 Hi, I'm Zor. Welcome to Unisor Education. I would like to just present a very, very simple problem related to energy and momentum of life. Now, this problem is part of the course called Relativity for All presented at Unisor.com. The website Unisor.com contains a couple of prerequisite courses. One is called Mass for Teens and another is Physics for Teens. Now, the Relativity for All is slightly more advanced course. Now, we were discussing all the different Relativity aspects and obviously, very important aspects were energy and momentum of objects, basically. So, this little problem is based on whatever already had been covered before about what exactly is a kinetic energy of moving object and what's the momentum. And I just would like to apply these two characteristics to light. Now, just a couple of more words about the website Unisor.com. Well, first of all, it's totally free. There are no advertisements, no strings attached. You don't really have to sign in if you don't want to do it in some kind of an organized fashion under a supervision of somebody, et cetera. So, then you will need just to enter name and password. Other than that, if you just do it yourself, you don't even have to do that. It's relatively comprehensive courses as rigorous as possible because I'm a mathematician by education and I would like everything to be very, very rigorously proven, if possible. In physics, it's not always possible. Obviously, experiment takes a lot of weight, actually, in all the physical research. But theory is very, very important and it's all based on mathematics, obviously. So, you really have to know mass before you study physics more or less seriously. The website contains more than a thousand lectures. Every lecture has notes, very detailed notes, which are basically like a textbook. So, you have a video presentation of the lecture and you have notes for this particular lecture at the same time on the screen. All right. So, I do prefer actually you to use the Unisor.com website because it contains all these notes, et cetera. The YouTube, for example, where you can find these lectures, it's not as user-friendly in this particular case. And what's important, the website basically is driven by menus, which means you have a sequence of lectures, what follows, which, et cetera, et cetera. You can always switch in YouTube or somewhere else that's not available. Okay. So, back to problems. First of all, let me just explain what exactly have been covered before and how would I like to present it in reference to light. First of all, we have already derived the expression for kinetic energy of moving object. If its rest mass is M0, then its kinetic energy is described by this formula. That's basically a previous lecture. I think previous one before that. Also, we were discussing the momentum of moving body and momentum is this. So, M0 is a rest mass in the system of coordinates, which is basically tied to this particular object where this object is at fixed position. U is speed in this particular object. Now, obviously, in case of momentum, U is a vector and P is a vector. But right now, it's not really important. Let's just consider only one dimension. So, it's moving with some kind of a constant speed, let's say, and then that would be its momentum. So, the difference between relativistic momentum, for example, and classical momentum. Classical momentum is this. Just mass times speed and this particular factor. So, one over this square root is called the Lorentz factor and it's called gamma. So, existence of this Lorentz factor is what's really very characteristic for relativistic mechanics. It's actually present in most of the formulas in some way or another. So, what's my problem right now? I would like to find out what is this ratio, kinetic energy to momentum, for light. So, what's the problem about light? Why do I really specify it as a problem? Why don't I just divide one by another? Well, the problem is that, number one, light has no rest mass. S is zero. Light rest mass is zero. That's number one, number one problem. So, it's like zero divided by zero. That's undefined. Another problem is speed of light is C. So, if U is equal to C, I have C square over C square which is one, one minus one is zero and I have to divide by zero. So, none of these formulas really have some sense in case of a light. Now, so why do I would like to calculate this one? Well, because if I will start really calculating this, it might be, you know, a little bit easier to do something. And let's just think about how can we do it. Well, first of all, mass here, here and here will just cancel each other. So, we have dealt with mass. Doesn't really depend on mass at all, this ratio. So, we can simplify our problem. Another simplification is the following. If this can be expressed in this way, K is equal to m C square times gamma which is one over the square root, right? Minus one. And P is equal to m zero U times gamma. So, if I will divide one by another, what will I have? Well, mass is definitely cancels out. But here, now U is equal to C, right? For speed of light. So, it would be C square divided by C, so it would be, now, how about this? Gamma minus one divided by gamma. Now, gamma is equal to zero, so I cannot really divide by zero. But what I can do, I can divide gamma minus one by gamma and I will have one minus one over gamma, right? Now, one over gamma is square root of one minus U square C square. So, in case if U is equal to C is just equal to zero, so this one is equal to zero. And what do I have? I have a very simple equation. K over P is equal to C. So, that basically is the formula which I wanted you to know about light. Well, usually the specification of kinetic energy is not emphasized for light because light has only kinetic energy, basically. So, and usually we use the letter E for general energy. So, E divided by P is equal to C or E divided by C equals to P. In any way, this is the formula. Now, this formula was very interesting, which we have derived from relativistic expressions. It was actually obtained at the end of the 19th century, something 1800, as a consequence of research done by the physicist by the name Pointing. So, it was actually based more on electromagnetic properties of light and Maxwell equations. So, it kind of predates, this formula predates the relativistic theory. So, from the relativistic standpoint, we still come up with the same thing. Now, why? I mean, for a very simple reason, because relativistic view is completely in accord with Maxwell equations. So, Maxwell equations internally are not really classical. It's more relativistic physics, because the classical, for example, blue land transformation, they are not invariant relative to Maxwell equation. So, Maxwell equation are a leap forward from the classical physics to relativistic physics without even knowing about this. It's very important. So, as I was saying, this is basically derived a long time ago, and right now we have come up to the same formula, but from a different aspect, from the aspect of purely relativistic standpoint. But, again, as I was saying, it was first derived from classical, well, not from classical physics, I would probably say from Maxwell equations in the theory of electromagnetic fields. That was a really very kind of a simple thing to basically explain, because everything is already here, and to get to these little arithmetic is not really difficult at all. On the other hand, you can consider this to be maybe a little bit less contradictory if instead of m0 is equal to u and u is equal to c for life, speed of light. You can consider u as a variable which tends to c. So, we're talking about some kind of an object, the speed of which is changing, and it's approaching c. Now, if it's changing and approaching c, we will just have to have a limit of this, gamma minus 1 divided by gamma, which now gamma is not equal to 0, obviously. But even in this case, it's really very simple because the limit of that thing, gamma is actually going to infinity, right? If u is equal to c, that would be 1 minus 1, so it's 0 in denominator. So, what is the limit of gamma minus 1 divided by gamma if gamma tends to infinity? Well, that's a simple mathematical thing, and obviously you can do it dividing member by member by gamma. So, it would be 1 minus 1 over gamma. Now, if gamma goes to infinity, that thing goes to 0, obviously, and you have 1 as a limit. So, basically I did exactly the same here, I just didn't really use the word limit, and that was kind of obvious that 1 over gamma goes to 0. So, that's a very simple problem, but it helped to derive a very important formula, which I did not really derive from the electromagnetic field standpoint from the Maxwell equations, from this theory which was developed by pointing. Maybe I will have to return back to physics 14's course and add this particular piece, so I will have it from there as well, but I'm not sure if I will have time for this. But in any case, we have derived the formula, it's the right one, so the energy divided by speed of light gives you the momentum. It actually combines energy and momentum for light into one very important formula. Okay, that's it for today. I suggest you to read the notes for this lecture, and take the whole course. I mean, if you found one particular lecture, it doesn't really stand by itself, because it's always using something else which was before it, and there will be other lectures which will follow it. So, it's very important for you to take the whole course to be more or less equipped with the knowledge. Okay, that's it. Thanks very much. I'll see you the next time.