 Hi, I'm Zor. Welcome to Unizor Education. Today we will solve a couple of problems related to vector and scalar fields. Obviously, it's all preparation for Maxwell equations. Now this lecture is part of the course called Physics for Teens presented on Unizor.com. I do suggest you to watch this lecture from the website because every lecture including this one has textual supplement. It's like a textbook basically, so you have parallel video and text. Something I explained a little better. Sometimes maybe my text explains it a little better. It happens too. There is a prerequisite course on the same website called Maths for Teens. Something like calculus and vector algebra are absolutely mandatory to understand whatever we are talking about. So I do suggest you if you don't remember certain things, you can always refresh from Maths for Teens course on this website. Now everything on the website is completely free. There are no advertisement. Sign-on is optional. I mean it's for certain specific purposes if you want to supervise the education. Other than that, you just do whatever you want on the website. Go to anywhere. But it's more important actually to go with a course because the course has menus. The lectures are in certain logical order, obviously. Like if I'm presenting certain problems right now, it's obviously the problems about something which have been just recently presented to you. All right, so back to problems. First of all, maybe a couple of very important reminders about what we are talking about. We will talk about operators of gradient, of scalar field, three or two dimensional depending on, which is basically a vector by partial derivatives by corresponding variables, arguments. Sometimes we say nabla F, where nabla is triplet of operators of partial differentiation. dp dx, dp dy and dp dz. So this looks like a multiplication of vector dp dx, dp dy and dp dz. by a scalar f. I mean it looks like it, so that's by the research. Then we're talking about divergence, which is about the vector field. So if you have a vector field, which has three components. fx, fxyz, fy and fz. Then divergence looks like a scalar product of vector of nabla, which is dp dx, dp dy and dp dz times f, which is actually dfx by dx, dfy by dy and dfz by dz. So I'm just reminding something, which we have been presenting in the previous lectures. And then there is a curl, which is actually looking like a vector product, also on the vector field, defined also on the vector field. I don't want to put a big formula, I will actually in one of the problems, but for two arguments, for the two-dimensional, it's actually very easy thing to do. The curl at every point, it's dfy by dx minus dfx by dy. I do remember that. So the curl on the two-dimensional field, actually, it's a function, basically, which defined for every point. Well, it's kind of a scalar, but we usually associate a vector perpendicular to the surface of the two-dimensional plane. It more or less resembles the angular velocity, if you remember. We have something, which is angular speed when something is rotating. But then we usually associate the vector along the axis of rotation, which is magnitude, which is equal to angular speed, and we call this vector angular velocity. So this is also curl is kind of a constant, which we associate with a vector perpendicular to this particular two-dimensional surface. In three-dimensional, it's a little bit more complex to define, so it's just delayed. Okay, so this is just preamble before we present any kind of problems. Okay, now let's talk about problems. Problems are relatively easy, and in some cases it's just plain technicality, I would say, manipulating this formula. But in one case, there is something important. Okay, the first one, the first problem is easy. Now, you remember in one of the previous lectures, we have introduced the concept of circulation of the field. Now, circulation of the vector field, let's talk about two-dimensional case. So we have a plane, and we have a vector field on this plane, defined, a vector defined at every point. Now, let's say we have certain trajectory from A to B. Well, basically, the circulation is an integral of this force by the lengths of these each particular segment where the force is defined. Force is different, right? So in every case, it's different vectors. So we divide it in many, many different cases, different intervals, and within each interval, we assume that the force is the same. So we multiply the vector of force. We usually use the scalar product, so the projection onto this particular segment. And we multiply by the lengths of the segment, and then we summarize them together. Then we go to the limit when the number of segments is increasing and their lengths is decreasing. It's basically integration. The only thing is, when we were integrating, let's say, the area under the curve we divided in many tiny segments. And we multiply the value of the function by this segment, gauging the area of this little almost rectangular piece. And then summarize them together, then go to a limit. Now, obviously, assume that the limit exists. There are certain conditions when it exists, which physicists usually don't even think about it. But in any case, this is curvilinear integration, if you wish. So that's how it's basically done. Now, in physical sense, this is basically work. Because whenever I'm multiplying the constant force by the lengths of the segment during which this force is acting, I will get work. That's the definition of work. So basically, we're trying to find out what kind of work is needed if we move one particular object from here to there. If we know the force, either the force helps us or we go against the force, whatever it is, depending on the direction. So our work will be positive or negative from our standpoint based on whether we exhort this work or we use the force to do that for us. But in any case, with a plus or with a minus, it's basically a work. Because we have on every particular little segment which we assume to be straight, the force is directed this way and the segment is directed this way. So we have a projection of the force by the segment and multiply the projection by the lengths, which is actually a scalar product of this vector and this vector. We did address this before. And then we introduced the concept of conservative field. Conservative field is if the work from here to here depends only on endpoints. It's not important how we move, important is from where to where. And examples of these conservative fields are gravitation and electrostatic. We are familiar with that. Now what's interesting is, okay, now we're talking about the first problem which I would like to solve. The problem is if the field is conservative, two-dimensional field, conservative field. So basically work against the force from A to B depends only on A and B doesn't depend on the trajectory. Now my question is, well actually the problem is, curl of two-dimensional conservative field, vector field. We have to prove that this is equal to zero. Okay, now let's recall how curl was defined in two-dimensional field. That was exactly in the same lecture where I introduced the circulation and conservative fields. The curl was defined the following way. If you have a point, let's make some kind of a closed path around this point. We're talking about two-dimensional case. Now let's just calculate this integral around this path which is circulation. Now we calculate this and then we have area of this. So we divide circulation by area. Then we will start shrinking our path around our point to this particular point. Point is x, y. If this ratio has a limit and in most good cases, smooth functions, etc., etc., it does have a limit. This is called the curl. Now, how can I prove that the curl of two-dimensional conservative field is equal to zero? Well, that's kind of very easy because if you have this particular property, that our work which we perform moving from A to B is independent of the path. So let's just take two points, let's say A and B. This is a closed path. So moving from A to B this way, we have exactly the same result as moving this way. Because it's conservative field. Now what if I move not from A to B on this particular segment but from B to A? Well, I will have the same but with a minus sign. Because the vector field is exactly the same, but all these little intervals are in the opposite direction. So I will have the same with a minus sign which means if I will go circularly, I will get zero. So circulation of conservative field around closed paths from A to A is always equal to zero. Which means this is equal to zero and that's why curl is equal to zero in this case. So for conservative field, any circulation around closed path is zero and that's why curl is zero. So that was very easy problem number one. Now problem number two also related to two-dimensional vector field is the following. Again, we are considering conservative two-dimensional field. Again, for example, gravitation field or electrostatic field. Now, apparently, if you remember from the electricity part of this course or from anywhere else, in both gravitation and electrostatic cases, well, let's talk about gravitation case. There is something which is called potential energy. Potential energy. Remember when you are lifting a stone from the surface of the earth, let's say, it has potential energy and if you let it go, it will fall down, converting potential into kinetic energy and the kinetic energy when it hits the earth is basically converted into heat or something else. Now the concept of potential energy is very interesting. Let's calculate the potential energy. First of all, let's talk about defining what is a potential of, let's say, gravitational field. Again, that's something which were addressed before in the corresponding part of this course, but I'll just remind you. If you have a unit of mass, let's say in the C system of units, it's one kilogram. The potential gravitational point at some point, x, y, let's consider my mass is at zero and the mass is m. Now, what is a potential? Potential is amount of work which is supposed to be performed by moving this unit of mass from infinity to point x, y. This is the definition of potential. Now, it's a function of x and y. It's a scalar function. Now, the force which exists at any point x, y depends on the distance from the source of gravity. Now, let's, for simplicity, use even one dimension. So we are moving along the x-axis. So this is my x, this is my infinity, and I'm moving from infinity to x. Now, the force of gravity is, that's the Newton's law of gravity divided by square of the distance, right? This is the gravitational constant. This is the mass of the source of gravity. Our unit is at point x, so the distance is obviously x. So one over x squared, that inverse proportional to square of the distance. We know about that, right? Now, so this is the force. Okay. If this is the force, how can I calculate the work to bring from infinity to the point x my unit mass against this type of a force? Well, obviously we have some small interval from x to x plus delta x. We assume that the force is exactly this. Don't need to access. So this is my interval. So we assume that the force is exactly the same because interval is very small. Which means if I will multiply times delta x, I will have work which is supposed to be performed on this particular small piece. And if I will integrate it from infinity to point x, let's use a different variable here just not to confuse. Let's say u. Okay. If I will integrate this, I will have the work, right? Now, what's integral of, indefinite integral of one over u squared is minus one over u. Remember? Because the derivative of one over u is minus one over u squared, right? So we should not really miss gm. And then we have to put it from infinity to x. So if I put x, I will have minus gm divided by x. If I will have infinity, I will have zero. So this is my potential. That's interesting actually. You see, if this is my potential, I use capital U. The derivative of potential of x is equal to function of x, right? Derivative of gm divided by minus gm divided by u is gm divided by u squared. Or x doesn't matter. So what's interesting is that derivative of the potential is the force. I will use this particular consideration to basically prove now what is gradient. Gradient is the derivative of the scalar field by arguments, right? We were just talking this in the beginning of this lecture. So my point right now, and this is my second problem. If you have a conservative field, like a gravitation field, two-dimensional, let's talk about two-dimensional, it's easier. If you have a conservative field, then this vector field is a gradient of some scalar field, which is called a potential, which means it's df by dx comma df by dy. So I have to prove that for any conservative field, there is some kind of a scalar function of the same two arguments, which is the scalar field in the same two-dimensional space, gradient of which, which means a vector of partial derivatives, is equal to this particular vector. I use p and q instead of fx and fy just for convenience purposes. So p of x, y, it's a projection of vector x onto x-axis and q is fy. Just easier to manipulate. So we have a vector field with components p and q, x-projection and y-projection, and I would like to find the function f, gradient of which is equal to this particular vector. And this function is actually called a potential of this vector field. Now, go back to gravity. In the case of gravity, we were talking about potential being such exactly a scalar field, because when I was taking the derivative of the potential, I got the force. Well, it was one-dimensional case, but it just brings me to a certain logical point. What if, now what is a potential? Potential is work for a unit of mass, right? Now, what if I will use the property of this field being conservative, which means the work against this field would be the same regardless of the trajectory. What if I will use this work as my candidate for such a potential function? Or a scalar function? Well, let me try. I mean, it's an idea which we have to try. So how can I try? All right, fine. So let's say I have a point x, y, and I would like to have some work which I have to exhort or the field has to exhort by moving into this point from somewhere else. Well, obviously the easiest point is from the origin of coordinate, and I have to find an easy way to reach this point because the field is conservative, so I can choose any way and calculate work. So I will choose this way. I'll go this way and I'll go this way. So this is point x, 0, and this is point 0, y, right? So I will choose trajectory from this and then from this. Why does it work? Why is it easier? Well, because in this case I can disregard my y component of the vector field and in this way I can disregard my x component. So it's kind of easier. All right, so let's just think about what happens if I will calculate the work. If I'm moving from here to here, my vector field is equal p of x, 0. But you know what? I would use again some other arguments because I will integrate. Let's call it u. So I will take a very small interval from u to u plus delta u and the work during this interval is equal to delta u, right? The force, well, x component of my vector field because it's along this x-axis times the length, force times the length. Thanks. That's the work. Now I will integral it. Instead of delta I will put d and I will integrate it from 0 to x, right? That's my work on interval a, b. Great. Then I will use my second from b to c. So it's from b, c. It's integral from... Now I will integrate from 0 to the y. My variable of integration I will choose v, let's say. Now I have to really take into consideration only a vertical component of my vector field which is q. So it's q. Now the first component is always x because I'm always on the abscissa x. Along the y-axis it will change. So I will put v here and dv from 0 to y. So some of these is my work. Well, I've got everything I wanted. All I need now is to check if this is really a function, a scalar vector, gradient of which is equal to my initial... That's my initial vector field, right? That's my force. It has an x component and it has a y component. Now if I will use this sum and take a gradient of this, we have to have this. So partial derivative of this by x should be equal to this and partial derivative by y should be equal to this. Well, let's check. Am I wrong? Am I right? All right. So partial derivative of sum is sum of partial derivatives. Now what's interesting is now integral, if I will have partial derivative by x by upper limit of the integral, it's basically this function at... Okay, the indefinite integral is p of u0 itself and all I have to do is put limits. Now, why? Well, consider what's derivative. Derivative is first I have to... Now this is a function of x, some kind of a function of x. So I have to put f plus delta x minus f of x divided by x. That's what derivative is. Increment of function divided by increment of argument. What is increment of function from x plus delta x? Well, the increment of this function is p of x0 times delta x. Right? Consider this is... Forget about the second argument, it's always 0. So it's function of one argument and let's say you have an area under this function. This is your u, this is your function, this is from 0 to x. Now you increment x plus delta x and now it's a new integral. So new area. So what's the increment of this whole integral? Well, it's f of x times delta x. So basically integration by top limit of integration gives you the same function. Then you divide it by delta x because this is the derivative so you have to increment the function which is f of x times delta x and divide by delta x so f of x remains. So this is... So this is p of... Okay, this is p of f0 minus p of 0, 0. Something called this. Just p of x. p of x like this. I don't have to calculate this. It's already definite integral. I don't have to go to indefinite integrals. Yeah, that's it. It's a function only of x, exactly like this. So under integral. Great. Now derivative again by x of the second of integral from 0 to y, q of x v dv. Now we are differentiating by x which means y is not involved and v is not involved. So it's basically I can go and change the order. First I have to derivative and then integration. We can change the order of differentiation. Why? Because again integration is by v from 0 to y. It has nothing to do with dx so I can insert dx inside. Great. Now what is... Okay, I don't need this anymore. What is dq by dx? Okay, that's important actually. If you remember again in the lecture about circulation and conservative fields and introduction of curl, I had the expression, the differential kind of expression for curl of this field f at x, y with these components being equal to d f y by dx minus d f x by dy. Well in this case instead of dy I'm using q and instead of f x I'm using p. And the curl of this field is 0 because the field is conservative. That's my first problem which we started this lecture. Which means the dq by dx is equal to dy, dp by dy. So instead of this I can put, instead of this, I can put integral of 0 to y, dp of xy point dy because they're equal. The difference is equal to 0 instead of this. But instead of y I have to put obviously v in this case. dv. So I'm using explicitly using conservativeness of my field. And that's extremely important otherwise I couldn't really change dq by dx to dp by dy. And what's the result of this? Well this is differentiation and this is integration. They are reverse each other, right? So I know that indefinite integral from this is equal to p, x, v. And now I have to put it in, this is formula Newton's Leibniz formula. Indefinite integral is equal to p, x, v. So I just put the limits which is p of xy minus p of x0. So this is my second integral along the BC. Now if I will add this to this, this is plus p of x0, this is minus, so the only thing which is remaining is this. So I have proven that derivative by x, partial derivative by x, from that formula which I had sum of two integrals, wab plus wbc. So I have proven that partial derivative of this by dx is equal to p of xy. Okay, now how to prove that partial derivative by dy is equal to q, exactly the same thing. I can actually leave it to you as your home exercise if you wish. But it's exactly the same thing, exactly the same manipulation with integrals. So again, I would like to add that forget about the technicality, the idea was that in gravitational and electrostatic field potential which we know as being a work being done on unit mass or unit of static electricity to bring it from infinity to some point, this function potential energy which you have after that is really something which the derivative of which or partial derivative by each argument gives you the force, gravitation force or electrostatic force, the Newton's law or Coulomb's law. So I'm just using this idea and check if it works. And then everything else is just technicality manipulation with integrals. It's very important and by the way I was reading a few articles about how to prove that conservative field has always some kind of a function gradient of which is equal to this vector field and the function is called potential. And they were only manipulating integrals without really talking about where the ideas are coming from and I didn't really like it. I mean maybe I didn't find the better way and that's why I have decided to approach it this way. I think it's much more natural. You know about something physically and you know that potential is the source of, you can restore the vector field from the potential. And that's what it is. Now I have a third problem which is purely technical and now it's a three-dimensional but it's easy. Here it is. Coral of gradient of some scalar field is equal to zero. Well, in the previous problem we talked about the conservative field being always a gradient of something which we called potential. So this looks like we are talking about conservative field which is a gradient of this. So the gradient of this scalar function gives vector. So it's a vector field now and now the curl. So the vector field seems to be conservative and we talked about the conservative field in two-dimensional case. That's the first problem in today's lecture and the curl is equal to zero because the integral along the pass, closed pass was equal to zero. Now how is it in three-dimensional case? Well, in three-dimensional case there was already a lecture before which found the expression of this three-dimensional case. So if you have a field f which has three components fx, fy and fz. Now each of them is function of three arguments now. Then the curl of f is equal to, that's kind of difficult. So I put it before z by dy minus df y by dz. That's my first component. Second component xz fx by dz minus fz by dx and the third component df y by dx minus df x by dy. Now that was in one of the previous lecture when I was talking about curl in three-dimensional case. So it has three components. Now we have to use this fact. This is a gradient of f which means it's df by dx, df by dy, df by dz. So fx is this, fy is this and fz is this. That's what it means. So my field is not just any field curl of which I'm trying to find out. My field is gradient of some scalar field. That's why my first coordinate is equal to partial derivative of the potential by x. Again this is a potential. Second by y and third by z. Okay let's substitute it. So instead of fz I will put df by dz. So it's df by dz and now I have to differentiate it by y which means dy, dy minus df y is df by dy. But then I have to differentiate it again. Second it's a mixed derivative. So that's my first component of the curl. And look at this. They are not different. The only difference is that I'm differentiating first by z and then by dy and then by y and here by y and then by d but it's the same function x, y, z. And under normal conditions it doesn't really matter how you differentiate first by x and then by y or by y and then by z or first by z and then by y. So that's equal to zero. Exactly the same thing for the second and the third component. Well that's it. So the curl of gradient is always zero. Because the function as a result of the gradient the vector field is conservative and the curl of conservative field is always zero. So if you will take some kind of curl well around any point you can do this circulation and shrink it etc. If you will do the curl for any conservative field like for instance gravitation field or electrostatic field you will get curl equal to zero. Okay and by the way if you have a non-equal to zero curl it means that the field is not conservative. So it's not like gravitational field. Well that's it for today. I suggest you to read the notes for this lecture. This is the same thing which I was writing but probably in a little bit more orderly and more complete fashion. I didn't really jump over certain like saying this is exactly the same. I just put it in writing that this is the same. Well other than that you got it. So thanks very much and good luck.