 All right, first bad news. The final exam will be on December 19th. Exactly one month from today. Well, that's good news in some sense, right? You still have one month. And the exam will be from 12.30 to 3.30 in the afternoon at Hearst Gym, which I think is very appropriate because, you know, perhaps I put some physical exercises on some of the problems. Shoot some hoops and run a couple of circles. You have a question? So again, you can find this information online. I don't remember, but I hope it's not swimming pool. Although for this big class, perhaps, that's the only place on campus where we can fit everybody if we drain the water from the pool. So last time we had this discussion about this Thursday, next Thursday. So this is next month, okay? December, not this month. All right. But in the meantime, we continue with vector calculus. And today actually, so we're kind of getting to more and more interesting stuff. And today we will do some preparatory work for a generalization of a Green's theorem which we discussed last time to the three-dimensional case. Come on, guys, let's focus, okay? So this three-dimensional generalization of a Green's theorem is called Stock's theorem. And it is actually very useful in many areas of science and engineering. I'll talk a little bit about Maxwell equations today because we'll finally have all the tools needed to actually formulate Maxwell's equations which govern electromagnetism. But let me start by drawing an analogy. And I guess this has become sort of a prevailing theme in this course in the last few weeks. And I have to say that mathematics is really about finding and exploiting analogies. There are a lot of things which are parallel and kind of the same ideas and same patterns play out in different domains in slightly different ways. And if you learn how they play out in one place, you can actually gain some insights about how they play out in other areas. So case in point is this type of theorems which we are discussing right now which as I said already many times take the kind of general form integral over some domain over derivative of some differential form and here there's a boundary of this domain and then here you have this form itself. So this is a guiding principle. And what we are doing now is we are finding concrete realizations, incarnations of this general principle in different dimensions. So let's see what we have found so far. So first of all, we have worked with the case when the dimension of D is one. In other words, this object D is a curve, okay? So there are, this tells us the dimension of this object of the curve, right? But we should always remember that geometric objects, they live in some space, right? So there is some ambient space and there is a geometric object itself and we have to distinguish between them. So if you are talking about a curve, that's a one dimensional object. It could live in a one dimensional ambient space or it could live in two dimensional ambient space or it could live in three dimensional ambient space. And I would like to distinguish between these cases now, okay? So let's first look at the case when the curve is in the ambient space is R. That is to say just a real line. So here is a real line. And in this case, the one dimensional object which we will be integrating over will be just an interval. And generally it could be a union of several intervals but in that case the integral will be just a sum over those integrals. Therefore, without loss of generality, we might as well assume that our domain is a single interval. So that would be an interval from A to B. So now we are in the realm of one dimensional calculus and in the realm of one dimensional calculus, this formula, this general formula takes the following shape. So omega first of all is just a function. And so on the right hand side, we're simply evaluating the function at the end points. Whereas on the left hand side, we are differentiating the literally the derivative of this function over this interval. So that's the formula that we are talking about in this case. And that's the simplest case really because this corresponds to the smallest possible dimensions of all objects involved. Our domain of integration is one dimensional and it lives in the one dimensional ambient space. So next, we generalize this. So let me do it like this. So here, I will do it like this. Next we generalize it and we consider ambient space which is two dimensional, okay? So now we are talking about the more general curve on a plane. So a curve on the plane would look something like this. So there is some, and now that's some wiggling room, so we can play with it. It doesn't have to be a segment of a straight line because now we are on the plane. Okay, in this case, we also have a formula and that's what we call the fundamental theorem for line integrals. In this case, again, we have a function F. We have two end points. I will now denote them by capital A and B because here A and B really corresponded to some real numbers. And now this points, this A and B, they represent points on the plane. So these are not numbers anymore, right? And so in this case, on the right, we just evaluate our function at those two points. So it's very similar. And on the left, we are taking the line integral over this curve. Let's denote this curve by C as we always do. And here we'll have the line integral of nabla F dr, a line integral of the gradient vector field, right? So that's what we have in the case when the domain is one dimensional. There is actually one more. There will be one more square when the ambient space, so I would have to draw it here, right, the ambient space will be three dimensional. And actually we stop there. I don't have to go any lower, because in this class, as I said many times, we only talk about lines, planes, and three dimensional spaces. We don't talk about four dimensional spaces, for example. And we'll talk about three dimensional spaces later next week and the week after next. For now, I just wanna focus in this line integrals. I would like to just look at R and R2. So now I would like to make an analogy. And my analogy will be with the case when the dimension of D is two. In other words, now the domain D in this formula on the left, didn't put a bracket, nobody told me. Do tell me if you see something odd on the blackboard. So the domain now is two dimensional, okay? So now if it's two dimensional, we certainly cannot fit it in a one dimensional space. So the simplest example here for the ambient space is not R, but R2. So the ambient, let me abbreviate this, ambient space is R2. That is really the simplest example, right? We don't have smaller, we cannot put a two dimensional object into a smaller dimensional space. And so what is it going to look like? The analog, what's the analog of each of this form, you know, all the formulas, the pictures and the formulas. What's the analog of this formula? The analog, or first, what's the analog of this picture? The analog of this picture is the following. It's just a domain which looks something like this. We'll have some corners, it doesn't matter. And I mean really the interior here, the shaded region. So this is D, this is now D. And this domain has a boundary, which I would like to do, maybe I'll use a different color for it. So let's just call it B of D, the way we did last time, B of D, the boundary. So maybe to make it more, to make the picture more analogous, I'll put yellow here also. That's the boundary. So what I'm saying is if the first piece of the analogy is that this red interval is analogous to this red domain. And these two points, these two yellow points are analogous to this curve, right? Is that clear? Yes? Okay, so that's the first thing. So clearly there is a simple geometric analogy, we just bump all dimensions by one. Whereas here the domain was one dimensional and the boundary was zero dimensional. Now the domain is two dimensional and so the boundary is one dimensional. Okay, just bump everything by one. Now the formula. The formula is the formula we learned last time. In this formula on the right hand side, on the right hand side, we have the line integral of a vector field. So we can write it like this, p dx plus q dy over the boundary, right? That's the right hand side. Because now this omega is actually a vector field and we take the line integral of this vector field over the boundary. The boundary, remember, is oriented according to a rule, special rule, goes counterclockwise. Okay, what about the left hand side? The left hand side now is going to be a double integral because we are going to integrate over D. And so it is going to be the integral over D, double integral of the expression, which at first looked mysterious, but then we discussed last time what the meaning of this expression is. So that's the left hand side. So what I'm saying now is that this formula, the analog of this formula in this column is the left hand side of this here, right? And the analog of the right hand side of this formula in this column is this right hand side. So this is very important to realize. Because once we see a general pattern with all of these formulas, they start making a lot more sense. If we don't see this analogous, it looks like a collection of different formulas which seem to be totally unrelated. But in fact, what I'm trying to explain is the fact that actually they're very closely related. You see, okay, so this is Green's theorem. This is Green's theorem, which we discussed last time. So we have found these three corners, right? Of this diagram, of this picture. And as always, when you do an IQ test, we talked about this before. You're asked to complete the diagram, right? That's what we are going to do. Not, we will not do it everything today, but we will start laying the foundations for proper understanding of what should go here. And that's what's called Stokes's theorem. So, but we can use this picture to see the different elements of what's to come in this corner, okay? So we should just try to generalize by using analogy between the left and right, and also kind of see the progression from the first line to the second line. So what should this theorem be about? I mean, we know, we see that there must be some result here as well. So what should it be about? Well, first of all, we are still within this column, which means that the dimension, I will use now this board because I don't have enough space to work this out here. So I'll just think of this as a magnified corner of this blackboard. So we're still in the second column, and that means the dimension is two. So on the left, we're still going to integrate over two-dimensional object. But now the ambient space should be three-dimensional, right? Why? Because I want to establish an analog of what I had here when the ambient space was R2. The ambient space was R2, whereas the object was a curve. So the gap between the dimension of the object and the dimension of the ambient space was one. And so we want the gap to be one here as well. That means two-dimensional object inside three-dimensional space, okay? So what kind of object, let me give you an example of such an object. Well, the simplest example of such an object would be, would be a plane, a plane in three-dimensional space. And here is our favorite example of a plane. This is our, you see, this is why, by the way, I erased that, but there is a big game this weekend. So of course, I hope everybody knows about this and will come to support Brokliak. So anyway, this is our plane. So you see, if we were in the top column, so to speak, in the top column, our two-dimensional domain is confined within a given plane. So let's say that plane was this vertical plane. So our domain would be part of this, right? And the plane is rigid. So if we are within that plane, that's all we can do. But now we say, okay, there is actually three-dimensional space out there. So let's look at more general two-dimensional objects within that three-dimensional space. And the first thing we can do is we can take, for example, this same plane and just rotate it like this. And we can also move it like this, right? So this way, we can actually get any plane. And then we can take a piece of that plane. So that would be a more general object than before. Slightly more general object because it would not fit in the original two-dimensional plane, which was just this. So now it would say this one would not be exactly like this. Would not be exactly like this one, right? So this is a very small generalization, but generalization nonetheless. If we were in this column, if we were still in this column, that would correspond to just restricting our attention to line segments, but line segments which can go anywhere on the plane. Whereas before we had a line segment confined within particular line, okay? But of course, here there is a lot more to curves. We don't just consider line intervals, we also consider this wiggly curves like this, right? So what's the analog of that kind of object in this context? So that would be something which does not fit even in a rotated plane, right? So for instance, of course the simplest example is a sphere. Or a sphere is kind of difficult to draw. So what I'll draw is an upper hemisphere. Let me draw an upper hemisphere. And let me use red. So this is upper hemisphere. You see, I mean the, it's like this. So it doesn't fit in any plane. It doesn't fit in this plane. I'm drawing it on this plane, but it doesn't live on this plane, right? Unlike this one, which actually fits in a particular plane. So, right? So this is what I'm trying to say is that this is the analog, this is a proper analog, two dimensional object analog of a curve on the plane. And it also has a boundary, also has a boundary. Now I know that there's one confusing point here. Oftentimes people get confused why do we assign to this upper hemisphere or to sphere itself dimension two and not three. And this is kind of common confusion about how we count dimensions. And I said it before and I'll say it again. You have to distinguish between the geometric object and the ambient space. Here I'm drawing a sphere, think of a dome, a roof, a dome shaped roof like this. It lives in a three dimensional space, but it doesn't mean that it itself is three dimensional. It is two dimensional because if you have a little bug on this, living on this dome, the bug can only go in two independent directions. There are two independent, two only two degrees of freedom so to speak in which the bug can move, not three. I mean, it doesn't go inside or outside. If it lives here, then that's the way it is. So, a more mathematical way to think about it is the following. You can actually think of flattening this dome, this half dome, onto the disk which lies at its foundation, at its base. And so we can actually identify points on the dome with the corresponding projections onto this flat disk. So there is actually one-to-one correspondence between the dome, the hemisphere, and the disk. Now the disk surely is two dimensional. So because we established such a one-to-one correspondence, that also is an indication or actually is a proof of the fact that this is also two dimensional. So this is all to say that it is this kind of objects that we will have to integrate in this corner on the left. So on the left we will be integrating over things like a sphere, like a sphere or upper hemisphere. Okay, and on the right, what are we supposed to integrate on the right? Well, you see now we actually, so to understand this picture, we looked at this analogy with analogy on this side crossing from right to left. Now to understand what we're going to be integrating, we cross from bottom to top. You see, if we were in this column, the right-hand side actually stays the same. The right-hand side stays the same. We just take the values of the function at the endpoints here and here. The difference is that these are the endpoints of an interval on the line. And these are the endpoints of a curve on the plane. But structurally it's the same. So that's why we should expect that going from top to bottom within this column, we should also be integrating very similar objects. So what are we integrating here? Here we are integrating a vector field which has components P and Q, and we integrate it over the boundary. So we should do the same thing here. So this will be an integral of a vector field over the boundary. So I will now write it out in more detail. And so basically I have already assembled the formula that we need, the stock's theorem. I have already assembled most of the elements. The only thing which is missing right now is this, the question mark here. What should appear here? And that should be the analog of this expression, dQ dx minus dP dy. So that's the only thing which is missing right now. So, but let me write this, let me write out the right-hand side in more detail first before I explain what that question mark stands for. So now F is going to be a vector field in R3. We shouldn't be scared of that. I mean, we understand that vector fields can be on a plane or it can also be in three-dimensional space. Remember, we talked about wind maps. Well, if you think about wind map, wind map is two-dimensional. So a wind map would represent a vector field on a plane. But what does it mean? It actually means that we have to say where exactly do we measure the wind vector for this map? And usually we just measure it on the surface, right? But in fact, if you look at the entire three-dimensional space, each point in that three-dimensional space has some wind vector. Even now in this classroom, there is some mass winds. Winds of change, as they say. So you can sense some winds, right? But the wind actually depends, it lives in a three-dimensional space, not just a two-dimensional space. And so that's a vector field in a three-dimensional space. That's totally natural. So that's what F is going to be because now that we live in a three-dimensional space, let's look at the most general vector fields in a three-dimensional space. There's no reason to confine ourselves to vector fields in a two-dimensional space. So that vector field now has three components. There's some p, which is a function of x, y, z times i, plus q, also a function of x, y, z, but let me not write it out just to save time, and r times k, where, remember the i, j, and k are the three basis vectors. This is i, this is j, and this is k. This is x, this is y, and this is z. Okay? So the difference is that in this corner of the diagram, we consider a vector field on the plane. That vector field, first of all, does not have a third component, so it only has p and q. And second of all, because it only lives on the plane, it does not have a dependence on z. So this is a more general situation in two ways. First of all, we have a third component, r, which corresponds to the third direction, k. And also p, q, and r all depend on x, y, and z. Okay? So then what is f dr? Well, dr, as always, is just dx times i, plus dy times j, plus dz times k. And so you just take the dot product of these two guys, and you find pdx, plus qdy, plus rdz. Just like before. Before, we did not have r. So that's why what we got was just pdx, plus qdy, which you see on the upper right corner. Now we'll have a slightly more general expression, pdx, plus qdy, plus rdz. So let me now write down the theorem which we would like to formulate in more detail. In that theorem, we will have on the right hand side a line integral of pdx, plus qdy, plus rdz, where p, q, and r are three functions, three functions of x, y, z, of x, y, z. But in fact, it is better to think about them as components of a vector field, components of one single vector field. And we have, we are going to integrate over the boundary over surface in three dimensional space. And a very good example of this is upper hemisphere. So that's just a good picture for it, okay? But of course there are, you can always make something more fancy like this. So it looks like a little rabbit. You see what I mean. Okay. And that's the right hand side. And on the left hand side, we will have a double integral because we're going to integrate over the rabbit's head or over the upper hemisphere. I want to stay politically correct, so I hope I don't offend any rabbits. So let's just, let's just do an upper hemisphere. Just to be on the safe side. So we integrate over. Maybe I should emphasize that here we're integrating over the yellow thing. So maybe I should erase it now, right? So actually the whole thing. Now I'm really politically correct. And here I'm integrating over the red thing, of which the yellow is the boundary. So in fact, now I don't even have to, to make it a dotted line. It is, we can actually see the entire boundary. So boundary is here. Okay, is that clear? So that's exactly the same as in previous cases. And the only question that remains is what do we have here? What should we integrate here? And that should be something which is the analog, analog of the expression dq dx minus dp dy. So you can ask, why not just put the same expression? dq dx minus dp dy. That worked in the case when we had a flat region, which fit entirely in on the plane. Well, there is a problem with that expression. The problem with that expression is that it does not depend on r at all. Only depends on p and q. That's why it worked when we were on the plane. On the plane, our vector field only has two components, p and q. So such an expression makes sense. It depends on both of them. But now the right-hand side depends not only on p and q, it also depends on r. So surely this r should appear on this side, right? Because of course, if we change r, we will get different answers on the right. So it's impossible to have a formula where on the right you have r, but on the left you don't have r. So it has to be something like this, okay? It has to be something like this. And that's what we're going to find out. What exactly is it? So here's the plan. I will have to explain what kind of integrals. First of all, we'll have to learn how to integrate over general surfaces, general surfaces in R3. And the second is that we will have to find an analog of this expression. And the analog of this expression is what's called curl. So just define, explain what is a curl, curl of f, which also can be written as a cross product between nabla and f. And finally, we will put everything together and we will write the left-hand side of this formula. We will have to write the following thing. It will be double integral over d of what we'll call curl or nabla f like this, dot ds. And that will be the answer. That's the answer which will go instead of the question mark over there and that will complete our diagram. But you see, we can't do it right away now because there are a couple of things which are missing. First of all, we don't know yet how to integrate over surfaces. What we do know is how to integrate over flat surfaces. But see, there's a big difference between a region like this and a region like this, which is what we are talking about now, okay? Or think about this curtain. You can bend and shape it in any way you like, unlike a flat region like this. For flat regions, we know how to integrate. For this curvy regions, we actually do know something because remember at the very beginning of this course, we talked about surfaces of revolution and about areas of such surfaces. So we actually know a few examples of formulas for surface areas. And what we'll do, we'll just generalize that. First of all, to get surface areas for general surfaces, but also to get surface integrals of vector fields which are like line integrals of vector fields that we found for curves. So we'll now proceed to develop this and to kind of laying down the foundations for explaining what the left-hand side is. And so the first thing we'll do, what I would like to do is to talk about the curl because that's sort of the easier part. This kind of algebra, more kind of algebraic part. So what is the curl? Any questions about this, by the way? So speaking of the exam, I'm actually going to put some materials on the course homepage by the end of this week. I've been asked by some students to put the review problems and things like that before Thanksgiving so that you guys have more time to work on this. So I'll do it this week. All right, so first, so we talk about curl. So here F is again a vector field like on that board with components P, Q and R. And curl is just what you would get by taking the cross product formally between this vector field and nabla. What do I mean by nabla? By nabla, I mean a vector which has components d, dx, d, dy and d, dz. So we just follow the general rule. How to do cross product? To do cross product, we have to make this diagram like this where in the first row, we put i, j, k. In the second row, we put the components of the first of the two vectors involved. In this case, it's going to be d, d, dx, d, dy. And d, dz. And then we'll have P, Q and R which are the components of the second vector involved, right? That's just the usual formula for the cross product except the novelty is that, well, one of this is actually bona fide vector field. The other one is kind of strange looking expression. It has partial derivatives as components. But nevertheless, we can assemble this diagram, this picture, matrix really is a proper name for it and take its determinant, right? So what is this determinant equal to? Just the usual rules apply. In front of i, we should have d, dy of R minus, well, here we should multiply them in the order from the second row to the third row. So it would be d, dz of Q. So let me write this down. This means dR, dy minus dQ, dz times i. You see what I mean? Just this minus this. But you have to, the derivative goes first. It wouldn't make much sense if we wrote Q times d, dz. So, but d, dz of Q makes sense as a partial derivative of Q. And then we proceed in the same way. So the next one will be dR, dx. Minus dP, dz, and that's the j component. And finally, we will have, for this one, we'll have dx of Q minus dP, dy. And so what I'd like to say is that actually this expression should be thought of as the analog of the expression dQ, dx, minus dP, dy. So let me convince you that this is a good analog for that expression. It's a good analog for this expression. Well, first of all, let's try to reduce this formula to the two-dimensional case. The two-dimensional case can be obtained from the three-dimensional case by first of all setting R to be equal to zero and also saying that P is a function of X and Y and Q is a function of X and Y only, right? If we do that, then our vector field PQR will actually become just a vector field PQ on the plane. So let's see what will happen with this formula. If indeed we say that R is zero, P depends on the X and Y, and so does Q. Well, first of all, because R is zero, all terms involving R will disappear. Second, because I said that P does not depend on Z, this will disappear. Q doesn't depend on Z, this will disappear. So the first two summands in this formula will disappear and what will we be left with? We'll be left with this. So in other words, we get the old expression DQDX minus DPDY, but now we actually, before we thought about it as a function, but now we think about it as a vector. We think about it as the third component of a vector. So in other words, in space we arrange it like this. Our vector field F is a wind map on the X, Y plane. And the curl of that vector field actually happens to be going in the vertical direction, direction of K, and its magnitude is equal to DQDX minus DPDY at each point. So it's a slightly different perspective on this expression. Before it was a function, now it's a vector, but always pointing in the Z direction in the three-dimensional space. But this is already the first indication that this is a good candidate for it. For replacing this formula in the three-dimensional case. In a sense, you see, it's very symmetrical because each time what happens is, let's say we take K and we get an expression which involves derivatives of the remaining two variables, remaining two components, P and Q. So K gives us P and Q and we take derivatives, cross derivatives for the remaining two coordinates X and Y. And likewise here, for example, here's J, but we work with the things which correspond to X and Z. Here it's I, but we work with things corresponding to Y, X and Z and Y and Z here. So that's the first reason why it's a good replacement for it. But there is actually a better reason because actually this expression played a very important role when we talked about conservative vector fields. And that's the next thing that I want to explain. Remember, conservative vector field is a vector field which is a gradient of a function. And these are very nice vector fields. For example, for these vector fields, it's very easy to calculate line integrals. We just have to know the values of the function at the end points and take the difference. So conservative vector field, vector field, I recall. First of all, let me recall what happens in R2. In R2, it means that your vector field, which is P times I plus Q times J is the gradient of a function, for some function, for some function, okay? And if a vector field is conservative, we can use the fundamental theorem for line integrals to evaluate line integrals of such vector field. And we saw examples of how nicely this works. So it is good to know when is a vector field conservative, when is F conservative, conservative. And we had a very simple criterion. We said that it is conservative, F is conservative, if and only if the following formula holds, dQ dx is equal to dP dy, right? Remember, now there was a caveat. There was a subtle point here, which was that actually, that's not always true. It is true if your vector field is defined over a simply connected domain, because we didn't really talk about the reasons for this, but actually there's a very simple geometric reason. The point is that if you have a non-simply connected domain, in other words, for example, you have something like this, or even just remove one point. And your vector field is only defined here and it cannot be extended to inside. What happens, what could happen is that when you take integrals over, say, circles or some curves going around this region, you could get something non-trivial. Even if this equation is satisfied, you could get something non-trivial. But we know that the vector field is conservative if and only if its line integral over closed curve. No matter whether it goes around some hole, it doesn't go around some hole. Over all closed curves should be zero. So the problem is that this equation almost guarantees that such integral is at zero. What it surely guarantees that such integral is at zero when you can contract your curve, your closed curve to a point. Here, you cannot contract to a point. And that's why this argument breaks down. So this is only true if your vector field is defined on a simply connected domain. Now, what are the simply connected domains? In this, if you look at all the, say, homework exercises that we have, 99% of the examples are vector fields which are actually defined everywhere on the plane. They don't have any poles, any singularities, right? So in fact, this is the main, in this course, we mostly focus on such vector fields. So let's assume for now that the vector field actually is defined everywhere on the entire plane. And the plane is simply connected. Non-simply connected things are things like you have to remove some points. That would mean that your vector field actually has poles. You have something like one over x squared plus y squared in a denominator. So let's suppose to simplify matters that f is actually defined everywhere. It's defined on the entire plane everywhere. Everywhere. It's defined, defined everywhere on R2. Then this issue does not arise and this is actually true statement. It would not have been a true statement had I not made this assumption that the vector field is defined on a simply connected domain. So then this is actually very simple criterion. But what is this formula? What does this formula say? It says that the difference between these two guys is zero. Let me erase this. We will remember this. Suppose that it's defined everywhere. So this is the criterion. And now we can appreciate more, even more this strange looking expression dq dx minus dp dy. This strange looking expression tells us whether our vector field is conservative or not. If it is zero, this vector field is conservative provided it's defined everywhere. And if it's not zero, then for sure it's not conservative. You see, that's exactly this expression that we're talking about. That's the expression which appears in Green's formula. Okay? And if this formula is correct, is this formula satisfied? Then we actually have an algorithm how to find, how to find F. We talked about it before and you must have done some exercises doing this. And you see the point is that you can try to apply this algorithm in the case when this formula is not satisfied but you'll get stuck. You'll see that it doesn't work. The algorithm doesn't work. You will not be able to find a function F. So the first step when you're asked whether the vector field is conservative or not is to compute this expression and see if it is zero or not. Okay? So the reason why I'm saying all this is because the notion of conservative vector field makes sense not only for vector fields on the plane like this, but also for vector fields in the three dimensional space. And if I want to prove to you or convince you that this curl now in the three dimensional case plays the role of this expression, that would be a very simple test. Would I be able to use this expression to test as to whether or to see as to whether my vector field is conservative or not? So that would be the second argument in favor of this expression. So let's talk about vector fields in the three dimensional space. Now in R3, in R3, we have a vector field which is PI plus QJ. And now there is a third component which is RK. And what we would like to do is we would like to apply the same notion. Well, a vector field is called conservative again if it is equal to nabla F. You see nabla F, the gradient, makes sense on the plane and in space. It's just that now it is going to have as components, it's going to have dF dx i plus dF dyj plus dF dzk. In the two dimensional case, we would not have this term. And now we have this term, but this notion makes sense anyway in both cases. And now we can again ask the question, when is F conservative? And the answer is given by the following theorem. Let's suppose again that F is defined everywhere is well defined. Well defined means that it does not have poles. So for example, if you have a vector field which you may have seen minus yi plus xj divided by x squared plus y squared. This is what makes it ill defined or not well defined because this gives you in zero at when x and y are equal to zero zero, you get a pole. You cannot evaluate this vector field at this point. So this vector field is not defined on the entire plane. But for example, vector field like this is certainly defined everywhere on the plane. That's what I mean when I say that F is well defined. Defined everywhere in Rc. So I assume that if we assume that then saying that F is conservative is equivalent to saying that the curl is zero. And that is a justification for saying that that's another justification for saying that the curl is a good replacement for this expression. Which in fact, which in fact can be found as one of the components, this one for the curl. And this is actually proved. This is actually proved by using stock theorem which we are now trying to establish. So this is to tell, this was all this long discussion was to convince you of the importance and usefulness of the strange looking expression. Because you probably thought on Tuesday that this looks strange. Well now this looks even more strange, right? So like what does it mean? We'll talk more about the meaning of this later. But at least now you see that it is useful. It is a very functional thing. It is, first of all, it reduces to the old expression for when our vector field actually is a vector field on the plane. And second of all, it can be used as a criterion or for a criterion as to whether the vector field is conservative in the three dimensional space just the way we used this criterion on the plane. So let's see how it works in practice. Let's suppose you are given a vector field now in the three dimensional space and you're asked, is this vector field conservative? And if so, find the potential function. Find the function F for which this vector field is, of which this vector field is the gradient. So here's an example. Suppose your vector field is like this, e to the zi plus j plus x e to the zk. So the question is, is it conservative? Well, first of all, let's see. It is well-defined everywhere. There are no poles, there are no singularities. We can evaluate, we can find the value of this vector field for any x, y, z. No expressions like one over x squared plus y squared. So great, we can then apply this theorem. This theorem says that if F is well-defined everywhere, it is to find out whether it's conservative, it's sufficient to just calculate this. So let's calculate the curl. Oh, we just write i, j, k, d dx, d dy, d dz. Then we have e to the z1 and x e to the z. So we start calculating. What do we find? So here you have to take the derivative of respect to y, partial derivative with respect to y of this expression. It does not depend on y, so that's zero. Minus d dz of one, that's zero two. So zero. Good, next. You have j, let me put minus. d dx of x e to the z, that's e to the z. Minus d dz of e to the z, also e to the z. So that's zero also. Plus k times d dx of one, that's zero. Minus d dy of e to the z, that's zero also. So zero k. So this is indeed the zero vector. Please note that I put an arrow over the zero. Remember we had a conversation about this a while ago about different zeros, right? A curl is not a function, it's a vector field. So we cannot, if you put it like this, if you put zero like this, that would mean, that would be okay for a function. But saying that it's zero is more than saying that the function is zero. It means that three functions are zero. Not just one, but three functions, the components in front of i, j, and k. And that means that this is zero as a vector. It has all components zero. That's what this criterion means. So criterion is actually a collection of three equations, not just one, but three equations. Okay, great, so this is satisfied. So it is conservative, so answer is yes. Okay, a follow-up question. Follow-up question is if you're so smart and you know it's conservative, find the function for which it is, the gradient, right? And that's really very easy to do. We just follow the same algorithm that we used in the two-dimensional case. So find f such that f is equal to nabla, okay? So what we do is we start doing anti-derivatives. And the first time you take anti-derivatives is back to one of the three variables, and then you continue. So we have to make a choice. So let's just do this three-fourth way. Let's take anti-derivative with respect to, of the first function. Take anti-derivative of e to the z with respect to x. So what do we get? We get x times e to the z, but that's not all. We have to add a constant when we take anti-derivatives. But now this constant is not really a constant. It is a constant with respect to x, which is a variable under which we differentiate. So this constant actually could depend on y and z. So it's not really a constant. It's a function of y and z. That's what the first step of the algorithm tells us. Now we're going to differentiate this with respect to y. Differentiate, well, let's just say, take d dy. Take d dy. Well, this will be zero. And this will be d c d y. We are supposed to get the second component of our vector field, which is one. You can see I wrote the second component is j. It means one times j. So this should be one. If this is one, it means that c is equal to y plus another constant. Let's call it c one. But so if we were working with a vector field in two variables, vector field on the plane, which depends only on x and y, that's where we would stop. But now we have a third variable. So a priori, this c one could depend on the last variable, which remains, which is z. So we have to just make one more step in this algorithm. Otherwise, it looks exactly the same as before. So now we should take d dz of this, of what? We have to assemble our, what we've learned so far. But we've learned so far is that f is equal to this, but this c is equal to this. So that means that we can replace now this c in this formula by y plus c one of z. And now we have to take the derivative of this whole thing with respect to z. So we find x e to the z, the derivative of this is zero plus c one prime from z. And we should compare it to the expression which we were given. We were given x e to the z. So that means that c prime is actually zero, which means that actually c prime is a constant. Sorry, c one prime. So c one prime, c one, not c one prime. C one prime is zero. C one is actually a constant. Is an honest constant. It doesn't have any hidden dependence on anything. So the answer is, the answer is that f is number of the function which we found, which is x e to the z plus y plus c one. But this is actually constant. Okay? Any questions? Yes. Yes. Let me switch the board. I think the derivative of, first I have to reload f to use all the information I found so far. Right? I have found that it is equal to this from the previous step, where this is already just a function of z. Because I found that c of x, this was c of y z, but we have found that it's y plus c one of z. So I already put this back for the function f. Right? And now, I think the derivative of this with respect to z. And this is what I get. And now I have to compare it to my component, the third component of my vector field, which is x e to the z. Right? And the very top row in front of k, you have x e to the z. So I say this is equal x e to the z. So I can cancel out these guys. And I end up with c one prime is zero. That means c one is actually a constant. It actually does not depend on z. And that gives me the answer. Okay? All right. Any other questions? Okay. So this is how it works. This actually works in a very similar way. And this was all to convince you that of importance of this curl. And this is the expression which we will use for which we will use to establish the stocks formula that elusive formula which appears in a lower right corner. If you think, if you thought this was too many formulas, there's actually one more, which is called divergence. And divergence we won't need until the last lecture. But since it is in this chapter of the book, I guess the idea being, let's just put it all on the table. All the derivatives that we have. And let's look at all of them at the same time. Okay? So we might as well, I might as well write a formula for divergence. So divergence is also an operation on vector fields in three dimensional space, which we can think of as a dot product, a dot product as opposed to a cross product with nabla. So in other words, it is dp dx plus dq dy plus dr dz. So it's a very interesting operation. It takes the vector field and it spits out a function, not a vector field, but spits out a function. And the nice thing about it, so far it's not clear what this is good for. But here is one result which is, which might convince you that it is important, which is that if we take the divergence of a curl. So let's say we have some vector field F. Let's first apply to it curl, that is to say cross product with nabla. That's what's given by this formula. So we get some expression, right? And let's take now the divergence of the result. In other words, you substitute these three components, this, this, and this into this formula. These are not the P, Q, and R of the original F, but these are P, Q, and R of the curl, okay? So you'll get actually double derivatives. So it looks like a really ugly expression, but actually it turns out to be zero. So that's the good thing. So this is something, this is kind of a kryptonite for curl. That's what kills curl. So curl looks very complicated, but there is a nice formula which actually kills curl. And now we can actually assemble all of this, all of this operations we've learned up to now. And now we can actually see that there is some system to this, that it's not random. So let me explain this. We have learned, we have learned three different operations. We have learned three different operations. The first one was the gradient. The gradient goes from functions to vector fields. You have a function F, you get a vector field nubla F, right? That's the first operation we've learned. The second operation which we've learned today goes from vector fields to vector fields, all in R3, all in space. And that's the curl. It takes the vector field and it sends it to its curl, which maybe it's better to call it curl to kind of emphasize that it's different from this guy. Even though I really like this notation nubla cross F, but for the purpose of this diagram, maybe I'll just stick to curl to emphasize the difference. And now we've learned one more, which is the divergence. And divergence now goes from vector fields to functions. So you start with functions, go to vector fields and vector fields to vector fields. And there is another operation which goes from vector fields back to functions. So this one takes the vector field and it maps it to divergence. So three different operations. And now we've learned a very interesting aspect of this. Since you have these three different operations, you can apply two operations one after another. You can start with a function and you can apply the gradient, you get this vector field. Because it's a vector field, we can apply to it the curl, right? And what do we get this way? Zero, right? This is exactly the theorem which I have formulated. In other words, if you apply this operation twice and you take curl of nubla F, you get zero. This is this formula. Because again, curl is the same, I recall, let me maybe write it one more time. So divergence we can write like this and curl we can also write as cross. This is just notation, the same notation, two different choices of notation for the same thing. So you see, apply these two operations one after another and you get zero, but zero is an error. That's interesting, right? What about if we apply this one and then this one? So that means take a vector field, take its curl and then apply the divergence. Also get zero. That's this formula right here. It's the same as writing it, you can write it like this. This is zero without an error because it is a function. It's a function zero, it's not a vector zero. So you see, we have three different operations and these operations have this property that if you apply two of them in sequence, you'll get zero. I would like to contrast that with something we discussed last time about taking boundary. You have geometric object, if you have a geometric object like a domain we discussed, then we can take its boundary, right? We can take its boundary. So we just get this and let's apply boundary one more time. Is there a mistake? What if you take the gradient of divergence? I see, I see. So in other words, you wanna apply it like this. Well, you get what's called Laplacian, so I think. No, Laplacian is the other way. Sorry, see the point is that actually, that would be like going up, that would be going from like kind of going back to the bottom to the top. And in fact, it should be, we should only be going down. So it doesn't look, it doesn't look right. But it's a good idea. In a way it's a good idea, but it's important here to go down. Just like for boundaries. So see what I'm trying to explain now. When you, the point that I'm trying to explain is that this property that you have an operation which kind of goes, makes you go one step. And if you take it twice, you get zero. And what I'm trying to explain is it's exactly like taking the boundary. When you take the boundary, you go, you lower dimension by one. And you see, and if you take boundary one more time, you get nothing. So this, by the way, is another zero. This is empty. This means empty set. For our purposes, it's like zero, right? It's nothing. So the point is, if you think about, this is a very intuitive concept, boundary, right? Any geometric object has a boundary. And then you can, once you realize that there is such a thing as boundary, and you just start thinking about it. It's like, what is a boundary of a boundary? Why not? If you take a boundary, why not take boundary of a boundary? And at first it looks like a good idea, but then you realize that actually, it always gives you empty set. So there is this very interesting geometric structure taking the boundary. And it has this very interesting property, which mathematicians call nilpotency. It's a nilpotent operation. Nilpotent meaning that if you square it, you get zero. So now you try to look for such an operation algebraically in the algebraic world, in the world of functions and vector fields. And this is what you find. You find that there indeed exists such operations, which have the same nilpotent property, that if you square it, if you apply it twice, you get zero. And this is a very important aspect of this formulas that we are trying to establish. The formulas relating integrals in different dimensions because this is what I call in this guiding principle. This is what I call D. You see, I explained already many times that the general guiding principle we are pursuing here has to do with integrating over domains and boundaries. And then here you have some algebraic object and it's derivative. So going from left to right, you take the boundary, just like this. And going from right to left, you take the derivative. And now here is a derivative that I'm talking about. Here I lay it all on the table. If you start with a function, when you are doing, when it's sort of going from zero to one, you take the gradient, then you take the curl, and then you take the divergence. And this operation D, which kind of an algebraic operation, it lives in a different world. But turns out that first of all, there is a trade-off between these two, which gives us this beautiful identity. And also it has exactly the same property as B. In other words, D squared is zero, just like B squared is zero. So there is this parallel analogy between the geometric world and the algebraic world in which taking boundary on the geometric side corresponds to taking the derivative in this sense on the other side. And this formula is just the expression of the fact that one operation is kind of dual to the other. So in some sense, they are one and the same, okay? So that's what these formulas are all about. They look complicated, they look very abstract, but in fact, they are all part of a very conceptual phenomenon, very conceptual thing, and very important phenomenon in mathematics. And as a bonus, we can now write down the Maxwell equations. Because Maxwell equations use, so this is a curl, this is divergence. And now I can write down Maxwell equations because these equations use curl and divergence. So in the Maxwell equations, Maxwell equations are the equations which govern electromagnetism, which govern the behavior of electric and magnetic fields. Well, you know that there are, first of all, there are electric fields. For example, if you have charged particles, they will, if they have opposite charges, they will attract, if they have the same charges, they will repel each other, right? That's electric field. So if you have, if you put a charge somewhere, if you put, for example, the nucleus of an atom consists of some protons and neutrons and protons have positive charge, so neutrons have neutral charge, but protons have positive charge, so they all, they create a field which would grab and attract electrons which are negatively charged particles. So it's a very important field. There's also magnetic field. That's the field that you have when you have a magnet and you have a, someone told me a story that they drop the key at night in a paddle and it's like, what do you do? And there is a very nice solution. You take a big magnet and you put it in the water and boom, you got your key, see? So that's magnetic field. So that's also very important. And not just for finding keys, I suppose, right? So now you look for equations which govern electric and magnetic fields. And here are the equations which were written in 19th century by Maxwell as well as other people. So the, usually the notation we choose is like this. E for electric field and B is for magnetic field. This are just vector fields, just like the kind of vector fields which we talked about, like on that board, here is a vector field. Electric and magnetic fields are just, vector fields like this. But they change, you know, they change in space and they also depend on time. And these equations describe how they depend on space and time. And the equations look surprisingly simple, deceptively simple, perhaps I should say. But let me assume that there are no charges and currents. So it's kind of equations in the vacuum, if you will. Just to simplify, just to give you a flavor of what these equations look like. No charges or currents. So the equations just involve the divergence and the curl. So that's the first equation. See, this is the divergence of E is zero. And you have divergence of B is zero. And then the curl of E is minus one over C dB dt, the derivative with respect to time. And the curl B is one over C dE. Now, what is C? C is the speed of light, speed of light, which is approximately equal to 300,000 kilometers per second. So it's very fast, but it's not infinite, it's finite. So this is very important actually. So you see, so these are the equations. We wouldn't be able to even understand, to read these equations, to even understand these equations if we did not define the curl and the divergence. What's perhaps more important is that these methods, these formulas that we are proving now, this guiding principle and various incarnations that we are talking about. By using those, we can actually derive very important consequences. For example, the Gauss law, the Ampère's law and things like that. But if you just look at this, and it's really, I think it's amazing that such complicated interaction, which governs basically the entire universe, there's two forces in the entire universe, can be summarized basically on half of the blackboard in this very neat and beautiful way. And just looking at these formulas, you already see some very important things which actually were, which were sometimes the cornerstones of physics of the 20s, maybe 21st and 22nd and so on centuries. The first thing you see is that these numbers, you have this number which doesn't see, speed of light does not depend on anything, right? It's just a constant here, which is actually an incredibly powerful statement because in our real, in kind of, in our everyday life, we are used to the fact that velocity or speed depends on who the observer is. If you are standing here and there is a bicycle going in this direction, right? Doesn't matter, okay? So let's say with some speed, right? This is the speed which you see, but if you are going on a bicycle in this direction or if you are walking with some speed, right? You will see, you will observe a different speed. But if you are with speed of light, it's not like this. Speed of light will appear in the same way to this observer and to this observer. You don't add the velocities one way or the other. It is constant. This is what Maxwell's equations indicate. And this was one of the first arguments for Einstein to create his special relativity theory. He said, look, these equations work. So what they show us is that the speed of light is constant in all inertial coordinate systems. That was one of the first steps in creating special relativity. I'm telling you all this to convince you that the kind of stuff we are doing is not just a bunch of formulas that you need to memorize, but actually this stuff makes sense and this stuff is very important, okay? So we'll continue on Tuesday and go bears.