Consequences of Relativity: Study with Video Lessons, Practice Problems & Examples

Time dilation and length contraction are key consequences of special relativity. Time dilation occurs when time is measured differently in moving frames, leading to the concept that the dilated time (Δt') is always greater than the proper time (Δt₀), expressed as γ1 = 11-u2c2. Length contraction states that lengths measured in a moving frame are shorter than in the proper frame, with the contracted length (L') being less than the proper length (L₀) as L₀γ1.

Time Dilation

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. Besides that, it was done in a 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. Okay? 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. Okay? Remember that s primeprime, 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 2h, just twice the distance that it had to go from the bottom to the top. And so the time measured is going to be the distance divided by the speed, right, which is 2hc because light is traveling at c, 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 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. Okay? Let me minimize myself here really quickly. This height is the same. It's still just h. 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 2 times h. It's traveling this hypotenuse, which by definition has to be longer than h, that hypotenuse doubled. So it's traveling to l, not 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 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 2hc. The time measured in s is now going to be 2lc, 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 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 is 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. An outside observer would see the ball go up and then 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, 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, s prime. Where gamma is something called the Lorentz factor, and it's this denominator right here. 11-u2c2. Don't forget that u 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. Okay? In this case, the time in s sorry. Let me do this. Yeah. Okay. 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. 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 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. Okay, 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. Okay? 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. Okay, given our particular choice of 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, spaceships have to travel faster than 11.2 kilometers per second in order to escape the Earth's 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 gamma times the proper time. Now, if we look at Let me minimize myself really quickly. If we look at the spaceship right here, right 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. Okay? That's moving at this velocity u relative to s. Okay? 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 ticking relative to the clock in the ship, that's going to be the dilated time. Okay. Oftentimes, 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. Okay, so let's just look at gamma, basically. 1-u2/c2×deltatnot. 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 the 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 negative 8. So look at what you're doing. You're doing 1 minus 10 to the negative 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. Okay? Or maybe 0.9999-999 something. Alright? But it's most likely just gonna 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. Okay? 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.

Time Dilation for a Muon from the Atmosphere

Hey, guys. Let's see this problem. Okay? We have muons, which are very, very tiny charged particles similar to an electron but heavier. They're emitted from very high up in the atmosphere when high-energy particles from the sun collide with the atmosphere. So if here's the earth's surface, there's a bunch of right atmosphere, just air molecules, high-energy particles coming from the sun collide with atoms inside the atmosphere and produce muons. They're 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 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 light. But in the muon's own frame, s prime, right, where that frame is moving at 90% speed of light, the muon is static. Okay? 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. Okay. Because the event that we are interested in is the decay of the muon. Okay. 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. Okay? Now what exactly is that amount of time?

Well, Δt' = γ ⋅ Δt ₀ = 1 1 - u 2 c 2 ⋅ Δt ₀ . Okay? 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 2 c 2 is the same as u divided by c squared. So if we plug in 0.9c divided by c, you'll see that those speeds of light cancel. So this is just going to be 1 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. Okay? 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 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. Okay? 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. 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 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 the dilated time because we're measuring it in the lab frame, not in the proper frame. Okay? Alright, guys. That wraps up this problem. Thanks so much for watching.

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.

• A. 1.00000000033 s
• B. 0.9999999993 s
• C. 0.9999999997 s
• D. 0.0052 s

Length Contraction

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, 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 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. Right? Time is going to be longer. What length contraction says is if you measure the length in the proper frame, the length in 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 your contracted lengths, your non-proper lengths, should always be less than the proper lengths. Okay? Now, in order to understand where length contraction comes from, we need to imagine measuring a rod in two different ways. Okay? First, we're going to 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. Okay? Because if the clock is stationary in the lab frame and the rod is moving past it, that's the same in the lab's 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 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, 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. Only 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, the 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 it 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. 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 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, 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: L'=L0γ. And remember that because the Lorentz factor, γ, is always going to be larger than 1, the contracted length \( L' \) is always going to be less than the proper length \( L_0 \). Okay? This is the opposite logic for time dilation: in dilation, you get this equation with γ in the numerator. Since γ is always greater than 1, dilated time is always larger than proper time. For length contraction, because γ is in the denominator and γ is always larger than 1, 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 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? Right? That doesn't make any sense. So that 100 meters 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. Okay?

First, let's just solve for γ: 11-u2/c2. And like most problems, \( u \), the speed, is given in terms of the speed of light. Right? 10% speed of light means that \( u \) is 0.1 times \( c \). So this is one over the square roots of 1 minus 0.1 squared, and this is going to be 1.005. Okay? And then this leads us to the conclusion that the contracted length, which is 100 meters over γ, is actually going to be 99.5 meters. Okay? So half a meter short, shorter than it was. 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. 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.

Length Contraction for a Muon from the Atmosphere

Hey, guys. Let's do this problem. Okay. 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, 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 muon's rest frame, they last 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 traveling 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. Okay? 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 the muon's perspective. So from the muon's perspective, 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 muon. So how much of this atmosphere is going to pass the muon before it decays? Okay. 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 a chunk of atmosphere that is that's going to pass the muon before it decays. Right? That's just going to be the speed that it's going. So it's vt= 0.9 × 3 × 108 × 2.2 × 10-6 meters. The question is, is this the proper time or is this the contracted length? This is 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 at 90% speed of light. So how far would we measure the muon 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 the muon is moving through. So the proper length, 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 L0= 11-0.92 × 594 m. Now the Lorentz factor gamma, we got in the previous problem, and it was equal to roughly 2.29. Okay. Multiplying, Okay? So that is how far the muon will travel in the lab frame before decaying. 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 dilation 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 muon is traveling at 0.9 times the speed of light. So what time would we measure in the lab frame before it decays? Right? This is the dilated time because in the rest frame of the muon, we measure the proper time. In s prime, we measure the proper time in this case. So this is going to be gamma times delta t naught 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? Right? That's once again going to be velocity times time. So that'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 negative 6 seconds and that's going to be 1350 meters. 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 L0. 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 it 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 carried enough significant figures, these should be exactly equal. Not off by 10 meters, but that's because of rounding error. Alright. 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. Alright? Alright, guys. Thanks so much for watching. That wraps it up for this 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

Proper Frames and Measurements

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 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. Okay? 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 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, a ship passes Sally who's standing on the earth's surface watching the ship fly past her. Right? 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 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 going to be the shorter length, the proper length is going to 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 meters 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 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. 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 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. Okay? 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 problems with time dilation, you're basically going to be comparing 2 clocks. That's always the scenario. When we saw a problem with a 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 2 clocks. One clock moved with the muon and measured 2.2 microseconds. 1 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 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 2 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 far away. The time that the astronaut measured was 5 years. Oftentimes, the way that these problems are phrased are going to specifically, give some sort of preference to one of the 2 people. In this case, the preferred person is the astronaut. Right? 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, right, that time measured by the astronauts, 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. Behind. Okay? 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. Okay? 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.

Here’s what students ask on this topic:

What is time dilation in special relativity?

Time dilation is a phenomenon predicted by Einstein's theory of special relativity, where time is observed to run slower for an object in motion relative to a stationary observer. This effect becomes significant at speeds close to the speed of light. The relationship between the proper time (Δt₀) and the dilated time (Δt') is given by the equation:

${\mathrm{\Delta t\text{'}}}_{}=\gamma {\mathrm{\Delta t}}_0$

where $\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$ is the Lorentz factor, u is the relative velocity, and c is the speed of light.

What is length contraction in special relativity?

Length contraction is another consequence of special relativity, where the length of an object moving at a significant fraction of the speed of light appears shorter to a stationary observer. The proper length (L₀) is the length measured in the object's rest frame, while the contracted length (L') is measured in the moving frame. The relationship is given by:

${\mathrm{L\text{'}}}_{}=\frac{L_0}{\gamma }$

where $\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$ is the Lorentz factor, u is the relative velocity, and c is the speed of light.

How does the speed of light affect time dilation and length contraction?

The speed of light (c) is a fundamental constant in special relativity and plays a crucial role in both time dilation and length contraction. For time dilation, the time measured in a moving frame (Δt') is related to the proper time (Δt₀) by the Lorentz factor (γ), which depends on the speed of light:

${\mathrm{\Delta t\text{'}}}_{}=\gamma {\mathrm{\Delta t}}_0$

Similarly, for length contraction, the contracted length (L') is related to the proper length (L₀) by:

${\mathrm{L\text{'}}}_{}=\frac{L_0}{\gamma }$

In both cases, the Lorentz factor (γ) is given by:

$\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$

where u is the relative velocity. As u approaches c, γ increases, leading to more significant time dilation and length contraction.

What is the Lorentz factor and how is it used in special relativity?

The Lorentz factor (γ) is a key component in the equations of special relativity, describing how time and space are affected by relative motion. It is defined as:

$\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$

where u is the relative velocity and c is the speed of light. The Lorentz factor is used to calculate time dilation and length contraction:

1. Time Dilation: ${\mathrm{\Delta t\text{'}}}_{}=\gamma {\mathrm{\Delta t}}_0$

2. Length Contraction: ${\mathrm{L\text{'}}}_{}=\frac{L_0}{\gamma }$

As the relative velocity u increases, γ increases, leading to more pronounced effects of time dilation and length contraction.

How do proper frames and non-proper frames differ in special relativity?

In special relativity, a proper frame is a reference frame in which an object or event is at rest. Measurements made in this frame are considered 'proper' measurements. For example, the proper length (L₀) of an object is measured in the frame where the object is at rest. Conversely, a non-proper frame is a reference frame in which the object or event is in motion. Measurements made in this frame, such as the contracted length (L'), are affected by the relative motion. Proper time (Δt₀) is the time interval measured in the frame where the event occurs, while dilated time (Δt') is measured in a frame where the event is moving. Understanding the distinction between these frames is crucial for correctly applying the concepts of time dilation and length contraction.