Okay, so here's another example, now the example involves neutrons. So, in the very, very first chapter, we learned about fission, nuclear fission, where we can send a slow neutron to split atoms into two part. Let me ask Melody, do you remember how many neutrons were created when you send one neutron to uranium atom? I don't actually remember, sorry. Okay, so let me remind you. It creates more than two neutrons, it's between two and three, usually 2.6. So, it means if you cut atoms by one neutron, you create more cutters. So, if this is the case, what will happen? I think that might be dangerous for a reactor, because if you create more and more of a driving force like a catalyst then the reaction will proceed faster, and then you might have a runaway reaction. Exactly, you will have a runaway reaction or in other words it will be avalanches. So, it is important to see how we can use a break like neutron absorber to make sure you have no runaway situation. So, in this case, we're not going to cover that topic. We're going to cover diffusion of neutrons, where carbon or graphite can be a very good materials where nutrients are not being scattered, so they are not being absorbed. So let's take an example of diffusion of neutrons in a material like graphite, as written here carbon doesn't absorb slow neutrons, and the question would be what is the average flow of the nutrients in graphite? So let's take a look, and if you think of this example, you may remember we have solved the equations for uniform sphere of charge, with charge density being the same everywhere or you may remember the distributed mass of earth, and if we make an elevator or escalator that connects from one part of the surface to the outer part of the surface, how the gravitational field will vary. So those kind of problems we already have solved. We can use those knowledge to solve this problem. So, let's derive the capital N of x, y, z, which is the density of the neutrons or number of neutrons in Delta v at point x, y, z. As you can see it, this will be density, because if you multiply by the volume you got the number. So this will be number per unit volume, and let's say let j be the flow vector, and j sub x its x-component, and the net number of neutrons to pass in unit time per unit area perpendicular to the x-direction will be described by the Fick's first law, where you can see J sub x is equal to minus D times round N over round x, and you know D stands for the diffusion constant. In terms of mean velocity, mean free path l between scattering, where you see the diffusivity is one-third of lv, and it's one-third, in this case because we only are considering one direction out of three. So the vector equation for J will be J equals minus D times Del N, which is gradient of the density of neutrons. So again, what does this equation mean? So, Melody, can you remind our students what this equation is telling us? It kind of looks like the equation we used before where there's a driving force, and then some sort of gradient will be equal to the flow, so I'm guessing this is sort of similar. Exactly, so the gradient of the density and putting minus there will be the driving force. So it will be the steepest downhill direction of the density where the neutron will flow, and D will be your constant thermal conductivity where it was related to the kinetics. So, both two should work in hand to make a large flow. So let's take a look at the comparison between neutron diffusion and electrostatics. So if there are sources in the volume that generate S neutrons per unit time in a unit volume, then the net flow out of Delta V will be described like this, and again if it is a steady state, we will remove this term which is related to the change of the neutron density in a given volume. If you do that, you will see that the divergence of D Del N is equal to minus S, and if you compare this with the electrostatic equation, you can replace one by one. So in this case, Melody. What does N correspond to, what is it? Also corresponds to the potential. Exactly. So, if we solve the equation for potential of charged sphere, then you can imagine or you can already know, now we can also get the neutron density as a function of the distance in this case. So indeed, these are two schematics as you can see, this is the uniform sphere of charge, and describing the potential, you will see you have quadratic function here and then you have 1 over r function. For the neutron case, we will guess it will have the same shape of curve. So we will solve what is the density of neutrons everywhere, what is the ratio of neutron density at the center to the nutrient density at the surface of the source region using this analogy. You may say, if you are creating neutrons at the same rate everywhere, the density of neutron should be uniform. That might be a naive guess, but in fact, you can already see here that's not the case. So here, we're going to first revisit the problem of electric potential inside and outside of a uniformly charged sphere, as you can see here, where density is the same and see how the potential varies as a function of distance. Before solving this, let me ask my teaching assistant, Melody, what type of strategies or methods we can use to tackle this problem? Okay. So, we're looking at essentially two different regions here, one is within this sphere and one is outside of the sphere. So within this sphere, it depends on radius, because as you expand the radius, the amount of charge inside is increasing. However, outside of the sphere, there are no additional charges added. So, it only depends on what's inside the sphere. Exactly. So like Melody said, we will rely on the fact that outside the sphere, if we make a Gaussian surface larger than the uniform sphere of charge or uniformly charged sphere, then we can think this as a point charg., then the relationship will be 1 over r, the potential, but if we use Gaussian surface smaller than this sphere, then the charge that the Gaussian surface can contain depends on the radius of the Gaussian surface. So we will have totally different r dependence. So using those conditions and method of Gauss law, we are able to find the potential outside will depend on 1 over r like a point charge, and as you can see, you will be able to get Rho a cubic over three Epsilon not, as a coefficient. For points inside as mentioned before, the electric field will be linearly increasing as a function of radius because of the reasons Melody just mentioned. So then we can integrate it to get a quadratic form of r dependence with a constant, and at the interface, at the surface because these two should have the same value, we can get the unknown constant, C, using that boundary conditions. When we do that, we find that the C is equal to Rho a squared over two Epsilon not. So if we put that back into our equation, the potential inside will have a quadratic function that is described here. So you can see you have a very smooth continuous function where the first part have a quadratic dependence, and then the second part have 1 over r dependence. So, neutral density inside and outside a source can have the same solution because the equation that governs the phenomenon it has the same form as electrostatics, so you just replace one by one. For example, the potential by nutrient density, and then you can get the same type of formula and then plot the density as a function of distance. As you can see, at the center, you have the largest neutron density. As you go to the surface, it decreases as a function of r. So, a uniform source doesn't produce a uniform density of neutrons. That's what we learned from this equation, and also, we can know the ratio of N at the center to that at the edge which is 3 over 2 which is 1.5. So you have 50 percent more nutrients in the center when compared with the surface. All right, let's take another example. So in this case is it's a rotational fluid flow, the flow past a sphere. So imagine in your bathtub, you somehow dropped a bead a glass bead, and then you wonder what would be the velocity vector around this bead. If that's really something you want to solve, then you can solve it using the Poisson equation that we learn with some assumptions. So let's consider an example of dry water which is an incompressible. Non-viscous and circulation free liquid, and we represent the flow by giving the velocity vector v of r as a function of position r and if the motion is steady, v is independent of time. So you can see the divergence of density times the velocity vector is equal to minus round rho over round t, but because the density doesn't change as a function of time, it should be zero. So in this case, you can see the divergence of velocity will be zero, and if there is no circulation of velocity like in the drain where you have a whirl-type of vector, then it will be zero. So then you can see this is really like electrostatics, and we already learned if a vector field has circulation-free condition, we can make a relationship with a potential. So, we can have a relationship of v equals minus del phi and therefore if we put it back, it becomes like a Poisson's equation with net charge density zero. So, with that in mind, let's take a look at this picture again. Imagine, instead of ball moving, the water is moving. Everything is relative. So, in this way, we can imagine this in a better simpler term. Now, as you look at this field line, what do you think? Can you think of a situation in electrostatics that is similar to that, say you have uniform electric field and you are putting some sphere inside, and do you think you can create this kind of situation? Let's ask my teaching assistant Melody. So, Melody, can we make this kind of situation using say metal sphere or dielectric sphere? I think maybe this would be the case where the dielectric constant is zero again. I'm not sure. Yeah, and that's a very good guess. So dielectric constant zero doesn't exist. So, then you may ask, what if I have metal sphere? What happens? How does the electric field change? So let me ask Melody. If I put metal sphere inside uniform electric field, what happens? Yeah. So, if you fit the metal sphere inside there, then the charges will accumulate, and they would create an electric field within the sphere that cancels out. So, if I have a metal sphere, as Melody mentioned, you cannot have electric field inside the metal sphere. So, the field that want to penetrate will be stopped here. It will not be able to penetrate further because you are creating a situation where you have a plus charge accumulated, minus charge accumulated, so you annihilate. Then you have penetrating electric field from here. Now, if I have a field from a sideway, and if I create a lateral field here, what happens? It doesn't feel happy. So, it will make sure any field that borders will only come in normal direction, and the same goes here and then same goes there and same goes there. So you will see the shape of this field is only saying that you don't have any field inside the ball that's the same. But outside you have a different boundary conditions where in this case, you only have normal component,. In that case, you only have in-plane component. So, how can we understand this? So, as Melody already mentioned, this can only be solved with unusual conditions, where you have almost zero dielectric constant or negative dipole moment, which we will discuss shortly. So, how does water flow past a sphere at rest? As we mentioned, it is convenient to describe what happens in a frame of reference fixed in the sphere. It this frame, we're asking the question, how does water flow past a sphere at rest when the flow at large distance is uniform? So we want to get a quantitative description for the velocity field that is to say an expression for the velocity at any point P. So, you see here, any point P and I want to know the velocity factor. So let's see how we can solve this. So, we are solving the Poisson's equation with boundary conditions, where there is no flow in the spherical region inside the surface of the ball and the component of v normal to the surface of a sphere must be zero at r equals a. So, the round phi over around r is always zero at the surface, and the flow is constant at large distance, meaning round phi over around z is constant, v0. So along the z direction, you have v naught, and along the r direction, you have zero at the surface. Now, there is no electrostatic case which corresponds to this problem as we just discussed. Maybe a sphere of dielectric constant zero in a uniform field will mimic this situation as Melody rightfully guessed. So, let's take a look at the situation where we have zero dielectric constant sphere in a uniform field. Without a sphere, we know that the electrostatic potential will be a function of E naught z where you can see it's constant and only a function of z. We have analyzed the case of a dielectric sphere which has a uniform polarization inside it, and the field inside such a polarized sphere is a uniform field of rho delta d over 3 epsilon naught. We used a situation where you have uniformly-charged spheres with plus charges and minus charges and displace it by delta d. The field outside is the same as a field of a point dipole located at the center. So we know the phi of dipole, phi sub dipole is equal to 1 over 4 pi epsilon naught times p.e r over r square, and that can be rewritten in this form, and we can guess the solution is a superposition of a uniform field plus the field of a dipole. So here's our guess. The phi is the superposition of this one plus this one. I'm going to revisit the dipole and an electric field inside the sphere where I just mentioned we have displacement of two uniformly-charged spheres with opposite sign of charge and because of the displacement, the same point will have the different Gaussian sphere for each sign of charge. If we use that to our advantage, we'll be able to calculate the electric field by superposition and here we've got the answer which is minus rho times delta d over 3 epsilon naught. So, continue our discussion. If r is much larger than a, which is where we have uniform field, then as we mentioned, the phi is equal to minus epsilon naught z. That's boundary condition number one. How about boundary condition number two? Since z is equal to r cosign theta in this spherical coordinate, r cosine theta will be z. We can replace z by r cosine theta, and if we do that, we come up with this equation, phi is equal to minus E sub zero r cosine theta plus p cosine theta over 4 pi epsilon naught r squared. Now, if I take a derivative of this phi with respect to r and put minus in front of it, then you can see this is the equation we get and for the boundary condition number two where it is stated the normal component of the velocity should be zero, then we have to solve this equation. Of course, when r is a for all theta, then the only answer we can get is p the dipole moment is minus 2 pi epsilon naught a cubic times E naught, meaning if we only have negative dipole moment. In that case, we will be able to solve this problem. So, a sphere moving through a dry water is similar to have potential over zero dielectric constant sphere in a uniform electric field, which we can describe the potential as in this case. The solution of the fluid problem can be written simply by replacing the parameters one by one for the solution we got for the sphere, and this will be the equation. So you will be able to plot the velocity field around the sphere moving through a dry water using the fact that this potential, if you take the gradient del and minus in front of it, will give you the velocity as a function of distance. So this could be your homework to take a look. So, we're going to take another example here, illumination, and you may ask how many more examples do we have to cover? So, I'm going to ask my teaching assistant Melody. So how many more do we have? I think we only have this one. Exactly. So this is the last one. So stay tuned and stay with us until the end of this lecture. So, this example is also relevant to design our room where we want to have uniform lighting, especially for our kids or even for ourselves. When we are studying, we don't want the variation of light intensity interfering our reading. So, but you cannot think of a ceiling covered fully with the lighting source. So you may have to have some compromise. So, we will use Fourier transform as well as the equations that we learn to see what would be the realistic design of lighting source to make uniform lighting on your desk. So, you know if we have a point source like the sun, and this is our earth surface, it's not flat, but we can approximate flat if it is a small town, then you know the intensity of the light normal to the surface is equal to S over r squared er.n. So here the S is the strength of our light source and r is the distance and this is the odd factor, n is the surface normal so you know what it means. Now, here we assume that the light sources are incoherent sources with intensity always add up linearly. So, what if my light source is coherent, what will happen? So, let's ask Melody, what will happen if I have coherent light source that has the same frequency and when they are added, what happens? They add with each other. Exactly. So, they add with each other and depending on the face, if I have two waves of the same face, then I will have a constructive interference, the amplitude of my light wave will be doubled and the intensity will be four times, because intensity is a square of the amplitude. However, if I have two wave with the same frequency coherent, but however, it is out of phase like this, then you will have zero amplitude. So, it depends on the phase of your light, if they are coherent. However, in case of incoherent as written here, you can just simply add the intensity up. We also assume that the light energy arrives at the top of an opaque surface without any sources below the surface, so you can imagine your desk is an opaque surface, I don't think you have a mirror type of desk or fancy, shiny desk when you are studying. Now the problem here, what is the widest spacing b from fluorescent tube to tube that we should use if we want to surface illumination to be uniform to say, within one part in 1,000? One part in 1,000 is like what? 0.1 percent, so it's very small. So, imagine you have a room and this is the ceiling. This is the ceiling, this is your room and you're going to put fluorescent light on top of your ceiling, on your ceiling and you want to make sure, on your desk which is just below here, the intensity of light is uniform to 0.1 percent. So how many lightings do I have to put, how many fluorescent bulb do I have to buy? That's a relevant question. So, let me ask Melody her guess. So, imagine the distance to your ceiling is like let's say, two meters from the desk which is high ceiling, then what would be the distance between the fluorescent bulb that you want to place? I want to say it would be okay if it was less than two meters, but I'm not entirely sure. Okay. So, let's see if her answer is correct. So, here is the steps that will take to find the answer. Number one, we will find the electric field from a grid of wires with a spacing b, each charged uniformly. We'll compute the vertical component of the electric field, and we will find out what b must be so that the ripples of the field are not more than one part in 1,000. So, before, we saw that the electric field of a grid of charge wires could be represented as sum of terms, each one of which gave a sinusoidal variation of the field with a period of b/n, where n is an integer. An amplitude of any one of these terms is given by F_n is equal to A_n times exponential to the power of minus 2pinz/b. We need consider only n equals one because the terms for n larger than two is much smaller than n equals one. Since we only need A_1, we can estimate that it's magnitude is roughly the same as that of the average field. Now, the exponential factor would then give us directly the relative amplitude of the variations. If we want this vector to be 1/1,000, we find that b must be 0.91z. So, we can calculate that putting in to this term that is 0.91z. What does it mean? It is the distance between the fluorescence bulb will be smaller than the distance to the ceiling, so she was right. You can have smaller than two meters, and to be exact, an exact calculation shows that A_1 is really twice the average field, so the exact answer is b equals 0.8z. So 80 percent of two meters, so it would be 1.6 meter apart. Now, let's discuss the underlying unity of nature. We have just started with the sentence that the same equations have the same solutions and we discussed how this is the case for such variety of phenomena in nature. Melody has read rightfully discussed that it comes from our approximations, our geometry, as well as, the way we solve the problem. So, in learning electrostatics, you have learned at the same time how to handle many subjects in physics, and that keeping this in mind, it is possible to learn almost all of physics in a limited number of years. That's fantastic. However, a question surely suggests at the end of such a discussion: why are the equations from different phenomena so similar? We may say, "It's the underlying unity of the nature or nature." The underlying unity might mean that everything is made of the same stuff, and therefore obeys the same equations, but we're not dealing with the same stuff as we know, electrostatics, electrostatic potential, temperature, displacement of membrane, density of particles, they're all different. Then why do we simply have this underlying unity? It comes from the fact that we're using differential equation which is just an approximation, and we use only second order type of differential equations, and we assume that the neutrons are smoothly distributed in space, for example, for density of neutrons case, which is not always true. As long as things are reasonably smooth in space, then the important things that will be involved will be the rates of change of quantities with position in space, and the derivatives must appear in the form of gradient or divergence, and because the law of physics are independent of direction, they must be expressible in vector form. Before going further on these points, you may say, "Oh, can it be true that all things are reasonably smooth in space?" So, let me take an example of like atom. You have plus, protons, neutrons and electrons, cloud, electron cloud, and then you have another atom to your neighbor. Then the electric field distribution will not be smooth, why? Why is the electric field discontinuous? So, let me ask Melody. At the center of your atoms or at the center of charge particles, what happens? In terms of the atom, there's both a proton then the neutron. Yes, or let me ask it this way, if you have electrostatic potential is phi, then you have 1/r dependence. What happens if r is 0? Then it goes to infinity. Exactly. So, at each point of point charge you have infinity divergence. So, it is extremely hard to make smooth. So, in order to make this approximation you need to scale up, you need to have a bigger scale like in a macroscopic world where quantized world looks to be continuous. So, that's one of the caution that you need to have when you use these kind of governing equations. You see, the equations of electrostatics are the simplest vector equations that one can get which involve only spatial derivative of quantities. Therefore, any other simple problem, or simplification of a complicated problem must look like electrostatics. What is common to all our problem is that they involve space and that we have imitated what is actually complicated by a simple differential equation. So, that's how we use the same kind of equations for different kinds of phenomena. So with that, we'll wrap up our 11th lecture and we'll see you again. Bye.