 Hi, I'm Zor. Welcome to Unisor education. We continue talking about conservation laws in theory of relativity. So the last lecture was about momentum and we basically came up with a formula for momentum in theory of relativity, relativistic momentum, which is slightly different, well, not slightly different. than the definition of momentum in Newtonian mechanics. And we needed it to basically preserve the integrity of conservation law of momentum. And I will use this formula today. Now, today we will talk about energy, primarily about kinetic energy. We will gradually go to conservation of energy. So today we will talk about kinetic energy and basically, again, we will come up with a formula which defines kinetic energy in theory of relativity. So the today's lecture is about relativistic kinetic energy. Now this lecture is part of the course called relativity for all presented on Unisor.com. I suggest you to watch this lecture from this website because every lecture has very detailed notes like a textbook. And then it's a course, basically, which means I always depend on something which I have already covered before in most of the lectures. Like today, for example, we will use the formula which I have derived yesterday for momentum. Now the same website contains prerequisite course called mass for teens. And for example, today I will use calculus and you definitely have to know calculus. You have no vector algebra to study physics. So the website is totally free. No sign-in is necessary, although you might. No strings attached. No financial involvement, etc. completely free. No advertisement, by the way. No ads. So all concentrated on studying. All right, so let me start. So today we're talking about kinetic energy in primarily relativistic kinetic energy. Now let's start from force. According to second law of Newton, the force is mass times acceleration. Acceleration is first derivative of the velocity. Actually, these are vectors. But we will probably consider only one direction, also it doesn't really matter. So it's m times dU by dt. So U is velocity. And dU by dt is first derivative, which gives acceleration. Now in Newtonian mechanics, mass is constant, so I can put it under differential. I will put it here. And this is momentum as defined in classical Newtonian mechanics. M times U, mass times speed is momentum. Now, today this is very important. Again, I will refer to the previous lecture about momentum. We are talking about relativistic momentum. And that's why retaining this formula, I have to express f force in terms of relativistic momentum. And that would be relativistic force. So momentum as a function of t by the way of the time in zero relativity is okay, first of all, I put dependence on time. And obviously speed depends on time. And therefore momentum depends on time. When we are talking about force, we no longer talk about constant speed. It's not a uniform movement. It's accelerated movement. And not necessarily a acceleration is supposed to be constant because f is not supposed to be constant. f can be variable as well. So this is the definition where gamma is one over square root one minus U squared divided by C squared. And this is function of time. Now, this is no longer a vector in this particular case. You can consider this as a scalar product of vector by itself or basically the length of this vector squared. So that's what my expression for momentum in theory of relativity is. And I have to use it here basically to define relativistic force, right? So now force is equal to as a function of t, as a function of time, it's equal to derivative by time of this expression, which is m0 U of t I can put vector here and here plus vector square root of one minus U of t squared divided by C squared. C is the speed of light, of course, you remember that, right? So that's my expression for the force. Great. Now, from the force I have to somehow get to energy. Now, the energy, if you remember, is definitely related to work which the force is doing. So let's just assume that we have an object. This is x direction y z. I'm not talking about y or z dimensions this is only one-dimensional movement. The force, this is the object at t is equal to zero at the moment at the origin of coordinates. Now, this is location of object at time t and the force is acting only in the direction of x-axis. So my movement is always along the x-axis. So that's simpler. So if I have some kind of force and some kind of speed with which we are actually moving as a function of time, of course, speed for example, if force is constant then acceleration is constant and speed would be gradually increasing with uniform acceleration. But that doesn't necessarily have to be constant. It doesn't really matter. Okay, so what happens during certain period of time? Let's say I have this period of time, capital T. During this time the force is acting and the object at time zero was in the beginning of this and the origin of the coordinate and then it's moving during this thing. What is the work which my force is doing? Well, let's just do very simply. Let's just have a tiny t plus d t. So a tiny piece of this trajectory from t to t plus d t. During this time obviously as usually done in calculus and physics we are assuming the force during this infinitesimal period of time is constant and it's equal to f of t. Then we will multiply it by the length of this piece which is actually d x. It's differential of the length so it's from t to t plus d t. It's d x. And we also assume that the speed during this period of time, infinitesimal period of time is constant and equal to u of t. That's the speed. And now we can calculate the differential of work which is done only during this period of time. So this differential of work which is actually amount of kinetic energy which our object gained during this period of time because the force is moving it faster and faster. So the work which force is doing is exactly equal to increment of the kinetic energy. So it would be d k if you wish. So what is it equal to? Well, it's equal to force at this time times distance d x. Okay, now we know what it is. We know the description of the force which is d by d t m0 u of t divided by square root 1 minus u square of t divided by c square. Now what is d x? Well, if I know the speed then would be speed times differential of time, right? Speed times the... during this time. This speed is constant. So u times d t would be my d x piece of the trajectory during which the force is acting. Okay, so what do I have to do now is to find out the total kinetic energy which the object moving along x axis from 0 to time is equal to t. Well, I have to just integrate it. So kinetic energy and we can actually use something like subscript 0 times capital T. It's equal to work with the same subscript from 0 to time c. It's equal to integral from 0 to capital T of this thing. Okay? Well, don't get scared, actually. That's something relatively simple. So let me just get the middle part. We don't really need it. So we are talking about integral of this from 0 to capital T and that would be my kinetic energy which object has gained during the time period from 0 to capital T. Okay? That's what it is. Now, first of all, my notes for this lecture are very detailed about how to take this integral but I'll just spend a couple of minutes basically explaining what it is. Now, this dp dt is related to this fraction. Okay? Now, recall the integration by parts. You remember that if you have two functions f of x times g of x and you want to take the derivative, it would be derivative of f times g plus derivative of g times f. Now, if you want to integrate it, it would be sum of integrals. Okay? Now, if you want to consider this thing, I have to put brackets around it. So this is integral of derivative and what is integral of derivative? It's the original function, right? Original function is the product of f times g. But the only thing, if my integral is definitive, let's say from a to b and this is probably dx and this is from a to b and this is from a to b and this is dx and this is dx. So I will probably have to put it from a to b. So it's the original function in these limits, which means this function of b minus this function of a. Remember the forward Newton labels? Okay. And that's equal to, again, integral of a, b, f of x, g of x dx plus integral of b, f times g times dx, from which follows that this thing, and that's where the formula of integration by parts is, integral from a to b of f dx dx is equal to f of x times g of x in the limits from a to b minus, so this one is equal to this minus this minus integral from a to b of f of x, g of x, the, yes, derivative by dx. We will use this formula because it looks exactly like what we have. Derivative times the function. Derivative times the function is equal to this function times this function in the limits which I was indicating, minus integral of derivative of this function times this. We changed the derivation, right? Okay, so I did it quite accurately in my notes, and I will try to do it, I'm not sure I have time, I will check. Right now, but it's really very, very straightforward. So I will do just one interaction here. So this k from zero to capital T is equal to, I have to take this function under the derivation times this function, which is m zero times u square of t, u and u, that's why u square divided by square root 1 minus u square of t divided by c square in the limits from zero to t, minus integral from zero to t. I have to take this function as is and derivative of this. So it would be m zero u of t divided by square root minus u square of t divided by c square times u d t. Okay? All right. Now, this thing is du. Differential of the function is equal to derivative of that function times the differential of argument, right? Differential of f of x is equal to derivative times differential of argument. So I can replace this with du. Now what I will do, I will consider my u of t as substituted independent variable. You can do this integral m zero u divided by square root 1 minus u square divided by c square du. And the only thing which I have to do, I have to change the limits of integration from zero to u of t, the maximum value, okay? Now this integral, again, can be simplified even further because instead of u times du, I can put d of 1 minus u square divided by c square and I need some differential of this. Well, one doesn't matter. There is a minus, so I have to change the sign of the whole thing and then I have to, it would be 2 times u. So I have to multiply it by c square and I have to divide it by 2, I think so. Am I right? Yes. So with this multiplier, I can replace du on d of this expression. Why this expression? Because it's exactly the same as this expression. And again, I can take the integral. Now, using this technique, now the integral of d, let's say, whatever, some kind of dz, whatever, by square root of z. I mean, you know what that is actually, right? It's the derivative of square root is 1 over 2 square roots, right? So you will easily take this integral and again, I don't want to go into all the details because I might make a mistake and it will take the time. But in the notes, I really do it all very accurately and I will just give you the result of this. The result of this is, whenever you will do this and this and all the manipulations, the result of this is that it's equal to a very interesting formula. It's m0c square divided by square root of 1 minus u of t square divided by c square. u of t is that limit of time, minus muc square. This is a very interesting formula. This is the result of this thing, basically. Now, let's analyze it. First of all, you see the resemblance, well, not resemblance, but part of this formula is something which looks very familiar. The most famous formula in physics is that energy is equal to m0 times c square, where m0 is a rest mass. By the way, I used m0 everywhere here just to make sure that we are talking about rest mass because, again, in the last lecture, I was talking about so-called relativistic mass, which is mass divided by this square root, which is basically gamma, the Lorentz factor. So, I don't want to get involved in this. That's why I'm always using the rest mass. The mass of the body at rest in the system of coordinates where it's at rest. Okay. So, it looks, again, as part of that thing. And what's interesting thing is that what's really very, very tempting to say that this is the energy of the body, which is basically at rest. This, the same thing with the Lorentz factor, the same thing, the energy of the body, but it's not at rest, it's moving, which means it has both energy, which it was before, which means energy concentrated in the object at rest plus kinetic energy. And that's why the difference between them, between the total energy and the energy of the object at rest, is really the kinetic energy which we have just derived. It's tempting to say right now, and obviously physicists were thinking about this way, I am not going to prove anything right now. I will leave it for the next lectures. But basically that's exactly where we are moving. We are moving towards the total energy and the energy of the body at rest. And obviously the difference would be the energy of the movement itself. Movement adds the kinetic energy to the energy at rest and that's why I have a total energy. But, again, right now I'm not talking about this in any details, just mentioning this. Now, another thing which is very important is, now you remember that the kinetic energy as defined in Newtonian mechanics is this one. Right? By the way, I put T here. Now T is the limit of the time I'm calculating and we are talking about the energy of the body accumulated up to this time. But T can be anything. So basically whenever you have just any kind of speed u, which is the limit speed, so we are considering that in the beginning the body was at rest and then during some time I can put any value. It accumulated certain speed up to this value and this value would be basically the basis for calculating the kinetic energy. So I can just drop this dependence on time just saying that whatever the speed is at the very end, at the very moment when we are interested in how much energy kinetic energy the body has, that's the speed which we have to use. So it's obviously depending on something, on time or maybe on distance with you, but whatever it depends on doesn't really matter. Whatever the point in time and space where we are measuring the speed would be the total kinetic energy using this formula, which object actually is in possession of. Okay, now the question is how they are related. Well, they must be related if the speed of the object is really very, very small relative to speed of light and that's absolutely true. Now how can I prove it? Well, to prove it is very easy actually. You have this gamma thing, this is gamma, right? So the whole thing can be gamma is m0c squared minus m0c squared which is gamma minus 1m0c squared. That's what it is, right? Now, if you will take this gamma which is 1 over square root of 1 minus u squared divided by c squared and represent it in the Taylor series by the powers of u over c, it makes a lot of sense because in Taylor series you will have this thing as first member in first degree then the second degree, the third degree and the greater the power you raise u squared divided by c squared the smaller it becomes, right? So we are assuming that u over c is very small that's why I will just use whatever I have done before I am very lazy, I did not really calculate it myself I took it from the internet so this is my first, this is my second and I think this is my third 5 u to the sixth I think it's 16 c to the sixth you see u squared is very small now this one would be even smaller and this one would be even smaller, etc so what usually people do in these cases when you want to approximate you basically cut off the tail considering it's really very, very small much smaller than this one and you take only this one so if you will replace gamma with this what we will have? you will have gamma minus one so this one goes out so you have only one u squared c squared times mc squared what do you have? one half mass times square so indeed if u is small relative to c if the speed of the object is really small relative to speed of light then this formula and this formula are very close to each other so this can be used as an approximation in smaller speeds of objects and that's all I wanted to talk about today so it's a kinetic energy I have the formula for kinetic energy and we have proven that this formula is corresponding to Newtonian when the speeds are really low so I do suggest you to read the notes because the notes are much more detailed as far as the integration is concerned but this is kind of technical details I'm more interested in the approach rather than technicalities which are just basically a matter of your basic knowledge of calculus and nothing more than that okay, so that's it, thank you very much and good luck