 Hi, I'm Zor, welcome to Unisor Education. Today we will talk about momentum, relativistic momentum. We are talking about theory of relativity, right? So relativistic momentum. Well, I think I have to spend some time just discussing purely philosophical issue about momentum. Now, you remember the second Newton's law. Force is equal to mass times acceleration, right? Now, what's acceleration? Acceleration is basically the rate of changing the speed, or mathematically speaking it's the first derivative of speed by time. So it's m times dv. I'll use the letter u just in case, because I'll do something else. By dt, by time. So u is speed, a is acceleration. Well, m is a constant in Newtonian mechanics, right? So I can put dm times u by dt. I put constant into the differential, no problem here. Now, what is this? Mass times the speed. Well, basically this is definition of something which is called momentum in Newtonian mechanics. And I'll use the same term, momentum. So let's talk about what is actual momentum. It looks like force is equal to dv by dt. Rate of change of momentum, right? Mu. So it means that the faster momentum is changing, the stronger the force, right? Or it can be presented in slightly different term. The larger momentum, the larger change of momentum requires the larger force. Now, obviously looking at this particular formula, you see that the mass and speed are linearly contributing to momentum in Newtonian mechanics, right? So it means that if I will increase mass by a factor of 2, my momentum increases by the factor of 2. Which means the strength needed to change momentum twice as fast needs to be twice as strong. So the momentum is well interpreted as something like a measure of resistance of the body, of the moving body to change its movement. So the larger the momentum, for example, the larger the mass, the more difficult it is to move this mass from its trajectory. Same thing with speed. The faster the object is moving, more difficult it is to divert this object from its trajectory. So that's just general consideration about momentum. Now, another consideration, and it was in the previous lecture about Neuter's theorem. The conservation of momentum is a fundamental property of our space. The uniformity of space is sufficient to prove that momentum is supposed to be preserved. Similarly, the uniformity of time was sufficient to prove the conservation of energy. And uniformity of direction inside the space was a sufficient condition to prove the angular momentum conservation. So that was in the previous lecture. So in any case, right now we're talking about momentum, and momentum is supposed to be conserved if there are no outside influences. So let's say we have certain closed system, there are no other object but these couple, and they interact somehow. So before interaction and after interaction momentum is supposed to be the same. That's what conservation of momentum is. Okay. Now, let's switch to relativity. And I would like momentum to be conserved in theorem of relativity. Now, what's the difference between theorem of relativity and Newtonian mechanics as far as momentum is concerned? In Newtonian mechanics, momentum is this. So what I will try to do is I will try to find out if momentum is preserved, if I will calculate momentum using this particular formula. It will appear that it does not, which means I have to change the formula. And then I will suggest you how to change the formula to retain the conservation, the law of conservation of momentum. So that's the plan for today. Now, what I'm going to do is I'm going to talk about certain experiment, one particular experiment, and demonstrate using this experiment that this formula is not really good for definition of the momentum in relativistic mechanics. Okay, so let's do it. Here is the experiment which I would like to conduct. And then I will use the another reference frame, because the momentum is supposed to be preserved in any reference, if it's preserved in one reference frame, it's supposed to be preserved in another reference frame, right? Of course, the frames are in actual. So what I'm going to do is the following. Let's say I have two spherical objects, this one and this one. Now, this one goes this way, this one goes this way. That's where they are, touch each other. And I assume that this is the horizontal line and this is the x-axis. And this will be my y-axis. So I'm talking about two dimension, not three dimensional space, okay? Easier this way. By the way, if I will use only one dimension, I will not be able to demonstrate what I want to demonstrate. But in two dimension, I can. Okay, so what happens then? I assume this is an ideal elastic collision, which means it will go this way and this will go this way. So if I assume that the speed is the same by magnitude, but different in direction, opposite in direction. Now, if it's ideal collision, then I will have this u as a minus and this would be, you know what? I will probably better use coordinates. So if this coordinates would be increasing, so it's vw, this is the x coordinate of the speed and this is the y coordinate of the speed. Both are positive because x is positive and y is positive. Now, the coordinate of this one would be minus v, minus w. That's the vector of speed of velocity actually of this object. Now, this object will have positive v, negative w and this would have negative v, positive w, right? So these are my vectors of velocity. So before we have vw, minus vw, after the collision, this one, let's call this one green and this one will be blue. So the green one will go this way and this way, the blue one will go this way and that way. Okay, so this is my experiment and let's call this frame a beta frame. Now, is the momentum conserved in this particular case? Well, before it was mass times this vector and this one was mass times this vector. If I assume that the masses are the same, these will nullify each other and the total momentum would be zero, right? If you have two balls going into each other with the same mass and directly opposite velocities equal by absolute value, obviously the total momentum is equal to zero. So momentum is basically a vector, right? So the x component of this vector in this particular case would be what? m times v plus m times minus v. m times v, m times minus v. That would be my px and my py would be mw plus m minus w, zero, zero. This was before collision. Now, what's after the collision? After the collision, I will have a very similar situation. I would have px equals to mv plus m minus v, right? And my py would be m minus w. That's the green one plus mw. That's the blue one. And this is obviously zero as well. Before zero, after zero, both x and y components are equal to zero. Everything is great. Fine. Let's wait. So what's my next step? Well, I know that classic definition of the momentum works in beta frame. Now, what's important about theory of relativity? Always we are trying to approach the same thing from another inertial frame and see if the same laws will be held. I mean, that was actually the original idea from which we have derived that the time is supposed to be diluted, et cetera. I mean, all these theory of relativity properties which we have derived, we have derived because we are comparing how the system behaves or how the parameters of system behaves in different inertial systems. Okay, so this is great for this particular inertial system. Now, let's assume our beta system is moving relative to alpha frame. Okay. Now, if I have alpha frame and then I have beta frame moving, let's do it in a simple case. It's moving along x-axis with, let's say, speed s. So beta is moving along alpha x-axis with speed s. Now, if there is some object which is moving with some speed, let's say, dw, these are two components. And I basically ignored the z component. Now, then there is something which we can calculate what would be the speed or components of the velocity of the subject in the alpha frame. So that was a subject of one of the lectures when I was discussing Einstein's view on theory of relativity, how velocities are added to each other if you have velocity in one system and that system is moving relative to another system. Okay, so I will just write down the formula which were straight from that lecture. And all I can say that if these formulas looked unfamiliar, I do suggest you to go through that lecture. Okay, so u alpha x is equal to s plus u beta x divided by one plus s times u b x divided by c square. And u alpha y divided by square root of one minus s. Okay, these are two formulas. Now, obviously, x and y components are components of velocity. Alpha and beta means this is speed in beta frame, this is speed in alpha frame. So u beta x is the x component of velocity in the beta frame and this is x component velocity in alpha frame. Same thing, y. This is y component of velocity in beta frame and this is y component is in alpha frame. Pay attention, this is component x component. So the x component depends on an x component. y component depends on both y and x component. Why? Because the whole thing is moving along the x axis. x axis is a special direction and that's why it participates here. Now, again, these formulas were derived in the lecture about adding the velocities. There are two lectures actually, adding x velocity only where I derive this formula and then adding y and z velocities when I derive this one and similar for z. So both lectures, I do suggest you to refresh them. Otherwise, just use these formulas as given, basically, right? So I have these formulas. Now, using these formulas, I would like to calculate the components of velocities of all these things in the alpha frame and then I will do exactly the same as did before. I will check if my momentum is preserved. Now, so using these formulas, what can I say? Okay, I will use the table. This is speed. This is before. This is after. Collision. Okay, my green object. Green x coordinate. So x coordinate was a green object before collision was v, right? So I will put, instead of u, beta x, I will put v and I will get what? s plus v divided by 1 plus sv divided by c square. Now, same thing after was exactly the same. It's also v after the collision. Now, the blue x component, the blue x component was minus v, so it would be s minus v divided by 1 plus c square and exactly the same here. The blue x component is minus v after the collision as well. Okay. Now, what I will do next is I will do the y component. So, to save space, I will just wipe out this and put the y component. So, green y component. Okay, y component before collision was w, right? So, the y component I should calculate from this. So, it would be w square root y minus s square divided by c square and here 1 w. So, it's plus s w. No, it's a, sorry, it's x component. So, it's v. So, sv divided by c square. After the collision, my green object changes the direction from w to minus w, right? That's the y component, which means it would be minus w square root 1 minus s square c square divided by 1 minus sv, sorry, v is still plus, so it's plus divided by c square. And similarly, I can put the blue. Whatever the formula is, I don't want to spend too much time on this, but I have put all this in writing in notes for this lecture. So, notes for this lecture are very, very detailed and I do suggest you to read them after you watch this lecture. So, I have this table and what I will do next is I will check if this formula represents the momentum which is preserved, which means if before collision it was preserved, then in after collision it's supposed to be preserved. And, well, look at this very, very simply. For x direction, this will be preserved because the x coordinate is not changing. I had before collision v and after the collision v, same thing there, but which means that the sum of momentum, this plus this plus momentum of green plus momentum of blue before the collision, and it was, by the way, zero, would be after the collision, the same thing. It was zero in beta frame. In alpha frame it will not be zero, but it will be something. But again, sum of momentum of this guy, so I have to use these formulas, sum of momentum of this guy before collision and this guy together should be equal to sum of this plus this, which means that momentum is conserved, whatever it was, will be after the collision. And again, as I'm saying, the x component will be preserved, y component because of this denominator will not be preserved. And that's the major obstacle which physicists actually faced in the beginning of the theory of relativity until they basically came up with the solution. So what is the solution? Well, you know, time dilates, length is changing when you switch from one system to another. Well, momentum is also changing. There is some kind of a factor which you have to multiply or divide whatever to basically bring it into the proper balance before and after. And, well, let me just tell you the result. The result is that you have to really take into consideration this particular factor and the proper definition in theory of relativity for p is mass times u. And then you have to divide by something which is already quite familiar to you because it's basically the same factor as used. Now, this is vector and this is the length of this vector basically, right? So I can put this one or I can put u scalar product with the same u whatever it is. So square root of one minus u square divided by c square defined this way. And then going through all the calculations, you will see that obtained speeds, this table which I was just trying to basically to show you. So if you will fill up this table completely then you will use this formula to calculate the momentum before some of these two and after some of these two you will have exactly the same result. Now, for strange reason I did not really see in the textbooks a real derivation, how to check basically that this formula really works. So I decided to do it myself. And in the notes I have a link to another page which basically has all the details step by step, very detailed proof that this formula is working. It's nothing but algebraic transformation of the original formula. But it's a large formula because all these square roots, etc. So they're definitely kind of bulky, I would say. So I put it in a separate page and they put a link onto separate page into notes for this lecture. So for those people who are really curious whether this really works, if you don't take just my word, you can just follow the calculations yourself. And again I really don't like when you have something in the textbook. From this follows that and basically in between they skipped all the intermediary manipulations. I do not skip it. So the page which contains all these verification that this formula really works are very detailed, so considered to be a very nice and complete proof. So this is the formula. So now let's think about what exactly is happening. First of all, obviously it increases with increasing of the mass, as we expected. Now if the speed is increasing in magnitude, let's say, what happens? Well this is increasing. Now this is increasing, so this is decreasing. The denominator decreasing, increases the whole fraction even more. So we have a kind of logically correct formula, which means that increasing of mass and increasing of speed really increase the momentum, which means we increase the resistance of our body to change its trajectory. It's more difficult to move it from whatever, wherever it's moving. It's difficult to move it from it to divert it. You need greater force. Okay, so that's fine. Now here we have two different opinions. One opinion is, let's just consider this. And they call this thing as a relativistic mass. So they're saying, you know, with this increase of the speed, the mass is increasing. Well, u is increasing, so 1 minus u over 6, but it's decreasing, so the whole fraction is increasing. And they call this thing a relativistic mass. Well, another group of physicists do not like it this way. They're saying mass is a mass. And basically the mass is a measure of energy which contains in the matter. Remember the very famous formula which we will try to address in the next lectures. So mass and energy is basically the same thing, so don't make this relativistic mass the issue. Just consider the whole thing as a momentum. This is a momentum. Mass is a mass, and they usually use subscript zero, which means mass at rest, or rest mass of the body. Mass as it is in the reference frame where the body is at rest. So this is the mass of the object. Don't, you know, mess with the thing. Don't mess with the mass. And the whole thing is a momentum, including this particular factor. Look, I'm not going to express any kind of feelings towards or against it, whatever it is it is. The formula basically speaks for itself. So you need some kind of a mass multiplied by real speed and divided by this square root. So that's the real definition of relativistic momentum. And as I was saying, through straightforward calculations you can prove that this really preserved before and after the collision, you have some of all momentums before and after the collision is preserved, the law of conservation of momentum. Well, that's it for today. I just wanted to present you this concept of relativistic momentum, which is this formula where M is actually the rest mass. And I do suggest you to read the notes. They are, I would say, lengthy in this particular case, because I'm talking about momentum, etc., but still it's very important to read it. And for those who are really curious in calculations, and again, it's just plain algebra, goes through the calculations just to be comfortable that whatever people are saying that, okay, this is the right formula. Yes, this is the right formula and there is a proof of it. That's it. Thank you very much and good luck.