 Hi, I'm Zor. Welcome to InDesire Education. We continue talking about a curl of a vector field. Previous lecture was about two-dimensional curl of the vector field, of two-dimensional vector field. And that's kind of easy, because if you can remember, we were using, let's say, this whiteboard as a two-dimensional surface and we can install a tiny paddle wheel somewhere and rotation of this paddle wheel was basically a characteristic of the vector field, which is defined at every point of this surface. So if it turns this way, it's curl well, negative. If it turns that way, it's curl positive. And basically the intensity of the spinning is a quantitative characteristic of curl at any point wherever we put this wheel. And it was easy because it's flat surface. So there is only one orientation of the this paddle wheel, which we can arrange. Just put it in such a way that the axis of rotation is perpendicular to the plane. In three-dimensional space, it's kind of more difficult because no matter how we put this, it's still kind of different orientation of this wheel and probably we have to put it in some way that rotation is most intensive and that would be the curl. Sounds reasonable, actually. So anyway, that's what will be a subject of today's lecture. This lecture is part of the course called Physics 14's, presented on Unisor.com. I suggest you to watch this lecture from this website because it's a course on the website. Physics 14's is a course. So there is a menu, sub-menu, etc. So everything is logically related. There are notes for each lecture. So not only the video presentation, but also you can just read it as a textbook parallel to the video. There are problems, there are exams, which you can take in any number of times you want and the site is totally free, no advertisement, no strings attached. Okay, so let's go back to three-dimensional curl. So first of all, we have to have a three-dimensional vector field. So this is a vector, which is defined at any point as code in its x, y, z in space. Now every vector in three-dimensional space can be represented by its three components, which I call vx, vy, and vz. At the same time, if you have three components, we can have three unit vectors on each axis. Usually the terminology is i for x vector, j along the y-axis, and k along the z-axis. So we basically represent a vector, this vector, with these three components as sum of unit vectors along the axis, multiplied by corresponding components. This is a plain vector algebra, and we are not talking about the details of this. It's supposed to be known. If it's not, there is a parallel course, prerequisite actually course, called mass proteins on this website, and there are vectors explained there. Okay, also the same mass proteins has calculus part where you can read about partial derivatives, which will be used here. Okay, so that's done. Now, now we have to place our paddle wheel at point x, y, z, and wait. After some time, it will probably orient itself properly, probably when its axis is parallel to the most important kind of direction of the vector field, and it will turn or not turn, and that would characterize the curl. So we're waiting until it basically start spinning or not spinning in a stable kind of way. And that would be our curl of the vector field, but now we have to quantitatively evaluate it. Now, first of all, if this wheel has basically established its position and speed of rotation, how can we characterize the position of this wheel? Well, if you remember some time ago when we were talking about rotation, we characterized speed of rotation with a vector, which is directed along the axis of rotation. So let's say if this is speed of rotation, we associated it with a vector, magnitude of which is basically equal to a speed of rotation and the direction of this vector, one or another side, was such that from the top of this vector, we see the rotation counterclockwise. That's basically a vector. So we can characterize the spin, the rotation of this probe wheel by a vector which is along, which is directed along the axis of finally established the axis of rotation. And the length, the magnitude of this vector would be equal to the speed of rotation. And the gain direction would be in such a way that from the top of the vector you see the rotation counterclockwise. That's basically our agreement. It's convenient because one vector would be basically completely describing the rotation of this pedal wheel at any point in space. Well, let's call this vector K. And by the way, it can also be represented as any vector as a sum of its component. So it's Kx. I assume x, y, z in the parenthesis. So my question right now is, basically my task is to evaluate this vector if I am given this vector. Kind of complex. But we can always try to simplify any problem by dividing it into smaller, simpler problem which we have already solved before. That's exactly how we're going to do. Think about it this way. If my vector of wheel rotation, basically it's a curl, it's a curl vector, if my curl vector has three dimensions and this vector has three dimensions, what I'm kind of assuming would be a natural thing to represent the x-component of the curl, x-component of the rotation. x-component of rotation. This one. As caused by projection of the v vector which is somewhere, let's say, projection onto the y-z plane. So again, rotation around the x-axis is rotation within y-z plane. So I assume that it's a lip of face. I mean, yes, obviously you have to really kind of understand it and agree because it's my assumption which happened to be actually corresponding to the real thing is, obviously experiments can be arranged, etc. that the x-component of the curl of rotation around the x-axis. It's basically a rotation of this vector projected on the y-z plane. So basically I can assume that x is a constant. It doesn't really matter what it is and consider only a function of two arguments, kind of, which basically in a correct representation would be v y of x, y, z times j plus v z of x, y, z times k. So j is along the y, component of vector v along the y-axis. v z is a component along the z-axis and sum of this and this is basically a projection of whole vector onto y, z plane. So we project this to this point y, z and the vector would be somehow, I don't know how, it's the end of this projection. So that would be the projection of the vector v of x, y, and z in three-dimensional space onto y, z plane. And as vector v is rotating in space in different places, I mean the whole vector field, v has certain rotations, certain curls in certain spaces or not half whatever it is. Its projection on the y, z-axis would have basically some kind of a rotation which causes the x-component of the rotation. This is probably the most important part of this lecture to understand this. So again, vector in three-dimensional space is projected on two-dimensional space y and z and if there is any kind of a rotation at point x, y, and z, any kind of a, if the wheel would spin in this particular position in this particular, in some space, then projection of this vector will also have certain wheel and in this case, if we put this wheel onto y, z, its axis would be perpendicular and it will cause basically only the rotation of the curl vector around the x-axis, only the x-component of rotation around the x-axis. This is the most important and that you really have to kind of feel and understand because after that basically everything is easy. How easy? Very easy. It means that kx of x, y, z is a curl of this vector and we have already learned from the previous lecture that rotation in the 2z, which is represented by components according to the coordinates, is the minus dvy of x, y, z by dz. We have already established this in the previous lecture. In the previous lecture we had x and y, x-axis doesn't really matter. In this lecture we have right now y and z only because we started from the x and y and z was the plane of rotation. We will move to y and z obviously. So we have already determined the x-component of the curl of our vector. Now obviously if I want to do this, I have to multiply it by i and I have to multiply this by i and that's what this particular thing is. Absolutely similarly we have ky equals to d so it's xz plane, right? So it's x of x, y, z by dz minus d z dz by dx and finally ky z that's the curl of projection of rotation onto x, y, z. That's the formula which was in the previous lecture dy of x, y, dv, y. dv dv, y at x, y, z by dx minus d vx at x, y, z by dz by dy, sorry. Basically that's it. We finished. We have determined three components of the curl of the vector field. So the vector field has three components. We have basically considered every pair of components which is basically a projection onto corresponding plane. So dy and dz is projection onto yz plane which causes a rotation around x x and z plane is rotation around y and x and y projection is rotation around z. That was exactly what was determined in the previous lecture. We were using x and y. The only problem which I have right now so this should be multiplied by j and that gives me this component and this is supposed to be multiplied by k. This is by j and this is by k and that gives me this component. The only problem is it's kind of cumbersome. I mean people, mathematicians and physics don't like long formulas so I would like to simplify it and it's actually the result is amazingly simple. Here is the simplification of this. Let's say we have two vectors vector p which can be represented as three components p x, p y and p z and vector q which has its own three components. I would like to express in terms of p x, p y, p z, q x, q y and q z vector product of p times q vector product, cross product. Two vectors in three dimensional space everything is supposed to be very simple, right? So three components and three components and multiply each by each and we'll have a result. Now the only thing is I would like to point out that we will end up with multiplication well, constants p x, q x, q y, p y etc will be multiplication, simple. What about the vector multiplication? Do you remember something like first of all is anti-commutative so if you change the order a times a cross product b is minus b cross product a I hope you remember that from the vector product from the cross product and also if you have two vectors which are perpendicular to each other let's say a and b the result of their vector product is the vector perpendicular to both of them which magnitude is equal to times this times this and the direction would be some kind of rule if you will put from if you will rotate from a to b it would be this direction towards us from the port so based on this there are some rules about i and j and k multiplication now multiplication of vector by itself they are basically collinear so sign of angle between them is zero so that would be always zero as well as j by j zero and k by k is zero now as far as i times j it would be a vector perpendicular to both now this is a long x-axis this is a long y-axis so the perpendicular to both will be z-axis and the magnitude would be the product of the magnitude right each magnitude is one it's unit vector so that would be k and then j times k would be i and k times j times i would be j I mean we can always consider all the different combinations if we want k times j it would be minus i etc. so using all these rules which are quite simple I will write the result and if you want intermediate results it's all written in the notes for this lecture but anyway the result is the result is p y qz minus pz q y i plus analogous j plus analogous k and what I would like to say I don't want to spend too much time just rewriting the problem is that if px if p is a nubla nubla is dpdx dpdy dpdz same as previous lecture three components dpdx dpdy and dpdz a triplet of operators nubla and q is our vector v at x y z with components v x y z then this is exactly this which means that I can write a very simple expression for my vector product of nubla times v this is exactly curl of v so this is the simplest formula possible and all these complication results results in this particular formula again I skip a couple of intermediary steps they are absolutely trivial I just don't want to waste my time but in your time but all these results are written on the notes on this website notes for this lecture so I suggest you just to go there well basically that's it we have determined how in a simple form express the curl of our vector field at any point and well you know that my favorite model right now of the vector field is the wind basically you have the wind you can always if you know the velocity of each molecule at each place at some moment in time you can determine where exactly the tornado might be actually located I don't know but maybe those guys who are predicting the weather maybe they're doing something similar to this I don't know okay that's it for today thank you very much I suggest you to read the notes for this lecture they are more detail in some cases than whatever I presented on the board but basically along the same line so thanks very much and good luck