 Hi, I'm Zor. Welcome to Unisor Education. During the previous few lectures, we have discussed scalar and vector fields, and we were talking about certain characteristics of these fields, like gradient for scalar field and for vector fields, we were talking about divergence and curl. And curl was in two-dimensional and three-dimensional case. All this was basically a preparation for Maxwell equations of electromagnetic field, because these operators are very much involved in expression of these Maxwell equations in contemporary form. Maxwell himself, by the way, did not use something like a nabla symbol. He was using straight partial derivatives. And in our case, we were trying to kind of shorten the formula, and that's why we have used this very convenient symbol nabla, which can be applied as a multiplication to a scalar field or as a scalar product of a vector field or a vector product of a vector field. Now, again, before getting involved in Maxwell equations, let's just do a couple of problems related to fields. We will basically calculate all these characteristics like gradient, divergence, and curl, just to be a little bit more familiar with numbers, with fields, how they feel, etc. Now, this lecture is part of the course, Physics 14, presented on Unisor.com, and I suggest you always to watch the lecture from the website, because it's a course, which means there is a menu, there is a sub-menu, so you have to go, for instance, for this lecture to a Physics 14 course on this website, and there are other courses like Mass 14s, for example, and the part of the course related to whatever I'm talking about is called Waves, and then the sub-menu would be field waves, and among field waves, you will see this particular lecture, which is field problems, basically. And there are other lectures as well as the ones which we were talking before, and the ones which we will be talking as far as Maxwell's equations are. Okay, so I have three problems relatively straightforward, but again, the purpose is to get familiar with fields and calculations of these characteristics of the fields. Okay, the first problem is about gravity. So we have a gravitational field. Let's consider a planet Earth on a macro scale. So we are always considering these masses as point masses, and obviously the gravitational field is the field where you have the forces radially directed towards the source. So if this is the source, mass m, so all the gravitational forces are this type, towards the center. This is not exactly towards the center. Okay, now according to the Newton's law, if we have a mass m, then the force of the gravitational field would be g times m times m. This is gravitational constant. This is mass of the source of gravitation. This is mass of the probe object. And it will be inversely proportional to square of the distance, right? And I'm talking about x, y, z as coordinates basically of this point, considering my center mass is at point 0, 0, 0, at origin of coordinates. And this mass is at x, y, z. So this is something which I'm sure we all remember from physics, from elementary physics. Now maybe a little bit less familiar or maybe you don't remember it well. For this gravitational force there is a concept of potential, gravitational potential. Now what is potential? Potential of the field is amount of work which is needed to bring the unit mass from infinity to point x, y, z. Now this was addressed actually in the part of the course, the same course, physics for genes in the part called energy. And within that part there was a topic potential energy and within it there were some lectures related to gravity but I do suggest to maybe to review these lectures if you don't remember what is this gravitational potential. But in any case the explanation of this is amount of work per unit of mass to bring it from infinity to a particular point. So m is equal to 1 and this unit of mass needs a certain amount of work. Now in this particular case I put minus in front of this amount of work because if not we have to do this work. Its gravitational field does it for us. The reverse operation from the point to infinity would be on us. That would be positive but the same value basically. So its g times m and it was inversely proportional to square root. To a distance basically. Square root of x squared plus y squared plus z squared is a distance to this point. I mean if we are talking about let's say another variable called r for example distance it would be minus gm divided by r. Now again to derive this again you can go to previous lecture which I mentioned about energy. I consider this to be a known fact. So let's talk about gravitational field. This is a function which is a potential of this field at any point. Now what is this particular field in our language which is related to the previous lectures? It's a scalar field. This is a scalar. Now this f is a vector because it has a magnitude and a direction. Now this is amount of work. So it's a scalar. So we have a scalar field. Okay great. So we have a scalar field. Now what exactly do we need? This problem is what is the gradient of this field? Very simple problem. Now we remember that gradient is nabla u at x, y, z. Where nabla is d by dx, d by dy and d by dz. So let's assume the vector multiplied by this constant. That means we have to really produce vector which is this. Where u is basically a function of three arguments. And we have this function. Okay all we need to do is basically to take partial derivative by x by y and by z. That's it. Now thankfully they are symmetrical so we will take only for x for example and then we will extend it to y and z by symmetry basically. Minus. Okay. So what is the derivative of this? Well we can always rewrite u at x, y, z as minus gm and x squared plus y squared plus z squared minus one half, right? One divided by square root is the minus one half exponent. Okay. That's easier to take the derivative, right? So the derivative would be let's say by x. By x. So minus gm retains, these are just numbers, scholars. Now if I have something in some exponent, so that would be exponent times this in an exponent minus one, right? Times derivative of inner function, inner function by x only so we ignore y and z so it would be 2x. Derivative of x squared is 2x. Same thing by the way. The exponent and then we reduce exponent by minus one. It's all related to x to some exponent a if you want to d by dx it would be a times x minus one. So you have to remember this from the calculus. Which is equal to 2 and 2 minus and minus so it would be g times m times x squared plus y squared plus z squared minus 3 times x. Okay, great. Now, so this is our x component. Now d u at x, y, z by dy would be gm x squared plus y squared plus z squared minus 3 seconds times y. And d u at x, y, z by dz would be gm. Nothing to do with general motors. That's the gravitational constant times mass times x squared plus y squared plus z squared minus 3 seconds times z. Great. So we have three components. This is our vector of gradient. Interestingly, what is the length of this vector? Well, the length of this vector is if we will take square of one, square of two and square of three, the third, we add them together and have a square root of this, right? Okay, so square of this is g squared m squared x squared and x squared plus y squared plus z squared minus 3, right? That's square of this. Same thing for y squared, same thing for z squared. This remains exactly the same. If I will add them together, I will have basically g squared m squared and this minus 3 can be factored out. And I will have x squared plus y squared plus d squared in the first degree, right? So I have minus 3 as a degree of this sum and plus 1 degree of this sum. So the result would be g squared m squared x squared plus y squared plus z squared minus 3 plus 1 would be minus 2, right? Now the square root of this and what do we have? We have gm times x squared plus y squared plus z squared minus 1, which means 1 over, basically, right? So let me write it down this way. It's more familiar. And what is this? If I will add m here, that would be gravitational constant times mass of the source, mass of the probe, divided by square of the distance. This is a gravitational force, remember, of x, y, z. That's what we started with. This is the Newton's gravitational law. So what's interesting is that the gradient is basically a force of gravitation for a unit mass. So if m is equal to 1, gravitational field as a scalar field has a gradient. And the gradient is basically, first of all, it's a vector, right? Because it has x component, y component and z component, which means it's directed regionally, right? Because this is the same. This multiplier is the same and only x, y and z is. So basically it's like x, y, z times something, times the length of it. So the gradient is directed exactly towards the center of mass of the source of the gravitation. And its magnitude of this vector, gradient is a vector. So its magnitude is equal to the force by unit mass. Very interesting. So potential is a scalar field. Gravitational force per unit of mass is actually a gradient of potential of the field. Okay, that's the first problem. Next, I'm just giving you not a practical field, but just a field for educational purposes. Okay, this is a vector field now, x, y, z. And it has three components. For every point x, y and z, I have a vector. Vector has three components, x, y and z. So component of the vector, projection of the vector on the x-axis is zero, on the y-axis is zero. So its only projection on the z-axis is equal to basically the distance from the origin of the coordinate. Okay, my first problem is just describe it. I started to describe this field verbally. And how can we just draw the picture, maybe? Well, these are coordinates x, y, z. So for every point x, y, z, what is this vector? Well, projection on x and y should be zero, right? So it should be directed only along the z-axis. So all my fields are, all my vectors are directed, well in this case vertically. If I consider z to be a vertical and x and y to be horizontal plane. Okay, now what's interesting is that the vectors which are close to x, y plane which have zero z coordinate. So vectors here are zero. A little bit higher than that, they are small. And the higher they are, the stronger they are. Because their magnitude is z. So the higher we start, the longer the vector will be. So they're all parallel and their length is proportional to the height of the point where they start from x, y plane. Okay, so that's a kind of verbal description of the field. Now, what is the front surface? Front surface is where all these vectors are the same. Okay, the answer to this question is, well any plane parallel to x, y plane would be a front surface because all the vectors originated on any point here will have the same z coordinate and that's why they will have the same magnitude. So they're all the same because direction is all the same. It's all vertical parallel to z axis and their lengths would be always the same because they have the same z coordinate of origination of the vector. Next, what is divergence? Okay, divergence. That's something calculable. The divergence of vector v of x, y, z is equal to nabla scalar product with vector which is d0 by dx plus d0 by dy plus dz by dz which is equal to 1 this is 0, this is 0, this is 1. This is my x component, vx if you wish, this is vx, this is vy, projection on the y and this is vz. So divergence is always 1. That's interesting. Now, why is it positive? You remember we were talking about divergence, let's say, of the wind. We were talking about wind coming from the higher pressure to the lower pressure. Here we have exactly the same except it's from the low to the high. So we have certain direction. Masses are moving in some direction when the air is concerned. If there is a wind with nonzero diversion, it means it's not constant everywhere. It's maybe stronger in one case and bigger from another. We really have a source of higher density going into, molecules are going into lower density. So there is certain source of something. Now here we also have something, from nothing we have something. So it's just reverse project from the low to high. But anyway, there is a movement, there is a real movement. It's not just like a wind going with a constant speed everywhere in the same direction. That's not the case. Then the diversion would be 0. In our case it's positive. From something we get something more basically. That's what it is. Is there any place where the diversion is equal to 0? No, because it's always equal to 1, it's constant. And the next is curl. Curl is a vector, which is a vector product. Of nabla and v. So what is the vector v? v is 0 times i plus 0 times j plus z times k. If i, j and k are unit vectors along the axis. What is nabla? I will also use d by dx i plus d by dy j plus d by dz. Now, this is not the real multiplication. But again we are using symbolics of vector algebra to basically have our lives easier, so to speak. Now, what is the vector product of these two things? Well, if you remember, we were actually calculating this previous lecture. And it's a vector with the components which I will just try to write from memory. But in any case, if you will go to the textual part of this particular lecture, it's all derived in more details. So basically it's like this. So you have x component, y component and z component, right? So for z component, I do remember it's d of dy by dx minus dx by dy. And this is my z component. It's like that's multiplied by k. Now, what is in this particular case? dy, y is zero. So this is zero. And x by y, it's also zero. So this is zero. Okay, now I will just substitute circularly. So it would be dz by dx minus dx by dz times j would be equal to dz by dx. It's zero. And dx by dz, now this doesn't depend on x, it depends only on z. That's why the partial derivative is zero. Now, in this case, it's just straight zero, so it's also zero. And the third one would be dz by dx by dy minus dz... Oh, I'm sorry, I think this would be y. x, y, yes, that's y. Is that the same thing? z by y is zero, of course. Yeah, it's still zero. And here we have dx by z minus dz by dx times i. And it's also zero because this is zero and z doesn't depend on... This is v. So it's all zero. So the curl of our field is equal to zero. Now, it's kind of obvious because think about this way. If I will have this pedal wheel positioned anywhere, you see all these are parallel. And whenever I put it anywhere, if I put it in such a way that pedals... If I put my pedal wheel with the axis parallel to z, now my pedals would be this way, right? So it will not push anything. It will not spin. If I will turn, let's say, put it parallel to z. If I will put axis parallel to x or y anywhere, my forces will be the same on all plates. Left and right of the axis. So again, since they are the same on this plane, then if I put axis this way, my pedals would be, let's say, horizontal. But on both sides of the axis, that will be the same force. So this is a field, a vector field, which will, if these vectors are forces, they will always be the same regardless of how I put my pedal wheel. And that's why pedal wheel will not be spinning and that's why the curl is equal to zero. That's it. Third problem. It's a simple, actually, kind of field, obviously, specifically for educational purposes, just to go through some calculations. And the third problem will be similar, just different field. I will use a two-dimensional field, minus y square root of x square plus y square x. Now, the question is why did I put it this way? What is the length of this vector? It's square of this plus square of this, which will be y square plus x square divided by x square by y square, so the length is always one. So I wanted to have a field, which all the vectors at all points will have the same magnitude, which is equal to one. Secondly, let me just have a vector xy multiplied by v. Now, xy is a radial vector. Now, if I will multiply it, it's a scalar product. It's x times minus xy divided by square root plus y times x divided by square root, whatever, x square plus x equals to zero, right? Minus xy plus yx. Scalar product of this vector of a unit length is zero with a radial, which means it's perpendicular. I'm having a field, which is always at any point equal to the same value by magnitude, but it's always perpendicular. That's my field. So it's always of the same length, but always perpendicular. So it's kind of circular field, right? Okay, so I probably will have a curl not equal to zero, but let's do it step by step. So first of all, it's verbal description. All the vectors are perpendicular to the radius towards the point where the vector is originated, and all of the vectors have the same length. So that's the verbal description of the field. Now you can, well, it's a two-dimensional, so you can imagine it's on this surface of the board, right? So all vectors are kind of radial, but again, the length is the same. Okay, now, is there something which can be called a front line? Front line, well, in a three-dimensional we're talking about surface. In two-dimensional we're talking about front line. That's the set of points where the vectors in each point, all the vectors, are exactly the same. Well, the answer is any line which is going from the center would be such a front line, because all the vectors will be perpendicular to this line, and all of them have two lengths of one. So this is the front line, so to speak. So the whole, like, animal vector field, if you wish, are moving like this. Okay, fine. Now, what's the divergence? Okay, divergence is a scalar product of nabla and our vector. Okay, so we have to calculate derivative of this by x plus derivative of this by y. Okay? All right, fine. So my vx is equal to minus y times x square plus y square minus one-half. And I have to partial derivative by x is equal. Minus y is just a multiplier because we are deriving by x, partial derivative, minus one-half. That's the degree exponential, x square plus y square minus three second, which is one-half minus one, times two x. Which is two and two goes plus and plus goes, which is xy times x square plus y square minus three second, right? Great. Now, what if I take the y component of my vector by dy? Again, x, that's basically x times x square plus y square in the minus one-half. But I am deriving by y, so x is just a multiplier. And then I have to multiply by this x square plus y square to the minus one-half by y. So that would be x square plus y square minus three second and one-half and two y. Which is equal to minus one-half, sorry, minus one-half. Which is equal to xy, yeah, two and one-half goes minus one-half. Minus remains, it's important. Why is it important? Because if I will add them up, that would be zero. So divergence is equal to zero. Well, again, some explanation. Remember the divergence of the wind would be not zero if the wind goes from a higher to a lower. So in the area where it's higher pressure, the higher density it would be positive. And in the area where it's lower, it would be negative. But in between, where the air is just moving without any kind of modification in the magnitude, if they are far away, these points of lower and higher pressure, then somewhere in between, divergence would be equal to zero. Because it's just moving the masses without modification of their density. Here we have exactly the same situation because all the vectors, they are of different direction, I understand that, but they all have the same magnitude equal to one, right? And that's probably why we have this particular result. Does the direction make some influence on the divergence? I don't know. Probably it does, because in my case, everything is symmetrical. Everything is really circular. Now, if my vectors would be of the same length, but not such a nice symmetrical direction, maybe in some cases we will have divergence greater or less than zero, but not in this case. In this case, we have everything very, very smoothly moving without any change of direction, without any change of magnitude, sorry. So maybe if I will do it somehow differently, and you might actually try to create some kind of a field where the magnitude of all the vectors is the same, but divergence would still be somewhere different from zero. That's an interesting question. Okay. Now, where is divergence equal to zero everywhere again? And I do attribute this to constant magnitude and relatively symmetrical kind of always circular direction. And the last thing is let's calculate the curl. Okay, the curl. Curl is a little bit more involved, but what we have to do is exactly the same as I did in the previous problem. So, again, it's ddy by dx minus dvx by dyk. That's the z component. That's what I remember. Everything else are circularly changed the variables. I'm sorry. There is no k here. Oh, I forgot. This is a two-dimensional, not three-dimensional. So, basically, that's it. That's all we have. This is the curl in two-dimensional case. If you remember, I started with this when I was trying to explain the curl, and then I moved to three-dimensional where I had this whole ijk, et cetera. In two-dimensionals, that's it. So, this is my dx. This is my dy. All right. Let's just calculate it. dvy by dx is equal to minus y stands like a multiplier because it doesn't depend on x. x plus y squared in the power of minus one-half. So, I have minus one-half. x squared plus y squared minus three-half and two-x. Actually, I have already done it for dy. That's this one. No, I'm wrong. I'm wrong. So, it's x divided. So, I'm partially driving by x. So, it's not that simple. So, it's u times v. Okay, uv is equal to uv plus uv. Remember this formula for derivatives? Okay. So, in this case, my u is x and my v is x plus y squared to the power of minus one-half. Okay, so derivative of x by x is one times this thing without change plus x is without change. And now derivative by the second function. So, it's minus one-half. x squared plus y squared minus three-half times two-x. Which is what? Okay, let's just leave it as is. Now, my dvx by dy would be similar. So, first I have the top one. So, minus stays. So, it's one times x squared plus y squared to the power of minus one-half. So, that's the one thing. And then minus y minus one-half x squared plus y squared minus three-half times two-y. So, this is the difference between them. Now, I have to take and that would be a curl. So, what's the difference between them? Now, two and two would be actually out. So, the difference would be this is x squared. Now, if I would change sine to a minus would be plus here. Now, minus and minus and then I would change the sine so it would be minus here. So, two and two can go. Minus stays. So, if I would write this. So, this is minus and this is minus. This is plus and this is plus. So, I'll have two x squared plus y squared minus one in the power of minus one-half. And here I have x squared plus y squared with a minus sign. So, it would be x squared plus y squared in the first minus three seconds. So, again it's minus one-half, right? x squared and y squared would be x squared plus y squared in the first degree. And this is minus three seconds. The degree we will add and that would be this. So, it's two minus one, it would be only one. So, that would be my curl of v. That's interesting. Now, first of all, obviously it's the length of the distance basically from zero to the point of interest, right? So, the curl is diminishing. It's inversely proportional to a distance. The further we are, the less curly basically the field becomes. And that's kind of obvious because since our field is circular, the further we are, since vectors are of the same lengths. Remember we started from the same length so this is equal to one. So, they are more, the further they are from the center of the circle, the closer they are to being parallel to each other because on a small radius, on a small radius, two vectors are on this particular angle between themselves. So, that's kind of more curly than if I would take a bigger radius, they are much more parallel to each other. So, that's why the curliness of this particular vector is diminishing the further we are from the center. And the proportionality is exactly like a distance from the center. Well, that's it for today. It's just small exercises in differentiating basically. So, you have to remember what is the gradient, which is multiplication by NABLA, which is divergence, which is scalar product by NABLA, and what is the curl, which is a vector cross product with the NABLA. That's it. Everything else is technicality. Please read the notes for this lecture. It's on the Unisor.com. And, well, basically that's it. The next would be related to Maxwell's equation. So, I will use this apparatus to present the Maxwell's equation. That's it. Thank you very much and good luck.