So part two of the Michelson-Morley experiment. And what we're going to do here is we're going to set it up actually using an airplane example. And the reason for doing that is it's just a little more intuitive for us. Sometimes, when we do it with things we're not used to thinking about so much, for example, light waves. Just the strangeness of the situation can give us some conceptual problems or challenges. And so we'll just do it with an airplane here. The analysis though will work exactly with light waves, and I'll explain why in just a second. Let's set this up. So imagine we have an airplane, we're going from A to B and then back again, so round trip. First leg up to B and then back down to A, we'll say our plane has some velocity, v sub plane or we'll probably just use v sub p for the plane velocity. And then we have a wind involved. Actually, we'll just do no wind at first, but then we're going to turn on the wind, as it were, and the wind's going to go directly in the face of the plane on the first leg of the trip and then it'll be with the plane, helping the plane on the return leg. So clearly on the first leg, when there's wind involved, the wind will retard the plane's velocity, so it'll be going a little slower. And just as, you know, if you've traveled by plane before, sometimes the pilot would say, hey, we have a tailwind today so we're able to make the trip a little faster than normal. Sometimes we're battling head winds and, therefore, we'll be a little slower. That's what going on here. So, on the first leg from A to B, battling head winds and so it's going to be slower that they're, the effective velocity of the plane will be the velocity of the plane minus the velocity of the wind. And then on the trip back down from B to A then the plane will have added velocity, because of the tail winds, so it will be the velocity of the plane plus the velocity of the wind. Before we get into that though let's do the no wind case, in other words, how long will it take the plane to make a round trip from A to B and then B to A back again. Well we know that time traveled is just distance divided by velocity, again remember if you are going 60 kilometers, right, let's just say 120 kilometers at 60 kilometers an hour. 120 kilometers divided by your velocity, 60, takes you two hours to get there. Or miles per hour if you want to do miles per hour. So with no wind then, the total time, is going to be, The total time of our trip. For one leg it's the distance divided by the velocity. So it's just D divided by v sub p. So it's going to be D divided by the velocity of the plane and then of course on the return leg there's no wind involved. Same thing plus D divided by the velocity of the plane whatever it happens to be. And so clearly we can just add that together, we get 2 D over v sub p. Okay? So that's the case. If there's no wind, we just need to know the distance. We need to know the velocity of the plane. And a simple calculation to give us the total time it takes for the round trip for the plane. Now let's turn on the wind, okay? So we've got the velocity of the wind coming here. And actually, at this point, for those of you who want to do a little quantitative challenge. Or even just conceptually for everybody to think about this. You think, okay. On the trip up, the wind's going to be against the plane. So it's going to retard. It's going to take a little longer. On the trip down though, the wind's going to be helping the plane. Shouldn't it cancel out? In other words. Shouldn't you get the same answer for the total time? Definitely you're going slower on the first leg of the trip. But then you're going faster on the second leg of the trip. So, should it not just cancel out? And we get the same result. As we will find, it's not quite that simple and so you might think about why that might be. You can think about just in the qualitative conceptual fashion. But you can also think about it, for those of you who want to say pause the video now and see if you can work this out. I'll get you started by saying the velocity on the first leg of the trip is going to be the velocity of the plane minus the velocity of the wind. So that's the net velocity on the first leg and then we have our same equation involved here: distance divided by the velocity and the velocity of the second leg of the trip is going to be the velocity of the plane plus the velocity of the wind v sub p plus, v sub w. We use vw for the wind. Equations the same. You just have a different velocity on each leg now. The total time is going to be add those two things up and get the total time. So again if you'd like the little challenge and see if you come up with what we're going to come up with here in a minute. So we can do it correctly. Then go ahead, pause the video and try that and then come back and we'll look at the details here. Okay, so the A to B trip, the time. Let me, what range? So here is the time for that again is going to be the distance D divided by the velocity and the velocity going against the wind is v sub p minus v sub w. So it's going to be D over v sub p minus v sub w, just same equation as before, just different velocity this time, and then the B to A trip is going to be distance D, but velocity in this case is v sub plane plus velocity of the wind. So here, we've got the time equals D v sub p plus the velocity of the wind. And the total time is then just add those two things together. So that total time is going to be D plus v p minus v w plus D v p plus v w. Okay and now what we'd like to do is this is where more of the algebra comes in. We'd like to simplify this a little bit. See if we can get it into a nicer form, such that we can really compare it to this. because remember what we're maybe thinking here, is that it should cancel out. Shouldn't this, it doesn't look quite like that, but shouldn't it maybe end up sort of like that? Well, it's going to end up not quite like that, but almost. So, let's see if we can do that. So, to do that, just to make some more room here, we're going to I'm going to bring this up here. We'll leave this up here, because we want that result there. So we're still working with the wind stuff here. Okay, so we're saying here the total time, just rewriting this thing here, is going to be D vp minus vw plus D, vp plus vw. And at this point, we've done the analysis, now we just have to crank through the algebra a little bit and see what the result is. So this is one of the reasons we did the math reminder here, we have two fractions. We need to get, we want to add them together so we can combine them. We need to get common denominators, and the little trick we use is we multiply each in the form of one. But with the other denominator. So here's what we're going to do, we're going to say this equals, D over v p minus v w times the other denominator in the form of one v p plus v w over v p plus v w like that. Okay, so I'm just taking this first term here multiplying by one but in a special form here and do the same thing for the second term. Plus D over v p plus v w because that's this term here. And now multiply it by one in the form of the other denominator v p minus v w all over v p minus v w. So I have not changed anything. I can also multiply anything by one if I want, and the results will still be the same. But now you see in the denominator I've got vp minus vw times vp plus vw over here. Same thing over here but just in opposite order there so let's put it all together and see what happens in the numerator. So we are going to get, here we get D times v p plus v w, all over v p minus v w, v p plus v w, actually I really should extend this over, because I am putting these two terms together. They both have the same denominator, that's this thing. Just in a different order there so I can flip it around. The second part of the numerator is plus d times vp -vw. Yes I just combine them together, I haven't done anything else besides that here. But now I can look at the top and I can say, okay I've got D times vp and D times vw, I multiply that out and over here I've got D times vp and I've also got D times -vw. So over here I've got a D times vw. Over here I've got a D times -vw. They cancel out. I'll have D vw minus D vw and then we'll left with D times vp here and D times vp here. So that's two of them, so I can write this as, okay squeeze it in down here. Two times D times vp all over vp- vw times vp + vw. And we're going to bring it up here in a minute as well. Okay, so see what we did? We just said okay, here was our total time. That plus that, for each leg. Put them together for a common denominator. And then noted that in the numerator, some of the terms cancel out. So I've got D times v w, and then minus D times v w. So they go to 0, and I'm left with this. Okay, so now let's, we'll bring this back up here. We'll leave that, because that was our starting point. Okay so what we've got so far then is 2 D vp over (vp- vw) (vp + vw). So that's what we've got so far. Now let's again crank some more algebra here. Let's multiply this out in the denominator on the bottom here. So we'll leave the top as it is for a minute. 2 D v p. I've got a v p squared. V p times v p is v p squared. Minus v w v p plus v w v p. So those terms will cancel out. -vwvp here and I've got vp times vw + vw times vp. And then -vw squared. If you've done algebra recently, you recognize immediately what we're going to get here and that is this. vp squared- v w squared on the bottom there. Okay, so [SOUND] okay, now what. Well again, going back to the math review we did in the last video clip. We're going to factor out in the bottom here, we're going to factor out a v p squared. And so it's going to look like this. We got 2Dvp, we haven't done anything on the top for a couple steps here, but on the bottom, we're going to write this as vp squared times 1- vw squared over vp squared. And just pull out the vp squared term here so that gives me the one there. And then on the other term, v p squared times that would give me the v w squared again. So they're identical there, they're equal. And the reason I did that is now I can cancel this v p on the top. Got a vp on the top. I've got a vp squared on the bottom. And so this thing then equals, we'll write it up here, 2Dvp, oops, sorry. Not the vp, that's what we're trying to get rid of. 2D on the top. That's what I've got left on top, on the bottom I've left one v p here. So I've got a v p. And then the 1 minus v w squared over v p squared. And we'll rewrite this in just a second but let's see what we've got there. And another reasons we did this is now we see hm. Going back to the total time with no wind, it was 2 D over v p. I sort of almost have that. I have a 2 D on the top in the numerator, a v p on the bottom, but I've got that extra term there. And so let's just rewrite it so we can look at it closely here. So after a few steps of algebra there we find the total time for with wind is going to be 2D over vp, and then this second part of the term over here is which is one over one minus vw squared over vp squared. Like that. And all I did to go from here to here is just write this part of the denominator as a separate factor. And the reason I did that is to show very clearly I had the 2D over vp or 2D over vp, but it's not the same. I have an extra factor in here. So it turns out when I've got the wind involved going up and going back down even though it's herding going up, headwind, tail wind going back, it doesn't cancel out exactly. It actually is a difference with a wind involved such that you do not get your no wind time of travel for the round-trip you get a slightly different time. Note also that, and this is a good thing to check in these cases, you say, okay this is the case with the wind involved. This equation should be true if there is no wind, right? Does it give us the right result? Well, if there is no wind, the velocity of the wind is 0. So this is 0 and I get 1 minus 0 here. Velocity plane of something. And then I get one divided by one, so that's just one. And I get 2D over vp, the answer I got with no wind. So this is a more general equation I have derived, it works with wind but it also works if I've got vw equals zero for the wind in there. So a somewhat surprising result because again intuitively thinking shouldn't it cancel out? But let's think about this in terms of doing conceptual ideas like this or conceptual problems like this, sometimes it helps to take an extreme case. So let's take an extreme case here. Let's say the plane velocity is 300 kilometers per hour. Sort of a slow plane actually as well. Or 300 miles an hour if you want that. Just or whatever you want to use for the velocity of the plane. And let's say the wind, okay, so this is going to be vp. Let's say the wind is 299 kilometers per hour for the wind, very strong wind of course. Tornado wind really, but again, it's just a thought experiment. Like many thought experiments Einstein did. Maybe not quite sophisticated as some of his, but it's a thought experiment. Let's think about what happened here. Okay just intuitively. So going up the effective net velocity of the plane would be 300 minus 299, it would be one kilometer per hour. Okay if it really could fly like that. So it would be a very slow trip going up. And then on the way back down the velocities would add, would be 599 kilometers per hour. So about twice as fast. But by doing that, you can see it's going twice as fast going down, but it's not just going half as fast going up. It's going about 299 times slower than it did before. So it's not an equal situation in terms of going against the wind and coming back with the wind. You can double your speed, in this case, going with the wind. But you're slowed down many more times than just the doubling you get. Much less than half the speed. And therefore you see that yes there is a significant difference when the wind is involved. So the point being here, how does this apply back to the Michelson-Morley experiment? Well in this case essentially this would work for light as well if there is a light medium. If either exists, okay so in other words it would be like sending a light beam against the either wind. And then sending a light beam with the either wind. And if you could somehow see that difference in speed and detect it then you could actually prove they either exist or investigate the properties to that either a little bit more. Now the problem is that the velocity of the Earth through as it rotates around the sun is about 30,000 meters per second or something like that. And that's still much much less than the velocity of light. And so the effect notice that when the velocity of the wind here is much less than the velocity of the plane, there's not much effect on the travel time and that's the case with, even though the Earth moves around the sun at a relatively fast speed for most, what we're used too, certainly. That velocity of light is so much greater than that, this is a very small factor in here and this is essentially one and therefore it's very difficult to see an effect, it does exist, it's very tiny. But with the experimental techniques they had, they weren't able to do it just from that alone. So they actually added on to this idea. And here's what they did. And that'll be the subject of our next, part three for our video clips here. But essentially they said okay, we'll do one up and back like this against the wind. But let's analyze the case where it's cross wind. So if the airplane or the light beam is going this way and the wind whether the regular wind or the ether wind is going cross ways against it. What happens in that situation? So we'll analyze that and then compare this situation with that situation and see what we can get out of it.