we can calculate the time it takes for the one-way trip: Time = Distance / Speed Time = 12 light-years / 0.6c Time = 20 years (for one-way trip)
Step 1/5 First, we need to calculate the time it takes for twin A to make the round trip to the star and back. Since the star is 12 light-years away and twin A is traveling at $0.6c$, the time it takes for the round trip from Earth's perspective is $\frac{2 \times 12}{0.6} = 40$ years.
One of a pair of 24-year-old twins takes a round trip through space while the other twin remains on Earth. The traveling twin moves at a speed of 0.87c for a total of 38 years, according to Earth time.
First, let's calculate the time it takes for Twin A to make the round trip at 0.6c. The distance to the star is 18 light-years, so the time it takes for Twin A to reach the star is 18 light-years / 0.6c = 30 years. Since Twin A is traveling at 0.6c, time dilation occurs.
For twin A I need to know how long it takes B to complete the round trip. time = d/r = 24LY/0.6c = 40 years Twin a would send 39 pulses since B would arrive at year 40. For twin B, the universe is moving past him at 0.6c so I need to compute the length of the trip according to his point of reference.
Twin A is moving at a speed of 0.6c (60% of the speed of light) with respect to Twin B who is stationary on Earth. The star that Twin A is traveling to is at a distance of 12 light years. Therefore, the total time Twin A experienced the trip is calculated using time dilation in special relativity, i.e. Time_A = Time_B/sqrt(1 - (v^2/c^2)).
Question: Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals does A receive? (3) A. 16 B. 20 C. 32 D. 40 E. 50. Show transcribed image text. Here's the best way to solve it.
So the star six light-years away to the homebody appears to be only 4.8 light-years away to the traveler at a speed of 0.6c. Therefore, to the traveler, the trip to the star takes only eight years ...
The total travel distance and round trip will be 24 light years totally travel distance, let that be will not so for a due to the length contraction, the total distance will be L. That is a loss, not root 1 -3 square by c square. El norte is 20 food light use. This is route over 1 -0.6 squared. This is equal to 19.2 ideas.
Science. Physics. makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals d. makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth.
Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals does A send during the trip? А. 16 В. 20 С. 32 D. 40 O E. 50. Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth.
Science. Advanced Physics. 2. Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. (a) How many signals does A send during the trip?
Physics. Physics questions and answers. Twin A makes a round trip at 0.6c to a star 12light-years away, while twin B stays on the earth. Each twin sendsthe other a signal once a year by his own reckoning. (a) how manysignals does A send during the trip?
Tiwin A makes a round trip at $0.6 c$ to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning.
VIDEO ANSWER: Due to the length contraction, the total distance will be L, which is a loss. El norte uses 20 calories per light use. This route is over 1 -0.6…
Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals does B send during the trip? Twin A makes a round trip at 0.6c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal ...
Solution: Given: A and B are twins. A makes a round trip to star while B stays on Earth. Each twin sends the other a signal once a year by his own recoking. v = speed of twin A = 0.6c s = distance between Earth and …. Question 17 Twin A makes a round trip at 0.6c to a star 18 light-years away, while twin B stays on the earth.
Twin A makes a round trip at 0.6 c to a star 12 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. Each twin sends the other a signal once a year by his own reckoning.
Twin A makes a round trip at 0.6c to a star 18 light-years away, while twin B stays on the earth. Each twin sends the other a signal once a year by his own reckoning. How many signals. does B send during the trip. Expert Solution. This question has been solved!
VIDEO ANSWER: I would like to send a letter to a student. We have a question and answer about the round trip from Earth Point Oh view as the first signal is sent. It's Mhm. The first signal is sent one year after the launch and the…