Any questions?  We'll get to that one later. Do you really don't have an astronaut brother in a lot of ways? No, because he quit and is now teaching at the Naval Postgraduate School. He's been up on two shuttle missions and won six months stay on the space station, which meant once, every hour and a half, he was closer to me than he ever has been any other time to do. It's only like 260 miles overhead. Anyway, that's my trial. Life. What do astronauts do with their astronauts? Drag, mostly. I pick up girls in bars, I think, and put on weight. Only Joe looked at that. Only Joe has a question about that swinging chair thing, man. All right, Joe. There you go. Check it out there. All right. Any other open questions, though, before we get going here? All right. What we're looking at, if you remember, are different types of forces because our number one tool right now for solving kinetics problems is Newton's law. F equals ma. By f, I'm being a little bit loose. I mean, the sum of the forces. A lot of times in our problems, there's just one. But if there's more than one, it's no big deal. We add them all up. It's all the forces together. And whatever's left over, if anything, that will determine if there's any acceleration. So we looked at some tributal force yesterday. Did we talk about spring force a little bit? I think I did just real quick. And there's a business in the book about it. Everybody's played with springs before.  You just may not know how to quantify it. It's pretty straightforward. So we'll spend a little bit of time on some other things as we go here. So what we're going to look at today, well, let's do a little force experiment here and see what's going on if we can all figure it out. Very simple setup. I've just got a little block of wood that's on the table. I attach a string to it. We know what strings do. They only pull. And then I've got a scale here to help me determine what's going on. So I'm going to put it there. Let's see. What's this reading? This reads in Newtons. So I'm pulling with pretty much all my might right now. Three Newtons. I should probably push my sleeve up so you can see the veins. No, you feel to see from there where he was. He could actually see it pushing out against my shirt. Oh man, that shirt ripped open. I'm going to lose my eyesight or something. So about three Newtons. So let's see. Let's draw a free body diagram of that setup. The string is pulling on this little block with the weight on it by about three Newtons. We read that right off the scale. Any other forces on it? Who's going to leap forward and take the easy one? Who did that? Nobody? Nobody wants to put somebody on there? Yeah, well wait. See, you got to be smart here and always jump in and get the easy one right off the bat. I've got about 2,000 grams on there. And then that then weighs maybe 100 or 200 grams or so. Any other forces? The normal force. Otherwise, the weight would accelerate it down. So there's got to be a normal force. In fact, in this case, we figure unless there's some other vertical forces that we don't have up there, and I can't imagine what they would be, then in this case, the normal force equals the weight. Not always. We need a free body diagram to check to make sure whether it does or not. But in this case, the weight and the normal force happen to be equal in opposite. So I'm pulling on it to the right with three Newtons, but it was not accelerated. The only way that can be, oh, all the way up to four now, it's not accelerating. Listen, if it's not accelerating, if I sum the forces in the x direction, we'll let that be the x direction, that be the y direction. As usual, if we don't have to specify it, we'll just take it as that, I guess. There happens to be no x acceleration. Well, wait, how do I know there wasn't any x acceleration? I mean, did I measure it? Did I run a little ticker-tape thing through it? Wow, that high-sounding. Because it was just sitting there, and that's all it was doing. If it was accelerating, we'd see it do something else a second later. That's what Mike meant, but he wanted to sound more important than that. So the forces then must also sum to zero. Here's our first indication, at least in a quantifiable term, what you all know to be going on anyway. In fact, I think I heard someone say it. What's the force that's missing? Clearly, we don't have all the forces. Because I don't have any force up here that balances this one. There's got to be another one that balances, at least one more, that balances this force is equal and opposite in the other direction. Somebody said it. Who said it? Friction. It's friction. Obviously, friction is pulling back in that way. We typically, just to remind this, it's a different force, and so we don't just treat it like other ones because we're going to have to do some other stuff with it here in a minute. Give it that little sort of script F for friction. Now, the interesting thing is, it appears to be adjustable. See, I pull with three Newtons. The friction must be three Newtons. I go up to four Newtons. The friction must be four Newtons. Five. I try to trick it. I go back down to four. It catches that. I try to trick it. I try to distract it. I say, hey, look over there. To five, it knows. It's reading my mind. It knows what I'm doing. If we graph this, let's see. Here's the force I'm applying. And I tried three Newtons. I couldn't get much below three Newtons because then just the weight of the scale itself tends to pull down too much. There's not even enough tension in the line to hold that. I have the force I apply across there and the friction I calculate there. I'm not measuring the friction. I'm calculating it. All I can do is sum the forces in the x direction. I know what I'm doing and I know the friction's opposing it. If I do three Newtons, then friction does up to four Newtons. And if it doesn't know, I was doing four Newtons. Here, you know what? It's peaking. So I'm going to cover you. Okay, Joey, get ready. Damn it, Newtons. It knows what I'm going to do.  I can't get you guys to know what I'm talking about half the time and this thing's reading my mind. It's going to hate that. It knows what I'm doing. One for one. What's the slope of that line? One. That slope of that line is perfectly one. Not sort of one. Not occasionally not one. Not one for me and 1.1 for you. Because if that slope was anything other than one, then either this force would be bigger than this one or this one would be bigger than that one. And that thing would start to accelerate. It's perfectly one. It knows exactly what I'm going to do. Even everywhere in between. Because this is an analog scale. Everybody know what that is? Analog scale. Analog is a scale that can read any possible value. Whereas a digital scale could only read certain values. This is an analog scale. Not only do I get three Newtons, four Newtons, five Newtons, I can hit absolutely every little miniature tiny bit of a Newton in between. Six. A new record. Guys, things are reading my mind. It's perfectly reading my mind. It's really reading my mind. It knows exactly how much force I apply. It won't apply anymore. And it won't apply any less. So I go to... Ah! Something happened there. Something where did it happen, Joey? It happened about seven. Well now we got things going. Now things are sliding a little bit. There's a lot of shock all over here. So now, some time or other, I get to a point where I can actually pull it hard enough and it starts to accelerate now. You know what? I think we had a goober right there stuck on. But you know what? That's important to us. We need to know why it moves and why it doesn't. So now, it looks like we can't hit five or six anymore for some reason. Now it starts to move. Unfortunately, I don't have this table I want. There's five. Hit five a little bit. Oh, and then they even drop down. Got the five. Drop down a little bit. So clearly, there's a limit. And you know this. If you've ever tried to push a couch across the floor or something, you got to... If you don't push hard enough, it doesn't go. You got to push harder and harder until it starts to go. And then once you finally get it going, what happens? Huh? You ever move the couch? Yeah, it's actually... Once it gets moving, isn't it actually easier? Some, there's just too much garbage on the floor there. Sooner or later, it breaks free and the friction force actually drops down. And as we apply more force, now it starts to accelerate. That's much harder to work with because we have to test, well, how much force do I have to apply for constant velocity where I can have it moving but not accelerating or how much force do I have to apply to get it to accelerate? It's a much trickier thing. But what we find is once it's moving, if we could be sense enough to do that test, we find that the friction force would never change and we can very easily get above the friction force, give that thing some pretty good acceleration. So it appears that we've got two things going on. We had this region where nothing would move. That's when I couldn't pull it hard enough to get it to move, but it always pulled it back exactly what I did every time it pulled back against me. Then we had this region here where the friction force actually drops down. You know that from shoving couches across the floor or whatever it is you were trying to steal that day where actually things are moving. This is a static region. Static means no acceleration or no movement. This is a kinetic region where things are moving. So we did that before and then we got up to five and then it started moving and then things got a little goofy here on the surface somehow. I'm not sure why. Now we could get things moving. So let's see if we can try something out. Let's get a reading here with our new slippery surface, whatever it is happening. So about four and a half. Right, Joe? And then things start to move. I'm a little less there. Not big news, is it? Isn't it easier to slide? When you need to slide your couch across the floor, do you want your drunk uncle Earl sleeping on the couch when you do that?   I got to move the couch because the couch is moving. You don't want to push Uncle Earl across the floor because he's going to make it too hard to push that. So it seems like friction has something to do with the weight. Is that seem reasonable? If I put both of these on, what were we at with just this one, Joe? Four and a half? And then you see when it started to move, you see it drop down pretty hard. This is a real rough set. But now we're all way up to six or something. And here's the friction. Well, not the friction. The static friction is always one for one. No matter how much I'm towing, isn't it? Because I never have any acceleration. No matter what I'm trying to tow, I never have acceleration. Whether I'm here or whether I take that one off, as long as I'm not moving, I must not have any, it's always one for one. But here's like this highest point I have to pull. Might have something to do with the weight. Is that what seems like? The heavier it is, the higher I have to go, the higher the load I have to exert before it will finally break free. Does that seem reasonable? Maybe we draw it to something like this. Here's my force. There's friction force. If I don't have very much weight on there, just the one little one, then I don't have to pull very hard before it breaks loose. But if I have more weight on there, I've got the big one, or I've got the two of them together, it appears like I've got to go a little bit farther before it breaks loose. Before that, to the left of that peak, it's always one to one. Friction's always one to one with my pull, because that's the zone where it's not accelerated. So it's this peak here that seems to be of interest. How can we predict what that peak is going to be? Before that, there's nothing to predict. It's doing the predicting. It's one for one what I applied. I applied three, it applies three. I applied 3.269, it applies 3.269 back. But that peak, this, let's call it, let's call it the maximum static friction. Maximum friction, that's the maximum force I can apply before it finally breaks loose and is no longer in the static region, it's down in the kinetic region, because it's starting to move. If it's starting to move and it wasn't before it accelerated. It seems like that point would be of interest. Maybe it's not something you'd actually calculate when you're trying to move your couch, but if you're trying to calculate whether you've got a good design for snow tires or how much friction is in a machine in the bearings or in the belts or something when you're part of that engineering design team, it seems like that kind of thing might be of interest to you. What is this maximum static friction? If you can calculate it, then you can take it into your design calculations to do a better design. It seems like it has something to do with the weight, doesn't it? So we'll put a question mark there because it seems like it has something to do with the weight. Try it again. I've got the weight down, I'm there, and what we're trying to find out is what's the maximum force I have to apply to finally get the surfaces to break loose and the object starts to accelerate. It doesn't accelerate for long because I can't pull it very far before it's off the table. But once it's at that maximum amount I have to pull, I know that friction is at its maximum amount because until that point, they're one for one. So it's very easy for me to indirectly measure the friction. I'm looking to see when's the maximum static friction. Clearly, just double check. It's not accelerating up or down. There must be some normal force up that's equal to the weight. And I'm looking now what's the maximum force I apply because that's the maximum force that the friction reapplies back against me just until it takes me to break the surfaces loose. So now I'm going to do a test, a little bit of a test here. I just want to double check something. All right, here we go. Let's see if we can get some consistent results here, Joe. About eight. Okay, we'll try it again because it seems to act silly after moving about five and a half. There must be some something on the surface that if it sits there a while it kind of gloms on to it and then once I get it moving it isn't there so much. So somewhere between five and a half and six, right? And then it starts to move. Phil, you agree with that? Something like that unless we let it sit for a while. So I won't let it sit for a while. Five, five and a half. So somewhere around five and a half or six it'll start to move again. I'm going to do something. I'm going to press down on this. Now what's its weight? Has the weight gone up? How'd the weight go up? What's my fingertip weigh? It depends on whether it's just been in my nose or not. It doesn't weigh much. I'm pressing down. I can press pretty hard. What's the weight of this? Does my finger change the weight of that? I still want to ask. Answer my question. Don't pull an owl on me. What happens to the weight of this, the little board and the two masses there when I press down? What happens to the weight of those things? Nothing. Is there a new force on here that I need to draw in? Well, yeah, of course. It's my finger. So I'm pressing down on it with one finger worth of force. Maybe we'll call that FF. I'm going to give this an F applied, force applied. That's the pole there we've got going sideways. There's my finger pushing down. The weight doesn't change because that's a factor of the mass of the thing sitting there. That doesn't change. So why would its weight change? Does anything else need to change now? The normal force, it had better get bigger. It needs to be that big plus that big. That normal force has to equal the weight we had there, the force I'm exerting with my finger. Yeah, my finger has a little bit of mass, but not much. So now let's see what happens, Joe. Let's retest it since we see it gets funny when it sticks for a little bit. All right, so that's six and a half. Now I'll press down. Let's get here in a minute. I wish Malcolm was here. Well, what time will it be? Malcolm could do this, couldn't he? Malcolm lifts weights. He tells us that. It's probably got a half. Huh? It's probably got a half because of it. There's funny. That's like a world record if I ever saw one. So I don't think since the weight didn't change and the maximum force I had to apply is clearly, in fact, it's off the scale now. I still couldn't even get it to move. It doesn't appear the maximum friction has to do with the weight. It appears more likely. It has to do with the normal force because that increased and that did increase the maximum static friction I had to apply. And we can do a lot of other tests that are much more quantitative than that. I don't know how much force I pushed down with. Once the normal force went up. But if we could test this much more carefully we would indeed find that this maximum peak point here this maximum static friction would indeed be a factor of the normal force. In fact, they're directly proportional. So I could write instead of that little squiggle which means has something to do with I could actually write is proportional to. Have you seen that symbol before in math? Some of you have and some of you haven't, I bet. It means is proportional to. Means that this peak is a linear function of the normal force. If I double the normal force either by doubling the weight or putting the weight on and pushing down with the same amount of force equal to the weight either way if I double the normal force I'll double that peak maximum friction. It's tough to show with this little thing on a crummy tabletop but if we could do that test carefully that's indeed what we'd see. Anytime I have a symbol that means is proportional to I can take that out replace it with an equal sign and a constant. Then I can make a direct calculation between the two. It turns out we call that the coefficient. No, that's a mu. It's a Greek letter mu.  Our M came from that but that's a Greek letter mu. Remember we never write out these Greek letters. Reminds us that's the limit of the static region. So if I do anything to increase the normal force which means I could increase the weight that would increase the normal force or I could apply more force here downward myself that would increase the normal force. Anything, any combination of those is going to increase where this peak happens. The bigger the normal force the higher up this line goes. This one for one line just keeps going higher and higher and higher the higher the normal force gets. And in fact for static friction we could say static friction itself now notice I don't have any max written on there static friction is always less than or equal to mu s times M equal to, because we're at the maximum point thing than that we're down somewhere on this one for one slope this maximum static friction equal to this coefficient of static friction you got to figure we're going to have to know something more about that but for right now it just turns out that we do this measurement we set the normal force to something measure how much static friction we do that for a whole bunch of different normal forces measure the maximum friction force and then we can calculate that static coefficient. We could do that in the lab if we didn't have so many other busy things to do. The other region of interest I guess is this kinetic region where once it's moving it turns out no matter how hard you pull the friction force pulling back is pretty much the same it doesn't really change it wobbles a little bit but if I pull 8, 9, 10, 20 if I get Malcolm in here go to 25 if I get Malcolm and Samantha we go to 29 it doesn't matter the static or the kinetic friction will be pretty much the same no matter how much more I apply so we can also say there must be some kinetic friction that we could calculate it is also proportional to the normal force but it turns out the proportionality constant is different well you can see that because this friction force tends to drop down in fact in every situation I can think of mu k is less than mu s if nothing's accelerating yet if nothing's broken loose if we don't have this one surface sliding over the other there's going to be more friction than when we do get things moving and that's what you experience when you shove something across the floor try to get it refrigerator moving or something big effort to get it moving but then you try real hard not to let it stop again because now it's a little bit easier until you get it to where you're going you know that from pushing stuff across the floor so let's see maybe a little bit more word about these coefficients they're both very much the same in that they're essentially constant but there's some things we really need to understand with these coefficients let's be a little more thorough we'll say coefficients of friction whether static or kinetic there's a couple things that are true besides just the generality that the kinetic is generally less than the static coefficient of friction what you have to realize is when we talk about the difference between this static region and this kinetic region we're only looking at one thing we're not looking at the object necessarily itself and whether it's moving or not because I could do the same experiment by somehow holding this thing still and pulling the table underneath it and I would still get the same kind of friction measurements there if I could do that without much difficulty so it's not a factor of whether this thing is moving or not it's a matter of whether these two surfaces in contact are moving over each other it's this little wooden block moving over the tabletop that's important that's the difference between the static region when those two surfaces aren't moving over each other and the kinetic region when those two surfaces are moving over each other we don't look at the object and whether it's moving or not necessarily what we look at is the two surfaces in contact in fact this depends very very much on surfaces in contact remember I told you friction was a contact force we only have friction if we have two surfaces in contact if those surfaces are not moving and I mean not moving relative to each other not moving relative to each other I don't care which one's actually moving which object I don't care if one object's moving and the other isn't I don't care if they're both moving a little bit all I care is about what's happening if the surface is in contact that's where the friction is and that's where our concern is not moving relative to each other that's a static friction situation if they are moving relative to each not over other that's a kinetic friction situation that's the dividing line between these two regimes here you look at the two surfaces in contact where the friction is remember that the normal force is also a contact force with two surfaces in contact and in this case we're talking about the very two same surfaces it's the normal force at the surface where the friction is occurring that we're talking about here one other thing is very true about these coefficients of friction very much on what the surfaces are I'm trying to pull something across the floor and then I come in and well we even saw it change somehow here there's something weird that's on the surface but I'm really serious about this I have to clean these surfaces I have to clean them off of alcohol let them dry, be very careful where I touch them because all kinds of weird stuff can happen but if I grease the table if I let you guys borrow the table how many of you guys are living alone not living at home anymore Joey so if I let you bother this table because bachelors don't know how to clean anything I let you borrow this table for say a week or two and come back pretty greasy after that wouldn't it there'd be spaghetti chocolate milk yeah there we go and so we have entirely different test values for the maximum static friction and the kinetic friction it took to get that to move over so if I took the surface and I laid over it a real sticky foam rubber or something and then re-ran the test that'd be completely different again it very much depends upon what surfaces are in contact it's very different wood on fermica or in Joey's case wood on greasy fermica or say rubber on that pretty important one don't you think because that tells you how well your car tires stick when you're going around corners we know that if we don't have enough friction force on a corner you can't go around a corner where you're looking at centrifugal force the other day very much depends upon what the co-efficient friction or what the surfaces are that are in contact what would be nice then is if every time we need to do some kind of friction problem where we have some problem we understand friction to be a part of it it'd be very nice if we didn't have to re-test it every single time we have two surfaces in contact it'd be nice if somebody would test some surfaces say rubber on steel or tennis shoe rubber on ice or concrete on asphalt be nice if somebody would do that test for us and then just let us know what the coefficients are so we can do the calculation wouldn't that be nice well it turns out that a lot of those calculations have been done a lot of those tests have been run and this is the only way you can get these coefficients to do the test in your book on page 175 if you have your book if you don't, I can put it up on the screen why don't I go ahead and do that anyway trick you to one close enough nothing yet everybody wants to come up and gather around my monitor let's say you said I'm not going to camera is the light there in the right spot? the light is in your mouth no, the screen is there that's just how much light that is looking at you isn't it the lens is here so that's not the case I think we don't actually have this let's see where's my choice here it was blinking green before but now it's blinking green again there we go it was just sleep see if you don't bring the books I don't know how to do this kinetic and static coefficients so we see mu k that's connect what does that kinetic mean again, that k? what's that mean? what's moving? the two surfaces relative to each other I don't care what object is moving I care whether the surface is in contact or moving static coefficient means they're not moving the two surfaces are stuck to each other steel on steel well that's the kind of thing you'd want to know if you were building some machinery like this is what's going on with your pistons in the piston cylinders in your car these are actually kind of high numbers so you might want to put some oil in your car to decrease that a little bit that would be then lubricated steel on steel which would be a very different number copper on glass poison very important ones here rubber on concrete there we go at the bottom that's your car tires on the road surface we get pretty high coefficients of friction about the same but the static one is definitely definitely greater as most of you well that's why you don't want your tires spinning because the static coefficient is higher than the kinetic coefficient the tires are spinning your tires are spinning on the road surface that's a kinetic friction situation and typically the kinetic friction is less it's saying there are both about one but well maybe with concrete it is about that with asphalt it's very definitely true that the kinetic coefficient is less than the static coefficient and if you need other numbers than that there's a whole bunch of other publications that will tabulate a whole bunch of those things so you don't have to always look them all up alright so we need to do some problems with that so let's try a couple problems here another one with my car we like to drive in our car places so here's my car about a thousand kilograms am I in it or am I not in it a thousand kilograms I'm going about 30 meters per second I slam on the brakes lock up the wheels so I'm skipping them lock up the wheels because I panic when I need to stop I'm like you guys you know if I'm not squealing the tires when I stop I'm squealing the tires when I start I'm not going to meet the girl in my dreams so you've got to squeal those tires fellas so I lock up the wheels and I'm skidding coefficient of friction between my car tires and whatever surface I happen to be on let's say it's 0.8 kind of low much lower than rubber on concrete was so maybe I'm on a dirt road roads we have right now there's a lot of grit on them that's going to lower the coefficient of friction it's a lot easier to skid out when there's grit all over a road and of course I do that to bring myself to a stop how far do I travel in that skid until I come to a stop there's my question before we've done problems where somebody's moving and then they're not moving and we could figure out the acceleration or the distance or any of those type of things we needed so we're talking about what force is it that causes me to stop I've locked up the tires my tires are skidding over the road surface so there's a big friction force backwards that's what's slowing me down is that kinetic friction or static friction Joe says kinetic yeah my tires are skidding over the road the car is moving but that's not what matters what matters is what are the two surfaces doing with each other so my tires are skidding over the road surface that's a kinetic friction situation I'll skid before I come to a stop remember taking driver dead and they said stopping distance at certain speeds is so far as if you can calculate that you see a little boy run across the street chasing the ball I always do those calculations real fast but by the time I'm done I'm hidden there goes one more tuition fallar which is lost there how are we going to do this problem how are we going to get to this distance traveled we have initial velocity final velocity we're looking for distance that kind of sounds like the constant acceleration problems we were doing I told you this friction force when the two surfaces are sliding is fairly constant so if the forces are fairly constant the acceleration is fairly constant so we could do this as a constant acceleration problem so we have initial velocity final velocity we're looking for distance it's a constant acceleration problem how do we solve constant acceleration problems remember we've got those four equations but to work the four equations what do you need I already said constant acceleration you need three things there's five possible variables we've got three of them involved two we have one we're looking for so we need find the one thing we're looking for one thing that we know before we can do this is a constant acceleration problem remember there's only five possible variables in a constant acceleration problem we've got three of them so there's only two other possibilities what are they time how long is this acceleration going on or the acceleration itself we need one or the other of those then we know what equation you use we just look it up on our table and find the distance we traveled which one of those can we get and how time it takes to come to a stop or the acceleration we undergo to do this stop what good will that do because that's not either one of these when I say you're wrong I'm just asking you what good will that do I can find the force of friction I've got the canade but what good will it do time or the acceleration then we can do this then I can find the distance I got to find either one of these which one can I find now Mike was throwing out some stuff but I don't know that it was going anywhere I'm saying he's wrong I'm just not sure it was going anywhere and let's see some of the forces now let's do the usual let's call that x call that y so the sum of the forces in which direction why the x direction because we're looking for acceleration in the x direction this delta s is actually in the x direction here so it sounds like we at least ought to start there in fact maybe that's what we're looking for that would be our acceleration in the x direction which changes s to an x if you want no big deal it's just an arbitrary choice of a letter we've got the mass if we can find the acceleration we can do this problem so it all depends over here if we can sum the forces in the x direction what do we need before we sum the forces Phil now give me your usual answer we need to best sum the forces a free body diagram we don't have the right forces we don't have all the forces in the problem how are we going to sum the forces we can't just guess at them we got to think and make sure we got them all any forces any forces on the car who spoke up or who grabbed them Phil's starting to figure out how to avoid getting called on later jump in just say wait any time I ask a question just say wait and then go back to sleep any forces on that car got to be a normal force otherwise it would accelerate towards the center of the earth normal force between what two surfaces normal force is a contact force the tires in the ground there's only surfaces in contact in fact that's our surfaces of interest any other forces which direction opposite direction of the movement is the stationary ground the car tire remember the car tire is not turning it's moving but it's not turning so this surface is sliding over that surface friction opposes that so the friction will be back that way okay on it because this is a situation where two surfaces are sliding over each other yeah the car is moving and yeah the wheels moving with the car but the two surfaces are what we look at one surface is sliding over the other surface that's a situation of kinetic friction no other ways about it what other forces don't bother some of the forces until you're done with the free body diagram you don't know now remember it's got to be anytime we put a force up there you've got to tell me what's causing it Phil, oh you're awake again you're going to just say normal no you're going to say weight go back to sleep I can go put my hand on the previous movement well you could stick your head out you hit by a car tire for a second slide on the brakes so I can go back here before I put the brakes on I can roll down the back window stick my hand out the back touch the previous movement Phil you want to you want to just say weight and leave it at that don't forget no forces go on here unless you can tell me what I can touch that causes that force I can't touch previous movement I can't touch acceleration I can't touch circular motion like we were trying to do yesterday we wrestled with that a little bit forces are only caused by ropes surfaces people reaching in and pulling or pushing that's it sooner or later you're done you can't keep putting forces on forever sooner or later you're done we're done there are no other forces in this problem so some of the forces are in the x direction that's the x direction that's the force in the opposite direction so it's minus fk minus because I chose to the right positive and this force pushes to the left there aren't any other forces in the x direction times acceleration in the x direction well that's what we're looking for so we've got the mass friction force how big is it? if we can't figure that out we can't figure out this acceleration how big is that friction force? you have mu kinetic sure do friction mu kinetic right there so that might help a little bit minus still stays there minus mu k the coefficient of kinetic friction times and the normal force so I don't know maybe we're a little bit closer well we don't have the normal force where do you get the normal force? you get the normal force from your free body diagram there's no other place to get it in fact we need to sum the forces in the y direction well there is no y acceleration it's going on level ground it continues to do so so I know in this case n equals w if the ground isn't level n doesn't equal w so don't just assume that in this case n equals w for how we find w do I give it to you? we got the mass we don't have the weight mg equals m a it doesn't matter it does matter then in determining stopping distance if the mass doesn't matter what matters most is what kind of tires do you have on what kind of surface that's what matters the most oh also matters which planet you're on for some of you we're not sure where that is now we can figure out the acceleration now you can figure out the stopping distance so go ahead and finish the problem take a couple seconds to finish the problem we're going to have a calculator you called for help you called up your mom said mom calculate this quick figure out the acceleration then we've got a good situation here we don't need the time nor do we find it directly we can find the time once we found the acceleration but that's just an we need to find out what what was it that's more heading towards that little kid finish the calculation remember we're looking for delta s yeah you're saying oh oh his eye real big his throat is solid he can't move check the acceleration the acceleration is not right and your delta s isn't right maybe make those delta x since we did pick that at the x direction but we might as well be consistent especially since we're taking notes and chalk classes after this you have to drive home it's a long way so that's going to be what two stops, three stops maybe stop to pick up more of the pizza and you agree Mark, who would you check with nobody you know what to do here that I gave you that's on your calculator g is what it always is so you can find that then when you find that there's three things you know one thing you don't use your constant acceleration did it work who would you check with now you agree Tyler, you guys agree now Patrick almost what did you get for a just to make sure nope nope what there we go, got it all finally 8.5 what's the minus mean we picked the positive effect on his x the acceleration is negative to the left what we typically call deceleration now that you've got the acceleration is now just a constant acceleration problem that you can finish and we got for a stopping distance 57.4 meters which is lucky the kid was standing at 57.6 meters so scared to hell which is cool teach him a lesson don't run out the street alright we don't have much time I don't see you until Monday so I have to give you something that will consume at least several hours over the weekend see if I can keep you busy enough keep you out of jail so here's the problem there's a wall on the floor two kilogram box one kilogram box is tied to the wall with a titanium cable you are pulling on the bottom box with your full might I redlined the scale today 20 newtons I want you to find oh wait two more things coefficient of friction right there UK is 40 for any time two surfaces are in contact there can be friction UK is also box is doing otherwise we wouldn't have kinetic friction we'd have static friction because if it's moving it's moving over that surface and it's moving under that surface so find two things if you don't don't bother coming for the rest of the semester and that's the last week next week before spring break isn't it man yeah things are flying find the acceleration of the two kilogram box what's the acceleration of the one kilogram box zero it's tied to the wall and the tension in the cable back here titanium is expensive I don't want to buy a thicker cable than I need to got the problem that dude on your forehead when you come on Monday if I see it there you can come in however is there over forehead what are we going to do fifth thing yeah Allen's got a hat on you have bigger equations that's why you need to get into graduate school when your older equations get bigger because you have more forehead to put them on alright if you don't make your time with it when you're done I'll be in my office I guess


Hello everybody, my name is Daria Grudzień, I am a site reliability engineer and a co-founder of Cyber Magnolia and Women's Tech Collective in the Czech Republic. Hi all, I'm Aneška, I work as a project and community coordinator in Czecho-Digital and I am also one of the co-founders of Cyber Magnolia community. Hello there, my name is Karolina Surma, I work in software development. I am a part of Cyber Magnolia community. And together we would like to introduce you to our community and the vision behind it. There's two things we would like to achieve, one of them is to connect women working in technology regardless of experience with programming languages, seniority or a job description, because our community already includes people working as software engineers, web developers, QA, cybersecurity, site reliability, project coordination and more. And so the second thing we would like to do is to help build the first list of women tech speakers in the Czech Republic because as event organizers behind conferences like PyConCZ or the Fest, we really know that it's very hard to build diverse speaker lineups and this is something we'd like to help fix. This is where we come asking for your help. If there's a lady who's working, appreciate and would like to see her speaking about it, we would really ask you to point us to her by going to our website at cyber magnolia.com slash speakers where you will find a super short form where you can point us to the lady, give us her contact and tell us which fields you think she could talk about and we will be happy to follow up and start building that list. So that's the first thing. The second thing is if you are the lady working in technology, we would like to invite you to join us, make new friends, join our online meetups in the future offline meetups as well and chat about everything and anything technology related. You can do that also by signing up on our website at cyber magnolia.com slash contact and we're very much looking forward to meeting you. Lastly, if you're curious what we're up to, you're welcome to follow us on social media who are available almost anywhere as cyber magnolia, particularly on Twitter who are the Twitter girls. So come say hi, say hello, ask questions if you're curious and we're very much looking forward to meeting everyone in real life in the future. Thank you.


Hi and welcome to the session. I am Purva and I will help you with the following question. Prove the following integral dx upon x square into x plus 1 where limit is from 1 to 3 is equal to 2 by 3 plus log 2 by 3. Now we begin with this solution. We denote this left hand side by i. So we have i is equal to integral limit is from 1 to 3 dx upon x square into x plus 1. We prove this question by partial fractions method. So we consider 1 upon x square into x plus 1 and this is equal to a upon x plus b upon x square plus c upon x plus 1 where a, b and c are real numbers and we have to find out the values. This implies 1 is equal to a into x into x plus 1 plus b into x plus 1 plus c into x square and this is equal to opening the brackets we get a into x square plus a into x plus b into x plus b plus c x square. Equating the coefficients of x square, x and constant terms we get. Now equating the coefficients of x square on right hand side and left hand side we get a plus c is equal to 0. Equating the coefficients of x on right hand side and left hand side we get a plus b is equal to 0 and equating the constant terms on both the sides we get b is equal to 1. So we have got b is equal to 1. Now putting this value of b in the second equation that is equation 2 we get a plus 1 is equal to 0 which implies a is equal to minus 1. Now putting this value of a in first equation we get minus 1 plus c is equal to 0 which implies c is equal to 1. Now putting the values of a, b and c in this equation star we get putting the values of a, b and c in star we get 1 by x square into x plus 1 is equal to minus 1 by x plus 1 upon x square plus 1 upon x plus 1. Therefore we get integral dx upon x square into x plus 1 where limit is from 1 to 3 is equal to minus integral 1 upon x dx where limit is from 1 to 3 plus integral where limit is from 1 to 3 1 by x square dx plus integral limit is from 1 to 3 dx upon x plus 1. This is equal to now integrating the right hand side we get minus log x because integration 1 by 3 is equal to log x plus into minus 1 by x because integral 1 by x square is equal to minus 1 by x plus now integral 1 by x plus 1 is equal to log x plus 1. In each case we have limit from 1 to 3 this is equal to minus log putting upper limit in place of x we get 3 minus minus log putting lower limit we get 1 plus minus 1 upon putting upper limit we get 3 minus minus 1 upon putting lower limit we get 1 plus log putting upper limit 3 in place of x we get 3 plus 1 is equal to 4 minus log putting lower limit 1 we get 1 plus 1 is 2. So log 2 this is equal to minus log 3 minus into minus becomes plus log 1 minus 1 by 3 minus into minus becomes plus 1 plus log now we can write 4 as 2 square so we get log 2 square minus log 2. Now this is equal to minus log 3 plus log 1 now minus 1 by 3 plus 1 gives plus 2 by 3 and we can write this log 2 square as 2 log 2 minus log 2 and this is equal to minus log 3 now we know that log 1 is 0 so plus 0 plus 2 by 3 plus now 2 log 2 minus log 2 gives log 2. Now this is further equal to we can write this as 2 by 3 plus now log 2 minus log 3 can be written as log 2 by 3 and this is equal to our right hand side. Hence we have proved that integral limit is from 1 to 3 dx upon x square into x plus 1 is equal to 2 by 3 plus log 2 by 3 hope you have understood the solution take care and god bless you.


I'm very glad to have this opportunity to speak here. And as you can see later, the title of my talk is very much related to this workshop in the next slide. But before that, let me just mention that I will talk about self-similar, discretely self-similar, and their rotating versions. So these are special solutions of Navier-Store equations. So the connection to this workshop is that this name was learned from this workshop. So this name didn't exist for Navier-Store's community before I attended this workshop in 2009. So that's the connection with this workshop. I was very happy to find the name finally. And that talk is in cosmology, which I will explain later. So these are my two collaborators. Mikhail Kolopkov, he is a young guy in Russia based on the Biasq. He is an expert in new analysis. And Zagre Brescia, he is a postdoc at UBC. So he has several joint work with me. I can just online. So first I will give an introduction. So I will give three constructions for self-similar solutions. So the first one is all of them are based on a prior bounds. But the pros are different. So the first construction is based on herd estimate near individual time for local delivery solutions. So that was first done by GRSFRAC for self-similar solutions and then for myself for DSS solutions. And this construction works only in R3. And the same construction works only in half space. And the prior bounds are based on method of contradiction and reduction to a real field theorem for all the equations. So this is a classical method invented by LeRae. And so it's a joint work of Kolopkov and myself. And this works only for self-similar solutions, not for DSS solutions. And the third construction is based on a new explicit applied estimate. So it's a weak solution approach. And it works in every cases. So self-similar and discrete self-similar and also whole space and half space was done last year. And then we work out the rotating version this year. So my original title did not have this part. But CJ was, in my audience, three months ago in a talk in Harvard. So I decided to give this actual new part to make this more interesting to her. And finally, I will remark on the backward case. So here's the table. So with three methods, they are based on different things. The first method is based on non-field stock equation itself. The second one and third one are based on the LeRae equations. And so this is a, OK, so the applied bounds are put in different ways. So the first one is based on the graduate criteria. And so the bounds, we get is a point-wise bounds. The solutions here are strong. The second way is based on contradiction and Euler equation. It's H1 bound. And the last way is a explicit cut-off. And it's usual energy bound. OK, so introduction. So here's the usual non-field stock equations in R3 times positive time, with the usual data U0, divergent free. So here are the notation. And you drop a linear term, we call it a stock system. So it has two basic properties. The first property is the applied bound. The linear term is anti-symmetric.  So this is a usual applied bound for non-field stock equations. And here we don't see the linear term. So it's the same as the stock system. And it was based on this applied bound, the LeRae contracted solutions in whole space R3 for any L2 data and hope did it for domain case. And now open questions, since LeRae's work is uniqueness and regularity. And the second property of the equation is a scaling variance. So if I have one solution, and for any lambda greater than 0, I can define U0 superscree lambda to be given by this formula.  And related to this property is the ideal mild solutions. It's treated a linear term as a source term to a stock system and contract solutions by the Picard iteration. And so if U0 is in LQ, then that's a unique local in time solution called mild solutions. And for Q equals 3, this solution would be global if L3 no is small. And so now comes the self-similarity. So it would say it is self-similar if U and U lambda are equal for every lambda. So in this case, the value of U decided by its value at any time moment. And then the discrete version is that U equal to U lambda only for one lambda. This does not need to be for every lambda, but just for one lambda. So in this case, the value of U is decided by its value in this time interval t between 1 and lambda squared. So maybe draw a picture. So for self-similar solutions, the value of U is decided by the value of this time instance. And then the value of U at any later time is decided by a corresponding point at the earlier time. And then if it's discrete self-similar, then the value is decided in this time slice. So self-similar object appears in minimal surfaces and gas dynamics. And discrete self-similar object appears in factors. So a counter set is discrete self-similar with factor 3. And in cosmology, so here's a quotation from a survey paper. It says that there's evidence that discrete self-similarity occurs at the mass threshold for the formation of a black hole. So when I look at this mass threshold, I think about the work of Burrow and Kenick. But anyway, so I don't miss them in the talk here in 2009. And this is also basic. It's a Schaefer's example, which I will explain later. And so the forward and backward distinguishes positive time and negative time cases. And when it is stationary, we say stationary, there's no time dependence. So here, we'll take a parameter a to be 1 minus 1 and 0 correspondingly. So it's different from other equations discussed in this workshop. Navier-Stokes equation is time irreversible. So forward and backward are different. And so I want to talk about the decay of a typical solution. So if the self-similar, or if the u0 is self-similar, then it's very simple. u0 will be, if u0 is self-similar, then u0 of x will be equal to its value of uniball divided by its length. So it's mass homogeneous. And so it has to have this decay. And so the natural space for u0 is L3 weak. And the corresponding forward solution will satisfy this point-wise decay. C0 first goes to t plus x. So this t could be part of the negative. But here, I'm talking about the forward solution. And so because of this problem, u is already beating in time and L3 weak in space. So a backward solution, satisfyingly above bound, is called the type 1 singularity, if it is singularity. So the existence is an open question, except actually a similar case, which I did with Bob a few years ago. So I decided to highlight every person who is present in this workshop. So now here's the real equation. And it comes from the similarity transform. So for a equal to plus or minus 1, we define the similarity variables. So u xt now changes to capital U with variable y and s.  And y is x over root t. And the scalar time s is a log of t. Then this capital U will satisfy the following equation. And this box part is a new term compared to a log of t equations. So this comes from the time derivative of this guy in this factor and in this factor y. So that's always new. So the behavior of capital U as s goes to positive infinity in caused behavior of little u at infinity. So that means the time asymptotics when a is positive. And it's in caused behavior near a singular time if a is negative. And there's an important difference here. So little u is self-similar only if capital U is independent of s. And little u is discretely self-similar if only if capital U is periodic in s. So a little bit history. So the stationary equation was written down by the Ray in his 1934 paper. And then motivated by this problem, Gigan-Kong introduced similarity variables. And he used that to study nonlinear heated equations. And then it was extended by many people including Merle and Zag and also to energy critical wave equations by Koenig and Merle. And so now it's a bit about the Schaefer's example. So Schaefer in 85 has a paper, which is a factor field defined for negative time over nonlinear stocks equations with a singular force. However, this singular force is speed-reducing. So f dot u is non-positive point-wise. So normally speaking, this f should help to make u smooth. But this f u is actually a singular has a singularity at the origin.  So f has a difference getting low. So this was 85. So that was 30 years ago. So it was there. And so now the question about the existing problem for forward self-similar solutions. So for given self-similar initial data, want to contract the self-similar solution. And the natural space here is a low infinity in time. And then x should be L3 weak or some other critical spaces. And for small data, so when c star is small, and then we have this unique, we can use a unique existing theory for mild solutions. And so this has been done for many answers, including Giga, Miyakawa, Kato, Kanon-Plenchon, Paraza, and Kog-Tatalu in different spaces. So the theory for small data is very satisfactory. So this also includes discretely self-similar solutions and the rotating versions. So now the question is, what want to construct solutions for large data? What puts c star large? So when c star is large, we don't have an existing theory for mild solutions. And the usual low-week solutions does not contain such new data. And so a theory, due to the Mahi Newset, local data solutions is what is useful. But then this is a weak solution theory, so it has no uniqueness result. So for given self-similar data, we cannot guarantee that the solution itself is self-similar. So in 2012, Giga and the Sepharic contracted self-similar solutions for every self-similar initial data, even small or large, which is required to be locally heard continuous. So I highlight Giga because his work on wave equation was a topic of a technique. So what do we care about these guys? So here is one page notes on the reference to the non-uniqueness problem. So we say that the uniqueness problem for the Navier-Store equations for any L02 initial data is an open question. So here is one way to control non-uniqueness. So we consider a self-similar solution w sigma corresponding to initial data sigma times u0. So u0 is fixed. A sigma is available. So for sigma small, w sigma is unique. However, for one increase sigma, what might get by vacation? And by vacation has two different types at least. So if by vacation is a sudden not type, then we will get two self-similar solutions, w sigma and w sigma prime with the same initial data. And we could get the whole by vacation. And then the new solutions would be time periodic in the similarity variables in the linear equations. So they correspond to DSS solutions. And so these guys are L03 infinity valued. And just first show that if we have such solutions, we can prove non-uniqueness in the real-hope class by cut-off. So that's why they are interesting. But so far, it's not easy to prove by vacation. So these are introduction. So what do I do? Can I just? No, it refuses to echo. I don't believe it. OK, this is my part one. So I need to speed up. So it's a first construction using heard estimate in linear initial time. So the basis of GR and SPRIC is they have a local heard estimate for local linear solutions. So it's iterated in a framework of local linear solutions. So which you will give estimate here and also here. It's a short time estimate. But then by self-similarity, they have estimate below the parabola. And then for the middle part, they just use regularity CRV for stationary solutions. And then, the only way to get a prior estimate and they apply the rich auto theorem. So this I skip. For DSS solutions, the story is a bit different. Because in the middle, in this part, it's not stationary. I wanted to have regularity. So I had to require short time. So it cannot be arbitrary. So here is my theorem. I say that if lambda minus 1 is sufficient small, which says that this time interval is sufficient small, respect to the size of data. Or if u0 has some symmetry, then we have a regularity theory to guarantee bounds in this middle region. OK. The rest is the same, the rich auto theorem. And here it's important to note that the regularity gives us companies, which is essential for the rich auto theorem. And then the same construction, we want to work on R3 high space. In high space, there is no local solution theory. The difficulty is that it's very hard to infer the formula for the pressure from a solution if the solution is not very well localized. And then what we do is that we look at the regularity equation directly. And then we want to control the solution. And then so here, this equation is coupled with the boundary condition u0x. u0x is the Taiwan map of the usual data u0. So the usual data for original little u becomes the boundary data for the Ray's equation. And then u0x is declared as 1 or x. What we expect is that the difference here has a better decay is in L2. And so the theory says that, OK, so A is a Stokes operator in high space. For any self-similar, for any self-similar usual data u0, which is c1 local, satisfying the necessary condition, then there's a subunit self-similar model solution in this usual class, L3 infinity valued. And then the difference between the solution and the linearity part is decays in L2, like t to 1 quarter. And our proof actually would work for any call. So the proof works for R3 and R3 plus. If you work for any call, you can prove this data. This image says that for A to be the Stokes operator in omega, u0 is the Taiwan map of usual data, u0. 1, u0 to be L6, and it's gradient to be L2. So this is where we use that u data is c1. So L2 is a better decay than L3 weak. And I don't know any semi-group theory which guarantees a better decay for the gradient. So that's why we can only do this in high space, not in any other call. So this beta decay is essential for our prior estimate. We could do this. We could prove something actually much better using Solikov's point-wise estimate for green tensors in high space. So here's sketch. So we look at the equation for the difference. So the equation for difference is this one. And then the idea is the proof has three steps. So step one is to show that solutions to this equation with a parameter lambda multiplied to a right-hand side has a uniform bound, Cr. And we'll prove this by contradiction. I will mention contradiction later. And second step to show that in the previous step, the constant Cr is actually independent of r. And so if we can do these two steps, we can apply the rich-older theory to get solutions in a half ball r and then take limits r to infinity to get a solution in r3 plus. So that is the lowest method called invading domains. So the key here is step one, two, by basic contradiction. So suppose we don't have a uniform bound. We need a solution vk to have a h1 bound jk. And look at the normalized vk. And then we'll converge it weekly to a solution to the Euler equation, v h1 0 over step 2 h1 half space of Euler equation. But then with this important inequality, v dot greater v dot u0 is at least 1. And we need to show such solutions does not exist. And this problem is easy when the boundary omega is connected. And it's a difficult question if a boundary omega has at least two components. And so there is a design work in a DSS setting. So it does not work in a time periodic case for the Ray equations because the lag of time companies. So I mentioned earlier, for the Ray-Shadow theory, it's important to have time companies. So I'll probably hear the limit of vk hat in the time periodic case does not satisfy any equation because the time derivative just blows up. So it's related to this following open problem. So for any domain 2D, 2D is easier than 3D. But this 2D is open. So for any open domain with the most multi-connected boundary, and then with periodic boundary data satisfying the necessary condition, do we have a periodic solution? And so this is the open question. Part three is my third construction. So it's important to know that this is the weak solution theory. So my data would be quite general. My data is only required to be L3 weak, not heard continuous, or C1. So we start with the DSS initial data U0, which is divergent-free. But we don't require symmetry over small lambda minus 1. So we don't have regularity. So what can I apply the Ray-Schilder theorem? So what we do is we want to study the Ray's system and look for time periodic solutions. So I denote U1 to be the corresponding vector field after similarity transform of this linear solution. So this t, the periodic t, is equal to 2 log lambda. So it's a log of lambda squared. So here's a new observation. So here's an equation for the difference v, u minus u1. So this is similar to what I wrote in part two, which is the time derivative here. This is only thing, which is new. So you multiply this equation by v and integrate, from time 0 to time t. So in this way, this term will drop out. So formally, this left-hand side equal to this right-hand side. So this is quadratic. And this part is cubic in v and u1.  Because this term can be very, very large and cannot be controlled by this term. And there are examples showing that this term is bigger than this term. And what our new observation is that, unlike usual non-field story equations, we have this term here. This is only available for the linear equation, not for non-field story equations. Well, this new term could help us. And also, this u1 has a case like 1 over x. So these two combined enable us to control this term. So this is much better the situation for non-field stocks. So we'll replace u1 by a cut-off w. kasiya basically keeps the outside part for large radius u1. And then this correction term to make w diverges free. And this correction term is global. It's not localized. It's convenient this way. So w is divergent free. And its difference with u1 is coming from this part. It has a decadent y to a negative 2. So that's acceptable. So for error 3, OK, I can say that u0 is in error 3 in this annulus if u0 is in error 3 weak. That's a theorem for DHS factor fields. So if u0 is in this space, then we can prove that the LQ term of w for any q bigger than 3 is actually decaying when r goes to infinity. So when we take r large enough, w is small in LQ. So for a difference term, this v dot gradient v dot w turn is controlled by this. And then this term is not small.  So this is a new observation for the Ray equations. So again, this term is important. Without this term, here one can only take error 3 weak, not anything else. So here's our theorem last year. So for any divergence free in your data, which is DSS and in error 3, v lambda minus b1. So as I said, this is equivalent to that. It is in error 3 weak in the whole space. Then there's a local solution, which is also DSS with the same lambda. And the difference between u and the linear path in error 2 is decayed like t to 1 quarter. So what is the interesting of this solution is that, yes? That sort of position that replaces like the initial condition, right? You are saying like. That's the initial condition. You know when you write this, and you are also saying that u of u at equal 0 is equal to u 0, right? Let's give it by this, right?  It depends on which sense. You have to talk about your definition of initial data. So usually, this is a global norm. But for local solutions, for example, is that the converges error 2 in any compact set. So it's a weaker. It's OK to be a weaker. And for the theory of error 3 weak, it's usually weak convergence is sufficient. But here, this is stronger. This is error 2. So you have a strong information. Right. But it's so similar, or DSS. So any other question? I have recovered my time, so I should slow down. So that's an interesting scenario. So only in this part 3, this serve construction, the solutions could be singular. So the previous two constructions, the solutions are all have a point-wise bound. But this solution could have singularity. And suppose it is singular at a particular point, x0, t0. And then, by DSS property, it's also singular at all this risk-gilled spacetime point. So it's from time 0 to up to time infinity. So we say that uxt does not become regular for t sufficiently large. And its after-statement is called eventual regularity. So this says that it's possible for a solution to lose eventual regularity. So in contrast, there's a simple theory saying the following. So any local linear solution, any, is any. We don't have any structure, but we just need data to be in L3, R3, not L3 weak. Then this solution must have eventual regularity. So it's this borderline difference which allows the failure of eventual regularity. So at the beginning of my project, I was actually trying to prove the opposite. I want to show that any local linear solutions with new data in L3 weak has eventual regularity. And then I realized that it's not possible. Schedule proof. So this, so I need local linear solutions. So I need pressure. So this is usually the Ray multiplication. But I do this in the Ray equation side. I do multiplication here. And then I had to solve a periodic, time periodic solution from 0 to type of t. And then this is in a gargoyle approximation. And so the Ray equation has some damping. So the size of v at the type of t is less than some sigma, less than 1, theta less than 1 per constant. And then one can use for all of this point to construct a periodic solution. And then take a limit first in k. And then we still have epsilon, back to physical variables to recover the pressure p epsilon. And then take the limit. So here we could call slim, who is somewhere here. But there are many people who did these kind of things. So that was part c. So I talk about all three constructions. So the part four is a new part, notative versions. So here I need to fix my notation. My notations will be notation about the x-axis. So you just discuss. And so these notations, they commute with each other. So their derivative has a simple form. It's jr or equal to rj, j, just this matrix. So these kind of solutions was first proposed by Grisha Perlman for backwards solutions a decade ago as a candidate for singularity. He asked, does this kind of solution exist in the critical space? But they both couldn't solve it. So here I managed to set Yifam. So I was working with him on the new Schrodinger equations at the time. So I have set it all three organizers of this workshop in this transparency. Is there a fourth one? No. OK, very good. So here's the definition. We say u is a notative self-similar. If for some alpha real number, uxt is equal to this formula for every space point for any time and for any scaling factor lambda. So here the new part is this notation part, outside and inside. So you can see that first they have opposite sign here. And then they are both. So it's a number of alpha times the log of this scaling factor lambda. So we actually mentioned why log. So the choice that the angle theta of lambda is for alpha log lambda is natural because we want this to be true for every lambda. And so because of that, we need the sum of theta lambda plus mu has to be equal to theta of the product. So because then this gives us this formula here. And then notative self-similar solutions is always discrete self-similar. Just choose any lambda, such that alpha times log lambda is 2 pi times the integer. And if alpha is 0, it is reduced to self-similar. And then so this is a formula here. Notative self-similar. So this formula works for forward case and also for positive case, for backward case. So now for forward case, we say lambda to be t to negative 1 half. Then this u is given by this formula here now. So what it says is that, so this is time 1. So it's the same as this picture. The value of u is determined by its value at the end of fixed time, in particular at time 1. And now rotate it is quickly self-similar factor fields. So its definition is actually easier. So we need two numbers, lambda and theta. We require UST to be equal to this formula here. So we are theta inside and are negative theta outside. And this is true for ABXT, but only for 1 lambda. So there are some particular cases. If theta is 2 pi times the integer, or then r theta is just identity. So it's lambda dSs. If n theta is 2 pi times the integer, then u is dSs with vector lambda to power n, which is iterative formula n times. And then rotation part will return to identity. However, if theta over 2 pi is irrational, then u is not dSs. In this case, I think it's quasi periodic in the Ray equation side. And so for any time positive, we can choose tau between 1 and lambda squared, which is lambda to 2k times t. And then UST can be expressed by the value of u at time tau. So u is also decided by its values at the slice from 1 to lambda squared. So now, similarly, transform. So first, u0 is the rdss with vector lambda. And the phase theta, we have to choose the alpha. And this alpha is not unique. We can choose any alpha k for any integer k, just to be this guy. And then we define the new variables. So this part is the auto one. But now we do a rotation here, r, alpha s, outside of v. And then in the inner variable, we'll do an inference rotation. This is an inference rotation of y. And then the equation for original Laffy-Stoy equations, u is equivalent to this new equation. So this part is there for the Ray equation. And this box part comes from the rotation. So y is from outside. y is from inside. So at the beginning of my project, I chose z to be y. And thankfully, I thought that that's a rotation. That's good enough. But then the Launier terms does not change well. So they have to be coupled. Otherwise, Launier term is not preserved. And divergent free-taste is also not preserved. And this is actually quite natural. So I forgot to put in the slides the study for rotating fluid with a rotating obstacle. It's a very well-studied object. So rotating obstacle in fluid. So if you do Google search, you will see several papers on this subject. And there, I think they also have these two terms with different forms. But of course, they don't have these two terms coming from self-similarity. So the new terms, OK, they are new. But they are divergent free by direct computation. And the new terms are suckled to v in no two sense. And so it will have the same applied bounds as the Ray questions. And then RSSU correspond to stationary v. And RDSSU correspond to periodic v. And then so the proof itself is not very hard for this theory. So it's similar. So for any L3 week, usual data U0 in the whole space or in half space, if it uses RSS, then there's a RSS solution. This is a typo. This is RSS. U0 is RDSS. Then there's solution, which is also RDSS. So the only thing I require for data is L3 week. So that's everything on my work. And so I just give two pages like a new mark on the back work case. So what I know is that in the back work case, self-similar solutions do not exist. If U is L5 in time, time here is a negative L3 week. So this was done by my advisor and his advisor and my academic uncle in the L3 case. And then I did a case for L3 week. And then DSS solutions do not exist. If UT is L5 and L3 in space, so this is a separate work of Escalariza, Serakis, and Seferik. But then the existence of DSS solutions in this class is unknown. So here I also mentioned that Schaefer's example of single factor fields DSS. So these are basically what are known. And then the second page, the open problems, problem two. So problem one was the existence of periodic solutions for multi-connected boundary. So problem two is this. So existence of discrete self-similar solutions satisfying this critical bound. And the special case of problem two is the problem of Briecher-Perilman. He said that the existence of rotated self-similar solutions satisfying the same bound. I put question mark here because I didn't talk to him. So what I learned are a few sentences from Serakis. So this is my guess. Here are a few remarks. So are SS solutions you can deal with a stationary equation? So that is good. However, for this stationary solution, there's no maximum principle unlike the case studied by my family. So this is very hard. Finally, problem three is I think it's possible to improve Schroeder's example from DSS to self-similar. But I don't know how to do it yet. OK, and that's the end. Thank you. Right, do we have questions? I just wanted to go back to part one of your talk.  And you claim that you know it from the half space because you know the half space, but you don't know on the cone. So what are some difficulties when you're trying to show it on, say, certain types of cones? I say quarter space or? Yeah, I thought about a quarter space. However, Solonium of estimates for green tensors is already very difficult in half space. It's a, his estimate is not complete. His estimate is only for divergence free initial data, which has zero normal component. But then, so that does not apply to non-field storage equations because the source term V dot gradient V will be on the right hand side, and it's not divergent free. So even for half space, the available estimates are very limited. And I don't know any estimate of this sort in quarter space.  But at the beginning of the talk, you have mentioned that your BSS solution is related to the formation of black holes and some applications in cosmology, right? Yes. My question is, I noticed that we are looking at incompressible fluid with viscosity. And in an incompressible case, I think the speed of the sonic speed is infinite. Is this a conflict with the principle of general or special relativity? I think self-similar solutions are. Is there any some properties of finite speed propagation to be in the region of general or special relativity? So your question about general relativity. So general relativity is, in some sense, tiny reversible. But we have self-similar objects in many tiny reversible systems. So that's my question, but how is the relation of this? I say it's not really related to. So I mean the dissipation, the presence of dissipation is not a deciding factor for self-similar objects. Actually, OK, it may have less flexibility. But actually, without dissipation, there is a bigger chance of self-similar object. That's my opinion. And is there some possibility to regard, for example, the compressible fluid? Compressible fluid, yes. It's because compressible fluid in the region, you have finite speed of sun. I see gas dynamics is compressible, right? Is it to relate your rotating self-similar solution with this problem, with the obstacle? The problem you have on the board, the obstacle problem of the obstacle? This one is different, because for rotating obstacles, the actual is rotating. So they choose a coordinate fixed to the obstacle so that the domain becomes the exterior domain and the fixed. So it wasn't necessary for them. OK, so there is no direct relation, right? No, but the derivation of equation are the same. Thank you.


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Let us begin our electrostatics. So, now that we have had a reasonable amount of discussion on the vector calculus required for it, let me now go over to what we actually want to do. We are interested in talking about electric field. So, I will not waste a lot of time on some of these things because this is all very familiar to you, but it is a good idea to realize that physics is not in isolation. No matter what you are doing, whichever part of physics you are doing, you need a set of building blocks and the fundamental law of electrostatics is what is known as Coulomb's law and we all know that the Coulomb's law, I will not state the way you are doing it in school. They like charges, repel, unlike charges attract all those you know, but basically this is the idea that I will summarize the essential point of this. Supposing I have a charge q1 here at a position vector r1 and a charge q2 there at a position vector r2, so that the vector r is the relative position vector of q2 with respect to q1, then the force on the second particle due to 1 is given by 1 over 4 pi epsilon 0 which is a constant that comes q1 q2 the product of the charges divided by square of the distance between them and a vector r is there. Depending upon whether q1 q2 is positive or negative the product, I have either a repulsive force or an attractive force. So what one does very frequently is to write this instead of a unit vector and r square make this an r cube and a vector r this is the same thing really and that is the inverse square law. And since by Newton's third law the force due on 2 due to 1 is the same as force due on 1 due to 2, but with a negative sign I have f12 equal to f21. So that is my Coulomb's law and the Newton's third law, let us summarize this what is so special about Newton's law what are the points that one has to know first inverse square law it is a inverse square law force of course we already know that that the force varies inversely as the square of the distance like charges repel unlike charges attract is also something which you know. The third point which is an important point which should be made very clear it is a long range force. You see a long range force means it is a force which really never becomes 0 never becomes 0 no matter how far you keep one from the other it may become weaker and weaker, but never becomes 0. So inverse square law is a long range force another important point is the force is a central force meaning thereby that the magnitude of the force only depends on the distance between the two and the direction is along the line joining charges. So this is the central force and well it we have defined something called a permittivity of the free space which is an epsilon 0. Now I want to summarize the Coulomb's law force with a bit of a comparison of various types of forces. Now you see in the nature we only have four types of basic forces does any force that you think of can belong to one or the other of this category. Now in terms from weak to strong from weak force to strong force the weakest force is the gravitational force and taking some scale of one which is the strength of the strong nuclear force between nucleons attractive nuclear force between the nucleons. The gravitational force has a strength which is 10 to the power minus 39 or 38 times the strength of a nuclear force which is essentially a negligible force and its range is infinite as I told you it is a long range force. The next in order of strength increasing order of strength comes a weak nuclear force and that has a relative strength of 10 to the power minus 6 with respect to the strong force and it has a range of 10 to the power minus 18 meters. Remember gravitational force is infinite only ranged that is long range the nuclear force is very short range but not that short but it has a 10 to the power minus 18 meter. Now the real strong force with which we deal with leaving aside the nucleus is the electromagnetic force and the relative strength of it is about 1 by 100 actually a typical characteristic number is E square by C H cross which is 1 over 137. So that is the fine structure constant which is 1 over 137 but roughly less than 1 percent of the strength of the nuclear force and the strong nuclear force with the range of a water farming which binds the nucleons namely neutrons and protons together that has a strength of 1. So since we are talking about the fundamental four fundamental forces there is also a question of how do these things act. Now remember that I have supposing I put two charges now they have a force even when they do not have to be in contact in order that there is a force they have a force even when they are separated by a distance the question is how does one particle know that the other particle is around. So, as to either attract or repair. So, this is a question that has bothered people for a very long time that you know I mean. So, what actually happens now this is the reason why we are learning about the field what we say is this that let me give all my explanation with respect to the electric magnetic forces. So, what we say is that when you have a charge now the all the field all the space around that charge is a seat of electric field the region carries the information that there is a charge there. Now, so when another charge comes to that region it has it is of course, own region, but when another charge comes it gets influenced by this field. Now this is let me give you a very bad example similarity that is supposing you are in the in the presence of a person of very strong personality the fact of his presence is sometimes felt by you. Now as I said it is a bad example, but this is precisely what we are talking about we are saying the fact that one charge is there it creates its field and since it is the range of electromagnetic forces infinite it essentially means that this information is there all over the space. In the modern language what we believe is that these charges are or any objects where we are looking for forces they continuously exchange certain things between them and these exchange particles are bosons since you are all teachers I do not have to explain to you. So, the carriers of long range forces which are the electromagnetic forces they are massless and in this case it is photon and as you know photon does not carry any charge. So, that is all about the Coulomb's law now I want you to realize another very important concept. So, I start with the building blocks the first building block is the existence of Coulomb's law which is of course, an axiom this second axiom is that supposing you are looking for calculating the electric field at a point due to let us say multiple charges. Now the point is that how do what is the effect at a point due to the presence of multiple charges they remember I told you that each one of them has their own electric field. Now this is where we talk about what is called superposition principle what we are saying is this that suppose I have a charge q at let us say some point and I said already that all region around it is a seat of electric field. Now if I have an electric field whatever is the reason for of its creation supposing there is an electric field at a point p and that field is e if a charge q finds itself at the point p then it experiences a force q times e. So, basically the electric field is the force exerted on a unit charge at a particular point. So, what we are now saying is this that supposing there are charges q i's at different point let us suppose R i's are the positions at which the charge q i's are there then the electric field at a point p is simply given by a sum remember it is very important it is not given by product or division or by anything it is given by a vector sum of the forces exerted on that a part charge at the point p due to all the charges q i's. Now since it is an electric field at the point p I put a unit charge. Now this unit charge due to the presence of the charge q i at the point R i experiences a Coulomb's law force 1 over 4 pi epsilon 0 q i this is the relative position vector of the point p with respect to the position R i of the point where q i is there divided by of course R cube as I told you this is Coulomb's law. Now this is a very important concept we are trying to say that physical effects add up in other words if there is a force exerted on a charge due to one field is E 1 due to another field is E 2 then when they are simultaneously there it will be the vector sum of E 1 plus E 2. Now notice this is not something which you can prove it is a superposition principle which is axiomatic that I take Coulomb's law and the superposition principle as the basic building block of my electromagnetism. So having defined what is an electric field and having talked about the principle of superposition I need to know there is a question raised which I said I will take up this is almost now time come for discussing the electric flux and the Gauss's law. So remember we defined a flux of a vector field we said that if you have a surface then you define the flux as the surface integral of the vector field over it which we defined in a particular way. Now supposing this field now is my electric field now that this tells me that I can define the flux through any surface as given by this is the definition of flux the surface integral of the electric field over that surface and I have told you that the definition of ds vector is the outward normal. So let me now come to this question which came up the question was that why is it that when we put a charge inside I have a flux coming out positive or negative will depend upon the sign of the charge let us not worry about that then but on the other hand if the charge is outside I do not have a flux now let us look at it in a slightly different way. But before I do that let me explain the concept of what is known as a solid angle look at this supposing I have a surface here this is a surface and I am looking at a point p. Now I would say I will define that a surface makes a solid angle at a point p I will sort of explain it to you by comparison with something else. So let me come back and try to explain this. So let us first go to the definition of an angle ordinary angle how do I define an ordinary angle remember that this is the way we looked at it supposing this angle is theta. Now if this angle is theta how do I define this it is an ordinary angle. Now what we did is to say that look if this remember angle is a dimensionless quantity the you may sometimes measure it in degrees radians or whatever but angle does not have a dimension and the reason is this supposing I look at this situation where this length is L and this is let us say R then my L is given by R theta this was your definition of ordinary angle theta of course is in radians this is my definition of ordinary angle. So what did I do I said that look take a curve which is remember that this curve is such that these are two radial lines. So therefore these are perpendicular to it take a curve and then divide the length that you make between these two points by R and that will give you angle theta but supposing you had this a different length then also the definition is true accepting that you have to take the projection of this along the normal to this one. So this is basically and this is also the definition tells you that theta is dimensionless because L is a length unit R is a length unit I may have a unit but I may not have a dimension. Now the so far as this thing is concerned now the solid angle is concerned now it is a straight forward generalization of whatever I told you just now. So accepting that we had said that in defining a an angle we look at a curve and ask the what is the angle that is subtended between the two radial lines which intersect that arc. Now here what we do is this we go one step and we say supposing I have a surface. Now this surface at a point P makes a cone like structure remember the process is in three dimension earlier we are in two dimension. Now so what we said is in the earlier case an arc made an angle ordinary angle but in this case a surface is making a type of a cone angle and that is what I will call as the solid angle look at this picture. So what we are trying to say is this that here is a surface I have taken it to be small because most of our definitions are defined like this. So what we are saying is imagine not keep on drawing tangents from point P to various points on that surface and then you will get a cone angle here. Now remember I define the angle theta as the ratio of the length to the distance. Now in this case I define a solid angle as the area like in the other case I also take the perpendicular area that is the area which is normal to those and divide it by because of dimensional reason I do not divide by a length but I divide it by a square of a length. So once again if this surface is not directly perpendicular then you take its projection. In other words if the outward normal to this surface is making an angle alpha with the radially outward direction then you have ds perpendicular as ds cosine alpha divided by r square. So that is the solid angle that this surface element subtends at the point P. Tomorrow I will take up the questions that have come in because we will collect all the similar questions together but what I will do tomorrow is this. I will go repeat the idea of solid angle and then tell you that how the Gauss's law follows from our idea of what is an electric flux and the idea of a solid angle. The question that was asked is why is it that when we have a charge inside a volume we have an outward flux whereas why is it that if the charge is outside the flux through that surface is equal to 0. So since I have run out of time I will take over from there. Thank you very much.


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So with UNC we talk all the time about their high profile running backs Which makes sense because they're very good, but not the only skill guys in this draft from that team You've also got Diami Brown. What do you see in his profile? I see a big play waiting to happen in 2020 according to pro football focus Fifth and deep catches second and deep yards seventh and averaged up the target He averaged 20 plus yards per reception in each of his last two years So the very least he should be a vertical receiver at the next level and given his size Given his production profile. I think we could see something a little bit more than that Which is why I think his top comp is willful