36. Special Relativity

Lorentz Transformations

# Lorentz Transformations of Position

Patrick Ford

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Hey, guys, in this video, we're going to start talking about Lawrence Transformations. When we first introduced relativity, we talked about the addition of velocities or the addition of speeds, which was the way in Galilean relativity to take a measurement in one reference frame and move it to an inertial reference frame. Now, in special relativity, we can't do something as simple as addition to speeds as edition of speeds. But we can still move measurements from one inertial frame to another, using more complicated Lawrence transformations. All right, let's get to it. All right, So, like I said in Galilean relativity, the process of moving one measurement into another relatively simple Okay, the position in one reference frame, um, is easily relatable to the position in another reference frame, based on how quickly that reference frame is moving relative to the first. So in X, if you make a measurement of sorry s if you make a measurement of X, and as prime is moving at a velocity or speed you relative to, um s then ex prime is just gonna be X plus or minus You times t. It depends on the relative direction. Okay, but it's basically just the initial position in s plus how far the frame moves in your time measurement teeth. Okay, now it's much, much more complicated in special relativity, because this thing right here, time sort of messes it up for everybody. Because you can't just say that the time measured in one frame is equal to the time measured in another frame. You have to consider time dilation. It makes the complication the calculation. Ah, lot more complicated. That's what I said here. The time duration is not the same between two inertial frames. Okay, So, Lawrence transformations are what are going to allow us to relate these measurements in different frames, taking into account phenomena like time dilation and length contraction. Okay. Galilean relative Galilean Transformations like this one right here will always work if the speed is low enough. And that's something that we're going to see in the example below. Okay, um but they are not gonna work as you get closer and closer and closer to the speed of light. And once again, just like we've been saying the entire time we've been talking about relativity. This Onley works. If you are considering inertial frames, it does not work if you're talking about accelerated reference frames like non inertial frames. Okay, let me minimize myself here. So to use Lawrence transformations, we need Thio. Have the two things Okay first. Well, first of all, we need thio. Inertial frames here were saying s is the rest frame just like we've been using in all of our other videos. The convention that we've been using so far and as prime is theme moving frame and they are inertial. So the velocity you is constant. Okay, Now what we're choosing is that their origins are aligned. Okay? ATS the initial time. So this is a some time t the corn systems. They're gonna look like this. But originally, here's why. Prime, Here's the prime. Here's ex prime. This is at T prime equals T equals zero. We're choosing to a line their origins. Okay, Now what we need to do is we need to choose a particular orientation of our frames such that the boost, which is what we call this relative velocity. I've probably used that word before, but the boost is along some axis. Commonly. We are all we're going to choose the X direction if you look at the images here, the access along which the boost lies is the X axis. There is not going to be any relative velocity for the wide axes or for the Z axes. They're not gonna be any relative velocity. So, like we saw when talking about length contraction, there is not going to be any length contraction along those axes. Length contraction will only apply along the X direction because that's the direction of the boost. Okay, so now we can actually write down our transformations. Our coordinate transformations. Now, first of all, because there's no boost in the why or the Z direction, that means that there's no length contraction along those directions. So there's no change in those positions that why prime equals Y Z prime equals Z, right? If this is s and this is s prime and this is X, and this is ex prime where this is some speed, you you can see that if I were to measure something here that is going toe have a completely different measurement from the origin in s prime. Okay? Because as time goes on, ex prime is moving away. Started the origin of ex prime is moving away and there's also length contraction due to relativistic effects. But if I were to look at why versus why prime and let's say that this event occurred here where this was, why not weaken? See that? Because there's no relative velocity between the UAE directions that why not? Prime is going to be the same length as Why not? There's no relative velocity. There's no length contraction. There's nothing. And the same sort of thought applies to the Z direction as well. Okay, so Onley in the X direction or we're gonna get length contraction and for time. We're also gonna have time dilation, so we need to take that into account as well. The equation is gamma, where this is the same Lawrence factor that we've used times. Time minus you x over C squared. Where you remember is the boost. Okay, that is the boost velocity and then ex prime is going to be gamma again, right? This is special activity. Pretty much every equation is going to have a gamma in it, a Lawrence factor. And this is going to be X minus you t OK. So interestingly enough, if we look at our ex prime equation. It looks exactly like what we had for Galilean relativity, just with a factor of gamma in front of it. Because that gamma, when you multiply it out, you're going to get length contraction from the first term, and then you're going to get time dilation with the second term. That's basically what happens. The time equation is a little bit more complicated. The first term in the time equation is clearly just time dilation. But the second term there's no easy way to just look at it and be like, Oh, yeah, that second term is obvious. So I'm not gonna talk about that, Okay? Just accept the second term as fact it has to be there, and there's a mathematical reason why there's just not a good sort of intuitive reason why. Okay, so let's take a look at this problem. We have two frames SNS prime, which have their origins aligned at this time, right? That's one of those requirements, and we have a boost of 580 kilometers per second in the X direction. Remember, that's the second requirement, and by convention, these air typically going to have boosts in the X direction so that we can always use these above. Lawrence Transformations. What is the position and time in as prime? Okay, so first of all, the position in s prime is going to be a gamma. The position in S u times the time. By the way, this time, this position, this time, this position those are always measured in s just look at them. They don't have primes, so they're not measurements in s prime. So before we could do anything, we do need to know gamma. And remember, gamma is one over the square root of one minus. You squared over C squared. So this is gonna be one over the square root of one minus 580 kilometers per second is the same as 580 times. 10 to the 3 m per second speed of like three times. 10 to the 8 m per second squared. And this whole thing is gonna be 1. So if you look basically one, there's gonna be very, very, very little effects due to relativity because you're not really going that fast. Okay, so 1. X is going to be in units of meters just because our speed is in kilometers per second. So it's easiest to put everything in meters you could represent this is centimeters but then this would be 580 times 10 to the three to get 2 m times 10 to the to to get it to centimeters per second. And then everything gets all messed up Like it would just be easier if we stuck in meters for now. So 0.1 m minus the speed 580 times 10 to the three meters per second times. Five seconds is how long this occurs. Obviously this number is going to be way, way, way bigger than 10 centimeters. So this is going to be negative. 2,900,000 and five meters. Okay, notice that you times t let me actually write this in red just because it's easier Notice that you times t if you plug this and for you to send for tea is actually just 2,902,000 and 0 m. So you can see that the length contraction is 5 m. Okay? And we know that the length contraction should be very, very small because we're not actually going that fast. Okay? And now time is going to be gamma time measured in the s frame. The rest frame minus you. X over C squared gamma once again. Time measured is five seconds minus. The speed was 580 kilometers per second times. So times 10 to the 3 m per second. The position 01 m. You need to get this in the same units. Okay, so you wanna ultimate meters and then this is going to be three times 10 to the 8 m per second squared. This number right here is going to be really, really, really, really small. Okay, Right. It's 580 times 10 to the to, because that 0.1 drops x one by one, right 580 times 10 to the two divided by nine times 10 to the 16. So that number is basically zero. Okay, so what you're going to get is you're going to gamma times, the time interval, and that's just link. That's just time dilation. Right. So listen to being a 5.1234 five, 935 seconds. Okay, let me put that. But that so you can see that the time dilation is all the way out here at 10. To the negative. Six seconds. Okay, nine times 10 to the negative. Six seconds. Is the time dilation very, very small, as we expected, because gamma is very, very near one, because we're going pretty slowly. All right, so this is the basics of the Lawrence transformation. What you're essentially doing is time dilation and length contraction just at once. Okay? And it's often times much, much easier to relate these two because you don't need to worry about what's the proper frame. What's the non proper frame? You know? S you know s prime. You know the boost of s prime. You just plug in the numbers into equations and you're done. Alright, guys, that wraps up this discussion on Lawrence Transformations. We're gonna follow us up some practice problems. All right, I'll see you guys in another video. Thanks so much for watching

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