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.