Consequences of Relativity

Concept: Time Dilation

13m
Video Transcript

Hey guys, in this video we're going to talk about the specific consequences of special relativity starting with time dilation, alright let's get to it. Now time dilation is always introduced with the same thought experiment and it's pretty much the same thought experiment that Einstein used in 1905 to introduce the concept of special relativity the only difference is they didn't have lasers back then and we have lasers now with besides that it was done in a train imagine that we have a rest frame S which is shown immediately above me here my finger gets cut off so I can't quite point to it and then above that we have a moving frame S prime, so S prime is moving at the speed of the train so the train appears to be at rest in S prime and S is the lab frame so the train is just moving at whatever speed it's moving remember that S prime we're going to consider the proper frame because we are interested in what's happening inside the train in this instance and the lab frame as always is just the one that is at rest with respect to the earth so let's say that there was an observer inside the train so this is an observer inside S prime and they're watching as a ray of light leaves some source bounces off a mirror at the top of the train and then comes back down and is detected at the bottom and the distance from the floor to the mirror at the ceiling is some distance H. How much time is that observer going to measure passes well the distance traveled right was 2H just twice the distance that had to go from the bottom to the top and so the time it measured is going to be the distance divided by the speed right which is 2H over C because light is traveling at C the speed of light now what about the same experiment measured from an outside observer and observer in the lab frame while that observer isn't going to see light going up and down that observer is going to see light starting over here and traveling at an angle bouncing off the mirror and coming back down right because the train itself is moving so it's going to see light start here travel up and then travel down in this triangular direction.

Let me minimize myself here really quickly this height is the same it's still just H, but the light actually has to travel a further total distance right as we can see in this triangle right here that it's not traveling 2 times H it's traveling this hypotenuse which by definition has to be longer than H the hypotenuse doubled so it's traveling 2Lnot 2H. right and 2L is bigger than 2H. Now I do the actual math here just to show you what is involved in it but you don't need to worry about this it's just that this is clearly longer than H because it's H plus some number so far like I said here there's nothing strange about this this is just geometry now where the strangeness comes in is the next bit which is the fact that if we want to measure the time that the light took to travel that triangle we're going to need the speed of the light right however the speed of that light is the same in the lab frame as it is in the moving frame remember that we got that the time measured in S prime was 2H over C the time it measured in S is now going to be 2L. over C that same speed of light so you can see right away if the distances are different the amount of time that you measure to have passed has to be different and it has to be different because that freakin speed of light is the same in both frames this is why stuff gets so weird in special relativity it's because the speed of light has to be the same in both frames if you were just throwing a ball right inside the train the dude in the train was the ball up and down up and down that guy would measure some time it takes for the ball to go up and down an outside observer would see the ball go up and down and that would be a larger distance but a ball is not the same as light a ball's speed is not the same in both frames the speed is going to be different in both frames so if you did the same analysis what you would arrive at is the time measured in both frames is the same because the ball's speed is justifiably sorry not justifiably is different enough to compensate for that change in distance right that this distance right here is longer so this speed is going to be faster so the time measured is the same but that's not true for light.

Let me minimise myself. Everything here is all of the math required to come to the conclusion this right here to come to the equation that compares the speed of light measured in the moving frame to the speed of light measured in the rest frame just in case you need to know this for your class, but the important thing to take away is this sort of simplified equation that the time measured in S the lab frame is going to be something called gamma times the time measured in the moving frame S prime where gamma is something called the lorents factor and it's this denominator right here 1 over the square root of 1 minus U squared over C squared don't forget that U is the speed of the frame relative of the moving frame relative to the lab frame now once again I have been using from the beginning the concept of a proper frame and a lab frame in this case the time and sorry let me do this. So I did write it out like this so the time in S is actually known as the dilated time I'd wanted to talk about the proper time first but this is where we are it's the dilated time. and it's always going to be greater than the proper time, let me minimise myself here the proper time because the event remember that we were talking about was the laser going up and down inside the train that was the event we were interested in so when we are measuring the time taken for that light to travel from the laser at the floor of the train to the top and back that is the proper time so the time in S prime is called the proper time and the time in S is called the dilated time and if you look at gamma right here as U gets larger the denominator gets smaller and as the denominator gets smaller gamma gets larger so as U gets larger gamma gets larger so you are taking the proper time and multiplying it by a number greater than 1 so this dilated time is always going to be larger than the proper time. Now typically this is where the notation can get a little bit weird and we're going to continue using this notation from now on just because this is how people do it don't ask me why the dilated time is typically given by delta T prime now the reason why this can be weird is because the dilated time is actually the time in S not the time in S prime this notation delta T prime doesn't have anything to do with reference frame it doesn't have to do with S or S prime this is just the dilated time given our particular choice of S and S prime the time measured in S happens to be the dilated time and the proper time or the time it measured at rest with respect to the event right that is Tnot delta Tnot.

Let's do one quick example here. It just says space ships have to travel faster than 11.2 kilometers per second in order to escape the Earth's gravity this is called the velocity of Earth, we want to know can astronauts measure any noticeable amount of time dilation on a spaceship traveling at 11 kilometers per second basically. So the dilated time is going to be gamma times the proper time now if we look at. Let me minimise myself really quickly, if we look at the spaceship right here traveling fast away from the Earth's surface then the lab frame S is going to be the frame that an observer watching the spaceship leave the earth at rest on the earth is measuring right and then the astronauts inside the spaceship are going to be in a frame S prime that's moving at this velocity U relative to S now if we want to know how a clock inside the spaceship is ticking that's going to be the proper time and if we wanted to know how a spaceship sorry a clock on the earth is taking relative to the clock in the ship that's going to be the dilated time often times it's easy to remember that the moving clock measures time more slowly the moving clock will be the proper clock the stationary clock will be the dilated clock so let's just look at gamma basically 1 minus U Square over C squared times delta Tnot right this is going to be the square root of 1 minus let's just call it 10 kilometers per second so that's 10 times 10 to the 3 over 3 times 10 to 8 we can just call it 1 times 10 to the 8 because this number is actually going to be 0 anyway and then squared so what you're getting right here this number on the interior is going to be this is 10 to the 4 in the numerator 10 to the 8 in the denominator that's 10 to the negative 4 but you still have to square it so this whole number is going to end up being 10 to the -8 so look at what you're doing you're doing 1 minus 10 to the -8 and then you're squaring you're taking the square root of that if you plug that into your calculator it's going to tell you it's 1 or maybe 0.99999 something alright but it's most likely just going to tell you that it's 1 this means that astronauts traveling at this speed 11.2 kilometers per second do not notice any difference in time measured by their clocks relative to clocks on earth there is no noticeable time dilation for an astronaut on this ship leaving Earth at the regular escape velocity just 11 kilometers per second, so this wraps up our sort of introduction into time dilation and now we're going to follow this by some specific practice problems to get more comfortable with making these calculations. Alright thanks so much for watching guys.

Example: Time Dilation for a Muon from the Atmosphere

5m
Video Transcript

Hey guys, lets see this problem, we have muons which are very very tiny charged particles similar to an electron but heavier they are emitted from very very high up in the atmosphere when high energy particles from the sun collides with the atmosphere so if here's the earth's surface, there's a bunch of atmosphere just air molecules high energy particles coming from the sun collide with atoms inside the atmosphere and produce muons there given by the Greek letter mu those muons travel at 90% the speed of light and as measured in a lab right measured with respect to the middle muon it's going to decay at 2.2 microseconds right in their rest frame so that is the proper time because we are talking about muons moving at 0.9 the speed of light so this is a muon and this is measured in the rest frame so here's some dude at rest with respect to the surface watching a muon fly by at 90% the speed of lights but in the muons own frame S prime right where that frame is moving at 90% the speed of light the muon is static. The muon has no speed this is technically V prime it has no speed and it's going to decay in an amount of time of 2.2 microseconds this time is the proper time because the event that we are interested in is the decay of the muon the moving clock measures time more slowly so in the lab frame when this guy sees the muon fly by he's going to see the muon live for a longer time because he will be measuring the dilated time. Now what exactly is that amount of time well Delta T prime is going to be gamma times delta Tnot which is going to be 1 over the square root of 1 minus U squared over C squared times delta Tnot. Now most of these problems the speed U is going to be given in terms of the speed of light if we look at this term right here U squared divided by C squared is the same as U divided by C squared so if we plug N. 9C divided by C you'll see that those speeds of light cancel so this is just going to be 1 of minus 0.9 squared and this is typically how these problems are going to be given every now and then instead of giving it something like 0.9C they'll say like 3 times 10 to the 6 meters per second but most of these problems are going to be given in terms of the speed of light because it just makes the calculation more easy it just makes it easier.

So gamma the Lorentz factor if you plug this into your calculator you're going to get about 2.29 and this is times 2.2 microseconds which remember 2.2 microseconds is the proper time so the Lorentz factor says that the dilated time is 2.29 times larger than the proper time and this is going to be about 5 microseconds. So the time that an observer on Earth right in the lab frame measures for the muon to decay is 5 microseconds whereas the muon in its rest frame decays in 2.2 microseconds. This is a perfect example of what's the proper frame what's the lab frame what's the proper time what's the dilated time the event that we're interested in is the decaying of the muon and when we measure that time at rest with respect to the muon that's the proper time when we're watching the muon zip by the time that we're going to measure it taking to decay is going to be dilated time because we're measuring it in the lab frame not in the proper frame. Alright guys that wraps up this problem. Thanks so much for watching.

Problem: The international space station travels in orbit at a speed of 7.67 km/s. If an astronaut and his brother start a stop watch at the same time, on Earth, and then the astronaut spends 6 months on the space station, what is the difference in time on their stopwatches when the astronaut returns to Earth? Note that 6 months is about 1.577 x 107 s, and c = 3 x 10 8 m/s.

7m

Concept: Length Contraction

8m
Video Transcript

Hey guys, now we're going to start talking about the second consequence of the second postulate of special relativity which is length contraction. Alright let's get to it. Now because time is measured differently in different inertial frames so this is actually not its own consequence technically it is just a consequence of time dilation. Because time is measured differently in different reference frames, length is also going to be measured differently in different reference frames and this fact is known as length contraction. So we had time dilation which said that if you measure time in the proper frame time in the non-proper frame is going to be dilated. Time is going to be longer. What length contraction says if you measure the length in the proper frame, length in the non-proper frame is going to be contracted it's going to be shorter. So just be on the lookout for that that your contracted lengths your non-proper lengths should always be less than the proper lengths. Now in order to understand where length contraction comes from we need to imagine measuring a rod in two different ways.

First we're going to imagine measuring it in its proper frame which means at rest with respect to the rod, at rest with respect to the distance that we want to measure. Now because the frame that the rod is in is moving we want to imagine a clock that is stationary in the lab frame moving past the rod because if the clock is stationary in the lab frame and the rod is moving past it that's the same in the lab, sorry, in the rods frame in the proper frame as the clock which I'm holding in my right hand moving past that length and basically all we're going to do is we're just going to click the clock when we pass one end, let it pass the other end and click it off. So it's like a stopwatch when it clears the other end so we're just measuring how much time is elapsing as the clock passes and given that time, we will get some measured length based on how quickly the rod is moving. Now in the lab frame, instead of having a moving clock the clock is stationary remember that the clock was always stationary in the lab frame only when we are in the proper frame of the moving rod does the clock appear to be moving. Now the clock is stationary and the rod itself is moving past the clock so same exact idea, the rod is moving at the same speed U that the frame was moving, the proper frame, so this rod is going to pass the clock and we're going to click on when the rod just approaches the clock, start measuring time, click it off just as the rod leaves and we're going to measure a different time right because the time is different between the proper and the non-proper frame. We have time dilation so those two times that we measure have to be different. Now if you actually work through the equations you get that the length in the proper frame, remember the proper frame is the proper frame for the rod which means that the rod is at rest. The non-proper distance, the non-proper length is the one measured in this case in the lab frame and if you put them together you're going to get something that looks like this and if you use the time dilation equation you're going to end up with the proper length divided by gamma and remember that because the Lorentz factor gamma is always going to be larger than one, the contracted length L prime is always going to be less than the proper length L not. This is the opposite logic for time dilation because for time dilation you get this equation with gamma in the numerator since gamma is always greater than one dilated time always larger the proper time for length contraction because gamma is always in the, sorry, because gamma is in the denominator and gamma is always larger than one you always get a smaller non-proper length, a contracted length.

Very simple problem here to get us started in length contraction. A spaceship is measured to be 100 meters long while being built on Earth. That means that that is the proper length while it's being built on Earth we're assuming that the people who are building it and measuring it are at rest with respect to the spaceship. Why would they be building the spaceship as it flew by them? That doesn't make any sense so that 100 meter should be the proper length. Now if the spaceship were flying past somebody on Earth, they would measure the contracted length, the non-proper length of that spaceship because now that spaceship is moving past the observer at some speed. First let's just solve for gamma, that's 1 over the square root of 1 minus U squared over C squared and like most problems U, the speed, is given in terms of the speed of light. 10 percent speed of light means that U is 0.1 times C. So this is 1 over the square root of 1 minus 0.1 squared and this is going to be 1.005 and then this leads us to the conclusion that the contracted length which is 100 meters over gamma is actually going to be 99.5 meters. So half a meter shorter than it was. Basically half a percent shorter in length going 10 percent the speed of light which is very very very fast, you only get a half a percent of drop in length. Alright guys, that wraps up this video length contraction. It's not that big a deal it's actually much easier than time dilation because proper lengths are easy to recognize, it's just measured at rest with respect to the object and then applying length contraction super easy. Alright guys, thanks so much for watching and I'll see you guys probably in the next video.

Example: Length Contraction for a Muon from the Atmosphere

8m
Video Transcript

Hey guys, let's do this problem. We've already seen a problem basically the same as this with muons but we were looking at time dilation. Now we want to look at the length contraction aspect of it. So once again we have a bunch of atmospheric particles high up in the atmosphere that encounter these high energy particles emitted from the sun and every now and then and not every now and then it actually happens millions of times a second there is a collision that's going to produce these heavy particles that are like electrons called muons. Now in the muons rest frame they have a they last, I should say, 2.2 microseconds. They decay after 2.2 microseconds. We looked at how long the muons would last in the lab frame given the fact that they're travelling at 90% the speed of light. Now we want to look at how far they will travel in the lab frame but specifically we want to use length contraction.

I'll actually show after I solve this that you can use time dilation to arrive at the exact same answer, well roughly because of rounding errors, you would arrive at the exact same answer if you didn't have to deal with rounding because length contraction and time dilation are actually two different sides of the same coin. So let's look at this from the muon's perspective. So from the muon's perspective it's travelling, well sorry, it's at rest in a frame that's traveling at 0.9 times the speed of light so it's not moving but distance is rushing past it and by distance I mean atmosphere. So there's some amount of atmosphere right here that's rushing past the muon so how much of this atmosphere is going to pass the muon before it decays? That's pretty easy. The frame is going at 0.9 the speed of light, we know that it decays in 2.2 microseconds so let's just figure out how long chunk of atmosphere that is that's going to pass the muon before it decays that's just going to be the speed that it's going so it's 0.9 times the speed of light, 3 times 10 to the 8 times the amount of time that passes. Distance is velocity times time so this is going to be 2.2 microseconds and don't forget micro is 10 to the -6 and so this is going to be 594 meters the question is is this the proper time, sorry, the proper length or is this the contracted length? This is actually the contracted length because the proper length would be the one that we measure with respect to the Earth. We're talking about the Earth's atmosphere so if we are at rest with respect to the Earth we would measure the proper length of that atmosphere so this is actually the contracted length because the muon is not at rest with respect to the atmosphere, it's moving through the atmosphere at 90% speed a light so how far would we measure the muon travelling in the lab frame? That's actually the proper length, the lab frame represents the proper length because the lab frame is the one at rest with respect to that chunk of atmosphere that the muon is moving through. So the proper length sorry let me write out the length contraction equation. Length contraction says it's the proper length divided by gamma so the proper length is going to be gamma times the contracted length. This is going to be 1 over the square root of 1.1 minus 0.9 squared times 594 meters. Now the Lorentz factor gamma we got in the previous problem and it was equal to roughly 2.29. Multiplying these together you're going to get a distance of 1360 meters. So that is how far the muon will travel in the lab frame before decaying and this is found just using length contraction, no concept of time dilation was used here but like I said leading up to this solution we can still use time dilation and not worry about length contraction at all to solve this particular problem because in the lab frame, so this is S prime the moving frame which happens to be the proper frame for the time but the non-proper frame for the distance.

The lab frame is the proper frame for the distance but the non-proper frame for the time. So the muon is traveling at 0.9 times the speed of light so what time will we measure in the lab frame before it decays? This is the dilated time because in the rest frame of the muon we measured the proper time. In S prime we measure the proper time in this case. So this is going to be gamma times delta T not and we showed that this was about 5 microseconds in the previous problem so we can straight up just measure length in this case how far does it physically travel before it decays? That's once again going to be velocity times time so it's going to be 0.9 times the speed of light, 3 times 10 to the 8, times the amount of time that passes which is about 5 times 10 to the -6 seconds and that's going to be 1350 meters. So we have some rounding error between these but that's no big deal, the idea here is that time is proper in this frame, time is proper in this frame but length is non-proper. In this frame, length is proper by the way this is L not right that's the proper length but time is non-proper and so the whole idea is that length to get it to be contracted you have to take the proper and divide it by gamma. Time, to get to be non-proper, you have to take the time and multiply it by gamma and that divided by gamma and multiplied by gamma is going to cancel out when you compare the two results. So these should be if I cared enough significant figures these should be exactly equal not off by 10 meters but that's because a rounding error. Alright so in this particular problem we can easily see that length contraction is just a consequence of time dilation but in a lot of problems length contraction the equation is much easier to use. Alright guys, thanks so much for watching. That wraps it up for this problem.

Problem: In the following figure, a right triangle is shown in its rest frame, S'. In the lab frame, S, the triangle moves with a speed v. How fast must the triangle move in the lab frame so that it becomes an isosceles triangle?

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Concept: Proper Frames and Measurements

8m
Video Transcript

Hey guys, before moving on to the next topic I want to spend just a little bit more time talking about proper frames versus non-proper frames. Just to clear up any sort of inconsistencies that we have in those definitions. It's really really really important to know exactly what your proper frame is for the particular measurement that you're making and what the non-proper frame is for the measurement that you're making. We typically deal with lab frames and moving frames and sometimes the proper frame will be the lab frame and sometimes the proper frame will be the moving frame it just depends on what you are trying to measure. Now a proper frame is always a frame that is at rest with respect to, I'm just going to write here, something important. Now that seems extremely vague and the reason that's so is because it's extremely vague. Proper distance proper length is very very easy when you have an object the proper length of that object is always going to be measured at rest with respect to that object. So if you're talking about you know a spaceship which is the most popular example to use in special relativity because spaceships you can make go extremely fast then whenever you're measuring the length of that ship at rest with respect to that ship that's going to be a proper length. Let's say there was a person going very very very fast, if you wanted to know how the height of that person would change based on their speed you would need to know you would say that the proper height for that person, that proper length was measured when that person wasn't moving. So if we look at this example right here a ship passes Sally who's standing on the Earth's surface watching the ship fly past her. If Sally measures the length of the ship to be 100 meters is this the proper length or the contracted length? Now in some problems they will give you both lengths and they'll ask you to find the speed. We saw a problem with the triangle where that was true we were given both lengths and asked to find the speed. That's very easy because the contracted length is going to be the shorter length the proper length is going to the longer length but in problems like this where you are only given one length you need to know which is the proper length, which is the contracted length. In this case this 100 meters is the contracted length. Here's Sally and the ship is passing her at some speed and when that ship passes her she sees a length of 100 meters that ship is moving with respect to Sally so it's absolutely not the proper length that she is measuring. The proper length, this by the way we would call S, the lab frame. The proper length would be measured in a moving frame S prime and specifically a frame that moves with the ship. So the ship is at rest in this particular moving frame S prime and this length whatever that would be that would be the proper length. So it's really easy to understand for objects it's a little bit harder for times.

For time dilation it's often times more tricky to establish which is the proper time and which is the dilated time. Now like I had said the proper frame is always going to be the frame where the thing that you're interested in is at rest. So the proper frame is going to be the one at rest with respect to the clock you care about. Ran a little bit of space here. In problems with time dilation you're basically going to be comparing two clocks that's always the scenario. When we saw a problem with the muon we said that the muon took 2.2 microseconds to decay in its rest frame, how long is it going to take to decay in the lab frame? So that was essentially comparing two clocks, one clock moved with the muon and measured 2.2 microseconds, one clock was stationary while the muon flew past it and it measured a different time. The clock that we were interested, in the clock measuring the event that we cared about was the clock moving with the muon so that was the proper time and the lab frame time was the dilated time. If we look at this problem right here, an astronaut is leaving home on a long trip but before he goes he synchronizes a watch with his brother. These are very popular problems by the way. Allowing them to compare the amount of time that passes when he returns. During the astronaut's trip he measures himself to be 5 years older while his brother measures a different amount of time passing. Which of the two is measuring the proper time? Now one guy is stationed on Earth as the other astronaut or as the other brother, the astronaut is flying very very far away. The time that the astronaut measured was 5 years. Often times the way that these problems are phrased are going to specifically give some sort of preference to one of the two people. In this case the preferred person is the astronaut, the guy on Earth we're saying is just stationary the astronaut is flying away very very fast from that guy. All of this implies that the thing that we're interested in is the aging of the astronaut and then the brother on Earth is going to compare his age to the astronaut's age. So this would be the proper time, that time measured by the astronaut and then the guy on Earth would measure the dilated time so that when the astronaut came back he would be younger than his brother who stayed behind. So this is sort of how you want to approach these problems to figure out which is the proper time, which is the dilated time, which is the proper length, which is the contracted length. Like I said for length contraction it's very very easy, for time dilation it can be a little bit more difficult but you want to look at the problem and figure out what is the event that you care about which is the clock that you care about? If it's talking about something like a particle decaying like for the muon, it's very very easy because that's the thing that you're interested in the decay of the muon. For something like this it can be a little more difficult and a little bit more ambiguous but you want to figure out what is the event that's actually happening that you care about and in this event that is the astronaut aging. So that's going to be the proper time the brother is going to compare his time to the astronaut when he gets back and the brother at rest on Earth is going to measure the dilated time. Alright guys that wraps up this concept this, little last bit about time dilation and length contraction that we needed to talk about. Alright thank you guys so much for watching.

Consequences of Relativity Additional Practice Problems

A spaceship flies at a speed of 0.4c towards a stationary mirror. The ship sends a laserbeam at a wavelength of 500 nm towards the mirror, which reflects back towards the ship. What wavelength of light does the ship measure coming off of the mirror?

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It is known that the universe is expanding, so all stars are moving away from us, and the further away they are, the faster they are moving away from us. For convenience, there is a quantity known as redshift, z, that can describe how "far" an object is away from us. Redshift is given by the equation

1 + z = λ'/λo

where λo is the wavelength of light emitted at rest and λ' is the wavelength of light observered on Earth.

(a) Is redshift a distance measurement?

(b) If a star is supposed to emit light at a frequency of 4.57x1014 Hz, what would the observed frequency be if the redshift of this star were 0.1?

Watch Solution

Red light is emitted from a source at rest at a wavelength of 650 nm. How fast would this source have to be moving relative to you for you to see it as blue? Would it need to be moving towards or away from you?

Watch Solution

Samir (who is standing at rest on the ground) starts his stopwatch at the instant that Maria flies past him in her spaceship. She continues to fly her spaceship at the same constant velocity. According to Maria, at the instant that Samir’s stopwatch reads 8.0 s, Maria’s stopwatch reads 10.0 s. According to Maria, her spaceship is 100 m long (along the direction of motion). According to Samir, the length of Maria’s spaceship is

A. 64 m.

B. 80 m.

C. 100 m.

D. 125 m.

E. Not enough information is given to decide.

Watch Solution

An arrow of length 1.00 m flies by you at 0.8c. In the reference frame of the arrow, how long does it take to pass by you?

A. 0.75 ns

B. 1.25 ns

C. 2.5 ns

D. 4.2 ns

E. 6.9 ns

Watch Solution

An arrow of length 1.00 m flies by you at 0.8c. If you were to measure the time it takes to fly past you, what would you measure this to be?

A. 0.75 ns

B. 1.25 ns

C. 2.5 ns

D. 4.2 ns

E. 6.9 ns

Watch Solution

An arrow of length 1.00 m flies by you at 0.8c. You observe the length of the arrow to be

A. shorter than 1 m

B. 1 m in length

C. longer than 1 m

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Anna boards a spaceship and does an interstellar flight while her twin brother, Carlos waits on Earth. Anna’s ship travels at β = 0.80 and, according to Carlos, is gone for 10 years. According to Anna, how long was she gone?

A. 3.0 years

B. 6.0 years

C. 10. years

D. 16 years

E. 19 years

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A flight in a space ship going half the speed of light takes one hour as measured by a person in the airplane. How long did this flight take according to an observer on the ground?

A. 4.00 hr

B. 1.15 hr

C. Exactly 1.0 hour.

D. 0.500 hr

E. 0.563 hr

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A flight in an airplane going 300 m/s takes one hour as measured by a person in the airplane. How long did this flight take according to an observer on the ground?

A. 1.8 ns less than 1.0 hour

B. 3.6 ns less than 1.0 hour

C. Exactly 1.0 hour.

D. 3.6 ns more than 1.0 hour

E. 1.8 ns more than 1.0 hour

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In The Planet of the Apes, humans "traveled" into the future by hibernating on a spaceship traveling relative to the Earth for 120 years before returning. If the ship traveled at 99.99% the speed of light, how much time would have elapsed on Earth while they hibernated?

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Manufactured on Earth, a cube has a side length of 2.0 cm. As a spaceship flies past an observer on Earth, it fires the cube at a speed of 0.7 c relative to the observer on Earth. What volume for the cube would the observer on Earth measure?

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At what speed does time dilation have a "significant" effect? Assume that significant means a difference in 1% of dilated time and proper time.

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A subatomic particle called the pion has a measured lifetime of 4.4x10-8 s. If the lifetime of a pion measured at rest is 2.6x10-8 s, how fast is the pion moving in the lab frame?

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Muons are commonly produced in the upper atmosphere due to collisions of cosmic rays with the atoms in the atmosphere. The muons produced commonly travel downwards at speeds up to 0.99 c. If the half life of the muon, in the rest frame, is 2.2 μs,

(a) How long is the muon's half life, according to an observer at rest on the Earth's surface

(b) How far can the muon travel, according to an observer at rest on the Earth's surface?

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A spaceship passes an observer on Earth at a speed of 0.7 c. On the ship, an antenna has a length of 1 m and an angle of 15o measured from the surface of the ship. When passing the Earth-bound observer, what length and angle is the antenna observed to have?

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A spaceship passes an Earth-bound observer at a speed v. The observer on Earth measures the ship to be 300 m long, but an astronaut on the ship would measure it to be 400 m long. What is the speed of the ship, v?

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An astronaut travels at 0.8 c from Earth to a star that is 20 light years away from Earth, as measured from the Earth's perspective. If the astronaut is 25 years old when leaving Earth, what is the age of the astronaut when arriving at the star?

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Ted travels in a spaceship at 0.45 c relative to Alice, who's at rest on the Earth's surface. Ted travels for 20s, as measured on his watch. 

(a) Who measures the proper time, Ted or Alice?

(b) How much time elapses on Alice's watch during this motion?

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