Hello everybody. Welcome back to Electrodynamics and Its Applications. I'm with my teaching assistant, Melodie Glasser, and my name is Professor Seungbum Hong, so this will be the seventh unit of our lecture series. So, we have learned about electrostatic world, where either the charges are fixed in their positions and we know exactly the quantity of the charge, as well as the position they are occupying. Now in this lecture, we're going to divert our attention to some of the mathematics we can use to solve interesting electrostatic field in a given situations. If you remember, before this lecture, we'll learned about image charge method, which is kind of a backward solution method, where people have solved a lot of different kinds of configurations and then backwardly fit that to some of the interesting conductor configuration. So, as written here, we will describe some of the more elaborate methods for solving problem with conductors and we take up two examples, where the charge distribution is neither fixed nor is carried by a conductor, but instead, is determined by some other physical law. So, this will be a big part in the end of our lecture including the plasma oscillation, colloidal particle, salting out of colloidal particles, and also the grit or cif problems. Now, we are first deal with the direct method, evaluation of an integral where initial distributed charges are known, that was what we have done before and in this chapter we're going to use indirect method where we have already used the use of image, so this was already taken care of like sphere or parallel plates or cylinders. In this one, we also have indirect mathematical technique which I will describe shortly, complex number method where we can solve 2-D problems using the mathematics of functions of a complex variable. Okay. So, Melodie, can you help our students understand what we mean by complex variable, what is complex variable? So, the complex number i, is the square root of negative one, and basically if you're going to represent something in complex numbers, you can do it by a coordinate plane where your x is your real number, and then as you can see, over here there's a complex number multiplied by the quantity y, and then that's going to be on this axis here, which is your imaginary number. Exactly. Then, that squared is going to give you some real number that can be represented. Yes. So, exactly. So here's the. So, as mentioned by Melodie, complex numbers have two parts, a real part and imaginary part. An imaginary number is defined by this unit i, which is square root of minus one, and real numbers and imaginary numbers, they're orthogonal, and therefore we can use the fact that x and y are orthogonal in x-y coordinates, and visualize the numbers in the graph. So, that would be our plan zeta. This is zeta equals x plus iy, and we can use zeta as a single variable, and with that write the usual kinds of mathematical function of f of zeta. For example, the quadratic function here, zeta squared or other complicated functions where you can see two examples of which in this slide. Now given any particular f of zeta, we can substitute zeta by x plus iy, and we have a function of x and y with real and imaginary parts. So, let's take a look at this example of quadratic function: f of zeta equals zeta squared, equals parenthesis x plus iy parenthesis square. If we expand this, then it will be x square minus y square, which is the real part and plus two ixy, which is the imaginary part. We can rewrite that in this kind of notation where u of xy is the real part function, v of xy is the imaginary part function, and separate them out like here. So, from any complex function f of zeta, two new functions, u of xy and v of xy can be derived as you can see here, and in the later slides, we will show you that these functions have very interesting properties. Now, before going that, we will claim that for any ordinary function, the functions u and v automatically satisfy the relations written down here which is round u over round x equals round B round y, and round v or round x equals minus round u over round y. From this relationship, we can we can derive that they fulfill the Poisson's equation for neutral object in two-dimensional space. So, that's why you see here, round square u over round x squared plus round square u over round y square equals zero, and the same applies for v as well. Okay? So, we're going to prove what we just claimed for any ordinary function using the mathematical manipulation. So, you can see f of x plus iy equals u of xy plus i times v of xy, where we have two functions instead of one. So, we're going to separate out each term and use the chain rule to prove this. But before doing that, let me ask Melodie, what is chain rule? Okay. So, we have the chain rule actually written right here, and what the chain rule implies is that you can break derivatives up, so they become easier by replacing some component with another variable. So, in this case, we have the derivative x taken in terms of some component z, and then if you multiply that by the derivative of z in terms of x, then you actually get the function in term taken into derivative of z, and sometimes it's easier to break the derivative off in this manner. Exactly. So, let's take a look at here. Look at this equation round f over around x equals round f over around z, so z is another parameter that we introduced to make it easier to solve. Then we need to put round z over around x to make it equivalent. Now, looking at this equation, look at this, round z over around x is one. Why is it one? Because, z, we defined is equal to x plus iy. Right? So, if I do partial derivative of z with respect to x as you can see here, this one becomes one. Right? So, because this is one, we can ignore this and we only have this. So round f over round x is really equal to round f over around z. Using the same logic, we can apply it to round f or round y, and in this case round z over around y will be i, because it's the imaginary part. So, it is i times round f or round z. Now, knowing this, now we can play with this round u over x plus i round v over round x, this is round f over around z. Why? Because round f over round x is equal to round f over round z. So, we replaced round f over around x by round f over round z. Right? Here too, round u over around y plus i round u over around y, we replace round f over round y by i round f or round z. Now with this two equations, if I do a subtraction or an addition, what happens is, subtraction over addition, by multiplying i to this equation, because then these two becomes equivalent, then we have real parts and imaginary part and by corresponding each part to each other, we come up with these two equations. Which is the same equations as we found in our slides, just one slide before this. Okay? Now. So, thus starting with any ordinary function, right? We can arrive at two functions, U of x, y and V of x, y, which are both solutions of Laplace's equation in two dimensions. We just proved that right? We just proved that. So, each function represents a possible electrostatic potential because they fulfill Poisson's equation. We can pick any function f over zeta, which represents two electric field problems. We can write down as many solutions as possible by just making a functions and find a problem that goes with each solution. So then we will come back to this quadratic function, where we had real part x squared minus y squared, and the imaginary part 2xy. If we plot that in this graph, if you plot that in this graph, 2xy will be this one or this one, the dotted function, and x squared minus y squared will be this one, the line that is connected. Now to see what problem the function U belongs to we solve for the equipotential surface by setting U equals A. So, we can vary A from zero to positive numbers or zero to negative numbers. If you see that, you can see here A equals minus one, minus two, minus three, minus four, here A equals one, two, three, four. So if it is positive you create curves that are symmetric with respect to y, if it is minus then you create curves that is symmetric with respect to x axis, okay? The fine details of the field near the point halfway between two equal point charge will resemble the potential of A equals one, positive numbers. In that case, the electric field will be repelling like this, so you can see the dotted lines will represent the electric fields of two positive facing each other okay. We can also take a look at the field inside at a right angle corner of a conductor like this. We just figured out that some of the battery society was to make trans like this for one electrode. When they do that what happens with the opposite electrode is you have lesser concentration and electric field close to the corner of the trans, while for the entrance part you have concentration of the field. That will influence the kinetics of lithium ion transport because lithium ion is ionized and they will have driving force from electric field, as well as, concentration gradient. Now, question. Melody, outside this conductor, if I have another plate here, and I apply field, what will happen to the tip here? Last lecture we learned that the electric field will be more concentrated at the tip, and actually you can see that right here where there's a greater collection of field. Exactly. Where these reaches. Exactly that's very good. So with the knowledge we acquired from our lectures we can now apply that to different configurations, right? Good. Another applications we can think of is quadruple lenses, where these are applied for focusing particle beams in an accelerator Earth like Argonne National Lab or even in Pong where they have accelerator facilities to accelerate charged particles like electrons. In that case, we can design the lenses following the contour of x squared minus y square, and apply voltage to control the driving force that will control the left and right and up and down position of the beam that is going through this plane, right? So, the solution corresponds to a hyperbola shaped electrode near right angle corner, that's another knowledge we can use to design quadrupole lens. In this slide we're going to prove the orthogonality between two functions U and V, one is real function the other is imaginary function. The curves for V equals constant or orthogonal to the ones for U equal constant, and we're going to use the mathematical trick again to derive that the slope of each curve is really orthogonal. The way we do that is to prove dy over dx at a constant U, times dy over dx at constant V, if we multiply them is equal to minus one. Some of you might not be familiar with this concept, so I'm going to give a very simple example. So, take a look at this: If you have a y equals let's say x squared, y equal x, right? Then Melody, what is the slope here? It's one. It's one right, now I'm going to draw a line that is orthogonal to this curve. What would be the function here? Y equals negative x. Negative x, so here what is the slope? Negative one. Negative one. If I multiply those slope what do we have? Negative one. Negative one, and that's an exemplary proof where the slope of those two curves if they're orthogonal, if I multiply it's minus one. Of course, this is not a full proof, but this is just to give you some easy feel how we prove it, okay? All right. So by doing this, we're now looking at this curve, the red curves always intersect the blue curves with dotted lines at a perpendicular angle, right? That's also a hint that if one is potential, the other is electric field, why? Electric field is always the gradient of the potential, and what is gradient Melody? The gradient is the tangent vector, right. It is the vector where the potential change the fastest. If you have a topography, if you're climbing a mountain, the fastest way to go is to go orthogonal to the iso, what do we call the isobaric? Iso, the same height. So the same height line, right? Yeah. You have to go perpendicular to that line to go fastest. So that's why it's always orthogonal. So, the electric field will be orthogonal to this equipotential line which is the fastest way to change the potential, okay? Good. Now, this was one example right? The quadratic function can be applied to a corner tip or could be quadrupole lens or two positive charges facing each other. Now the other functions can represent other situations, for example, the square root of zeta function can represents the edge of a conductor in 2D. This could be applied to graphene studies or cut coaches and I 2D materials. So any 2D materials where you have edges, we have electrostatic potential similar to this function. But of course, these days people use Matlab or Mathematica or Python to use the equations inside the machine and can graph this out. But of course, if you already know some of the tricks that we learned from this lecture you can verify whether those results spit out by machine are making sense or not, right? So, more examples here. This is the laudre list of the examples, like outside our rectangular corner in 2D you can use zeta to the power of 1.5. The field for a line charge, we have already thought about a line charge with infinite length, right? But with finite length in 2D you can use log logarithmic function. In 2D analog of an electric dipole where you have two parallel lines charged with opposite polarities are very close together, we can use this one over zeta function. So, there are list of functions people have already categorized and published and if you're more interested in those functions you can search for those literature. So from here, we're going to cover some of the examples where we can use other physics plus the electrostatic potential that we just learned from the previous lecture. So the first example will be plasma oscillations. So Melody, can you tell our students what plasma is? So, plasma is another phase in matter where there are neutral atoms, ions and electrons all coexisting. Exactly, and where can we find plasma in our nature? So, there are some examples in outer space but also in the atmosphere. Yes, we have ionosphere outside the atmosphere where we have plasma created by the rays from the universe, from the space. Because of this ionosphere, we can use electromagnetic wave to go through that sphere or reflect electromagnetic wave to reach the outer parts of the earth. We are going to tackle this problem using very simple 1D case mathematics, okay? So, physical situation where the field is determined neither by fixed charged nor by charges and conducting surface but by a combination of two physical phenomena, this will be our main focus. As I told you, the plasma oscillations will be the first example. In this case, the field will be governed by the equations from electrostatics, relating electric fields to charge distribution, and an equation from another part of physics that determines the position or the motions of the charges in the presence of field will be taken into account. In our case, we will think about the distribution of charge inside the plasma. Plasma oscillations is defined by where the motion of charge is governed by Newton's law. So in addition to electrostatic equation, we're going to use to Newton's law to describe the motion of our plasma, okay? So, here, we will start with some assumptions. The first is positive ions are much more heavier than the electrons, so that we neglect the ionic motion in comparison to that of the electron. So we will think the positive ions are fixed in their position. The density of positive ions will be n_0. The plasma will be electrically neutral, so the density of the minus intensity of plus will be the same initially, and we will only think about the one-dimensional motion in x-axis. This one where delta s is describing the motion of the electron cloud in terms of expansion of shrinkage will be smaller than the initial width of the electron cloud. Okay? The translational motion will be described by s here, and when it is moving to the right side by an amount of s, then the density of electron cloud will be denoted n, and this n, the density, will be determined by the volume of this electron cloud, which is delta x plus delta s. So, that's why n will be equal to n naught delta x over delta x plus delta s, because the total amount of charge will not change, only the volume will change. And if the volume changed, there is a density change, and thus this. If I do some rearrangement, then it will be n naught over n plus delta s plus over delta x. Because of this assumption, this will be a very small number, and if this is a very small number, we can use binomial expansion and our approximation to the first order, and if you do that, it is approximated to be n naught times one minus delta s over delta x. So, we can now see that average charge density at any point, rho, will be minus n minus n naught parenthesis q_e. Why? Because we have to calculate the difference between those two densities, right? Because we know this is n is equal to n naught times one minus delta s over delta x, if I replace this here, then we have delta s over delta x only. If we put that limit to zero, then it becomes derivative, ds over dx. So this is what we get. So the density at any point here will be equal to n naught q_e times ds or dx. Okay. So, we are going to use the Maxwell's first equation, divergence of electric field is equal to charge density over epsilon naught in one dimension. If we use the divergence in one dimension, we only have x component, so it's around dx or around x equals rho over epsilon naught. From the previous slide, we solved rho as a function of n naught, q_e in around s over around x. And if we solve this partial derivative equation, then we will have this e_x equals n naught q_e over epsilon naught times s plus a constant. So, for the initial boundary condition where s equals zero, e_x equals zero, also k should be zero, right? Now, we see electric field, which is the force per unit charge is linearly dependent on the displacement s. Here, I have a question for Melody. So, can we think of this as a spring mass system? Yeah. I think so, because the equation for the force is F equals kx, right? Yes. This is the distance and here is the spring constant, so we can think of this value here as spring constant, and then we have our distance as well. That's perfect, yes. So, mathematically, this situation where we have plasma oscillations in one-dimensional is equivalent to a spring mass systems, so we can just replace the parameters one by one. So, if I know the spring equation, we can use that to our advantage and solve it for the case of plasma oscillations. So, that will be a Harmonics oscillation because spring mass system is known for being a harmonic oscillator. So equation of motion will be F equals ma, that's the Newton's law, m_e times d squared s or dt squared which is acceleration, is equal to minus n_0, q_e square over epsilon naught s, so this will be the restoring force, this will be the equation of motion. If I solve this, we will have the displacement, we'll follow the harmonic oscillator behavior, which is e to the i omega p over times t, where omega p is the plasma frequency or characteristic number of the plasma where above this, the system will be frozen, below this, it can respond to the electric field stimulus, but at the right point of this, it will diverge, so we'll have maximum amplitude. So this is omega p square equals n naught q_e square or epsilon naught m_e. So, a disturbance of a plasma will set off free oscillations of the electrons about their equilibrium positions at the natural frequency omega p, which is proportional to square root of n naught. So, you can see, the more denser electrons we have in our plasma, the higher the frequency of the resonance, and if we just perturbed the plasma at it's natural state, it will oscillate at this frequency. Okay? All right. So, here we're going to think about phenomena related to the natural resonance and what we are doing with this phenomenon of plasma frequencies. So, one example would be the radio wave communications. Let's say, we have earth here, we have ionosphere here, we have North Pole station, South Pole station, and we have a satellite communications. So, in order to communicate with satellite, we need to use higher frequencies, so the ionosphere is frozen, so higher frequencies than the plasma frequency. But if I want to reflect this and reach the station in the South Pole, than we need to use frequency smaller than or lower than plasma frequencies. Now, I'm going to ask Melody, why does the case? So, in the first case, because it's above resonance, there's going to be no movement, so the wave can just get through the atmosphere. However, below resonance, the contact with the ionosphere will cause vibrations, I think, and that will send the wave back. Yes. So to complement Melody's explanation, let's take a look at this one. This is the interface between atmosphere and ionosphere, and you can see the incoming radio wave which is the red curve, will oscillate the charge inside the ionosphere, which will emanate waves with the same frequencies symmetrically to inbound and outbound. As you can see, in inbound, you have destructive interference which will cancel out, and outbound, you see they have the same phase but different directions, so they don't interfere with each other. In this way, you can have a very good reflection. Now, so plasma oscillations can occur in this big scale like ionosphere, but also can occur in a very small-scale, like in metal. According to quantum mechanics, a harmonic oscillator with a natural frequency, omega p, has energy levels separated by the energy increment of haba omega p. So, if one shoots electrons through an aluminum foil and measures the electron energy on the other side, one find the electrons sometimes lose the energy haba omega p to the plasma oscillations. Okay? That's the quantum excitations of the plasma oscillations in the metal. So, that's a small universe inside a solid metal. Okay? So, the same physics, the same equations can be applied to different scales. Okay.