 Hi, I'm Zor. Welcome to Nezor Education. The previous lecture on the item which I called Einstein View was about adding velocities. So basically we had one frame and then another frame which is moving along x-axis of the first frame. And there is a velocity of the whole frame and there is an object which is moving in the same direction along the axis with some speed view. And I have derived the formula of speed of this object in this main system, so to speak. I call it alpha. Now what's important here is that the whole beta system is moving along x-axis and the object itself is moving along x-axis. And that's how the law of addition of speeds is basically derived here, relative in relativity, theory of relativity. Now, what if the object moves in some other direction, not exactly along the x-axis? Well, obviously we can split the speed into three components. Component along x, along y, and along z, and do the calculations separately. And that's what this lecture is about. I would like to talk about directions along the y-axis and along the x-axis, how these speed additions actually look like in formulas. So that's what this lecture is all about today. Now, this lecture is part of the whole course, actually, which is called Relativity for All. And that's why this particular item I called Einstein's View is, before that was Galilean View, then there are some other items, etc. So it's the whole course. It's presented on unizord.com, totally free website, no advertisement, no strings attached. And what's important, the same website contains two prerequisites, I would say, courses. Mass for teens and physics for teens, basically directed to the advanced high school level, maybe a little bit more than that. Now, the relativity is probably living more advanced course. I still try to do it as understandable for everybody who is familiar with mass and physics of high school course. So don't hesitate to take this course because it's nothing special, basically. Okay, what else? Well, the prerequisite courses are, they contain lots of problems to solve, exercises. What's important is that every lecture, it's not only the video lecture, as you might actually catch on YouTube or somewhere else, but also there is a full textual part you will see on the same screen. This is the video and this is the textual part. Textual part is basically like a textbook. So it's not like notes, some very short notes. No, it's a full explanation which is as good as the, or maybe even better than whatever I'm talking about during the lecture. I might forget something. So basically I do recommend you to use the website Unisor.com for both the video and the textual part. And I don't know what's better to read first and then which lecture, which lecture and then read first depends on you. But if you have something like a problem solving lecture, it's better to read it first without solutions, which is sometimes presented, sometimes not. And try yourself all these problems whatever is presented and only then watch the lecture where I usually present the solution. Okay, back to theory of relativity. So let's just assume as before that we have an alpha frame that is supposed to be alpha. So we have alpha frame and we have a beta frame which is moving along the capital X, capital Y, capital Z. Lowercase x, lowercase y, lowercase z. So it moves the beta frame is moving with the speed V along the x-axis of the alpha frame. Now at the moment time equals zero, which is the same for both systems. We have them completely coinciding. But generally speaking the beta frame is moving retaining the parallelism of the axis. So it's very very natural kind of arrangement of two frames moving relative to one against another. First they coincide at moment t0, but basically speaking it moves with a constant uniform speed V along the x-axis. Now based on that we have already derived so-called Lorentz transformation of coordinates where speed of light is involved etc. So if you missed that particular lecture it's one of the previous within the same Einstein's view item in the course. So I assume that Lorentz transformation is a known thing and I will definitely use it. Now I assume we have some kind of a point object which is moving with certain speed and direction uniformly moving. Let's just simplify our job uniformly moving along some direction in beta system. The beta system is the one which has lower case coordinates and it's moving relative to the alpha system with the speed V. So my question is how the speeds of this particular object in the beta system, if we know them, how they will transform if I view the same object from the alpha system. Obviously using Lorentz transformation of coordinates. So it's a very simple exercise, the only thing is it involves just very accurate writing if you wish and some derivative basically. So the task is obvious. I know the movement of this thing in the beta system and the speed, vector of speed in the beta system looks like u, beta, x, u, beta, y and u, beta, z components, constant components because this is a uniform movement. So this is the vector of movement x, y and z coordinates in the beta system. What I do need, I need to find out alpha components of the same vector, question mark. That's my task. And again, if it's only movement along the x axis, so if u, beta, y and u, beta, z are equal to zero, in the lecture which precedes this in the same Einstein's new topic, I have derived the value of u, alpha, x. And you can basically go there and find it out. But I will probably repeat it very briefly here as well. So now I'm in a more general case when these are not equal to zero. So the question is, what should I do? Well, the answer is pretty simple and it's exactly the same approach as I used in the lecture about only x coordinate not equal to zero. So first of all, we have to write the equations of motion in the beta system. Now, I think I have to assume that the position of, again, for simplicity, the position of this object at time zero is in the origin of coordinate. That just makes my job easier, but it doesn't really make any kind of simplification. So what is my movement of the object in the beta system? What are x, y and z coordinates? Well, if I know that it's a uniform movement with components of the speed, constant components of the speed are these. So basically this is u, v, x times t, u, v, y times t, and u, beta, z times t. That's what uniform movement is, constant speed times the time, and that gives me the coordinates. Great. Now, let's go back to Lorentz transformation of coordinates. So I will transform x, y and z from beta system to alpha system knowing the formulas for Lorentz transformation, which I'm going to write here. So I will write first the basic formulas, and obviously I would like to use my previous c plus v square c square. So this is how time is transformed from lowercase t to capital T, from beta system to alpha system. Now the x coordinate, this is capital X, which is alpha x, is x plus vc divided by square root minus square c square. So the denominator is the same everywhere. Y is equal to y and z is equal to z. Since my movement of the beta system is along x axis, y and z coordinates are exactly the same of any point in beta system as in alpha system. But time and x coordinates are obviously changing, and these are the formulas. Now, using this, well let me just substitute this to x, y and z here. And I will have basically the representation of x, y and z and time t of alpha coordinates as functions of only one parameter, the time, because the speed is constant. So the same thing would look like t is equal to t plus v. v is the speed of the beta system relative to alpha system along x axis times x coordinate, which is u beta x times t divided by this square root. Now x would be equal to x, instead of x now I can put explicit u beta x times t plus vt divided by the same square root. Now y is equal to y, which is u beta yt and z is equal to u beta zt. So these are positions x, y and z and the time t expressed in terms of lowercase t time in beta coordinates. Now, what should I actually find out? I need to know the components of the speeds in the alpha system. So I have to find basically u alpha x and what is basically the speed? Speed is the first derivative of coordinate by time, which is supposed to be dx divided by dt, right? That's the derivative of position by time and all of them are in alpha system. That's why this is alpha component of the x. So this is x component, y component would be dy by dt again, all capital and z component would be dz for dt. So that's what I have to find out. How? Very, very simply, again using exactly the same method, though just in the lecture about x coagulate only without y and z. So what I did was, I was using simple formula. It's a compound derivation. If you have a compound function, let's say square root of 2 plus x cube, how do you find its derivation? First, it's supposed to be this one as some kind of a y. So it's square root of y. First, you have to derive by y, which is derivative of square root as I think is 1 over 2 square roots. Maybe it was a minus sign. And then derive multiplied by derivation of y by x, which is derivation of derivative of 2 plus x square is 2x. So that's how you do it. This is a compound kind of a function. This is exactly the same thing. So if x is a function of t and capital T and capital T is function of lower case t, the derivation by lower case t is derivation by capital T and then derivation of t by t. From which I can have this one as dx divided by dt is equal to this divided by this, right? So dx by dt divided by dt by dt. Both of them are derivative by lower case t. And we can very easily have it because all of these functions are linear function of lower case t. Okay, so that's why we can just put very simply dx by dt is equal to... Okay, derivative of capital X by lower case t is this u beta x plus. It's linear function by t. Derivative of this divided by square root of 1 minus v square minus v square. And this is what we have already obtained in the lecture when only x component was involved. So this is a known formula that was a lecture, preceded lecture to this one. Now dy by dt is equal to... The derivative of this by dt is just u beta y divided by dt. Wait a moment, I think I forgot to divide. Okay, I forgot to divide. Yes, of course. Sorry. Yes, I did obtain this formula but I did it incorrectly. You have to have this one, which is basically what I just said, divided by derivative of this one. So the square root will cancel out. So it will have u beta x plus v divided by numerator. Sorry, which is 1 plus v times u beta x. By the way, this is supposed to be divided by c square, right? Yeah, of course, I just missed it. Okay, divided by c square. That's the formula which was derived in the previous lecture. Sorry about that. Now this one, again dx per dt is this one divided by this square. Okay, the square root will always cancel as far as I understand. So you will have this one divided by this. The square root will always cancel out when we divide one derivative by another. Now dz per dt would be, again, dz by lowercase t, which is u beta z, divided by, I think I'm right. So let me check again. dx by dt is supposed to be dx by dt, which is u b x plus v divided by square root. This is a constant. It's not dependent on t. So t is only the numerator, so it will be some of this divided by square. Divided by derivative of capital T by lowercase t. So square root also will be here, and here it will be cancelled, and u one plus this divided. Yes, so that's correct. Now, y is different. y doesn't have this square root, so there is nothing to cancel. So we will have this in the numerator, which is this, and then the whole, both numerator and denominator from t. The numerator will be in the denominator of this, which is derivative by t with the one plus v divided by square. And the denominator of this will go to numerator. It will be turned around because it's a division, and you will have here, and the similar thing with this. So these are the formulas of transformation of coordinates from beta system to alpha system. For any object which is moving with constant speed with all three components, x, y, and z. This one is for x, it's special, because it's exactly the same direction as beta system is moving. These are look-alike, basically, and these are transforming the y and z components of the beta system into y and z components of alpha frame, alpha system. So it's not easy. I mean, the formulas don't really look easy. By the way, in Galilean system, this would only be for x, because in the Galilean system you can probably consider speed of light to be infinite, and then these formulas will be converted into Galilean. If c is infinite, the whole thing is disappearing, this is disappearing, this is disappearing, and you will have that the x component is the sum of x component of the object and the beta system plus speed of the beta system itself relatively to i, and y and z components would be exactly the same, because this would be zero, so it's just one, and this would be zero, so it's just one, so you will have only. So the y and z speed will be the same, and the x component will be just the addition of the speed. That's the Galilean transformation. That's the Galilean principle of adding the velocities. But in the relativity, in the theory of relativity, it's much more complex because the c is not infinite. Speed of light is the maximum, possibly achievable, and to basically accommodate this property of the universe, the formulas are getting a little bit more complex. However, there is one simple case which I would like to point out. What if the object is moving in the beta system only along the y axis? So u beta x is equal to zero and u beta z is equal to zero. What happens then? Well, u beta z is equal to zero, then this is zero, and what's important is that dx would be g. In this case, it would be equal to only v. There is no x component moving only along the y axis. Now, what would be dy v dt? That's important. That would be u beta x would be equal to zero, and this component disappears and only the numerator, u beta y times square root of 1 minus y square, that's important. This multiplier is important, and dz is equal to zero. So if the movement is only along the y axis, then the coordinate of the x would be only related to the movement of an entire beta system, which is moving along the x system. Now, y direction would have this particular multiplier as a correction factor. So the speed will be smaller in alpha than it is in beta along the y axis. So along the beta y axis it's only u beta y, but along the alpha system, the coordinates would be slightly less. And the greater the speed of object, sorry, the greater the speed of beta system relative to the alpha system, the greater to zero will be this particular expression. So the perceived by alpha observer speed along the y axis of the object would be diminished with faster moving beta system. The faster it moves out, the smaller seems to be from the alpha observer the speed along the y axis. That's very important. Now, if vice versa, the z component is not equal to zero, only z and all others, it would be basically the same thing. This would be zero and this would be u beta z multiplied by the same multiplier. So that's how speed is transformed from one frame to another. Obviously frames are considered to be inertial and the beta is moving relative to alpha. So from the beta speed we are converting to alpha speed. And these are formulas, this one which was before the previous lecture, but it's the same methodology that's why I decided to put it in. Okay, that's it for today. I never remember these formulas to tell you the truth. Whenever I need it I just go to either my lecture or any article, but anyway it's important to understand what actually is going on and these particular cases when only one code in it is used along the movement. So these formulas are much simpler for y and for z and we might actually use it for some other purposes. Okay, I do suggest if you didn't read the textbook for this because it doesn't contain some small errors which I make on the way. Other than that, good luck. Thank you.