[MUSIC] Hello, everybody. Welcome back to our lecture Electrodynamics and Its Applications. My name is Seungbum Hong, and to my right side. >> I'm Melodie Glasser. >> Mm-hm. So we will cover the third part of our lectures, mainly line integral of vector fields. So as you can see from this slide, the vector integrals start from the line integral of del five, which is the gradient function of a scalar field. So let's get some understanding of the significance of the derivatives of fields. Then we will have a better feeling for what vector field equation means. So the meanings of the divergence and curl operations can be best grasped via vector integrals and equations relating such integrals. So you may remember in our previous lectures that in order to describe the correct characters or characteristics of vector field. We need number one, flux, and number two, circulation, and those characteristics are best described by divergence and curl. And in this lecture, we're going to show you how those meanings can be very clear by using the vector integrals. Now the equation we just study are mathematical theorems, which will be useful not only for interpreting the meaning and the content of divergence and the curl. But also in working out general physical theories. So it's not just limited to electrodynamics. However, the theorems are not very useful for solving problems, except in very simple case. So this would be the pros and cons of our lecture today. So we will start with the first theorem, where you will see the line integral of the del psi function. And see how this line integral along arbitrary path connecting those two starting point and end points will result in the state function. So Melodie, can you explain to our students how we can understand the equation in this box? >> Sure. The integral is basically just taking small infinitesimal changes and then seeing how some property, in this case, psi is dependent or changes along that small tiny section. >> Yeah. So as Melodie pointed out, the integral on the right side is really summation of infinitesimal sized segments. And we will see, depending on which way you take, it can be a long path or a short path. But however, no matter which path you will take. They will all cancel out and you will only get the results that are always the same. And we will prove it using mathematics. In order to visualize what we are doing, here is the schematic diagram of connecting the starting point 1 and the end point 2 with arbitrary gradient function delta psi. Where the tangential component will be most important when we calculate the line integral here. So let's revisit what we mean by line integral. So the value of the function at the Is segment will be denoted by a subscript here, I. So we can equate this integral by summation function in using arithmetic of the function sub i, which means the function value at each segment, i segment. And the length of that segment, delta psi, which have the same links and its very small. And as you can see, we are putting the links of del psi approaching zero, so that means it is infinitesimal. So if I replace the function by del psi, which we just discussed, then you will have an equation that shows delta psi sub t, which is tangential component of the delta psi times del sub I. And you do the summation along the path gamma. And the line integral here will be the limiting value of the sum as we add more and more segments by reducing the links of each segment, okay? So here is the proof mathematically and before explaining the proof to you, I'll revisit the equation that we just learned previously. The temperature change or temperature difference between two neighboring points is equal to gradient of temperature. That the relative position vector that connects the two points, right? So Melodie, do you remember what we learned? Or what the meaning of the gradient of T means using this equation? >> Yeah, it's the direction in the highest rate of change, I think. >> Yes, so as you can see, del T is the direction of the temperature change, the maximum temperature change and the magnitude. Because the maximum delta T will always be the case when delta T and delta R are in the same direction, right? If they are in perpendicular direction, the change in temperature will be zero, right? >> Yeah. >> So that's what we learned. And we can expand this equation based on the definition of the dot product. So dot product between the del T and delta R is round T over round x times delta x, plus the round T over round y, plus run T over run z Delta Z, right? And if we make a generalization here, so we are going to put del precise equals del precise dot delta S, right? And if we do it segment by segment, so starting from the first segment where delta psi 1 is equal to psi of a minus psi of 1 is equal to del psi at first segment, dot delta s, the first segment. And if you do that segment by segment, till the last segment, and if you arrange it in this way, we can see all the other components will be canceled out and what are left, Melodie? >> Just the outside boundaries. >> Exactly, the starting point and the end point. So no matter which way you connect those two points, the intermediate values will always cancel out. So the equality doesn't depend on how the points a, b, c are chosen, and it doesn't depend on what we choose for the curve [INAUDIBLE] joint one and two. >> Mm-hm. >> So it's state function. So that's the end of our proof, right? >> Mm-hm. >> So in a nutshell, again, if we think about this theorem, The line integral of the gradient of this color function psi, along any curve from 1 to 2, is equal to the psi of 2 minus the psi of 1, which, you can see, they are all state functions. All right, so we will then deviate from our electrodynamics and discuss a little bit about the thermal transport, okay? And in this case, we will think about the flux of a vector field. We have already discussed the flux in circulation, but now we are revisiting this with a specific example of thermal transport. So imagine you are traveling to some place with your friends on a beach and have a campfire, all right? Then, from your experience, you know if you're close to the fire you'll have more heat. If you're further away you will have less heat, right? And we can formulate a mathematical equation to describe what is happening when you do that. So imagine you have fictitious boundary that will enclose the campfire. And if you think about the total heat flow outward through this fictitious boundary S, then you can find it by doing a line integral of the heat flow vector h and the surface normal vector of that fictitious boundary. So S equals integral of h dot n, times da, over the surface, which is a closed surface, right? >> Right. >> And if I expand the meaning of flux, right, it will be surface integral of the normal component of any vector that could be a component for vector field, right? So if we think about the electric flux, flux of electric field, how would we define that? >> I'm not sure. >> You're not sure? Okay, so that's good, so let's take a look at this box. So flux of electric flow through fictitious boundary S, which encloses a charge, is equal to electric field dot n, like here, h dot n times da. In this case, there's nothing physical flowing out. >> You see? So mathematically they look similar, but physically they have different meanings. In this case, you have a real physical quantity that is going out as a function of time. But in this case, it is just a static field emanating from charge. >> Also, there's no time involved. >> Exactly. >> in this scenario. >> Exactly, that's true. Now, let's think a little further from here. What if the heat is conserved? Maybe from thermodynamics, or other fields of science, you may have learned conservation of energy and especially conservation of heat energy. If the heat is conserved, then what we can say is the total heat flow outward through this fictitious surface will take away the existing heat inside the surface. And mathematically you can formulate an equation like this, where S = h dot n times da over the closed surface, is equal to -dQ over dt, where Q is the internal heat energy inside the fictitious boundary. So what it means is the rate of decrease of the heat inside the fictitious boundary is equal to the total heat flow outward through that boundary, okay? Now we're going to cover a very important theory based on what we've just learned. So we will use the concept of flux, which is the normal component of our vector field integrated over a closed surface, and see how the flux of an arbitrary surface can be evaluated based on a very simple principle. So here think of an arbitrary surface, and this looks like a potato from Idaho, so I sometimes tell my students this is Potato Law, okay? And imagine now you are going to assess, let's say, the heat flux, the flux of heat out of this, as we mentioned before. Then, if I cut this potato into two, then we will create a new surface or interface where the flux across this boundary should also be considered. However, if you think about those two components, they will share this interface as their surfaces, but the direction of normal vector will be opposite. So the flux through this surface will have the same magnitude, but opposite side. So if I sum the flux from each component, it will be equal to the flux out of the original potato, right? >> Mm-hm. >> So we can do that proof using mathematics here, where you can see a new interface part is denoted by the integral of the vector filled C dot n1, where n1 is the surface mole vector of this part, the left part, part where n2 is the surface normal vector of the right part, right? So if we do addition of those two they will cancel out, because they have the same magnitude but opposite plurality. So what happens is the flux through S1, the newly created component, and flux through S2, if we sum them up, will be equal to the flux of the total entity. So what does it mean? So for any way of dividing the original volume, it is true that the flux through the outer surface, which is the original integral, is equal to a sum of fluxes out of a little interior pieces. So this is wonderful, why? Because then we can use the simple theory of differentiation. Meaning, for example, if you had the arbitrary shaped potato and if you cut them into a cube, then you can do a very simple and nice mathematical analysis. Then, what you can do is to put one of the cube out, Assess the flux of this cube, and do integration, all right? We can use the theory of integral and just evaluate the flux out of our arbitrary surface. So this is what we will do in the next slide. So we will discuss the Gauss's theorem using the simple cube. So Melody, instead of the simplicity of this shape of a cube, what is good about cube? >> It's symmetrical. >> It is symmetrical, it is symmetrical in three directions right? >> Mm-hm. >> Meaning, if I just think about one direction, say x direction, then we can immediately understand about y and z, right? >> Mm-hm. >> So here we will only thing about the surface 1 and surface 2, which has a surface normal along the x direction, okay? And the flux through S1 is equal to integral of C dot nda over S1. Where because the right direction is positive, the surface mole, which is pointing to the left, will have a minus sign. So that's why we put minus, right? dydz will be the area of this infinitesimally small cube, right? Cx will be the x component of the C vector. So in a nutshell, it will be approximated by -Cx(1) because we're thinking of surface 1, times delta y delta z. Now, if we think about flux through S2, which is the adjacent surface, they have the surface. This has in the surface null to the right, which is positive, so we don't have minus sign in front of it. So this will be Cx(2) delta y delta z. Now, we are going to use a first order approximation of our function where Cx(2) is approximated by Cx(1) + round Cx over round x times delta x. And if I replace this in this equation and if we sum this up, what remains is round Cx over round x times delta x, delta y, and delta z. And this is the volume of the cube that we are now discussing, right? And as we have solved for 1 and 2, right? What would happen to y and x, Melody? >> Well, because they're symmetrical they should have a very similar appearance. >> Exactly, so if you look on this slide, the next slide, you can see from S1 and to S2, we just solve it. S3 and S4, it just changes the subscript from x to y. S5 and S6, you just change the subscript to z. And if you sum them up you will have this form for an infinitesimal cube, right? So the flux out of a cube = (rounded Cx over round x + round Cy over round y + round Cz over round z) delta x delta y delta z. And if you remember this, this was just the formula for divergence of a vector field. Then you can make it neater by using the vector formula (del dot C) del V. So looking at this formula, what can we know or what is the meaning of the divergence of a vector field, Melody? >> So it's the flux- >> Mm-hm. >> Over the volume is equal to the divergent of the vector field. >> Exactly. So what Melody told us right now is the fact that if you're relearning this equation, then the flux, which is D dot nda over delta V, which is the volume of the cube is equal to the divergence of the vector field. So this is a point function, right? Divergence of the vector field is a point function. And so at each point, we can define the flux out of this point per unit volume, per unit volume. So it's flux per unit volume, all right? So it means, if I integrate this over an arbitrary space, then we can evaluate the flux out of that arbitrary volume, right? So this is really important, so divergence of a vector field is the flux at that point per unit volume. So in this page, we're going to discuss one more time about the heat transport, to see how the theory we learned, the divergent theorem, is connected to heat conduction case. So we'll revisit the problem of outward heat flow from a body placed in a system where the heat is conserved. And that was an example we just discussed before about the campfire and fictitious boundary that encloses the campfire. In that case, total heat flow outward through the boundary S is equal the surface integral of h dot nda. Which is equal to minus dQ over dt, where the Q represents total heat inside the boundary. Now, Melody, what would happen if we shrank the volume in this to a very little cube? >> Thank you. So, if you have a smaller cube, then from the previous slide we learned this equation. So we can now write this is the differential form with the dels, which is more convenient later because we don't need boundary conditions. So we set this del h times delta V equal to this differential. And if we think of a very small cube, then you have a uniform density inside, or in this case a density of heat. So you can change this into a q, which represents the heat per unit volume and multiply it by sum delta V. And then you get this equation right here, which is the native differential of the heat density multiplied by the volume. >> Very good. So from here, you can see that you have the common factor, delta V, on both sides, left and right. So we can remove it and derive the conservation law in a differential form. So this was a conservation law in an integral form, this will be in differential form. So now you can define at each point for the conservation of heat energy. And as you can see, the negative of the heat testing change as a function of time is equal to the diversions of the heat flow vector, okay? Now we're going to use Gauss' Theorem, or Divergence Theorem, to prove the heat conservation law in integral form from differential one. So this will be a mental exercise to go back and forth what we just learned it for. So let's start with the Gauss' law which states the flux of a vector field from and arbitrary shaped volume, where the boundary is the closed surface denoted by S. That is equal to the volume integral of divergence of this vector field across the volume, or including the entire volume. Now, from the previous slide, we learned that the energy conservation, heat energy conservation can take a differential form where- dq over dt = del dot h, right? So we are going to use this inside the Gauss' law, or Gauss Theorem. And as you can see here, the C is replaced by h. So you will see we're just repeating the first part here. And then we are going to replace the divergence of vector field by this equation for energy conservation. And if I do that, instead of the divergence, we will have dq over dt. Then we can change the sequence of integration and differentiation. So d over dt will come out, and then it will be the volume integral of the charged heat density in this arbitrary shape volume. And that will really give you the total heat, right? Total heat, so again, you can see the total heat flow outward through S = integration of h dot nda = dQ over dt. Which is the integral form of heat conservation, so that will be end of proof. So what we do in this kind of mental practice, we can go back and forth and corroborate what we just learned, right?
