 Hi, I'm Zor. Welcome to a new Zor education. Today, we will talk about how to add velocities. So if you have, let's say, a person who is moving in one inertial reference frame, and then the whole reference frame moves relative to another one, how to add these two speeds to get his speed in the in the second reference frame. Now this lecture is part of the course called Relativity for All presented at Unisor.com. The same website Unisor.com contains two prerequisite courses, Mass 14s and Physics 14s. Both are definitely needed to study relativity. The whole website is totally free. There are no advertisements. Every lecture has textual notes, which basically like a textbook. So you have the advantage of seeing the lecture and reading the same material as in the textbook. There are exams, there are problem solvers, etc. You can use it at will, no restrictions, no signing in, so basically it's all open. Okay, back to adding velocities. So let's start with Galilean transformation and Galilean approach to mechanics and physics. For example, if you have, let's say, a platform which is moving relative to Earth, let's say, with a speed V, and then you have a person here who is moving along the platform in the same direction with the speed U. So what is the speed of this person relative to Earth? Well, in Galilean transformation system, they're just adding the speeds. It's very easy to derive. Now, in relativity, in special theory of relativity, it's basically wrong because the completely different transformation of co-ordinates exists. It's a Lorentz transformation rather than Galilean transformation. And in addition, I mean the most important part, the Lorentz transformation specifies that the speed of light, for example, is the same and that's the maximum possible in all systems, in all reference frames. So we will start with Lorentz transformation and we will try to derive this law of adding velocities from explicit Lorentz transformation formulas. So let me just remind that we have come up with Lorentz transformation as being, so V is the speed of beta system relative to alpha system. Alpha system is something which is inertial and beta system is also inertial and basically the approach is exactly the same. Let's consider that the beta system moves with certain speed V relative to alpha system. And then there is an object whatever, which is moving with speed U beta relative to beta system. Our task is to find what is the speed relative to alpha system of the same object. Same as this previous example with a platform and the person, so U beta would be the speed of the person relative to platform and U alpha means speed relative to the Earth's, okay? So that's how we have our problem stated and let me just write the second equation and I'm only talking about movement relative to the x-axis, just to simplify the number of variables. So no y's, no z's, only x and time t. Alright, so for x it will be x plus V times t divided by the same square root of 1 minus v squared divided by c squared. So c is the speed of light. V is the speed of beta platform relative to alpha platform and U beta is the speed of the person or whatever object within the beta system. U alpha is the speed relative to alpha system. Okay, so lowercase x, lowercase t are coordinates of the space along the x-axis and time in the beta system and capital T and x are in the alpha system. Okay, so far so good. I have defined everything whatever we need it. Now, what do we need basically? We need U alpha, which is dx of times t by dt, right? That's what we need. The speed of the object that's x-coordinate of the object relative to alpha system. That's why we differentiate by capital T. What is U beta? U beta is differentiate of lowercase x of lowercase t by lowercase t. So this is what we have. This is what we have to get, right? Okay, so from here on it's just pure mass and very easy mass. Just don't forget such a very simple thing. First, we can say that differentiate x by dt by d capital T is sorry. Yeah, we can say dt is equal to differentiate by d capital T times dt by dt. This is the composite function. x as function of lowercase t can be basically considered as function of capital T and then T as capital of as a function of lowercase t. From here, we solve it for this guy dx by dt is equal to dx by dt divided by dt by dt. Now, why is it easy? Because what I will do is I will just substitute this and this into this formula and to find out what's x as function of lowercase t. All I have to do is basically substitute x. What is x? As a function of lowercase t. If my platform is moving with a constant speed v, then this is v times t, right? Well, let's consider that at the moment t is equal to 0, we are at the beginning of coordinates. I forgot to tell it, but it's kind of natural condition. So in the beginning of the time, both beta and alpha reference frames are coinciding. And then basically beta is moving with a constant speed v relative to alpha. Then, obviously, if this is... Oh, I'm sorry. Not this one. If this is the movement of the object, then it's speed v. That's uv. Yes, obviously. So this is the law of movement of the object in the beta system. Speed times time. And obviously if we differentiate it, we will get the derivative which is equal to u beta, right? So what I will do, I'll just substitute it to here and to here. And I have x and t, capital X and capital T, as function of lowercase t time only. So that means I can both... I can calculate this and this. Alright, so let's do this trick. Again, as I was saying, it's just pure math. Calculus is more precise. So my t is equal to t plus v divided by c squared times x, which is u beta t divided by square root of 1 minus v squared c squared. My x is... Substitute this one. So instead of x, it's u beta times t plus v times t divided by the same square root. So now I don't need this one. So all I have to do is differentiate this, differentiate this by t. And these are linear functions. So there is nothing actually complicated. You just basically factor out t and whatever remains would be the multiplier, which is the derivative. So d t by d t is equal to 1 plus v divided by c squared times u beta. Beta. It's not beta. This is beta. Divided by square root and dx by dt is again t factors out. Whatever remains is u beta plus v divided by square root. So all I have to do is divide one by another to get the speed. Okay, so if I divide dx by dt by this one, well square root goes out, obviously. So my d capital x with d capital t is equal to u beta plus v divided by this thing. 1 plus v c squared u beta. And this is the final speed of the object relative to alpha frame, which is u alpha. So that's how we obtain the combined, the composite speed. So the speed local within the beta system is u beta, u beta. The speed of the whole beta system relative to alpha is b, which brings us to this formula as the speed of this object relative to alpha. Okay? Now just to check, we know that we cannot exceed the speed of light, right? What if you beta is equal to speed of light? What happens in this case? What will be the result? I mean, we know that we can't really exceed speed of light and if something is moving with the speed of light in one system, it's exactly the same speed of light in the another system. Even if these two reference frames are moving relative to each other, speed of light is constant. Well, let's just check it out. So if ub is equal to c, what will be this formula? Well, u alpha is equal to c plus v divided by 1 plus v divided by c square and multiplied by c square. So it's c, which is equal to c plus v divided by 1 plus v c, if I will multiply it by c, what I will have c times 1 plus v over c would be c plus, so this will be canceling out and I will have c. So if u beta is equal to c, u alpha is equal to c as well. So our principle of constancy of the speed of light is preserved. So basically as an exercise, you can just make sure that this particular equation, if u beta is less than speed of light, then u alpha also should be less than speed of light, but greater than u beta. So you can check these inequalities are actually held as an exercise. Well, that's it. I mean, that's basically the formula which I was trying to present to you. This is how the speed of one object in one system is added with the speed of the whole system, speed of object added with the speed of the entire system and that's what we get. By the way, if the v is very, very small and u beta is very, very small relative to c, then as you see, this would be very, very close to one and the whole thing would be very close to just the sum which is the Galilean transformation of the Galilean speed addition rule, right? So that's why, unless these speeds are really significant, and obviously until the end of the 19th century we did not have significant speeds, people just were completely satisfied with the rule where both speeds are added to each other and that was fine. That basically corresponded to experiments with the precision these experiments were conducted. But obviously with increasing our precision, working with things which are much faster than whatever was before, people did confirm that this is exactly the case. So experiments show that this is the correct formula. Okay, that's it. I suggest you to read the notes for this lecture. So you go to unizord.com, the course is called Relativity for All from the front menu and it has certain submenus. This goes to Einstein view and within this Einstein view you will find the Erich velocity lecture. That's it. Thank you very much and good luck.