 Hi, I'm Zor. Welcome to a new Zor education. We continue talking about vector fields. We have already covered gradient of scalar field, which produces the vector field of gradients. We have also covered divergence. Today we will talk about the characteristic, which is called circulation of vector field. Well, obviously you understand that circulation is somehow related to circular character of the vector field. Like a tornado, for example, if our vector field is three-dimensional velocities of each molecule of the air, just as an example. Anyway, this lecture is part of the course called physics for teens. That's the wave part of this course, and in particular part of the topic field waves, because we are gradually moving to the waves of the fields. But we have to know the characteristics of fields, including the circulation, for example. Now it's presented on the website called Unisor.com. The site contains also a prerequisite course, which is called mass routines. I do recommend you to go through material presented in the mass, because we definitely use a lot of math in the physics course. Definitely calculus, definitely vector algebra. Mathesis sometimes. Now the the course contains basically video-recorded lectures like this one, which you are looking at. And every lecture has a textual complement, supplement. It's basically like a textbook. So every lecture has like a chapter of the textbook, which is related to that particular material. There are some problems solved. There are exams. The site is completely free. There are no advertisement, no strings attached. You don't even have to sign in if you don't want to. Exams, you can take exams as many times as you want, until your score is 100 percent, basically. Alright, so let's return to circulation. First of all, I would like to talk about circulation in two-dimensional case, because the subsequent lectures will cover two-dimensional curl of the vector field, which is basically a subsequent topic after the circulation. And then three-dimensional, which basically is reduced to three two-dimensional cases. So one three-dimensional is three two-dimensional cases. Basically, that's how I approach this. So today we'll be talking about two-dimensional vector fields, which means that every point on the plane has a vector coming from it into some direction of some magnitude. That's what basically vector field is. So obviously this vector at point xy, this is a vector. So at every combination of xy, which are coordinates on the Cartesian plane, we have a vector which has certain direction and certain magnitude. And obviously this vector has projections on x-axis and on y-axis. So the vector has two components, obviously. So this is x-component, projection turned to x, and this is y-component, projection on the y. Okay. Now we are assuming that the vector field is not changing with the time right now. So it's a stationary vector field. I mean, all the vectors can be either the same, maybe or different or whatever, but they are not changing with the time. Now, now we would like to introduce the concept of time, but not to the changing of the vector field, but to the movement within this vector field. So let's assume that there is a steady wind, which is blowing into some directions, not necessarily in one direction. Maybe there is some kind of small whirlpool or whatever, doesn't really matter, but it's there. And let's assume that we have a surface of the lake with paper ships on it, or some particles, whatever, and the wind is blowing, and we would like actually to move our paper ship from one place to another, along certain paths, and I'm interested in work, which I'm supposed to do. Now, if I'm working against the wind, so I'm having certain, I have to exert certain effort to go against the wind. Now, the wind is known, there is F, this is basically the force of the wind, which I have to overcome, and I know my trajectory, so in theory I can calculate the work, which I'm supposed to do. If, on the other hand, I'm moving with the wind, then there is also work, but it's kind of negative. So the work in this particular case becomes positive or negative, depending on whether I'm going against the wind, or the wind helps me. How can I calculate this amount of work to basically make some kind of a judgment about, well, which way to, which paths to take from A to B? We will consider three different cases of what kind of a wind actually we have, what kind of vector field we have, and the first and very simple situation is when the wind goes straight along one line, so all these vectors are equal in both direction and magnitude, and my path, my trajectory, is a straight line. So I'm moving from A to B and the wind goes this way. Doesn't matter actually. Where, as long as we are considering the case when all these vectors are exactly the same, same direction, same magnitude, and I'm moving along the straight line. Well, let me just start from a simpler case. What if I'm moving from A to B exactly opposite to the wind, exactly opposite to the force? Well, then I have to exhort efforts, my force supposed to be basically compensated. It's supposed to compensate the force against which I'm going, so I have to exhort force the same F as the wind. These are all F and I have to cover the distance from A to B, and the work is basically force times the distance. We all know that, right? Now, if my path is not parallel to direction, there is a slight modification to this, cosine of alpha, cosine of angle. Why? Because, well, if in this particular case, my force goes this way, the force can also, can be represented perpendicular to my path and parallel to my path, right? So these are two components. Now, the components which is perpendicular to my path doesn't really resist my movement, but this component, which is exactly opposite, it does. So I have to really take only this component into consideration, and this component is equal to, if I would take the vector F, and I would take the vector AB from A to B, the way how I'm moving actually, then their scalar product would be exactly the multiplication of this force, the projection, by lengths of this vector, and the proper angle will be, because the scalar product is magnitude of this times magnitude of this times cosine of angle between them, so that would be basically my work, which is the same as this one. So this is the simple case, and it's just elementary problem in mechanics. Whenever we were talking about dynamics in the part of mechanics of this course, even I, we can actually take a look at the lectures of this particular course. That's where it's all explained. Okay, so the first case is simple. That's uniform vector field and straight line path. Now, just as an example, what if my force is perpendicular to my path? Well, then obviously this is equal to zero, because I don't have to really, my, my, my, wind does not resist my movement against this particular path. Now, if my force is directed this way, it helps me. So, then basically I will have the negative negative scalar product. Well, it obviously depends on how we count the the angle. So the cosine of zero is equal to one, and the cosine of 180 is minus one, so it all depends. So the sign is kind of not as important right now, but the magnitude is definitely correct in this particular case. Plus or minus depends on whether I'm going from A to B or from B to A and how I calculate the angle from from the vector to from one vector to another or vice versa. So sign is basically not that important. In any case, this is basically the correct representation of the work, which is supposed to be done either by me, who is moving these paper ships, or by wind itself. So it depends on who I am actually, which side I am representing. Okay, fine. So that's a simple case. Now, let's consider we have a different case. We still have a straight line movement, but our forces, our our wind, it's not, so this is again from A to B, but our forces are different, and they still have to basically calculate work, which either I'm supposed to do or the wind is doing, whatever, moving the paper ship from one location on the surface of the lake to another. Okay. Well, how do we approach it? Well, first of all, for simplicity, I will use the system of coordinates. This is y, and this is x. Doesn't really matter. I mean, it's all up to me. I know my paths from A to B, so I established this particular type of coordinates. Now, in this coordinates, every vector has some kind of magnitude and direction. Okay. Now, again, obviously every vector can be represented as this type of thing. Now, as I'm moving from A to B, my y-coordinate is always zero, right? So I don't really have to talk about any f of x, y. I'm basically talking about f of x, zero. So vectors which are at this particular point, they are moving in some direction. So I'm interested to analyze these, because all these are not important. I don't go through these points. I'm going only through these points, and the wind in these points is important. Okay. So now, this thing, this vector, each vector, again has two components. It has f of x component, and it has a y component. Obviously, depending on the coordinate x, zero, x, zero. Okay. So this is my x component. From this vector, for example, this is my x component, and this is my y component. This is fx, this is fy. And again, since fy is perpendicular to my paths, I don't care about it, and I'm only involved with fx, because only fx goes either helping me or resisting me, right? Okay. Now, how can I basically calculate the work which is supposed to be done here? Well, I'm doing exactly the same as I've done with whenever we were approaching integration. If you remember, if you would like to have some kind of an area under the curve, we break the argument into small pieces, x0, x1, etc., xn. Then we consider this to be almost like a rectangle with the height being equal to f of xi, right? This is xi. And then we will just multiply fxi times delta xi, and then we sum them together. And that will be, this is f of xi. This is delta xi in between these. And then we will summarize them, and that would be the total area, and then we will go to a limit when the number of these points is increasing to infinity with each one shrinking to zero. That's exactly the same approach which we will take here. So, there is nothing new here. What I will do here is I will break from a to b, this would be my x0, it would be my xn, and I will break them into equal parts. Each one is equal to delta x. Let's say it's equal. It doesn't really matter, but let's, for simplicity, consider it's equal. All right? So, this is my xi, this is my xi plus 1, this is my xi minus 1, right? Now, at each interval from xa minus 1 to xi, I consider that the strength of the horizontal component of my wind is not really changing. Now, it's basically something which mathematicians are much more sensitive than physicists. Physicists always assume that something like a smoothness of anything which they're dealing with exists. Mathematicians are more sensitive, they're basically asking, is it a smooth function or something like this? Smooth in terms of existing some kind of limits, etc. So, we do assume, as physicists do, that we can consider the wind is not changing on this small infinitesimal, ultimately it will be infinitesimal segment. So, during this particular piece of the pass, the wind is exactly the same. And it's equal to f of fx of xi comma zero. That's the horizontal component, the projection of the x-axis of the vector x at point xi. So, what we have to do is, now we have this projection of the vector, we can multiply it by delta x. So, we have the distance which is delta x, it's between xi minus 1 as xi. We have the strength of the force which is this one. And then we will summarize it, i from 0 from 1 actually, from 1 to n, to get all the work done on each particular segment of our pass. And then we will take the limit when delta x goes to zero. That's the approach. And if I know this particular function, if my vector field is given as a function, and by the way, this is the function of one argument right now, right? It's a function of one argument of x. So, basically the result would be integral from a to b, from zero to b, whatever, of function fx of x zero dx. So, if my function fx is given, and it is given because my field, vector field is given, which means I know this at every point, x and y. So, that's why I know projection onto x-axis. And I can take this integral. So, the straight pass is fine. Now, let's go to a curve pass, my third option. So, I'm trying to do it like in a sequentially more difficult cases. Now, speaking mathematically, we are talking right now about introduction into integral along a curve, basically, which is very much similar to regular integral. But in any case, we just have to really spend some time. So, we're talking about a non-uniform vector field f of x, x, y, and we are talking about direction which is not really straight. This is a, this is b. Now, how can I calculate this particular work? Okay. Now, I think it would be easier if my way from a to b is measured in time. Now, my field remains the same within the time frame. But my movement, so my position is changing. So, somehow, I actually have to have my position, my position is x, y. And I have to have my position specified as some kind of function of time. That what defines the curve. Now, define the straight line, I define basically by a and b. And this is a straight line. So, b has certain lengths and basically that completely defines my pass. If pass is not straight, something like this, I have to define it somehow. So, this is how I define it. So, my position, let's call it r of t vector has two components, my abscissa coordinate as the function of time and my position, my ordinate position. So, that's what defines my pass. Okay. Now, I will do exactly the same thing as before, but instead of dividing a straight line into pieces, I define time into pieces. Let's say I'm at a at t is equal to zero and I'm at b at t equals to some kind of a t. So, I divide time segment from zero to t into small segments from t zero, which is zero to t one, then from t one to t two, etc., from t n minus one to t n, which is my end of time. And let's assume that I divide in such a way that delta t is constant. So, difference in time between t two and t one, same as from t three to t three, t three, etc., so all the different intervals are the same. Now, what am I buying by doing this? Well, now I have positions. Now, this is position at t one, this is position at t two, this is position at t i. This is position of t i minus one. Now, I'm making my delta t is so small that this particular segment of my curve is practically almost approximately straight. Now, if I'm talking about work, I have to have the force at this point and again assume that the force is the same during this period of time. So, the force which is acting at this point is exactly the same as this and this and this. Again, mathematician would probably ask, okay, is that true that the function is smooth enough that whenever we are making small enough distance in time between these two positions, we can assume approximately that the wind will be basically the same along the same straight line. So, I'm basically reducing my micro problem into a certain number of micro problem. Now, every micro problem is exactly like the first one which I was talking about when I have a uniform vector field and straight line pass. So, this is a straight line pass approximately and this is a uniform vector field approximately. So, if I will multiply my lengths by projection, now, if this is, let's say, my force at this particular point. So, I'm representing it as perpendicular to this and parallel to this segment. Obviously, perpendicular is not involved at all. So, I have to find projection on the segment itself. Okay, how can I characterize this particular segment? I need its direction and its length, right? Okay, fortunately enough, physics has its own approach. There is a concept of a speed. Now, speed, velocity, velocity is a speed and direction. So, basically, we're talking about velocity. So, the velocity of my movement at this point, and at this point, we are assuming that delta t is small enough, is always a vector which is tangential to my segment itself. No matter how curvy my way is, velocity is always tangential to my curve. Again, if you remember, if you are moving along the circle, velocity is always tangential to a circle. So, it doesn't really matter how you move. It's always, because velocity is actually calculated based on position at this point, position at this point, and then you have a difference, and you reduce the difference between this point and this point, and basically divide it by the length, divide by time, and that will give you the tangential length. And velocity is, as a vector, is a derivative of position, this is of xy, and this is of xy. So, at any point, velocity is a vector which is x of t, y of t. These x and y are two components of my r vector, the position vector, and their derivative, which is the prime, derivative is basically the vector of force. So, that's what my velocity is, okay? So, I have a vector which is definitely related to this segment since it's tangential, and the delta t is very small, so the points are very close to each other, and obviously the closer points are to each other, the closer the tangential line is to the segment between these two points. When this point is moving towards this one, this segment is directed more and more as a tangential line. That's the definition of the tangential line. Okay, so as you see, we do have some mathematics in this, so we need to understand these simple concepts. Okay, so what's the length of this segment? If I know the speed and I know the time covered from one point to another, my length is v of xy times delta t. Well, the length of this, that's my length of the segment, and direction is actually the direction of the vector. So, I know all the components I need. I need the projection of my vector f onto the direction which is defined by v, so I need f of xy. I have to multiply it as a scalar product by v of xy. This is a scalar product, and that scalar product will give me magnitude of this as projected onto this direction times magnitude of this, which is actually what I want, magnitude of projection of f onto v times the length of this particular vector. All I need is multiplied by delta t, right? So, delta t times v will give me the lengths, this times this will give me this and the projection, so I have my work. This is i's, right? So, I assume this is i, this is i, this is i, and this is i. This is the piece of work which I am actually performing during this. What should I do then? Well, obviously I should summarize it, which is actually and go to a limit when delta t goes to zero, right? When these points are closer and closer to each other, which basically gives me integral from zero to t f of xy a scalar product v of xy dt. Or alternatively it can be expressed differently from zero to t f of xy dr of xy. Well, they are obviously xy, x of t y of t. That's position. Or simply dr times f. That's how in the simplest form it actually expressed. So, this is basically not a simple integral, it's integral along a curve. But it's approached exactly the same as integral of any regular functions. Well, why did they do all this? Why is it called circulation? Okay, basically this particular integral, the last one, is called circulation from point a to point b. It's the definition basically. What's more interesting is that there are certain vector fields when their circulation from a to b would be equal to circulation from a to b by any other pass. So, no matter how we move from a to b, the result, work which we have to really spend, will be the same. Now, these fields are called conservative. For instance, gravitation field and electrostatic fields are possessing this particular quality and they're called conservative. So, what's interesting is, and I'm not going to prove it right now, but maybe I will do it as a problem later on, later on that any conservative field f of x, y is actually a gradient of some scalar field. You remember what gradient is? That was a couple of lectures before. That's the operator df of x, y to dx, df of x, y to dy vector. So, if I have a scalar field, which is function of two arguments basically, there the partial derivative by first argument and partial derivative by second argument is a gradient. Physical sense, physical meaning of the gradient is which direction it changes the fastest, these values. So, it's a vector of fastest change. So, any conservative field is actually a gradient of some function which is called potential. Remember, gravitational potential, electric potential for electrostatic field. From these fields, we can always get the force. And again, maybe we'll just have a couple of problems later on solved in this particular case. Okay, now, if this is true, then what's interesting is that if I'm moving on any closed circuit, let's say, from A to A, in the conservative field, my work, my circulation should be equal to zero. Right? Why? Because I'm just choosing any other point from here to here supposed to be the same from here to here. But integral from here to here is opposite with a sign to move in an opposite direction because I'm moving in my line, my old little pieces are in a different direction. So, from here to here, and then from here to here, which is the same as this one was an opposite sign, would give you zero. So, that's an interesting fact, just by itself. So, if the field is conservative, then the circulation is equal to zero. Well, actually, because of this type of property, the name circulation comes because we're always going around the circle. But in some cases, for some fields, it's not zero. And let's analyze these cases. So, we're talking about something new. Now, imagine you have some kind of a field, and you have to move around some kind of a closed path. Okay, that's some work which is probably not equal to zero if the field is not conservative. But what is the relationship of the work which I'm spending here and the area of this inside this circle? We're talking about two-dimensional. Everything is in two-dimensional. So, this is area. Okay, what's interesting is that if I would like to know the circulation property in one particular point of the field, what I can do is I can surround this field with some paths, have the circulation calculated, and have the area calculated. Now, obviously, if I'm squeezing, if my path is shrinking to the point, both my work is changing down, reducing probably to zero, and my area is reducing to zero. And what physicists have introduced, well, I'm not sure actually who introduced it. They have introduced an interesting characteristic which is called curl. Curl is actually of the field f of x, y. This is a characteristic of the field at the point x, y. And it's a limit of circulation along a closed path divided by area when I am shrinking, shrink my circle around it. Now, when I'm saying circle, it doesn't necessarily mean that it is circular. If the field is smooth enough, and again, this is, mathematician is talking to me right now, you still have to prove that this limit exists, because maybe it depends on how exactly the path is formed. Is it a circle? Is it a rectangle? Is it something else? Will this limit exist? And again, if it does exist, maybe it would be different depending on what kind of a shape of the circular path, or any path, or rectangular path, or any closed path would be. So, right now, I'm completely leaving these issues out. These are relatively complex mathematical issues. But we assume that the field is smooth enough, so that this particular limit indeed exists. And now, I will just go through the calculation of this limit, how it looks. And again, all we need for this, from the mathematical standpoint, a relative smoothness, well, differentiability, if you wish, of the field. So, the field is supposed to be differential, by x, by y, by whatever. But let's try to basically calculate this thing, assuming it exists. All right, what we do is, we will choose a particular path, a rectangular path around my point, this is my point, p of x, y. And I'm choosing the delta x here, delta y here, and I will squeeze these sides of this rectangle. So, I will calculate my circulation around this particular path. And then I will squeeze down to zero delta x and delta y, and see what happens. Well, let's see what happens. If the middle point of this rectangle is x, y, then this point is x minus delta x, y minus delta y divided by two divided by two, right? Half. But this is half of delta x. This point is x plus delta x divided by two y minus delta y divided by two, right? This is x plus delta x divided by two y plus y delta y divided by two, and this one x minus delta x by two, and y plus delta y by two. Now, I assume that this is a very small rectangle. So, the force within each side of this rectangle, my force f of x, y, is the same. And again, that because my force field is really smooth enough, so whenever these points are close enough, there is no change of the vector of force on this segment. All right, so let's just calculate the work which is required in this particular case. My f of x is, has two components, f x and f y. f x would be parallel. f y would be parallel to this, right? So, whenever I'm moving along this side, I have to take into consideration only f x because f y would be perpendicular. And whenever I'm moving along this side and this side, I'm taking into consideration only f y component. All right, so if I move from here to here, my component is equal to f x. Let's take this one, for example. I'll take in the middle. That would be easier, okay? So, the component of this, f x component of this force at this point, this point has what corrosion as a what? x would be the same and y would be minus delta y divided by 2x, y minus delta y divided by 2 times. And this length is delta x, right? Okay, go this way. The component would be f y of, this point has x plus delta x and y, x plus delta x of 2y times delta y. Now, this way. Now, mind you, here we move from here to here. And here we move in the opposite direction. So, that would be with a minus sign. If I'm using the same component, it would be minus sign. So, minus sign would be what? So, plus, plus, minus f x at this point. It's x, y plus delta y over 2 times delta x and again minus here. And again, I'm moving against delta y's that way. So, I'm moving against, so it's minus sign. f y of x minus delta x over 2y delta y. Okay, we've got our formula. Now, it's simple. Remember this thing from calculus. Okay, so this is my a, this is my b. Now, the difference between them is this one, right? So, if I take the line which is parallel at some point c, then the difference between this and this would be equal basically to b minus a times tangent of this angle. And tangent of this angle is the derivative at one middle point. So, whenever my points a and b are very close to each other, I can say that this is approximately times b minus a. But since points are very close to each other, I can put a b or a, it doesn't really matter what it is. When a and b are very, very close. So, basically that's the definition of the derivative. So, from this simple calculus case, we can combine this and this. What is this? Well, both of them are delta x and both of them are f x. So, it would be f x. Now, the first argument is the same x. Second argument is y plus delta y divided by 2. No, the first is minus. Minus f x x y plus delta y divided by 2. Now, and both are multiple by the x. Now, let me just talk about the y component. The x component is fixed because right now where x is here and x is there. This is the difference between them. What's the difference is? Well, that's exactly the same as when I was talking about f of b minus f of a. b is this, a is this, and it's equal to the difference between them, which is delta y with a minus sign actually times derivative. In this case, partial derivative by y in some middle point. Middle point between these two is y. So, basically I can say that this is equal to f x of x y d by dy times the difference between arguments b minus y, b minus a, which is delta y and delta x. But I think it's with a minus sign, right? Because this minus this would be equal to delta y minus delta y. All right, fine. Now, very similarly, if I will take this and this, I will have df y of x y by dx times delta x delta y. Exactly the same consideration because y and y is the same and the difference between arguments is delta x. That's why I multiply by delta x and delta y itself. So, this difference between these two is my circulation. And they have to divide it by the area, right? Divide by the area. Circulation divided by area. That's my curl definition. Area is delta x times delta y. So, from this, I came to a formula that circulation divided by area when area is shrinking to a point x y is equal to df y x y dx minus df x dy. Because this is area which is cancelling when I divide the circulation by area. So, this is basically a final formula which we will return in the next lectures related to curl of the field. Curl is very important. For electromagnetic field, the Maxwell's equations related to electromagnetic field, they contain the curl. And that's basically why I have introduced the whole thing. This nubbler symbol with gradient and divergence. And now we have the curl. And what's interesting is that the curl of this f of x y of vector is equal to nubbler vector product with vector f of x y. Now, this will be explained later on. And that's what actually makes nubbler very important symbol in physics. So, you have gradient, you have divergence, and you have curl of the field expressed in nubbler. So, now it's vector product, the cross product. Okay, that's it. I would suggest you to read the notes for this lecture. And as I said, that you go to unizor.com. The part is called waves. And a particular topic is field waves, which contains this lecture and others about gradient, about divergence, and about Maxwell's equations. That's it. Thank you very much and good luck.