 In the last lecture, we had introduced the subject of vector calculus and we discussed fields in particular scalar and vector fields. We also brought out the meaning of gradient of a scalar field and I emphasize that the gradient of a scalar field is a vector field. Today, I would be spending a little more time in discussing special functions of vector fields or operators which act on the vector fields. But before that, I would introduce what are known as the line and the surface integrals. We are all familiar with normal integrals. As you are all aware that an integral is always regarded as a sum. Now, what we want to do is this that let us first talk about a force which is acting on a particle and is in one dimension. We are aware that integral f d x if it is a one dimensional problem gives me the work done in let us say moving the particle from a point a to the point b. Now, in computing this work done, we do a simple integration. However, suppose my force acts on a particle and takes it along a curve. It could be two dimensional or it could be even a three dimensional path. It could be an open curve like what is being shown here or it could be a closed curve for example, like this. Question is how does one compute the work done in taking the particle let us say from a to b in this case or in a circuit in the other case. Now, we will represent the work done let us call this curves as c, c standing for curve. It could be a closed curve or an open curve. The work done as we are aware is given by dot product of f with d l where d l is along the path. Now, in particular the vector d l is directed it is an infinitesimal element directed along the tangent to the path. For example, d l is directed like this here d l is directed like that there. Now, so what we want to do is to generalize the concept of computing work in one dimension to determining how much of work has been done when a force acts on a particle and takes it from one point to another or in a circuit. Now, the point is in general the this integral that is integral of f dot d l from a to b depends on not only on the end points as we are accustomed to while we compute integrals, but it also depends on details of the path taken. Of course, there are exceptions to this for instance if the force is a conservative force exact details of these we will talk about later in the day to day, but if it is a conservative force for instance a gravitational force or electrostatic force then it can be shown that the work done is path independent. However, in such a case it would always depend upon the end points. Now, let us look at what how does one write this. So, what I have is along the curve f dot d r let us say because d r is a an element along the curve and what we do is this that let me write this as f dot d r by d t d t where I have introduced time as a parameter time as a parameter and you notice that this is nothing but the dot product of force with the velocity with the velocity and I take a time integral. So, in other words what I have done is to parameterize that with time. Now, along a curve along a curve I always have a single parameter and this is what we are going to exploit in computing the work or in general an integral like f dot d l along a given path. Now, let me start with an example. So, for instance I have given here a force which is written as f is equal to now I have taken in two dimension one could easily extended the curve to three dimension, but let me illustrate this by taking a force whose x component is x square minus y square and whose y component is two times x y. Now, what I am going to do is to compute the work along three different paths in both the cases I am going from a point which is origin let me draw it here again origin to a point two one. So, supposing this is one along the y axis and this is the point two along the x axis. So, from origin to the point p which is the point two one first I will calculate the you know if this is the force the line integral of this force along a straight line joining the point o to b o to p. Now, notice since this straight line joins origin two two one I can easily find out what is the parameter or what is the equation of this line. The equation of this line obviously is x equal to x equal to 2 y alternatively y is equal to x by 2. Now, let us look at how does one compute this? Remember integral f dot d l if you like is f x d x plus integral f y d y. Now, f x is x square minus y square, but we have seen that y is equal to x by 2. So, therefore it is x square minus x square by 4 d x the limit of integration over x is obviously from 0 to 2. I write the y part again. So, f y is given to be 2 x y, but x is 2 y. So, as a result I get 4 y into y is y square d y and limit on y is 0 to 1. You compute this, this is this trivial integration you will get this to be equal to 3 by 2 plus 4 by 3 which is equal to 17 by 6. So, that is the work done along this line. Now, let us compute the work done along line along this curve 2. Now, I have taken the curve 2 to be a parabola I have taken the curve 2 to be a parabola and my equation of the curve 2 is y is equal to x square by 4 I have given this. You can see that the it is of course an equation to a parabola and when x is equal to 0 y equal to 0 and when x is equal to 2 y is equal to 1 because it is 2 square by 4. Now, how does one do this for the same force field or same f the vector field. So, once again I will compute this. So, this is now 0 to 1 I have got x square I have got y square, but y is equal to x square by 4. So, y square is x to the power 4 by 16 d x plus 2 times integral x y. Now, x is 2 root y. So, I have got 2 times 2 square root of 2 root y and I had already a y there d y and limits are once again the x limits are from 0 to 2 and the y limits are from 0 to 1. So, if you compute this now this very straight forward x square is x cube by 3 x to the power 4 is x to the power 5 by 5 which gives me x to the power 5 by 80 limits are from 0 to 2 and this is 4 integral of y to the power 3 by 2. So, I get y to the power 5 by 2 divided by 5 by 2 limits are from 0 to 1 and you can easily compute this. This happens to be 34 by 15 plus 8 by 5 and is equal to 58 by 15. Now, you already noticed something that the value that we got for in the first case was 17 by 6 whereas, the value that we got in the second case is 58 by 15. Now, you can see that the value of the line integral indeed depends upon the path that we have chosen. In the third case, I am taking a path from origin to again to 1 by a little complicated path. I am first going from here 0 0 to 0 1. So, let us draw that path here. So, I want to go to 2 1 and I am going by a broken path like this in 2 steps. So, this is 0 1 this is 0 0. So, I go like that first like this then like this. Recall my force is x square minus y square x component and the y component is 2 x y j. Now, look at this. When I go from origin to this point, let us call this O P Q. So, as I go from O to Q, my x remains 0 the value of x does not change. Now, if x remains 0 further along this path along this path d x is equal to 0. So, notice that if I now compute this integral from O to P integral of f x d x that is x square minus y square d x that gives me 0 because d x is 0 f y d y namely 2 x y d y gives me 0 because on this line along the entire path my x is 0. So, therefore, contribution to the line integral from the path O P. So, let me write that as integral I 1 is equal to 0. In the second path I go from P to Q. So, I will go from what I meant is I go from O to P and then I will go from P to Q. Now, when I go from P to Q, when I go from P to Q, on the P to Q my value of y remains constant y is equal to 1. Now, once y is equal to 1, d y of course, becomes equal to 0 because constant have differentiation as equal to 0. So, since d y is equal to 0, I am only going to have to compute f x d x. So, this is simply 0 to 2 x square minus y square which is 1 d x and this is equal to x cube by 3 minus x evaluated from 0 to 2 which is 8 by 3 minus 2 which is equal to 2 by 3. So, you notice one thing that I calculated this integral f d l f dot d l in three different paths 1, 2 and 3 and every time I got a different result. Now, I will talk about a slightly different technique now. I will return back to the same integral with a slight modification as I go along. Now, look at this field, remember we are talking about a vector field. Now, this vector field is f is equal to minus y i plus x j. Remember last time we gave a prescription for drawing the vector fields, catching the vector fields. So, you could see this vector field, this is and I am going to compute the integral of this vector field in along a quadrant of a circle, along a quadrant of a circle taking me from this point that is 1 0 2 0 1 in the anticlockwise fashion 1 0 2 0 1 in the anticlockwise fashion. Now, if you look at the sketch of the vector field that I have made, the sketch of the vector field that I have made you notice that in this quadrant, in this quadrant the because x is positive, because x is positive there is always a positive y component of the force field of the vector field. Now, as a result loosely speaking these vectors are all directed somewhat in the upward direction, very loosely speaking. Now, another thing that you notice is this that the vectors are directed in the upward direction and I am going to compute my line integral along this path and in this path the direction of d l is like this. So, therefore, the angle between the direction of the vectors the which we have shown in this sketch and the direction of the tangent to the curve always makes an acute angle. Now, since it makes an acute angle f dot d l which is the magnitude of f times magnitude of d l times cosine of the angle between the two remains positive. Now, it so therefore, what I expect it is a purely going by geometry I expect this line integral to be of positive value. Here I will use a slightly different technique the technique is that along the curve I need to have a parameter I have already said that a along a curve I have a single parameter. Now, notice that if I have this curve which is quadrant of a circle. Now, I can use a parameter like this using this angle theta as a parameter. Now, if theta is a parameter notice that in term this is a quadrant of a circle of radius 1 because I have said the equation is x square plus y square equal to 1 and I need to compute minus y d x plus x d y. In terms of this theta since the radius of this circle is 1 I can write x is equal to cos theta and y is equal to sin theta. So, d x is minus sin theta d theta and d y is cos theta d theta. So, I notice that this integral now which is over the curve C minus y is minus sin theta d x is another minus sin theta d theta. So, minus minus plus sin square theta d theta plus integral again over the same curve x which is cos theta d y is cos theta d theta. So, I get cos square theta d theta. Now, I can simply add them and get it as sin square theta plus cos square theta which is equal to 1 of cos d theta. So, which is nothing but integral of d theta and the limits on theta we know is from 0 to pi by 2. So, as a result the result of this integration is simply pi by 2. So, this is the parameterization. Now, in general as we said in the beginning the line integral depends on both the path of integration and the n points. Now, if the line integral f dot d l depends on the path of integration and if f is a force field then we say that this force is a non-conservative force. That is a non-conservative force is one where the line integral depends upon the path of integration. As we had seen in the first example that we gave you that there were three paths along which we calculated the line integral and found the values to be different. However, f dot d l only depends upon the n points supposing I am going along some path from a to b and it does not depend upon exactly which path do I take. I do I take this path or for example, that path or this path I do not care result is always the same. In other words this line integral in such situation depends only on the two end points a and b. Such a force is called a conservative force such a force is called a conservative force and you can see that if I have a closed path supposing I am calculating the line integral like this and closed path integrations are written like this with a circle over the integral sign it is also called contour integral closed contour. For a conservative force for a conservative force you can show that this is equal to 0. Now the proof is really straight forward supposing this is the path you are going starting from the point a you are returning back to the point b. Let me take any path arbitrary point b on the way. So, you notice my integral is from a to b by path 1 then once I reach b I am following a segment of the path which is 2. So, we will say b to a by path 2 now what you have said is this that this integral from a to b depends only on the end points depends only on the end points. So, in other words this integral f dot d l so a to b f dot d l along path 1 can be written as something like phi b minus phi a just I evaluate this and then subtract from the value of that integral standard definite integral result at b minus result at a. When I am going on path 2 I am going from b to a f dot d l now once again we have said that it depends upon only the end points. So, as a result this is going to be phi at the upper limit minus phi at the lower limit which is b. Now, add them up obviously the integral over f dot d l over the contour is equal to 0. So, for a conservative force the integral line integral from a to b depends on only the end points and if you take the force field through a closed contour the line integral is identically equal to 0. Now, let us spend a little more time on the conservative force field the reason is that the field which is of importance to us in this course namely the electrostatic field is a conservative field so is gravitational field. Now, suppose I have a force field vector field which can be written as a gradient of a scalar field force field can be written as a gradient of a scalar field. Now, remember we had said the gradient of a scalar is a vector now I am not saying that all force fields can be expressed as a gradient of a scalar field. But supposing it can be done now let us see what it means what it means is this supposing I am going from a to b again I have to calculate f dot d l. Now, since f is a gradient you remember the expression for the gradient of a scalar function in Cartesian and we had seen that this is equal to I partial phi with respect to x plus j partial phi with respect to y plus k partial phi with respect to z dotted with dotted with d l which is obviously I d x plus j d y plus k d z. Now, take the dot product remember I dot I is 1 j dot j and k dot k are also 1, but I dot j I dot k etcetera are 0. So, what I am left with is a to b d phi by d x partial phi partial x d x plus partial phi with partial y d y plus partial phi with partial z d z, but this quantity d phi by d x d x d phi by d y d y plus d phi by d z d z is the differential of the function phi itself. So, in other words this is from a to b just d phi because this expression includes changes due to variation in x y and z. Now, this is the an exact differential. So, as a result the integration is very trivial it is phi b minus phi a. Notice what we have done we have said or we have proved that if a force field is expressible as gradient of a scalar function then integral f dot d l only depends upon the two end points. In other words such a field is conservative field such a field is conservative field. Now, we make a statement at this stage that if I have a conservative field of force for instance an electric field or a gravitational field I can always express the force as a gradient of some function. Generally because of various reasons instead of writing f as equal to gradient of phi we define f as equal to minus the gradient of phi and this phi is then given a name called potential this is conventional, but in principle a mathematician would call phi as a potential whether you have a minus sign in front of it or not, but we will of course in electrostatics we will be using this convention namely force is negative gradient of a potential function and the potential is a scalar function the potential is a scalar function. So, this is about line integrals. So, we have seen line integrals in general depend upon the path that we take in some cases when they do not we call the vector fields as non-conservative field in which case the line integrals only depend on the end points line integrals only depend upon the end points. So, line integrals was essentially one parameter integral though I am not saying it is in one dimension, but I go along a line I go along a curve. So, there is a single parameter in such a situation. Now, let me now extend this let me now extend this to define what is known as a surface integral. Now, in surface integral we will write this surface integral as integral of f dot n I will explain what it means d s. Now, in order to understand the surface integral look at this situation let me take an open surface for example, a hemisphere which is an open surface a better situation is to have a net for example, a butterfly net as it is called you notice that there is a rim here there is a rim here and there is a net which is bounded by this ring and this is an open surface in the sense that this surface is there only to one side of the boundary. Now, this is exactly the picture that we gave in the other one. So, I have an open surface and so what we do is that whatever is the bounding curve of the open surface. So, imagine I have a cup imagine I have a cup as hemispherical cup or even an ordinary cup it does not matter. Now, this is my bounding surface this is my bounding surface and a curve which binds it again going back to my butterfly net this is what it is this is the rim this is the rim and this net is the surface. Now, when that happens we say that direction of this n it is a unit vector which is perpendicular to the surface. Now, it is not a constant vector it changes from point to point it changes from point to point. So, for example, you are at this point. So, draw a perpendicular at that point and draw it outward. Now, what is meant by outward because after all it the outward needs to be defined the. So, what we do is we adopt a convention which is called a right handed corkscrew rule which is with which you are all familiar that is if I am rotating or if I am moving suppose this curve is being turned along the direction of a corkscrew then the direction in which head proceeds is the direction of my normal to the surface the outward normal and. So, that is why I said a butterfly net is very clear. Now, there is some issues connected with it. Number one that I need to have what are known as two sided surface in order that my definition of a surface integral makes sense I need a two sided surface. What is a two sided surface for example, an open hemisphere it has an inside surface and an outside surface you take a cup you have an outside surface and an inside surface. Now, on the other hand there are certain surfaces which are one sided surface and for example, one of the examples is what we call as a mobius strip it is diagrammatically shown here, but let me try to explain what is a mobius strip. Now, I have this strip of blue strip of paper I have taken a small strip with it. Now, notice that if I just fold it like this making it a ring then I have two sides in other words supposing I am on the upper surface and I want to go to the lower inside surface. Then I have to always cross an edge I have to always cross an edge and this is known as a double sided surface. This is true of for example, if I take this glass at this moment there is water in it, but this is the outside surface and there is an inside surface of the glass. If I want to go from the outside to the inside I will need to cross this ring I will need to cross the edge, but on the other hand please look at it this way supposing I take this again this is diagrammatically shown here supposing I take this strip again and instead of folding it like this I will give it half a turn I will give it half a turn and fold it in this fashion fold it in this fashion that is the same side does not get stuck I will get. So, I have actually already stuck it with a glue and this is the type of picture that I get. Now, notice what is so special about this this special about this is I can come from any point on this surface and traverse both sides both inside and outside by ever having to cross this ring. So, in other words at a given point the outward normal is not uniquely defined the outward normal can be both outward like this or inward on the inner surface this is a one sided surface. So, for such surfaces this example this is called a mobius strip my definition of a surface integral is not valid. So, let me talk about normal surfaces which are two surface two sided open surfaces and in going from one side to the other side I need to pass through a rim need to pass through an edge well here the same thing the this is the hemisphere surface is A and the rim of the hemisphere is the curve C and I have to sort of define that if I am going in a particular direction then the direction in which the head proceeds will be the direction of the normal outward normal. So, that is about two dimensional surface integrals. Now, one can now go and if I now remember the surface integral is dot product of a vector field with the outward normal and integrated over the surface this is usually called a flux of the force field F through this surface flux of the force field F through the surface. Now, we can now extend this concept to a three dimensional integral, but this is much simpler because the integral is simply a volume integral over a scalar function is phi and. So, this is nothing, but normal triple integration which you are familiar with. So, volume integral phi d v v here is the volume is nothing, but triple integral over x y z of phi which is a function of x y and z now I can define another type of triple integral or volume integral which is F d v that is an integral of a vector field over a volume. Now, this is to be interpreted as actually three integrals because I can write my F as i F x plus j F y plus k F z d v. So, this is nothing, but i integral F x d v now remember F x is a scalar function plus j F y d v plus k F z d v each one of them we have just now learnt what it means because this is just like any other volume integral of a scalar function which is nothing, but a triple integral. So, therefore, this is i times 1 triple integral plus j times a second triple integral plus k times a third triple integral. So, this is all about the line surface and the volume integrals. Now, let me concentrate a little more or spend a little more time on vector field once again. Now, I define what is known as divergence of a vector field the reason for the name will become clear in a little while. Now, I define divergence of a vector field as a limit of the surface integral per unit volume I will explain this with figures as I go along as the volume enclosed by the surface becomes smaller and smaller becomes infinitely smaller. Now, let us look at this relationship a little more carefully and now divergence of the vector field is written as a dive dive F one also writes it for reason to become clear as del dot F both of them are identical ways of expressing divergence. So, what I have said is this that now this is a point relationship this is a point relationship in all the words you define divergence of a vector field at a particular point. Now, supposing around that point around that point I think of a small volume infinitely small volume described by a surface d s. Now, so what we do is this take this volume to be infinitely small let us call it delta v. So, I define the del dot F as limit of delta v going to 0 of the surface integral F dot n d s. So, I have taken a closed volume very small closed volume at the point where I am computing the divergence of F. So, that is that is why this circle it circle essentially means it is closed. So, over a closed surface I have taken and divided by the volume enclosed by that surface take this as the volume becomes smaller and smaller I reduce that volume essentially to that point and that is why I say it is a point relationship. Now, this picture is what explains what happens now supposing I have a volume bigger volume. Now, what do I do let us consider a volume now I can think of splitting that volume into a large number of infinitesimally small volumes. Now, let us look at what am I trying to do in the process now look at this supposing I have a volume of this type this is a rectangular parallel pipe and I have a volume and notice I am at this moment concentrating only on two surfaces. Though it has six surfaces I have a upper surface whose direction is outward normal is this direction and a lower surface the this surface the bottom surface whose outward normal is directed like this. Now supposing this surface I am to split into two smaller surfaces. So, in other words this big surface this big volume gets cut into two I take a section and just split it now when I split it remember nothing has happened to the direction of the normal to this surface nothing has again happened to the normal to the bottom surface, but once you cut it in this section in this section the normal is directed like this and in this section the normal is directed like that I do a mental exercise. So, if I just cut it since the vector field is the same over this surface which I am going to cut then the contribution due to the surface integral from this surface in this surface because they are directed oppositely will become 0. So, what it means is that if you have a volume if you have a volume and you split that volume into in. So, supposing I have a volume like this and I split it up into smaller volumes etcetera. Now I can reintroduce talk about the same argument once again and say that the contribution from things which are inside the pairs will always cancel and what will I be then left with I will be left with only contribution from the outside surface only contribution from outside surface well. So, let us now try to compute we have said that divergence of a function is the ratio of the limit of surface integral over the vector field at a point I am saying it is a point because I have an infinitesimal volume there bounded by an infinitesimal surface. I divide it by the volume and I let the volume and the surface go to 0 and this ratio is finite and that quantity I call as the divergence of the vector field at that point. So, what is this divergence of a vector field? This picture here sort of gives you an idea of if here we have drawn the pictures of the vector field you notice as the name suggests the field gives an idea of how much the vector field diverges from a point. The situation becomes clear when we look at for example, a velocity field in a vector we know that if there are if I look at a closed region then if the fluid is flowing in to a point then of course, the density inside would increase and similarly if the fluid is flowing out the density would decrease. So, here is an example where the field is spreading out from the origin. So, here the divergence is positive in this case however the divergence is negative this is the physical meaning of a divergence of a vector. So, to summarize what we have done today is we defined a line integral and emphasize the fact that in general the line integral of a vector field could be path dependent. In case it does not depend on the path it is called a conservative field and we have seen for a conservative field I can write the force as a gradient of a potential we defined a surface integral and also defined the divergence of a vector field through a ratio of a surface integral of a closed small surface at a point with the infinitesimal volume that it encloses. Next time we will be obtaining an explicit expression for the divergence in Cartesian coordinate system.