===Middling===
===Middling===
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 – \frac{v^2}{c^2}} = (10\,years) \sqrt{1 – \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years – 9.17\,years) = 0.83\,years</math>.
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 – \frac{v^2}{c^2}} = (10\,years) \sqrt{1 – \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years – 9.17\,years) = 0.83\,years</math>.
Chaz Sporrer Fall 2024
Short Description of Topic
The Main Idea
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.
A Mathematical Model
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance [math]\displaystyle{ D }[/math] away at a velocity [math]\displaystyle{ v }[/math]. The twin traveling on the spaceship does not immediately have a velocity [math]\displaystyle{ v }[/math], but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth “appears” to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with [math]\displaystyle{ \Delta t }[/math] being the time for the round trip according to the twin on earth, and [math]\displaystyle{ \Delta t_0 }[/math] being the time according to the twin on the spaceship.
Time Dilation Formula: [math]\displaystyle{ \Delta t = \frac{\Delta t_0}{\sqrt{1 – \frac{v^2}{c^2}}} }[/math]
A Computational Model
To more easily see how this works, consider again the situation where one twin travels a distance [math]\displaystyle{ D }[/math] and back at a velocity [math]\displaystyle{ v }[/math]. The time [math]\displaystyle{ t }[/math] seen by the twin on Earth equals [math]\displaystyle{ \frac{2D}{v} }[/math], where [math]\displaystyle{ 2D }[/math] equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.
The formula for length contraction: [math]\displaystyle{ \Delta L = \frac{\Delta L_0}{\sqrt{1 – \frac{v^2}{c^2}}} }[/math]
Where [math]\displaystyle{ \Delta L_0 }[/math] is length measured by inertial frame, and [math]\displaystyle{ L }[/math] is length measured by observer moving at velocity [math]\displaystyle{ v }[/math] relative to the inertial frame. Thus, time [math]\displaystyle{ t_0 }[/math] according to twin on spaceship is [math]\displaystyle{ \frac{2D \sqrt{1 – \frac{v^2}{c^2}}}{v} }[/math]. Thus you can say [math]\displaystyle{ \frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 – \frac{v^2}{c^2}}}{v}} }[/math]. Rearranging, you get [math]\displaystyle{ t = \frac{t_0}{\sqrt{1 – \frac{v^2}{c^2}}} }[/math], which is the equation for time dilation.
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
Question 1: If there are two twins, and twin 1 travels a distance [math]\displaystyle{ d }[/math] away from twin 2 at a velocity [math]\displaystyle{ v_1 }[/math] and then travels back to twin 2 at a velocity [math]\displaystyle{ v_2 }[/math], which twin will be older?
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.
Middling
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed [math]\displaystyle{ 0.4c }[/math] ([math]\displaystyle{ c }[/math] = speed of light), and the distance between Earth and planet B is [math]\displaystyle{ 2 }[/math] light-years.
Solution: First, the distance of a light-year is equal to [math]\displaystyle{ (1\,year) \times c }[/math]. So time [math]\displaystyle{ t }[/math] to travel a round trip of [math]\displaystyle{ 4 }[/math] light-years = [math]\displaystyle{ 4 \times c \times (1\,year)/0.4c = 10\,years }[/math]. Using time dilation formula and rearranging for [math]\displaystyle{ t_0 }[/math] (time according to twin 2), we get [math]\displaystyle{ t_0 = t \sqrt{1 – \frac{v^2}{c^2}} = (10\,years) \sqrt{1 – \frac{(0.4c)^2}{c^2}} = 9.17\,years }[/math]. This means we can subtract [math]\displaystyle{ t_0 }[/math] from [math]\displaystyle{ t }[/math] to find how much older twin 1 is than twin 2. [math]\displaystyle{ (10\,years – 9.17\,years) = 0.83\,years }[/math].
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 [math]\displaystyle{ (9.17 years/2) = 4.59\,years }[/math] to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of [math]\displaystyle{ 10 }[/math] signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where [math]\displaystyle{ f_0 }[/math] is the frequency observed by twin 2 and [math]\displaystyle{ f }[/math] is the frequency observed by twin 1.
Relativistic Doppler Effect: [math]\displaystyle{ f_0 = f \sqrt{\frac{1 – u/c}{1 + u/c}} }[/math]
While traveling to planet B, [math]\displaystyle{ f_0 = (1/year) \sqrt{\frac{1 – 0.4c/c}{1 + 0.4c/c}} = 0.655/year }[/math]. On way back (towards Earth), [math]\displaystyle{ f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 – 0.4c/c}} = 1.53/year }[/math]. So on trip to planet, he receives [math]\displaystyle{ (0.655/year)(4.59\,years) = 3.01 }[/math] signals. On way back, he receives [math]\displaystyle{ (1.53/year)(4.59\,years) = 7.02 }[/math] signals. This means that over the entire journey, twin 2 receives [math]\displaystyle{ 7.02 + 3.01 \approx 10 }[/math] signals.
Difficult
Question 3: twin 1 travels to a planet [math]\displaystyle{ 10 }[/math] light years away and then back to Earth at a speed of [math]\displaystyle{ 0.8c }[/math]. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of [math]\displaystyle{ 0.999c }[/math]. Which twin is older when both twins have finished their journeys and by how much?
Solution: The time that passes on Earth during twin 1`s journey is [math]\displaystyle{ \frac{20\,light-years}{0.8c} = 25\,years }[/math]. The time that passes in twin 1`s reference frame is [math]\displaystyle{ t_0 = t \sqrt{1 – \frac{v^2}{c^2}} = 25\,years \sqrt{1 – \frac{(0.8c)^2}{c^2}} = 15\,years }[/math]. Twin 2 will leave [math]\displaystyle{ 10 }[/math] years after twin 1`s journey began, and it will take [math]\displaystyle{ \frac{10\,light-years}{0.999c} = 10.01 }[/math] Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is [math]\displaystyle{ t_0 = 10.01 \sqrt{1 – \frac{(0.999c)^2}{c^2}} = 0.45\,years }[/math]. In Earth years, there is still another [math]\displaystyle{ (25 – 10 – 10.01) = 4.99\,years }[/math] until twin 1 returns. So, during the [math]\displaystyle{ 25 }[/math] Earth years, twin 2 aged [math]\displaystyle{ 10 + 0.45 + 4.99 = 15.44\,years }[/math]. Because twin 1 aged [math]\displaystyle{ 15 }[/math] years during the [math]\displaystyle{ 25 }[/math] Earth years, twin 1 is [math]\displaystyle{ (15.44 – 15) = 0.44 }[/math] years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.
Connectedness
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.
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History
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.
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