 Hello, let us recapitulate what we did in our last lecture. We had given a general definition of proper time interval between two events. When we had first introduced the concept of proper time interval, it was in the context of the use of time dilation formula. At that time, we told that if two events occur in a given frame in the same position, then the time interval between these two events, as measured in that frame of reference, will be called proper time interval. Of course, this particular definition sort of assumed that it is possible to find out two events, which occur in a, it is possible to find out a frame of reference in which the two events occur at the same position. Here we have seen, when we are defining space-like and type-like events, that it need not always be so. If we are restricting our speeds to up to the speed of light, then it is not necessary that if we take any two arbitrary events, I will be able to find a frame of reference in which these two arbitrary events would occur at the same position. But then, we discussed in our last lecture that we can still give a more general definition of proper time interval, even in such cases, where it is not possible to find a frame in which the two events occur at the same position. Only thing in this particular case, the proper time interval would turn out to be imaginary. And of course, the way we have now defined the proper time interval, it turns out to be a forescaler. It means it does not change when I change the frame of reference. So, what we have said that this particular time interval, proper time interval can be defined even when the events are space-like separated. And remember, when the events are space-like separated, it is not possible to find a frame in which the events occur at the same place. So, even in such cases, a proper time interval can be found out by finding out delta x, delta y, delta z and delta t. Only thing for space-like intervals, space-like separated events, the proper time interval will turn out to be imaginary. While for time-like separated events, it will turn out to be real, because if the events are time-like separated, then it is always possible to find a frame of reference in which the two events occur at the same position. So, it is possible to make a measurement in that particular frame of reference, of time difference between these two events, which will turn out to be real time interval and that is what is the proper time interval. Then of course, we have given certain examples to mention about how the proper time interval turns out to be same, just to convince ourselves that whatever we are trying to say is correct. Now, let us go a step further. We had defined a position four vector, then we had defined a displacement four vector. Now, let us try to define a velocity four vector. The question is that why we are at all talking about four vectors? One thing is of course, I have mentioned that it is convenient. Any new quantity that we define is because it makes further mathematics or further physics simple. Also, here we are doing with a specific aim of looking at the new definition of momentum, so that I showed that conservation of momentum in a frame of reference implies that conservation of momentum is same unchanged in any other frame of reference. It becomes a universal quantity and advantage of dealing with four vectors is that certain amount of universality is maintained in these things. So, now let us go to a velocity four vector. Now, we imagine that two events that we had talked about earlier, those two events are not arbitrary events, but they are specific events related to displacement of a particle in a real space. So, there is one particular particle which is found at a given position, this I call as one event, then the same particle gets displaced and at a different time is found at a different position in the same frame of reference, then I call this as the second event. So, the two events that I am talking are not really now arbitrary events, but the events that are related to the displacement of a single particle of a particular particle in a given frame of reference. So, let us imagine that a particle is located at a position x 1, y 1, z 1 at time t 1. So, in a given frame S, where we have erected our axis and we have our watch, we find that at a particular time t 1, the particle is located at a position x 1, y 1, z 1 and at a later time t 2, the same particle is now found at a coordinate x 2, y 2, z 2. So, these become the coordinates or position coordinates of the two events which we are talking about relating to the movement of this particular particle. And of course, we assume that both these observations are being made in the frame in a given frame S. Now, I take the difference between these two and if I call delta x as x 2 minus x 1 as we have been calling earlier. Similarly, we call delta y as y 2 minus y 1 and delta z as z 2 minus z 1 and assume that the time that has elapsed between this particular two events is very small. Then, in the limit delta t tending to 0, if I divide these quantities by delta t, I will get the instantaneous velocity of the particle. Let me again remind you that this particular particle need not be moving with the constant velocity. The constant velocity restriction we have put only on the frames, between the relative velocity between the frames, because that is why we are talking of the inertial frames. Now, both these inertial observers could be observing one particular particle which may or may not be having acceleration. So, on the particle which is being observed by the two observers in the two frames of references, that particle need not be moving with the constant velocity. So, I have to talk about the instantaneous velocity and this aspect we have discussed even earlier that the instantaneous velocities will be given by u x in the limit delta t tending to 0 delta x by delta t. Similarly, u y in the limit delta t tending to 0 delta y by delta t in the limit u z in the limit delta t tending to 0 delta z divided by delta t. So, this is how an observer in the frame of reference S would measure the components of the velocity. Once he knows the time difference between these two events and the position difference between these two events, then by dividing by appropriate by the time as measured by in his, by him in his own frame of reference. Remember, we have always been emphasizing this aspect that a observer has to be consistent. All the measurements have to be made in his or her own frame of reference. So, if I am talking of delta x, this delta x is in his or her frame of reference. Similarly, delta t is also in his or her frame of reference. Now, let us try to talk about the four vector language. We have earlier seen that delta x, delta y, delta z and I c delta t, these are the components of the four vector. What it means that if I go to a different frame of reference, delta x will be different from delta x measured in S, which I call now as delta x, delta x will be different from delta x. Similarly, delta t will be different from delta t, but they will transform using the same transformation matrix that we have written earlier. Then only they qualified to be called as the components of the four vector. So, we realize that delta x, delta y, delta z and I c delta t are the components of the four vector. Now, if I divide these things or multiply these things by something which is a constant, which is not a frame dependent quantity, then that quantity will also remain or will also turn out to be a components of four vector. So, if for example, if we have matrix here and I am multiplying two matrices to get something here, here you have a1, a2, a3, a4, here you have a1, a2, a3, a4 and this is the transformation matrix, which is used to transform this to this. If these are the components of four vector, if I multiply this by a constant c here and this also I multiply by a constant c, c times a1, c times a2, c times a3, c times a4 will automatically give me c times a1 prime, c times a2 prime, c times a3 prime, c times a4 prime provided this c and this c are same. It means once I change my frame of reference, this c should not change. So, if it is a four scalar, then if I multiply the components of a four vector by a four scalar, multiply or divide by a four scalar, then that quantity remains. The components still, even after multiplication will remain the components of a four vector. But if I look at the velocity definition, what I have done, I have divided in all to achieve this particular velocity, I have divided this delta x by delta t. I have divided this delta y by delta t. I have divided this delta z by delta t. Now, delta t is not a frame independent quantity. We have seen from Lorentz transformation that if I change my frame of reference, delta t would become different. The time interval between these two events will turn out to be different. Hence, though I agree that delta x, delta y, delta z and I c delta t are the components of the four vector. But once I divided by delta t, delta t not being a four scalar, being something which depends on the frame, once I divide these things, I will not get what I will get will not be the components of a four vector. Hence, division by delta t to these equations will not let them remain a four vector. And that is how we had defined our velocities. The fact will means that u x, u y, u z and I c cannot be transformed to another frame using the transformation matrix. Hence, u x, u y, u z and I c are not the components of a four vector. They will not transform using the similar equation. This you might have even noticed earlier that when I transform the velocity, the velocity transformation equations that I was getting, they are very different from the normal Lorentz transformation equations. They appeared very, very different. It is basically because u x, u y, u z and I c, they do not transform by the same transformation matrix because they are not the components of a four vector. Now, because I want certain amount of universality to be built in all my definitions because eventually from velocity I have to go to the momentum. Let us see how can I make this particular quantity or how can I define a velocity four vector which will transform the way I want all the transformation of four vectors to be. And if you realize, the answer is very simple. If instead of dividing delta x by delta t, I would have divided delta x by delta tau, I would have realized that delta tau is a proper time interval and this would not change when I change my frame of reference. So, instead of dividing delta x by delta t, delta y by delta t, delta z by delta t, if I would have divided delta x by delta tau, delta y by delta tau, delta z by delta tau, then I would be getting them to be the components of four vector. So, what I would should see? I should take delta x by delta tau, where delta tau is the proper time interval. Similarly, delta y by delta tau, delta z by delta tau and of course, I see delta t by delta tau. If I would have defined them in this particular fashion, then resultant four numbers or four variables will really be the components of four vector. Hence, I define a velocity four vector using delta tau and not using delta t. Therefore, that is what I have written here. If we divide by four scalar which is delta tau instead of delta t, we shall be able to retain them as components of four vector and eventually, I have to take the limit of delta t tending to 0, which effectively means delta tau also tending to 0. Then, the velocity four vector is defined as limit of delta tau tending to 0 delta s. This is what we have used for four vector divided by delta tau which I can write as d s d tau using exactly the similar type of technology that we use in differential calculus. So, we agreed that the velocity four vector is defined as d s d tau rather d s d t. Now, my question is that if I am in a given frame of reference, I directly measure only delta t. If I have to find out how or I have to find out the components of the velocity four vector of a given particle, I have to go and find out delta tau. So, let us look at the way I can find it out. It is essentially very simple, but let us formalize it. So, what I should do if you remember in u x, I have divided delta x by delta t. What I wanted is that it should have been divided by delta tau. So, whatever velocity component or instantaneous velocities that I measure for the particle in a given frame of reference, if this u x I multiply by d t by d tau, then I should be able to get the velocity component of the velocity four vector. Just to make it clear, my u x was delta x by delta t. In the limit of course, delta t tending to 0. Now, I want to take first component of velocity four vector. So, let us say a 1. I want this to be divided by delta tau. So, a 1 is defined as delta x by delta tau. What I can do? I can write this as delta x divided by delta t into delta t divided by delta tau. This quantity is u x. So, I can write this as u x delta t by delta tau. So, it means all I have to do is take my velocities u x, u y, u z and multiply them by delta t by delta tau, which of course, in the limit differential limit means I multiplied by d t by d tau. So, each the components which we have set u x, u y, u z and i c, if I multiply by d t by d tau, then I would be able to convert them into components of velocity four vector. Now, let us evaluate this quantity delta t by delta tau. Once I evaluate the quantity, I will be able to eventually find out a method of how to find out in a given frame of reference the components of velocity four vector. Now, we have seen by the definition of delta tau the proper time interval between these two events. This is the general definition, which is under root of delta t square minus delta x square plus delta y square plus delta z square divided by c square. Now, we realize that delta x in a frame of reference s is actually the motion of the particle. We have said earlier that the two events that I am talking are related to the motion of a single particle. Therefore, this delta x essentially is the displacement of the particle along the x direction. Similarly, delta y is the displacement of the particle along y direction delta z is the displacement of the particle along the z direction. If I would have divided this by delta t, I would have got this as u x or other u x square. If I would have divided by this delta t square rather, I would have got as u y square. So, I can write this delta x square as u x delta t square u y as delta y as u y delta t square and delta z as u z delta t square. I can take delta t square common out of this. Then I will get in this bracket u x square plus u y square plus u z square, which eventually means u square. So, this whole quantity can be written as minus u square delta t square divided by c square. If I take this delta t common, other delta t square common, because of this particular under root, this will become delta t. And in the bracket, what will be remaining is 1 minus u square by c square under root. So, what I will be having is delta t tau is equal to delta t 1 minus u square by c square. Now, I define a quantity gamma u as, let me put three lines just to say that this is definition 1 minus u square by c square. Therefore, delta tau becomes delta t by gamma u. There is a little chance of confusion here. So, let us be clear. Let us revise our ideas, because as we will be seeing that we have so many u v's and gamma, gamma u's etcetera are all appearing. Remember earlier, we had talked of only one single gamma. This gamma was equal to 1 upon under root 1 minus v square by c square. So, our gamma was 1 upon under root 1 minus v square by c square. This v was a relative velocity between the frames. In fact, this is the only gamma that we have been using in the present context. In fact, if you look at the transformation matrix that we have used for transforming a four vector, they are also used gamma. This gamma is something which depends on v and v is relative velocity between the frames. This v is constant, because the two frames that I am talking are inertial. Therefore, this v has to be constant. So, this is the gamma that we have been talking so far. But when we use velocity transformation, at that time we have said that we might have to handle with three different speeds, three different velocities. So, you have one particular particle which is being observed by two observers, one observer sitting in S frame, another observer sitting in S frame of reference. Both these observers are observing one particular particle and that particular particle need not be moving with constant velocity. Now, the relative velocity between S and S, the frames in which our observers are sitting, for that we have reserved the symbol v and using that v I define gamma. Now, an observer sitting in S frame observes this particular particle and finds its instantaneous speed as u. Of course, u in fact, you need not be even constant as we have said. Now, whatever is the value of u using that particular value of u, that observers uses exactly a similar expression and finds a different gamma that we will call as gamma u. So, this gamma u depends on u which is the speed of the particle minded as being observed in frame S and this is the instantaneous velocity. Therefore, gamma u need not be constant as a function of time while gamma is constant as a function of time. Now, the same particle is also being observed by another observer in S frame which is moving with relative speed v. Now, when that particular particle observes that particular particle and at a given instant finds its speed as u, then using that particular u, that observer at that instant of time calculates gamma using that particular u in his or her own frame of reference that I will call as gamma u prime. So, gamma u prime is 1 upon under root 1 minus u prime square divided by c square. So, as we had 3 different speeds that we were talking, speed v relative velocity between the frames, speed u, the speed of a particle being observed in a frame S and a speed u prime, the speed of a particle as being observed in frame S prime. Similarly, using these 3 speeds, I can define 3 different gammas. One gamma relating to v for which I am not putting any subscript, another is gamma u which depends on u and a gamma u prime which depends on u prime. Again, as I have mentioned u and u prime need not be constant. Therefore, gamma u and gamma u prime need not be constant as a function of time. So, I realize that if I am sitting in frame S, then if I have to calculate delta tau, then if I know the instantaneous speed of the particle, means u, I can evaluate using that particular instantaneous speed of the particle gamma u which is a very similar expression as gamma except that v is replaced by u. Then this t tau will turn out to be delta t divided by gamma u. So, delta tau will be delta t divided by gamma u. And all that I have to do is in velocity expression, I have to multiply by an appropriate constant so that this particular thing gets converted into a component of 4 vector. So, this is the way I would write the components of the velocity 4 vector. This is the definition that we had given as d S d tau. This I have expressed in terms of dt because the advantage of expressing this in terms of dt is that the first 3 components will directly turn out to be the instantaneous speed of the particle in S frame of reference. But in order to make this delta dt here, I will have to multiply it by gamma u. Therefore, the components of the velocity 4 vector have to be derived first by taking the standard velocity components and then multiplying them by gamma u. And of course, the fourth component I c will also get multiplied by gamma u. I have just expanded this particular thing. The components of the velocity 4 vector will thus be given by I have multiplied by gamma u dx dt, dy dt, dz dt, d dt of I c t. d dt of I c t if I take it just becomes I c. So, the components is gamma u, u x, u y, u z, I c. It means this I have to multiply by gamma u in order to convert them to a velocity 4 vector. Now, if these are velocity 4 vectors, they must transform by using this particular transformation equation. That is the way we have always been saying that the components of any 4 vector, if I know in a given frame of reference, if I multiply this by standard 4 by 4 matrix, then the resultant that I will be getting will turn out to be the components of the 4 vector in S frame of reference. Here, I have to use gamma, which is dependent on the relative velocity between the frames. As far as velocity 4 vectors are concerned, the components are gamma u, u x, gamma u, u y, gamma u, u z, gamma u, I c as we have just now mentioned. Now, if an observer tries to measure the speed of the same particle in his or her own frame of reference, the way he or she would calculate the components of the 4 vector would be, first calculate the velocities, components u x, of course, the velocity in that particular frame would be u x prime, u x prime, u y prime, u z prime, for c does not change. So, fourth component is I c. Then, he or she has to multiply it by gamma u prime, because the speed has changed in his frame. So, using his speed, he will have to calculate his gamma u, which will be different from the gamma u in S. So, that is why I have called this as gamma u prime. Therefore, what according to him would be the components of the 4 vector will be gamma u prime, u x prime, gamma u prime, u y prime, gamma u prime, u z prime, gamma u prime, I c. These gammas depend on u prime, the instantaneous speed of the particle. These gammas depend on again u, which is the instantaneous speed of the particle in S frame of reference. Transformation matrix depends on gamma, which is the relative velocity between S and S prime frame of reference. So, there are 3 gammas, which one has to be clear that we are using an appropriate gamma in order to work out this particular equation. Now, let us try to see whether if I multiply this particular matrices, I get back my velocity transformation, because that is what it should happen. I know that by a totally different method, I have found out a velocity transformation. In principle, if I expand this particular matrix, I should be able to get back my velocity transformation. So, let us try to do this particular thing. Small amount of mathematics, so let us just try to see it. Let us look at this particular equation here. First, I look at this particular component here. This component, the first component will be given by gamma multiplied by this first component plus 0 multiplied by this component plus 0 multiplied by this component plus i beta gamma multiplied by this component. So, eventually, these 2 components do not lead me to anything because they get multiplied by 0. So, this will be gamma times this, i beta gamma times this. This must be equal to the first component. So, let me first write the first component. So, gamma, gamma u, ux plus i beta gamma times the fourth component. The fourth component is gamma u ic. So, this becomes my first equation. Let us look at the second thing here. As far as second equation is concerned, it is simple because this is equal to this multiplied by the first component, 1 multiplied by the second component, third multiplied by the third component here, fourth multiplied by the fourth component. So, this is just gamma u prime, ui prime is equal to gamma u uz. So, let us write again here. Similarly, we can write for the z component because it is exactly similar. Now, let us try to write for the fourth component. For the fourth component, we have gamma u prime ic should be this multiplied by the first component, 0 multiplied by the second component, 0 multiplied by the third component. So, that does not give me anything plus gamma multiplied by gamma u ic. So, let us write this equation as the fourth equation minus i beta gamma times gamma u ux plus gamma gamma u ic. So, this is what I get after opening up this particular matrix. Let us see whether this particular matrix yields me the velocity transformation. What I have done just written the same equations which I have written here, gamma u prime is equal to ux prime multiplied by ux prime is equal to gamma gamma u ux plus i beta gamma gamma u ic. These two are just equality, gamma u prime ic is equal to minus i beta gamma gamma u ux plus gamma gamma u ic. The equations look complicated, but let us just try to work out. Let us start with the fourth equation. This is the equation. Let us first look at this particular equation and then whatever I get, I will substitute this here. In fact, from this equation I will get gamma u prime in terms of gamma u. Let us realize that this ic, there is i here, there is i here. So, all these i's will cancel. I will let this c also cancel. There is a c here. So, that c also will get cancelled. There is no c here, but there is a beta which is v by c. So, once I cancel c, this will become v by c square. So, what I will get is gamma u prime is equal to gamma gamma u. Gamma gamma u, in fact, I can take common. So, I will get 1 minus ux. There is a ux here and this will become v by c square. So, this is what happens to this fourth equation. This is what I have written here to be more careful. Gamma u prime ic, this is just a repetition of the same equation. I have taken i gamma gamma u common. I get, there is a c here. So, this gets c. This term I am writing first. This term I am writing the second. So, this becomes, I have already taken i gamma gamma u common. So, this becomes beta times ux and I realize that beta is equal to v by c. So, once I write in terms of, when once I try to cancel this particular c and i, then I get gamma u prime is equal to gamma gamma u 1 minus v by c square ux. So, this is an equation which essentially sort of relates gamma u prime in terms of gamma u. This is the equation which I will now put into the other equations and try to work it out. So, let me first write this equation here. Gamma u prime is equal to gamma gamma u 1 minus v ux by c square. Let me put it back into the first equation. This was my first equation. All I am doing is putting for gamma u prime gamma gamma u 1 minus v ux by c square, which we have just now seen from the fourth equation. So, this gamma u prime has been replaced by this quantity multiplied by ux prime should be equal to gamma gamma ux ux minus v. Just stop. Repeat this transparency. So, let us substitute this value of gamma u prime in this particular first equation which I have written here. So, this is the same equation which I had obtained after expanding the matrix transformation matrix. Now, this for this gamma u prime I substitute gamma gamma u multiplied by 1 minus v ux upon c square as we have just now seen from the fourth equation. And this ux prime remains ux prime. On the right hand side, I take gamma gamma u out. So, this becomes gamma gamma u and in bracket what is remaining is this ux, then this i square becomes minus 1. So, this becomes, there is a v by c here, there is a c here. So, this c cancels with this c. So, what remains here is just v. So, you get here as gamma gamma ux ux minus v. This gamma gamma u cancels from this side. I can find out for ux prime, ux prime would be this divided by this. So, you get ux prime as ux minus v divided by 1 minus v ux upon c square which as we know which we have derived earlier directly by the use of Lorentz transformation, the first equation of velocity transformation, the transformation of the x component of the velocity of the particle. Now, let us look at the second equation. This was my second equation which was gamma u prime u y prime is equal to gamma u u y. I do exactly the same thing for gamma u prime. I substitute the same expression which I have obtained from the fourth equation which is gamma gamma u multiplied by 1 minus v ux divided by c square. Of course, u y prime remains here, gamma u into u y. Gamma u here cancels with this gamma u. I can write u y prime is equal to u y divided by this quantity and of course, one of the gamma remains here because this gamma does not cancel here. So, I get u y prime is equal to u y divided by gamma 1 minus v ux upon c square which as you would have realized is the standard transformation of the y component of the velocity. Similarly, I could have substituted this into the third equation and shown that this turns out to be the transformation for the z component of the velocity. So, I can derive the velocity transformation from the transformation of velocity 4 vector. Let us relook at the fourth equation and let us just see what it gives. I have expanded it. For gamma u prime, I have written the actual expression of gamma u prime which is 1 upon under root 1 minus u prime squared divided by c square. Here, there was a gamma. So, for this gamma I have written as 1 upon under root 1 minus v square by c square. There was a gamma u here. For this gamma u, I have written this as 1 minus u square by c square and this gets multiplied by 1 minus v ux by c square as we have just now seen. So, all I have to do is to take, write this particular expression in terms of c square, readjust these terms and you will get comparatively a simplified expression which I can interpret slightly in a different fashion. That is why I am discussing little more about this last equation. So, this equation can be very easily shown to be written in terms of 1 minus u prime square by c square and writing in this particular fashion by taking the inverses to a small amount of simplification by multiplying by c square or c 2 power 4 or whatever it is depending upon the cases and eventually you will land into this particular equation. From this equation, I find out something which is interesting which I would like to point out. One is that if I assume, for example, that we have already said that I would not expect the relative velocity between the frames to exceed c. So, this quantity for any value of v less than c is always going to be positive. In the denominator, we have something which is square which is always going to be positive. There is no imaginary number involved here. Here, you have c square minus u square. Here, you have c square minus u prime square. So, if u is less than c which is what we expect, then this quantity is positive. So, this quantity will also turn out to be positive which it means shows that if in any frame u is less than c, in any other frame u prime will also turn out to be less than c because this quantity has to be positive. Nothing surprising, but let us little extend a little bit the argument. If u happens to be c, if there was a particle which was at all found to be moving with the same speed as c, about this we will discuss little later, then this particular quantity will be 0. This particular quantity will also be 0. It means if a particle is found to be traveling with the speed of light, then in any other frame of reference also that particular particle will found to be traveling with the speed of light. And remember, this is something which we have earlier said, there is nothing special about light as far as transformation equations are concerned relating to particle or light. If light velocity is same in all the frame of reference, if at all we find a particle which also travels with the same speed as speed of light, it is possible, then in every other frame of reference that particular particle will also be found to be moving with the speed of light. Now, let us extrapolate a little further. If at all we are able to find a particle which travels with speed greater than speed of light, then this quantity in this particular bracket will be negative, then here also should be negative. It means in any other frame of reference also that particular particle will turn out to be moving with speed greater than speed of light. So, speed of light is something which is in between any particle, if at all we are able to find a particle which is travelling with the speed greater than speed of light, in every frame this will move with the speed greater than speed of light. If it is found to be travelling with speed of light in every other frame of reference, it will be found to be travelling with speed of light. If in a frame it is found to be travelling with the speed less than speed of light in every other frame, it will be found to be travelling with the speed less than speed of light. This is what I have summarized in next few transparencies. We can see from this equation that if u is less than c in a particular frame, it would be same in any other frame also. If u equal to c in a frame, it would be same in other frames also. Also, if it happens that if u is greater than c in a frame, the same would be true in other frames also so long as v is less than c. That is what we have said earlier. One curiosity we may have, we have always said that the length of four vector is always same in all the frame of reference, all the frame of references. Let us just evaluate the length of the velocity four vector and see whether it really turns out to be same in all the frame of reference because see like for the displacement four vector, we had taken that particular four vector and look at its length and said that this proper time interval is same in all the frame of reference. Similarly, the length of the velocity four vector should also be same in all the frame of reference. Let us just evaluate the length of the velocity four vector. This is what I have done in this particular transparency. This is the first component. This is the second component. This is the third component. Fourth component was gamma u i c. I have taken the length. I have a square and because of this i, this sign becomes negative. So, the length of velocity four vector will be given by gamma u u x whole square, gamma u u i whole square, gamma u u z whole square minus gamma u c whole square. I can take gamma u square common. If I take gamma u square common, I will get u x square plus u i square plus u z square minus c square. u x square plus u i square plus u z square is u square. So, this whole thing I can write as u square minus c square. So, this is what I have written here. u square minus c square. I have expanded gamma u square which is 1 divided by 1 minus u square by c square. If you look at this particular denominator, this you can write as c square minus u square divided by c square. So, this c square will come into the numerator. Here you will have u square minus c square divided by c square minus u square. So, this will become minus 1. So, the whole quantity will become minus c square. And indeed, this is going to be same in all the frame of references because I know c is independent of frame of reference. It is same in all the frame of reference. So, obviously, the length of the four velocity four vector is really a four scalar. It does not depend on the frame of reference that you have used to describe this particular four vector. Now, let us take some example. Let us take one particular example. And this example is essentially similar to the one we have done in our last class. Essentially, the numbers have been picked up from the same thing. So, we do not have to describe another different problem. So, we just take numbers from an earlier example. And let us try to work out and find out the velocity four vector. So, we know how we are doing it. So, this is the one which we have done it earlier, that there is a particular particle which is falling along the tower, etcetera, all those things. But that is not all that important. As far as we are concerned, let us treat this as a particle which is falling vertically downwards. So, it has a velocity along the minus y direction. And the velocity that we had evaluated at that time was minus 0.8 c. Let us try to be little more careful about our symbols which we have not always been earlier, because we have not described these things so much in detail. So, let us call this as u, because this is the velocity, this is the particle speed as being observed in S frame of reference. And for that, we have reserved the symbol u. We have another frame of reference S, which is moving relative to S. This, for this relative velocity, I have to reserve the symbol v. So, v turns out to be equal to 0.6 c. This same particle is being observed by this particular observer. And we had done a velocity transformation last time and found out that it has a x component of the velocity, which is minus 0.6 c and a y component of the velocity, which is minus 0.8 c divided by gamma, which is appropriate gamma, which turns out to be equal to 1.25 in this particular case. Because this is the particle speed as being observed in S frame of reference, I have reserved the symbols U. Being the particle speed, I have put U, not v. And because this is being observed in S frame of reference, I am putting U prime. So, you have Ux prime, Uy prime. So, these are the two frames of reference. This is the relative velocity between the frames. This is particle. This is particle velocity as being observed in S. This is the particle velocity as being observed in S frame of reference. Remember, of course, in this particular example, this U, we had assumed to be constant, because we wanted also to transform to U frame of reference. But now, so long we are talking only between S and S prime, this speed did not have been constant. So, this could then be a instantaneous U. This could have been an instantaneous U prime at a given instant of time. But of course, in this particular example, we have taken the case when this particular particle velocity is also constant. Now, let us first look at the frame S and try to calculate the components of the velocity 4 vector. I have just put the same thing here. In S, Ux is equal to 0, Uy is equal to minus 0.8c and Uz is equal to 0. First thing that I have to do is to calculate gamma U, because just Ux, Uy, Uz is not enough. So, I calculate gamma U using these velocity components, the velocity of the particle. So, gamma U is 1 upon under root 1 minus Ux square plus Uy square plus Uz square divided by C square. It essentially means, because all other components are 0, only 0.8c, we have used this particular gamma value for 0.8, this turns out to be equal to 5 by 3. So, once I know gamma U, I can very easily calculate the components of the velocity 4 vector. Remember, my velocity components were 0 minus 0.8c and 0. So, I have to multiply them by gamma U. Of course, 0 multiplied by gamma U, it still remains 0. So, minus 0.8c, I have to multiply by 5 by 3, which is the gamma U, which I have just now calculated. 0, nothing happens. I see multiplied by 5 by 3. If I just simplify it, I get the components of the velocity 4 vector of that particular particle as 0 minus 4 by 3c, 0, 5 by 3, Ic. These are the components of the velocity 4 vector of the particle as seen in S frame of reference. Let me try to find out for S frame of reference. This particular transparency is superfluous, because this we have done earlier. Calculate the x component of the velocity and y component of the velocity of the particle in S frame of reference. This, in fact, I have even written in the figure that Ux is minus 0.6c, Uy is minus 0.8c divided by 1.25. I calculate gamma U. So, the observer sitting in S finds Ux to be this much and Uy, in fact, I have put the number here, this much. So, I can calculate that particular observer will calculate gamma U in his or her own frame of reference. This also we had done in the last lecture. This gamma U will turn out to be 1 divided by 0.48. So, the components of the velocity 4 vector in S frame of reference, he has to take the velocity components. After taking the velocity components multiplied by gamma U to convert them into the components of the velocity 4 vector. So, this was Ux. I have multiplied by gamma U, which is 1 divided by 0.48. In fact, I have put it now 0.8c divided by instead of 1.25, I have put 5 by 4 here, divided by gamma U, which is multiplied by gamma U, I am sorry, gamma U prime, which is 1 divided by 0.48. Third has anyway 0. Fourth is Ic, multiplied by gamma U prime, which is 1 divided by 0.48. If you sort of simplify this, this becomes minus 5 by 4c, minus 4 by 3c, 0, 1 divided by 0.48 Ic. So, according to S observer, the components of the velocity 4 vector are given here. Now, if whatever I am saying is consistent, then if I use the transformation matrix for transforming the velocity 4 vector, these 4 components must be obtained by from the components in S frame of reference. Let us just try to see it. We see that the components are different in S frame of reference. However, they should transform as per the 4 vector transformation equation. This is the 4 vector transformation equation. These are the components in S frame. These are the components in S frame of reference. This equation must be true if I whatever I am saying is correct. Remember here, what is appearing is gamma and what is appearing here is beta. This gamma depends on the relative velocity between the frames, which is 0.6c. For 0.6c, I get 1.25 here. So, this is 1.25. This I get as 0.6. So, let me expand this particular thing, writing these numbers. So, all that I have done is expanded this particular matrix by writing the value of gamma which turns out to be 5 by 4. Here, we had 0.6 multiplied by 5 by 4, which gives me 3 by 4. Here, I had minus i beta gamma. So, I get minus i 3 by 4. This is just same gamma 5 by 4. You can very easily see that this equation is satisfied. For example, if I take the first component 5 by 4 multiplied by 0, 0 multiplied by 0, 0 multiplied by 0 here. I am sorry, 0 multiplied by minus 4 by 3c, 0 multiplied by 0. Then, i 3 by 4 multiplied by 5 by 3, i c. If I multiply these things, what I will be getting? 3 will be cancelling. I will be getting 5 by 4. i square will be minus 1. So, I will get minus 5 by 4c. This is same as this. This is same as this as expected. Similarly, minus i 3 by 4 multiplied by 0, 0 multiplied by minus 4 by 3c, 0 multiplied by 0. It means the fourth component is just 5 by 4 multiplied by 5 by 3, i c. You can see very easily that this will give you 1 divided by 0.48. So, this equation is satisfied. These are the components of the velocity 4 vector and s frame. These are the components of the velocity 4 vector and s frame of reference. This equation does follow the same transformation equation that I expected it to follow. So, this is what I have written. This can be easily seen that this equation is true by expanding the matrix equation into four equations. So, in the end, I will just show this particular figure to make things about the gamma is very very clear. So, remember we had put here u y, here we had put u x prime, u y prime. If I use this particular u to calculate gamma, I will get gamma u. If I calculate these particular values of u to calculate gamma, I will get gamma u prime. And if I use this particular value of v to calculate gamma, I will get gamma. I will just summarize whatever we have said today. We have defined the velocity 4 vector. Then we gave an example to show how to evaluate these components in a frame and how do they get transformed into another frame. Thank you.