36. Special Relativity

Lorentz Transformations

# Lorentz Transformations of Velocity

Patrick Ford

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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.

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