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

2

Problem

In a lab frame, S, an object crosses a distance of 15 m in 10 s. In an initially aligned frame S', moving at 1000 km/s in the x-direction relative to S, how far a distance does the object have to travel, and in what time does it travel the distance?

A

0.99998889 s

B

9.9998889 s

C

1.0000056 s

D

10.000056 s

3

concept

Lorentz Transformations of Velocity

13m

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Hey, guys. So far we've talked about Lauren's transformations for individual positions. So if an object Zach, position X at time T in s. Where will it be? In s prime. What position? What time? Now we want to talk about how we use Lawrence transformations for velocity. And these are much, much more commonly used than just pure Lawrence. Transformations for position. Okay, so this is what you're much more likely to be tested on when you actually see questions about special relativity. All right, let's get to it. So remember, for Galilean relativity, it's easy to find a speed in some moving frame relative to some rest frame or vice versa. You just have simple addition of velocities. If a car is when we have 10 miles an hour and you throw a ball out the window at 20 miles an hour Ah, guy on the street sees the ball moving at 30 miles an hour. Okay, Addition of velocity is very simple. It's much, much more complicated when talking about a relativistic problem because of things like lame contraction and time dilation. Okay, remember the two things that we need for a Lorentz transformation? We need to have two frames, okay? With their axes and origins aligned at Time t zero, This actually isn't that important for velocity. Okay, this was incredibly important for position. But as long as if this is true, right, learning transformations will work for a position and velocity. Okay, The second one is definitely true. We need to choose a direction for the boost. And typically you're always going to choose that to be in the X direction. Okay, so these equations, they're gonna be written assuming that there is a boost in the X direction. All right, Now, what is the X component of the velocity on to be in s prime in the moving frame? Okay, well, it's going to be the X component in the rest. Frame s minus the speed of the boost divided by one minus the x component of the velocity times you divided by C squared. OK, this is exactly the Lorentz transformation of velocity along the X direction. What about the Why in the Z direction? Remember that positions don't transform unless the boost is going in those direction. There's no length contraction. I could just as easily have written this as delta y Delta Z. So now these air specifically distances, specifically lengths because there's no length contraction. You might want to say that there's not gonna be any change in the velocity, but bear in mind that the velocity is not just the distance. Right. Velocity is going to be that distance over time, right? And while the distances are going to remain the same, the time intervals will not remain the same, right. There is definitely going to be time dilation. And because of time dilation, you are going to change the denominator. It's very easy to see that you should have some sort of gamma term in the denominator in order to transform these. It's not going to be as simple as that, but you could see that it should have something to do with that. Okay, now the velocity in the Y direction in the prime frame is going to be the velocity in the Y direction in the UN prime frame, divided by gamma times this same term. Okay, it's very, very important to recognize that that velocity term in the denominator is in the direction of the boost its Vieques. It's not V y. Okay, and we get basically the same equation for Z. We're gonna get a gamma in the denominator one minus V x, you over c squared. And once again, we have that X component of velocity in the denominator there. We do not have the Z components. Okay? Those denominator terms are incredibly important because off time dilation. Okay, let's do a problem right here. Let me minimize myself to get out of the way. Okay, So a spaceship is passing the earth at five C from an observer on this ship. Right. So this is in s prime A missiles fired forward at 0.1 at sea. According to an observer on Earth. How fast is the missile moving? This is going to be our V X that we want to find. So we're going from a sorry hit my knee against the table. We're going from Vieques, Prime to Vieques. So this is from s prime to s. So this is sort of like the opposite transformation of what we've been doing. But don't worry, Doing the inverse transformation is really, really easy. And it uses the same equations they're just going to be. It's changed in one respect. So if this is s okay, and this is s prime moving forward at a speed. You This is absolutely equivalent. Okay, To s moving backwards at a speed you and s prime being stationary, right? These two absolutely equivalent So we can say that the transformation from s prime to s is going toe Look the exact same. Let's not see, that's a you as the transformation from S T s prime, the one that's been negative, there's just going to be one major difference. The one major difference is we're no longer dealing with you Now. We're dealing with negative. You write the direction is opposite because the direction is opposite. We're gonna pick up a negative sign. So this was negative. Now it's going to be positive this term was negative, but that you is going to become negative. So this is going to be positive. And this is exactly the Lorentz transformation we're going to use to go from s prime to s. Okay now, we were told that the missile, which is our thing that's moving, is moving at 0.1 c and the frame right, the frame is going to be the ship in this case is moving at 0. C. Right, This guy is you divided by one plus 0.1 C 0. c. Oversee squared. This is why it's really nice to use everything in terms of C because those were going to cancel. And if you plug this into your calculator, you will get 0. c. Okay, Now, this answer makes perfect sense. If we were just going toe, add the velocity is we would have gotten not 0.57 c, but 0.6. See, the problem with just adding the velocity is that as to get higher and higher and higher, you will eventually cross the speed of light. Right? If the ship fired the missile forward 2.6, the speed of light relative to its captain, it's going at 0.6 ships going at 0.5. If you just add those two numbers, that's 1.1 times the speed of light that violates special relativity. That cannot be it. So it has to be the addition, but slightly less and how slightly less it is Depends on how fast you're going right. The faster and faster and faster you're going the Mawr and more and more or less you're actually going to be. Then just the addition. What you would expect it to be. Okay. Another Lorentz transformation problem. Ah, spaceship passes the earth at 0.7 times. This to be the light from preserver on this ship. Missiles fired laterally. Okay, so here is the Observer on Earth. Here's the ship moving at a speed of seven C when it fires a little missile laterally at point to see relative to the ship. This point to see is the prime. But now this is actually the ex prime, right? Assuming that the boost is in the X direction, I'm going to assume that this is the X direction and that this I'll call the Y direction, actually, So this is V Y. Prime. This guy is still you right in the X direction. And what we're looking for is how fast is the missile? We're looking for V y non prime right in the s frame. So remember what's really important. Well, first of all, let's do the same thing to our Lawrence transformation equation for the Y component, as we just did for the X component. This was originally the X Times you over c squared But remember when going backwards from s prime toe s we're gonna take you and make it negative You So instead of this being a negative here, it's now a positive right away The thing that you have to remember Oh, sorry. This is also times gamma one plus the ex prime, You over C squared right off the bat. I think that you have to remember that you cannot say is that the velocity in a perpendicular direction to the boost is going to be the same because it's not because of time violation, it will be different. Okay, let's find what Gamma is really quickly. Let me minimize myself for this. Gamma is gonna be one over the square root of one minus. You squared over C squared the speed of the frame right we were told was seven c. So it's gonna be seven squared. Okay. And sorry. I wrote the wrong number in my calculator and my calculator. I used 0.2 times the speed of light. It's not point to because the frame is not moving at point to the speed of light. The frame is moving 0.0.0.7. The speed of light. So this is actually 1.4. Okay, We're not going to use 0.2 times the speed of light because that is the speed of the missile, not the speed of the frame. Okay, now, what about this term right here? Well, that is definitely zero, right? The missile has no component in the X direction. Imagine if it was fired at an angle. Well, then it would have a component in the X direction and a component in the Y direction. But that's not the case here. It's on Lee fired in the Y direction. So this term is absolutely zero, and all we're going to get is the speed in the Y direction right, which we were told was 0.2 times the speed of light divided by gamma, which were just calculated to be 1.4. And that is going to be a 0.14 times the speed of light. Okay, so the speed that you measured is definitely reduced because the time interval that you measure in the non proper frame which is on earth, the time that you measure is going to be longer than the time that you and measure in the proper frame on the ship. And because that distance in the Y direction the same. But you measure a longer time, you have a lowered velocity. Okay, So velocities acts absolutely do change in directions perpendicular to the boost. But it's actually a pretty easy change. All right. Okay, guys, Thanks so much for watching. And I'll see you guys in another video. Hopefully shortly.

4

Problem

In a particle accelerator, a neutron is traveling at a speed of 0.7 c, as measured by you in a laboratory. This neutron decays (becoming a proton), ejecting an electron. If you measure the electron’s speed to be 0.5 c, traveling in the same direction as the neutron, what was the relative speed between the electron and neutron when the neutron decayed? Was the electron ejected forward or backwards relative to the neutron’s motion, as “seen” by the neutron?