[MUSIC] Hello, everybody, welcome to Electrodynamics and Its Applications. This will be lecture 11. I'm with my teaching assistant, Melodie Glasser. My name is Professor Seungbum Hong. So here we will first discuss about the meaning of the governing equations to describe any natural phenomenon. So we will start with this sentence, the same equations have the same solutions. So you may wonder what that means. And as we have learned, the equations for metastatic world, now we are going to see if there is a way we can use the same equation to describe other phenomenon. And to our surprise or to your surprise, you will see the equations for many different physical situations have exactly the same appearance. And as we have studied one subject that is metastatics, we immediately have a great deal of direct and precise knowledge about the solutions of the equations of another. Then we may wonder, how come such a diverse phenomenon can have the same equations as well as the same answer? So let me ask my teaching assistant about her opinion on that. So Melodie, can you comment on that? >> Well, I think one of the reasons so many different physical situations have the same sort of solutions is because, first of all, we can simplify many situations. So we often make assumptions. There's uniform density of some quantity like charge or heat or something. And the second one is just geometry because a lot of situations we examine have circles or squares, and because of that, the equation starts to look kind of similar. >> Exactly, and if I complement her comments, we can also think that we use a lot of approximations like we only considered up to a second order derivatives and high order terms which we ignored. So you can imagine for partial derivative and partial differential equations, we will have similar types of solutions. But that doesn't mean neccessarily that all the natural phenomena work under the same mechanism. Okay, so let's take a look at the equations that we learned from electrostatics and how we can apply that to other fields. So you can see here the first equation from Maxwell equations. It describes that the kappa times electric field. That's considering the matter in our world. The divergence of this field is equal to the free energy density divided by [INAUDIBLE] and in electrostatic world, we don't have circulation. So the curl of electric field will be 0 and we already derived from these two equations that we can come up with Poisson's equations, where you can see divergence of kappa del phi is equal to minus rho phi or epsilon 0. And so if you see some similar type of equations from other fields, then we can use all the solutions that we have struggled to learn throughout the lectures and see the meaning of the solution in each field. Now, let's revist to the flow of heat. We have used a lot of the analogies between thermal transport, as my teaching assistant Melodie has explained before. And we will revisit it and see how that is, again, related to the equations from the electrostatics. So let's call s the heat energy produced per unit volume per second by the source at the point of interest, and h, which is here, the amount of heat energy which flows per unit time through a unit area perpendicular to the flow at the same point. Then we have proven that the divergence of the heat flow vector is a function of s, which is the rate at which the heat is generated at the source point, and du / dt will be the rate at which the internal energy will be decreasing. So as you can see, if we think about the steady state heat flow where you have no change of heat inside, then du / dt should be 0 because there is no decrease of energy. So del.h = s. So let me ask my TA, Melodie, to recap what it means to have the heat flow vector as a function of the thermoconductivity and gradient of temperature. So Melodie, what does this mean? >> So basically, the temperature gradient from a heat equation like this is the driving force and then the thermal conductivity I kind of think of as a measure of how easy it is to transfer heat. So if you have a very high temperature gradient or a very high difference between the two temperatures, that means the heat conduction will go faster. And then if the thermal conductivity is also high, that means that the material is very susceptible to changes in temperature as well. >> Exactly. So the gradient of temperature is the driving force and kappa, which is the thermal conductivity, is related to kinetics. And the gradient of temperature has the direction normal to the temperature isothermal lines where the change of temperature is the steepest. So you can imagine the heat flow along the steepest gradient, the steepest hill, downhill of the temperature world. And even though you have huge driving force, if you have low thermal conductivity, the thermal transport will be limited by the thermal conductivity. So that's a very good discussion. So let's take a look and compare electrostatics versus steady state heat-flow. So as we mentioned, for steady state heat-flow, the diversions of K del T will be equal to- s, and, as we can see from our electrostatics, the delta K del phi, where K is the electrostatic constant, is equal to- low subs free or epsilon naught. So we can see the formalities. They are exactly the same. So you can just replace each parameter to each case. And then use the solution you're familiar with to understand the phenomena that you are not familiar with. So this is a very powerful tool. So let's take a look at an example. So imagine you have a pipeline. A pipeline where you can say, transport heated water. Right? Or yes, heated water. And inside, you have another pipeline that you have a heater. So if you heat this inner part, you can imagine the heat will flow in a radiant direction and make the outer part warm. So then you can put this under your floor, so then you can have a very warm apartment say, for instance, in Korea. So in that case, you may wonder, what is the temperature distribution and the rate of reaction? How can I solve this? And in fact, you can also ask, what is the rate at which the heat is lost from a length L of the pipe? So how fast is the heat being transported? Across this radial direction. So those questions can be answered in a different way. And one thing that you may remember is to use the Gauss law to solve this problem. And let me ask my TA, Melody, what Gauss law is. >> So the way we learned about it was the integral of E.ds is equal to the sum of charges divided by epsilon not. >> Exactly, so what welearned was if we have a charge inside a closed volume, the flux of electric field out of that charge is exactly the charge divided by epislo non, right? So we can use the same analogy here, if we use absorption of surface, which is cylindrical, because we know this is cylindrical symmetry then the flux of the heat flow vector around the cylindrical surface, which is Gaussian surface, will be equal to the heat generated per unit time inside this volume. So you can say 2 pi rL, which is the area of the cylinder, times H, which is the heat flow vector Will be the flux, will be equal to G, which is the heat generated inside the caution surface. So then h = G over 2 pi r L, and since we know the magnitude is equal to the h flow of vectrates, so it's equal to -K Gradient of T, in this case, the gradient is along the radial direction. So it's dT/dr. So we can equate these two, and if we equate these two we wil have this partial differential equations, where dT/dr is equal to minus G over 2 pi KLr. So by comparing between the answers for heat flow and electrostatics you will get a better feeling or better grasp of how we can use one equation to another phenomenon. So as you can see if I divide the infinitesimal increments from dt and dr to both sides. And do integration for T1 from T2, which is from the inner pipe to outer pipe temperature. Then it will be the biggest of inner part pipe to the outer pipe. So when you do this integration, this is the temperature difference and on the right side you will have logarithmic term because 1 over r will be integrated to logarithmic term. So if I solve for g, then g becomes 2 pi k l over t 1 minus t 2 over log b over a. And you see this result corresponds. This result corresponds exactly to the result for the charge of a cylindrical condenser. So imagine you have a plus charge and minus charge across this pipe. So the charge accumulated at one end of the electrode will be depending on the potential difference between those two as well as the lengths of the pi and the radius ratio of the pi as well as the permittivity. So without going further into doing the same kind of calculation, you can just replace the parameters. So I'm going to show you a real example in the research that my student, [INAUDIBLE] Kim Under supervision. The same kind of calculation for spring type energy harvesters. Where you coat a PVFTRFE what is a polymer on top of a spring structure. And then you deposit electrodes on top of it. So the process is like dip coating. When you dip coat your ice cream, vanilla ice cream into chocolate sauce, then you have chocolate layer, and if you it again, you have another layer. This is how we did. And for the [INAUDIBLE] we did like barbecue. So we rotate it, this spring, while we do position. So in that way we are able to make a cylindrical type of pipeline of two electrodes. Which are sandwiched by dialectic. So if that's the case, then we can use the equations that we learned from this course. And that's what he did and you can see how much charge will be accumulated on the outer electrodes and we could use the knowledge from this course to get the answer. So melody, so in your research, do you see you can use the same type of equations in your for example. >> Yeah, I think so. >> Yeah, so she's working on a multi-layer that could mimic the tectile information. So for that also we need to know how much charge is accumulated. On one surface or the other, and I hope that she can use this equation in our research as well. >> Perfect. >> Okay, all right, now let's take a look at our very interesting sample, Heat source near the ground. Imagine, somehow, you detonated, say, TNT Underground, to make a tunnel, right? If you do that, you want to know, what is the temperature distribution underneath. Maybe for safety, or maybe for designing the destruction of the area of interest. So if that's the case you will ask, what is the temperature distribution near the surface? And as you can see air is a very good insulator, and that's why we use air or vacuum in a tumbler, to make heat preserved inside the volume of interest. So you can think is nearly heat conductivity with zero, kappa is almost zero. If kappa is almost zero, it means you cannot have any heat transfer normal to the surface. So we'll only have parallel component near the surface. So in that case, you can imagine, how can I solve this? If you looked at this picture at a glance, maybe it may remind you of some different kind of distribution. So I'm going to ask my teaching assistant, Melody, what this picture reminds her of. >> Yeah, so in an earlier lecture, we talked about using the mirror to find equations. So I remember we did it with a negative charge and a positive charge. So the field lines would kind of look like this. But this look kind of different with the same pattern. So I'm guessing, instead, we have maybe two of the same charges, and therefore There would be a field line going towards it instead of away from it. >> Exactly. If you have two charge with the same sign as Melody mentioned, you can imagine that they are confronting each other, and because of that the normal component will annihilate. And only what will be left is the, so the normal component will be annihilated and only you have parallel component near the surface. So you can make a guess or maybe this is similar to plus and plus charge distribution. And we will try to prove that, that that's the case. So this will be a special topic with the title of image charge with unusual boundary conditions. So it's very unusual. So it's like an electrostatic problem with two materials with different dielectric constants kappa or on opposite sides of plan boundary. And you can see that the physical condition is that the normal component of h equal vector on the surface is 0. Since we have assumed that there is no heat flow out of the block. So you can ask, in what electrostatic problem do we have the condition that the normal component of the electric field E analog h is 0 at the surface? And there is none. And when we say none it is considering the case where we have electroneutral situation. Electrically neutral situation. So for physical reasons, there may be certain restrictions in the kinds of mathematical conditions which arise in any one subject, as we just mentioned and discussed. So for example, there's no material with a dialectric constant zero, whereas a vacuum does have zero thermal conductivity. And there's no electrostatic analogy for a perfect heat insulator, but we can still use the same method by imagining what would happen if the dielectric constant were zero. And in fact, this kind of thinking can give you some motivation to think about metamaterials or artificially structured materials to have effective dielectric constant of zero, or even negative, right? So, but for macroscopic world, dielectric constant is never zero. And we might have a case where there's a material with a very high dielectric count constant so that we could neglect the dielectric constant of the air outside. So this is kind of contrasting the dielectric constants across the surface or interface. But in this case, we are going to use a little bit different way, so we will use unusual image charge method. We'll try to pick an image source that will automatically make the normal component of the field zero at the surface. And image source of the same sign and the same strength, so this is very important. So this is what Melody already mentioned. So what if we have the same sign of charge? Then and we have them placed at an equal distance from the interface. Then the temperature everywhere is the same by a direct analogy as the potential due to two equal point charges, because we know that from their competition, the normal component will all vanish. And the only one that is left is the horizontal or in-plane component. So if that's the case, we know the potential of a point charge follows 1 over r rule. So T, which is temperature, in that esthetics, it would be phi is equal to G over 4 pi kappa r. So you see this 1 over r. And because we can do super position of two potentials, you will see the temperature for a point source with its image source is G over 4 pi K r1 + G over 4 pi Kr2. So, this will be the answer to the question we already asked. So this is the temperature distribution, and we can know by putting the numbers inside this equation. All right, so let's move onto another example. So let me ask my teaching assistant Melody, whether she has an experience to play with a drum. >> Yeah, of course. >> So when you hit the drum, what happens? >> So when you hit a drum- >> Uh-huh. >> It kind of looks like this picture here, where the membrane of the drum begins to vibrate at whatever, based on the force you hit it at, it vibrates at a certain frequency. And then it generates sound waves, which you can hear. >> Exactly. So when you hit the drum, as Melody mentioned, you will have deformation at certain frequencies, which is characteristics of this structure, as a function of geometry. And as it vibrates it will give you a sound, right? But in this lecture, we will only consider a static case where you are hitting the drum with constant force, but that's it. So in that case you can think of a constant deformation, right? If you poke the drum membrane with a stick from inside you will have a protrusion in some point. And we will see how the protrusion will make surface profile or topography, based on what we have learned. All right, so here the question is, in that case can we describe the shape of the surface, right? So here we are going to learn a few terminologies that are important to understand the deformation of membrane. So we will start with surface tension, which is denoted by tau in the membrane, and we will assume the deflections of the membrane are not too much. Meaning we will have very small, gentle deformation. And in that case, the way you can visualize the presence of surface tension is to make a small cut using a razor blade on the membrane, and see how the cut behaves. And as you see, if you poked the membrane into the normal, to this surface, you see it will be widened in the direction perpendicular to the interface. And in that case you can see it is counteracted with the surface tension which is the force perpendicular to the interface per unit lengths. So you can see force per unit length perpendicular to the direction of force. That will be the surface tension. And it holds together the two sides of a hypothetical cut in the membrane. Now let's do some force analysis on a small piece of a membrane, and you see here in the schematics where it is a part of a membrane, and we are looking at a cross-section of the drop, right? And you can imagine you have a stick here where you are poking upward, and then you can imagine that this will be the shape of the deformation. And in that case, if it is in equilibrium then the reaction force will be the same with the opposite side of the direction, right? So we'll kind of analyze this situation using the force order approximation, where you can see in x direction and y direction, which is perpendicular to this one. You will see the first term refers to the surface tension that is to the left of the stick. And you see tau 2 delta y is the length along the y, times sine theta 2 Will be the vertical portion of this surface tension minus tau 1 delta y sine theta 1 will be from the right-hand side of this surface tension, and again, will be the portion of the vertical sine. And you will see that delta f will have a negative value. And if we assume it is a very small distortions, that is to say, that theta is close to 0, then sine theta will be close to tangent theta, because cosine theta will be close to 1. So then, we can think of this as the slope, instead of sine theta, so we can replace that by the derivative, du over dx. And if we do that, you will see you can replace this equation to this form, where you see in the bracket, you have tau 2 times round u of round x of 2- tau 1, tau 1, round u of round x, sub 1 and then you can multiply by delta y. And if you take a look at this one, this is exactly the definition of the partial derivative with respect to x. So you will have this form of equation. So if you do this kind of analysis along another direction, which is perpendicular to the x, you will find that you will have another term that is similar to the one, but with change of the parameters from x to y. So you will have these two inside a bracket with delta x, delta y, being the area of this infinitesimal part of the membrane, okay? And you see, if you divide the force by the area, and if you define that by a small letter f, what is f? So let me ask Melody, if you divide the force by the area, what does it become? >> Pressure. >> Yes, it becomes pressure. So it will be in Newton per meter squared or Pascal unit. So, as you can see, because this is pressure, you will divide delta F by the area of this infinitesimal part of the membrane. And if you do that, you will be able to see this pressure is equal to minus divergence of tau delta u. And you can see, from the form of the equation, it is very similar to the electrostatic equation. So let's take a look. So 2-D mechanical version of Poisson's Equation, you see is this one, for the membrane. And this is for metastatics. And you can see that they are one to one. So let me ask Melody where the tau corresponds to. So, Melody, where does tau correspond to in metastatics? >> It corresponds to kappa. >> Exactly, so tau corresponds to kappa. What about u? >> That would be the potential. >> Exactly, so u, which is the height information, is to pull of the information. u is potential information, right? So then, with the stick, if I move the sticks with certain displacement, then you can think of this as a pi voltage to this part, right? And then if you apply voltage, the height distribution along the membrane will be the voltage distribution along this direction. So that's how you can visualize the voltage distribution, for example, of a pi actual cable using a drum and a drumstick. So that's the way people have done. And also by rolling a ball from this part to this part you will see how current will flow, okay?