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6 - 32 - What Happens With Perpendicular Velocities- (12-09, Low-Def) (1)

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  In a previous video clip we looked at combining velocities, really adding velocities is what we were doing when we had motion in the x direction. So remember we used the Lorentz Transformation and you got an equation like this and what I've done here is added these x subscripts to the velocity is, use of r remember is the velocity in the rocket frame, our example was Bob, here, going with some velocity, and Alice is here, say, get her in in the picture, going with some velocity, v, with respect to Alice's frame of reference, and shot off in an escape pod in his direction of motion, with velocity u sub r compared to him. And we discovered, then, that the velocity that Alice would observe. The escape pod to go follow this equation here. And we did a few examples with that just to check out how it might work. And in particular we saw that even though the two velocities here, the velocity of Bob with respect to Alice, of his spaceship, and then plus the velocity of the escape pod. If they were both, if they both added up so they were greater than c, just by adding them together, this equation would be such that the result would never be greater than c, in terms of what Alice observed, in terms of the velocity of the pod. What we want to do now is do another case, where, what happens if Bob shoots it up maybe straight up, or just perpendicularly off to the side here? And, figure out, what would Alice see in that situation? And interestingly enough, this will bring us back to the light clock that we are talking about, recently. So, here's a situation. Bob shoots it up at some velocity u sub r, and we'll use a subscript y to in, emphasize this is just in the y direction, now. You can say it's the z direction but we'll choose the y direction for, for that direction. And of course definition of velocity, delta y sub r over delta t sub r. This is for Bob, now, he says, how far does it go in the y direction? And what's the elapsed time for how far it's going in that direction? That's going to be the definition of velocity.  And so now let's write down a couple of equations, actually, for u sub ly. That's what we're after, here. What is, what does Alice observe the velocity in the y direction to be? So U sub ly is going to be delta yl over delta tl, just by definition of velocity. And this is why I, as a reminder I added the y and z components for our Lorentz Transformation, remember they don't transform at all.They're the same. And so delta y sub l is going to be the same as delta y sub r. So this is going to be delta y sub r over delta t sub l. But that does change, of course, by our equation we used before. Remember we could use deltas in here to indicate change of quantities. So we'll write that as delta y sub r on the top, and then delta t sub l, remember, using the equation here, is going to be gamma times delta t sub r plus v over c squared, delta x of r. Right? Again, these are the quantities that that, that Bob is measuring. And we, we're assuming everything is constant velocity here. For those of you who've had some physics, you say, well what if it's accelerating? Well we're not dealing with that case here. And therefore we can define velocity as simply change in, not change in direction, but the direction, elapsed direction if you want, the direction covered divided by the elapsed time. So, delta y sub r over this quantity, here. Again we'd like to simplify it. We'd like to be able to get some of the u sub r rocket frame y components in here. And so let's pull out this delta t sub r, sort of like we did before, and see what happens here. So we have delta y sub r on the top still. And, got a gamma here. Gamma's don't cancel this time. And we're going to pull out a delta t sub r from that expression. So we've got delta t sub r, times 1 plus v over c squared, delta x sub r. Over delta r sub r. Okay, so delta r sub r times 1 is the delta t sub r. Delta t sub r plus this thing gives us that because the delta t sub r's would cancel there, of course.  But then, this is a little strange here because, notice that we're, we're doing an equation for, for the y component but we've got some x components in here as well. Remember what this was? This is just the velocity of the pod and the x direction in the rocket frame. Okay. So we could write, in fact just, you know, to remind ourselves here. U sub r x equals delta x sub r over delta t sub r by definition. Have a good delta there. that's an even worse one, there we go. Okay, now, so, but in our example here, note what happens up for Bob, not what happens but what is happening for Bob here. There is no velocity of the pod in the x direction. Remember Bob's stationary as far as he's concerned. He's shooting the pod straight up. As far as he's concerned, it's straight off to the side. And therefore, there is no velocity in the x direction for this example. So, in this example, this equals zero. No velocity of the pod in the x direction as far as Bob is concerned. And so this is nice, cause this whole turn then here. Is zero for us. So this thing here turns into zero for this case. And look at what we're left with here. Delta y sub r divided by delta t sub r, and also divided by gamma. Delta y sub r divided by delta t sub r. That's the velocity in the y direction for Bob, in the rocket frame. U sub ry. So this simply becomes U sub R in the y direction divided by gamma. Okay, now we're, we're going to come back to that in a minute, but let's, let's do one other thing here too just to make sure here, because, you know, for Bob it's going just straight up. Right? But for Alice, Bob is moving that way. So, Alice may, and probably will and actually will, actually see, will observe that the pod is, is going to go up at an angle because you do have the velocity of motion here of the spaceship involved. And so let's see if we can get that in here as well. So, we'll squeeze it in right here.  We just did that. And what we're going to do here make a little room. So lets do u sub lx. In fact, there's the formula you want to use right here. Don't have to rewrite it. We already wrote it. And so in this case, u sub rx. Remember we just finished talking about it. I just erased it. That's zero. Right, for as far as Bob's concerned, no velocity in the x direction. So I get 0 plus v all over 1 plus, where again u sub rx is 0 to Bob in this example. Therefore this whole thing is 0 so we get 1 plus 0. And in fact we did this example. When we were talking about the transformation equations here for the velocity. This just gives us v. So there are two components as we talk about the velocity here for Alice. One component Is just the velocity v in the x direction because that's just the motion of the spaceship here, and the other component velocity in the y direction for Alice is use of our y over gamma so it is, whatever velocity Bob sees there shooting up, then that's going to be divided by, divided by gamma. Note one very interesting thing here. Gamma is in the denominator, so as the velocity of light gets, not velocity of light, the velocity of the pod shooting up here gets faster and faster close to the speed of light. Remember what happens to gamma, gamma gets close to infinity. And therefore this whole thing no matter how big this velocity is. This is like something divided by a very, very, very big number. This goes to zero. So this approaches zero, as v approaches c or gamma. Another way to say it, gamma approaches infinity. So that's very interesting, because what that's saying is no matter how fast Bob shoots up his escape pod If his relative velocity is close to the speed of light, Alice sees almost no upward velocity at all. This is Alice's upward velocity in the y direction that she's observing.
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