 Hi, I'm Zor. Welcome to Unisor Education. The title of this lecture is Speed of Light, but actually the light would be subject of one of the very last couple of statements in this lecture. This lecture is part of the course called Physics for Teens, presented on Unisor.com. I do suggest you to watch this lecture and all other lectures of this course from the website rather than from, let's say, directly from YouTube or somewhere else, because the website is basically, it's a driver. I mean, it has a menu, it has chapters, if you wish, parts, chapters, whatever. Now this lecture is part of the chapter called Field Waves, and it's important to understand all the material presented in this chapter. So if you go to Unisor.com to Physics for Teens course, it will be the part of the course which is called Waves, and within that part there is a chapter, Field Waves. So everything called lectures. There are lectures about gradients, diversions, curls, the Maxwell equations. I mean, all this information is really necessary. You have to know all the stuff before approaching this particular lecture. So it's all in the menu, in the logical order, so any subsequent lecture is using the material from the previous, etc. Okay, so that's a preamble. And obviously, you do need to know math, in particular partial derivatives which are used in the whole, basically, chapter dedicated to Field Waves. You need to know vector algebra. I mean, it's all part of the physics which physics cannot live without, contemporary physics. Well, actually, not contemporary, only, not only contemporary. Whatever I'm talking about today and whatever I was talking about in all the lectures which belong to the Field Waves, it's as it was at the end of 19th century. The development of the 20th century, which are related to theory of relativities and quantum physics, etc., not yet happened. So James Maxwell, who basically was the author of so-called Maxwell equations described in the previous lectures, four of them. He was basically a very knowledgeable guy. He knew the physics as it was at the time, knew the results of experiments, and knew his math, by the way. Especially, again, vectors, derivatives, calculus, basically derivatives, partial derivatives, integration, etc. So, right now, we are not really moving beyond the end of 19th century, believe it or not. Okay. So, the previous lectures were dedicated to four Maxwell equations which describe electromagnetic field. So, we're talking only about electromagnetic field. Now, in this lecture, we will talk about a very, very simple electromagnetic field. Not anything like in general. Not very, very simple. Now, the very simple electromagnetic field has very simple both components, electrical components and magnetic components. So, electrical components, electric intensity E is a vector, and we will assume that this vector basically is changing in one particular direction along the x-axis. Now, the magnetic field intensity, magnetic components of the electromagnetic field intensity, that's vector B, and that will be always perpendicular to the electric intensity. So, if you remember, if we are changing the electric intensity vector value, the changing electricity produces changing magnetic field, changing magnetic field produces changing electric field. So, I assume that this is a simple, and it's changing along the x-axis, up and down, or whatever, changing. What's important is changing. And then, the B vector would be perpendicular to it, always. And the direction of the propagation of the electromagnetic oscillations is always perpendicular to both. And I assume the direction would be along the z-axis. So, I would draw it in this particular way. So, that would be my, let's say, this would be my x-axis, that would be my y, and this would be my z propagation of waves. So, x is E. So, E is, I assume this is something like a sinusoidal, but it doesn't really matter. What matters is it only changes the x-component, y and z-component are not changing of this field. Now, the perpendicular to E would be, how should I explain it? Something like this. And the, no, I think, yeah, well, something like this. Yes, this is along the x, and this is along the y. Well, I have a much better picture in the notes for this lecture. So, I do suggest you, when you will go and you will open the right side of the screen in the Unisor website, it has notes, and there is a nice and pretty picture of this. And so, the propagation of the waves would be in this particular direction. I think I probably do it slightly differently. I think I should do it something like this. Like this. Yeah, so, this is B, and this is E vector. E goes up and down. B goes along the y-axis. And the whole thing is propagating in this particular direction. So, the changing electric field produces changing magnetic. Magnetic produce changing electric, etc., etc., and that's how it's propagating. Now, so, this is how electromagnetic oscillations are actually going. Okay. Now, let's talk about Maxwell equations and mathematics of this. Now, if vector E is, has only x component, I can write it as x component. Now, what it depends on? It changing as we move along the z-axis. And at any particular point, it's changing with time. So, basically, it's a function of two arguments, time and z components, how far our electromagnetic field propagated. Now, two other components are zero. There is no component which is changing along the y direction or along the z component. So, it's not changing this way and it's not changing this way. It's changing only this way. Now, speaking about B, now, I'm talking it's only changing along the y-axis. You see, this is the B component. So, it's changing up and down within the y-axis. And obviously, it depends again on time and how far our field propagated. But it's x component is zero, it's y component, it's the y of t and z. And the z component is also zero. So, now, the four equations which we have learned about before, four Maxwell equations, contain a lot of variables, basically, right? I mean, it contains two vectors, E and B. And it contains all components of each of them. Each of them has three components. It's basically like six unknown variables. Now, in our case, we have only two variables because we have simplified the complexity of the electromagnetic field which can be produced by many different sources of moving electrons somewhere. Like, take a sound, for example. Sound has a huge number of sources of electric oscillations which produce magnetic oscillations which is electromagnetic field. And obviously, it goes to all different directions and it's a combination, it's a superposition of many, many different fields. Now, we're talking about one simplest electromagnetic field which has only one component along the x, which is electricity and only one component along the... So, now, we have two functions, basically, of two arguments. And we have four Maxwell equations which mix them together. E depends on B, B depends on E, etc. Now, we would like to solve it somehow. We are not going to solve it in a general case. Number one, only in this case and even in this case, we will restrict ourselves to one particular solution which will be proven as a solution to the Maxwell equations. So, you remember that if you have a system of equations, how can you basically solve it? Well, you have to reduce the number of equations and number of arguments, right? So, that's exactly what I will do. I will try to reduce system of whatever number of variables with whatever number of equations with single equations. And each equation will contain only one particular unknown. And if I will be able to solve this equation for this particular unknown, that would be basically a solution. So, I need an equation for E and equations for B separately, not together as they are in the Maxwell equation. So, I would like to do this mathematics. It's not a trick actually, but it's some calculations are involved. So, just follow me. So, first of all, what we will do, we will recall the third Maxwell equation which is nabla E is equal to minus dB or dt. Okay. So, if you don't remember this Maxwell equation, please go to the earlier lecture in the same chapter which I have. It's called third Maxwell equation where this is basically explained. Well, I do suggest again to know very well the whole chapter which is called field waves with all these nabla, gradient, curl, diversion, etc., etc. So, this is all must, otherwise you will not understand it. But if you do know that, this is a simple stuff really. So, just do it slowly, thoroughly and try to go through all field waves, lectures so that you understand all the material. Our purpose is not really to like force this knowledge upon you. I would like you to understand how it's not really very complicated if you go with methodically and step by step. So, don't jump. This is the third equation. I will take it as granted because it's already being discussed. Now, I will apply this to my concrete vectors. But I would like to really open this up, nabla vector product with a vector. Let me just remind you what is a vector product of operator nabla. Nabla is basically a pseudo vector. It's d by dx, d by dy, d by dz. It's three operators of differentiating by partial differentiating by x, by y, and by z combined into a pseudo vector and multiplication in this particular case. It's a vector product. It's a cross product. I can really component by component actually convert it into something which is like normal operator and normal algebraic operations. And here is how. If you have, let me just put it somewhere here, if you have a vector, you know what, just for savings of real estate, I will put it this way, dy zero. And here I will put my general nabla some vector d, which has three components, vx, vy, and vz. In general, this is the following. It's dz component by y minus dy component by dz. And this is my component of the x-axis. Plus, component of the y-axis would be dx by dz, d dx by dz minus dz dz by dxj. This is my i, j, and k are unit vectors along the x, y, and z-axis. Axis and vx, vy, and vz are components of my vector v. And the third component would be dy, dy by dx minus dvx by dy, and k. Now, if this looks like Chinese for you or something like that, again, go back to one of the previous lectures where it's all explained in the details. So, I'm just taking this as granted. Okay, so this is general for vector v having vx, vy, and vz components. And the vector actually v is, in this particular case, it's a general vector, which depends on t, x, y, and z. It's a vector field, which means for each space coordinates, it has a value, and it's changing in time. So that's what vector field actually is, a changing in time vector field. Okay, now, I will use this big formula for our e and our v. Now, ex is the only non-zero component of the vector e. So if I need to multiply pseudo-vector nabla by vector e, I have to really use this formula, but have in mind that vy and vz are equal to zero in this case. Ex is the only thing which is not equal to zero. So, basically, I have this thing where dx is actually participating and this one. So if e is substituted instead of v here, then this components are zero. Now, this component still has the x component, right, vx, but it's partially differentiating by y, but we know that ex doesn't really depend on y. It depends on t and z time and the direction of propagation of the electromagnetic oscillations. The vector e does not change along the y axis. So this is also equal to zero. So the only thing which I have is this one. So this thing on the left is equal to very simple, this d ex component of tz by dz. That's what nabla times e in this simple case and it's equal to minus d again. Now, v is the vector which has only y component. So, yeah, I forgot to put j here. This is my unit vector along the y axis, right. Now, in case of db by dt, b also has only one component and it happened to be the component along the y axis. So I can convert this into dby of tz by dt along the same y component. So this is a vector, but since the vector has only one coordinate not equal to zero, which is y coordinate, then the, okay, here it is. b is equal to bx times i plus by times j plus bz times k, right. That's how my vector is represented along the coordinates. These are unit vectors and bx, by and bz are components, x, y and z components of the vector. But this is equal to zero and this is equal to zero, right. So I have to differentiate, when I differentiate it by time, I differentiate only this by time. Now, this is a constant vector and this is the constant which is changing, right. So that's why I put this. Constant goes outside of the differentiation. Again, it's all simple if you know the basic manipulations with vectors and derivatives. Now, I have a very simple equation because right now what I can do, I can just do this. Now it's not a vector, it's a scalar equation. Well, it's relatively simple, fine. That's one equation. Now let's consider the second equation, the fourth. The fourth maximal equation says the following. You know what, I will write this equation here again for the sake of real estate. So I have dx of tz by dz equals minus dby of tz by dt. And I will have this on three. Okay, now I will consider the fourth maximal equation which says the following. Again, nobler times b vector is equal to mu epsilon dE by dt plus mu sigma p. Okay, it's a little bit more difficult, a little bit more complex than the previous one. And here is the reason for complexity. First of all, there are two different multipliers. These are constants actually. Mu is magnetic permeability of the media where electromagnetic field is propagating. Absolutely, it's electric permittivity of this media. Now it's different, it's constants. For every part of media, for vacuum, for glass, for metal, we have some constants which have been measured and known. These are constants. Now mu is the same as this one, magnetic. Sigma is conductivity of the medium. So it's basically in reverse of the resistivity. So we all know about the resistance, but this is inverse conductivity. Again, this was explained in details. And this particular member, it was added by Maxwell on the top of whatever. This is part of the Emperor's law, but Maxwell had to add this component to make the whole thing play together in these four Maxwell equations. Now, that we are talking about the medium, we are talking about vacuum. Again, we are simplifying. We are not talking about propagation of electromagnetic field in metals or somewhere else. We are talking about vacuum. We are simplifying the story. Because again, my purpose is for you to understand the mechanics, the basic mechanics of this. Obviously we can make it more complicated, but this would be more, I would say, quantitative complication rather than qualitative. So let's just simplify again our picture. And this is vacuum. Vacuum has no electrical conductivity. So this goes down. There is no such member in our case. Since we are talking about vacuum, and vacuum has special, I will put zero here. These are known constants of magnetic permeability and electric permittivity of vacuum. And then it looks more or less like this one, right? Okay. Now I will do exactly the same. I will simplify this, I will convert this into something similar to this. How should I do it? Again, nabla times vector v. Vector v has only y component. So in this general description of nabla times vector v, I should use only the y component. So this component stays, right? So I will have minus dby of tz by dz minus. That stays times i. Where is y? Okay, here is again. But now here, similar to the previous case, my by component, it does not depend on x. It depends only on time and the space coordinate z. So this is also equal to zero. So there is nothing else. So on the left, I have only this component. And on the right, I have equals mu zero epsilon zero. Now dE by dt. Same thing as before. Since E has only one non-zero coordinate along the x-axis, I can put very similar to the previous case. G E x of tz by dt times unit vector along the x-axis. Same thing because E is equal to E x times i vector plus E y times j plus E z times k. But these are equal to zero. So I have only one E x times i. And differentiating would give me this. Now let me wipe out this. I don't need my original Maxwell equations. Let me wipe out i and i. And I have only these two kind of lookalike equations. Now what's the problem? Because it's a system of two equations with two unknown functions. By the way, it's not unknown variables. It's unknown functions because E x and d y are both unknown functions of two arguments t and z. And it's a system. And these functions are mixed together. So I would like to do, I would like to separate them into separate equations somehow. Well, let's just talk about E components, the electrical components. Well, how can I do that? Well, actually it's very simple. I will differentiate this by z and this by z. So what happens? And this by t. Yeah, this by z, z per dz. And this d by dt. Let's see what happens. Let's talk about this particular component and this particular component. If I would differentiate this by z, I will have minus dby of tz by dt dz. And I could put two. Right? It's a second derivative, partial derivative. The first derivative was by dt. And now I have a second derivative by dz. So that would be it. What will be here? First was dz, again minus db of tz by dz. And the second one by dt dt. And again, from the theory of partial differential equations, we know that the order of differentiation for most smooth functions is non-important. As long as they're differentiable, the order of differentiation dz by dt or dt by dt and then by dt is the same. So these are equal. Which means this will be equal to this one. And that will be my equation which contains only e. So this would be d2 ex of tz by dz and dz. So it's dz square. That would be on the left. And the right would be nu 0 epsilon 0 again d2 ex of tz by dt and dt. So it's dt square, second derivative. And this is the equation which I am looking for. It contains only one function. It's a differential equation with one function. And don't forget that all of this was done at the end of the 19th century by, in this case, by James Maxwell. So this is the equation which describes the function, only one function of two arguments, time and position. And I can basically solve it. Now, I'm not going to solve it in general. However, I would like to point out that this equation, which is basically, by the way, called wave equation, it describes many different sinusoidal like oscillations. And here I would like to remind you a particular sinusoidal oscillation, which we were already considering before oscillation of the rope whenever you are moving it up and down at one end, and the waves are going all the way. It was all described in the same, by the way, chapter filled, not filled, the oscillations of waves, yes, in the same chapter waves. And if you will take a look at this, you will find the following solution I came up with. It's a times sine of omega times t minus z over b. Okay, now z is a space coordinate whenever we are make this wave oscillations along the z axis. t is time, v is the speed of propagation. Now, if you will take a look at this, if you have z, if the source of oscillations is at z is equal to zero, okay, this is my z axis. If this is the source of oscillation and I'm moving the rope up and down, okay, so z divided by v is basically a time delay. How much time is needed for a wave to reach the point z if the speed of propagation is v. So if there is no such thing, it describes that z is equal to zero, right, and this is a simple sign. This is amplitude and this is the frequency basically. This is the position, the height, if you wish, of moving my end at point z is equal to zero. And it's basically the same, but it was a time delay and that's why I have introduced this. So the derivation of this formula is really trivial and again it's one of the previous lectures. So I will use it and I will check if this is a solution. So instead of solving this equation, I will just check if simple oscillations, as we already kind of researched, would be a solution. Obviously the real course of physics in universities, etc, will contain a general solution, etc, etc. That's not the purpose. My purpose is to prepare you for understanding of that more complex things and why I'm doing this exactly for this simple case. So instead of solving this differential equation, I'll just check that something which seems to be most natural is indeed a solution. Let's try it. So I need two derivatives by z and two derivatives by t. Okay, by z. First derivative, d by dz. Okay, let's call it e of tz because I'm going to use it for this e. So e of tz is equal to the first derivative. So a goes as it is, sin goes into cosine of omega t minus z over v times derivative of the inner function. We are doing it by z. So it's minus omega over v. Omega over v. Right? That's inner function. Omega t would not be dependent on z and omega z over v would give me this. Okay, the second derivative, d2 e of tz by dz square is equal to, okay, it would be, again, a. It would be minus sin of omega t minus z over v times inner function minus omega over v and this omega over v. So plus and minus minus square omega square over v square. Right? And minus we will put in front. So a is amplitude, omega is frequency, v is speed. So that's this part. Okay, now let's talk about this part. Second derivative by this. Okay, so the first derivative, d e of tz by dt square equals, okay, cosine would be minus. So it's minus here a minus sin of omega t minus z over v times the derivative of inner function, which is just omega. The second derivative would be e of tz dt square equals, okay, cosine would be minus. So it's minus here a sin of omega t minus 0 over v omega and one more omega square. And this is supposed to be equal to this times mu 0 epsilon 0, mu 0 epsilon 0. So this is equal to this. Is it possible? Of course it is possible. The only thing we need is that this, so if 1 over v square is equal to mu 0 epsilon 0, my equation would be satisfied. So we know mu 0 and epsilon 0 and James Maxwell knew it at the time. So he can always calculate what is v. v is equal 1 over square root of mu 0 epsilon 0, right? From here. So he did this calculation. And what did he receive? He received that this speed is about 300,000 million, billion meters per second. Approximately. The exact number is in my notes. It's 299, blah, blah, blah. And lo and behold, this is exactly the same speed which was measured when physicists were measuring speed of light. And that was a kind of a wow moment for James Maxwell. At this moment he suggested that maybe the light is electromagnetic oscillations. And that was his, I would say, extremely useful gas and suggestion, because it was obviously true. It was numerous number of times. It was supported by experiments, etc. So that is the result of all these calculations. And that's why I called this lecture speed of light. Because this is exactly how speed of light was found theoretically. And it was, speed of electromagnetic oscillations was found theoretically. And it was absolutely the same. I mean, to their level, 19th century level of precision, it was the same as the speed of light which they have measured independently. And that was basically the most important kind of result of this. They understood the nature of the light. And that's the end of this lecture. And I think it's the end of this chapter of this course of physics. This is what I wanted to end up this field waves with. Again, there are other solutions to these differential equations. But what's important is that the most simple one really is part of the solution. And I do recommend you to read the notes for this lecture. And there is a nice pretty picture, actually, how these waves are propagating in space. And don't forget that this is the simplest way of simplest kind of electromagnetic field. Obviously, the real theory is much more complicated. But I just wanted to prepare you for something which you might or might not learn in universities. At least you know that this is, well, it's not a proof, but it's a very, very solid assumption that the light is the oscillations of electromagnetic field. With this, thank you very much and good luck.