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Length Contraction for a Muon from the Atmosphere

Patrick Ford
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Hey, guys, let's do this problem. Okay? We've already seen a problem basically the same as this with neurons. But we were looking at time dilation. Now we wanna 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, and not 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 Mulan's. Now in the Mulan's rest frame, they have a, um they last, I should say, 22 microseconds. They decay after 22 microseconds. We looked at how long the yuan's would last in the lab frame, given the fact that they're traveling at 90% the speed of light. Now we wanna look at how far they will travel in the lab frame, but specifically we want to use length contraction. OK, 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 them. Yuan's perspective. So from them yuan perspective, right? 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 mu on. So how much of this atmosphere is going to pass the muan before it decays? Okay, that's pretty easy. The frame is going at the speed of lights. We know that it decays into two microseconds. So let's just figure out how long chunk of atmosphere that is that's going to pass the muan before it decays, right? That's just going to be the speed that it's going. So it's 0.9 times the speed of light three times 10 to the eight times the amount of time that passes. Right distance is velocity times time, so this is going to be 2.2 microseconds. And don't forget Micro's 10 to the negative six. And so this is going to be meters. The question is, Is this the proper time, or is this the sorry the proper length? Or is this the contracted length? This is actually 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 atmosphere. So this is actually the contracted length because the muon is not at rest with the strength of the atmosphere, it's moving through the atmosphere at 90% speed of light. So how far would we measure them? You on 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 them you want is moving through. So the proper length Sorry, 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 gonna be one over the square root of one point one minus 10.9 squared times, 594 m. Okay, now the Lawrence Factor Gamma. We got in the previous problem, and it was equal to roughly two point Okay, multiplying these together, you're going to get a distance of 13 m. Okay, so that is how far the neuronal travel in the lab frame before decayed. 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 violation 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 yuan is traveling at 0.9 times the speed of lights. So what time would we measure in the lab frame before it to case this is the dilated time? Because in the rest frame of the yuan, we measured the proper time in s prime. We measure the proper time in this case. So this is gonna be gamma times Delta t not. And we showed that this waas thought about five 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 it's going to be a 0.9 times the speed of light three times 10 to the eight times the amount of time that passes, which is about five times 10 to the negative six seconds. And that's going to be 13 50 m. 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 l not 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 divided by gamma time to get 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 had carried enough significant figures, these should be exactly equal, not off by 10 m. But that should because a rounding error. All right, 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. All right. All right, guys, Thanks so much for watching that wraps it up for this problem.