Consequences of Relativity - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Time Dilation

Video duration:

12m

Play a video:

Hey, guys, In this video, we're gonna talk about the specific consequences of a special relativity, starting with time dilation. All right, 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. Besides that, it was done in the train. Okay, 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 as prime is moving at the speed of the train, so the train appears to be at rest and s prime and s is the lab frame. So the train is just moving at whatever speed it's moving. Okay, 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. Okay, 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. Okay. And the distance from the floor to the mirror at the ceiling is some distance. H Okay. How much time is that observer going to measure passes. Okay, well, the distance traveled, right was to H just twice the distance that it had to go from the bottom to the top. And so the time measured is gonna be the distance divided by the speed. Right? Which is to h oversee, because light is traveling at sea the speed of light. Now, what about the same experiment? Measured from an outside observer, An observer in the lab frame. Well, that observer isn't going to see, like going up and down that observers 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 gonna see light start here, travel up and then travel down in this triangular direction. Okay, let me minimize myself here really quickly. This height is the same. It's still just a church. Okay, 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 to times H. It's traveling this high pot news, which by definition has to be longer than H that high pot news doubled. So it's traveling to L not to h right into l is bigger than two h. 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 per se. It's just that this is clearly longer than H because it's h plus some number. Okay, 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 to h oversee. The time measured in s is now going to be to l over See that same speed of light so you can see right away that 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 freaking speed of light is the same in both frames. This is why stuff gets so weird and 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 trains throwing the ball up and down, up and down that guy would measure some time. It takes for the ball to go up and down on outside observer, would see the ball go up and then down, and that would be a larger distance. But ah, ball is not the same as light. Ah, balls. 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. Now, let me minimize 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. Okay, just in case you need to know this for your class, Okay? 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 as prime where gamma something called the Lawrence Factor. And it's this denominator right here. One over the square root of one minus. You squared over C squared. Don't forget that you is the speed of the frame relative of the moving frame relative to the lab frame. Okay, Now, once again, I had been using from the beginning the concept of a proper frame and a lab frame in this case, the time and s Sorry. Let me do this. Yeah. Okay. So I did write it out like this. So the time and s is actually known as the dilated time. I had wanted to talk about the proper time first, but this is where we are. It's the dilated time, okay? And it's always going to be greater than the proper time. Let me minimize 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're 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 and s is called the dilated time. And if you look at Gamma right here. As you gets larger, the denominator gets smaller and as the denominator gets smaller, gamma gets larger. Okay, so as you gets larger, gamma gets larger. So you are taking the proper time and multiplying it by a number greater than one. So this dilated time is always going to be larger than the proper time. Okay, now, typically, this is where the notation could 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 an s not the time and 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, Okay? Given our particular choice off s and s prime, the time measured in s happens to be the dilated time, okay? And the proper time or the time measured at rest with respect to the event, right? That is t not right? Delta t not. Okay, let's do one quick example here. It just says, um, spaceships have to travel faster than 11.2 kilometers per second in order to escape the earth. Gravity. This is called the escape Velocity of earth. Okay, 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 a gamma times the proper time. Now if we look at mineralized myself really quickly if we look at the spaceship right here, right? Okay. Traveling fast away from the Earth's surface than 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. Okay, that's moving at this velocity. You relative toe s. Now, if we want to know how a clock inside the spaceship is ticking, that's going to be the proper time. Okay? 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. Okay, Often times it's easy to remember that the moving clock measures time or slowly. The moving clock will be the proper clock. The stationary clock will be the dilated clock. Okay, so let's just look at Gamma. Basically one minus. You squared over C squared times Delta t not right. This is going to be the square root of one minus. Let's just call it 10 kilometers per seconds. Let's 10 times 10 to the three over three times. 10 to the eight. We could just call it one times 10 to 8 because this number is actually going to be zero 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 four in the numerator. Tend to the eight in the denominator. That's 10 to the negative four. But you still have to square it. So this whole number is going to end up being 10 to the negative eight. So look at what you're doing. You're doing one minus 10 to the negative eight, 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 one okay. Or maybe 0.999999 something. All right, but it's most likely just gonna tell you that it's one. 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 in a foreign astronauts on this ship leaving Earth at the regular escape velocity just 11 kilometers per second. OK, so this wraps up our sort of introduction into time dilation, and now we're gonna follow this by some specific practice problems to get more comfortable with making these calculations. All right, thanks so much for watching guys.

2

example

Time Dilation for a Muon from the Atmosphere

Video duration:

4m

Play a video:

Hey, guys, let's see this problem. Okay? We have mu ons which are very, very tiny charged particles similar to an electron, but heavier. They're admitted 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 right atmosphere, just a Ramallah cules. High energy particles coming from the sun collide with atoms inside the atmosphere and produce Mulan's. They're given by the Greek letter mute those Mulan's travel at 90% the speed of light and, as measured in a lab right, measured with respect to the mu on, it's going to decay a 2.2 microseconds right in their rest frame. So that is the proper time, because we're talking about Mulan's moving at 0.9 the speed of light. So this is a mu on, and this is measured in the rest frame. So here's some dude at rest with respect to the surface, watching Um, yuan fly by at 90% the speed of light but in the yuan's own frame as prime right where that frame was moving at 90% speed, the light, the Muan is static. Okay. The muan has no speed. This is technically the 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. Okay, because the event that we're interested in is the decay of them Yuan. Okay, The moving clock measures time Mawr Slowly. So in the lab frame. When this guy sees them, yuan fly by. He's going to see them. You on live for a longer time because he will be measuring the dilated time. Okay, Now, what exactly is that amount of time? Well, Delta T Prime is going to be gamma times Delta t not. Which is going to be one over the square root of one minus. You squared over C squared times. Delta t not. Okay. Now, most of these problems the speed you is going to be given in terms of the speed of lights. If we look at this term right here, you squared. Divided by C squared is the same as you divided by C squared. So if we plug in 0.9 c divided by C, you'll see that those speeds of light cancel. So this is just gonna be one minus 10.9 square. And this is typically how these problems they're going to be given every now and then instead of giving it something like 0.9 c, they'll say, like three times 10 to the 6 m per second. But most of these problems, they're going to be given in terms of the speed of light because it just makes the calculation more easy. It just makes it easier. Okay, so gamma. The Lawrence 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 Lawrence Factor says that the dilated time is 2. times larger than the proper time, and this is going to be about five microseconds. So the time that an observer on Earth right in the lab frame measures for them you want to decay is five microseconds, whereas the mu on in its rest frame decays in 2. microseconds. Okay, so 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 them yuan. And when we measure that time at rest with respect to the mu on, that's the proper time. When we're watching them, yuan zip. By the time that we're going to measure it, taking to decay is going to be the dilated time. Because we're measuring in the lab frame, not in the proper frame. Okay, Alright, guys, that perhaps with this problem, thanks so much for watching.

3

Problem

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 10^{7} s, and c = 3 x 10 ^{8} m/s.

A

1.00000000033 s

B

0.9999999993 s

C

0.9999999997 s

D

0.0052 s

4

concept

Length Contraction

Video duration:

7m

Play a video:

Hey, guys. Now we're going to start talking about the second consequence of the second part of special relativity, which is length contraction. All right, 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. Okay, 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. Okay, so we have time dilation, which said that if you measure time in the proper frame time and the non proper frame is going to be dilated, right, time is going to be longer. If what length contraction says is if you measure the length and the proper frame lengthen, the non proper frame is going to be contracted, it's going to be shorter. Okay, so just be on the lookout for that that you're contracted lengths. Your non proper length should always be less than the proper lengths. OK, now, in order to understand where length contraction comes from, we need to imagine measuring a rod in two different ways. First, we're gonna imagine measuring it in its proper frame, which means at rest with respect to the rod. Okay, 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 a stationary in the lab frame and the rod is moving past it, that's the same in the labs. Sorry. In the rods frame in the proper frame as the clock right, which I'm holding in my right hand moving past that length. Okay. And basically all we're going to do is we're just gonna 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. Okay? 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 Onley When we are in the proper frame of the moving rod, does the clock appear to be moving? Right now, the clock is stationary and the rod itself is moving past the clock. So same exact idea. The rod. It's moving at the same speed you that the frame was moving. Um, the proper frame. So this rod is going to pass the clock, and we're gonna click it on. When the rod just approaches the clock, start measuring time, click it off just as a rod leaves. And we're gonna measure a different time, right? Because the time is different between the proper and the non proper frame. Right? 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 and the proper frame remember, the proper frame is the proper frame for the rod, which means that the rod is rest at rest. Okay, the non proper distance, right? 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 Lawrence 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 Okay. This is the opposite logic for time dilation because for time dilation, you get this equation with gamma in the numerator. Since GAM is always greater than one dilated time always larger than proper time for length contraction. Because gamma is always in the did not sorry, because gammas in the denominator and gamma is always larger than one. You always get a smaller, non proper length, right? A contracted length. Okay. Very simple problem here to get us started in length contraction. A spaceship is measured to be 100 m 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 right? That doesn't make any sense. So that 100 m 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 off that spaceship. Because now that spaceship is moving past the observer at some speed. Okay, First, let's just sold for gamma. That's one over the square root of one minus. You squared over C squared. And like most problems you, the speed is given in terms off the speed of light, right? 10% speed of light means that U is 0. times. See? So this is one over the square root of one minus 0. squared and this is going to be 1005 Okay. And then this leads us to the conclusion that the contracted length, which is 100 m over gamma, is actually going to be 99. m. Okay, so half a meter short, shorter than it waas, right. Basically half a percent shorter in length, going 10% the speed of light, which is very, very, very fast. You only get a half a percent of drop in length. Okay, Alright, guys, that wraps up this video on length contraction. It's not that big a deal. It's actually much easier than time dilation because proper lengths are easy to recognize its 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.

5

example

Length Contraction for a Muon from the Atmosphere

Video duration:

7m

Play a video:

Hey, guys, let's do this problem. Okay? We've already seen a problem basically the same as this with neurons. But we were looking at time dilation. Now we wanna 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 Mulan's. Now in the Mulan's rest frame, they have a, um they last, I should say, 22 microseconds. They decay after 22 microseconds. We looked at how long the yuan's would last in the lab frame, given the fact that they're traveling at 90% the speed of light. Now we wanna look at how far they will travel in the lab frame, but specifically we want to use length contraction. OK, 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. Okay, so let's look at this from them. Yuan's perspective. So from them yuan perspective, right? It's traveling 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 right, and by distance, I mean atmosphere. Okay, so there's some amount of atmosphere right here that's rushing past the mu on. So how much of this atmosphere is going to pass the muan before it decays? Okay, that's pretty easy. The frame is going at the speed of lights. We know that it decays into two microseconds. So let's just figure out how long chunk of atmosphere that is that's going to pass the muan before it decays, right? That's just going to be the speed that it's going. So it's 0.9 times the speed of light three times 10 to the eight times the amount of time that passes. Right distance is velocity times time, so this is going to be 2.2 microseconds. And don't forget Micro's 10 to the negative six. And so this is going to be meters. The question is, Is this the proper time, or is this the 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, right? 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 the strength of the atmosphere, it's moving through the atmosphere at 90% speed of light. So how far would we measure them? You on traveling in the lab frame That's actually the proper length, right? The lab frame represents the proper length because the lab frame is the one at rest with respect to that chunk of atmosphere that them you want 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 gonna be one over the square root of one point one minus 10.9 squared times, 594 m. Okay, now the Lawrence Factor Gamma. We got in the previous problem, and it was equal to roughly two point Okay, multiplying these together, you're going to get a distance of 13 m. Okay, so that is how far the neuronal travel in the lab frame before decayed. Okay? 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 violation and not worry about length contraction at all to solve this particular problem. Because in the lab frame. So this is s prime, right? The moving frame, which happens to be the proper frame for the time. Right. 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. Okay, so the yuan is traveling at 0.9 times the speed of lights. So what time would we measure in the lab frame before it to case this is the dilated time? Because in the rest frame of the yuan, we measured the proper time in s prime. We measure the proper time in this case. So this is gonna be gamma times Delta t not. And we showed that this waas thought about five 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? Right, That's once again going to be velocity times time. So it's going to be a 0.9 times the speed of light three times 10 to the eight times the amount of time that passes, which is about five times 10 to the negative six seconds. And that's going to be 13 50 m. Okay, 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, right time is proper in this frame, but length is non proper, right? 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 divided 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. Right. So these should be. If I had carried enough significant figures, these should be exactly equal, not off by 10 m. But that should because a rounding error. All right, so in this particular problem, we can easily see that length contraction is just a consequence of time dilation. Alright, but in a lot of problems length contraction, the equation is much easier to use. All right. All right, guys, Thanks so much for watching that wraps it up for this problem.

6

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?

A

0.16c

B

0.84c

C

1.84c

D

1.19c

7

concept

Proper Frames and Measurements

Video duration:

7m

Play a video:

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. Okay, just to clear up any sort of inconsistencies that we have in those definitions. Okay, 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're trying to measure now. Ah, proper frame is always, ah, frame that is at rest with respect to I'm just gonna right here something important now that seems extremely vague. And the reason that's so is because it's extremely vague. Um, proper distance. Proper length is very, very easy. When you have an object, the proper length of that object is always gonna be measured at rest with respect to that object. Okay, so if you're talking about, you know, a spaceship, which is the most popular um, 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, right that proper length was measured when that person wasn't moving. Okay, so if we look at this example right here, ah shit passes Sally, who's standing on the earth's surface watching this ship fly pastor, right? If Sally measures the length of the ship to be 100 m, 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 this speed. We saw a problem with a 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 gonna be the shorter length. The proper length is gonna be 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 m is the contracted length. Okay, here's Sally and the ship is passing her at some speed. And when that ship passes her, she sees a length of m. 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 as prime and specifically a frame that moves with the ship. So the ship is at rest in this particular moving frame as prime and this length, whatever that would be, that would be the proper length. Okay, so it's really easy to understand. For objects, it gets a little bit harder for times. Okay. For time dilation, it's oftentimes mawr tricky to establish, which is the proper time in which is the, um, 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 out of space here. Okay, In, um, 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 Milan, we said that the Mulan took 2.2 microseconds to decay. And it's rest frame. How long is it gonna take to decay in the lab frame? So that was essentially comparing two clocks, one clock moved with them yuan and measured 2.2 microseconds. One clock was stationary while the muon flew past it and 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 mu on. So that was the proper time. And the lab frame time was the dilated time. If we look at this problem right here on astronaut is leaving home on a long trip. But before he goes, he synchronizes the watch with his brother. These were very popular problems, by the way, allowing them to compare the amount of time that passes when he returns during the astronauts trip, He measures himself to be five years older while his brother measures a different amount of time passing. Which of the two is measuring the proper time? Okay, 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 five years. Oftentimes, the way that these problems were phrased are going to specifically, um, give some sort of preference to one of the two people. In this case, the preferred person is the astronaut, right? The guy on earth we're saying is just stationary. The ash not 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 astronauts age. So this would be the proper time, right that time measured by the astronauts and then the guy on Earth would measure the dilated time so that when astronaut came back, he would be younger than his brother who stayed behind. Okay, so this is sort of how you want to approach these problems. Thio 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 violation. It can be a little bit more difficult, but you wanna 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 them yuan, it's very, very easy, because that's the thing that you're interested in the decay of them yuan. For something like this, it could 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 gonna 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. Okay, Alright, guys, that wraps up this concept this little last bit about time dilation and link contraction that we needed to talk about. Alright, Thank you guys. So much for watching.

Do you want more practice?

We have more practice problems on Consequences of Relativity