Hey, guys. Let's do this problem. Okay. We've already seen a problem basically the same as this with muons, but we were looking at time dilation. Now we want to look at the length contraction aspect of it. So once again we have a bunch of atmospheric particles high up in the atmosphere that encounter these high energy particles emitted from the sun. And every now and then, it actually happens millions of times a second, there is a collision that's going to produce these heavy particles that are like electrons called muons. Now in the muon's rest frame, they last 2.2 microseconds. They decay after 2.2 microseconds. We looked at how long the muons would last in the lab frame given the fact that they're traveling at 90% the speed of light. Now we want to look at how far they will travel in the lab frame, but, specifically, we want to use length contraction. Okay? I'll actually show after I solve this that you can use time dilation to arrive at the exact same answer. Well, roughly because of rounding errors. You would arrive at the exact same answer if you didn't have to deal with rounding because length contraction and time dilation are actually two different sides of the same coin. Okay? So let's look at this from the muon's perspective. So from the muon's perspective, it's traveling, well, sorry, it's at rest in a frame that's traveling at 0.9 times the speed of light. So it's not moving, but distance is rushing past it. Right? And by distance, I mean atmosphere. Okay? So there's some amount of atmosphere right here that's rushing past the muon. So how much of this atmosphere is going to pass the muon before it decays? Okay. That's pretty easy. The frame is going at 0.9 the speed of light. We know that it decays in 2.2 microseconds. So let's just figure out how long a chunk of atmosphere that is that's going to pass the muon before it decays. Right? That's just going to be the speed that it's going. So it's vt= 0.9 × 3 × 108 × 2.2 × 10-6 meters. The question is, is this the proper time or is this the contracted length? This is the contracted length because the proper length would be the one that we measure with respect to the earth. Right? We're talking about the earth's atmosphere, so if we are at rest with respect to the earth, we would measure the proper length of that at 90% speed of light. So how far would we measure the muon traveling in the lab frame? That's actually the proper length. Right? The lab frame represents the proper length because the lab frame is the one at rest with respect to that chunk of atmosphere that the muon is moving through. So the proper length, let me write out the length contraction equation. Length contraction says it's the proper length divided by gamma, so the proper length is going to be gamma times the contracted length. This is going to be L0= 11-0.92 × 594 m. Now the Lorentz factor gamma, we got in the previous problem, and it was equal to roughly 2.29. Okay. Multiplying, Okay? So that is how far the muon will travel in the lab frame before decaying. Okay? And this is found just using length contraction. No concept of time dilation was used here. But like I said, leading up to this solution, we can still use time dilation and not worry about length contraction at all to solve this particular problem. Because in the lab frame, so this is s prime, right, the moving frame, which happens to be the proper frame for the time, right, but the non-proper frame for the distance. The lab frame is the proper frame for the distance, but the non-proper frame for the time. Okay? So the muon is traveling at 0.9 times the speed of light. So what time would we measure in the lab frame before it decays? Right? This is the dilated time because in the rest frame of the muon, we measure the proper time. In s prime, we measure the proper time in this case. So this is going to be gamma times delta t naught and we showed that this was about 5 microseconds in the previous problem. So we can straight up just measure length in this case. How far does it physically travel before it decays? Right? That's once again going to be velocity times time. So that's going to be 0.9 times the speed of light, 3 times 10 to the 8 times the amount of time that passes, which is about 5 times 10 to the negative 6 seconds and that's going to be 1350 meters. Okay? So we have some rounding error between these, but that's no big deal. The idea here is that time is proper in this frame. Right? Time is proper in this frame, but length is non-proper. Right? In this frame, length is proper. By the way, this is L0. Right? That's the proper length. But time is non-proper. And so the whole idea is that length to get it to be contracted, you have to take the proper and divide it by gamma. Time to get it to be non-proper, you have to take the time and multiply it by gamma. And that divided by gamma and multiplied by gamma is going to cancel out when you compare the two results. Right? So these should be if I carried enough significant figures, these should be exactly equal. Not off by 10 meters, but that's because of rounding error. Alright. So in this particular problem, we can easily see that length contraction is just a consequence of time dilation. Alright. But in a lot of problems, length contraction, the equation is much easier to use. Alright? Alright, guys. Thanks so much for watching. That wraps it up for this problem.
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35. Special Relativity
Consequences of Relativity
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